fluid mechanics and hydraulic machinery
DESCRIPTION
Fluid Mechanics and Hydraulic MachineryTRANSCRIPT
E310/1 1
FLUID MECHANICS AND HYDRAULIC MACHINERY
UNIT I
PREPARATION OF FLUIDS
Aims: This unit aims at cursory review of properties of fluids like, viscosity, Bulk
Modulus, Vapour pressure, surface tension and capillarity.
Objective:
1 . To distinguish between dimension and Unit.
2. To define various properties.
3. To distinguish between Ideal fluid & Real fluid Newtonian and non Newtonian fluids
4. Practical application of the above preparation in solving problem.
1.1 Introduction: Fluid Mechanics is the science of mechanics, which deals with the
behaviour of fluids at rest or in motion. Both liquids and gases are called fluids. fluid is
defined as a substance which is capable of flowing and deforms continuously under a
shear stress however small it may be.
Before review of the various properties of fluid, dimensions and units of
measurement are discussed here.
1.2. Dimensions and units: Any physical quantity can be expressed in four fundamental
dimensions namely, mass(M), Length(L), Time (T) and temperature (θ) . But in
incompressible fluid flows the 3 dimensions M,L, and T are adequate to define the
physical quantities.
The standards to measure the above dimensions of the physical quantities, are
called unit of measurement. There are 4 systems of units as given below.
1.Centimeter - Gram - second (C.G.S)
2. Meter - Kilogram - Second (M.K.S)
3. Foot - Pound - Second (F.P.S)
4. International standard system (S.I)
(latest system)
In this present treatment of the subject , F.P.S is not used and mostly S.I. System is
followed.
E310/1 2 The dimensions and units of various quantities useful are given below.
S.No. Quantity Dimension C.C.S M.K.S S.I
1. Mass M Gram (m) m.slug Kg(m)
2. Length L C.m metre metre
3. Time T Second second Second
4. Velocity L/T = LT-1 cm/s m/s m/s
5. Acceleration LT-2 cm/s2 m/s2 m/s2
6. Force MLT-2 Dyne Kg(t) Newton
7. Pressure ML-1T-2 Dy/cm2 Kg(f)/m2 N/m2 Pascal
8. Mass density ρ= ML-3 gm(m)/cm3 m.slug/m3 kg (m)/m3
9. Specific
weight
w= ML-1T-2 Dy/cm3 kg(t)/m3
(9.81N)
N/m3
10 Viscosity ML-2 T-1
FTL-2
Dy -s/cm2
(poise) Kg( f ) − sec
m 2
(98.1 Poise)
N.Sm 2
(10 poise)
11. Kinematic
Viscosity
L2T-1 cm2 /s
(stokes)
m2/s m2/s
12 Workdone
Energy
ML2T-2 Dy-cm Kg(t) -m N-m
(Joule)
13 Power ML2T-3 Dy-cm/s kg(t) -m/s
75 kg(f) .m/s
= 1H.p
N-m/s= J/S=
Watt
1000w
=1k.w
14 Surface
Tension
MT-2 Dy/cm Kg(f)/m N/m
Self Assessment Questions
SAQ 1. Define a fluid
SAQ 2. Distinguish between Unit and Dimension
E310/1 3 SAQ 3 Derive the dimension of power
1.3 Fluid Properties:
Density or mass density or specific mass = ρ It is defined as the mass per unit volume of
fluid. It is denoted by ‘ρ’
Unit C.G.S gm(m) /Cm3
M.K.S. m slug / m3
S.I kg (m) /m3
For water at 40 c Unit of mass density
C.C.S M.K.S S.I
ρ 1g(m)/cm3 10009.81
= 101.94mslug m 3 1000kg(m) /m3
Specific weight is defined as the weight of the fluid per unit volume . It is
expressed as ‘w’ or ‘λ’
Unit is given as
C.C.S M.K.S S.I
Unit Dy| cm3 Kg(f)/m3 N/m3
For water at 40 c, the Unit is given by
C.C.S M.K.S S.I
Unit 1gm(f)/cm3
981 dy/cm3
1000 (f)/m3 9810N/m3
=9.81K.N/m3
Further the following relationship of units are also useful in the further study of Fluid
Mechanics.
Kg(f) = 1m slug ×1m/s2 = kg(m) ×9.81m/s2
N = (9.81) Kg (m) ×1m/s2 = kg (m) ×1m/s2
∴1m.slug = 9.81 kg (m) = 9810g(m)
1Kg(t) = 9.81N = 1000g(t)= 981×105Dy
1Kg(m) = 1000g(m)
Dyne = 1g(m) ×1cm/s2
N= 1000g(m) ×100cm/s2 = 105 Dynes
1g(f) = 1g(m) ×981cm/s2
=981 Dynes
E310/1 4 Thus Kg(f), g(f) are called engineering or gravitational units, where as Dyne, Newton are
called as absolute units.
Specific Volume: is defined as volume per unit mass. It is reciprocal of mass density.
It is denoted by υ = 1ρ
Unit CC.S M.K.S S.I
Cm3/g(m) m3 /m.slug m3 /kg(m)
For water at 40 c , ν = 0.001 m3 /Kg(m) in S.I System
Specific Gravity is defined as the ratio of specific weight of any fluid to the specific
weight of standard fluid which is taken as water (for liquids) at 40c it is denoted by ‘s’.
SAQ 4. What is the specific volume of a fluid whose density is 90Kg(m)/m3.
SAQ 5 10 m3 of fluid weighs 90 KN. find its specific gravity
Viscosity is defined as the property of the fluid with which it offers resistance to
shear force. It is due to both cohesion and adhesion of the fluid. It is also known as
Dynamic Viscosity,. Absolute Viscosity, Coefficient of viscosity . As force is involved in
this phenomenon, it is called as Dynamic Viscosity. It is represented by ‘µ’.
Newton’s law of Viscosity:- It states that the shear stress is proportional to the velocity
gradient normal to the direction of flow of fluid ( in rate of angular defer motion)
Thus ταdϑdy
or τ = µ dϑdy
where τ is the shear stress.
consider a fluid of viscosity in between two parallel plates at distance ‘y’ apart as
shown in figure. The lower plate is at rest and the upper plate of plane area ‘A’ is moved
by a Force “F” with a velocity ‘v’. Due to adhesion the particles at the Solid boundary
are adhered to them and so the velocity of the particles at lower plate is zero and that of
the particles at upper plate are moved by a velocity V. In between the particle move
linearly from zero velocity to v velocity, setting by a velocity gradient vy
ordϑdy
in the
direction
E310/1 5 normal to direction of plate movement. Suppose if the force is increased, then the velocity
gradient also is increased. by experiments it is proved that the shear force F αAvy
for
number of fluids.
ieFA
= ταvy
=dvdy
Where dv is change in velocity in dy as the force F is increased over the place A, the dv/dy
also is proportionately increased.
Thus τα dνdy =
dθdt
or τ = µ dudy
This equation is called Newton’s Equation of viscosity
Also = µ =τ
dϑ dy
So Viscosity is also defined as ;shear stress per unit velocity gradient in normal direction
to the movement of plate.
Dimension of Viscosity is obtained as
µ =FA
yv
=FL2
LL T
=FTL2 (force l-length- time)
or M.LT −2
L2 .LT
=MLT
C.C.S M.K.S S.I
UNIT Dy − scm 2 poise
Kg( f )− sm 2
98.1 poise
N.sm 2 10 poise
For Water 200C Viscosity = 0.01 poise (one centi poise)
E310/1 6 Based on ;the shear stress and the respective velocity gradient in normal direction,
fluids are classified as discussed below.
i).Ideal fluids and Real fluids
we know that τ = µ dϑdy
if µ= 0 τ = 0
This means that when viscosity is zero , there will be no shear stress to oppose the fluid
flow and the fluid particle ‘slip’; over the solid boundary. This leads to Uniform velocity
distribution over the cross section al flow. Such fluids are called as Ideal fluids for which
viscosity is zero and there will be no loss of energy, in the fluid flow from point to point.
On the other hand in fluids if viscosity is not zero and fluid particles satisfy ‘no-slip’
condition over solid boundary, then small fluids are called as real fluids. So for real fluids
viscosity is not zero ;and the shear stress opposes the fluids flow to make the velocity
distribution to vary from zero at solid boundary to free stream velocity to set up a velocity
gradient. In the real fluids there will be loss of energy in fluid flow from point
to point
The ideal fluid is represented in the graph shown in picture on ‘x’ axis for which shear
stress = 0.
Newtonian and non-Newtonian fluids: If fluids like air. water. kerosene. glycerine etc.
obey the law of viscosity and the shear stress in these fluid is proportional to velocity
gradient. Then such fluids are called as Newtonian fluids. If the shear stress is not
proportional to velocity gradient, then these fluids are called as Non-Newtonian fluids as
shown in figure.
E310/1 7 Plastic and Thixotropic fluids
Some fluids flow after certain yield stress. Thus plastic will flow with linear
relationship shear stress and velocity gradient, after certain yield stress. Thixotropic fluids
after yield stress, flow with a non linear relationship between shear stress and velocity
gradient. printers ink is called as Thixotropic fluid. These fluids are represented in the
figure.
Kinematic Viscosity is defined as the ratio of dynamic viscosity for mass density of the
fluids. It is represented by. ‘ν’.
∴υµρ
=
In this unit only Kinematic of fluid, like length, Time are involved and so it is called as
Kinematic Viscosity.
Dimension of Kinematic Viscosity
υµρ
= = × = −MLT
LM
L T3
2 1
Unit of Kinematic Viscosity
Unit C.G.S M.K.S S.I
Cm2/s stokes m2/s m2/s
Kinematic Viscosity of water at 200 c is 0.01 stokes.
SAQ 6. Define ideal fluid
SAQ 7 Why shear stress is zero for a fluid at rest
SAQ 8 Why Viscosity is called as Dynamic Viscosity
SAQ 9 Distinguish between ideal and real fluids
SAQ 10 The Viscosity of a fluids 0.09 poise Determine its Kinematic Viscosity if
its specific weight is 9Kn/m3
SAQ 11 What are the unit and dimension of velocity gradient
SAQ 12 Why the ratio of viscosity to mass density is called as ‘Kinematic
Viscosity’
SAQ 13 (a). of velocity gradient in normal direction is 50m/s/m . Determine the
rate of angular deformation
E310/1 8 (b) If the rate of deformation is 1 rad/s find its velocity gradient.
SAQ 14 If the Viscosity is o/.s/m2 compute the angular deformation for a shear
stress of 0.1N .S/m2.
Worked Out Examples
Worked Example (1) Two parallel plates are 2 cm part. This space is filled with glycerine
Find the force required to drag this plate of area 0.5 sqm at a speed of 0.6m/s when it is
placed at 0.75cm from the lower plate Viscosity of glycerine is 8.0 poise.
F = µAϑ1y1
+1y2
810
× 0.5 × 0.61001.25
+1000.75
=51.2 N
Worked Example (2) A piston of 496 mm dia and 150 cm long slides vertically down a
cylinder of 500 mm dia. If the clearance is filled with lubricating oil of Viscosity 5×10-2
poise find the terminal speed of the piston The weight of piston is 5N.
Sol:
p= µ vy
× A
E310/1 9 A = ΠDL =Π×0⋅496×0⋅15=0⋅233m2
∴5 =5 × 10−2
10
×
v0.002
× 0.233
∴v= 8.5 m/s
Worked Example (3) A 15 cm dia vertical cylinder rotates at 100 r.p.m concentrically
inside another cylinder of 15.10 cm dia. Both the cylinders are 25 cm high. If the
viscosity 8.0 poise is filled in between the space in between the cylinder, find the power
absorbed in overcoming the shear resistance.
sol
Ω =2ΠN60 −
=2 × Π ×100
60= 10.47
velocity = ϑ =ΠDN
60=
Π ×15100
×10060
= 0.785m / s
Shear force =τ×area
µ vtΠDL
=8
10×
0.7850.05100
× Π × 0.15 × 0.25 = 148N
Torque T = F×r = 148 ×0.15
2= 11.1Nm
Power absorbed = T ×Ω = or F×U =116 Watts
Worked Example (4) If the velocity distribution of a fluid over a plate is given by
u =3/4 y-y2, where u = velocity in m/s at a distance of ym above the plate, determine
shear resistance at y = 0.15 m from the plate . viscosity = 0.00833 poise
E310/1 10 sol
Given u =34
y − y2
∴dudy
=34
− 2y ;dudy
=34
− 2 × 0.15 = 0.45m / s / m
y = 0.15
So shear resistance = τ = µ dudy
=0.00833
10× 0.45
= 0.375 ×10-3 N/m2
Surface Tension: The surface of contact between a gas and a liquid or between two
immiscible liquids acts as a film capable of resisting small tensile forces. It is due to
cohesion between the liquid particles at surface. This phenomenon is known as surface
tension which is due to the imbalance of the intermolecular attractive forces between
liquid particles at the interface. This imbalance of forces causes the film or interface to
behave like an elastic film under tension.
So surface tension is defined as the force required so keep unit length of the film in
equilibrium condition. It is denoted by ‘σ’.
C.G.S. M.K.S S.I
Unit Dy/cm kg(f)/m N/m
For water with air at 18 0c surface tension σ = 0.074 N/m (S.I)
= 75 Dy/cm (C.G.S)
Effect of surface tension : The internal pressure inside a fluid drop is increased due to
surface tension.
Determination of excessive pressure inside a drop
Consider a drop of a diameter ‘d’ is divided into two halves. Now consider the force that
are acting on it to keep it in equilibrium condition as shown in figure
E310/1 11 The external force and internal force acting on the drop in horizontal direction. are given
by pressure intensity and projected area of the drop in vertical plane
ie ΠD 2
4Pa and
ΠD 2
4pi respectively
Now consider the forces in equilibrium condition i.e. the ∑F in horizontal direction = 0
∴ΠD2
4pi − ΠD2
4pa − ΠDσ = 0
(Pi-Pa)/ΠD /2
4= Π /Dσ
Pe = 4σD
is the expression for the inside excessive pressure.
SAQ 15 Find the excessive pressure inside a cylindrical jet of water of diameter ‘d’
SAQ 16 Find the expression for excessive pressure inside a soap bubble of dia d.
(hint there will be two surfaces) in contact with air.
SAQ 17 If the dia of drop is increased so 4d from d, find the excessive pressures
ratio inside the drops.
Worked Example (5) Air is introduced through a nozzle into a tank of water to produce
a stream of bubbles of 2 mm dia. calculate the excessive pressure inside the bubble over
surrounding water pressure. Surface tension of water 74 dy/cm
Pe=4σd
=4 × 74 ×10−5
10−2 ×1000
2
=2×74=148N/m2
Capillarity We know that cohesion is intermolecular attractive force between
molecules of the same liquids, where as Adhesion is the attractive force between solid and
liquid or between two different liquids.
When a liquid wets a surface, adhesion is greater than Cohesion and so the liquid rises in a
capillary tube immersed in the liquid. Similarly when liquid does not wet a solid,
Cohesion is greater than adhesion and so the liquid drops in side the capillary tube. This
rise or fall of liquid in capillary tube is known as capillarity which is due to with adhesion
and cohesion of the liquid.
Derivation of an expression for capillary rise or fall :Consider the rise or fall of liquid
in the capillary tube in water and mercury as shown in figure. Water wets the tube
surface where mercury does not.
E310/1 12
In case of water which wets the surface of the tube, the adhesion is more than the
cohesion and so the density of water decreases in the tube at the liquid surface. So to
make the pressure over the liquid surface equal, the water rises in the tube till the pressure
is equal in side and outside over the surface.
In case of mercury , cohesion is more than adhesion, as it does not wet the tube
There fore the density of liquid increases inside the tube near the liquid surface and so to
make the pressure in side and outside of the tube to be equal over the surface, the liquid
falls in the tube.
So consider the equilibrium condition between the forces in vertical direction on
the liquid column in the tube.
upward force = downward force
due for surface tension due to weight of liquid
ie //Π /dσ cosθ =/Π /d 2
4h × w
h =4σ cosθ
wdWhere ‘θ’ is the angle of the surface tension with vertical
For pure water ‘θ’ with glass = 00
For mercury ‘θ’ with glass = 1400
SAQ 18 What is the inclination of surface tension with glass tube with vertical for
pure water.
SAQ 19 Determine the capillary rise in two parallel plates ‘d’ apart immersed in a
liquid.
SAQ 20 Capillary rise in 5 mm dia tube is 0.6 c.m. Find the same in 10 m.m tube.
E310/1 13 Worked Example (6) What should be the minimum dia of a glass tube used to measure
water level if the capillary rise is to be limited to 1mm. Surface tension of water = 75
dy/cm. θ = 0
h =4σ cosθ
wd= 0.1 =
4 × 75 ×1981 × d
∴d= 3cm
Vapour Pressure When a liquid with a free surface is combined in a closed vessel, the
vapour molecules from the free surface will escape into the space above free surface in the
vessel. The escape of vapour molecules continue till the saturation of the space within
vessel with vapour molecules. This vapour exerts a pressure on the liquid surface which is
called as partial vapour pressure.
If, in any fluid flow, the fluid pressure in dropped to its vapour pressure at that
room temperature, then the fluid starts to vapourise. This vapour and the liberated
dissolved gases form as vapour bubble which grow in size. After some time these
bubbles move to higher pressure zone, where they collapse (decrease in size due to high
pressure) releasing very high pressures which are nearly equal to 20 atmospheric pressure.
This high pressure erode the solid boundaries by repeated blows. This phenomenon is
called as Cavitation. So care should be taken in design of Hydraulic machines to avoid
cavitation which causes damage of solid surface of machines.
Mercury vapour pressure is very low and so it is an excellent fluid for manometers.
S.I M.K.S
Unit N/m2 Kg(f)/m2
For water at 200c , vapour pressure = 1.23 K Pa(abs)
For pure water v.p = 18 to 24 Kpa(cabs)
SAQ 21 Which fluid is good for manometers
Bulk Modulus:
When ever a pressure is applied over a fluid in a container it will be compressed .
Compressibility of a fluid is defined as change in volume under a pressure It is the inverse
of Bulk Modulus of Elasticity ‘K’
which is defined as
K∆p−∆v
v
=stressstrain
E310/1 14 C -ve sign indicates the decrease in volume due to increase in pressure.
Unit N/m2 (S.I) or kg (f) /Cm2
K for water = 2.06×109 N/m2 at N.T.P
K for air = 1.03 ×105 N/m2
ie K of water = 20,000 × k of air
So water is considered to be incompressible as its K is very high. That means
the density of incompressible fluid remains constant whatever may be the pressure over it.
Air is compressible fluid.
Summary
1. Fluid is defined as a substance which is capable of flowing and deforms
continuously under a shear stress however small it may be
2. In fluid mechanics dealing with in compressible fluid, 3 fundamental
dimensions namely Mass (M), Length (L) & Time (T) are used to express
any physical quantity.
3 Units are the standards of measurement of the above dimensions.
4 3 systems of units namely C.G.S, M.K.S, and S.I. are used in present
treatment of the subject. But mostly S.I system which is the latest system is
followed
5 Kg(f) = 9.81 N ; 1M slug = 9.81 kg(m)
6. Viscosity is defined as the property of fluid with which it offers resistance to
shear force. Its unit in C.G.S is poise
7 Kinematic Viscosity is the ratio of viscosity to the fluid mass density
8 Ideal fluid is that for which viscosity and shear stress are zero.
9 Surface tension is defined as the force per unit length of the film to keep it in
equilibrium condition.
10 Vapour pressure is the partial pressure exerted by the vapour on the fluid.
11 Compressibility of fluid is the change in volume of fluid under a pressure and is
the inverse of bulk modulus of a fluid.
Answers to self assessment questions:
3. ML2T-3 4.0.0011m3/kg(m) 5. s= 0.91
7. dv/dy = 0 10. v =0.098 stokes 11. m/s/m/ -7-1
13. tan θ = 250, θ=890 .7, dθ/dt = 0.498 rad/s
E310/1 15 Vel gradient = 1.55 m/s/m
14. tan θ = 1, θ= 450 , 15. 2σd
16.pe = 8σd
17. p1
p 2
= 4 18. θ=0 19. h = 2σ cosθ
wd
20. h2 = 0.3cm
EXERCISE 1.(1) Of 5.27 m3 of a certain oil weights 44 KN, calculate the specific weight,
mass density and specific gravity of oil (8349N/m3 , 85109 Kg/m3 , 0.85)
1.(2) A certain liquid has a dynamic viscosity of 0.073 poise and specific gravity
of 0.87 compute the kinematic viscosity of the liquid in stakes of SI.I
system (0.08 stakes, 0.083x10-4m2)
1.(3) of the equation of a velocity distributing over a plate in given by v=2y-y2,
where v is velocity in m/s at a distance ‘y’ in from the solid surface,
determine the shear stress normality at 7.5 cm and 15cm from it.
Given m =8.6 poise.[0.175 Kg(f)/m2 , 0.162 Kg(f)/m2 , 0.149Kg(f)/m2 ]
1.4 Two large place surfaces are 20mm apart and the gap contained oil of
dyamic viscosity 0.6 poise. A this plate of 0.5m2 surface area is to be
parallel although the gap at a constant velocity of 0.6 m/s. The plate is
placed at 8mm from one of the surfaces. Find the face required to parallel
the plate. [F=3.75 N].
1.5 A flat plate of area 0.15m2 is to be tossed up an inclined plane of slope 1
vertical to 3 horizontal on a thin layer of oil m=0.75 poise of thickness
0.2mm. Of the weight of the plate in 250N , estimate the force required to
pull the place at 1.6m/s velocity [F=979N].
1.6 A piston of 7.95 cm diameter and 30 cm long works in a cylinder of 8.0 cm
diameter. the annular space of the piston is filled with an oil of viscosity 2
poise. Of an axial load of 10N is applied to the piston, calculate the speed
of the piston(v=16.68 cm/s)
1.7 A 90mm diameter shaft states at 1200rpm in a 100mm long cylinder
90.5mm internal dia mater. The annular space in the shaft and cylinder in
filled with oil of viscosity 0.12poise. Find the power absorbed to overcome
shear friction. (P=434 w)
E310/1 16 1.8 A tube of internal diamter 2mm is dipped vertically into a vessel containing
mercury. The lower and of the tube is 2cm blow the mercury surface.
Estimate the pressure of air inside the tube to blow a semi spherical bubble
at the lower end. Surface tension of the mercuty is 0.4N/m. [∆P=3.468
Kg/cm].
1.9 A soap bubble 51mm in diameter has an internal pressure in excess of the
external pressure of 0.00021Kg(f)/m2 . Calculate the tension in the soap
film [0.0134kg(f)/m].
1.10 Calculate the capillary rise ’h’ a glass tube of 3mm diameter when
immersed in water at 20oC. The surface tension at 20oC is 0.075kg(f)/m.
What will be the percentage increase in the value of ‘h’ if the diameter of
the tube in 2mm [10mm,50%].
1.11 By how much does the pressure in a cylindrical set of water 4mm in
diameter exceed the pressure of the sorrounding atmosphere of
σ=0.0075kg(f)/m, [3.75kg(f)/m].
Reference books 1. Fluid Mechanics and Hydraulics & Hydraulic Machinery P.N.Modi & S.M.Seth 2. Fluid Mechanics, Hydraulics & Hydraulic Machines K.R.Arora 3. Fluid Mechanics by K.Subramanyam
***
FLUID MECHANICS AND HYDRAULIC MACHINERY
UNIT II
HYDROSTATICS
Aim:
The aims of this unit are to define pressure at a point, the measuring of pressure head,
different standard practices of expressing the pressure, the principle of pressure
measurement by differential and micro manometers and fluid pressure over curved
surfaces.
E310/1 17 Objectives
1. To define fluid pressure and explain the meaning of a pressure head.
2. To explain atmospheric pressure gauge pressure and their relationship.
3. To explain differential and micro manometers and obtain expressions for differential
pressure.
4. To obtain expressions for hydrostatic pressure over curved surfaces which are
submerged, in horizontal and vertical directions and then to find resultant force over the
curved surface
5. To apply the above in solving the practical problems.
2.1. Introduction:
In Units (2) and (3) the fluid at rest is considered . As there is no motion of fluid
tangential forces are zero. So hydrostatic forces of fluid at rest are normal for the solid
boundaries .
In this Unit the fluid pressure is defined. The relationship between absolute atmospheric
and gauge pressures is explained. Differential and micro manometers are explained which
are useful to measure differential pressures. These manometers are useful for measuring
devices.
Knowledge hydrostatic forces on curved surface is useful in design of crest gates of
spillways etc.
2.2 Fluid Pressure is always normal to the surface. It is also called a pressure intensity.
Definition of fluid pressure ‘p’ It is defined as the weight or normal force of fluid
on unit area
p =PA
where‘p’ is normal force
P = Wt of fluid column over the area
= Volume ×Specific weight
= Ah × w
So P = w /Ah
/A= wh
∴p=wh
∴ Pressure at any depth ‘h’ is given ;by ;the product of ‘depth and specific weight of the
fluid’. It is same in all directions in a fluid at rest.
E310/1 18 Unit is N/m2 or Pascal (S.I)
Kg(f) /cm2 (as kg(f)/m2 is too big)
From the above expression we have
h =pw
which is called as pressure head as its dimension is only ‘L’.
2.3 Atmospheric pressure is the normal pressure excerted by atmospheric air over the
surface. It is measured by a barometer as shown in figure . In the top of tube of
barometer the pressure acting by air is zero as there is no air in that space. So this is
taken as absolute zero. This refers to complete vacuum.
At M.S.L the standard atmospheric pressure ‘p’ is 76 cm of mercury
=76
100×13.6 ×1000 Kg(f)/ m2
=76
100×
13.6 ×1000100 ×100
Kg(f)/cm2
= 1.03Kg(f)/cm2
=101KN/m2 or 101Kpa
So standard atmospheric pressure is given ;by 76 cm of mercury, 1.03 kg(f) /cm2 ,
101Kpa. The atmospheric pressure head is given by
ha =pa
w=
1.03 ×104
1000= 10.3 m of water
Local atmospheric press ure will vary from place to place, which can be measured by the
barometer.
SAQ 1. Find the magnitude of standard atmospheric pressure in meters of water.
SAQ 2 Pressure head at a point is 20 m of water . Find the pressure in Kpa.
SAQ 3 Pressure at a point is 98.1 Kpa . Find the pressure head in metres of liquid if
specific gravity 0.9.
E310/1 19 SAQ 4 The pressure at a point in a fluid at rest is same in all directions. True/False.
SAQ 5 The atmospheric pressure will vary with altitude. True/False.
SAQ 6 Local atmospheric pressure is constant at any place. True /False
SAQ 7 Atmospheric pressure depends on Viscosity . True /Flase.
2.4 Absolute Pressure and Gauge pressure
Absolute Pressure When a pressure is measured above absolute zero (as datum) it is
called as absolute pressure.
Gauge Pressure If the pressure is measured above or below atmospheric pressure
then it is called as gauge pressure. It is so because , when the gauge which measures the
pressure, is opened to atmosphere, reads zero.
Of the pressure is above atmospheric pressure it is called as positive pressure or gauge
pressure. If it is below atmospheric pressure it is called as negative pressure, section
pressure or vacuum pressure. of the pressure below atmospheric pressure, is expressed as
vacuum, then ‘-ve’ sign is not necessary for the value.
Relationship between absolute, atmospheric and gauge pressures
All the pressures are shown in figure. From this figure the relationships are obtained as
given below.
Absolute pressure = Atmospheric pr+ Gauge pr
” = Atmospheric Pr - Vacuum Pr
So absolute pressures are obtained by adding gauge pressure or deducting vacuum
pressure to/from atmospheric pressure respectively.
2.5 Measurement of Pressure
Introduction
We know already that pressure at a point can be measured by manometers. Simple
manometers were already known by you, which are used to measure the pressure at a point
E310/1 20 in a fluid. Now let us discuss about differential and micromanometers to measure the
differential pressure between two points in the same pipe line or in two different pipes.
Differential manometer This is used to measure the difference which is considerable,
between two points in same pipe or in two different pipes.
It consists of a glass ‘U’ tube. The two ends of the ‘U’ tube are connected to the two
points between which the difference of pressure is to be measured. The ‘U’ tube contains
in its lower portion a heavier manometer fluid than the fluid in the pipe and immiscible
with the fluid in the pipe
Expression for differential pressure
Consider a differential ‘U’; tube manometer as shown in figure. Let it be connected
between two pipe lines in which a liquid of sp gr sl is flowing. Now let us calculate the
pressure heads in terms of liquid flowing , at each interface in the ‘U’ tube. We know the
pressure will be more, below a column of manometric head is converted to liquid head as hsm
sl
. With this idea, let us consider the pressures from point (1) to point (2)as shown
below .
p1
wl
+ x + h
is the pressure head at A
p1
wl
+ x + h − hsm
sl
is the pressure head at B
In the same way, finally we have
E310/1 21 p1
wl
+ x + h − hsmsl
− y =p2
wl
This is known as gauge equation of the manometer in
terms of ‘m’ of liquid.
i.e. p1 − p2
wl
= hsm
sl
−1
− x − y( )
h(sm
sl
−1) − z
m of liquid
If z= 0 , ie the points are at the same level
Then p1 − p2
wl
= hsm
sl
−1
m of liquid.
p1 − p2
w= h(sm − sl) m of water
Qwl = sl×w
Here ‘h’ is to be measurable with accuracy, otherwise errors will be involved in pressure
measurement. In some cases ‘h’ may be too small to be measure accurately. Then
inverted ‘U’ tube manometer is t o be used.
Inverted ‘U’ tube manometer
This is used to measure small differential pressures. It is an inverted ‘U’ tube manometer
as shown in figure. The manometric fluid which is in the upper portion of ‘U’ tube, is
lighter than the fluid flowing in the pipes. An air cock is provided at the top to expel
entrapped air.
E310/1 22
Now the gauge equation is written as follows in terms of head of flowing fluid. p1
wl
− z − /y − h + hsm
sl
+ /y =p2
wl
p1 − p2
wl
= h 1 −sm
sl
+ z m of liquid
Now Qwl = slw, and if z = 0 , then we have
p1 − p2
w= h(sl − sm) . m of water
Micromanometer is used to measure small differential pressures very precisely .It
consists of two transparent reservoirs of enlarged section at the top of the two limbs of
‘U’ tube manometer as shown in figure. The manometer contains two manometric liquids
of different sp.gr and immiscible with each other.
Let tube CS area = a
E310/1 23 and reservoirs C.S area = A
When the micromanometer is connected to the points (1) and (2) , the fluid of sp gr s2 is
dropped by ∆2 in the reservoir connected to point (1) and rises by the same amount in the
another reservoir connected to point (2). Due to this drop, the fluid of sp gr s, in the ‘U’
tube is dropped by h/2 in the tube connected to point (1) and rises by the same amount in
other tube connected to point (2).
∴ The volume of fluid ∆z ×A= ah/2
or ∆z =aA
h 2
Now let us write gauge equation for the micro manometer as shown below in terms of
heads of liquid flowing in the pipes. p1
wl
+ x + ∆z + ( /y − ∆z)s2
s3
−h2
s2
s3
− hs1
s3
− ( /y −h2
+ ∆z) − ( /x − ∆z ) =p2
wl
p1
wl
+ 2∆z − 2∆zs2
s3
+h2
+h2
s1
s3
− hs1
s3
=p 2
wl
(p1 − p2 )wl
=hs1
s3
−hs2
s3
− 2∆z(1 −s2
s3
); as 2∆z =aA
h
we have (p1 − p2 )
wl
= hs1
s3
−s2
s3
−
aA
h(1−s2
s3
)
= hs1
s3
−s2
s3
−aA
+aA
s2
s3
h =s1
s3
−s2
s3
(1 −
aA
) −aA m of liquid
(p1 − p2 )
w= h s1 − s2 (1−
aA
) −aA
s3
m of water
If A is large aA
is too small and so can be neglected
Then (p1 − p2 )
w= h(s1 − s2 ) m of water
Here h = 2(∆z aA
)
This means by making A large ∆z is magnified to ‘h’ which is considerable.
E310/1 24 SAQ 8. Absolute pressure at a point is 2 m of water. Obtain vacuum pressure in k Pa.
SAQ 9. The pressure of a fluid at rest is always normal to the surface (true /false)
SAQ 10 Convert 2 m of water into mercury height
SAQ 11 Distinguish between negative pressure and vacuum pressure.
SAQ 12 Gauge pressures are measured by taking atmospheric pressure as datum.
(True/False)
SAQ 13 Convert 10 cm of mercury deflection in a differential manometer into a fluid of
sp. gr 0.9.
SAQ 14 The differential pressure head in 10 cm of water. Find the deflection of kerosine
in inverted ‘U’ tube manometer when fluid of sp. gr 1.6 is flowing in pipes.
SAQ 15 If the differential pressure is 20 k Pa determine the deflection in micro
manometer in m of water in which there is mercury in lower portion, fluid
of sp gr 1.6 in reservoirs and kerosine of sp gr 0.9 in the two pipes. The
reservoirs are too large.
SAQ 16 Very small pressure differences are measured accurately by micro manometer
(True /False)
Worked Example (1) Calculate the pressure at A in KN/m2 and m of water at the bottom
of the vessel shown in figure.
sol:
Total pressure at A
p = 0.8×9810×1.2
+0.95×9810×1.6
E310/1 25 +2×9810
=9417.6+14911.2+19620.0
=43948.8 N/m2 = 43.9488 KN/m2
Pressure head in meters of water = 43948.8
9810= 4.48m
Worked Example (2) Find the differential pressure between A and B in the figure
shown
Sol:
By gauge equation we have pA
w1
+ y − 25 ×0.961.2
+ (25 − /y) + 40 =pB
w1
pA − pB
w1
= 250.961.2
−1
− 40
=-45cm of liquid
Worked Example (3) Find the pressure difference of water in the pipes A and B as
shown in figure
E310/1 26 sol:
By gauge equation we have pA
w−15 −15 ×
13.61
− 15 =pB
wpA − pB
w= 15(13.6 + 1) + 15
` = 2.34 m of water
(pA-ps) = 2.34 ×9810=22955 N/m2
=22.955 K Pa
Worked Example (4) A micromanometer is shown in figure
Find the displacement of surface of separation when pressure over C is greater than B by 1
cm of water. Given C.S areas of tubes = 0.25 sq cm and of bulbs = 10 sq cm respectively
.n B side water if filled as C side red liquid of sp gr 0.9 is filled
sol:
By writing the gauge equation we have
pc
w+ (Zc − ∆z + h)0.9 − h − zb − ∆z =
pB
wpc − pB
w= 1cm = h + /zb − ∆z − (/zc − ∆z + h)0.9
But in the initial condition,
We have Zc ×0.9 = Zb ×1
∴pc − pB
w= 1cm = h + ∆z − (h − ∆z )0.9
but ∆z ×A = h×a, I e ∆z= h × a
A= h ×
0.2510
=h
40
E310/1 27
so pc − pB
w= 1 = h + h ×
140
− h −h40
0.9
=h 1 +140
−1 × 0.9 +140
× 0.9
1 cm = h0.1 +
1.940
= h × 0.1475
∴h = 6.78 cm
Pressure and curved surface
Introduction: We know already the total pressure acting on a plane vertically submerged
in liquid in give by
p= wx ×A
That means the total pressure over a vertical plane sub merged in fluid is given by the
product of pressure at centre of gravity of the plane from free surface and its area.
Also the centre of pressure in given by h =IGG
Ax+ x from the free surface. with this
understanding the expression for total pressure acting on curved surface submerged in
liquid.
Derivation of expression for total pressure.
Consider a curved surface submerged in a liquid of sp wt ‘w’ at a depth as shown in
figure. Let dt be an elemental area in the surface. The depth of liquid over this area is h.
Then the total pressure due to this depth of liquid over the elemental area is
dp = dA×wh
If this force is resolved into horizontal and vertical components as given below
E310/1 28 dpH = dpsinθ= dAwhsinθ = wh(dA sinθ)
Where θ is the inclination of dp with vertical. dpH is the horizontal component and the
pH vertical component dpv is given by
dpv = whdA cos θ
By integrating over the area, the above expressions we have pH = whdAcurvedarea
∫ sin θ
pH = pressure acting over vertical projection of the submerged curved area as dA sinθ
leads to vertical projection of the area dA.
This pressure acts at the C.G of the vertical projection of curved area.
Similarly , the vertical component is given by pv = whcurved area
∫ dA cos θ
As dA cosθ leads to horizontal projection of the area, wh dA cosθ leads to
weight of liquid over the area dA. So the integration leads to weight of liquid over the
curved area up to free surface. Therefore vertical component
pv = weight of liquid over the curved area upto free surface . It acts at the
C. G of the weight of liquid over the curved area.
Finally p = pv2 + pH
2
and its in clination with horizontal is given by’α’ = tan -1 (pv /pH ). The point of action of
p over the curved surface is obtained by extending the line of action of p through the point
of inter section of pv & pH to meet the surface at ‘O’ as shown in in figure..
Suppose the curve is as shown in figure. Then the vertical pressure is the imaginary
weight of liquid in the portion AOB.
E310/1 29
SAQ 17. The vertical pressure over a submerged curved surface is equal to the weight of
the liquid over the curved surface True /False
SAQ 18 The point of action of the vertical pressure over submerged curved surface
passes through the C.G of the weight of liquid over the surface. True/False
SAQ 19 Water is standing over a cylinder of 2m dia. Find the vertical force and its
point of action.
Worked Example (5) A 600 sector of 4 m radius is as shown in figure with one of
its edges horizontal, hinged to piers of spill way. Find the magnitude and direction of the
resultant force if the length of the gate is 3m. water is standing upto top of gate.
sol:
The horizontal force pH = ωA x
where A = BC×3
= 4 sin 60 ×3
∴pH = 9810 ×(4sin 60 × 3) ×(4 sin 60)
2
= 176.6 K N
E310/1 30
pV = Imaginary Wt of water on the curve A C which is in upward direction.
= (Area of gate portion in ABC) ×w×3
=( area of sector - ∆G area of BOC) w×3
=(ΠQ 2 ×60
360−
12
× 4cos 60 × 4sin 60) 9810×3
=(Π ×16
6− 3.464) × 9810 × 3
= 144.6KN
Total Pressure = p = pv2 + pH
2
= (176.6)2 + (144.6)2 K.N = 228.3KN
Its inclination with horizontal = α = tan −1(pv
pH
)
tan-1 (176.6144.6
= 0819
∴ α= 39.30
Worked Example (6) A cylindrical gate 2m dia is kept ion a floor with its
longitudinal ax is horizontal. The depth of water on one side is 2m and another side in 1
m. Calculate the resultant hydrostatic pressure on the gate and the minimum weight of it
so that it will not float away from the floor/ m length.
Sol:
E310/1 31 The horizontal forces are
pH1 = wA1 y1
9810 × (2 ×1) ×22
= 19.62KN
pH2 = WA2 y2
= 9810×(1×1)12
= 4.905 KN
∴The net horizontal force = pH1 - pH2 = 14.72 KN
The vertical forces are
Pv1 = ΠR2
2×1
9810 = 15.4KN
acts at x1 =4R3Π
=4 ×1
3 × Π= 0.42
Pv2 =ΠR2
4×1
9810 = 7.704KN acts at x 2 =
4R3Π
= 0.42
Total upward force = 15.4+7.704 = 23.104 KN
Resultant force = 14.722 + 23.12 = 27.4KN
Its inclination with horizontal = tan −1 (23.1
14.72)
=1.57, α = 57.50
The minimum weight of the cylinder not to be lifted by vertical forces 23.104 k.N
Summary
1. Fluid pressure is defined as weight of normal force of the fluid over unit area. It acts
normal to the surface
p = wh, pressure head = h = p/w
2. Absolute pressure = Gauge pressure + Atmospheric Pressure
Atmospheric pressure - Absolute pv = vacuum p2
3. Vacuum pr and -ve pr are same in magnitude, but vacuum pr will have no ‘-ve’ sign.
4. The differential pressure between two points as measured by differential ‘ U’ tube
manometer as ` (p1 − p2 )
w= h (sm -sl ) m of water
if it is inverted ‘U’ manometer (p1 − p2)
ω= h (sl -sm)
E310/1 32 m of water, which is useful to measure small pressure differences.
5. The small differential pressure is measured micromanometer very precisely as
(p1 − p2)
w= (s1 − s2 (1 −
aA
) − s3aA
) m of water
a - C.S of tubes
A- C.S of bulbs
= h(s1 -s2) m of water if A is too large
6. The horizontal pressure over a submerged curved surface is the total pressure over the
vertical projected area of the curved surface
pH=ωAx , Where is the C.G of the area A - which is the projected area in vertical
direction
7. The vertical pressure over a curved surface area which is submerged in the weight of
liquid over the surface up to free surface. If acts at the C.G of the weight of liquid over
the surface.
8. The resultant pressure over the curved surface is p = pH2 + pv
2
or its inclination with horizontal
α = tan −1( pv pH )
Answers for SAQ
1. 10.3m 2. 196.2 Kpa 3. 11.1m
4. True 5. True 6 False
7. False 8. 81.38 Kpa 9 True
10 .14.7 cm 12. True 13. 1.41 m of liquid = 1.27 m of water
14 14.28 cm 15. 17 cm of mercury 16 True
17 True 18 True 19. pv = 15.4 K N
C.G 4R3Π
= 0.42 m from centre of cylinder.
EXERCISE
2.1 For the given closed tube, the pressure at tip ‘A’ is 12cm of mercury vaccum.
Then determine the levels of liquid, in the tubes E,F & G from base
[38.96m, 27.168m, 18.48 m]
E310/1 33
2.2 A pipe containing water at 172 KN/m2 pressure in connected by a differential
gauge to another pipe 1.5m lower than the first pipe and contains water at high
pressure. The mercury deflection in the manometer is 7.5cm. Find the pressure in
the lower tube.[196KN/m2].
2.3 An inverted ‘U’ tube in connected between two pipes A and B. B is 60cm above
A. Water is flowing in A ∝ B. the deflection of manometer liquid specific gravity
0.9 is 50cm. Pressure in A is greater than B. Determine the pressure
difference.[0.058kg(f)/cm2.]
2.4 In the left tank shown in figure, the air pressure is -0.23m of mercury.
Determine the elevation of the gauge liquid in the right hand column at a, if the
liquid in the right and tank is water [94.62m].
E310/1 34 2.5 A micro-manometer consists of two cylindrical bulbs A and B each 1000sq.mm
(cross) sectional area which are connected by a ‘U’ tube with vertical limbs each of
25 sq. mm cross-sectional area. A liquid of sp.gr 1.2 is filled in A and another
liquid of sp gr 0.9 is filled in B, the surface of separation being in the tube attached
to B. Find the displacement of the surface of separation when the pr on the surface
B is greater than that in A by an amount equal to 15mm head of water.
[42.6mm].
2.6 For the container shown in figure find the resultant force on the hemispherical
bottom. [Rv=52.46KN,RH=0]
2.7 A radical gate retains 5m of water above the rest of a dam as shown in figure. Find
the resultant force on the gate per metre length.[R=124.4KN α=10.24o with
horizontal].
2.8 For a cylindrical gate 4m long shown in figure calculate the resultant force due to
fluid pressure. [R=219.64KN α=40.340 with horizontal].
E310/1 35
2.9 A sector gate of radius 4m and 5m long controls the flow of water in a horizontal
channel . Determine the total thrust on the gate. [ 205.7KN, θ=170-30’ with
horizontal]
2.10 A cylinder of 1m diameter supporting water is shown in figure. Find the reactions
at A and B if the cylinder is 5m long and has a weight of 6000kg(f).
2.11 A quadrant of a cylinder 3m long is shown in figure. Find the horizontal and vertical
forces FX=18000kg(f). F1=21424.2 kg(f)]
E310/1 36
2.12 Find the horizontal and vertical components of the force an the gate per metre
length and the resultant force and its direction. Shown that the resultant force
passes through the hinge.
References
Same as unit(1)
****
FLUID MECHANICS AND HYDRAULIC MACHINERY
UNIT-III
BUOYANCY AND FLOATATION
Aims: To know about the buoyant force, centre of Buoyancy, meta centre, meta centric
height and stability of floating bodies. Finally to gain ability to design a floating
body to support a given load.
Objectives:
E310/1 37 (1) To determine the buoyant force and the centre of buoyancy.
(2) To determine the conditions of stability
(3) To define metacentre, meta centric height and to determine the metacentric height
theoritically
(4) To determine the meta centric height experimentally
(5) To apply the above, in solving the practical problems.
3.1 Introduction
We know already that Fluid static deals with fluids at rest and so there is no motion
nor any shear forces between any two layers of fluid. So only forces involved are pressure
forces and body forces.
In the previous units it was shown that the pressure at a point in fluid at rest is
p=wh, where
h=the depth of fluid above that point.
This concept is useful in this unit .
Using Archimede’s principle the force of buoyancy is determined.
3.2 Buoyancy
When a body is immersed in fluid either wholly or partially, it is subjected to an
upward force which tends to lift the body up. This forces is in opposite direction to the
gravity force which is the weight of the body. The tendancy of the liquid to cause the
buoyant force over a floating or submerged body is called as buoyancy. The point of
application of the buoyant force on the body is called as centre of buoyancy.
Determination of buoyant force
The buoyant force over a submerged or floating body can be determined by
Archimede’s principle. This states that when a body is immersed in a fluid either wholly
or partially, it is buoyed or lifted up by a force which is equal to the weight of fluid
displaced by the body.
To obtain the buoyant force consider a wholly submerged body as shown in figure.
E310/1 38
The horizontal force that are acting on it are equal in magnitude and opposite and so the
net The horizontal force that is acting on the body is zero, as they are acting on the vertical
projection of the body. Now to find the resultant vertical force over the body, consider a
vertical prisum of ‘dA’ cross sectional area, out of the body.
The downward force over the area=p1dA
The upward force over the area= p2dA
So the net upward force = p2dA - p1dA
=(wh2-wh1)dA
=wydA=wd A
Where elemental volume of prisum=dv =ydA and wdv is the weight of the prisum of fluid
which is acting upward over the body.
∴ dFB =Elementar buoyant force = wydA=wdv
By integrating this over the body, we have
FB=Buoyant force ∫wdv =w v
This shows that the weight of the fluid displaced by the submerged body which is
equal to Buoyant force which acts vertically upward through the centroid of the volume of
fluid displaced. this centroid is equal to centre of Gravity of the body.
E310/1 39 If the body is immersed in two immiscible fluids of specific gravity ω1 and ω2 as shown
in figure, the buoyant force
=FB=w v1 + w2 v2
The centre of buoyancy coincides with the respective Centre of Gravity of the areas of the
body in the respective fluids.
When a body is floating as shown in figure the buoyant force FB=w v , whre v is
the volume of fluid displaced by the body. As the body is in equilibrium condition,
Upward force = down ward force
is Buoyant force = weight of the body
FB=W.
The line of action of these forces must lie along the same vertical line to make the
net moments about any axis zero.
Principle of floation: It states that the weight of a floating body in a fluid is equal
to the buyoyant force which is equal to the weight of the fluid displaced by the body.
When a body is immersed in a fluid if the buoyant force is more than the weight of
the body, then the body will rise up till the weight of the body is equal to the buoyant
force. If the weight of the body is more than the buoyant force, then it will sink.
SAQ (1) Buoyant force is equal to weight of body . True/False
SAQ (2) If the volume of water displaced by a body is 10m3. find the buoyant force.
SAQ (3) If the buoyant force more than the weight of body, then the body will rise
till its weight is equal to buoyant force. True/False
SAQ (4) Buoyant force in sea water is more. Why?
SAQ (5) 0.2 m3 body weighs 1.5KN in water. Determine its weight in air
E310/1 40 Worked Example(1) A boat when moved from sea water to fresh water sinks by 5cm. A
man weighing 686.7N gets out of the boat to keep it to its original
level. Find the weight of boat. Sea water w=10.06KN/m3.
Sol:
Let ‘W’ be the weight of boat
W=FB=v1w1
We know
W-686.7=∀1w2
is v1w1-686.7=∀1w2
v1(w1-w2)=686.7
∴ v1=686.7
(10.06 − 9.81)1000= 2.8m 3
So weight of boat and man
W=w1v1=2.8x10.06KN=28.15KN
∴weight of boat
=W-686.7
=27.485 KN
Worked Example (2)A rectangular wooden barge without a deck is 6m long, 3m wide
and 1m deep weighs 26KN. find the load that it can carry when water level is 0.6m above
bottom.
Sol:
Buoyant force FB
=6x3x0.6x9810=105.95 KN
∴Total weight = FB
W of boat +load = FB
E310/1 41 load = FB - weight of boat
=(105.95-26)KN=79.95KN
3.3 Metacentre and Metacentric height
When a floating body is in equilibrium condition the two forces that are acting
over it are (1) weight of body ‘W’ acting through its centre of gravity and (2)
Buoyant force FB equal to weight of fluid displaced by the body, which acts
vertically upward through the centre of buoyancy. Therefore these two forces must
be equal is
W=FB. Since there is no moment about any axis of the forces acting over the
body, which is in equilibrium condition, the two forces must lie on the same
vertical line as shown in figure(1)
Figure(1) Figure(2)
Now if the body be titled by an external force through a small angle ‘θ‘(angle of heel), the
buoyant force is changed and its line of action shifted to B2 from B1 as the value of water
displaced is more towards B2 while the G is assumed to be the same, even though a small
shift may be there.
The new buoyant force acts vertically upwards through B2 which meets the
extended line GB, at M. This point of inter section is called Metacentre and the distance
from Metacentre to centre of Gravity of the floating body, GM is called Metacentric height
as
θ→LO.
SAQ(6) Define metacentre and Metacentric height.
E310/1 42 3.4 Stability of floating bodies(or submerged bodies)
Consider a floating body as shown in figure(1) be tilted by a small heel angle ‘θ‘
by an overturning moment in clockwise direction. Now the buoyant force is shifted to new
Figure(1) Figure(2)
position B2 which is acting upward. This force forms a couple weight = w GMSinθ
anticlockwise direction. This couple restore the body to its original position. So this
condition is called stable equilibrium condition.
Stable Equilibrium condition: A floating body is said to be in stable equilibrium
when a small angular displacement of the body sets up a couple with its weight and
buoyuancy in tilted position, that tends to oppose the displacement and brings back to its
original position. Thus it may be stated that a floating body, if the metacentre lies above
its centre of Gravity G, is BM > BG will be in stable equilibrium condition. In this
condition the couple formed by the weight of body and buoyancy in new position will be
in opposite direction to the angular displacement and tends to bring back to its original
position. Here it may be noted that more portion of the body is to be submerged.
Unstable equilibrium condition:
Consider a floating body as shown in figure, be tilted through a small heel angle
‘θ‘ by a overturning moment. Here the couple formed by the weight of the body and
buoyant force in new position will be in the same direction of the overturning moment.
E310/1 43
Therefore the body will not be brought back to original position. This condition is called
as unstable equilibrium condition. Thus it may be stated that a floating body, if the
metacentre lies below the centre of Gravity G, ie BM < BG , will be in unstable
equilibrium condition.
Neutral equilibrium condition:
However if the metacentre coincides with the Centre of Gravity is BM = BG ,
then the floating body is said to be in Neutral equilibrium condition and it will occupy new
position, because the restoring couple formed by weight and buoyancy in new position, is
zero(GM = 0) .
Examples of forces causing overturining moments are wind forces, wave action, pressure
due to tidal or river current, shifting α(oa) in ship from are place to another.
So the ship is be to designed such that the metacentre is always well above the
Centre of Gravity, by lowering the centre of gravity by adding weight(by loading the ship
with ballast)
3.5 Determination of metacentric height
(a)Experimental determination:
E310/1 44
Consider a ship floating in the water as shown in figure. A plump bob is suspended from
P on the centre line of ship, ‘l’ above scale.
Let the weight of ship =W and weight of moving load = w(Magnified)
Now the movable load is shifted to a new position by a distance ‘x’ from the centre
of ship. This causes an over turning moment about ‘Μ’ = wx, in clockwise direction
through a heel angle ‘θ’. The plumb bob also shifted on the scale by ‘d’ giving tan θ =l/d,
where ‘θ’ is equal to heel angle.
In the new position, the buoyancy is shifted to B2 and the new centre of gravity is
G2 . The tilt of ship causes a restoring couple in anticlockwise direction =W GM tan θ.
∴ we have
WGM tanθ = wx or
GM =wx
w tanθ=
wxlwd
E310/1 45 Analytical method to determine metacentric height:
Consider the ship floating in the water as shown in figure be tilted by a small heel
angle ‘θ’. Due to this tilt, AOB wedge emerges from water causing -ve buoyant force
while COD wedge sinks into water causing +ve buoyant force which is equal to AOB
wedge. This causes an overturning moment as out ‘Μ’ due to the shift to Buoyancy B1 to
new position B2. To determine the overturning moment consider an elemental volume of
dx width at ‘x’ from the centre of ship.
Weight this elemental volume = wdx × xθ × L
Where L is the length of ship
= wθLxdx
The overturning couple of the elemental volume due to Buoyancy
= (wθLxdx )2x
Now by integrating this, the overturning couple is obtained as = ∫ wθLxdx 2x
= wθ∫ 2Lx2 dx = wθ ∫2(Ldx)x2
The integral value ∫2(Ldx)x2 is the double moment of the area(Ldx) which is equal to
moment of Intertia Ixx of the ship about its centre of gravity.
∴ overturning couple = wθIxx
Due to the tilt the restoring couple due to shift of buoyancy from B1 of B2 is
W BΜ tanθ (about M) = W × B1 B2. Note the weight of ship is equal to Buoyant force
and further as there is no change in volume of water displaced, the buoyant force at B1 and
E310/1 46 B2 are equal. This moment is the moment of the upward force F B1at B1 about Μ. This is
same as the moment of Buoyant force F B2 about B1 which oppose the overturning
moment.
∴ WBM tanθ = wθIxx as θ is small tanθ = θ
∴ BΜ = wIxx/W
But we know W = FB = ωv , Where v is volume of water displaced.
∴ BΜ = /wIxx
/wv=
Ixx
V
Metacentric height = GΜ= BΜ - BG
SAQ(7) What is the condition for stable equilibrium
SAQ(8) Loading the ship with ballast leads to what type of condition and why?
SAQ(9) A rectangular block 2m long 1m wide and 1m deep sink, into water by
0.6m deep. Find the weight of block.
SAQ10) Determine the distance between centre of Buoyancy to Meta Centre of a
wooden block 12 m long, 1m wide and 1m deep if it sinks by 0.6m.
Worked Example(3) A cylinder of dia 30cm having specific gravidy 0.8 floats in water.
What is the maximum permissible height of cylinder so that it can float in stable
equilibrium with its axis vertical.
Sol:
Volume of water displaced = V = πd 2
4× h1
we know
E310/1 47
BM = I / V; Ixx =πd 4
64
=πd 4
64×
1πd 2
4× h1
=d 2
16h1
But weight of body = Buoyant force ie
πd 2
4h × sw = πd 2
4h1 × w
= sh = h1
substituting this in above equation we have
BM =d2
16sh
From figure we know that OB = h1 / 2 = hs / 2
and OM − OB = BM
is OM = BM + OB
=d 2
16sh+
s2
for stable equilibrium, M should be at higher level than G,
so OM>OG = h/2 d 2
16sh+
s2
> h/2
or h2−
hs2
<d 2
16sh
h2(1-s) < 2d2/16s
∴ h < 2d2
16sx 1
(1 − s)
< 2 × 302
16 × 0.8−
1(1− 0..8)
= 26.52 cm.
Worked Example(4) A cone floating in water with its apex downwards has a diameter (d)
and vertical height ‘h’. If the specific gravity of the cone is ‘s’ prove that for stable
equilibrium.
h 2 <14
(d 2s
13
(1 − s13
) and Sec2θ > h/y
E310/1 48 where y is the depth of submergence
Solution:-
Let the depth of immersion = y
W weight of cone = ws(13Πd2
4h)
FB = Buoyant force = w(13Πd 2
4y)
we know that W = FB
is ws(13Πd2
4h) = w(
13Πd 2
4y)
d2 hs = d12y (1)
also tanθ/2 = d1
21y
=d2
1h
∴ d1 /d = y/h (2)
By (1) and (2) we have
(d1
d)2 = (
yh
)2 hsy
is y3 = sh3
y = hs1/3
OG = C.G of cone from apex = (3/4)h
OB = Centre of Buoyuancy from apex = (3/4)y
BG = OG − OB =34
(h − y) =34
(h − hs13 ) =
34
h(1 − s13 )
Now BM =I
VI =
Πd14
64; V=
13Πd2
4y
∴ BM =Πd1
4
641
13
Πd12
4y
=3d1
2
16y=
3d12 y
16h 2
∴ BM =3d 2hs
13
16h2 =3d2s
13
16h
For stable equilibrium BM > BG
3d2 (s13 )
16h>
34
h((1− s13 )
E310/1 49
h 2 <14
d 2s13
(1 − s13 )
To find Sec2θ > h/y
d = 2h tanθ
d1 = 2y tanθ
BM =3
16d1
2
y=
316
z2 y2 tan2θ =34
y tan2θ
BG =34
(h − y)
For stable equilibrium
BM > BG
3/4 ytan2θ > 3/4(h-y)
y sec2 θ > h
∴ Sec2 θ > h/y
Worked Example (5) A rectangular pontoon 10m long 7m broad and 2.5 m deck weighs
70 tonnes. It carries on its upper deep a boiler of 5m diameter weighing 50 Tonnes. The
centre of gravity of the boiler and pontoon may be assumed to be at their centres of figure
and lie in the same vertical line. Find the meta centre height.
Solution:-
W = Total Weight = 70 + 50 = 120 Tonnes
Let the height of immersion = y m
FB = Then buoyant force = 7 x 10 x y x 1 = 70 y Tonnes
But W = FB
E310/1 50 120 = 70 y
y = 120/70 = 1.171 m
To find Centre Gravity of the combined figure
Let Centre Gravity of the combined figure about base be = OG
Then tasking momenty of the weight about base
120 x OG = 70 x (25/2)+50(2.5+2.5)
∴ OG = 2.81m
Centre of buoyancy above base = OB = y/2 = 1.171/2 = 0.855 m
∴ BG = OG − OB = 2.81− 0.855 = 1.955m
But BM =IV
=
112
×10 × 73
120= 2.38
QFB = V .w
120 = V ×1
V = 120/1 = 120m3
Meta Centric Height = MG = BM − BG = 2.38 - 1.955 = 0.427
Worked Example(6) A rectangular pontoon weighting 240 Tonnes has a length of 20m .
The centre of Gravity in 30cm above the centre of the cross section. The meta centric
height is 1.33m when the angle of heel is 10o. The free board is not to be more than 0.67m
when the pontoon is vertical. find the breadth and height of the pontoon if it is floating in
fresh water.
Solution:-
Volume of water displaced = W/w = 240/1 = 240m3
OG = Centre of Gravity above base = H/2 +0.3
OB = Centre of buoyancy above base = (H - 0.67)/2
E310/1 51 ∴ BG = OG − OB = (
H2
× 0.3) −(H − 0.67)
2= 0.635 m
BM =IV
=
112
× xB3 × 20
240=
B2
144
MG = BM − BG1.33 = ((B2 /144) - 0.635)
∴ B = 6.5 m
Volume of water displaced =
240 = B x (H - 0.67)20
240 = 6.5(H-0.67)20
H = 2.51 m
Worked Example (7) A battle ship weighs 130 x 103 KN. On filling the ship’s boats on
are side at a distance 10m, from centre, with water weighing 600KN the angle of
displacement of the plumb line is obtained as 2o16’. Determine the meta centric height.
Solution:-
Given w= 600 KN x = 10m
W = 130 x 103 KN
tanθ = d/l = tan 2.27o = 0.04
GM =wx
W tanθ=
600130 × 103 ×
10.04
= 1.15m
Worked Example (8) A cone of wood floats in a fluid of specific gravity 0.9 with its apex
downwards. If the specific gravity of wood is 0.6 and weight of it is 300N. find the
weight of steel .If specific gravity 7.6 which should be suspended with the help of a chain ,
so that it can be submerged.
Solution:-
E310/1 52 V = Volume of cone = W/w
= 300/(0.6x9810) = 0.0051m3
FB = Buoyant force
= Volume of water displaced = w x V
=0.9 x 9.810 x 0.051 = 0.45KN
Downward force = upward force
W+ T = FB
∴ Tension in volume
= FB - W
=0.45KN - 0.3KN
=0.15KN
Again (T + Buoyant force over steel ) = weight of steel
But Buoyant force of steel
= Ws
ωsteel
w liquid
Volume of steel x wliquid
(Ws/7.6x9.81) x 0.9 x 9.81 = (Ws x 0.9)/7.6KN
∴ T +( Ws x 0.9)7.6 = Ws
Ws(1-(0.9/7.6)) = T = 0.15KN
∴ Ws = 0.17 KN
Summary
(1) Buoyant force is equal to weight of fluid displaced by the floating body and acts at
the centroid of the volume of fluid displaced.
FB = w V ; Where V = volume of fluid displaced
(2) Buoyant force is equal to the weight of floating body ie FB = W. The line of action
of these two forces will lie on the same to vertical line because these two forces
keep the body in equilibrium condition and so their moment about any axis is zero.
(3) The Metacentre :- When a floating body is tilted through a small heel angle θ, the
line of action of buoyant force through new buoyancy will meet the line external
through the centre of gravity and original buoyancy, at a point. This point is called
Metacentre and the distance of it from the centre of gravity of the body, is called
as metacentric height. is MG as θ →0.
E310/1 53 (4) The floating body will be in(a)stable equilibrium condition if M is above centre of
gravity G.
(b) unstable equilibrium condition ] if M is below G
and (c) Neutral equilibrium condition ] if M = G(M coincides with G)
(5) Metacentric height GM is determined by a keeping a wt ‘w’ at a distance u from
the centre of slip and knowing the heel angle ‘ ‘ by the plumb bob as
Gr = ωx/W tanθ ; where W = weight of ship
(6) Analytical method to determine the metacentric height is as follows .
BM (Distance of buoyancy to Metacentre) = I/V
Where
I = moment of intertia of the body about liquid surface
V = Volume of liquid displaced
Then Metacentric height = GM = BM − BG
BG is the distance between Buoyancy and centre of gravity of the body.
Answers to SAQ (1) True
(2) 98.1KN
(3) True
(4) Because specific weight is more for sea water
(5) 3.46 KN
(6) Stable condition
(7) 1200 Kg
(8) 0.14m
EXERCISE
3.1 A cylinder has a diameter of 0.3m and a sp.gr of 0.75. What is the maximum
permissible length in order that it may float in water with its axis vertical. Obtain
E310/1 54 an expression for the length in terms of diameter ‘d’, sp.gr of cylinder s and sp.gr
of liquid
[0.245m,
dso
8s(so − s) ].
3.2 A conical buoy floating with its apex downward in 3.5m high and 2m diameter.
Calculate its weight if it is just stable when floating in sea water weighing
10055N/m3.[29.127KN].
3.3 A solid concial float of wood weighing 7.5537N/m3 is to float in a liquid weighing
9417.6N/m3. Find the least apex angle in order that it may float with apex down
and its axis vertical [30050’ ]
3.4 A buoy carrying a beacon light has the upper portion cylindrical 2.5m diameter
and 2m deep. The lower portion which is curved displaces a volume of 0.4m3 and
its centre of buoyancy is 2.5m below the top of the cylinder. The C.G of the whole
buoy and beacon in situated 1.5m blow the top of the cylinder and the total
displacement is 19.62KN. Find the matacentric height of the sp. weight 9 sea
water is 10055N/m3.[0.505m]
3.5 A rectangular pantoon weighing 1716.75KN has a length of 20m. The Centre of
Gravity is 0.3m above the centre of cross section and the metacentric height is to
be 1.25m when the angle of heel is 90 .The free board must not be less than o.6m.
when the portion is vertical. Find the breadth & the height of the portion if it is
floating in fresh water .[5.792m, 2.11m].
3.6 A ship displace s 4150 Tonnes when floating in a sea water with its axis vertical
and a weight of 410 tonnes is placed on the centre line. Moving this weight by 1m
towards the hole of the deck causes a plumb hoh suspended from a 2.75m long
string, to move by 30cm. Find the metacentric height of the unloaded, ship.
3.7 A floating buoy in the sea water is floating upright by a submerged weight of
concrete attached to the bottom of the buoy. How many m3 of concrete weighing
22759 N/m3 must be provided to get a net downward pull of 2217N from the
weight., [0.1745m3].
3.8 A hollow cylindrical vessel of diameter 2m and 3m high weighs 29.43KN and its
C.C1 is at the mid point of the longitudinal axis. Shown that it will not float in sea
water with its longitudinal axis vertical. Find to what depth the inside of the buoy
is to be filled by concrete of sp.wt 22.563KN/m3 to give a depth of immersion of
E310/1 55 b2.5m and what is then the meta-centric height sea water weighs 10.055 KN/m3.
[0.6984m, 0.572m].
3.9 A rectangular pontoon 6mm by 3m plan floating in water has a depth of
immersion of 0.9m and is subjected to a torque of 7848N.m about longitudinal
axis. If the centre of gravity is 0.7m up from the bottom, find the angle of heel
[4050’]
References
Same as unit(1).
***
FLUID MECHANICS AND HYDRAULIC MACHINERY
UNIT-IV
KINEMATICS
Aim: To Know about the flow concepts like velocity, accelerations both tangential and
normal, potential theory of fluid flow in which the velocity potential and stream
functions. Finally to know about flow net and practical applications of the above
theory in solving the problems.
Objectives:
(1) Velocity of fluid in explained
(2) To explain convective & local acceleration and to arrive expressions for the same
(3) Then to obtain expressions for normal and tangential accelerations
(4) To define velocity potential and stream functions and to explain their uses in fluid
Mechanics
(5) To explain equipotential lines and stream lines, then flow net and its uses in fluid
mechanics and irrigation.
4.1 Introduction:
Types of flow, stream line, path line and streak lines are known already.
Now the acceleration is defined and the expressions for total acceleration is
arrived. This is necessary in further studies in deriving Euler’s and energy
equations.
E310/1 56 Velocity potential and stream functions are introduced which are useful in
describing the kinematics of fluid flow.
Based on these functions flow net is explained which is vary useful in solving
many problems in irrigation and potential theory.
4.2 Velocity of fluid
Consider the fluid flow occupied by the space (x,y,z) at a point as shown in figure.
Let the particle moves in time ‘dt’ sec through a distance ‘ds’ in the space. Then
the velocity of the fluid in the flow field is
V= L
t dsdt
as dt → 0= V(x,y,z,t)
If it is resolved in x,y,α,z co-ordinates,
Then velocity in x direction =ϑ=Lt
dxdt
as dt → 0= u(x,y,z,t)
Velocity in y direction = v = Lt as dt → 0dydt
= v(x,y,z,t)
and velocity in z direction = w = Lt as dt → 0dzdt
= w(x,y,z,t)
E310/1 57 Thus V = iu+jv+kw
Where i,j,K are unit vectors parallel to x,y,z axes respectively.
4.3 Acceleration in fluid flow
Acceleration in fluid flow is defined as the rate of change of velocity of the fluid.
is a = Lt
dvdt
as dt → 0This acceleration can also be resolved into the 3 components
ax,ayaz in x,y,z directions respectively as given below.
ax = Lt as dt → 0du(x, y, z,t)
dt= Total derivative of u
= ∂u∂x
dxdt
+∂x∂y
dydt
+∂u∂z
dzdt
+∂y∂t
u∂u∂x
+ v∂u∂y
+ w∂u∂z
+∂y∂t
(Connective accl.) (local acceleration)
In the above expression the sum of 3 terms (u∂u∂x
+ v∂u∂y
+ w∂u∂z
)indicates the
change in velocity taking place in space and so it is called as connective
acceleration. And ∂u∂t
is called as local acceleration as the change occurs at a
particular point in ‘dt’ seconds.
Similarly the accelerations in y and z directions are given by
ay = Lt as dt → 0dvdt
=u∂u∂x
+ v∂u∂y
+ w∂u∂z
+∂u∂t
and az = Lt as dt → 0dwdt
= u∂w∂x
+ v∂w∂y
+ w∂w∂z
+∂w∂t
The above expressions are arrived for unsteady non uniform flows.
∴ Total acceleration = connective acceleration + local acceleration
4.3 Tangential and normal acceleration
Consider an unsteady and non uniform flow of fluid along a curved stream lines of
radius ‘r’ as shown in figure.
E310/1 58
At A let the velocity be V and in
‘dt’ seconds after travelling ‘ds’ distance the fluid velocity is V+dv. As shown in figure at
point B, the magnitude and direction of ‘dv’ is obtained graphically by deducting of ‘V’’
from V+dv vertically. It is than resolved along tangential and normal directions to the
stream line, as dvs and dvm respectively.
Further we know that the tangential and normal velocities can be represented as vs(s,n,t)
and vn (s,n,t,) and then the total accelerations along tangential and normal directions are
represented by
Tangential acceleration = as =Lt
dvs
dtdt → 0
=∂vs
∂sdsdt
+∂vs
∂ndndt
+∂vs
dt
as =Vs∂vs
∂s+ vn
∂vs
∂n+
∂vs
∂tNote Vs =V along the stream line.
Similarly , the normal acceleration an is represented by
an =Lt
dvn
dt=
∂vn
∂sdsdt
+∂vn
∂ndndt
+∂vn
∂tdt → 0
=vs
∂vn
∂s+ vn
∂vn
∂n+
∂vn
∂tAccording to the definition of stream line, the flow will take place along stream
line only, but perpendicular to stream line, the velocity is zero. as there is no flow in that
direction.
E310/1 59 ∴υn = 0
Thus as = ϑ s
∂ϑ s
∂s+
∂ϑ s
dt
and an =ϑ s
∂ϑ n
∂s+
∂ϑn
∂t
It may be noted here that though ϑn =0, dϑn is not equal to zero as it is the
component of ‘dv’ in normal direction.
If the flow is stready then
as = ϑ s
∂ϑ s
∂sand an = ϑ s
∂ϑ n
∂s
Now from the figure we have, from the triangle BCD, tanθ =θdϑ n
v(dvs is
neglected as it is small)
For he curve with radius of curvature ‘υ’ . ds = υθ is θ= ds/υ
∴ equating both expressions
dϑ n
ϑ s
=dsυ
dϑ n
∂s=
ϑ s
rSubstituting this expression in normal acceleration An, we have
an =ϑ s
∂ϑ n
∂s= ϑ s
ϑ s
r=
Vs2
r=
V 2
r
and as = V dvds
Q Vs = V
SAQ (1) In steady flow the velocity along a curved stream line of radius of curvature 3m,
is 3m/s and it changes over 1m distance of the stream line into 3.2 m/s. Find the
tangential and normal accelerations.
SAQ(2) In a converging cone a fluid is flowing with a velocity of 1m/s at inlet section and
3m/s at out let section under steady condition, over a distance of 2m. Determine
the tangential velocity.
SAQ(3) For steady uniform flow, the acceleration is how much?
SAQ(4) There is no flow perpendicular to stream line . True/False.
Worked Example (1).
E310/1 60 A sheet of water is flowing over a curved bucket with a radius of curvature of 10 m with a
velocity of 30m/s
Find the normal acceleration
an =v 2
r=
30 × 3010
= 90m / s 2
Worked Example (2) A converging cone is changing in diameter from 2m to 1m over a
distance of 2m. The discharge changes from 15m3/s at t = θ sec., to 75 m3/s at t= 3
seconds. Find the total acceleration at t = 0 at a section where the diameter is1.5 m.
Solution
We know that dia at a distance x is
dx = 2-(2-1)x/2
= 2-0.5 x
Ax =Π(2 − 0.5x)2
4
Vx =QAx
=15
Π4
(2 − 0.5x)2=
60Π(2 − 0.5x)2
dVx
dx=
60Π
(−2)(2 − 0.5x)−3 (−0.5)
VX dVx
dx=
60Π(2 − 0.5x)3 ×
60Π(2 − 0.5x)
=364.7
(2 − 0.5x)5
Local acceleration = dVx
dt=
1A
dQdt
=60
Π(2 − 0.5x)2 ×(75 −15)
3=
25.5(2 − 0.5x)2
when t = 0, total acceleration as
as =Vx dvx
dx+
dvx
dt
E310/1 61 =
364.7(2 − 0.5x)5 +
25.5(2 − 0.5x)2
Given dx =1.5m
∴1.5 = (2-0.5x)
x=1m
at t=0
as =364.7
(2 − 0.5 ×1)5 +25.5
(2 − 0.5 ×1)2
=47.99+11.33 = 59.32 m/s2
Worked Example (3)
In two dimensional flow the velocity components are as follows
u = 2xy+3t2 +6
ϑ=-2xy2-15t
Find the acceleration at a point (1,2) when t = 2 seconds
Solution:
ax = u∂u∂x
+ϑ∂u∂y
+∂u∂t
ax =(2xy+3t2+6)(2y)-(2xy2 +15t)(2x)+6t
Given x = 1, y = 2, t = 2
ax = (4+12+6)(4)-(8+30)(2)+12=-20m/s2
ay =u∂u∂x
+ϑ∂u∂y
+∂ϑ∂t
=(2xy+3t2 +6)(-2y)2 -(2xy2+15t)(-4xy)-15
=22(-8)-38(-8)-15
=-495m/s2
a= ax2 + ay
2
202 + (495)2
=495.4 m/s2
E310/1 62
θ = tan −1 ayax
= tan −1 1 − 495
20
θ = tan −1 (−24.75)= −870.68
4.4 Stream function and velocity potential function
For 2 dimensional flow a stream function is defined as a continuous function of the
space (x,y) such that the partial derivative of it with respective to any direction gives
velocity component 900 to that direction in anticlockwise direction.
ie u = −∂ψ∂y
ϑ = +∂ψ∂υ
Its value is constant along a stream line. ie when stream function is equated to a series of
constants, it yields equations of series of stream lines.
Along a stream line
ψ = constant
Qϑ = 0 =
+∂ψ∂x
= 0
∂ψ = 0
ie ψ = constant along stream line)
ie dψ =∂ψ∂x
dx +∂ψ∂y
− dy = 0 (1)
along a stream line
E310/1 63 But equation of stream line, we know as
ϑ/u = dy/dx ie ϑdx-υdy=0 (2)
Compare (1) and (2) , we have
ϑ=∂ψ∂x
and u =-∂ψ∂y
shows that ψ is constant along a stream line.
Further consider the discharge flowing through a stream tube with ψ , and ψ2 as boundary
stream lines, as shown in figure. Now consider the discharge flowing across the curve
ACB is given by
dQ = -udy+ϑdx = dψ (here flow is taken +ve in anticlockwise.)
∴ charge in the stream function is equal to the discharge flowing between the two stream
lines for which ψ 1, and ψ2 are the stream functions passing through the points A&B
respectively.
Consider continuity equation as given below for stready incompressible and continuous
fluid
∂u∂x
+∂ϑ∂y
= 0
Substituting the expressions for uαϑ in the above equation,
−∂ 2ψ∂x∂y
+∂ 2ψdy∂x
= 0
ie −∂ 2ψ∂x∂y
+∂ 2ψ∂y∂x
E310/1 64 This is true if ψ is continuous . So it can be stated that when fluid flow is continuous, ψ
also continuous which may be irrotational or rotational. So if it is be irrotational, we have
rotation about ‘2’ axis w2
w2 =12
∂ϑ∂x
−∂u∂y
= 0
ie ∂ϑ∂u
−∂u∂y
= 0
Substituting the expression for u ∂ ϑ we have
∂ 2ψ∂x2 +
∂ 2ψ∂y2 = 0
This is known a Laplace equation in ψ. So it replace equation in ψ is satisfied than the
fluid flow is irrotational, otherwise it is rotational only. So ψ exists both for rotational
and irrotational fluid flows.
Velocity potential ‘φ’ is defined in 2D flow as a continuous function of space (x,y) such
that its -ve a have partial derivative gives velocity in that direction.
ie −∂φ∂x
= u;-∂φ∂y
= ϑ
Substituting these in continuity equation for steady incompressible fluid flow we
have
∂u∂x
+dϑ∂y
= 0
∂ 2φ∂x 2 +
∂ 2φ∂y2 = 0
Which is Laplace equation in φ. So if Laplace equation in φ is satisfied the fluid is flow is
continuous and so ‘φ is continuous function. Further substituting the above expressions
for u & ϑ in condition for rationality we have
∂υ∂x
−∂υ∂y
= 0
ie -∂ 2φ∂x∂y
+∂ 2φ∂y∂x
= 0
E310/1 65
ie ∂ 2φ∂x∂y
=∂ 2φ∂y∂x
is true only and is continuous . That means, if φ exists it is
continuous, then it should represent only irrotational fluid flow. So φ exists only for
irrotational fluid flow, ie for ideal fluid flow. So ideal fluid flow theory is also called as
potential flow theory.
If the equation for velocity potential is equated to series of constant we have series
of equipotential lines, for which the velocity potential function is constant.
SAQ (5) Define ψ and φ functions.
SAQ(6) φ exists only for irrotational flow? True / False
SAQ(7) Irrotational flow is called potential fluid flow. True / False
SAQ (8) ψ = 2xy, find the discharge through the points (1,1) and (1,2)
SAQ(9) ψ = 3x2 -y3 , find u at (1,2)
SAQ(10) φ = x2-y2 find u at (1,2)
SAQ(11) If ψ exists it may be either rotational or irrotational. True / False
Worked Example (4)
Given φ = x2 -y2 , find the velocity at a point (1,2) and its direction.
u =−∂φ∂x
= -(2x)=-2×1=-2 units
ϑ = −∂φ∂y
=-(-2y)=2y=2×2=4 units
V = u 2 +ϑ 2 − 4 +16 = 4.47m / s
Direction with horizontal = θ =tan −1 ϑu
= tan −1 y
−z
= tan −1 −z( )
Worked Example 5
Given φ = x2 -y2 , find ψ and discharge through (1,1) and (1,2) point
Sol:
E310/1 66
−∂φ∂x
= u =−∂ψ∂y
ie ∂φ∂x
=∂ψ∂y
−∂φ∂y
= ϑ =∂ψ∂x
∂φ∂y
=−∂ψ∂x
φ = x2 -y2
∴∂φ∂x
= 2x =∂ψ∂y
Integrating w.r.to y, we have
ψ = 2x dy∫ = 2xy+const
=2xy+f(x)
∴ψ = 2xy+f(x)
Now differentiating ψ w.r.to x,
∂ψ∂x
= 2y+ f x( )( )
dx= −
∂φ∂y
)=-(2y)=2y
∴d(t(x)
dx= 0
f(x) = const
∴ψ = 2xy+constant
Now ψ 1 = 2 ×1 × 1 = 4
(1,1)
ψ 2(1, 2)= 2 ×1 × 2 = 4
∴dψ =dφ=ψ2 -ψ1 =4-2=2units
Worked Example (6)
Verify whether ψ = 2x2y2 is a possible irrotational flow . If so find velocity at (1,2).
Solution:
For flow to be continuous
∂ 2ψ =∂ 2ψ∂x ∂y
=∂ 2ϑ∂y∂x
∂ψ∂x
= 4xy 2 ∂ψ∂y∂x
= 8xy
∂ψ∂y
= 4x 2 y∂ 2ψ∂x∂y
=8xy
∴It is continuous and so ψ represents a possible flow.
E310/1 67 Now for irrotational flow Laplace.
equation in ψ to be satisfied.
ie ∂ 2ψ∂x2 +
∂ 2ψ∂y2 = 0
∂ 2ψ∂x2 = 4y2 ∂ 2ψ
∂y2 = 4x 2
∴∠p is ψ ± 0
So it is rotational flow only.
Vol at (1,2)
u =−∂ψ∂y
= −(4x 2 y)
=-(4×1×2)=-8 units
ϑ =∂ψ∂x
= 4xy2 = 4 × 1× 22 =16
V = u2 + ϑ 2 = (−8)2 + 162 =16.5 units.
Flow net
we know that
∂ψ∂s
= ϑ n = 0
Qψ is constant along ‘s’ direction
that means, there is no flow normal to stream line.
Also ∂ψ∂u
= −Vs
Similarly −∂ψ∂u
= ϑ n = 0
as φ is constant along u direction
−∂φ∂s
= ϑ s
Thus Ψ = different constants, gives a series of stream lines
and φ = different constants gives a series of equipotential lines, as already explained
E310/1 68
. Now let us consider the slope of these lines. The slope of φ lines is
= dydx
=∂φ∂φ
×dydx
=∂φ∂s
oφ oy =
−u−ϑ
=uϑ
Similarly the slope ofψ lines
dydx
=∂ψ∂ψ
×dydx
=∂ψ∂x
∂ψ ∂y =
ϑ−u
∴(slope of φ lines) ×(slope of ψ lines)
=uϑ
×−ϑu
= −1
This shows that the φ lines and ψ lines meet each other orthoganally at any point, giving
rise to square mesh as shown in figure.
This grid of stream lines and equipotential lines is called as flownet. Thus a
flownet is defined as a grid of stream lines and equipotential lines which are orthogonal
to each other at every point of intersection. It is applicable to ideal fluid flow. But except
in Boundary layer, it can be applied even to real fluid flows.
Uses of flownet
1. If a flow net is known for a flow field, then the velocity distribution over the entire
flow fields can be determined as follows.
E310/1 69
Between any two stream lines the discharge flowing perunit thickness of flow is known
as dψ=ψ2 -ψ1 etc. The mesh size at any two points are known as ‘du’ and du2 . Then
ϑ 1 =dψ
du1 ×1( )&ϑ 2 =dψ
du2 ×1( ) are known
In this away at any point the velocity is known in the entire fluid field for which the flow
net is known.
2 . Then by application of Bernoulli’s equation, the pressure at any point can be calculated
in the flow field.
2. Using flow net theory, the seepage through earthen dams and foundations can be
calculated as shown in figure. This is useful in the design of earthen dams.
3. The uplift pressures and exit pressure gradient downstream of irrigation structures like
weirs, barrages etch can be known. These are very useful in the design of the irrigation
structures, as shown in figure.
SAQ(12) Define flow net.
SAQ(13) The mesh size at a point in flow net is 1m where the velocity is 3m/s. Find
the velocity at another point in the flow net where the mesh size in 1/2 m.
SAQ (14) Mesh size is more, the velocity is less. Is it true? True/False.
E310/1 70 Summary
1. Expression for total acceleration in direction is written as
Lt
DυDt
=u∂u∂x
+ϑ∂u∂y
+ w∂u∂z
a1 24444 34444
curvature acceleration
+∂u∂tb
local acceleration
dsdt → 0
Similarly in other directions, y& Z the expression for
DvDt
&DwDt
can be written.
2. For steady flow, the tangential and normal accelerations are given by
as =vdvds
(tangential acceleration)
an = v2/r (normal acceleration)
where ‘r’ is the radius of curvature of the stream line and V is the Velocity of
flow .along the stream line.
3. Stream function ‘ψ’ is defined as function of space (x,y for 2D flow, such that
its partial derivative gives the velocity in 900 to its direction in anticlockwise
direction.
∂ψ∂x
= ϑ -∂ψ∂y
= u
4. Velocity potential φ is defined as a function of space such that its -ve partial
derivative in any direction given velocity in that direction.
ie −∂φ∂x
= u -∂φ∂y
= ϑ
5. Velocity potential exists for irrotational flow only
6. Flow net is a grid of stream lines and equipotential lines such that they meet
orthogonally at any point of intersection giving rise so square mesh.
ie slope of φ lines × slope of φ ψ lines = -1
7. Using flow net velocity distribution, pressure over a flow fluid can be determined.
It can also be used to determine seepage flow through earthen dam through
foundations, uplift pressures below irrigation structures and exit pressure
gradient.
Answers for SAQ
E310/1 71 1.as =0.6m/s2 an =3m/s2
2.as = 1m/s2 an=3m/s2
3.ax =0 4. True 6. True 7.True
8.2 units 9.6m/s 10.-2m/s
11.True 13.12m/s 14.True
EXERCISE
4.1 The velocity along the centre line of nozzle of length L s given by v=4 t(1−x
2L)2
Where ‘v’ is the velocity in m/s , t = time in seconds, x = distance from inlet. Find
total acceleration when t=3s,x=0.5m,L=0.8m. [233.023m/s2].
4.2 A 90cm diameter pipe is reduced to 30cm diameter in a length of 1.5m. Water flows
through this pipe at a rate of 280 lps. If at an instant the discharge is found
to decrease at the rate of 60 lps estimate the total acceleration at a distance of
75cm from the 90cm diameter inlet.. [1.0953 m/s2].
4.3 In a steady flow two points A and B are 0.5m apart on a straight stream line of the
velocity of flow varies usually between A & B, what is the acceleration at each
point if the velocity at A is 2m/s and the velocity bat B is 6m/s.
[at A=16m/s2 , at B=48m/s2]
4.4 Check whether the following sets of velocity components satisfy the continuity
equation of steady incompressible flow.
(a) u=4x+2y-3, v=2x+4y+3
(b) u=4xy+y2, v=6xy+3x+2\
(c) u=2x2+y2, v=-4xy
(d) u=x3+y3, v=x-3x2y
(e) u=c(y2 − x 2 )(x 2 + y 2 )2 , , v=ϑ =
−2cxy(x 2 + y2 )2
(f) u=(3x-y), v=(2x+3y)
(g) u=A Sin xy v=-A Sin xy
[not satisfied -a,b,c,f,d ; satisfies -c,d]
4.5 Calculate the missing utilising velocity component, so that they satisfy the
continuity equation
E310/1 72 (a) u=2/3xy3-x2y ; (b)u=Ayex
v=? v=?
(c) u=2x2,v=xy2, w=1
(d) u=(2x2+2xy), v=? , w=(x2 -4xz-2yz)?
(e) u = 3x2 , ϑ=-x2y-yz-xy, w-?
(f) u=x3+y2+2z2, ϑ=-x2y-yz-xy, w=?
[(a) =v-1/6y4+xy2, (b) ϑ=-−Ay 2
2ex + f (x)]
(c) w=-4xz-xz2+f(x,y)
(d) -3yz2+f(x,z) (e) w=-(6xz+2xz2)+f(x,y)
(f) w=2x2z-z22-xz+f(x,y)
4.6 Determine which of the following velocity fields represent possible example of
irrotational flow.
(a) u=cx, v=-cy
(b) u=-cx/y, v=c logxy
(c) u=(Ax2-Bxy), v=(-2Axy+1/2By2)
[(a) irrotational,(b)(c) rotational]
4.7 Verify whether the following flow fields are rotational. If so determine the
components of rotation about verious axes
(a) u=xy2, v=zx, w=1/2yz2-xy
(b) u=xy, v=1/2(x2-y2)
[(a) Rotational w2 = 1/2 z(1-x)
wx=1/2(z2/2 -2x)
wy=1/2y(x+1)
(b) irrotational wx=wy=0
4.8 For the following sets of velocity components obtain the relevant stream
functions.
(a) u=6y, v=6x, (b) u=-Alnx, v=A(y/x)+x2
(c) u=y3/3+2x-x2y, v=xy2-2y-(x2/3)
[ (a) Ψ = 3(y2 − x 2 ) + c
(b) Ψ =y 2
12−
x 2 y 2
2+ 2xy +
x 3
9
E310/1 73
(c) Ψ =−Ay | nx
mx−
x 3
3+ c ]
4.9 Which of the following stream functions ψ are possible irrotiational flow fields.
(a) ψ=Ax2+By2 (b) ψ=Ax2y2
(c) ψ=A sinxy (d) ψ=Alog(x/y)
(e) ψ= Ay(1 −c
x2 y2 ) (f) ψ=(y2-x2)
[(a) & (d) - None, (e) & (f) are possible irrotational flows]
4.10 Verify whether the follwing functions are valid potential functions
(a) φ=y3-3x2y (b) φ=y4-6x2y2
(c) φ=x2-3x2y (d) φ=x3-y3
[ (a) yes., (b) (c) (d) - No ]
4.11 Determine the velocity potential functions for the following . Also determine the
discharge flowing between the stream lines passing through the points (1,3) and
(3,3)
(a) ψ=3xy (b) 3/2(y2-x2) (c) ψ= x2-y2
[ (a) φ=3/2(y2-x2) 24 units (b) φ=-3xy Zero (c) 8 units ]
4.12 Calculate the velocity at the point(3.3) for the following stream functions
(a) ψ=-xln x y + x (b) ψ=1/2(y2-x2)+xy+6
[ (a) v=2.42units (b) v=6 units ]
4.13 Determine the stream functions, ‘ψ‘ for the following velocity potential functions
φ. Calculate the velocity at (2,2).
(a) φ=x + y (b) tan-1(y/x) (c) φ=A x/ (x2+y2)
[ (a) ψ=y-x + constant (b) ψ=-1/2 log(x2+y2)+constant
v=√2 units v=1/√13 units
(c) ψ=-A y/((x2+y2)+constant)
v =A/√169 ]
***
E310/1 74 FLUID MECHANICS & HYDRAULIC MACHINERY
UNIT-5
DYNAMICS OF IDEAL FLUIDS
AIM:
To list the various forces causing the fluid flow, to identity the important fluid
forces viz. body forces and surfaces in deriving Euler’s equation of motion and to derive
Bernoulli’s equation of motion by integrating Euler’s equation. Finally to apply
Bernoulli’s equation in solving fluid flow problems.
OBJECTIVE:
1) The various forces causing the fluid flow are listed and name the two forces
namely, body forces and surface forces
2) To formulate the Euler’s equation
3) To derive the Euler’s equation along a stream line considering ideal flow.
4) To derive the Bernoulli’s equation by integrating the Euler’s equation along stream
line. Here the assumptions necessary to derive this equation, the modifications and
limitations are discussed .
5) Kinetic energy correction factor is derived.
6) To derive momentum equations based on Newton’s 2nd law of motion.
7) To apply the above equations in solving fluid problems.
5.1 INTRODUCTION
In dynamics of ideal fluid flow the viscosity is neglected and so the loss of energy
due to friction is zero. The various fluid forces are listed to derive Euler’s equation and
hence. Bernoulli’s equation based on Euler’s equation using Newton’s 2nd law of motion
The importance of Bernoulli’s equation is discussed by applying it to solve many fluid
problems. Along this equation continuity equation is also useful in solving the problems.
The third important equation in fluid mechanics is momentum equation. This is
also very useful in solving the fluid problems, When Bernoulli’s equation along with
continuity
E310/1 75 equation can alone not solve the fluid problems. This equation is very useful in Hydraulic
machinery.
5.2 DIFFERENT FORCES CAUSING THE FLOW:
The different forces causing the motion are listed below.
Fp Pressure force which is a function of fluid motion.
Fg Gravity or body force is due to weight of fluid flowing
Fo Viscous force is due to viscous shear action of fluid. It depends on
viscosity of fluid. So for ideal fluid it is zero.
Ft Turbulent force is due to the turbulence of fluid. In turbulent flows due to
intermixing of fluid particles, these forces are developed. It exists only for
real fluids and so are zero for ideal fluid flows.
Fe Elastic force is due to elastic compression of the fluid. This force is
involved only in compressible fluid flows.
Fσ Surface tension force is due to surface tension of the fluid, Which is
extremely fluid film at interfaces between two different fluids.
We know that by Newton’s 2nd law that the sum of all external forces acting as the
fluid is equal to the rate of change of momentum in the direction of forces. That means
∑F = Mass × acceleration.
In majority of flows surface tension forces, elastic forces are negligible and so neglected.
Thus Fp + Fg + Fϑ + Ft = ma , is known as Reynolds equation because turbulance of fluid
is involved in this equation. If this is neglected then.
Fp + Fg + Fϑ = ma is called as Navier-Stoke’s equation because viscous forces are
involved in this equation. The above two equations will refer to real fluid flows. If
viscous forces are neglected then we have
Fp + Fg = ma is known as Euler’s equation for ideal fluid flows where velocity is
zero.
E310/1 76 5.3 EULER’S EQUATION OF MOTION:
Now let us know about the two forces causing the fluid flow in derivation of
Euler’s equation for ideal flow.
Surface forces are due to (stress x surface area). Further these forces are
subdivided into normal forces called pressure forces and tangential forces, which are due
to viscous shear force. As viscosity is zero for ideal fluid, the tangential forces are zero
and so surface forces in ideal fluid flows, are only pressure forces which are normal forces
obtained by (pressure x C.S area of flow)
Body force is proportional to volume of body, which is due gravity i.e. weight of
fluid is considered as body force.
Now consider a small elemental fluid prisum of C.S area dA and length ‘d s’ in
fluid flow along a streamline as shown in figure for steady incompressible, ideal fluid
flow. As the flow
Fig 5.1
considered is ideal, mean velocity is considered over the cross section of fluid flow. The
particles move ‘ds’ distance in time dt seconds.
We have by Newton’s 2nd law that
∑ Fs = m × as along streamline
Fp + Fg = m × as
The sum of pressure force and body force along streamline = mass × acceleration along
streamline.
Fp = pdA - p +dbds
ds
dA =
−dpds
dsdA
E310/1 77 Fg = sPdAds
Where ‘S’ the component of body force along streamline
∴−dpds
dsdA - sρdAds = ρdAds × as
Now dividing the above equation throughout by mass of the fluid, we have −1ρ
dpds
- S = as
which is known as Euler’s equation along a streamline
Body force along ‘S’ direction is the weight component of weight of fluid along
streamline.
i.e. Fg = ρgdAds cos θ
but dzds
= cosθ
∴Fg = ρgd A.ds. dzds
= sρd A.ds
i.e. S = gdzds
Further for steady flow acceleration along stream line is
as =Vdvds
+Avdt
; Where dvdt
= 0 for steady flow ∴ as =Vdvds
where ‘v’ is the
velocity of fluid along steam line over the cross section, which is uniform so
as =d v 2 2( )
dssubstituting the above in the Euler’s equation we have
−1ρ
dpds
−gdzds
=d V 2 2( )
ds
i.e. 1ρ
dpds
+gdzds
+d V 2 2( )
ds= 0
Derivation of Bernoulli’s equation
Now integrate the above equation with respect to s, we have energy equation,
because Force × distance = work done
Thus 1ρ∫
dpds
ds +gdzds
ds +d V 2 2( )
ds∫∫ ds= constant
E310/1 78 i.e.
1ρ∫ dp + gdZ + d V 2 2( )∫∫ = C
As the flow is incompressible, ρ is constant along the flow, so it is taken out of integration
as
1ρ
dp + gdZ +∫∫ d V 2 2( )∫ = C
i.e. pρ
+ gz + V2 2 = C
or pg
ZV
gC
ρ+ +
21
2
i.e. pw
ZV
g+ + =
2
2constant
This is called as Total head at a point on the stream line or total energy per unit weight of
fluid.
Total head = pressure head + datum head + velocity head
The above equation is called as Bernoulli’s equation which states that for steady
incompressible continuous fluid flow the total head is constant at any point along its flow.
This equation in the conservation of energy of fluid, as discussed below.
Consider pw
= pressure head
We know the work done by pressure is given by
p × a× v =pw
wav = pw
( weight / sec )
∴pw
is (W.D / sec ) per unit weight of fluid = pressure energy / unit weight = pressure
head = mkg/kg work done by elevation of the particles above a datum is given by potential
energy = Mass × g × Z.
Where Z is the elevation of the fluid particles above the datum W.D by elevation of fluid
= (ρAV) gZ
ω AVZ = W/ secxZ
∴Datum energy = W/sec xZ
ρPotential energy or Datum head = Datum energy / unit weight = Z m-kg/kg
E310/1 79 work done by movement of fluid is
Kinetic energy
= 12
mass × V2
=12
× ρAV × V2
=12
×wg
AV× V2=V 2
2g( w/sec) = m.kg/sec
or kinetic energy / unit weight = V 2
2gm.kg/kg
∴ + +
W
pw
ZV
g
W
2
2= constant
or Total energy / unit weight = constant
i.e. Total head = constant
i.e. pw
ZV
g+ +
2
2= constant which is called as Bernoulli’s equation.
ASSUMPTIONS
In the derivation of Bernoulli’s equation.
(1) Fluid is assumed to be ideal and friction less is µ = 0 and hL = 0.
(2) Fluid flow is steady flow , so that local acceleration is dVdt
= 0.
(3) Fluid is incompressible ρ is constant along its flow.
(4) Fluid flow is continuous, i.e mass rate flow is constant over the cross
section of flow.
(5) Fluid flow is considered as ideal and so mean velocity is considered over
the cross sectional flow. This mean velocity is constant over entire C.S. of
flow.
LIMITATIONS
(1) It is applicable to only ideal fluid only .
E310/1 80 (2) It is considered as ideal and so velocity over the C.S is considered as
constant which is not true in the actual fluid flow.
(3) It is applicable to steady flow only
(4) It is applicable to particles moving along a stream line, but total head is
varying from stream line to stream line.
(5) Energy added or subtracted from the control volume, is taken as zero.
MODIFICATIONS
(1) For real fluids there will be loss of energy due to fiction, which is to be
considered in the equation as
Z1 + + = + + +−
pw
Vg
Zpw
Vg
hL1 1
2
22 2
2
2 2 1 2where hL1−2
is the constant
of the energy from (1) to (2)
(2) Between two streamlines the centrifugal head added is to be considered.
(3) The energy added or subtracted is to be considered in the equation as
Z1 +pw
Vg
1 12
2+ + Hadded / unit weight = Z2+
pw
Vg
2 22
2+ (Pump)
Z1 +pw
Vg
1 12
2+ - Hadded / unit weight = Z2+
pw
Vg
2 22
2+ (Turbing)
(4) The velocity distribution over the c-s is not uniform. ∴ Kinetic energy
correction factors are to be considered in the equation.
Z1+pw
Vg
Zpw
Vg
hL1
112
22
22
2
2 2 1 2+ = + + +
−α α
where ‘α1’& ‘α2 ’ are called as energy correction factors
KINETIC ENERGY CORRECTION FACTOR
E310/1 81 In the derivation of Bernoulli’s equation mean velocity which is uniform over
entire c-s flow is considered. But in actual fluid flow , it is non uniform as shown in the
figure.
Fig 5.2
Thus kinetic energy calculated on the basis of mean velocity is less than the actual kinetic
energy. So a correction factor is introduced to get actual kinetic energy as αV 2
2gwhere ‘α‘
is the kinetic energy correction factor which is derived as shown below.
Kinetic energy based on
uniform velocity distribution = 12
( mass ) V2
=12
2wg
Avv
=12
3wA
Vg
The actual kinetic energy is calculated based on the actual velocity distribution over c-s
flow, is calculated as shown below
consider an elimental are dA on which the velocity of flow is v
∴dr = mass flowing through this elemental area = wg
dAv
K.E of this mass = 12
2wg
dAv v
×
And by integrating this over the complete c-s of flow, we have
Actual K.E = 12
12
2 3
A A
wg
dAv vg
v dA∫ ∫× =ω
.
E310/1 82 Thus the cube of mean velocity over the area is less than the cubes of instantaneous
velocity over the area. ∴ The kinetic energy of the flow based on mean velocity is to be
multiplied by a factor ‘α‘ to correct the K.E. This factor is known as kinetic energy
correction factor is given by
α = =∫ ∫
12
12
3
3
3
3
wg
v dA
wAv
S
v dA
AvA A
Here α >1.0
For turbulent flow α = 1.03 −1.06 and for laminar flow α = 2
In turbulent flow α can be taken as 1.0
SAQ(1) In Euler’s equation only pressure forces and body forces are considered in
steady flow condition.
True/False
SAQ(2) In I-D flow mean velocity over the C.S is considered
True/False
SAQ(3) In Euler’s equation the terms are per unit mass
True/False
SAQ(4) Bernoulli’s theorem studies the conservation of energy
True/False
SAQ(5) In Bernoulli’s theorem the terms are energises per unit weight of flow per
second
True/False
SAQ(6) Bernoulli’s theorem is applicable to ideal steady incompressible continuous
flow
True/False
SAQ(7) The unit of term is Bernoulli’s theorem is, m, N - m/N, Jouls/N
True/False.
SAQ(8) Bernoulli’s heads are also called as energies/unit weight
True/False.
SAQ(9) Is Bernoulli’s theorem applicable to rational flow along a stream line
Yes/No.
E310/1 83 SAQ(10) Is Bernoulli’s theorem applicable to real fluid flows
Yes/No.
SAQ(11) Why the flow in the derivation of Bernoulli’s theorem is to be steady ?
SAQ(12) Why flow is to be incompressible in derivation of Bernoulli’s theorem?
SAQ(13) Why the kinetic energy correction factor is greater than 1?
SAQ(14) Given the total head at a point is 10m of water and at another point the
piezometer head is 8m. Find the velocity at second point?
SAQ(15) If the total head at (1) is 10m and at (2) is 8m, Find the head loss and
direction of flow?
SAQ(16) If the total head at (1) is 10m of water. Find the pressure at another point in
kpa, Which is 2m above point (1) if the velocity head at this point is 1m?
SAQ(17) A point in a pipeline which is in the form of a syphon , is 2m above the
water level in a tank . The velocity head in the pipe is 1m of water. Find the
pressure at that point?
SAQ(18) Find the maximum elevation of the summit of a pipe line which is in the
form of a syphon, above water level in the tank, if the velocity head in the
pipe is 1m of water. Take vapour pressure as -8m of water?
WE(1) A Pipe line carrying oil of specific gravity 0.9 changes in diameter from 20
cm at A to 50cm diameter at B which is 4m above A. If the pressure at A and B are 100k
E310/1 84 pa and 60kpa respectively and the discharge flowing is 200 lps, determine direction of
flow and loss of energy.
SOLUTION
Fig 5.3
Velocity at A = ( )
2001000
40 2 2×
π .
6.34m/s ;VA
2
2g2.066m
Velocity at B = ( )
2001000
40 5 2×
π .= 1.0185m/s ;
VB2
2g=0.052
Now applying Bernoulli’s theorem between A and B taking datum through A, we have
Total head at A = ZA +Pw
Vg
A A+ = +××
+2
20
100 10000 9 9810.
2.066 = 13.4m of oil.
Total head at B = ZB +Pw
Vg
A A+ = +××
+2
24
60 10000 9 9810.
0.052 = 10.85m of oil.
Energy at A > Energy at B ∴Flow is from A to B and the loss of head.
= HA - HB
= 13.4 - 10.85
= 2.55m of oil.
WE(2) A conical tube 1.5m long is fixed vertically with upper end which is smaller in dia
in a pipe line. When a liquid of sp gr 0.9 flows down the tube at a of 294 l.p.s the velocity
at upper and lower ends are 4.5m/s and 1.5m/s respectively. The pressure head at upper
E310/1 85
end is 3m of liquid. The loss of head in the tube is 0.3 ( )ϑ ϑ1 2
2
2−
g, where ϑ1 and ϑ2 are
velocities in upper and lower ends respectively. Calculate the pressure head in Kpa at
lower end and diameters of the tube.
SOLUTION
Fig 5.4
Discharge = Q = 295
1000m/s
a1 =Qϑ1
=2951000
×1
4.5=0.0655m2
d1 = 28.9cm
a2 =Qϑ2
=295
1000×
11.5
= 0.1967m2
d2 = 50cm
Now apply Bernoulli’s theorem between A & B taking datum through B,
ZA +VA
2
2g+
PA
ω= ZB +
VB2
2g+
PB
ω+ hAB
1.5 + 4.5( )2
2 × 9.81+ 3 = 0 +
1.5( )2
2 × 9.81+
PB
ω+
0.3 4.5 −1.5( )2
2 × 9.81
1.5 + 1.03 + 3 = 0+0.115 + PB
ω+ 0.138
PB
ω= 5.53 - 0.253 = 5.28m of liquid
PB = 5.28 × 0.9 × 9810 = 46.617 kpa
E310/1 86 WE(3) A 5m long vertical conical pipe with upper and lower diameters as 0.3m and 0.6m
respectively is submerged in water in a sump, such that the depth of water above the lower
end is 1.75m. If the pressure at upper end is 30cm of mercury vacuum and the loss of head
in the pipe is 1
10of K.E at the upper end, calculate the discharge and pressure head at
lower end. The water in the sump is exposed to atmosphere.
SOLUTION
Apply Bernoulli’s theorem between A and B taking datum through A
Fig 5.5
PwA + ZA +
Vg
Pw
A B2
2= + ZB +
VB2
2g+ hf A. B
Pw
Vg
Vg
Vg
A A B B+ + =−
××
+ + +02
30100
136 98109810
52
012
2 2 2..
= - 4.08 + 5 + 1.1 VB2
2g
a = aA vA = aB vB
VA =aBvB
aA=
0.30.6
2
× vB =vB
4
To find PwA , apply Bernoulli’s theorem between (A) and free surface, take datum through
‘A’
E310/1 87
0 + Pw
Vg
A A+2
2= 1.75 +0 + 0
Pw
Vg
A A+2
2= 1.75 , substituting this in above equation
1.75 = - 4.08 + 5 + 1.1 VB
2
2g
VB2
2g=
0.831.1
= 0.7545
VB = 3.87m/s
Discharge = Π 0.3( )2
4× 3.87 = 0.272m3/s = 272lps
PwA = pressure at A = 1.75 -
VA2
2g= 1.75 −
VB4
2×
12g
= 1.75 −3.87
4
2×
119.62
= 1.708
m
WE(4) A 2.5cm dia pipe is fixed to the end of a 10cm dia pipe discharging water from a
tank to atmosphere. The depth of water in the tank over the centre line of the nozzle is 4m.
The loss of head in the pipe is given as 20v2
2g, where v is the velocity in the pipe. Also
determine the pressure at the base of the nozzle.
SOLUTION
Fig 5.6
Apply Bernoulli’s. theorem between 1 and 3 datum through 3
E310/1 88
Pw
ZV
gPw
Zg
ghL
11
12
33
32
2 2 1 3+ + = + + +
−
0 + 4 + 0 = 0 + 0 +V3
2
2g+
20v22
2g
But by continuity equation, we have
A2 V2 = As V3
V2=A3
A3
V3 =2.510
2
× v3 =1
16v3
Substitute this in the above equation, then
4 = V3
2
2g+ 20 ×
116( )2
V32
2g= 1.07813
V32
2g
V3= 8.53m/s ; V2 =1
16× V3 = 0.5332m/s
∴Discharge Q = Π4
0.025( )2 × 8.53 = 4.19lps
Loss of energy = hL1−3 = 20 ×V2
2
2g= 20 ×
0.5332( )2
19.62= 0.29m
Now apply B. Theorem between 1 and 2
4 02
202
2 22
22
= + + +Pw
Vg
Vg
4 =( )P
wVg
Pw
Pw
2 22
22
2212
210 5332
19 620 3+ = + × = +
..
.
∴Pw2 4 03 3 7= − = .
i.e. P2 = 3.7 × 9810 = 36.3 KN/a2 (kpa)
WE(5) A siphon consisting of a pipe of 2cm dia is used to empty water from a tank which
discharges into atmosphere at a point 3m below the water level in the tank. The highest
point in the siphon is 2m above the water level. Determine the discharge and pressure at
highest point the siphon. The losses unto the highest pt is 0.5m and unto the end of pipe is
1.5m
E310/1 89
Fig 5.7
SOLUTION
Apply Bernoulli’s Theorem between 1 and 3 take datum through 3
Z1+Pw
Vg
ZPw
Vg
hL1 1
2
33 3
2
2 2 1 3+ = + + +
−
3 + 0 + 0 = 0 + 0 +V3
2
2g+1.5
∴V2
2
2g= 1.5;V3 = 2 × 9.81× 1.5 = 5.425m/s
Discharge = ( )π πdV lps
2
32
4 40 02 5425 17× = × =. . .
Apply Bernoulli’s Theorem between 1 and 2 datum through 1 V2 = V3 = 5.425m/s
0 + 0 + 0 = Pw
Vg
Z hL2 2
2
22 1 2+ + +
.
=( )P
wP2
225425
19 622 05 15 2 5+ + + = + +
..
. . .ω
Pw2 = - 4m of water
P2 = - (4 × 9810) = 39.24 kpa vacuum
E310/1 90 WE(6) A closed tank in which water is filled partly and the space above it is under
pressure. A hose pipe of dia 5cm connected to a tank discharges water to the roof of a
building of 6m high above water level in the tank. The frictional losses are taken as 3m of
water. Determine the pressure of air in the tank be maintained to deliver 15l/s to the roof.
SOLUTION
Apply Bernoulli’s Theorem between 1&2 datum through 1
Fig 5.8
Pw
Vg
ZPw
Vg
Z hL1 1
2
12 2
2
22 2 1 2+ + = + +
−
Pw
Vg
1 22
0 0 02
6 3+ + = + + +
but V2 =( )Q
a2
2151000
10 05
47 64= × =π . . m/s
∴Substituting V2 is above equation, we have
( )P
wPw
m12
17 6419 62
9 9 2 98 1198= + = + =.
.; . .
So P1= 11.98 × 9.81 × 1000 = 117kpa
E310/1 91 WE(7) Water flows from a tank into a pipe at the rate of 60lps as shown in figure. At ‘P’
a head of 20m is added by external device, to the fluid. The pipe is 10cm in dia. Determine
the pressure at B if the loss of head is 2V 2
2g, Where V is the velocity in the pipe.
SOLUTION Velocity in the pipe
Fig 5.9
=Qa
=60
1000×
1Π4
0.01( )2
= 7.64 m/s ; V 2
2g=
7.64( )2
19.62= 2.975m
Now apply Bernoulli’s Theorem between A and B. Take datum through ‘P’
Pw
Vg
Z HPw
Vg
Z HA AA ad
B BB LAB
+ + + = + + +2 2
2 2
Where Had is the energy added.
0 + 0 + 3 + 20 = PwB + 2.975 + 5 + 2 × 2.975
23 = PwB + 3 × 2.975 + 5 =
PwB + 13.925
∴ =PwB 9 075.
PB = 9.075 × 9810 = 89.025 kpa
5.4 MOMENTUM EQUATION (impulse momentum equation)
E310/1 92 INTRODUCTION
Momentum equation is derived based on Newton’s 2nd law. It is another
important equation in fluid mechanics. Along with continuity equation and energy
equation, it is used to solve many fluid problems which are not solved only continuity and
energy equation.
DERIVATION OF MOMENTUM EQUATION
Momentum equation states that the external forces acting on a fluid are equal to
rate of change of momentum in that direction
( )
a Fd mv
dtΣ =
i.e. Σ F × dt = d( mv )
Where Σ F × dt is called as impulse.
Now consider a steady I-D flow through a bend as shown in the figure.
Fig 5.10
between 1 -2. Let the particles move in ‘dt’ time through the distances of ds1 and ds2 from
1 and 2 respectively and occupy 3 and 4 positions.
Here the particles in the portion 3 -2 is common before and after ‘dt’ sec and so the
momentum of the mass in this common portion gets cancelled in the calculations of
change in momentum of the mass before and after ‘dt’ seconds. So to calculate the
momentum of the mass before ‘dt’ sec, it is enough to consider the mass in 1 & 3 and 2 &
4 as the mass in the portion 3 & 2 is common.
Thus the momentum of the mass before ‘dt’ sec is called as ‘Initial momentum’ given by (
in x direction )
E310/1 93
= × ×wg
A d Vs a1 11
Where A1 is the cross section area of flow at (1) and V1 is the mean velocity over the C.S
at (1) After ‘dt’ sec, the final momentum in X derivation is
= = × × ×wg
A d Vs x2 22and
Where A2 is the C.S of flow at 2 and V2 is the mean velocity at 2. So change in
momentum in dt sec is.
= Final momentum - Initial momentum
= wg
A ds Vwg
A ds Vx x2 2 2 1 1 1−
i.e. Rate of change of momentum is
= −wg
Adsdt
Vwg
Adsdt
Vx x22
2 11
1
=wg
A V Vwg
A V Vx x2 2 2 1 1 1−
where A2 V2 = A1 V1= Q = Discharge ∴ change in momentum per second is
( )wQg
V Vx x2 1−
This is equal to external force in the direction
i.e. ( )ΣFwQg
V Vx y x= −2 1
Similarly in y direction
( )ΣFwQg
V Vy y y= −2 1
and the resultant force on the bend is
ΣF = ΣFx2 + ΣFy
2 which is inclined to horizontal
byθ = tan −1 ΣFy
ΣFx
ADVANTAGE OF MOMENTUM EQUATION :
E310/1 94 When the details of fluid flow with in a control volume are not known and the
external forces acting over the fluid with in the control volume are known, then
momentum can be applied to this flow and solve the problem along with the other two
equations called 1 continuity equation and 2 energy equation. this equation is very useful
in hydraulic machinery
MOMENTUM CORRECTION FACTOR ' β' .
The same analysis which is followed to calculate the kinetic energy correction
factor, is now also considered to calculate momentum correction factor. Momentum over
the cross sectional area of fluid based on mean velocity is
Mass × velocity
= (ρ A× V ) × V = ρ AV2
But based on the actual velocity distribution over the cross sectional area of flow is
calculate d as follows over ‘dA’ area,
= elemental mass × V
= ( ρ dA v ) × v
Now if this is integrated over the area of cross section of fluid flow, the momentum of the
mass flowing over the cross sectional area of flow is
= ρdAV2
A∫
= ρV 2
A∫ dA
In these calculations, it is clear that the square of mean velocity over the area is
less than the square of instantaneous velocity over the area. So to get the actual
momentum based on mean velocity it is to be multiplied by a factor ' β' called as
momentum correction factor.
i.e. β × ρAV2 = ρV 2
A∫ dA
β =V 2
A∫ dA
AV 2
Q AV2 less than V 2
A∫ dA
E310/1 95 β is always greater than ‘1’. For laminar flow through pipe β = 1.33 and Turbulent flow
through pipe β = 1.01. So in turbulent flow through pipes β is almost equal to ‘1’ and so
this factor is not considered in the momentum calculations in turbulent flow i.e. β =1 is
taken.
SAQ(19) In momentum equation change in momentum takes place in the direction of
forces. True/False.
SAQ(20) There is no need of estimation of forces with in the fluid in control volume
in application of momentum equation.
True/False.
SAQ(21) For steady uniform flow, calculate the external forces acting on the fluid.
SAQ(22) 1kg(m) /s changes its velocity form 3m/s to 33m/s in a nozzle. Determine
the force acting on the nozzle.
SAQ(23) The momentum of a jet 30N.s is brought to zero by a plate perpendicular to
the jet, then calculate the force of jet on the plate.
SAQ(24) Given the mean velocity over 1m2 cross sectional area of flow is 2m/s and
the integral volume of squares of velocities over the area is 4.4, calculate
the momentum correction factor.
WE(8) In a 450 bend of cross sectional area 1m2 is gradually reduced to 0.5m2 area. Find
the force required to keep the bend in position if the velocity and pressure at 1m2 section
are 10m/s and 300 kpa respectively
E310/1 96 SOLUTION Velocity at 2 is
Fig 5.11
V2 =a1v1
a2
=1 ×10
0.5= 20m / s,
V22
2g= 20.387m
V12
2g=
10( )2
2 × 9.81= 5.097m
Pw
m1 300 10009 810 1000
30=××
=.
Now apply Bernoulli’s Theorem 1 and 2
Pw
Vg
Pw
Vg
1 12
2 22
2 2+ = +
30+5.097 = Pw2 + 20.387
∴P2
ω= 10.2 ; P2 = 10.2 × 9810 = 100 kpa
Now apply momentum equation, we have in ‘x’ deviation,
P1a1 - Rx - P2 a2 cos 45 = ( )wag
V Vx x2 1−
Here R is the resultant force required to keep the bend in position. This is resolved into
horizontal and vertical components as Rx × Ry
So 300 × 1 - Rx - 100 × 0.5 × cos 45 = 9.819.81
× 1 ×10( ) 20 cos45 −10( )
Rx = 223.22 KN
in y direction, P1a1 sin θ + Ry - P1a1 sin45 = ωas
V2 sin 45 − V1 sin 0( )
E310/1 97 0 + Ry - 100 × 0.5 × sin45 = 10 (20 × sin45-0)
Ry =176.78 KN
∴ The resultant force
R = RX2 + Ry
2
= 223.22( )2 + 176.78( )2
= 284.7 KN
Its inelination = tan-1 Ry
Rx
α = 38o.37
WE(9) A 2.5 cm dia nozzle is fitted to a 4.0 cm dia pipe. If the discharge is 15 lps.
Calculate the force exerted by fluid on nozzle tends to tear it of the pipe .
SOLUTION
a1 =Π4
0.04( )2 = 0.00125m 2
The velocity in pipe = 0.015
Π4
0.04( )2= 11.94m / s
The velocity through the nozzle = 4
2.5
2
×11.94 = 30.57m / s
V12
2g=
11.94( )2
2 × 9.81= 7.27m;
V22
2g=
30.57( )2
2 × 9.81= 47.63m
Now apply Bernoulli’s Theorem to fluid pressure at base of nozzle as
Pw
Vg
Vg
1 12
22
20
2+ = +
Fig 5.12
Pw
Vg
Vg
m1 22
12
2 247 63 7 27 40 36= − = − =. . .
E310/1 98 P1 = 40.36 × 9.81 = 395.9kpa
Now apply momentum to find the force an nozzle as
P1a1 - Rx = ( )wag
V V2 1−
395.9 × 0.00125 - Rx =9.8109.81
×15
100030.57 −11.94( )
Rx = 0.7765 KN = 776.5N
WE(10) A vertical jet of water issuing from a 5cm dia nozzle with a velocity of
10m/s supports a horizontal flat plate at a height of 3m.
Find the weight of plate Neglect losses
SOLUTION
Fig 5.13
Q = a1× v1=Π4
0.05( )2 × 10 = 1.96 ×10−2 m / s
To find velocity at 2 apply Bernoulli’s Theorem between 1 and 2
ZPw
Vg
ZPw
Vg1
1 12
22 2
2
2 2+ + = + +
0 + 0 +10 2
2 × 9.81= 3 + 0 +
V22
2g;V2
2
2g= 5.0968 − 3( )
V2 = 5.0968 − 3( )× 2 × 9.81 = 6.414m / s
To find W, apply momentum equation
W = ( )wQg
V0 2−
E310/1 99 =
98109.81
×1.96 ×10−2 −6.414( )
weight of plate = 125.94N
SUMMARY
(1) Forces causing the flow of fluid are listed as Fp,Fg,Fv,Ft,Fe and F
(2) Out of these, FP and Fg are the two forces causing flow in ideal fluid. Fg are called
as body forces which are due to weight of fluid. FP pressure forces are due to
surface area. The pressure forces are further divided into tangential forces and
normal forces. But in ideal fluid tangential forces are zero. So pressure forces are
given by = (pressure × C-S of flow) = normal force.
(3) In Euler’s equation. Body forces and pressure forces are considered as it is applied
to steady incompressible ideal fluid flow.
(4) 1ρ
dpdx
− S = a, is the Euler’s equation along stream line.
(5) Then Bernoulli’s equation is derived by integrating Euler’s equation along the
stream line as
H = ZPw
Vg
+ +
2
2= constant at any point.
Where Z = datum head = datum energy / unit ωt of fluid
Pw
= Pressure head = pressure energy/ unit ωt of fluid
V 2
2g= Velocity head = K.Energy/ unit ωt of fluid
The sum ZPw
Vg
+ +2
2is also called Total head ‘H’ or Total energy / unit weight of
fluid flowing where w= weight of fluid flowing = wQ. This Bernoulli’s Theorem
states that for ideal incompressible steady and continuous fluid flow the total head
and velocity head is constant at any point along its path
(6) Pw
Z+
is called as piezometer head
E310/1 100 (7) In real fluid losses are to be considered in the application of Bernoulli’s equation
as
H1 = H2+HL, where HL is the loss of head between 1 and 2
(8) Momentum equation states that the external forces acting an a fluid flow are equal
to rate of change of momentum in that direction , i.e.
( )ΣFwQS
V VS S S= −2 1
(9) The advantage of momentum equation is that when the information about the flow
process with in a control volume is not known then such problems can be solved
by momentum equation by knowing the external forces acting on the fluid.
ANSWER FROM SAQ
(1) to (8) True (9) Yes (10) No (14) v = 6.26m/s
(15) 2m, 1 and 2 (16) 68.67 kpa (17) -3m of water (18) 7m
(19) True (20) True (21) Zero (22) 30N
(23) 30N (24) β 1.1
UNIT 5
EXERCISE
(5.1) A pipe line carrying oil of sp gr 0.877 changes in size from 0.15m dia at A to
0.45m to B. Which is 3.6m above A. If the pressure at A and B are respectively
90.252 KN/m2& 59.841 KN/m2. The discharge is 145 lps. Then determine the
cross head and direction of flow.
[ HL= 3.32m of oil A →B]
(5.2) A pipe line carrying water changes in area from 0.031m2 at A to 0.124m2 at B
which is 3m above A’. If the velocity and pressure at A are 5m/s and 100KN/m2,
find the velocity and pressure at B. Neglect losses
[ 1.25m/s & 82kpa]
(5.3) A 60cm dia pipe is supplying water under a pressure head of 40m to a Hydraulic
Machine (T). Then water flows out of it through a 45cm dia pipe at a rate of 0.8m3/s
E310/1 101 into atmosphere. Neglecting the losses determine the energy consumed by the
machine.
Fig 5.14
[ 38.71m ]
(5.4) Water enters a Hydraulic machine at A where the dia of pipeis 40cm, under a
pressure head of 30m. Then water flows out of the machine through tapering tube as
shown in figure at a rate of 0,5m3/s. At a point B which is 2m below the point A, the
dia is 60cm and pressure head is -4m. Determine the power developed by the machine
if the efficiency of the machine is 90%
Fig 5.15
[ 161.45 kw]
(5.5) A conical pipe has diameters 0.40m and 0.80m at its two ends. The smaller end is
2m above the larger end. For a flow of 0.5m3/s of water the pressure at the lower
end is 10kpa. Assuming a head loss of 2m and kinetic energy correction factor
α=1.1 and 1.5 at smaller and larger ends respectively estimate the pressure at
smaller end
[P17.144 kpa]
E310/1 102
(5.6) A pipe line has the following data at its two section’s A and B
A B
Diameter 30cm 45cm
Elevation(m) 10.0 16.0
Pressure kpa 40.0 30.0
Kinetic energy correction
factor
1.08 1.25
Assume a loss of head equal to 20 times the velocity head at A. Calculate the
discharge of water flowing from B to A
[Qa = 147.5 l/s]
(5.7) Water enters a hydraulic machine ‘P’ at 70 lps through A and delivered by it through
B to higher level. A mercury manometer connected between A and B shows a
deflection of 40cm. Determine the head added by the Hydraulic machine.
Fig 5.16
[Hp = 6.193m]
(5.8) A 20cm dia pipe leading water from a reservoir ends in a nozzle of dia 10cm at
elevation of 900m. The water level in the reservoir in 100m. The loss of head in
the pipe is 12 times velocity head in the pipe. Calculate the discharge.
E310/1 103
Fig 5.17
(5.9) A pipe line delivering water from a reservoir is shown in figure. A hydraulic
machine P is adding energy to the flow. The rate of discharge into atmosphere at
out let is 4.5lps. Assuming loss of head as 2 times the velocity head in the 20cm
pipe and 10 times the velocity head in 15cm dia determine power delivered by the
machine.
Fig 5.18
(5.10) Calculate the discharge through the pipe line shown in figure.Also find the
pressure at A and B. The depth of water in the reservoir in 5m. Neglect losses.
Fig 5.19
[ 0.0194m3/s, 0.769 kg./m2, 0.231kg/cm2]
(5.11) Velocity distribution in a pipe is given by V = Vmax 1 −r 2
R2
at any radius ‘r’
from the centre of the pipe. Vmax= Max velocity at centre of pipe R= Radius of the
E310/1 104 pipe. Determine kinetic energy correction factor and momentum correction factor
β
α = 2,β =43
(5.12) A 20cm dia pipe has a 90o bend in the horizontal plane. Oil of sp gr 0.8 is flowing
at 150lps in the pipe. At in let the pressure is 0.5m of Oil . Find the resultant force
exerted by the oil on the bend
[F = 1612.5 N inclined at 4.190to horizontal]
(5.13) A 60o reducer bend at the end of a pipe 30cm dia, discharges freely through 20cm
dia into atmosphere at 10m/s. The bend is in vertical plane and the centre of exit is
60cm above the centre line of pipe. The weight in side the bend in 700N.
Determine the force required to keep the bend in position
[ R= 4593 N, θ = 131.95o with horizontal]
(5.14) Water flows through 180o vertical reducing bend shown in figure. The pressure at
in let is 20kpa and the discharge is 0.4m3/s If the bend volume 50.8m3. Calculate
the force required to hold the bend in place.
Fig 5.20
[ F = 10854N at 136.19o to
horizontal]
E310/1 105 (5.15) A sluice gate in a rectangular channel carrying water is so opened to create depths
of flow of 1.70m and 0.25m on u/s and d/s of the gate respectively. The discharge
intensity is 1.3m3/s per meter width. Estimate the force per meter on the gate
[ F = 8.09 KN]
(5.16) A tank and a trough are placed on a trolley as shown in figure. Water issues from
the tank through a 4cm dia nozzle at a velocity at 5m/s and strikes the trough
which turns it up by 45o. Determine the compression of the spring if its stiffness is
2kg /cm
[1.129cm]
Fig 5.21
References same as unit 1
***
FLUID MECHANICS & HYDRAULIC MACHINERY
UNIT-6
AIMS
The aim of this unit is to know about some more fluid flow measuring devices like
pitot tube, elbowmeter, Trapezoidal notch, their principle of measurement and uses using
Bernoulli’s Theorem.
OBJECTIVES
(1) To know about the meaning of pitot tube, static pressure, stagnation pressure and
dynamic pressure heads.
E310/1 106 (2) To derive an equation for velocity of flow by pitot tube by applying energy
equation and to know the application of pitot tube in measuring discharges
flowing in open channels and through pipes.
(3) To understand the meaning of elbow (bend) meter and to obtain an expression
for discharge flowing in a pipe of elbow meter.
(4) To review wiers and notches .
(5) To explain about end contractions velocity of approach and ventilation of wiers.
(6) To derive an expression for discharge flowing in open channels for Trapezoidal
notch.
(7) To explain about cippoletti weir. which is a special type of Trapezoidal weir.
(8) To know how to solve practical problems using pitot tube, elbowmeter and
Trapezoidal notch in measuring the rate of flow.
6.1 INTRODUCTION
The rate of fluid flow in open channels or through pipes is very much useful in
solving many practical fluid flow problems like in irrigation and water supply etc. The
discharge in open channels or through pipes can be measured by various measuring
devices. In previous courses some of such devices like venturimeter, orificemeter. flow
nozzle meter, orifices, mouth pieces rectangular notch and triangular notch are known.
Now in this unit still some more useful measuring devices like pitot tube, elbow meter and
trapezoidal notch are to be studied.
6.2 PITOT TUBE
Before deriving an equation for velocity of flow in a pipe, let us consider flow of
fluid with a velocity over a cylinder as shown in figure
Fig 6.1
E310/1 107 When the flow approaches the body the velocity tends to be zero at S1andS2, Which are
known as stagnation points. At this stagnation point all the kinetic energy is converted to
pressure energy. If Bernoulli’s equation is applied between 1 and S1, we have
Pw
Vg
Pw
s+ = +2
20
or P + ρv2
2= Ps
Here P is called as static pressure
Ps is called as stagnation point
and ρv2
2is called as dynamic pressure
At stagnation point, the dynamic pressure is converted into equivalent pressure.
With this understanding let us study about pitot tube. Derivation of equation for velocity
of flow by pitot tube
Pitot tube is a bent tube used first in 1730 by French scientist Henri Pitot to
measure velocity of flow in a river. So it is named after him. Let the pitot tube be kept in
the fluid flow as shown in figure in open channel and pipe
Open Channel Flow Pipe Flow
Fig 6.2
Let at point 1 the undisturbed mean velocity of flow is Vm/s and the static pressure head
isPw
Now applying Bernoulli’s Theorem between 1 and 2 taking datum through them.
E310/1 108
Pw
Vg
Pw
s+ =2
2, neglecting losses
V
gPw
Pw
Hs2
2= − =
V = 2gH is the ideal velocity
or Actual velocity = Va = C 2gH = 2g H − hf( )Which is less than ideal velocity
where hf = loss of head
so the coefficient C < 1.0
i.e. C = 0.98 nearly in actual case to account for loss of energy. The actual value of ‘C’ can
be determined by calibration.
If the differential pressure head is more, then a differential momentum may be
used to measure the differential pressure as shown in figure
Fig 6.3
By gauge equation we have
Pωl
+hsm
sl− h− =
Psωl
Ps − Pω l
= H = hsm
sl
− 1
H = 12.6h (m of water) if
sm = 13.6 (mercury)
sl = 1.0 (water)
Pitot - static tube
It measures both static pressure and stagnation pressure as shown in figure
E310/1 109
Fig 6.4
The central tube measures the stagnation pressure and the outer tube with holes
around the periphery as shown in figure, measures static pressure.
Prandtl pitot tube
Is a pitot - static tube as shown in figure with standard proportions which gives
accurate results. The blunt nose is so designed that the loss of head due to turbulence is
negligible and so the coefficient ‘C’ is almost equal to 1.0
Fig 6.5
Pitometer
Is a combination of both static pressure tube and stagnation pressure tube as shown
in figure which gives more differential pressure head than the pitot - static tube. It consists
of two pitot tube, one facing upstream and another facing down stream.
E310/1 110
Fig 6.6
V = C 2gh1
Here the value of ‘C’ is less than that of pitot static tube. i.e. C is about 0.8 to 0.85.
IDENTIFY THE FOLLOWING WHETHER “TRUE” OR “FALSE”
SAQ(1) Pitot tube measures total head.
SAQ(2) Pitot static tube measures both static and stagnation pressures.
SAQ(3) In pitot tube dynamic pressure is equal to kinetic head.
SAQ(4) The difference of static and stagnation pressure heads measured by pitot
static tube gives the dynamic head which is equal to velocity head.
SAQ(5) Pitot tube gives velocity at a point or discharge flowing in the pipe.SAQ(6)
What do you understand by C=1?
SAQ(7) The difference between static and stagnation pressure heads is 100cm of
water. what is the velocity C=1.
SAQ(8) The mercury deflection connected to pitot static tube gives 10cm of
mercury. Find the velocity of water C=1
SAQ(9) If velocity of oil of sp gr 0.9 at a point is 5m/s, find the deflection in
mercury differential manometer connected to a pitot static tube C=1.
WE (1) A pitot tube is inserted into an air stream where the static pressure is
100kpa. If the water deflection in the differential manometer is 30cm, determine the
velocity of air. Given C=1 and density of air = 1.22 kg/m3
SOLUTION
Differential head
E310/1 111
H = 0.3 10001.22
−1
= 0.3 ( 818.67 )
= 245.6 m of air
V = 0.98 2 × 9.8 × 245.6
= 68m/s
WE (2) Carbon tetrachloride of sp gr 1.6 flows in a pipe. A differential manometer
attached to a pitot static tube gives a deflection of 100mm of mercury. Determine the
velocity of flow C=1.0
H = hSm
Sl
−1
=
1001000
13.61.6
−1
= 0.75m
V = 2g × 4
= 19.62 × 0.75 = 3.84m/s
WE (3) In a 12cm dia pipe line water is flowing. Given the diameter of the pipe the
pitot static tube gives a deflection of 5cm and 15m of water at centre point and at 1.5cm
from the wall respectively. Determine the discharge C=1
Fig 6.7
a1 =Πd1
2
4=
Π 0.06( )2
4= 2.83 × 10−3 m 2
V1 = 2g × 5 = 9.9m / s
∆Q1 = a1V1= 0.028 m3/s
E310/1 112 a2 =
Π4
( 0.122- 0.062) = 8.48×10-3m2
V2 = 2g × 1.5 = 5.43 m/s
∆Q2 = 0.046m3/s = a2v2
Total discharge Q = ∆Q1 +∆Q2
= 0.074m3/s
6.3 ELBOMETER OR BEND METER
We know already that by creating a pressure difference between two points in a
pipe flow, by reducing the cross sectional area of the pipe, the rate of flow through it can
be measured, viz. venturimeter, orificemeter and flow nozzle meter by using the equation
Q = cd a1a2
a1 − a22
2gH or cd a1
a1
a2
2
− 1
2gH
Where a1a2 are the cross sectional area of pipe throat/ orifice/ nozzle. respectively
H=Pw
ZPw
Z11
22+
− +
; the piezometric head difference. Similarly based on
the above principle rate of flow through a pipe can also be measured by a bend or elbow as
derived below. And it is to be noted that bend meter is very simple and no construction of
meter at all, except pressure tappings at inner and outer walls of the bend or elbow as
shown in figure. In the bend the pressure difference is due to centrifugal force on the flow.
The flow in the bend is free vertex flow.
Derivation of equation for discharge
The particles in a bend move on the curved stream lines in a free vortex. Therefore
the velocities at 1 and 2 are
E310/1 113
Fig 6.8
V1 =C
R − r( )V2 =C
R + r( )Q We know in free vertex Vr = C at any radius ‘r’. Where C = a constant
Now applying Bernoulli’s Theorem between 1 and 2, We have
Pw
ZV
gPw
ZV
g1
112
22
22
2 2+ + = + +
H =Pw
ZPw
ZV V
g2
21
112
22
2+
− +
=
−
But V1=c
R − r( ) V2=c
R − r( )
H =c2
2g1
R− r( )2−
1R + r( )2
=c2
2g4Rr
R − r( )2 R − r( )2
=c 2
2g4Rr
R2 − r 2( )2
i.e. C = 2gHR2 − r 2( )2 Rr
To obtain an expression for discharge through the bend consider a stream tube of thickness
dx at a radicus x .
E310/1 114
Fig 6.9
The mean velocity over the C.S stream tube is
Vm =c2
R − x( ) R + x( ) =C
R2 − x2
Then the discharge through this stream tube is
dQ = 2ΠxdxVm
= 2Πxdx ×C 2
R − x( ) R + x( )
= 2Πxdx ×C
R2 − x 2
Now by integrating this equation the discharge through the bend is obtained as
Q = dQ xdxC
R xo
Q
o
r
∫ ∫= ×−
22 2
Π
= 2ΠCx
R xdx
o
r
2 2−∫
Q = 2ΠC R R xr
− −
2 20
= 2ΠC R − R2 − r 2[ ]
E310/1 115 Substituting the value of C, from the previous equation we have
Q = Π 2gHR2 − r2( )
R2R − R2 − r2
= Πr 2 2gH Rr
2
−
R− R 2 − r 2( )R2
The actual discharge is
Qa = cd A 2gHRr
2
−1
R − R2 − r 2
R2
Qa = cd k1 A 2gH where k1=Rr
2
−1
R − R2 − r 2
Rr
= KA 2gH
Where K = Cdk1= Bond meter constant . But by experiments it is shown that
K1 =R
2a
and K = 0.55 - 0.88
SAQ(10) In bend meter, the flow is subjected to centinfugal force. Is it True/False?
SAQ(11) In bend meter the flow is free vertex flow. Is it True/False
WE (4) Oil of sp gr 0.9 flows in a horizontal bend meter of 10cm dia of the
pressure difference between inner and outer walls is 10 KN/m2, find the discharge. Given
bend meter constant as 0.6.
SOLUTION
Q = KA 2gH
H =10 ×10000.9 × 9810
= 1.132 m of oil
A = Π4
0.1( )2
Q = 0.6 ×Π4
0.1( )2 2 × 9.81×1.132
E310/1 116 = 0.0222 m3/D
6.4 TRAPEZOIDAL NOTCH
We know already that the discharge flowing in open channel can be measured by
weirs and notches viz. Rectangular notch and triangular notch.
The equation for discharge for rectangular notch is
Q = Cd23
L 2gH 3 / 2
and for triangular notch
Q = Cd8
152g tanθ / 2H5 / 2
Where L = Sill length of rectangular notch.
H = Head over sill of notch.
θ = The vertex angle of triangular notch.
End contractions
When the sill length of rectangular notch is less than the width of channel, then the
nappe will contract giving rise to two end contractions as shown in figure 6.10
Fig 6.10
The effective sill length Le is
Le = ( L - 2 × 0.1 H )
where and contraction = 0.1 H If there are ‘n’ end contractions then the effective sill
length is.
Le = [ L - ( n× 0.1 ) H ]Velocity of approach
E310/1 117
Fig 6.11
Let Q be the discharge flowing in the open channel, then the velocity of approach with
which the particles are approaching the notch is.
Va =Q
B Z + H( )
where B = With of channel
Z = Height of notch sill over the bed of channel.
H = Head over sill
ha = Velocity head = Va2/ 2g
Then the equation for rectangular notch considering both end contractions and velocity of
approach is
Qa = Cd23
L - h × 0.1 × (H +ha)( H+ha)3/2- ha3/2
and for triangular notch considering velocity of approach is
Qa = Cd8
152g tanθ/2 ( H+ha)5/2-ha5/2
Steps Involved in finding discharge considering velocity of approach
1) First neglect the velocity of approach and determine the discharge to 1st
approximation.
2) Now in the 2nd trial, find velocity of approach as indicated already using the 1st
approximate discharge. Then using this velocity of approach 2nd approximate discharges
is determined.
E310/1 118 3) Again is 3rd trial, the above procedure is repeated to find the refined discharge. These
trials are repeated till the difference in the final discharge and the previous discharge is
negligible.
Ventilation of weirs
In case of suppressed rectangular weir where the width of channel is equal to sill
length of rectangular notch, the space below the nappe is hollow containing air as shown
in figure.
Fig 6.12
As the flow takes place the air inside this hollow space is carried away by the
flowing water, making the pressure inside the space is vacuum, because both ends of this
space, are closed by the walls of channel. Thus there is no way for the air to enter in for
this space below the nappe. So as the pressure below the nappe is vacuum, the nappe will
cling as shown in figure due to atmospheric pressure over the water. This clinging nappe
leads to error in the discharge measuring about 6 to 7% more than that of force nappe. To
avoid this error, air is supplied through end walls as shown in figure by providing two
numbers of holes of 25mm diameter in the end walls.
E310/1 119 Fig 6.13
Discharge of Trapezoidal notch
The Trapezoidal notch is a combination of rectangular and triangular notches. The
equation for discharge for this notch is
Fig 6.14
Q = Discharges through( rectangular notch + Triangular notch )
Q = Cd1
23
L 2gH 3 / 2 + Cd2
815
2g tanV2
H 5/ 2 of Cd1= Cd2
Q = Cd 2gH 3/ 2 23
L +8
15tan
θ2
H
Cippoletti weir
It is a special trapezoidal notch in which the loss of discharge due to end
contraction, of rectangular notch is made up by adding half triangular notch to each side of
rectangular
notch. That means the discharge lost by end contractions is equal to extra discharge
through triangular notch portions at the two ends. Finally the discharge for this cippoletti
weir is equal to the discharge of suppressed rectangular notch. Therefore the discharge
through cippoletti weir is
Q = Cd23
L 2gH3 /2 = Cd23
(L − 0.2 × H ) 2gH3/ 2 + Cd815
2g tanθ2
H5/ 2
i.e. Discharge lost through end contractions = Discharge added by triangular notch
∴ Cd23
0.2H( )H3/ 2 2g = Cd815
2g tanθ2
H5 / 2 14
= tanθ / 2
So the side slopes of this notch is 1 Horizontal to 4 vertical.
This trapezoidal notch with side slopes 1 in 4 is called as cippoletti weir whose discharge
is.
E310/1 120 Q = Cd
23
L 2gH 3 / 2 . Which is same as that for rectangular notch without end
contractions.
SAQ(12) Trapezoidal notch discharge is sum of both rectangular notch and triangular
notch discharges. Is it True/False.
SAQ(13) A trapezoidal notch with side slopes 1 in 4 is called as cippoletti weir. Is it
True/False.
SAQ(14) Given Q = 0.1m3/s in a channel of 0.5m wide. The height of sill of a notch
is 0.1m and head over the notch is 0.1m. Then find the velocity of
approach.
WE(5) The discharge flowing over a rectangular weir under a head of 0.4m is 5m3/s
without end contractions. Find the side slopes in order to increase the discharge by 5%.
Take Cd = 0.6. Find the increase in discharge for side slope of 45o with vertical.
SOLUTION
Q2 = Q - Q1= 0.05 × 5 = 0.25 m3/D
0.25 = 0.6 8
152g tanθ / 2 × H5/ 2 = 0.1434
tanθ / 2 = 1.743
θ/2 = 6.16o or slope is 1 to 0.5737
Increase in discharge for sides slope of 45o with vertical is
Q2 = 0.1434 tan θ/2 = 0.143 × 1m3/s
∴ increase in discharge = 0.1434
5= 2.868%
SUMMARY
1. The equation for velocity by a pilot tube is V = C 2gH , where H is the
differential head between stagnation pressure and static pressure.
2. The equation for discharge by an elbow meter is
Q = KA 2gH
where K = meter constant = 0.55 to 0.88 and K = Cd K1
E310/1 121 where K1= R / 2d , R = radius of curvature of bend centre line and d = diameter
of pipe.
3. The equation for discharge for trapezoidal notch is sum of discharges rectangular
and triangular notches.
i.e. Q = Cd1L3
L 2gH3 / 2 + Cd28
152g tan
θ2
H5/ 2
4. Cippoletti weir is a special trapezoidal notch with side slopes 1 H to 4V and the
discharge is equal to that of rectangular notch without end contractions
i.e. Q = Cd23
L 2gH 3 / 2
5. The discharge for rectangular notch considering end contractions as velocity of
approach is
Q = Cd23
L − 0.1h H + ha( )[ ] 2g H + ha( )3 / 2 − ha 3/ 2 where ha = Va2/2g
6. Ventilation of nappe is necessary in case of suppressed rectangular notch to avoid
cligging of nappe. Further cligging of nappe leads to 6 to 7% more in discharge
estimation meter the same head.
Answer to SAQ
1-4 True 5. velocity
7. V = 4.43m/s 8. V= 5.58m/s 9. h = 9cm 10-13 True
14. Va = 1m/s
Exercise
(6.1) The velocity of an oil of sp gr 0.9 was measured by a pitot-static tube. The tube’s
coefficient ‘C’ is 0.98. Calculate the velocity if the deflection in the mercury
manometer is 6cm
[ Vo= 4.0 m/s]
(6.2) A pitot tube was arranged in centre line as shown in figure. Calculate the velocity
if the coefficient is 1.0
E310/1 122
Fig 6.15
[ Vo = 3.851m/s]
(6.3) Kerosene of sp gr 0.81 is flowing in a pipe and the static pressure at a point is 3kpa
of the stagnation pressure of a pitot tube ( c=0.99) inserted in the pipe at centre line
is 4kpa, find the velocity.
[ Vo = 1.56m/s]
(6.4) At the summit of a siphon in a water pipe the centre line velocity is 2.5m/s and the
pressure as 3m of water (vacuum) If a pitot tube (c=0.99) is inserted into the pipe
centreline at this section, what will be the stagnation pressure in kpa (abs).
Atmospheric pressure = 101kpa
[ Pst = 74.81kpa(abs)]
(6.5) A trapezoidal sharp-crested weir has a base width of 1.2m and side slopes of 1.5
horizontal-1vertical. Calculate the discharge over the weir for a head of 35cm.
Cd=0.62
[ Q = 0.614m3/s]
(6.6) Estimate the head over a cippoletti weir of base width 0.9m required to pass a
discharge of 600l/s. Cd = 0.63
[ H = 50.45 cm]
E310/1 123 (6.7) A trapezoidal notch is to be designed to pass a discharge of 1.0m3/s at a head of
0.8m over the crest of 0.5m3/s at a head of 0.5/m. If Cd = 0.7, then Calculate the
base width and the side slope of the notch.
[L = 0.643m θ =2.96o]
(6.8) Find the discharge through a trapezoidal weir with inward sloping sides as shown
in fig if Cd = 0.6
Fig 6.16
[ 210l/s ]
***
FLUID MECHANICS & HYDRAULIC MACHINERY
UNIT VII
FLOW THROUGH PIPES - TURBULENT FLOW
Aims: In this unit the aims are (1) to distinguish between Laminar flow and turbulent flow
by Reynold’s experiments, (2) to derive an equation for loss of energy in turbulent flow
through pipes, (3) to distinguish between flow in series pipes and parallel pipes and (4) to
study the flow in branch pipes.
Objectives:
1) To explain the Reynlod’s experiments to distinguish between laminar and turbulent
flows by flow patterns and a graph between loss of head and velocity on logarithmic scale.
(2) To define and derive Reynold’s number and its use in flow through pipes.
E310/1 124 (3) To explain fluid friction laws for laminar and turbulent flows.
(4) To derive an expression for loss of head in Turbulent flow through pipes.
(5) To analysis flow through series pipes.
(6) To explain about hydraulic gradient and energy live and
(7)To study flow through parallel pipes and branch pipes and to apply the above in solving
practical problems.
7.1 Introduction
When fluid flows under pressure in a closed conduct, then this flow is called as
pipe flow in which fluid run full. Due to viscosity of fluid in rear fluids there will be
opposition to fluid motion, which leads to loss of energy in fluid flow. Its determination is
very important in design of water supply system, carriers of water in irrigation etc. So in
this unit we will study how to determine the loss of energy in laminar and turbulent flows
and how to make use of this loss of energy in solving fluid problems.
7.2 Reynold’s experiments
Osborne Reynold’s was the first man to demonstrate an experiment in l883 to
distinguish between Laminar and Turbulent flows. He maintained a constant water level
in a glass tank as shown in figure.
He allowed a dye of same specific gravity as that of water, from a small tank into a bell
mouthed glass tube as shown in figure. A control valve is provided to the glass tube to
control the fluid flow in the glass tube.
E310/1 125 At low velocities, the dye appeared as parallel straight lines, indicating that the
fluid particles move in parallel layers. This flow is called as’ Laminar flow’ because the
fluid flow is taking place in layers, one layer sliding above and the layer, as shown in
figure.
When the velocity of flow is increased the particles slightly deviate from the
parallel paths but the viscous force makes the particles to move back to their original path,
leading to wavy paths of motion. This flow is called as transition flow.
When further the velocity is increased the colour diffuses into the water , making
water completely colourful.
This indicates the zig zag motion of fluid particles and this flow is called as turbulent flow.
In this turbulent flow, some particles move fast, some slow, some dash against another
leading to mixing of fluid particles. This mixing leads to loss of energy.
The velocity at which the region of flow changes, is called as critical velocity.
In the above experiments it was observed by Reynolds that the motion of water
was governed by relative magnitudes of Inertia force and viscous force. The ratio of these
two forces is called Reynold’s number, which is non dimensional.
Thus Reynolds’ no= Rn =Inertia force
Viscous force
E310/1 126 Inertia force = Mass × acceleration
= lL3 × LT2 = lL2 L2
T2 =lL2 V2
Where l-is mass density,
L-Characteristic length
V- Velocity of fluid
viscous force = τ × Area
= µ VL
× L2 = µVL
Where τ = shear stress = µ VY
µ = Viscosity of fluid
∴RN =ρL2V2
µVL=
ρVLµ
=VLυ
Where υ =µρ
For pipe flow let L=D-Dia of pipe
So RN =ρVD
µor
VDυ
At low velocities, the viscous force is predominant and so in laminar flow, this
force controls the particles. So move on parallel paths, where as in turbulent flow the
inertia force is predominant and so the particles deviate from the parallel paths leading to
mixing of the particles.
Thus in laminar flow the Viscous force is more so Reynolds’ number is less, where as in
turbulent flow the Viscous force is less and so Reynolds’ number is more.
To obtain the range of Reynolds’ number to distinguish between laminar and turbulent
flows, experiments on flow through pipe were conducted in which the loss of head over a
E310/1 127 length ‘L’ is measured for different velocities from zero to maximum possible. as
indicated in the diagram. The loss of energy per unit length is taken on ‘Y’ axis and
velocity on ‘X’ axis in logarithmic scale and a graphs drawn between loss of energy and
velocity of flow as shown in figure.
Again the loss of energy is measured from maximum possible velocity to zero
velocity by decreasing the flow rate. The trends shown in the figure.
From the graph it is clear that
(1) At low velocities the head loss hL V (velocity)
Upper A or even B, in which the regime of flow is laminar flow. At ‘A’ the flow is
definitely laminar even there is some disturbances in the flow. Because these disturbances
are damped off by viscose force which is more in the case of laminar flow. So ‘A’ is
called as a stable critical point at which regime of flow is changed.
The flow can be maintained laminar without any least disturbance in the flow upto
B at which the regime is changed. This point is unstable as the least disturbance will
cause the regime of flow changed. So this point is called unstable critical point.
E310/1 128 From the graph it is clear that when the velocity is going on increased gradually,
the regime of flow changes from laminar to transmission at B, and into turbulent at C. In
turbulent flow the trend of loss of energy Os non linear and so hL ϑυ Where υ = 1.72 to 2
when the flow is gradually decreased from maximum to minimum, the trend of loss of
energy is not the same as in the previous case, but deviates as shown in fig. 1 in which the
flow changes from turbulent to transition at ‘C’ and transmitted to laminar at ‘A’.
At point A the RN= 2000, which is known as lower critical point, while at point B,
which is unstable the RN can be maintained even upto 12000- 14000 without any
disturbances. This point is called as upper critical point. The practical value of Rnat
which the negative of flow is changed from transition to turbulent is 2700-4000. At these
point, the velocities are called as critical velocities. In between 2000-4000 the flow is
transition flow.
Thus from the graph it is clear that when RN ≤2000, the flow is always laminar and
when RN ≥ 4000, the flow is turbulent in a pipe and for RN from 2000 to 4000 the flow is
transition.
SAQ(1) for laminar flow in a pipe, the RN is≤ 2000. Is it true or false ?.
SAQ(2) The lower critical point is stable point Is it true or false?
SAQ(3) Determine the lower critical velocity in a pipe of 2cm dia given the velocity of
water=0.01 stokes.
7.3 Laws of fluid friction.
From the experimental observations the laws of fluid friction in laminar and
turbulent flows are given below.
(i) Laws of fluid friction in laminar. flow
The fluid frictional resistance in laminar flow is
(a) proportional to velocity of flow
(b) proportional to surface area which is in contact with fluid.
(c) Independent of the nature of the surface of contact, since a thin fluid layer adjacent to
the surface, known as laminar sub layer, submerges all the projections of the surface.
(d) Independent of pressure
(e) is affected by temperature,
since it depends on viscosity, which in turn varies with temperature.
(II) Laws of fluid friction in Turbulent flow.
E310/1 129 In turbulent flow the fluid frictional resistance is
(a) Proportional to Vn where =1.72to2.0
(b) Independent of pressure.
(c) Proportional to density of fluid
(d) Proportional to surface area in contact
(e) Dependent on nature of surface in contact
(f) Slightly affected by temperature.
III Froude’s experiments
From the experiments conducted by Froude, to investigate, the frictional resistance
for turbulent flow on different surfaces, the following conclusions are drawn.
(a) Frictional resistance varies approximately as square of velocity.
(b) The frictional resistance varies with the nature of surface.
(c) The frictional resistance per unit area of the surface decreases as the length of board
increases, but is constant for long lengths.
So the frictional resistance ‘F’ in turbulent flow is given by
F=f1 A Vn where n =2.0
Where f1 is the frictional resistance per unit area at unit velocity.
7.4 Equation for loss of head in Pipes in which the flow is turbulent
Consider a steady and uniform turbulent flow in a pipe of cross sectional area ‘A’
m2 and ‘L’m long in which fluid flows with a mean velocity Um/A
Let p2 and p2 be the pressures at (1) and (2) respectively.
Now applying Bernoulli’s theorem between (1) and (2), we have
p V L
gZ
p V Lg
Z hf1 1
12 2
22 2ω ω+ + = + + +
But Z1= Z2 , v v1 2=
∴(p1 − p2 )
ω= hf =loss of head due to friction
E310/1 130 Further for steady flow
∑ =F 0 because acceleration =0
i.e., (p1A-p2A) = Frictional resistance
=f1xSurface area xϑ n
(By Froude’s experiments)
=f1 p x L x vn
Where p= Wetted perimeter
(p1-p2) =f1 pLA
vn =f1MLv n
Where m= Hydraulic mean depth
=pA
=π
πD
D2 4=
4D
Take n=2.0
Then (p1-p2) = f1 X4D
Lϑ2
i.e.,(p1 − p2 )
ω= hf =
4 f 1
ωLϑ 2
D=
8 f 1Lϑ 2
p2gD
ie hf=fL
gDf
fϑρ
2 1
28
; ( )=
Where ‘f’ is called as friction factor or coefficient of friction
The above equation is called as Darcy- Weis bach equation
This loss is more in magnitude and so it is called as Major loss of head.
7.5 Minor Losses
Minor losses are those which are due to sudden change either in magnitude or in
direction. The magnitude of these losses is less than the major loss due to friction and so
there losses are called as minor losses. These are to be considered when the lengths of
pipe are short. These losses are due to
(a) Sudden expansion(ϑ1 −ϑ2 )2
2g
(b) Sudden contraction=0.5v2
2g
(c) entrance = 0.5v2
2g
E310/1 131
(d) exit = v2
2g
(e) graded expansion or contraction = k( )V V
g1 2
2
2−
(f) bend kv
g
2
2
(g) Obstruction like value ,pipe fittings = kv 2
2g
Derivation of loss of energy due to sudden expansion.
Consider a sudden expansion as shown in figure from A1to A2.
Let p1 and p2 be the pressures and ϑ1andϑ2 be the velocities at (1) and (2) respectively
Now applying Bernoulli’s theorem between (1) and (2) we have
p V
g
L1 2
2ω+ + Z1=
p V Lg
2 2
2ω+ + Z2+ hLSE
Here when the cross sectional area of the pipe is suddenly enlarged from A1 to A2,
the fluid separates from the boundary at sudden expansion leading to deceleration of fluid.
This leads to folders formation which consume certain energy of fluid. After some
distance of travel those eddies die away giving rise to certain heat energy which is not
important in incompressible fluid. So the energy consumed by the eddies is a loss of
energy.
So hLSE=P1 − p2
ω+
V12 − V
2
2
2g
Further applying momentum equation over fluid in between (1) and (2) ,we have
p1A1 + p0 (A2- A1) - p2 A2 =ωQg
(V2-V1)
E310/1 132 Where Q= a1v1=a2v2
But by experiments it is shown that p0≈ p,
So (p1-p2) A2=ωA v
g2 2 (v2-v1)
i.e., (p1 − p2 )
ω=
v v vg
2 2 1( )−, substituting this in the above equation, we have
hLSE =v v v
g2 2 1( )−
+v v
g1
22
2−
=2 2 2
22 1 2 1
22
2v v v v vg
− + −=
( )v vg
1 22
2−
Loss of energy due to sudden contraction
Here loss of energy due to sudden contraction is also due to sudden expansion as shown in
figure
As the flow takes place from A1 to A2 the fluid contracts to a minimum cross sectional
area Ac and Ac
A2
=Cc called as coefficient of contraction Cc=0.62.
So loss of head due to sudden contraction is
HLSC=( )v v
g2 2
2
2−
But v a a vC C = 2 2
vaa
vCC
=2
2
=vC
C
C
HLSe = (vC
v XgC
22
2 12
− )
E310/1 133
=v
g CC
22
2
21
1( )−
=v
g ov
g2
22 2
2
2162
1 0 3752
(.
) .− =
But the actual value of this loss is
HLSe =0.5v
g2
2
2( as the loss of energy due to contraction is not considered here)
Loss of head at entrance
It is also similar to sudden contraction and so HLen=0.5vg
2
2
Loss of head at exit
It is similar to sudden expansion
ie Hlex =( )v v
g1 2
2
2−
But v2≈0
∴ Hlex =v
g1
2
2
7.6 Flow through series pipe line
(a). Between two tanks
Consider the flow of water from upper tank to lower tank through a series
(compound) pipe line as shown in figure.
E310/1 134
The total energy line is drawn taking into consideration of entrance loss hLen=0.5v12/y, loss
of head due to friction in pipe(1) =hf1=f1L1v1
2
2gd
loss of energy due to sudden contraction hLSe=0.5v2
2
2g, loss of head due to friction in pipe
(2) =f2 L2V2
2
2gd2
=hf2, loss of head due to sudden expansion =hLSE =(V2 − V3 )2
2g, loss of head
due to friction in pipe (3) = f2 L3V3
2
2gd3
= hF3 and loss of head at exit =hLex=V3
2
2g, respectively
from A to B, as shown in figure. From this total energy line the respective velocity heads
viz, V1
2
2g,V
g2
2
2,V
g3
2
2
in pipe (1),(2),(3) are deducted, the Hydraulic grade line is obtained which is parallel to
Total energy line which is a line joining piezo meter levels along the pipe line.
The flow through series pipe line can be solved by applying Bernoull,s’ theorem
between (A) and (B) above the datum.
Let
V1,V2,V3 - Velocitiesd1,d2,d3 - diametersL1,L2,, L3- LengthsL1,L2,, L3- Lengths
of pipes (1), (2) , (3), respectively
Then ZPw
Vg
pw
BV
ghA
A A B BL+ + = + + +
2 2
22
2
E310/1 135 Z Z h h h h h h hA B Len f Lse f LsE f Lex+ + = + + + + + + + + +0 0 0 0 1 2 3
( )Z Z HVg
f L Vgd
rVg
f L vgd
V VgA B− = = + + + +
−0 52 2
02 2 2
12
1 1 12
1
22
2 2 22
2
2 32. . ( )
+ +f L V
gdv
g3 3 3
2
3
32
2 2
and by continuity equation
Q= a1v1=a2v2= a 3 v 3
Using the above equations the problems of flow through series pipe line can be solved ,V13
(a) of H is given, and if it is required to find Q, then the unknowns V1,,V2, ,V3 can be
determined using the above 3 equations.
(b) of Q is given, the unknowns, V1,V2 V3 and H can be determined from above 4
equations.
(c) of the pipe lines are large the major loss due to friction is more when compared to
minors losses and so the minor losses in that case can be neglected , without any serious
error. Then the solution of the problem is simple.
(4) Flow through series pipes discharging into atmosphere
Consider the flow through series pipes discharging into atmosphere as shown in
figure.
The energy line and Hydraulic grade lines are also indicated.
This problem can be solved by applying Bernolli’s theorem and continuity equation as
explained below.
Applying Bernaulis Theorem between (A) and (B) we have
Z o o o oV
gh h h h h hA
BLen f Lsc f LsE f+ + = + + + + + + + +
2
1 2 32
E310/1 136
( )V
gVg
f L Vgd
Vg
f L Vgd
v Vg
f L Vgd
B B B
B
212
1 1 12
2
22
2 2 22
2
22
3 32
20 5
2 20 5
2 2 2 2+ + + + +
−+
. .
and by continuity equation we have
Q = a1v1 = a2v2 = a3v3
By using the above equation the flow through series pipe line discharging into atmosphere
can be solved as explained in series pipe line between two tanks.
7.7 Flow through parallel pipes
Let there be ‘n’ no of parallel pipelines in between two tanks as shown in figure.
Now applying Bernoulli’s theorem, between A and B over the datum, we have
ZPw
Vg
ZPw
Vg
hAA A
Bw B
L+ + = + + +2 2
2 2
ZA + 0 + 0 + = ZB + 0 + 0 + hL
ZA - ZB = H = HL = hf1 = hf2 = hf3 =--------=hfn
and Q = Q1 +Q2 + Q3+ --------------------------Qn
Using the above equations the flow through parallel pipe lines can be solved.
Here it can be noted that,
(a) in case if series pipe line loss of head is additive and discharge is constant, where as
(b) in case of parallel pipe lines, discharge is additive and Loss of head is same for all
parallel pipe lines.
7.8 Flow through branch pipes
Consider flow of fluid from tank (A) to tank (B)and tank (C) which are at lower
levels than tank (A), as shown in figure.
E310/1 137
Now applying Bernoull’s theorem between (A) &(D), (D)&(B), (D)&(C), We have
Z Upw
Z hAd
d f+ + = + +0 1
pw Z Z hd
d B f+ = + 2
p w Z Z hd c fθ + = + 3
Q1 =Q2+Q3 if pw
Z Z or Zd b cφ +
>
(Q1+Q2)=Q3 if (Zb)> pw
Zdφ +
The un knows Pa/w v1, v2, and v3 can be obtained by solving the above equations.
Further if fluid flows from A to B and C,. we have
ZA= ZB+hf1+hf2 - along route A to B
ie (ZA-ZB)=hf1+hf2
and ZA= Zc+hf1+hf3 along route Ato C
ie (ZA-Zc)=hf1+hf3
and φ 1 = φ 2+φ 3
Using these equations, without knowing Pd/w, the pressure at junction the flow through
branch pipes can be solved if routes of flow are known.
Identify the following as true or false.
SAQ (4) Loss of head in Laminar flow in proportional to velocity.
SAQ(5) Loss of head in Turbulent flow is proportional to v4
SAQ(6) Loss of head in Laminar flow depends on surface roughness
E310/1 138 SAQ(7) Loss of head in Turbulent flow is dependant on surface roughness
SAQ(8) Loss if head in pipes depends on surface area
SAQ(9) Minor losses are due to change in velocity and direction of flow
SAQ(10) Velocity head in a pipe line is 0.5m, then find the loss of head at entrance.
SAQ(11) In problem (10), find loss of head at exit.
SAQ(12) Hydraulic grade line is always falling in the direction of flow
SAQ(13) Total energy line is always falling in the direction of flow
SAQ(14) The level difference in two tanks is equal to total loss of head
SAQ(15) In series pipe line discharges are additive
SAQ(16) In case of parallel pipe line the loss of head is same for all pipes.
SAQ(17) Tank A and B are above tank C then QA+QB=Qc
SAQ(18) Friction factor is non dimensional
WE (1) Two reservoirs are connected by a pipe line which is 15 cm in dia for the first 6cm
and 25 cm in dia for the remaining length of l5 cm. There is a valve in the 25 cm dia pipe
line. When the valve is completely opened a discharge of 0.112 m3/s is flowing in the pipe
line. Determine the level difference considering minor losses also. The energy loss at
valve is equal to entrance loss of head.
Applying Bernaulic Theorem between (A) and (B) taking datum through (B) , we have
0 + 0 + H = 0 + 0 + 0 + Total Flows
H = ( )05
2 2 2 205
2 212
1 1 12
1
1 2 2 22
2
22
22. _ .V
gf L V
gdv v
gf Lv
gdVg
vg
+ + + + +
Q = a1 v1 = a2v2 ; v1 =2515
2 782
2 2
=v v.
So, H = 0.5 ×(2.78)2 vg22
2+
( )0 04 6 2 78015 2
2 78 12
0 04 150 25 2
152
222 2
22
22
22. .
.( . . ) .
..
× ×+
−+
×+
vg
vg
vg
vg
E310/1 139
H = 23.264 v
g22
2; a2 = ( )Π
40 25 0 0492 2. .= m
vQa
m s22
4 1120 049
2 28= = =..
. /
vg
22
2 = 0.265m
∴H = 23.264 ×0.265 =6.17m
_____
WE (2) Water flows through a 60 cm dia pipe line between two tanks whose level
difference is 20 cm of another parallel pipe of same length and dia ”d” is laid to reduce the
level difference to 10m. Find the dia of 2 nd pipe. Friction factor is same for both the
pipes. The discharge flowing is same in both cases.
Case(1) H1 =hf=
202 3 026 0 6
2 2
5= =fLV
gdfLQ
, . ( . ) (1)
Case (2) H2 = hf1 = hf2
102 3026 0 61
12
12
5=
=
fLd
vg
fL Q. ( . )
(2)
102 30262
22 2
5=
=
fLd
vg
fL Qd. ( )
(3)
Dividing (2) by (1)
1020
05 1
2
2
= =
.
ie Q1 = 0.707Q
But Q=Q1 +Q2 =Q-0.707Q= 0.273Q
Dividing (2) by (3)
1010
10 6
12
5
5
22= = ×
Q dQ( . )
d2
1
2 5
0 6
=
.
0 2730 707 0 6
0 68342 5.
. ..
d
=
=
E310/1 140 d = 0.41 m
WE (3)Two pipes of dia 30 cm and 60 cm are connected in parallel between two
reservoirs. The length and friction factor are same for both the pipes. Determine the level
difference if the total discharge flowing is 3 m3/s. The length of pipe is 100m and f=0.02
Solution:
hf1=hf2
fLQQ
fLQQ
12
15
22
253026 3 026. .
=
dd
2
1
22
1
5
5 22 3
=
= =
Q2/Q1 = 5.657
Q2 = 5.657Q1
But Q=Q1 +Q2 = 3.0
Q1 + 5.657 Q1 = 3.0
6.657Q1 = 3.0
Q1 = 0.451 m3/s
and Q2 = 3.0 - 0.451 = 2.549 m3 /s
so, level difference = hf1 = hf2
=0 02 100
3 026 0 312
5
.. ( . )
× × Q
= 271.99(0.451)2
= 55.32 m
WE(4) Two reservoirs have a difference of level of 10m are connected by a pipe line
which consists of 50 cm dia pipe 3000m long and then 2 parallel pipes of 25 cm dia are
laid to another length of 3000m. Find the total discharge if f=0.04
Solution:
E310/1 141
Route (1) & (2)
fLV
gdfL v
gd12
1
2 22
22 210+ =
0 04 30002 9 81 05
0 04 30002 9 81 0 25
1012
22.
. ..
. .× ×
× ×+
× ×× ×
=V v
v v12
222 0 8125+ = .
But Q1 = Q2+Q3 =
since hf2= hf3
Q2 =Q3
So Q1 =2Q2
Q Q2 1
12
=
( ) ( )Π Π4 4 2
5025 22
2
2 1
2 12
12d v d
v v=
=; ϑ
v2 =2v1
∴v12 +2(2v1)2 = 0.8175
9v12 = 0.8175; v1
2 = 0.0908
v m s1 0 0908 0 3= =. . /
So ,discharge Qd
v112
14=
Π( )
=Π4
(0.5)2 ×0.3
0.1963× 0.3
=0.0589 m3/s
E310/1 142 WE(5) A reservoir A is feeding two tanks, B and C. The level difference between A and
B is 20m and A and C is 30 m. The pipe line from A is 1500m long and 30 cm in dia upto
junction. From junction two pipe lines of 30 cm in dia feed the tanks B and C . Find the
discharge to each tank. Take f=0.04 in all pipes
Take f=0.04 for all pipes
Solution:
Route (1) and (2)
hf1+hf2= 20
Route (1) and (3)
hf1 +hf3 = 30
So 0 04 1500
2 9 81 0 30 04 15002 9 81 0 3
2012
22.
. ..
. ..× ×
× ×+
×× ×
=v v
( )1019 2012
22. h v v+ =
v v12
22+ = 1.962 (1)
and 10.19h (ϑ12 +ϑ2
2)=30
v12+v3
2 = 2.943 (2)
But by continuity equation we have
Q1 = Q2 +Q3
Π Π Π4
0 34
0 34
0 321
22
23( . ) ( . ) ( . )× = +v v v
i.e., v1 = v2 +v3 (3)
v1 = 1962 2 94312
12. . .v v+ −
Solving by trial and error
v1 = 1.37, v2 = 0.292 m/s v3 = 1.032 m/s
Q12
40 3 137 0 0707 137= × = ×
Π( . ) . . .
= 0.0968 m3 /s
E310/1 143 Q2 = 0.0707 × 0.292 = 0.02 m3 /s
Q3 = 0.0707 × 1.032 = 0.073 m3 /s Summary
(1) Reynolds number. is represented by RN =ρ
µvD
where, ρ is mass density
µ viscosity
υ Kinematic viscosity
(2) Initial Reynold’s number = 2000
below which the flow is laminar
(3) loss of head in turbulent flow = hfLV
gdf =2
2is known as Darcy = Weis Bach
equation where ‘f’ is called as coefficient of friction or friction factor.
(4) Pipes in series.
Energy difference H = total losses
Q1 = Q2 = ----------------- = Qn
(5) Parallel pipes
hf1 = hf2 ------------------ = hfn = H
Q = Q1 +Q2 --------------+Qn
(6) Branch pipes for tanks A,B,C
ZA = 2d+pd/w + hf1 where pw
d is pr head at junter
pw
Z Z hdd B f+ = + 2
pw
d +2d= Zc+hf3
QA= QB +Qc
Answers to SAQ
1 & 2 = True 3. 10cm/3 4. & 5 True 6. False
7. - 9. True 10. 0.25 m 11. 0.5 m 12. False
13 & 14. True 15. False 16. - 18 True
Exercise
E310/1 144 7.1 A compound pipe of total length 690 m connects two reservoirs. Water surface in
one reservoir is at elevation 601.2 and in the other at elevation 610.2. There is 300m of 30
cm pipe l50 m of 20 cm pipes, and then 240m of 25 cm pipe, 25 cm pipe is connected to
the lower tank. The changes are sudden (Cc =0.62 ) and friction factor of 30 cm, 20 cm,
and 25 cm are 0.02,0.022 and 0.0213 respectively. Determine (a) Discharge, (b) Various
losses and (c) draw the hydraulic gradient.
[Q= 0.076 m3/s, hLen = 0.0224,
hf1 = 1.19m, hLgc = 0.113m, hf2 = 5.0m hLse = 0.039m
hf3 = 2.52 m, Lhex = 0.12m]
7.2 A pipe 20 cm dia is 20 cm long and the velocity of the water in the pipe is 8m/s.
What loss of head would be saved if the central 6m length of pipe is replaced by30cm dia
pipe, the change of section being sudden (f = 0.04, Cc = 0-62)
[ 1.17m]
7.3 Two reservoirs are connected by a pipe 20 cm dia and 3000m long, the difference
in surface being 15m. Calculate the discharge in lps.
If a loop line 30 cm dia and 1200m long is connected to last 1200m of pipe,
calculate the increase in discharge in Lp on due to addition of loop line. Neglect all the
losses other than due to fraction.
F = 0.03
[1476 lpm, 386 lpm]
7.4 Two pipes with dia 2D and D are first connected in parallel and when a discharge Q
passes the loss of head is H1. When the same pipes are connected in series for the same
discharge, the loss of head is H2.. Find the relationship between H1 and H2. Neglect minor
losses. Both pipes are equal in length and have the same friction factor.
[A1 = 45.7 H2 ]
7.5 Two reservoirs are connected by three pipes laid in parallel, then dia are respectively
d, 2d, and 3d and they are all of the same length L. Assuming to be the same for all the
pipes, what will be the discharge through the larger pipes if that through the smallest is
one m3/s
[5.66 & 15.58 m3/s]
7.6 Three tanks A,B,C are connected by a pipe system. The elevation of A is 33.00 and B
is 39.88m. The discharge from A is 0.07 m3/s Determine the rate of flow into or from
E310/1 145 the reservoirs B and C, Find also the R.L of water level in the reservoir C, Given f =
0.030
Pipe length dia
A to junction 306 cm 30 cm
junction to B 250 cm 25cm
Junction to C 300 m 20 cm
[ Discharge from B = 0.03 m3/s , Q w to c = 0.1 m3 /s
RL of water level is C = 24.75 m]
***
FLUID MECHANICS AND HYDRAULIC MACHINERY
UNIT 8
Laminar Flow and Turbulent Flow
AIM: In this unit the aims are to derive expressions for velocity distribution and loss of
energy in laminar flow through pipes and parallel plate and to distribute between smooth
and rough pipes.
Objectives:
1. To derive expressions for velocity distribution over the cross sectional area of flow, loss of
head and shown stress at boundary for laminar flow through parallel plates, both plates at
rest.
2. To derive an expression for velocity distribution for laminar flow through parallel plates,
one plate at rest and other moving. This flow is called as couette flow.
3. To derive expressions for velocity distribution over the cross sectional area of flow , loss
of head and shear stress at boundary for laminar flow through pipe.
4. To distinguish between smooth and rough pipes in case of turbulent flow through pipes.
5. To derive a condition for the pipes to be hydraulically smooth and rough.
6. To derive expressions for velocity distribution over the cross-sectional area of flow,
friction factor for smooth and rough pipes.
7. To explain the useful moody’s diagram.
E310/1 146 To use the above in solving practical problems.
Laminar Flow:
8.1 Introduction:
We know in the Reynold’s experiments that in case of laminar flow the viscous forces
are predominant over inertia forces and so the layers slide one over the other. Due to
viscosity of fluid, there will be opposition to motion which leads to loss of head over a
length of flow. The movement of layers leads to shear stress in between layers, whose
magnitude varies from point to point. Based on these theories the laminar flow
in studied, between parallel plates and horizontal pipes. This analysis is useful in solving
the problems of laminar flow through parallel plates and pipes.
8.2 Laminar flow between parallel plates at rest:
Consider a steady and uniform laminar flow of fluid between two parallel plates at rest
‘B’ distance apart as shown in Figure. In this flow consider a small elemental volume of
thickness ‘dy’ at a distance ‘y’ from the lower plate, of ‘dx’ m long. Let the width of flow
be 1 unit length. The forces acting on this volume are pressure and shear forces as shown
in figure.
Fig.
As the flow is steady and uniform total forces acting on the volume
∑ F = 0
ie., pdy − p +∂p∂x
dx
dy − τdx − τ ×
∂τ∂y
dy
dy
= 0
E310/1 147
∂p∂x
dx dy =∂τ∂y
dy dx
or ∂p∂x
=∂τ∂y
But according to Newton’s law of viscosity
τ = µ dvdy
Substituting this in the above equation we have
dpdx
= µ d2vdy2
Integrating this equation w.r.t y we have
dvdy
=1µ
dpdx
y + C1
v =1
2µdpdx
y2 + C1y + C2
Using the boundary conditions
v=0 at y=0
v=0 at y=B , we have
0 =1
2µdpdx
× 0 + 0 + C2
∴C2=0
C1 = −1
2µdpdx
B
∴ v =1
2µdpdx
y2 −
12µ
dpdx
By
=1
2µ−
dpdx
By − y2
This equation gives velocity distribution over the cross sectional area of flow.
Velocity is maximum at y=B/2.
.
i.e. vmax =1
2µ−
dpdx
B2
y
=B2
8µ−
dpdx
E310/1 148 Mean Velocity:
To obtain mean velocity consider the elemental discharge through the volume as
dq = v dy
Integrating this
q =1
2µ−
dpdx
By − y2( )dy
0
B
∫
=1
2µ−
dpdx
By2
2−
y3
3
0
B
=1
2µ−∂p∂x
B3
6
=
B3
12µ−∂p∂x
The mean velocityV =q
C.S.Area flow=
qB ×1
V =B2
12µ−∂p∂x
then V =23
Vmax
Loss of heat:
From mean velocity we have
−∂p∂x
=
12µVB2
−dp( )p1
p2
∫ =12µV
B20
L
∫ dx
p1 − p2( )=12µVL
B2
∴hf =p1 − p2( )
w=
12µV LwB2
which is proportional toV .
Shear Stress at boundary:
We know the shear stress at y is given by
τ=µdvdy
E310/1 149
= µ ddy
12µ
−∂p∂x
By − y2( )
τ =12
−dpdx
B − 2y[ ]
τ = τ0 at y = 0, τ = τ0 at y=B and τ = 0 at y = B/2.
∴τ 0 =12
−dpdx
B at y = 0
=12
−dpdx
−B( ) at y = B
=12
−∂p∂x
0( ) = 0 at y =B2
This shows that shear stress is maximum at boundary and zero at the centre.
Further then −dpdx
=
12µVB2
∴τ 0 = ±12
12µVB2
B =
6µVB
Identify the following as true or false:
SAQ(1). ∂p∂x
=∂p∂y
is difficult for steady and uniform flow.
SAQ(2). Mean velocity in case of laminar flow, between parallel plates which are at rest is
2/3 Vmax
SAQ(3). The pressure difference over a length L for laminar flow through parallel plates at
rest which are ‘B’ apart, is 12µV L
B2
SAQ(4). The shear stress at boundary for laminar flow through parallel plates at rest is
±6µV
B
SAQ(5). In case of steady laminar flow through parallel plates at 10 cm apart the pressure
gradient in the direction of flow is -40 KN /m2 /m. Find the maximum shear stress.
WE(1) Between two parallel plates at 0.1 m apart laminar flow of nil of viscosity 2.453
NS / m2 is taking place with 58.872 KN / m2 pressure drop over 20m long plates. Calculate
the discharge per m width, the shear stress at boundary , mean and maximum velocity and
velocity at 0.02 m from the boundary.
E310/1 150
Solution:
−∂p∂x
=p1 − p2
L=
58.872 ×100020
= 2943.6N / m 2 / M
Discharge / m width = q = B 2
12µ−
∂p∂x
=(0.1)3
12 × 2.45358.872 ×1000
20
= 0.1 m3/p
Mean velocity V =q
B ×1=
0.10.1
= 1m / p
=B 2
12µ−
∂p∂x
=(0.1)2
12 × 2.45358.872
20
= 1 m / A
Max velocity Vmax = 3 / 2 Vmea = 1.5 ×1 = 1.5 m / D
Velocity at 0.2 m = 1
2µ−∂p∂x
By − y2( )
=1
2 × 2.453(2943.6) 0.1 × 0.02 − (0.02)2
= 0.96 m / D
Shear stress at boundary = τ0 = −∂p∂x
B2
= 2943.6 ×0.1 / 2
= 147.18 N2 / m2
8.3. Laminar flow through parallel plate one plate stationary and other plate at rest:
Consider steady and uniform flow through parallel plates at ‘B’ m apart, bottom
plate being stationary and tap plate moving with velocity V m/s. Let the width of flow be
1 unit length. Consider an elemental volume of ‘dy’ thick at ‘y’ from the bottom plate of
length ‘dx’.
E310/1 151
Fig.
This type of flow is called as couette flow.
The forces that are acting over the elemental volume are shown in Figure.
Now as the flow is steady and uniform the total force acting over it are zero as the
acceleration is zero.
∴SF = 0
i.e as in the previous case, here also we have
∂p∂x
=∂p∂y
∂p∂x
=ddy
µ ∂v∂y
= µ ∂2v
∂y2
Integrating this twice using the B.C
v =1µ
∂p∂x
y2
2+ C1y + C2
using the B.C v = 0 at y = 0
v = V at y = B
C2 = 0
C2 =1µ
−∂p∂x
B2
+VB
∴ v = −1
2µ∂p∂x
(By − y2) +
VB
y
Thus the velocity distribution in couette flow depends on both pressure gradient in the
direction of flow and velocity of plate.
Here the pressure gradient may be either +ve or -ve.
E310/1 152
Case(1) When dpdx
= 0
v= vB
y -Simple or plain couette flow in which velocity distribution is linear with y.
of V=0 , then velocity distribution is same as that of laminar flow through stationary
parallel plates. This shows that parallel plates. This shows that couette flow is a
superposition of simple couette flow and laminar flow through parallel plates at rest.
The non dimensional form of velocity distribution for general couette flow is
uV
=yB
+1
2µ−∂p∂x
By2 − y2
V
=yB
+B2
2µv−∂p∂x
1 −
yB
yB
Let B2
2µv−∂p∂x
= P
Then uV
=yB
+ P 1 −yB
yB
Now the velocity distribution for different values of P are shown in Figure.
Fig.
Case 2: ∂p∂x
=+ve i.e., P=-ve, i.e., increase in pressure in the direction of flow in this case
for some layers the velocity is -ve(opposite to V)
E310/1 153
Case 3: When ∂p∂x
=-ve , i.e., P = +ve there will be pressure drop in the direction of flow
and there will be no backward flow.
Shear distribution:
We know shear stress at any depth y is
τ = µ dudy
= µ ∂∂y
VB
y −1
2µ∂p∂x
By − y2( )
= µ VB
−1
2µ∂p∂x
B − 2y( )
= µ VB
+ − ∂p∂x
B2
− y
Which shows that P varies linearly with y.
At y = 0 τ01=µVB
+ −∂p∂x
B2
At y=B/2 τ =µVB
At y = B τ02=µVB
+ −∂p∂x
B2
− B
=µVB
− −∂p∂x
B2
τ = 0 at µVB
=∂p∂x
B2
− y
y =B2
+
µVB
−∂p∂x
E310/1 154
Fig.
At this y, the velocity will be maximum.
Identify the following whether true or false:
SAQ(6). In case of plain couette flow velocity distribution is linear with y i.e.,
v = V y/B.
SAQ(7). At B / 2 , in couette flow where B is the distance between two parallel plates, the
shear stress is µ V / B, V is velocity of Tap plate .
SAQ(8). Backward flow occurs in same layers of couette flow if ∂p∂x
= +ve.
SAQ(9). Couette flow is a superposition of simple couette and laminar flow through fixed
parallel plates.
SAQ(10). In couette flow V = 10u / s, the plates are at 10 cm apart, determine the distance
of maximum velocity from fixed plate if ∂p∂x
= -20 Kn/m2/m and viscosity µ = 2 µ-S /
m2.
WE(2). In the previous problem if the tap plate moves at 10 m ./ s, determine the
maximum velocity, its distance from the flow plate and the shear stress over plates and
velocity at 0.02m from fixed plate.
if −∂p∂x
= 20 Kn / m2 / m.
E310/1 155 Solution:
y =B2
+µVB
1
−∂p∂x
; −∂p∂x
= 20,000N / m2 / m
y =0.12
+2.453 ×10
0.11
20,000
= 0.05+0.0122 = 0.0622 m
Vmax = Vs
y +1
2µ−∂p∂x
By − y2( ) at y= 0.00622 m.
Vmax = 10 × 0.0622
0.1+
12 × 2.453
× 20,000 0.1× 0.0622 − 0.0622( )2[ ]= 6.22 + 9.588 = 15.8 m/s.
Vol at 0.02 m from fixed plate
v at y = 0.02 = VB
× 0.02 +1
2µ−∂p∂x
By − y2( )
=100.1
× 0.02 +1
2 × 2.543× 20,000 0.1 × 0.02 − 0.02( )2( )
= 2+6.29 = 8.29 m/s.
τ at y=0 =µVs
+ −∂p∂x
B2
=2.543 ×10
0.1+ 20000 ×
0.12
=254.3+1000 = 1254.3 N / m2
τat y=0.1 , =µVB
− −∂p∂x
B2
= 254.3 - 1000 = -745.7 N / m2
8.4 Steady Laminar flow through horizontal pipe Hagen- Porsenille law:
Consider a concentric cylinder of ‘r’ radius and ‘dx’ long as shown in figure in a
laminar flow through a horizontal pipe of diameter D.
E310/1 156
Fig.
By applying Bernoulli’s equation between (1) and (2) we have,
p1w
+ z1 + v12
2g= p1
w+ z2 + v1
2
2g+ hf
i.e hf =p1 − p2
w
Qv1=v2 , z1=z2
The forces acting on the elemental volume are shown in figure. Then for steady flow, ∑F
= 0 , Qacceleration is zero, we have
pΠr2 − p +∂p∂x
dx
Πr2 = τ2Πrdx
τ = −∂p∂x
r2
But for laminar flow we know that
τ = µ dvdy
where y = R-r is the distance form boundary.
So dy = - dv
i.e., τ = −µ dvdr
= −∂p∂u
r2
i.e., dv =12u
∂p∂x
r dr
Integrating and substituting the B.C v = 0 at r = R, we have
E310/1 157
v =
14µ
∂p∂x
r2( )+ C
and C = −1
4µ−∂p∂x
R2( )
∴ v =1
4µ−∂p∂x
R2 − r2( ) is the velocity distribution over the cross
section of flow, which is parabolic. Velocity is maximum at r = 0
Fig.
i.e., vmax =1
4µ−∂p∂x
R2
=1
16µ−∂p∂x
D2
Then V = vmax 1−rR
2
Mean Velocity:
To get the mean velocity, discharge is to be calculated based on velocity
distribution. So consider at radius ‘r’ a ring of thickness ‘dr’ as shown in Figure.
Fig.
The discharge passing through this ring area is
dQ = dav = 2Πrdr ×1
4µ−∂p∂x
R2 − r2( )
So by integrating we have
E310/1 158
Q = dQ = 2Πrdr 14µ
− ∂p∂x
R2 − r2( )
0
R
∫∫
Q = 2Π1
4µ−∂p∂x
R2 r2
2−
r4
4
0
R
=Π
8µ−∂p∂x
R4
=Π
128µ−∂p∂x
D4
So the mean velocity is v
v = Qa
= Π8µ
− ∂p∂x
R4 × 1ΠR2
=1
8µ−∂p∂x
R2
=1
32µ−∂p∂x
D2
So v =12
vmax
Fig.
Further v = v at a radius r given by
vmax2
= vmax 1 − rR
2
i.e.,rR
=12
∴ r = 0.707 R
E310/1 159 Loss of head:
We know that the mean velocity is given by
v =1
32µ−∂p∂x
D2
− ∂p∂x
= 32µvD2
−∂p( )p1
p2
∫ = 32µvD2
0
L
∫ dx
p1 − p2( )=32µvL
D2
orp1 − p2( )
w= hf =
32µvLwD2 is known as Hagen Poisemille equation.
But hf =fLv2
2gD=
32µvLwD2
Then f =64µρvD
=64Rn
where Rn =Raynolds no = ρvD
µ
So for laminar flow, the friction factor is 64Rn
Shear stress:
The shear stress at any radius ‘r’ is given by
τ = −∂p∂x
r2
So at boundary τ0 = −∂p∂x
R2
( )=
−
=
p pL
D
whL
Df
1 2
4
4
=wL
fLvgD
D2
2 4
E310/1 160
τρ
0
2
8=
f v
Then
τ0ρ
= vA = shear velocity
vA = vf8
Power required to maintain the laminar flow is
P = F × v
= τ × A × v
=whf
2D4
× πDL × v
= w πD2
4v
hf
= wQhf N-m / sec or Kg-m / sec
Also P = Force ×v
=(p1-p2)A ×v
=(p1-p2) Q N-m / sec or Kg-m / sec
SAQ(11). Determine the shear stress at wall in case of a pipe of 10 cm diameter in which
laminar flow is taking place, given the -ve pressure gradient in the direction of motion as
20 KN / m2 / m.
SAQ(12). Determine the maximum velocity in a pipe of 10 cm diameter in which laminar
flow is taking place . Given -ve pressure gradient in the direction of motion as 2 KN / m2 /
m and µ=2.5 NS / m2.
SAQ(13).A laminar flow is taking place at 2 lps in pipe of 0.01 m2 C.S. Determine the
maximum velocity.
SAQ(14). Find the velocity in a pipe of 10 cm diameter at a radius of 2 cm from centre, in
which diameter flow is taking place. Given max velocity as 0.5 m / s.
SAQ(15). Find the radius at which local velocity is equal to mean velocity.
SAQ(16).If Reynolds no =1280 , find the friction for laminar flow through pipe.
SAQ(17).Pressure drop over a length in a pipe in which the flow is laminar, is
2 KN / m2 .Find the power lost, if the ratio of flow is 2 lps.
E310/1 161 SAQ(18). Given the pressure drop over 20m of pipe is 200 KN / m2 in a pipe of 10 cm
diameter in which the flow is laminar. Find the shear stress at wall.
WE(3). Under oil of dynamic viscosity 1.5 poise and specific gravity 0.9 flaws through a
20 mm diameter vertical pipe .Two pressure gauges have been fixed at 20m apart. The
pressure gauges fixes at higher level reads 200 KN / m2 and that at lower level reads 600
KN / m2 .Find the direction of flow and rate of flow. Verify whether the flow is laminar.
Then find the power lost in overcoming the friction.
Solution Taking datum through A
HA = Total energy at A
= pAw
+ zA + vA2
2g
=600 ×10000.9 × 9810
+ 0 +vA
2
2g
= 67.95 +vA
2
2g
Fig.
HB = Total energy at B
= pBw
+ zB + vB2
2g
=200 ×10000.9 × 9810
+ 20 +vB
2
2g
= 42.65 +vB
2
2g
E310/1 162 Q VA = VB ,HA > HB
∴ flow is from A to B (upward).
Loss of head is hf = HA - HB
= 67.95 +VA
2
2g
− 42.65 +
VB2
2g
= 25.30 m.
Assuming laminar flow
hf = 25.30 = 32µLvwD2 =
32 × 1.510
× 20 × v
0.9 × 9810 × 0.02( )2
= 27.18 v.
∴ v =25.3027.18
= 0.93 m / s.
Q =π4
0.02( )2 × 0.93 = 2.92 × 10−4m3 / s
RN =ρVD
µ=
0.9 × 98109.810
×0.93 × 0.02
0.15
= 111.6 < 2000
So the flow is laminar .
Power lost = WQ Hf =0.9 × 9810 × 2.92 ×10−4 × 25.3
= 65.3 Watts.
8.5. Turbulent flow- Smooth and rough pipes
We know that for flow through pipes when Reynolds no is > 4,000 the flow is
turbulent. In this case analytical treatment is very different due to fluctuations in the
velocity in the turbulent flow and so in the deviation of velocity distribution over the C.S
of flow, experimental values are to be considered. Here due to mixing the velocity
distribution is almost uniform over the C.S of flow as shown in Figure.
E310/1 163
Fig.
Due to mixing there will be some change in momentum which offers shear
resistance to flow. In turbulent flow there will be fluctuations in velocity. Let in 2.D flow
the fluctuations in velocity be Vx and Vy in x ans y direction respectively. Now if a fluid
mass ρAvy while moving over C.S of flow ‘A’ in ‘y’ direction fluctuates with velocity Vx
in x direction. Therefore the momentum change is ρAVyVx which is equal to shear stress
over C.S area A.
∴Shear stress τ =Forcearea
=ρAvx vy
A= ρvxvy
The transverse distance in which the fluctuating velocity of a lamp is equal
to mean velocity is called as mixing length ‘l’ by Prandtle in 1925. This fluctuating
velocities are related to mixing length by Prandtle as follows.
vx = ldvdy
= vy
Where v is the mean velocity.
Here vy is of the same order as vx.
∴Shear stress in turbulent flow is ρvxvy
τ = ρl2 dvdy
2
Further Prandtle assumes the mixing length proportional to ‘y’ transverse distance
from boundary i.e.,
l ∝ y
or l = k y
Where k is called Karmen’s constant ‘Roppa’
So τ = ρk2y2 dvdy
2and for small values of y , τ = τ 0
E310/1 164
∴τ 0 = ρk2y2 dvdy
2
τ0p
= k2y2 dvdy
2
dvdy
= 1ky
τ oρ
= v0ky
where vo =Shear velocity = τ0ρ
Now by integrating we have
v =v0k
loge y + C -------------------A
which is a logarithmic velocity distribution in turbulent flow.
Smooth and rough pipes
In fluid flows over a boundary, Prandtle recognises that there would be a small
layer adjacent to the boundary in which velocity varies gradually from zero at boundary to
free stream velocity (undisturbed velocity), as shown in Figure.
Fig.
This layer is known as boundary layer. The thickness of this layer normal to the flow , at
which the velocity variation is with 1% of free stream velocity, is called as boundary layer
thickness ‘S’. When the flow in the boundary layer is of laminar nature , then it is called as
laminar boundary layer and if turbulent nature , it is called as turbulent boundary layer .
But very near to the boundary even in turbulent flow , in a very thin layer fluid can not
maintain normal velocity component and so the flow will be always laminar. Then this
thin layer is turbulent boundary layer, adjacent to the boundary is called as laminar
sublayer whose thickness is ‘s’.
E310/1 165
Fig.
But as shown in fig any boundary surface will have certain irregularities rendering the
surface a roughness with average height ‘k’ where these roughness projections ‘k’ are
complete by submerged by laminar sublayer then the flow is not affected by the surface
roughness. Such pipe is called a hydraulically smooth pipe . If the roughness projections
project into the flow by penetrating the laminar sublayer , then such pipe is called as
hydraulically roughness pipe as the flow is disturbed by the roughness projections.
Velocity distribution in smooth and rough pipes:
We have already obtained an expression for the velocity distribution over a cross
section of flow over a boundary , in turbulent flow is given by v =v∗k
loge y + c
But at y = 0 , v is equal to ‘-α‘ indirectly that the velocity must be zero at certain distance
y’ normal to the boundary as shown in fig.
Fig
i.e., at y = y’ v = 0
∴ = −
C
vk
yelog '
So the velocity distribution is given by
E310/1 166
v =v∗k
logeyy'
Of y’ is known for both smooth and rough pipes, then the velocity distribution is
known for turbulent flow through pipes.
Before arriving at this distance y’ , we have to remember that there is a laminar
sublayer in the turbulent flow , adjacent to the boundary , in which the flow is laminar
type. In this layer then the velocity distribution is parabolic . But as the layer is very thin it
can be considered as linear with y.
Fig.
As shown in fig there will be a transition layer normal to the surface in between
laminar flow , in which the velocity distribution varies gradually from logarithmic to
parabolic. But in the absence of clear cut demarcation between each zone, the
intersection point of these two velocity distributions can be assumed to be the laminar
sublayer itself.
Velocity distribution laminar sublayer:
We know that the shear stress at boundary in laminar flow is given by
τ0 = µ vy
E310/1 167 ie., v = τ0
yµ
= ρτ0ρ
yµ
where τ 0ρ
= v∗
vv∗
= ρv∗yµ
=v∗y2
Velocity distribution in smooth pipes
By Nikaradse’s(student of Parandtle) experiments on flow through sand coated
pipes , the parameter v yvs = 116. at y =δ’
i.e.,v∗δ '
v= 11.6
or δ ' =11.6v
v∗− − − − − − − (1)
and at y = y' ,v∗y'
v= 0.108
i.e., y' =0.108v
v∗− − − − − −(2)
But we have from equation (1)
vv∗
=δ'
11.6
substituting y = g this in equ (2) We have y’ =0.108 ×δ’ / 11.6=δ’ / 107
Substituting this value of y’ in the velocity distribution
v =v∗k
logeyy1
, we have using k = 0.4
v =2.30.4
v* log10v∗y
0.108v
vv∗
= 5.75 log10v∗yv
+ 5.75log10(9.254)
= 5.75log10v∗y
v
+ 5.5
This equation is known as Karman -Prandtl equation for velocity distribution for
hydrautically smooth pipes. The velocity distribution for smooth pipe may also be given
E310/1 168 by an exponential equation empirically for v*y / v between 70 and 700 as
vv∗
= 8.74v∗yv
17
Velocity distribution in rough pipes
Nikuradse and others found by experiments that the roughness height k ∝y| and
y|= k / 30.
Now substituting k = 0.4 and y| = k / 30 in v =v∗k
logey′y
we have
vv∗
=2.30.4
log1030y
k
= 5.75log10yk
+ 5.75log10(30)
vv∗
= 5.75log10yk
+ 8.5
which is known as Karman-Prandtl equation for velocity distribution in rough pipes.
Criteria for smooth and rough pipes
This criteria depends on relative magnitude of laminar sub layer and roughness
hight, i.e., k′δ
=k
11.6vv∗
=v∗kv
×1
11.6
i.e.,k′δα
v∗kv
By Nikuradses experiments it was found that when v∗kv
≤ 3 or k′δ≤ 0.25 the pipe is
hydrautically smooth, when 3 <v∗kv
< 70 or 0.25 ≤k′δ
< 60 then the flow is transition
and when v∗kv
≥ 70 or k′δ≥ 6.0 the pipe is called as hydraulically rough pipe.
Mean velocity for smooth and rough pipes:
To get the mean velocity discharge is to be calculated base on the velocity
distribution as explained below.
E310/1 169
Fig.
Consider an elemental ring of ‘dr’ thickness at radius ‘r’ in a pipe of radius R.
The discharging following in this elemental area is dQ = 2Πr dr v.
Q = 2πr v∗ 5.75log10v∗yv
+ 5.5
0
R
∫
is the discharge through smooth pipe.
But y = R - r.
∴Q = 2πv∗ 5.75log10v∗(R − r)
vrdr + 5.5rdr
0
R
∫
So mean v =Q
πR2 = v∗ 5.75log10v∗Rv
+1.75
i.e., v
v∗= 5.75log10
v∗Rv
+1.75
is mean velocity for smooth pipe.
Similarly using equation for velocity distribution for rough pipes , the mean velocity in
rough pipes is given by
v = 1πR2 2πrv∗ 5.75log10
(R− r )k
+ 8.5
0
R
∫
dv
vv∗
= 5.75log10Rk
+ 4.75
Subtracting these mean velocity distribution in the respective local velocity distribution we
have
v − vv∗
= 5.75log10yR
+ 3.75 is identical to both smooth and rough pipes.
At y = R , v = vmax
Then for both smooth and rough pipes we have
E310/1 170
vmax − vv∗
= 3.75
v − vmaxv∗
= 5.75log10yR
Friction factor for smooth and rough pipes:
By dimensional analysis we have fraction factor in turbulent flow is function of RN
and relative magnitude roughness height to diameter of pipe , i.e.,
f = φvDv
k
D
(a) For laminar flow f = 64 / RN for RN ≤2000
(b) For turbulent flow:
(i)Blasius developed
f =0.316
RN( )1/ 4 for RN =4000-105
(ii) For RN > 105 ‘f’ is developed as below. The velocity distribution for smooth pipes is
vv∗
= 5.75log10v∗R
v
+1.75
Substituting v∗ = v f8
in above
v
v f8
= 5.75log10v f
8 R
V
+1.75
1f
= 2.03log10 RN f( )− 0.91
But by Nikuradse experiments , it is corrected as
1f
= 2 log10 RN f( )− 0.8 for RN = 5 ×104 − 4 ×107
This is known as Kurmon-Prandtl resistance equation for smooth pipes, which is to be solved
by Trial and error .
Further Nikuradse gave another equation for the same as
f = 0.0032 +0.221
RN( )0.237
(c) For rough pipes friction factor is obtained as below.
E310/1 171 The velocity distribution for rough pipes in turbulent flow is
vv∗
= 5.75log10Rk
+ 4.75
Substituting v∗ = v f8
, in above we have
v
v fRk
8
5 75 4 7510=
+. log .
( )12 03 0 9110f
R fN= −. log .
which is corrected by Nikuradse’s data as
1f
= 2 log10Rk
+1.74
which is known as Kurman-Prandtl resistance equation .
Thus in smooth pipes ‘f’ depends on Reynold’s no. only where as in rough pipes it is
independent of RN but depends on (R / k) only. So criterion for smooth and rough pipes is
RN f
Rk
< 17 smooth pipes
> 400 Rough pipes
=17-400 Transition pipes.
For any commercial pipe , the friction factor ‘f’ can be obtained by L.F Moody, diagram
shown below on log-log scale , knowing RN and relative magnitude of radius of pipe ‘R’
or roughness height ‘k’ i.e., (R / k). The roughness height of commercial pipes can be
obtained by equating the loss of energy of both sand coated pipes and commercial pipes.
E310/1 172
Fig.
Identify the following as true or false:
SAQ(19) In turbulent flow the velocity distribution is logarithmic distribution.
SAQ(20). If laminar sublayer submerges the roughness , then it is called as smooth pipe, if it
projects out of the sublayer, it is called as rough pipe.
SAQ(21). In turbulent flow, the velocity is zero at certain depth from boundary for
logarithmic velocity distribution.
SAQ(22). The velocity distribution is zero at y| = S| / 107 where S| is the laminar sublayer
thickness, in smooth pipes.
SAQ(23). For rough pipes roughness value k = 3. y ‘ where y| is the depth of flow from
boundary at which velocity is zero.
SAQ(24). The criterion to distinguish between smooth and rough pipes, depends on relative
magnitude of laminar sublayers and roughness height.
SAQ(25).The difference of local velocity and mean velocity in turbulent flow is identical for
both smooth and rough pipes.
SAQ(26). Difference of Maximum and mean velocities in turbulent flow through pipes is
3.75×v* .
SAQ(27). Friction factor in turbulent flow is function of RN and R / k, relative magnitude of
radius and roughness height.
E310/1 173 SAQ(28). For rough pipes ‘f’ is independent of RN.
SAQ(29). For smooth pipes ‘f’ is dependent on RN but not on roughness height of pipe.
SAQ(30). Same pipe will behave smooth and rough if v is varied.
SAQ(31). Given v* =0.5 m / s in a turbulent flow v = 0.01 stokes, roughness height k=1.0 mm
determine whether the flow is smooth or rough.
WE(4) For turbulent flow in pipes , show that vmax
v= 1.33 f +1
Sol:
The velocity distribution for turbulent flow in pipes is
v − vv∗
= 5.75log10yR
+ 3.75 for both smooth and rough pipes.
Now at y = R , v = v max
sovmax − v
v∗= 3.75
substituting v∗ = v f8
vmax − v
v f8
= 3.75
vmaxv
= 3.75 f8 +1
vmaxv
= 1.33 f +1
But by Nikuradse experiments
vmaxv
= 1.43 f +1
WE(5) A turbulent flow of water is flowing in a pipe of 10 cm diameter with roughness height
0.5 mm with a mean velocity of 10 m / s. Given v = 0.01 stokes. Find whether the flow is
smooth or rough. If it is rough for what velocity it will behave like smooth pipe. f = 0.02 .
Find the decmeter of pipe.
Solution:
Criterion for smooth and rough pipe is v∗kv
E310/1 174
v∗ = vf8
= 10 ×0.02
8= 0.5m / s
So v∗kv
=0.5 ×100 × 0.5
0.01= 250 > 70
∴The pipe is rough pipes.
So for smooth pipe v∗kv
≤ 3
or v∗ =3 × v
k=
3 × 0.010.05
= 0.6 = 0.05 × v cm / s
v =0.6
0.05= 12cm /s.
Further for smooth pipes
RN f
Rk= 17
i.e., RN f = 17 × Rk
But for smooth pipe
1f
= 2.0 log RN f − 0.8
= 2.0 log17 × R
k
− 0.8
1
0.02= 2.0 log
17 × R0.05
− 0.8
Where R is in cm.
7.07 = 5.06 - 0.8 + 20 log (R)
= 4.26 + 2 log ( R )
log R = 2.8
R = 25 cm.
WE(6)
The velocities in a 30 cm pipe carrying oil are 4.5 m / s and 4.2 m / s on the central line
and at a radius of 5 cm from the axis. Calculate the discharge and shear stress at boundary.
Solution:
For smooth and rough pipes
v − vv∗
= 5.75logRy
E310/1 175
4.5 − 4.2( )
v∗= 5.75log
0.150.15 − 0.5
=1.01252.
v* =0.296 m / s = v f8
But mean velocity is related to vmax as vmax − v
v∗= 3.75 for both the smooth and rough pipes.
Now at y = R , v = vmax
So vmax − v
v∗= 3.75
Substituting v∗ = vf8
vmax − v
v f8
= 3.75
4.5 − v0.2963
= 3.75
∴v = 3.389 m / s.
Q =πD2v
4
=π4
0.3( )2 × 3.389
= 0.2395 m3 / s.
Further v* =v f8
0.2963 = 3.389 f8
f = 0.0612
Shear stress at boundary τ0:
v∗ = τ0ρ = 0.2963
τ0 = (0.2963)2 ×ρ
= (0.2963)2×1000 N / m2
= 87.79 N / m2
E310/1 176 WE (7). A 1000m long pipe line of diameter 0.3 m carries oil at the rate of 540 lps. of the
specific gravity of oil is 0.8 and Kinematic viscosity is 0.023 stokes, determine the loss of
head, max velocity and at 10cm from axis and the maximum roughness height uph which
it behaves like smooth pipe.
Solution:
v =QA
=540 ×10154
0.3( )2= 7.64 m / s.
RN =vDv
=7.64 × 0.3
0.023 ×10−4 = 9.97 ×105
For smooth pipe
1f
= 2.0 log10 RN f( )− 0.8
=2.0 log10 9.97 ×105 f( )− 0.8
=2.0 log10 f +11.2
Solving by Trial and error
f = 0.0116
For smooth pipe
RN f
Rk≤ 17
Rk
>RN f
17=
9.97 ×105 × 0.011617
= 6317.65
k =R
6317.65=
0.15 ×10006317.65
= 0.0237mm
Loss of head hf =fLv2
2gd=
0.0116 ×1000 × 7.64( )2
2 × 9.81 × 0.3
= 115.0 m
Shear velocity v* = v f8
= 7.64 0.0116
8= 0.29 m / s.
Maximum velocity for smooth pipe:
E310/1 177
vv∗
= 5.75log10v∗y2
+ 5.5
at y = 0.15 v = vmax
vmaxv∗
= 5.75log100.29 × 0.15
0.023 ×10−4
+ 5.5
= 5.75×4.278+5.5 = 30.0
vmax = v* ×30 = 8.7 m / s.
Velocity at r = 10 cm ,
y = R - r
=15 - 10 = 5
vv∗
= 5.75log100.29 × 0.05
0.023 ×10−4
+5.5
= 21.84 + 5.5 = 27.348
v = 7.9 m / s.
WE(8):
In problem (7) what should be roughness height . If the pipe behaves like rough
pipe. Then determine the maximum velocity and loss of head.
Solution:
RN fR
k
≥ 400
f ≥400 R k( )
9.97 ×105 ≥ 40.12 ×10−5 Rk
For rough pipes
1f
= 2.0 log10R
k( )+1.74
E310/1 178
140.12 ×10−5 R
k( )= 2.0log10R
k( )+1.74
Solving by trial, R / k = 363.3
k =0.15363.3
= 4.13 ×10−4 = 0.413mm
∴ f = 40.12 ×10−5 0.154.13× 10−4
= 0.1457
f = 0.0212
hL =fLv2
2gd=
0.0212 ×1000 × 7.642
2 × 0.3 × 9.81= 210.4m
Max velocity :
We know velocity distribution in rough pipe is
vv∗
=v
v f8
=v
9.64 0.02128
=v
0.393= 5.75log10
yk
+ 8.5
v = vmax at y = 0.15.
vmax
v∗= 5.75log10
0.154.13× 10−4
+ 8.5
= 14.72 + 8.5 = 23.2
∴vmax= 23.2 ×0.393 = 9.12 m / s
Summary:
(1). For laminar flow the velocity distribution is v =1
2µ−∂p∂x
By − y2( )for parallel plates at
rest.
The mean velocity =B2
12µ−∂p∂x
Max velocity = B2
8µ−∂p∂x
∴ v =23
vmax
Loss of head hf =12µvLωB2
E310/1 179 Shear stress at boundary
τ0 =12
−∂p∂x
B
(2) The velocity distribution in couette flow is
v =vyB
−1
2µ∂p∂x
(By − y2 )
Which is a super position of simple couette flow on laminar flow through stationary plates.
The non dimensional form of velocity distribution for general couette flow is
vV
=yB
+ P 1 −yB
yB
where P =B2
2µ2 −∂p∂x
where V is the velocity of top plate.
(3). For laminar flow through pipes the velocity distribution is
v =1
4µ−∂p∂x
R2 − r2( ) which is a parabolic distribution.
vmax =1
4µ−∂p∂x
R2
v =vmax 1 −rR
2
vav = 1
32µ−∂p∂x
D2
So vav = 1 / 2 vmax
Loss of head hL =32µvLωD2
and f = 64 / RN
Shear stress τ = −∂p∂x
r2
(4) Laminar sub layer thickness S’’ =11.6v
v∗
where v* = shear velocity = vf8
(5) Criterion for smooth and rough pipe is
(1) k / S| ≤0.25 smooth pipe
E310/1 180 k / S| ≥ 6.0 rough pipe
(2) v∗kv
≤ 3 smooth pipe
≥ 70 rough pipe
(6)
(a) Velocity distribution for smooth pipe v / v* =5.75 log10 (v∗yv
)+5.5
rough pipe = v / v* = 5.75 log10(y / k)+8.5 known as Kurman -
Prandtl equations.
(b) Mean velocity for
Smooth pipe v
v∗= 5.7log10
v∗Rv
+1.75
Rough pipe v
v∗= 5.75log10
Rk
+ 4.75
v − vmaxv∗
= 5.75log10yR
for both Smooth and rough pipes
and v − vmax
v∗= 5.75log10
yR
+ 3.75 for both smooth and rough pipe.
(7) Friction factor ‘f’
(a) smooth pipe
1f
= 2 log10(RN f ) − 0.8
(b) For rough pipe
1f
= 2 log10Rk
+1.74
(8)RN f
Rk
≤ 17 smooth pipe
≥ 400 rough pipe
Answers for S.A.Q.
1) - 4) True 5) 2.0 K N / m2 6) - 9) True
10) y = 0.051 11)500 n / m 2 12) 0.5 m / s
13) 0.4 m / s 14)0.42 m / s 15)r = 7.07 cm
E310/1 181 16) f = 0.05 17) 4 N m / s 18) τ0 =250 N/m2
19)- 30) True 31) Rough.
Exercise
8.1) An oil of specific gravity 0.92 and dynamic viscosity of 0.082 pose flows in an 80 mm
diameter pipe. In a distance of 20m the flow has a heat loss of 2m. Calculate (1) The mean
velocity , discharge velocity and shear stress at a radial distance of 38 mm from the pipe
axis and boundary shear stress.
[v =2.197 m / s , Q = 11.04 l/ s
v = 0.4284 m / s , τ = 17.114 Pa τ0 = 18.02 Pa]
8.2) What power will be required for kilometre length of a pipeline to overcome viscous
resistance to the flow of an oil of viscosity 2.0 poises through a horizontal 10 cm diameter
pipe at the rate of 200 l/ min? Find the Reynolds number of the flow if the relative density
of the oil is 0.92.
[ RN = 194.8 , P = 0.905 KW]
8.3) A flow of 60 L / s per meter width of glycerine of specific gravity 1.25 and dynamic
viscosity 1.5 poises takes place between two parallel plates having a gap of 25 mm
between them. Calculate the (1)maximum velocity (2) Boundary shear stress and (3)
Energy gradient
[ Vmax = 3.6 m / s , τ0 =864 Pa hf / L = 5.648]
8.4) Two parallel plates are placed horizontally 10 mm apart. The bottom plate is fixed and
the top plate is moved at a uniform speed of 0.25 m / s. The fluid between them has a
dynamic viscosity 1.472 N S / m2 . Determine the pressure gradient which corresponds to
the condition of zero discharge between the plates and the shearing stress at each plate.
[22.08 K N / m2 / m , -73.6 N / m2 , 147.2 N / m2]
8.5) A smooth pipe line 0.1 m in diameter and 1000m long carries water at the rate of 7.5 lps
of the kinematic viscosity of water is 0.02 stokes. Calculate the head loss , wall shear
stress , centreline velocity ,shear stress and velocity at 40 mm from the centreline and the
thickness of laminar sublayer.
[9.95m ,2.44 N/m2 1.15 m /s 1.95 N /m2
0.95 m /s ,0.47 mm]
8.6). A pipe of diameter 0.3m is to convey water at 40 oC at the rate of 200 lps of the power
required to maintain the flow in 100m length is 61.8 KW, Calculate the value of k, vmax .τ 0
and v*. Take v = 0.0075 stokes (at 40 oC for water)
E310/1 182 [0.57 mm ,3.4m / s ,23.18 N / m2 , 0.152 m /s]
8.7). A 300 mm diameter pipeline carries water at 20 oC with a mean velocity of 7.5 m / s .
The pipeline is new with no surface irregularities at the beginning , but it was found that
the surface irregularities grow at the rate of 0.075 mm per year. Find the number of years
after which the surface irregularities will affect the flow. Take v = 0.01 stokes.
[2.67 years]
8.8). Field tests on a 30 cm cast pipe carrying 0.25 m3/s of water [v = 1×10-6 m2/s] indicate
that the height of roughness projections has incurred to 1.5 mm after many years of
service. What increase in flow can be expected if the pipe is replaced by a new pipe with k
= 0.26 mm of the same diameter .
[40 lps].
*****
Unit 9
Open Channel flow Aims: The aims of this unit are to define open channel flow, types of flow, to explain velocity distribution over the cross section of flow, energy and momentum correction factors, to review velocity equations, in uniform flow and to analyse economical sections. Objectives :
1. To define open channel flow and compare it with pipe flow. 2. To explain types of flow in open channel. 3. To discuss about velocity distribution over cross section of open channel flow. 4. To calculate energy and momentum of fluid flowing in open channels and then to obtain their correction factors. 5. To review uniform flow and to obtain expressions for velocity of flow by chezy, manning and Bazio. 6. To define economical section and to obtain conditions for economical sections of rectangular, trapezoidal, triangular and circular open channels. 7. To apply the above to solve practical problems. 9.1 Introduction:
Like pipe flow, open channel flow is another important branch of Hydraulics, which is very useful to civil engineers. The flow of rain water in streams, rivers comes under open channel flow. In Irrigation the application of theory of open channel flow is very much necessary. So study of open channel flow is very important to civil Engineer without which he can not be a successful civil Engineer.
E310/1 183 9.2. Definition of Open Channel flow Open channel flow is defined as that flow which takes place in a passage with a free surface subjected to atmospheric pressure. Here the water surface is exposed to atmosphere.
Figure
When this flow is compared to the flow through a pipe it is seen as shown in figure. Here the bottom of open channel refers to centre line of the pipe, the free surface is similar to hydraulic gradient, and the total energy is same in both the cases over the datum. In open channels the bed will be slopped down towards the direction of flow, to make the gravity force (component of weight of water) to overcome resistance and to cause the flow of water. Channels are broadly classified into natural and artificial channels. Rivers streams etc are examples of natural channels. Artificial channels are made artificially to carry water with different cross sections like rectangular, trapezoidal, triangular, parabolic and circular sections. Closed conduits flowing partially, with free surface exposed to atmospheric pressure are also called as open channels. Under ground drains are the examples of closed conduit open channels. A channel which has the same shape of cross sections along its length is called as prismatic channel, otherwise it is called as non prismatic channel.
Identify the following as True or False SAQ1 When water surface in a passage exposed to atmospheric pressure it is called as open channel flow. SAQ2 In open channels the flow is gravity flow SAQ3 If the closed conduit running helpful it is called as pipe flow 9.3 Classification of Open Channel flow i. Steady and Unsteady flow If the flow characteristics like velocity depth do not vary with time at any given cross section of the channel, then it is called as steady flow
E310/1 184 ie
dvdt
= 0,dydt
= 0
In prismatic channels since cross section is constant along the length, then the flow is
steady if dydt
= 0
If velocity and depth vary with time at any cross section of channel then it is called as unsteady flow is
ie dvdt
≠ 0 and dydt
≠ 0
Flow in a river during rainy season is unsteady. ii. Uniform and non uniform flow (varied flow) When depth slope, cross section and velocity do not change along a given length, then the flow is called as uniform flow.
ie dyds
= 0, dϑds
= 0, dAds
= 0, dSds
= 0
So uniform flow occurs only in prismatic channels. If the above flow characteristics vary along the length of channel then it is called as non uniform (varied) flow.
ie dy/ds ≠ 0 etc Further varied flow is classified as gradually varied flow and rapidly varied flow. If the depth of flow varies abruptly over a short length of channel, then it is called as rapidly varied flow (R.V.F). Hydraulic jump is the example of R.V.F. iii. Laminar flow and Turbulent flow Flow in open channel can also be classified as laminar, transition and turbulent flows, like in pipe flow, based on Reynolds number which is given by
RN =ρVR
µWhere R = Hydraulic Mean radius
V = Velocity of flow By experiments it is shown that when RN = 500-600 , the flow is laminar in open channels and RN > 2000, the flow is turbulent flow and between 500-2000 the flow in transitional flow. iv. Subcritical, critical and super critical flow
Based on relative magnitudes of gravity force and inertia forces, the flow in channel can be classified as subcritical, critical and super critical flow. The relative magnitude of gravity force and inertia force is given by Froude Number which is square root of ratio of Inertia force to gravity force ie
E310/1 185
ie FN =Inertiaforcegravityforce
(ρL3 VT ) (ρL3 g)[ ]
12
= (ρL2 LT v
(ρL3g)
12
= ρL2 v2
ρL3g
12
=vLg
where L is taken is depth of flow = y
ie FN =vgy
If v< gy , FN <1, then the flow is called as subcritical or tranquil or streaming flow. If v= gy , FN =1, then the flow is called as critical flow.
and of V> gy FN >1, then the flow is called as supercritical, rapid or shooting or torrential flow. Hydraulic jump occurs when the flow changes from super critical to sub critical flow. Identify the following as True or False. SAQ4 During floods, the flow in a river is unsteady. SAQ5 In a prismatic channel ifdischarge is constant, then the flow is uniform flow SAQ6 When mean velocity is equal to gy then the flow is called as critical flow. Geometrical Properties of Channel section
y = Vertical Depth of flow T = Top width of flow A = Cross sectional area of flow P = Wetted Perimeter which is in contact with water R = Hydraulic mean radius or depth = A/P D = Hydraulic depth = A/T
Z = Section factor = A D = A AT = A3
T( )12 for critical flow
Z` = AR2/3 for uniform flow 9.4 Velocity distribution in open channel
Due to free surface and frictional resistance of the boundary surface, the velocity distribution over the cross section of flow is non-uniform. The velocity distributions as measured by pitot tube on various cross sections are shown in figure which are function of shape of section, roughness of channel and bends in the channel.
E310/1 186
The maximum velocity will occur at a depth equal to 0.05 to 0.15 y from free surface. The mean velocity can be computed from the velocity distribution and it is equal to local velocity at a depth o.6y from the free surface. A better approximation for the mean velocity is equal to average of velocities at 0.2 depth and 0.8 of depth from free surface.
ie Vmean = 12
(vel ab 0.2y+vel at 0.8y)
Due to non-uniform velocity distribution over the cross section, the computation of kinetic energy and momentum of flow based on mean velocity are to be corrected by multiplying those by ‘α’ kinetic energy correction factor and ‘β’ momentum correction factor respectively as shown below. α - Kinetic energy correction factor (coriolis coefficient) Kinetic energy based on velocity distribution =
------ do----- based on mean velocity = α12ρAv
v 2 = α
12ρAv3
∴α =ϑ 3
A∫ dA
Av 3 > 1
for turbulent flow α = 1.03 - 1.36β - Momentum Correction factor (Boussining co efficient)
β =ρϑdAϑ
A∫ρVAV
=ϑ 2 dA
A∫
AV 2
β = 1.01 - 1.12 for turbulent flow But generally for turbulent flow , these are taken as unity.
Identify the following as True or False SAQ7 Maximum velocity occur at a depth of 0.05 - 0.15 of depth from free surface. SAQ8 Mean velocity is equal to local velocity at a depth of 0.6 depth from free surface SAQ9 Kinetic and momentum correction factors are taken as unity in turbulent flow.
E310/1 187 9.5 Uniform flow in open channel
Review : We know already that the depth of flow wetted area, velocity and discharge are constant along the prismatic channel in case of uniform flow. So the water surface , the bed of channel are parallel to total energy line. Consider a steady and uniforms flow in a prismatic channel of length L is shown in figure
Figure
The forces acting are 1. Gravity force which ;is equal to the weight component in the direction of flow = w sinθ2. Frictional resistance given by τox PL where τo - shear resistance per unit surface area. 3. Hydrostatic forces p1 & p2 which are equal and opposite as depths y1 = y2 in uniform
flow Now as the flow is steady ∑F = m × a = 0 --- acceleration = 0 in steady flow ∴w sinθ - τ .PL + p1-p2 = 0w sinθ = τ.PL This shows that the gravity force is equal and opposite to frictional resistance in uniform flow w=wΑL
∴τ o =WALsinθ
PL= W
Ap
sinθ = ωRsinθ
But sinθ ≈(z1 − z2 )
l= s. bed slope = tan θ
for small angle ‘θ’∴ τ o = w R So
But τ0 = f8 ρv 2 by pipe flow analysis
ie τ 0 = wRSo = f 8ρv2
E310/1 188
v = 8wpf
Rs0 . = C Rs0
where C = 8gf
is called as chezy’s coefficient
and V= C Rs o is called chezy’s equation. for mean velocity in uniform flow in open channels. Here C is inversely proportional to ‘f’ Darcy weisback coefficient of friction.
The dimensions of C = L12 T −1 and so C is not constant and varies for each system of unit.
It is simple but its determination is difficult. Further by applying Bernoulli’s equation between (1) and (2) , we have Z1+ y1 + V1
2 /2g = Z2 + +y2 + V22 / 2g + hf
But y1 = y2 V1=V2 in uniform flow ∴ hf = Z2 -Z1
slope of TEL = z2 − z1
L=
hf
L= s f = so = sw
Thus all the slopes are parallel to each other in uniform flow. The depth of flow in uniform flow is called as normal depth ‘yn’
To determine C Empirical formulae have been developed to find C as given below. a. The Ganguillet - Kutter formula
Based on flow measurement, in open channels above two Swiss engineers proposed in 1869 an empirical formula to find C in M.K.S as
C = 23 +
0.00155s
+1n
1 + (23 + 0.00155s
) nR
where n- Kutter’s roughness coefficient = Manning roughness Coefficient n - depends on channel surface and its condition. The typical values of n for different surfaces are given below channel surface value of n 1. Very smooth concrete, planed wood 0.012 2. Ordinary concrete lining 0.013 3. B.W lined with CM
0.015
4. C.C finish. 0.015 5. Unfinished c.c 0.017 6. Neatly excavated rock 0.02
E310/1 189 7. Unlined earth channels in good condition 0.02 8. Rubble masonry 0.02 9. Rivers and earth Channels in fair condition 0.025 10. Earth channels with gravel bottom 0.025 11. Earth channels with dense weed 0.035 12. Mountain stream with rock bed 0.045 b. Bazin formula He proposed in 1897 the following formula to find C in M.K.S as
C =157.6
1.81 + mR
where ‘m’ is proposed by Bazin as below
channel surface value of mVery smooth cement , planed wood 0.11 Concrete, brick or unplan ed wood 0.21 Ashlar, rubble masonry or poor B.W 0.83 Earth channels in very good condition 1.54 ” ordinary condition 2.36 ” rough condition 3.17
c. Manning’s formula In 1889 he proposed the following formula to find mean velocity in M.K.S as
v= 1n
R2/ 3 S1/2 which is very simple and given satisfactory results and is widely
used in practice
Here C = 12
R1 6
Identify the following as True or False
SAQ10 Water weight component in the direction of flow is equal to frictional resistance in uniform flow. SAQ11 Chezy’s Coefficient is in verbally proportional to square root of Darcy weisback Coefficient of friction ‘f’ SAQ12 In uniform flow bed slope, water surface slope and energy line slope are all parallel to each other. SAQ13. Chezy’s coefficient C is dependent on surface of channel and its condition SAQ14 Chezy’s coefficient is inversely proportional to Manning’s roughness coefficient
Worked Example (1) A rectangular channel conveys a discharge of 10 m3/s . If the width of channel is
6m, find the depth of flow if C = 54.62 and bed slope = 1
5000Sol:
E310/1 190 Q = AC RS
1.0 = b × yn × 54.626yn
6 + 2yn
1
50006yn
3
6 + 2yn
=10
6 × 54.62
2
× 5000
y3n = 4.655 + 1.552yn
yn = 1.95
Worked Example (2) A discharge of 100 l.p.s is flowing in a rectangular channel of 60 cm wide with a normal depth of 30 cm. Find the necessary slope if C = 56
A = 0.6×0.3 = 0.18 m2
velocity = Q A =0.1
0.18= 0.555 m s
p = b+2d = 0.6+2×0.3 = 1.2 m
Hydraulic mean depth R = A p =0.181.2
= 0.15m
But ϑ = c Rs0.555 = 56 × 0.15 × 3
s =1
1500
9.6 Economical Sections Economical section is that when maximum discharge can pass through it for a given cross section, roughness coefficient and bed slope. By this definition it is clear that discharge is maximum for a given cross sectional area when velocity is maximum since by continuity equation Q = AV. Velocity is maximum
when R is maximum because V = C RS or 1n
R2 3S 1 2 ie R = A/P , R is maximum
whenP is minimum So for given slope and roughness value V is maximum when P is minimum. Therefore section is economical when P is minimum for a given cross section, roughness coefficient and bed slope . By making use of this condition expressions can be obtained for different channels whose sections are to be economical. a. Rectangular channel.
Consider a rectangular channel of bottom width ‘B’ and depth of flow as y
Then P = B+ 2y, A = By
So p = Ay
+ 2y which is function of y only
E310/1 191 If p is to be miximum
dpdy
= 0
ie dpdy
= −Ay2 + 2 = 0
A = 2y2
By = 2y2
B = 2y or y = B 2is the condition for rectangular channel to be economical Further Hydraulic mean radius R = A/P
R =By
B + 2y=
2y × y4y
=y2
ie R =y2
is another condition for rectangular channel to be economical
b. Trapezoidal channel
Consider a Trapezoidal channel with bottom width B and side slope Z : 1 as shown in figure with depth of flow as y Here the cross sectional area is constant The parameters involved are B,y, Z . So there will be three cases as explained below to get conditions for economical sections case (1) for given cross section are A, Z is constant and y is variable.
figure
A = y(B+YZ) and P = B+ 2Y 1 + Z 2
So B = Ay
− yZ
and then p2 =Ay
− yZ + 2y 1 + Z2
which is function of ‘y’ only
for P to be minimum dρ dy = 0
ie dpdy
= −Ay2 − Z + 2 1+ Z
2= 0
Ay2 + Z = 2 1 + Z
2
But substituting for A = y(B+yZ), we have
y(B + yZ)
y 2 + Z = 2 1 + Z2
B+2yZ = 2y 1 + z 2
ie 12
(top width ) = side slope is the condition for Trapezoidal section to be economical
E310/1 192 Further R = A/p =
y(B + yZ)B + 2y 1+ Z
2
But 2y 1 + Z2
= B + 2yZ So substituting this in above equation , we have
R =y(B + yZ)2(B + yZ)
=y2
ie Hydraulic mean radius = 12
depth of flow is another condition for trapezoidal section to
be economical. Further draw a perpendicular line OA from centre of Top width t side of the section as shown in figure.
figure.
OC = 12
top width = (B + 2yZ)
2sin θ =
yy 1+ Z
2 =1
1 + Z2
OA = OC sin θ =(B + 2yZ)
2×
11 + Z
2
But (B + 2yz)
2= y 1 + Z 2
ie OA = y 1+ Z1 + Z
2
2
= y
This means that Trapezoidal channel will be economical when a semi circle of radius ‘y’ with centre as mid point of top width will be tangential to both sides and bottom. So trapezoidal section will be economical for a given cross sectional area keeping B and Z as constant, when
1. 12
top width = side slope
2. Hydraulic mean radius R = y/2 3. Semicircle of radius y and mid point of top width as centre will be tangential to both sides and bottom. case 2. For given cross section B is constant here also we have
A = (B +zy)y p= B+ 2y 1 + Z 2
From (1) Z = A/y2 - B/y
E310/1 193 substituting this for p, we have
p = B+2y 1 +Ay2 − B y
2
dpdy
= 2 1 +Ay 2 −
By
2
+ 2y ×12
1
1 +Ay2 −
By
2 × 2Ay 2 −
By
−2Ay 3 +
By2
= 0
2 1 + (Ay2 −
By
)2
+ 2y
Ay2 −
By
−
2Ay3 +
By2
= 0
1 = ZAy 2
Z =y2
A=
y 2
(B + Zy)Y;
(B+Zy) = yZ
By
+ Z =1Z
,
By =
1Z
− Z =1 − Z 2
Z
ie By
=1 − Z 2
Z
are the conditions required for Trapezoidal section to be
&` A =y 2
Zeconomical when B is constant
case 3 Depth of flow y = constant here also we have A = (B+Zy)y A/Y-Zy=B p = B+2y 1 + Z 2
p =Ay
− Zy + 2y 1 + z 2
For p to be min dp/dz = 0
ie dpdz
= −y +2y
2 1 + z 22z = 0
2z = 1 + z2
1+z2 = 4z2
z= 13
or tanθ = 3
or θ = 600
So when depth is constant for a given cross sectional area, Trapezoidal section is economical when the inclination of slides with horizontal is 600.
c. Triangular channel
E310/1 194 Consider a triangular section with depth of flow as y and side slope Z=1. Let apex angle be 2θ.
figure
Then A =
12
2y tan θ y = y2 tan θ
p = 2y sec θ
but y = A
tanθ(fromA)
s0 p = 2A
tansecθ
for p to be minimum dp/dθ = 0
ie dpdθ
= 2 Asecθ tanθ
tanθ−
secθ2 tan 32 θ
sec2 θ
= 0
2 tan2 θ - sec 2 θ =0
sin θ =12
or θ = 450 ie Z = 1 So triangular section is economical when included angle is 900 i.e. side slope is 1:1 , i.e. half square on diagonal.
figure
Further R =Ap
=y2 tanθ2y secθ
=y sinθ
2
But sin θ=12
∴R =y
2 2
NOTE: Q = ACAp
s
E310/1 195 Q is maximum when A is ;maximum for a given p. Then the conditions will be same for triangular section to be economical as derived in the previous section , for a given p. d. Circular channel
In this case both cross sectional area and wetted Peri meter vary with depth of flow y. Hence in circular channels two conditions will be derived (1) for maximum dischargeand (2) for maximum mean velocity as arrived below. i. Condition for maximum discharge
Let in a circular channel of radius r, the depth of flow be ;y as shown in figure and ‘θ’ bethe angle subtended at centre by wetted perimeter. Then we have
A =Πr 2
2θΠ
− 2 ×12
2sinθ
22 cosθ
2
=r 2θ
2−
12
r 2 sinθ
=r 2
2(θ − sinθ )
p = r θ
Then Q = AC RS = CA Ap
s = c A3
ps
Then for given C & S, Q is maximum when A3/p is maximum
ie d A3 / p( )
dθ= 0
=p3A2 dA dθ − A3 dp dθ
p 2 = 0
but dA dθ =r 2
2(1− cosθ )
and dp dθ = rby substituting dA dθ , dp dθ in above equation we have
3p r 2
2(1− cosθ ) − Ar = 0
2θ-3θ cosθ + sinθ= 0ie = 3080
for maximum discharge depth of flow is y = r+r cos(180-θ/2) = r(1+cos 260)
= 1.8988 r ≈ 0.95D so the depth of flow is 0.95D for maximum Q and hydraulic mean radius is
R = A p =r 2
2(θ − sinθ) ×
12θ
E310/1 196 =
r(θ − sinθ )2θ
=r2
×1
308
=r2
×1
308180
× Π(Π
180308 − sin 308)
= 0.5733r 0.29D So for maximum discharge R = 0.29D The above conditions slightly vary if Manning’s equation is used for velocity i.e y = 0.938D Condition for maximum velocity
We have velocity = V = C RS= C A p s
So V is maximum when A/p is max
i.e d(A / p)
dθ=
pdA dθ − A dp dθp2 = 0
and substituting for dA/dθ as dp/dθ in above equation we have
rθ r 2
2(1 − cosθ ) −
r 2
2(θ − sinθ )r = 0
θ = tan θie θ = 257.50
So the depth of flow for maximum velocity is y = r+ r cos (180-θ/2) r[ 1+cos 51025’]
=1.626r = 0.81D So the depth of flow y = 0.81D for maximum velocity Further R = A/p
= r
2θ(θ − cosθ )
=r
2Π 257.5180
Π180
257.5 − sin 257.5
≈ 0.6086r = 0.3D is maximum velocity the hydraulic mean radius in 0.3D State whether the following are true or false.
SAQ15. When discharge passing through a given cross section is maximum, then it is called as economical. SAQ16. For a given cross section Q is maximum when p is maximum SAQ17 Width of a rectangular channel is 10 cm Find the depth of flow when it is to be economical. SAQ18. When for rectangular section hydraulic mean radius is equal to B/4, then it is economical.
E310/1 197 SAQ19. Half of Top width of Trapezoidal section is equal to side slope, then it is called as economical section.
SAQ20. When hydraulic mean depth is equal to 12
nthe depth of flow , then both
Trapezoidal and rectangular sections are economical SAQ21. The sides and bottom of a Trapezoidal section are tangential to a semi circle of radius equal to depth of flow and its centre being mid point of top width, then the section is economical . SAQ22. When depth of is kept constant, the trapezoidal section is economical if side slope is 600 with horizontal. SAQ23 When the slope of side with vertical is 45 0 then the triangular section is economical. SAQ24 In case of circular section analysis of economical section with cross section being constant is not possible. SAQ25 When depth of flow = 0.95 diameter, then circular channel is economical SAQ26. For maximum discharge to pass through circular channel the hydraulic mean depth is 0.29 diameter SAQ27 For circular channel Q is maximum when (A3/p) is maximum SAQ28 For circular channel velocity is maximum when (A/p) is maximum SAQ29 The mean velocity through a circular section is maximum when depth of flow = 0.81 diameter and hydraulic mean radius = 0.3 dia SAQ30 For a given cross section the economical section circular channel is semi circle. Worked Example (3) A lined rectangular channel with Manning’s n = 0.02, is 5m wide and the depth of flow is 2m with a bed slope of 1 in 1500 keeping the same rectangular shape of section wetted perimeter and slope, find the maximum extent increase in discharge. sol : A1 = 5×2 = 10 m2
p1 = 5+2×2 = 9 m
R1 = A1 /p1 =109
−1.11m
Q1 =12
A1 R1
23 so
12
=1
0.02×10 × (1.11)2 3 (
11500
)1 2
=13.8558 m3 /s For maximum Q, y = B/2 P = 2y+B = 2B 9 = 2B , B = 4.5 m, y = 2.25cm A = 4.5 ×2.25 = 10.125 m2
R =10.125
9= 1.125m
Q =1
0.210.125 × (1.125)2 3 1
1500
12
= 14.148m3 /s
∴ change in Q = 14.148 -13.8558 = 0.29242 m3 /s
E310/1 198 WE (4) Water is to flow in a channel at 12.5 m3 / s with a mean velocity of 1.25 m/s .Calculate the economical cross section of a). rectangular b). Triangular c). Trapezoidal and d). circular section. Which of these have least perimeter and maximum perimeter a). Economical rectangular channel
y = B/2, A = B× y = 2y × y = 2y2 ×
A =12.51.25
= 10 = 2y2 , ∴ y =102
= 5 = 2.236
So B = 2×y = 2 × 2.236 = 4.47 2m p = B+2y = 8.944 m b). Triangular section to be economical θ = 450
and A = 12
× 2y × y = y 2
10 = y2 y = 3.162m p = (2 × 2)y = 8.943m
c). Trapezoidal section to be economical side slopeθ =600 with horizontal
Z = 13
A= (B + 2zy) + B
2
y = (B + zy)y
(B + 2zy)2
= y 1 + z 2 = y 1+13
=2y
3
B + 2 ×13
y
2
=2y
3
B =23
y
So A = 23
y +13
y = 3y 2
10 = 3y2
y =10
3
12
= 2.4m
B =23
× 2.4 = 2.775m
p = 2y 1 + z2
+ B = 2y 1 +13
+23
y = 2 3y = 8.3m
d. Circular section is economical for a given cross section when Q = c A3 p s is maximum when p is minimum i.e
E310/1 199
P = dθ A= r 2
2(θ − sinθ ) =
d 2
8(θ − sinθ)
d =8A
θ − sinθ
∴ p =d2θ
=8A
(θ − sinθ )×θ
For P to be minimum, dp dθ = 0
θ − sinθ −θ
2 θ − sinθ(1− cosθ ) = 0
2(θ − sinθ ) − θ(1 − cosθ) = 0θ = Π
ie economical section is semi circle.
d=8A
θ − sinθ=
8 × 103.142 − 0
= 5.045m
p=rθ =5.045
2× 3.142 = 7.926m
∴The least perimeter is of circular one and highest perimeter is rectangular & triangular Worked Example (5) A power channel of Trapezoidal section has to be excavated through hard clay at the least cost. Determine the dimensions of the channel given Q = 14 m3 /s. Bed slope 1 in1500 and Manning’s n = 0.02
Side slope for best section is 600 with horizontal. So Z = 13
Trapezoidal channel to be economical we have
b + 2zd
2= y z 2 +1
b + 2 ×13
y
2= y y 3
ie b = 2∫3
y
and A = y(b+zy) = 3y2
Q = Aϑ = 14 = 3y 2 ×1
0.02y2
2 3 12500
1 2
∴y = 2.604m
b = 2∫3
y = 3.007m
Worked Example 6) A lined channel ( n = 0.014 ) is of trapezoidal section with one side vertical and the other on a side slope of 1.5 H to 1 v . If the channel is to deliver 9 m3 /s on a slope of
E310/1 200 0.0002 , find the efficient cross section which requires minimum lining. Find the corresponding mean velocity. Sol
figure
A = B +zy2
y
B = A y −zy2
and p = B + y + y Z 2 +1
=Ay
−zy2
+ y + y 1+ z2
=Ay
+ 1 −Z2
+ 1 + z 2
y
For p to be minimum, dpdy = 0
ie dp dy =−Ay2 + 1−
z2
+ 1+ z2
= 0
A = y2 (1− 1 −1.52
+ 1 +1.52
= 2.0528 y2
B =Ay
−zy2
=2.0528y2
y−
1.5 × y2
= 1.3028y
p = 1.3028 y +y+ 3.25y = 4.1056y
R = Ap =
2.05284.1056
y = 0.5y
Q =1n
AR2 3s 1 2
9 =1
0.0142.0528y 2 .(0.5y)2 3 (0.0002)
12
y = 2.062 m B = 2.687 m So A = 2.0528 y2 = 8.7282 m2
∴ Velocity v = QA
=9
8.7282= 1.031m / s
SUMMARY 1. Kinetic energy correction factor
α =ϑ 3dA
A∫
Av3 > 1, for turbulent flow
α = 1.03 - 1.36 and momentum correction factor
E310/1 201
β =ϑ 2
A∫ dA
Av 2 > 1
β for turbulent flow = 1.01 - 1.12 2. Chezy’s equation is v = C RS
C - Chezy’s coefficient R -Hydraulic mean radius S - bed slope for uniform flow bed slope = water surface slope = Energy slope
Manning’s formula v = 1n
R2 3S 1 2
n - Manning’s roughness coefficient and c = 1n
R1 6
4. Economical section is that when maximum discharge can pass through it for a given cross sectional area ie Q = Ac A pS is maximum when p is minimum for given A 5. For rectangular cross section it is economical when y = B/2, R = y/2
6. For Trapezoidal section it is economical when, for a given cross section A, with side slope constant
12
(Top width) = side slope
R = y/2 or when semicircle of radius y and mid point of top width as centre, will be tangential to sides and bottom. When bottom width constant , side slope variable, for given cross section, it is economical when
B y =1− z 2
zand A =
y2
zwhen depth y = constant,
it is economical when side slope θ = 600 with horizontal.
7. Triangular section is economical when the included angle is 900 , ie hole slope is 450 with vertical. 8. Circular section is economical when y = 0.95D for maximum Q R = 0.29D
y = 0.81D for maximum velocity R = 0.3D
and for given cross section A it is economical when it is semicircle Answers to SAQ
(1) - (16) -- True ; 17 - 5m
E310/1 202 (18) - (23) -- True ; 24 - False, (25) - (30) -- True Exercise
9.1 A trapezoidal channel base width 8 m and bed slope 1 in 400, carries water at 12 m3/s and side slopes are 1:1 , compute the normal depth. Take n = 0.025 [0.85 m ] 9.2 A trapezoidal channel has a bed width of 2.0 m side slope of 1.25 horizontal : 1 vertical and carries discharge of 9 m3 / s at a depth of ;2.0m . Calculate the average velocity and bed slope of the channel . Take n = 0.015 [ v = 1.0 m/s , s0 = 2.0554×10-4]9.3 A rectangular channel 3.0 m wide had a badly damaged living whose Manning’s ‘n’ was estimated as 0.025. The lining was repairedand it has now an n = 0.014. If the depth of flow remains the same as 1.3 m as before the repair, estimate the new discharge [Q = 4.894 m3 / s] 9.4 What diameter of a semi circular channel will have the same discharge as a rectangular channel of width 2.0m and depth 1.2 m ? Assume the bed slope & Manning’s n are the same for both the sections [ D = 2.396 m] 9.5 A triangular channel has a vertex angle of 75 0 and a longitudinal slope of 0.001 If mannigs n = 0.015 estimate the normal depth for a discharge of 250 lps in this channel [yn = 0.668m] 9.6 A rectangular channel [ n = 0.02] is 5.0 m wide and 0.9 m deep and has slope of 1 in 1600. If the channel had been designed to be of efficient rectangular section for the same wetted perimeter, what additional discharge could it carry [ ∆Q = 2.211
m3 / s] 9.7 A trapezoidal channel has one side vertical and the other side has a slope of 1.5 H ; 1V . This carries a discharge of 15 m3 /s with velocity of 1.5 m/s . Calculate the dimensions of an efficient section of this shape and also the bottom slope necessary to achieve this discharge [ n = 0.0130] [ y = 2.207n, B = 2.876m, s0 = 3.335 × 10-4 ] 9.8 Deter mine the dimensions of a concrete lined ( n= 0.014) trapezoidal channel of most efficient proportions to carry a discharge of 10.0m 3/s. The bed slope of the channel is 0.005 [ y = 1.25 m, B = 1.444m, side slope Z = 0.5773] 9.9 Determine the efficient section and bed slope of a trapezoidal channel ( n = 0.025) designed to carry 15 m3 /s of flow. To prevent scouring the velocity is to be 1.0 m/s and the side slope of channel are 1V : 2H [y = 2.463m, B = 1.163m, s0 = 4.735 × 10-4]
* * *
FLUID MECHANICS & HYDRAULIC MACHINERY
UNIT X
OPEN CHANNEL FLOW
E310/1 203 Aims:
The aims of this unit are to define specific energy, specific force and to derive conditions
for critical depth to compute critical depth and to apply this concept to channel
transistors.
Objectives:
1. To define specific energy and explain meaning of critical depth by means of specific energy curve.
2. To derive conditions for critical depth for (a). given discharge and (b) given
specific energy
3. To define momentum in open channel flow and obtain an expression for specific
force
4. To derive a condition for critical state of flow for (a) given discharge and (b)
given specific force.
5. To compute critical depth
6. To analyse critical flow in rectangular channels.
7. To apply the above to solve practical problems.
10.1 Introduction:
In the design of transitions in open channels, specific energy and critical depth concept is
very much necessary. Similarly the energy description below spillways, hydraulic jump is
utilised. So the conditions required for the foundation of hydraulic jump can be obtained
from specific force concept. In addition to it there are many practical applications of
hydraulic jump. So this unit is very useful for civil engineers in solving open channel
problems.
10.2 Specific energy critical depth
The concept of specific energy was first introduced by Bakhmeteff in 1912 It is defined
as total energy of flow per unit weight of water which is measured with respect to the
channel bed as datum. Thus datum head in Bernoulli’s total head is zero.
ie specific energy E = y +ϑ 2
2g, But V =
QA
so E= y +Q2
2gA2
E310/1 204 Thus for a given Q in a prismatic channel, specific energy is function of depth of flow
only. So the relationship between specific energy for a given Q, and cross sectional area
A, and depth of flow is as shown in the figure.
Since E = y + v 2
2g , the curve of specific energy is asymptotic to 450 line through
origin and x axis as shown in figure. In the figure it is seen that specific energy is
minimum at a depth of yc . This depth is called as critical depth. So critical depth ‘yc’ in
open channel is defined as that depth at which the specific energy is minimum for given Q
and A. Velocity at this depth is called as critical velocity.
When the depth of flow in the channel is increased velocity decreases for the same
discharge in same channel or vice versa. When depth of flow is decreased the velocity
increased. When depth of flow is more than the critical depth yc , the flow is called as
subcritical flow or tranquil flow and when the depth is less than critical depth the flow is
called as supercritical flow or rapid flow. Thus for any given specific energy E , there are
two possible depths y1, and y2 which are called as subcritical depth and y2 is called as
super critical depth. These depths are called as alternate depths.
For given Q the minimum specific energy can be obtained by differentiating E
with respect to y as follows
E = y +Q 2
2gA2
dEdy
= 1 +Q 2
2g(−2)
1A3
dAdy
= 0
But as shown in figure dA = Tdy
So dA/dy =T
substituting those in the above equation, we have
1=Q 2TgA3 or
Q 2
g=
A3
Tis the condition for critical flow.
E310/1 205 But v =
QA
hydraulic depth D = AT
Then we have Q2
A2
×
1g
=AT
v 2
g= D
v2
gD= 1 or
vgD
= 1
A at critical depth, Fr , Froude number = 1
So for sub critical flow Fr <1 Q D is more and v is less
for super critical flow Fr >1 D is less and v is more
Similarly for a given specific energy, we have
Q2
2gA2 = E − y (A)
Q = A 2g(E − y)
Thus for given E and A, Q is function of y only . So the relationship between Q and y is
obtained as shown in figure.
ie Q is maximum at a depth called as critical depth, which can be obtained by
differentiating Q w.r.to y as follows
dθdg
= 2g(E − y)dAdy
+12
A1
2g(E − y)2g(−1) = 0
2g(E-y)T-Ag = 0
2(E-y) = A/T
E310/1 206
But Q2
gA 2 = 2(E − y) from equation (A)
ieQ2
gA 2 =AT
or Q 2
g=
A3
Tis the condition for critical flow at which Q is maximum, which is same
as the previous equation, for critical depth.
The above curve can be obtained when water is controlled by a gate into a prismatic
channel. When the gate is closed discharge into channel is Q = 0 and depth of flow v/s of
the gate is maximum. As the gate is gradually opened, the depth of flow v/s of the gate
decreases and discharge is gradually increased, till it reaches a maximum value at which
the depth of flow v/s of gate is critical. Further opening of the gate will have no effect
over the flow and thus top portion of the curve is obtained ;and the bottom portion is
imaginary.
Similarly d/s of the gate Q = 0, y = 0 and depth of flow gradually increases till it
reaches a value at which the Q is maximum then this depth is called critical depth. After
words opening of gate will have no effect over the flow. Thus lower portion of the curve
is obtained and upper portion of the curve is imaginary.
From the curve it is seen for any discharge there are two possible depths, y1 and y2 . These
depths y1 is called as subcritical and y2 is called as super critical depth. These two depths
are called as alternate depths.
10.3 Critical flow in rectangular channel
consider a constant flow in a rectangular channel of B wide.
In this case specific energy E = y +v2
2g
Let q be the discharge per metre width
Then ϑ =qy
E310/1 207
So E = y +q2
2gy2
and for min E, dEdy = 0
ie dEdy
= 1 +q 2
2g(−2)y3 = 0
q2
gy 3 = 1
q2
g= yc
3
or yc = q g23 which is called as critical depth Thus at minimum specific energy the
flow is critical.
ie q2
(y2 )gy= 1
ϑ 2
gy= 1
ϑgy
= 1
ie Froude no Fr =1
suppose for E constant we have
q2
2gy2 = (E − y)
or q = y 2g(E − y)
for maximum discharge dq/dy = 0
ie dqdy
= 2g E − y +y
2 E − y(−1)
= 0
2(E-y) - y = 0
But from eqn (A) q2
gy 2 = 2(E − y)
∴q 2
gy2 = y
q2
g= yc
3 which is same as before
E310/1 208 So in rectangular channel , for constant discharge specific energy is minimum or for
constant specific energy discharge is maximum. For these two conditions the flow is
critical which is given by
q2
g= yc
3
Further 2(E-y) -y =0
(E-y) = y/2
E = y + y 2 =32
yc
ie critical depth yc =23
L5
At this depth velocity is critical velocity and bed slope is critical bed slope.
The above condition can be obtained by
Q 2
g=
A3
T
for rectangular channel T = B
∴Q 2
A2( )g =AB
= y
Q2
B2 y2( )g = y
q2
g= y3c which is derived as above
10.4 Momentum of flow in open channel-specific force
Consider a small reach of a channel as shown in figure
The forces that are acting on the fluid are
1. hydrostatic forces p1 and p2 where
P1 = wAz1 and P2 =wAz2
E310/1 209 z1 & z 2 = Centre of Gravity of area of cross section A1 & A2 from force surface
respectively
2. Water weight component W sin θ in the direction of flow where W = water weight in
the reach.
3. Frictional resistance due to surface in contact with water.
By Newton’s 2ns law, we have
∑f = rate of change of momentum per second
∴P1 -P2 +W sin θ -Ff =wQg
(v2 − v1 )
Here as the reach is small, we can neglect Ff =0 also as the slope is small, we can also
neglect the weight component in the direction of flow.
So p1 - p2 =wqg
(v2 − v1 )
wA1z1 − wAz2 =wqg
(v2 − v1 )
or qv1
g+ Az1 =
qv2
g+ Az2 = F is constant
ie qvg
+ Az is called as specific force which is constant . This specific force of the
water at any cross section is the force per unit specific weight of water.
In a prismatic channel for given discharge this specific force is function of depth ‘y’ . So
the relationship between specific force F and depth y is as shown in figure.
ie when the specific force is minimum the depth is called as critical depth. So for specific
force to be minimum we have dFdy
= 0
ie dFdy
= (−1)q2
gA2dAdy
+d Az( )
dy= 0
E310/1 210 where
dAdy
= T
and d(A z) u is the change in moment of Area A about force surface for a change in
depth dy which can be calculated as given below.
d(Az ) = A z + dy( ) + Tdy dy2
− Az = Ady +
T(dy)2
2
∴d(Az ) = Ady by neglecting small terms.
substituting these in above equation we have
q2
gA 2 T =Adydy
= A
q2
g=
A3
Tis the condition for critical flow at which F is minimum and the
depth of flow is critical depth yc
Further q2
A2 ×1g
=AT
= D
v2
gD= 1 or
vgD
= 1 = Fr
is at critical depth Froude number is 1 if Fr is <1, the flow is called as sub critical at which
depth of flow is more and velocity is less
If Fr is >1, the flow is called as super critical flow at which the depth of flow is less and
velocity is more.
Thus when flow passes from supercritical to sub critical through a critical depth, then
hydraulic jump force as shown in figure in which there is turbulent mixing of the fluid.
Due to this mixing lot of energy is dissipated.
So this concept is used for energy dissipation below spillways etc.
From the specific force diagram it is seen that for the same specific force these are two
possible depths y1 and y2.
E310/1 211 The depth y1 is called as super critical depth and the depth y2 is called as subcritical depth.
so these depths for which specific force is same is called as initial depth (y1) and
sequential depth (y2).
Now from specific force we have
qgA
F Az2
= −
or q = gA(F − Az )
so for a given specific force q is maximum when dqdy
= 0
ie ( )dqdy
gA F Az
F AzdAdy
Ad Azdy
=−
× − +−
=
12
1 0( )
( )
ie (F-A z)T = Ad(Az )
dy= A
Adydy
= A2
But (F-A z) =q2
gA
q2
gAT = A2
or q2
g=
A3
Tis the condition for critical flow at which depth is critical and q is
maximum
Thus for given specific energy or specific force, the discharge is maximum when the flow
is critical.
10.5 Computation of critical flow.
When the depth of flow is equal to critical depth, then the flow is called as critical flow.
This can be known by
q2
g=
A3
T
c
qg
= AAT
= zc
where zc is called as section factor for critical flow.
Thus for a given discharge zc is function of depth ‘y’ only.
So for different depths, the relationship between Z and y is shown in figure, where Z =
E310/1 212
AAT
, for different depths, Z can be called as A A T and plotted as shown in figure.
Now for a given discharging
Zc =qq
so from the graph, for this section factor Zc we have yc read. which is critical depth. Or
for a given depth of flow, the critical section factor Zc is known from the graph from
which
Zc =qc
g, the critical discharge ‘qc ’ can be calculated.
For a given discharge the slope of bed can be adjusted such that the uniform depth is equal
to critical depth yc . Or for a given depth of flow, the discharge and slope can be adjusted
such that the flow is critical . Then this slope is called as critical slope sc.
Thus when the flow is in critical state we have.
1. E = specific energy is minimum for a given discharge
2. F = specific force is minimum for a given discharge
3. Discharge is maximum for a given E or F
4. Froude number Fr = 1
5. For rectangular channels
yc =23
E or yc =q2
g
13
and ν c
2
2g=
12
yc or Ec = yc +ϑ c
2
2g= yc +
12
yc =32
yc
ie Ec = 32 yc
Identify the following as True or False
E310/1 213 SAQ 1. For critical flow at a given discharge Froude Number = 1 and E is minimum
SAQ 2. Discharge per unit width is 1m3/s. Find the critical depth in rectangular channel.
SAQ 3.. For rectangular channel critical depth is 1m. Find the maximum discharge
SAQ 4. For rectangular channel E is 1.5 m find the critical depth
SAQ 5 For a discharge 1m3 /s per metre width the critical depth is 0.465 m Find Froude
number.
SAQ 6. When qyc
= gyc , the flow is critical.
Worked Example (1) A rectangular channel with a bed slope of 1 in 200, carries a
discharge of 10 m3 / s , Find the critical depth if the width of channel is 2m.
q = 102
= 5m3 / s / m
yc =q2
g
13
=52
9.81
13
= 1.368 m
Worked Example (2) If the specific energy of flow in 5m wide rectangular
channel is 1.5 m, then calculate the maximum discharge
yc =23
E =23
×32
= 1m
but yc =q2
g
13
or qmax = gyc3( )
12 = 3.13m 3 / s / m
Worked Example (3) A trapezoidal channel with 5 m bottom width and 2H to 1v
side slope carries a discharge of 9.81 m3 / s. Find the critical depth.
A = (B+zy)y
T = (B+2zy)
Now q2
g=
A3
Tfor critical depth.
(9.81)2
9.81= 9.81 =
(b + 2yc )yc( )3
(b + 4yc )
9.81 13 (b + 4yc )
13 = (b + 2yc )yc
or yc = 0.59
E310/1 214 for different depths y = 0.1, 0.2, 0.3 , 0.4, 0.5, 0.6, 0.7. 0.8 find Z = A A T and plot a
curve between z & y as shown in figure
Then find zc =qg
= =9 819 81
9 81..
.
for this value from the graph find yc
10.6 Transitions
A transition is a small portion of a channel of varying cross section connecting two
channel sections. It may be sudden or gradual. It may be contracted or expanded.
It may be obtained either by
a. reducing and expanding bed width or
b. raising or lowering bed
c. by varying both bed width and bed elevation.
These transitions are necessary for measuring devices, to change the velocity to create
hydraulic jump to dissipate energy and to economise cost of hydraulic structures like
aquident etc.
The purpose of a transition is to minimise the energy loss by changing the hydraulic
conditions gradually .
10.6.1 Transitions with reduction in width
In this case the specific energy will be constant for a given discharge in the channel.
When the width is reduced the uniform depth y1 at (1) is reduce to y2 at (2) in subcritical
flow as shown in figure, as the velocity at (2) is increased. This reduction in depth at (2)
continues till it reaches a critical depth at which the discharge per unit width is maximum.
This critical depth yc 2 is less than y2..
If the width at (2) is further reduced beyond the critical flow, the depth at (1) is increased
changing the specific energy E1 at (1) to E11 such that the new depth yc2
1 at (2) will be
the critical depth for the new specific energy E11. The new critical depth yc2
1 is greater
than yc2.
E310/1 215
Similarly if the flow is supercritical flow the depth y2 at (2) is more than y1 at (1) for the
given E1 . As the width goes an reduced the depth at (2) ‘y2’ goes on increased, till it
reaches a critical condition at which the discharge qmax per unit width at (2) is maximum
and the depth yc2. If further the width is reduced beyond this critical flow, the head at (1)
is lowered to increase the specific energy E1 such that the new depth it (2) would be
critical depth yc2
1 for the new specific energy. E11 and the new critical depth Y1
c2 is more
than yc2.
Thus for a given discharge and specific energy, there is a limit for a reduction in width at
which the flow is critical and beyond this reduction in width causes a change in u/s depth
Here E1 = E2 = Ec =32
yc
and yc =q2
2
g
13
q2 = q/B2
10.6.2 Transition with raise in bed.
In this for the same discharge, the specific energy varies, when the bed is raised by ∆z as
shown in figure
E310/1 216
Here E1 = E2 + ∆z
Hence for a given q1 E reaches a minimum as bed gradually is raised. At this minimum
specific energy the raise in bed ie ∆Zmax is maximum possible. This can be obtained by
E1 = Ec + ∆z max
Ec =32
yc
and yc =q2
g
1 3
As in the previous section here also in case of subcritical flow the depth at (2) goes on
decreases till it reaches critical value yc2which is less than y2 which is less than y1, and
in
case of supercritical the depth ‘y2’ at (2) goes on increases till it reaches a critical value
yc2as shown in figure.
If further raise in bed causes the y, depth to raise or fall as the flow is subcritical
or supercritical.
E310/1 217 In case of both reduction in width and raise in bed is analysed by the same procedure.
Identify the following as True or False
SAQ 7 By changing hydraulic conditions the transition reduces loss of energy.
SAQ 8 For a given discharge in a channel the critical depth in the transition is same both
in subcritical and supercritical flows.
SAQ 9 The minimum width or maximum raise in bed of a channel corresponds to critical
flow.
SAQ10 If the specific energy in the channel is 10 m and the critical energy is 9.8 m, the
maximum raise in bed is 0.2 m
SAQ11 If the width is reduced or bed is raised beyond the critical flow, then the v/s depth
will be altered.
SAQ12 A rectangular channel 2.5 m wide, the specific energy is 10m . Then when the
width is reduced till critical flow is obtained the critical specific energy is 10m
SAQ13 The specific energy in a channel is 10.2 m. The maximum raise in bed is 0.2 m.
Then the critical depth is 0.67m.
SAQ 14 Critical velocity head is 12
critical depth
Worked Example (4) A 10 m wide is rectangular channel carries a discharge of 30 m3/s
with 1.66 m depth of flow. Calculate the maximum raise in bed and the corresponding
fall in water surface. If the raise is increased by 10% of the width at critical flow, find the
rise in water bed U/s.
q = 3010
= 3m 3 / s / m
ϑ 1 =3
1.66= 1.8m ϑ 1
2
2g = 0.166m
∴ E1 = y1 +ϑ1
2
2g= 1.66 + 0.166 = 1.826m
and yc = q 2
g
13
=32
9.81
13
= 0.97m
Ec =32
yc = 1.457m
∴ ∆zmax = E1 - Ec = 1.826 -1.457 = 0.36m
Drop in water level = y1 - (∆zmax + yc ) =vc
2
2g−
ϑ12
2g
E310/1 218 = 1.66-(0.36+0.97) = 0.32m
or0.97
2− 0.166 = 0.32m
raise in bed level =- 0.1 ×0.36 = 0.036
New raise in bed is 0.396 m
New specific energy U/s is E11 = Ec +∆z = 1.457+0.396 = 1.853m
E11 = 1.853 = y1
1 +32
y11( )2
× 2 × 9.81
y11 = 1.7 m
Worked Example (5) A rectangular channel 10m wide carries a discharge of
10m3/s with a depth of 0.8 m Find the minimum contraction in bed. If the bed width is
further reduced by 10%. What would be the rise in water level U/s.
v1 =qy1
=1
0.8= 1.25m / s
v12 / 2g = 0.0796 m
E1 = y1 + v12 /2g = 0.8 + 0.0796 = 0.8796 m
E1= E2 = Ec
∴ yc = 2 3 Ec =23
× 0.8796 = 0.5814 = q22 g( )
13
∴q2 = yc3g = 0.58643 × 9.81 = 1.4m3 / s / m
vc2 2g =
12
yc =0.5864
2= 0.2932m
vc = 2.398 m/s
10 =Bcyc Vc = Bc ×0.5864×2.398
∴ Bc = 7.11m
or q2 B2 = Q
∴B2 =Qq
m2
1014
714= =.
.
New width = 7.11 × 0.9 = 6.399m
q21 =
100.399
= y113
g = 1.562
∴ yc1 = 0.629m
Ec1 = E1
1 = 3 2 × yc1 = 0.9436m
E310/1 219
E11 = y1+v1
2/2g = 0.9436 = y 1 + q2
y12 2g
= y11 +
1y1
2 ×19.62
y11 = 0.86 m
Worked Example (6) A rectangular channel 5 m wide carries a discharge of 10 m3
/s at a depth of 1.5 m. If the bed width is reduced to 3m and a hump[ is constructed to
create critical flow determine the same.
ϑ 1 =qy1
=2
1.5= 1.53m / s
ϑ 12
2g = 0.0906m
E1 = y1 + ϑ 12 2g = 1.681m
q2 =10B2
=103
= 3.33m
yc =q2
2
g
13
=3.332
9.81
13
= 1.04m
E1 = ∆z+yc + vc2 / 2g
= ∆z+ yc +1/2 yc
1.681 = ∆z +1.5 × 1.04
∴∆z = 0.0545m
10.7 Summary
1. Specific energy E = y + ϑ 2
2gis the total energy with r.to bed as datum
2. The condition for critical flow is a prismatic channel is Q2 /g = A3/T where T =
top width of flow both for q constant or E is constant
At this critical flow , E is minimum when Q is constant and Q is maximum for
given E..
3. For rectangular channel critical depth = yc = q2 g( )13
where q = discharge / m width
and F r = ϑc2 / yc g = 1
Ec = 1.5 yc ; vc2 / 2g =
12
yc ; yc =23
Ec
4. Qvg
+ Az is called as specific force
E310/1 220 5. For given specific force Q is max at critical flow given by
Q 2
g = A3
T
6. The minimum width Bc of channel is given by critical flow at which
Ec = E1 = 1.5yc
and yc = qc2 g( )
13
where qc = discharge per unit width at (2) = Q /Bc
7. The maximum raise in bed is obtained by critical flow in
E1 = ∆zmax + Ec
Ec = 1.5 yc
yc =q2
g
13
8. Same analysis will be followed in case of both reduction in width and raise in bed.
Here to find the maximum raise in bed the critical flow condition is to be applied.
Answers to SAQ
1. T 2. 0.465 3. 3.13 m3 /s/m 4. 1.m
5. 0.465 6. to 9: T 10. 0.2m 11. and
12. T 13 0.67m 14 T
Exercise
10.1 A rectangular channel 7.5 m wide carries 12 m3 /s with a velocity of 1.5 m/s
compute the specific energy. Find the critical depth and critical velocity and the
minimum specific energy [ 1.18 J/N 0.639m, 2.503 m/s; 0.959 J/N ]
10.2 A trapezoidal channel having a bottom width 5m and side slopes 1:1 carries a
discharge of 12 m3 /s . compute the critical depth and critical velocity. If
Mannings n = 0.02 determine the bottom slope required to maintain the critical
depth [ 0.793 m, 2.61 m/s, 1 in 200 ]
10.3 A trapezoidal channel with a base width of 6 m and side slopes of 2h-1v conveys
water at 17 m3 /s with a depth of 1.5 m. Is the flow subcritical or super critical
[ Fr = 0.38, subcritical ]
10.4 A rectangular channel 2.0 m wide carries a discharge of 6 m3 /s. If the specific
energy is 1.8 m, calculate the alternate depths [ y1 = 0.625 m y2 = 1.626 m]
E310/1 221 10.5 corresponding to a discharge of 9 m3 /s, calculate the critical depth in triangular
channel of vertex angle 72 0 [ yc = 1.99m]
10.6 Calculate the discharge corresponding to a critical depth of 1.5 m in a
a. rectangular channel of 1.5 m wide
b. triangular channel of vertex angle of 1200
c. trapezoidal channel of bed width = 6.0 m and side slopes 1.5 h to 1 v
[ a. q = 8.63/m3/s b. 10.571 m3 /s c.. q = 42.08 m3 /s ]
10.7 A rectangular channel 2.4 m wide carries uniform flow of a 7 m3/s at a depth of
1.5m. If there is a local rise of 150 mm in the bed level, calculate the change in
water elevation. What can be the maximum rise in the bed elevation such that the
u /s depth is not effected. [73.3mm, 262 mm]
10.8 Uniform flow occurs at a depth of 1.5 m in a long rectangular channel 3m wide and
laid at a slope of 0.0009 . If Manning’s n = 0.015 calculate the width of
contraction which will produce critical depth without increasing the u/s depth of
flow. [ 2.08m]
10.9 A rectangular channel is reduced gradually from 2 m to 1.5 m and the floor is
raised by 0.25 m at a given section. When the approaching depth of flow is 1.5 m,
what rate of flow will be indicated if the flow at the contracted section is at
critical depth [3.95 m3 /s]
***
FLUID MECHANICS & HYDRAULIC MACHINERY
UNIT XI
DIMENSIONAL ANALYSIS
Aims:
The aims of this unit are to explain the use of dimensional analysis ,to define fundamental
dimensions and units, to explain Rayleigh’s method and Buckingham Theorem and their
applications.
Objectives:
The objectives of this unit are
1. To explain the use of dimensional analysis
E310/1 222 2. To define fundamental dimensions and units and to differentiate between unit and
dimension
3 . To explain Rayleigh’s method
4. Toexplain Buckingham Theorem
5. To apply the above in solving practical problems.
11.1 Introduction
For civil Engineers Dimensional analysis is very useful in arriving at an equation
governing a complex phenomenon where mathematical solution is not possible. In this
analysis it is possible to verify the homogeneity of an equation and to derive the non
dimensional numbers which govern the phenomenon
This is very useful in experimental investigation using model study. So let us understand
this clearly.
11.2 Dimensional analysis
In any phenomenon there are number of parameters which govern the phenomenon. In
some simple phenomenon it is possible to arrive at an equation governing phenomenon,
connecting the parameters by mathematical analysis. But it is not always possible in some
complex phenomenae. In this case dimensional analysis is a powerful tool to arrive at an
equation governing the complex phenomenon by connecting the parameters involved in it.
In this analysis it is also possible to know about the non dimensional numbers which
govern the phenomenon and how the constants in the equation will very with the non-
dimensional numbers. Using this analysis we can verify the homogeneity of an equation
and to convert an equation in one system of unit to another system of unit .
11.3 Dimensions and units
Any physical quantity can be expressed in terms of primary or fundamental quantities
Mass (M), length (L) Time (T) and Temperature (θ). Temperature is used in
compressible fluid flow.
The other quantities which are expressed in terms of the primary quantities are called as
secondary quantities viz. area (L2), Volume (L3), Velocity (L/T), Acceleration (L /T2),
some people use force (F) , instead of M, which is called as F.L.T. system instead of
M.L.T. system.
Unit is the standard measurement of the above primary quantities . There are absolute and
gravitational units. In absolute system absolute acceleration viz. 1cm /s 2 , 1m/s2 is used
E310/1 223 where as in gravitational units also called as Engineering units, acceleration due to
gravity is used. Thus is absolute units , say in C.G.S units, Mass = 1g(m), Length =
1cm, Time = 1sec, Force = Mass × acceleration = 1g(m)× 1cm / s2 = 1 dyne and in S.I
units , Mass = 1kg(m) , Length = 1m, Time = 1 second, Force = 1kg (m) × 1m / s2 =
Newton. In gravitational units Man = 1 Metric slug, L = Metre , T = second , Force
= 1 kg (m) ×9.81 m / s2 or 1 M.slug ×1m / s2
So Force = 1 kg (f) = 9.81 N
The dimensions and units of important quantities are given below
Quantity symbol Dimension
MLT
FLT
SI
Unit
MKS C.G.S
A) Geometric
Length L L L m m cm
Area A L2 L2 m2 m2 cm2
Volume V L3 L3 m3 m3 cm3
Curvature C 1 / L 1/L m-1 m-1 cm-1
Slope s1i M0L0T0 P0L0T0 m/m m/m cm/cm
Angle α 1θ M0L0T0 F0L0T0 radian or degrees
radian or degrees
radian or degrees
B).Kinematic
Time T T T S S S
Velocity
(linear)
V L/ T L/T m/s m/s cm/.s
angular velocity ω 1/T 1/T` rad/s` rad/s rad/s
Frequency n 1/T 1/T 1/s 1/s 1/s
Acceleration (linear) a L/T2 L/T2 m/s2 m/s2 cm/s2
Angular acceleration α 1/T2 1/T2 rad /s2 rad/s2 rad/s2
Gravitational
acceleration
g L/T2 L/T2 m/s2 m/s2 cm/s2
Discharge Q L3/T L3/T m3/s m3/s cm3/s
Kinematic viscosity v L2/T L2 /T m2/s m2 /s cm2/s
(stake)
E310/1 224 circulation J L2/T L2/T m2 /s m2 /s cm2/s
C). Dynamic
Mass M M FT2 /L kg m.slug g(m)
Force F ML/T2 F N kg(f) Dyne
Weight W` ML/T2 F N kg(f) Dyne
Mass density ρ M/L3 FT2 /L4 kg/m3 M.slug/
m3
g(m)/c
m3
sp weight w,r M/L2T2 F/L3 N/m3 kg(f)/m3
Dy/cm3
Sp gravity s M0 L0 T0 F0L0T0 ------- --------- -------
Pr intensity p M/LT2 F/L2 N/m2
(pascal)
kg(f)/m2
Dy/cm2
Shear stress τ M/LT2 F/L2 N/m2 kg(f)
/m2
Dy/cm2
Dynamic viscosity µ M./LT FT/L2 N.S /m2 kg(f).s/
m2
Dy.s/c
m2
(poise)
Surface Tension σ M/T2 F/L N/m kg(f)/m Dy/cm
Modules of elasticity E M/LT2 F/L2 N/m2 kg(f)/m2
Dy/cm2
Compressibility 1/E LT2 /M L2 /F m2 /N m2/kg(f
)
cm2
/Dy
Impulse, Momentum I,M ML/T FT N.S Kg(f).S Dy.s
Work, Energy W,E ML 2 /T2 FL Nm
Joule(J)
Kg(f),m Dy.cm
Torque Γ ML2 /T2 FL N.M KG(f).
m
DY.Cm
Power P ML:2 /T3 FL/T N.m/s
(Wallw)
Kg(f)-
m/s
Dy.cm/
s
Identify the following as True or False
SAQ 1 Complex phenomenon can be solved by Dimensional analysis
E310/1 225 SAQ 2 The non dimensional parameters governing any phenomenon can be known by
dimensional analysis
SAQ 3 The homogeneity of an equation can be verified by dimensional analysis
SAQ 4 Any physical quantity in one system of unit can be converted to other system of
unit by Dimensional analysis
SAQ 5 The fundamental quantities to express any physical quantity are called as
dimensions
SAQ 6 The standard measurement of any physical quantity is known as unit.
SAQ 7 The dimension of power is FL/ T
SAQ 8 The unit of pressure in S.I system is Pascal
SAQ 9 The unit of work is Joule in SI system
SAQ 10 The unit of power is SI system is Watt
11.4 Homogenous equations
Fourier’s principle of dimensional homogeneity states that an equation which expresses a
phenomenon of fluid flow must be algebraically correct and dimensionally homogeneous.
An equation is said to be dimensionally homogenous if the dimensions of the both sides of
an equation are the same. This can be explained as follows.
consider v = cϑ 2gH
LHs dimensions = v = L / T
RHs dimensions = 1 × L T 2 × L( )12 = L T
LHs dimensions = RHs Dimensions
So the above equation is homogenous equation
Similarly Q = 2 3 L 2gH 3 2 , hf = fLv2
2gd, hf =
32uvLwD2
are all homogenous equations.
But V = C Rs, Q = 1.84LH3/2
are not homogenous equations
Homogenous equations are independent of units and so these equations are the same in
any system of units where as non homogenous equations are to be modified from system
of unit to another system of units
E310/1 226 A dimensionally homogenous equation can be reduced to non dimensional form as
explained below
consider Q = cd 23
L 2gH 3 2 .
QL gH 3 2 = cd
23
2 = constant = M0 L0 T0
Two dimensionally homogenous equations can be multiplied or divided with out effecting
the homogeneity of the resultant equation. But it can not be so if those are added or
subtracted as the resulting equation will not be dimensionally homogenous.
In any dimensionally homogenous equation the number of dimensionless ‘Π’ terms is
equal to number of variables ‘n’ unions the fundamental quantities ‘m’ ie
No of Π terms = (n-m)
Identify the following as True or False
SAQ 11 Dimensionally homogenous equations are the same in any system of units/
SAQ 12 Non homogenous equations are to be modified from one system of unit to
another system of unit.
SAQ 13 The dimensions of both sides of an equation are the same in dimensionally
homogenous equation.
11.5 Rayleigh Method
This method was proposed by Lord Rayleigh in 1899. In this method, if there are ‘n’
variables like x1 , x2 , x3 -------- xn in any complex phenomenon, then dimensionally
homogeneous exponential form of relationship between any one of the variable s and the
remaining (n-1) independent variables can be obtained as shown below.
x1 = c [ x2a x3
b x4c --------------xn
n]
where a,b,c ----n are the exponentials of the independent variables and
C is a dimensionless constant .
Then the dimensions of each variable are substituted in the above equation. Let the
fundamental dimensions be m. Then by equating the exponent of each dimension on both
sides of the equation we get ‘m’ simultaneous equations connecting the exponentials of
the variables. Among the ‘n’ variable there will be (n-m) independent variables and m
repeated variables. The repeated variables are identified by experience and the exponents
of those variables are expressed in terms of the exponentials of independent variables.
E310/1 227 substituting these in the above equation the functional relationship can be obtained by
grouping the variables of the same exponential.
11.6 Buckingham Theorem
This theorem states that if there are ‘n’ variables involved in any phenomenon , which can
be described by ‘m’ fundamental quantities and these variables are related by a
dimensionally homogenous equation then the relationship among ‘n’ variables can be
expressed by (n-m) dimensional quantities and each quantity consists of (m+1) variables
where ‘m’ is the repeated variables and other one is independent variable in the (n-m)
variables. The repeated variables should contain all the ‘m’ fundamental quantities and
at the same time they should not form a non dimensional quantity. This can explained as
below.
Let (x1 x2 x3 - - - - - xn ) be ‘n’ variables in any phenomenon . then
x1 = f (x2 x3 - - - - - -xn )
or f1 ( x1 x2 - - - - - -xn ) = c
where ‘c’ is a dimensionless constant. Then the above phenomenon can be expressed by
(n-m) - non dimensional ‘Π’ terms as follows
f2 = ( Π 1 Π 2 - - - - - - - - - Π (n-m) ) = c1
Where each Π term consists of (m+1) variables, where ‘m’ being repeated variables
raised to some unknown exponentials which are to be determined and one more
independent variable which will be selected from the remaining (n-m) variables. as shown
below
Π1 = x1a1 x2
b2 x3c1 − − − − − − − − − xm
m1 xm+1
Π2 = x1a2 x2
b2 x3c2 − − − − − − − − − −xm
m2 xm+ 2
Πn-m = x1an− m x2
bn−m x3cn− m − − − − − xm
mm−m xn
By substituting the dimensions of each variables in the above Π terms , and equating the
dimensions of R.H.S the Π term to zero, ‘m’ simultaneous equations are obtained in terms
of exponentials. Then solving the above equations the un known ‘m’ exponentials are
obtained and then by substituting those in the aboveΠ term eachΠ term can be obtained.
Thus the governing equation of the phenomenon can be obtained by expressing any one of
Π terms as a function of the remaining Π terms as
E310/1 228 Π1 = f1 (Π2 Π3 - - - - - - - - - - - - - -Πn-m)
or Π2 = f2 (Π1 Π3 - - - - - - - - - - - - --Πn-m) etc
Identify the following as True or False
SAQ 14 When large no of variables are involved in a phenomenon , Bucking ham Π
Theorem is followed
SAQ 15 In Rayleigh method all variables which are treated as function of another
variable are raised to some exponentials.
SAQ 16 In Rayleigh method ‘m’ simultaneous equations are to be solved by determining
the exponentials of repeated variables in terms of exponentials of other variables.
SAQ 17 Rayleigh method is tedious one when large no of variables are involved.
SAQ 18 In Buckingham Π theorem each Π terms consists of m variables raised to
some unknown exponentials and one more variable out of remaining (n-m) variables.
SAQ 19 In Buckingham Π Theorem any phenomenon involving ‘n’ variables can be
represented ;by (n-m) Π terms where ‘m’ are fundamental dimensions.
SAQ 20 In Buckingham Π theorem a relationship between the variable involved in any
phenomenon can be evolved by expressing one of the Π terms as function of remaining
Πterms.
SAQ 21 Grouping of like exponentials is followed in Rayleigh method
SAQ 22 If the variables are wrongly selected the resulting functional relationship is
erroneous
SAQ 23 In dimensional analysis relation of variables depends on vast experience of
investigator.
Worked Example (1) Check the homogeneity of the following equations
Q = Cd a 2gH
Q = 1.84 LH 3/2
Sol: Lhs of Q = Cd a 2gH , dimension is L3 / T
RHs dimension = 1L2 LT 2 L
12
= L3 / T
∴Q = cd a 2gH dimensionally homogenous is dimension of LHs≠RHs
Now Dimension of LHs of Q = 1.84 LH3/2 is L3 / T
and the dimension of RHs = LL 3/2 = L5/2
E310/1 229 ∴ Q = 1.84 LM3/2 is non homogeneous as dimension of LHs ≠ RHs
Worked Example (2) Convert the equation Q = 1.84 LH 3/2 into FPS system
Q = CL H 3/2
C =L3
T×
1LL3 2 =
L12
T
∴Conversion factor from MKS to FPS is
C = (3.28)
12
1= 1.81
∴1.81 ×1.84 LH 3/2 = 3.33 LH3/2 is the equation in FPS system
Worked Example (3) Obtain drag force on sphere of diameter D in a fluid of density P,
and viscosity µ.
Let the velocity of sphere be ‘V’
FD =φ (D,V,ρ,M)
Then FD = C[ Da Vbρcµd]
substituting the dimensions of each variable we have
MLT -2 = M0 L0T 0 La L T( )b ML3
c MLT
d
Equating the exponential of each dimension of both sides, we have
-M- 1 = c+d
-L- 1 = a+b-3c-d
-T- -2 = -b-d
Let D, V, ρ, be repeated variables and os expressing a, b c in terms of d we have
c = 1-d
b = 2-d
a = 1-b+3c+d
a = 1-2+d+3-3d +d = 2-d
∴FD = c[D2-d V2-d ρ1-d µd]
c D 2V 2ρµ
ρvD
d
=c D 2v 2ρρvD
µ
− d
E310/1 230 ∴FD = cρv2 D2 f RN( ) = cDAρv 2 / 2
where A =ΠD2 / 4
∴FD = cDΠD2
8ρv2 , where c =
Π8
Worked Example (4) Solve the WE(3) by Buckingham method
Sol:
Let FD = f(Dv ρµ) or c = f1 (FD, D,V, ρµ)
Here n = 5
m = 3(M,L,T)
No of Π terms = n-m = 5-3 = 2 no
Let repeating variables be D,V, ρ
ie Π 1 = f1 (Da1Vb1 ρc1 FD )
and Π2 = f2 (Da2 Vb2 ρc2 µ)
Then substituting the dimensions of each variable in the Π term we have
M0 L0T0 = Π1 = f1 (La1 (l/T))b1 (M/L3)c1 MLT-2)
Equating the dimensions of RHS to 0 we have
-M- 0 =C 1 +1
-L- 0 = a1 +b1 -3c1 +1
-T- 0 = -b1 -2
∴c1 = -1
b1 = -2
a1 = -b1 +3c1 -1
= +2-3-1 = -2
∴Π1 = f1 (D-2 V-2 ρ-1 FD)
= (FD / ρV2 D2)
Similarly Π2 = M0L0T0 = f2 (Da2 vb2 ρc2 µ)
M0L0T0 = f2 La2 (L T )b2 ML3
c 2 MLT
Again equating the dimensions of RHS to 0 we have
-M- 0 = c2 +1
-L- 0 =a2 +b2 -3c2 -1
-T- 0 = -b2 -1
E310/1 231 c2 = -1
b2 = -1
a2= -b2 + 3c2 +1
=+1-3+1 = -1
∴Π2 = (D-1 V-1 ρ−1 µ )
=M
ρVD
= RN
−1
So the phenomenon can be represented by 2 Πterms viz.
C1 = f1 (Π1 Π2 )
c1 = f1( FDρv 2 D2 ,
1RN
)
f1 (RN) = FD / FD \ ρv2 D 2
or FD = ρv2D2 f (RN) = CDAρv 2
2
where A = ΠD 2
4
where CD is fn (RN)
11.7 Limitations of Dimensional analysis
1. It will not provide any clue to select correct variables involved in any phenomenon.
Experiments are necessary to verify the variables.
2. If will not give complete equation constants are to be determined by experiments.
Summary
1. Dimensional analysis is very powerful tool to solve complex phenomenon.
2. By dimensional analysis we can verify the homogeneity of equation.
3. By Dimensional analysis empirical equation from one system of unit to another
system of unit.
4. Fundamental Quantities are M.L.T system or F.L.T system which will express any
physical quantity.
5. Units are standard measurements of the dimensions.
6. There are absolute system of units viz. C.G.S system and S.I system and
gravitational system viz. M.K.S system.
7. In Rayleigh method homogenous exponential relationship[ between the variable can
be obtained by equating one of the variables to a function of other remaining
E310/1 232 variables raised to unknown exponentials . This method is ttedious when variables
are many
8. In Buckingham Π theorem any phenomenon can be represented by no of Π terms
equal to (n-m) where n = no of variables and m = Fundamental dimensions. The
relationship is obtained by equating one of the Π terms to a function of other Π
terms. Each Π term consists of (m+1) variables where ‘m’ are repeated variables
raised to some unknown exponential and the other independent variables will one of
the remaining (n-m) variables without exponential.
Answers to Short answer questions
(1) - (23) ---True
EXERCISE
11.1 The following is a formula for calculating the maximum flood discharge Q from a catchment of area A
Q = 7000A
A + 4Where Q is in Cu.Ft /s
A is in Sq miles
Convert this to M.K.S system in which Q is in m3 /s and A is in Sq. Km
[Q =123.1AA +10.4
]
11.2 The capillary rise h of a fluid of density ρ and surface tension σ in a tube of
diameter D depends upon the contact angle and gravity g. Obtain an expression for h
by Rayleigh’s method
h D = fσ
lgD2 ,θ
11.3 The critical depth yc in a trapezoidal channel depends upon the discharge Q, the
side slope of the channel, the bottom width B and the gravity g . Obtain expression
for yc by Rayleigh;’s method
E310/1 233
yc
B= f m,
QB2 gB
11.4 The stagnation pressure ps in an air flow depends upon the static pressure po , the
velocity V of the free stream and density ρ of the air. Derive a dimensionless
expression for Ps
pspo
= fp0
pv2
11.5 (a) The discharge Q over a v-shaped notch is known to depend on the angle of the
notch, the head of the water surface, the approach velocity v0 , and gravity g .
Determine the dimensionless discharge
(b) Q for any shape neglecting the velocity of approach, connecting viscosity and
surface tension
(c) Q for rectangular notch
(a) Q
H 2 gH= f θ,
v0
gH
(b) Q = H 5 2g12φ
υ
H3
2g1
2,
σH 2ρg
(c) Q = CdLH 3/2 ]
11.6 Obtain expressions for the resistance to motion of a motor boat, assuming it to be
(a) entirely due to viscosity (b) entirely due to waves
(c) due to viscosity & waves and (d) due to viscosity and surface tension.
(a) RρL2 v2 = φ
ρLvµ
, (b)
RρL2 v2 = φ(v2 gL)
(c) R
ρL2 v2 = φρLvµ
,v2
gL
(d)
RρL2 v2 = φ
ρLVµ
,σ
lv 2 L
11.7 By dimensional analysis show that ;the torque T on a shift of diameter d,
revolving at a speed N in a fluid of viscosity µ and mass density ρ is given by
T = ρd5 N2 φν
d 2 N
using Rayleigh’s method. Then show that power
p = (ρd5N3) φν
d2 N
E310/1 234 11.8 The resistance to the motion of a supersonic air craft of Length L, moving with a
velocity v in air of density ρ depends on viscosity µ & bulk modules of elasticity
K of air. Obtain using Buckingham ‘sΠ theorem, the expression for resistance R
R = ρL2v 2φµ
ρLvKρv 2
11.9 The torque T exerted on the shaft of a turbine is found to be a function of
diameter D, width B and speed N of the runner, densityρ and viscosity µ of the water
flowing in the turbine and difference of pressure P at in let and outlet of turbine. If p
= wH, where w = Sp Wt of water H is the head over turbine , show that torque T may
be expressed by
T = ρN 2 D5φ BD ,
µρD2 N
,gH
D2 N 2
11.10 Obtain an expression for Q through a pump or turbine or compressor which
depends o shaft work gH, Power P, speed of rotation N, diameter of impeller or
sunner D, mass densityρ, and viscosity µ by Bucking ham’s method.
[ QND3 = φ
pρD5 N 3 ,
gHN 2 D 2 ,
µρND 2
]
11.11 For laminar flow in a pipe the drop in pressure ∆p is a function of pipe length L, its
diameter d, mean velocity of flow V and viscosity of fluid µ. By Rayleigh’s
method obtain an expression for ∆p.
[∆p =µvd
φLd
]
11.12 The head loss hL dueto friction in a pipe is function of diameter D, lengthL, rough
ness magnitude ∑ of the pipe, velocity of flow V, the gravity g and fluid densityρ
and viscosity µ then find an expression for hL
hL D = fLD
;∑D
,gDV 2 ,
µρvD
11.13 The shear stress τ0 at the bed of a rough channel depends upon the depth of flow y,
velocity of fluid v, roughness height ∑ of the bed, fluid density ρ and viscosity µ
. Find an expression for τ0
E310/1 235
τ 0 ρv2 = fµ
ρvy,∑y
11.14 The terminal velocity of descent v of a hemispherical parachute is found to depend
on its diameter D, weight w acceleration due to gravity g, density of air ρa and
viscosity of air µ . Obtain an expression for V
[vgD
= fw
ρD3 g
,
µρD gD
]
11.15 The lift force F on an airfoil is a function of the angle of attack α velocity
of flow V, chord length C , span L , density ρ, viscosity µ and bulk modulus of
elasticity E obtain an expression for F
F
ρv 2c2 = fρvcµ
,v p
E,Lc
,α
***
FLUID MECHANICS & HYDRAULIC MACHINERY
UNIT XII
HYDRAULIC SIMILITUDE
Aims: The aims of this unit are to define similitude to explain geometrical kinematic and
Dynamic Similarities , to obtain Froude’s , Reynold’s , Mach’s, Wahers’ and Euler’s
numbers, to obtain model laws to explain types of models and to explain scale effects in
models.
Objectives : The objectives of this unit are
1. To define hydraulic similitude
2. To explain geometrical, Kinematic and Dynamic similarity
3..To obtain Froude’s Reynolds, Mack’s, Waher’s and Euler’s numbers and their
use in model study.
4. To obtain different model laws useful in model study.
5. To explain types of models and obtain model scales.
6. To explain scale effects in model study and to how to modify the results of
model in transferring to proto type , using scale effects
7. To apply the above to practical problems.
E310/1 236 12.1. Introduction
In complex phenomenon , experimental investigation is necessary to arrive at correct
selection of variables involved in the phenomenon and to arrive at the constants involved
in the equation.
In verification of designs , results on prototype is not possible as it is uneconomical and so
models are to be constructed and verify the designs by model results.
For this purpose model study is quite useful. So a thorough knowledge of hydraulic
similitude is necessary for proper design construction and operation of models.
12.2 Hydraulic similitude
Similitude is defined as the similarity between model and prototype. Then the model will
yield quantitative results about the characteristics of prototype . Compute similitude
between model and pro to type can be obtained if both are geometrically, Kinematically
and dynamically similar. These similitudes are explained as below.
12.2.1 Geometrical Similarity
If the ratios of corresponding linear dimensions like length, width, thickness etc., of the
model and prototype are constant, then geometrical similarity exists between model and
prototype. This means the model and prototype are identical in shape but differ in size.
Let Lr be the linear scale ratio then
Lr =Lm
L p
=bm
bp
=t m
t p
etc.
The Area ratio = A r =Lr
Lp
×bm
br
= Lr2
Volume ratio V r =Lm
L p
×bm
bp
×tm
t p
= Lr3
where subscript ‘m’ refers to model and
‘p’ refers to prototype.
12.2.2 Kinematic similarity
It indicates the similarity of flow pattern(flownets), between model and prototype ,. This
means Kinematic similarity between model and prototype exists when the flow paths of
homologous particles are those particles at similar points around model and prototype as
shown in figure.
E310/1 237
This can be obtained when the ratios of velocities and acceleration , of the homologous
particles at the corresponding points are same between model and prototype. For this
similarity, the model and prototype must be geometrically similar. So for Kinematic
similarity we have at similar points
Time scale ratio = (T1 )m
(T1 ) p
=(T2 )m
(T2 )p
= Tr
Velocity scale ratio Vr =(V1 )m
(V2 ) p
=(V2 )m
(V2 )p
=Lm
TmLp Tp
=Lr
Tr
acceleration Scale ar =(a1 )m
(a1 )p
=Lm
Tm2
Lp T 2p
=Lr
Tr2 =
vr
Tr
discharge scale ratio Q r =AmVm
ApVp
= Lr2 Lr
Tr
= Lr3 Tr
or Qr = Lm3 Tm( ) Lp
3 Tp = Lr3
Tr
12.2.3 Dynamic Similarity
Dynamic similarity means the similarity of forces at similar points of model and
prototype. So far dynamic similarity the ratio of all similar forces at homologus points
acting on fluid particles of model and prototype which are geometrically and
kinematically similar, are to be constant or equal. The forces that are acting on fluid
particle at homologus points of model and prototype may be one or combination of
Inertia force Fi
Viscous force Fϑ
Gravity force F gPressure force Fp
Elastic force Fe
Surface tension force Fs
E310/1 238 By Newton’s 2nd law we have The resultant force SF = m ×a
(∑ F)m
(∑ F)p
= ((rFv +
rFs + Fp +
rFc +
rFs )m
(rFv +
rFs +
rFp +
rFc +
rFs )p
=(m × a)m
(m × a)p
=(Fi )m
(Fi )p
as inertia force is equal to mass & acceleration.
Also we have
(Fi )m
(Fi ) p
=(Fv )m
(Fv ) p
=(Fg )m
(Fg ) p
=(Fp )m
(Fp ) p
=(Fe )M
(Fe ) p
=(Fs )m
(Fs ) p
or
(Fi )Fv
m
=(Fi )Fv
p
(Fi )Fg
m
=Fi
Fg
p
(Fi )(Fp )
m
=Fi
Fp
p
Fi
Fe
m
=Fi
Fe
m
Fi
Fs
m
=Fi
Fs
p
Thus complete similitude exists between model and prototype when they are geometrically
kinematically and dynamically similar. So in complete similitude the dimensionless
quantities are constant for both model and prototype .
12.3 Force ratios :
When a fluid mass is in motion inertia force exists . So in fluid flows some non
dimensional parameters can be developed which are going to be same for both model and
prototype as ratio of Inertia force to another predominant force in the fluid. viz,
Reynold’s number RN =InertiaforceViscousforce
12.3.1 Reynold’s number
In fluid flows when viscous force is predominant, this number is to be considered, which
is the ratio of Inertia force to viscous force
Inertia force is given by = mass × acceleration
= ρL3 × L T 2
E310/1 239 = ρL2 × L2 T 2 = ρL2v 2
and viscous force = shear stress × area of surface
= τ × A
= µ vL
× L2 = µVL
∴ Reynold’s no R N =InertiaforceViscousforce
=ρL2 v2
µvL=
ρLvµ
=vLυ
where υ = µ ρ
∴RN =ρvLµ
=vLυ
For flow through pipes, this number is very important.
12.3.2 Froude number ‘FN’
When gravity force is predominant this number is very useful as in the case of open
channel flows.
Froude number is defined as the square root of ratio of Inertia force to gravity force.
So gravity force = mass × acceleration due to gravity
= ρL3 × g
Froude no : ‘FN’ =ρL2 v 2
ρL3 g=
v2
gL=
vgL
This number is useful in the design of hydraulic structures, ships and in the study of
hydraulic jump etc. where gravity force is predominant
12.3.3 Euler’s number ‘EN’
When pressure gradients exists ie fluid flows i.e where pressure force is predominant then
this number is useful which is the square root of ratio of Inertia force to pressure force.
i.e. Euler’s Number ‘EN’ =ρL2 v2
p × L2 =vp ρ
12.3.4. Mach number ‘MN’
In the case of compressible fluid flows at high velocities like high speed missiles,
projectiles in air, etc, the elastic force is predominant and in this case. Mach number is
useful, which is depend as the square root of cauchy number which in turns is the ratio of
Inertia force to elastic force.
E310/1 240
So canchy’s no CN =P2 L2V 2
KL2 =V 2
K ρ
where K is bulk modules of fluid
and C = K ρ which the velocity of sound in that fluid
i.e. MN = CN =Vk ρ
=vc
Here when C = V, MN = 1 which is called as sonic flow
Similarly when V>C, M>1, which is called as supersonic flow and when V<C , M<1,
and so this flow is called as subsonic flow.
When M<0.4, the effect of compressibility of fluid can be neglected.
12.3.5. Weber number ‘WN’
In fluid flows when surface tension forces are predominant then this number is useful
which is defined as square root of ratio of Inertia force to surface tension force
WN =ρL2 v 2
σL=
Vσ ρL
where σ is called as surface tension force per unit length.
This number is useful in case of motion of small jets, droplets , formation of waves and
flow over weir under small heads etc.
For complete similitude the above numbers are to be same for both model and prototype.
Based on this fact the model can be planned, and the test results of the model can be
transferred to prototype.
In planning the models, only the important predominant forces are to be considered and
then less important forces are to be neglected.
12.4 Model laws or similarity laws.
In planning of models, complete similitude is to be considered. But it may not always be
possible because some less important forces in the phenomenon are neglected to make the
problem simple. Also in many practical cases only one force in addition to inertia force is
predominant. So taking this fact into consideration the following model laws are
developed.
12.4.1 Reynold’s Model law
E310/1 241 In fluid flows where viscous force predominant Reynolds number is same for both
model and prototype. So
(RN )Model = (RN)Prototype
∴ρm vn Ln
µm
=ρ p vp Lp
µp
i.e. ρr vr Lr
µ r
= 1
or vr Lr
υr
= 1
i.e. vr =υ r
Lr
=µr
ρr Lr
Based on this fact the other scale ratio can be developed as arrived below.
Let Length scale ratio = Lr
Area scale ratio = A r = Lr2
Volume scale ratio = L3r
Velocity scale ratio = vr =υ r
Lr
=µr
ρr Lr
Time scale ratio = T r =Lr
vr
=ρr Lr
2
µr
Acceleration scale ratio = ar =vr
Tr
=µ r
2
ρr2 Lr
3
Discharge scale ratio = Qr =Lr
3
Tr
=Lr µr
ρr
Force scale ratio = mass × acceleration
Fr = µr2 ρr
Work, Energy, Torque scale ratio = Force × distance moved
=µr
2 Lr
ρr
Pressure intensity scale ratio = µ r
2
ρr Lr2 =
ForceArea
power scale ratio = W.D / sec
=µ r
3
ρr2 Lr
E310/1 242 Mass scale ratio = ρr L3
r
Mass density scale ratio = ρr
specific weight scale ratio = Mass × g
L3 =µ r
2
Lr3 × ρr
Momentum scale ratio = Mass × velocity = µr L2r
12.4.2 Froude Model law
When gravity force is predominant Froude number is useful in planning the model
and so Froude Model laws are developed based on Froude number which is same for both
model and prototype where gravity force is predominant as explained below.
so (FN)m = (FN)p
i.e. vm
gm Lm
=v p
gp L p
or vr
gr Lr
= 1
i.e. Vr = gr Lr or when gr = 1 vr = Lr
Let Linear scale ratio = Lr
Area scale ratio = L2r
volume scale ratio = Lr3
velocity scale ratio = vr = gr Lr =Lr
Tr
Time scale ratio = Tr =Lr
vr
=Lr
gr Lr
=Lr
gr
Acceleration scale ratio = vr
Tr
= gr
Discharge = Lr
3
Tr
= Lr5
2 gr
Mass scale ratio = ρLr3
Force scale ratio = Mass × acceleration = ρr Lr3 gr = Lr
3υ r
Mass density scale ratio = ρr
Specific weigh scale ratio = ρr gr = υ r
pressure scale ratio = υr gr=ρrLrgr
Momentum scale ratio = ρ r Lr7/2 gr
1/2
E310/1 243 Impulse = ρr
1 2 Lr7 2υ r
1 2
Work , energy, scale ratio = L g Lr r r r`4 4ρ υ=
Power scale ratio = Lr7 2ρr gr
3 2 =Lr
7 2υr3 2
ρr1 2
Here is general gr = 1, This law is useful for flows over spill ways, flow through sluices
flow of jet from a nozzle generation of surface waves etc.
12.4.3 Euler’s Model law
When pressure forces control the flow this law is useful in which Euler’s number is same
for both model and prototype.
i.e. (EN)m = (EN)p
i.e. vm
pm ρm
=vp
pp ρ p
or vr
pr ρr
= 1
or Vr = prρr
12
This is useful for flows through pipes where the flow is turbulent in which the viscous
forces and gravity forces are completely absent.
In this case also using Vr , the other scale ratios can be developed is explained in
the previous articles.
12.4.4 Mach Model law
In a compressible fluid flows like flow of air in pipes, or motion of projectiles in air, this
law is very useful in which Mach number is same for both model and prototype. Thus this
law is useful in aero dynamic test, water hammer analysis etc.,
i.e. Vm
Kmρ r
.= or υr = Kr ρ r Based on this scale ratio other scale ratios can be
developed as explained already.
12.4.5 Weber Model law
When surface tension is the predominant force, then this law is useful in which Weber
number is same for both model and prototype.
i.e. vm
σ mρm Lm
=vp
σ pρ p L p
E310/1 244
or vr =
σ r
ρr Lr
Making use of this the other scale ratios can be developed from this. For flows over weirs
under low heads, very thin sheet of liquid flowing over a surface this law is useful.
Identify the following whether True or False
SAQ 1 To find constants involved in the equation and correct selection of variables
involved in a complex phenomenon , model study is necessary
SAQ2 To check the designs economically Model study is necessary
SAQ3. Complete similitude exists between model and prototype if both are
geometrically Kinematically and Dynamically similar
SAQ4 When the ratios of corresponding linear dimensions of Model and prototype are
the same then they are geometrically similar
SAQ5 At similar points the ratios of velocity acceleration of homologus particles are
the same, the model and prototype are Kinematically similar
SAQ6 For Kinematic similarity the geometrical similarity is a pre requisite.
SAQ7 For Dynamic similarity the ratio of all similar forces at homologus points acting
over a fluid particle of model and prototype are the same.
SAQ8 For Dynamic similarity the model and prototype are to be Geometrically and
Kinematically similar
SAQ9 In complete similitude the dimensionless quantities are constant for both model
and prototype
SAQ10 In Reynolds model law velocity scale ratio is Vr = µrρr Lr
SAQ11 In Froude Model law velocity scale ratio is
Vr = Lr if gr = 1
SAQ12 In Euler Model law velocity scale ratio vr = pr rρ
SAQ13 If Mach no = 1, the flow is called as sonic flow
SAQ14 If velocity scale ratio vr = σ r ρr Lr then this law is called as Weber
Model Law.
SAQ15 Linear dimension scale ratio in Froude Model is 1/4 , then find Velocity
of flow in model if velocity in prototype is 2m/s
Worked Example(1)
E310/1 245 A spillway 7.2 m high and 150 m long discharges 2150 m3 / s under a head of 4m. If a
1:16 model of the spillway is to be constructed, find the model dimensions, head over the
model and discharge in model.
Solution:
Lr = 1/16Tvr =1
16=
14
Qgr = 1
Then Lm
L p
=116
∴Lm =15016
= 9.375m
Qr =Lr5/2 =
116
5 2
= 9.765 ×10−4
∴Qm = Qr × Qp =9.765 × 10 -4 × 2150 = 2.0996 m3/s
Height of model = 7.216
= 0.45m
Head over model = 4
16= 0.25m
Worked Example(2)
Estimate the resistance of boat 100m long when moving in water with a velocity of 8m/s.
The resistance experienced by a model constructed to a scale of 1
100when moving in the
same water is 0.2N. Find velocity and length of model.
Solution
: Fr =ρr Lr
3
gr
Qgr =1, ρr = 1
Fr = Lr3
Fp =Fm
Lr3 = 0.2 ×1003 = 200KN
Vr = Lr =1100
=110
∴VM =1
10× vp =
810
= 0.8m / s
Lr =1
100∴ Length of model =
Lp
100=
100100
= 1m
Worked Example (3)
E310/1 246 A model of an open channel is built to a scale of
1100
. If the model has a Mannings
coefficient n = 0.013, find its value in prototype.
Sol: Vr = Lr (1)
But velocity v = 1n
R2
3 s1
2 and Rr =Lr
∴Vr =Rr
2 3
nr
=Lr
23
nr
(2)
by (1) and (2)
Lr =Lr
2 3
nr
∴ nr =Lr
23
Lr
= Lr1 6
∴nm =1
100
16
× np
ie np =nm
1 100( )1 6 = 0.013 ×1001
6 = 0.028
Worked Example (4)
A 1:6 scale model of an aircraft is tested in a wind tunnel. The drag force experienced and
power required by the air-craft to overcome the drag force in air are respectively 6420N
and 642 KW when moving at a velocity of 100 m/s. Find the drag force and power
required to overcome drag and velocity of model. Properties of air are same for both
model and prototype.
Sol: Lr =16
velocity scale ratio Vr =υ r
Lr
=1Lr
= 6
Qυr = 1
∴ vm = 6×up= 6× 100 = 600 m/s
Fr =µr
2
ρ r
= 1 Qρr = 1 & µ r = 1
∴ Drag force on model = Fr × Fp = 6420N
E310/1 247
Power required by model = µ r
3
Lr ρr2
×Power of prototype
=1Lr
× 642KW = 6 × 642KW
=3852KW
Worked Example(5)
Caster Oil ( s = 0.96, µ = 9.8s poise ) flows through a pipe of a 1 m diameter at the rate
of 5 m3 /s. If the head lost in a geometrically similar pipe of 5 cm diameter carry water ( µ
= 0.01 poise) is 0.035 cm of water/m length, calculate the head lost.
Solution:
Lr =5
100=
120
For dynamic similarity ρ
µr r r
r
v L= 1
ρr =1000960
µr =0.019.8
Vm
Vp
= Vr =µr
ρ r Lr
=0.019.8
×9601000
×201
=0.01959
Velocity in prototype Vp =Qp
ΠQ p2 / 4
=5
Π4
(1)2= 6.37m / s
Vm = Velocity in model = Vr × Vp
= 6.37 × 0.01957 = 0.1248 m/s
Discharge scale ratio = Lr µ r
ρ r
=1
20×
0.019.8
×960
1000= 4.898× 10-5
∴ Qm = Discharge in model = Qr ×Qp
= 4.8978 × 10 -5× 5= 2.45 × 10-4 m3/s
This is also = Vm × Qm
= 0.1248×Π4
0.05( )2
E310/1 248 = 2.45 × 10-4 m3 /s
pressure drop ratio = µ r
2
Lr2ρr
orρ r vr2
=(0.01)2
9.82 ×20( )2
12 ×960
1000= 0.01959( )2
=0.3998 × 10 -3
Pressure drop in model = 0.035100
× 103 = 0.35 Kg / m2 /m
∴ pressure drop in prototype = 0.35
Pr dropscaleratio
=0.35
0.3998 × 10−3 = 875.37kg / m 2 / 20m
=875.37
960= 0.912m of oil in 20m
∴ Pressure drop in on e metre = 0.912
20= 0.04559m
12.5 Types of Models
In general Hydraulic Models can be classified as undistorted and distorted models. A
model is said to be prefect or undistorted if it is similar to prototype i.e all the linear scale
ratios are same. The transfer of results from model to prototype is direct.
A model is said to be distorted when one or more characteristics of the model are not
identical to corresponding characteristic of the prototype. Distortion may be geometric
distortion, material distortion, distortion of hydraulic quantities or combinations of the
above. These models are used in river models and open channel models.
Distorted models are necessary for the following reasons.
1. To maintain accuracy in vertical measurement, otherwise if undistorted model is
designed, the depth of flow, as in the case of a river model, is very small which can not
be measured accurately. Similarly the slope of model channel is different from that of
prototype . So river models, open channel models etc., are going to be distorted models.
2. To maintain turbulent flow
3. To obtain suitable bed material and its adequate measurement
4. To obtain suitable roughness condition
5. To accommodate the available space, money, water supply and time.
Advantages of distorted models.
E310/1 249 1. Vertical exaggeration leads in river models steeper slopes, more depth of water,
magnification of wave heights . This leads to accurate measurement.
2. Since slopes are exaggerated turbulent flow is maintained in the model also otherwise,
it would be laminar flow.
3. On Distorted model sufficient tractive force can be developed which cause the bed
material in the model to move
4. Small models which are economical can be planned by distortion.
Disadvantages
1. Because of vertical exaggeration the velocity distribution and so kinematic energy can
not be reproduced.
2. Flow regime is upset because of steep slopes of the distorted model.
3. A model wave may differ from that of prototype in action.
4. Difficult to prepare river bends, earth cuts etc.
5. It causes an unfavourable psychological effect on the observer
12.6 Specific models
1. Submerged objects
In case of testing of aircrafts and submarines, the drag force is given by F = ρL2 v2φ(RN )
and for dynamic similarity
(RN) m = (RN)p
(MN)m = (MN)p
i.e. ρr vr Lr
µ r
= 1 and Vr =kr
ρ r
But on account of range of ρ,µ, k available for range of fluids, it is not possible to satisfy
both conditions. So as one force is taken as secondary force which resulted in scale
effects. Then scale effects are to be accounted in presenting the prototype performance.
Q(RN)m = (RN)p
we have Fm
Fp
= Fr = ρ r Lr2vr
2
But ϑ r =µ r
ρr Lr
E310/1 250
∴Fr = ρr Lr2 µ r
2
ρ r2 Lr
2 =µ r
2
ρr
which is same as force scale ratio obtained
by Reynolds model law
∴Fm =µr
2
ρ r
Fp
2. Ship models
It is a partially submerged body . The resistance experienced by it is sum of skin friction
and surface waves friction and pressure drag due to wave formation. But pressure drag is
negligible in case of stream lined body like ship.
∴ Total resistance of ship = R
R = Rw + Rf
where Rw = Resistance due to waves
and Rf = Skin friction
So total resistance = ρL2 v2φ(RN , FN )
For perfect similitude, we have
(RN)m = (RN)p
and (FN)m = (FN)p
i.e. ρr Lr vr
µ r
= 1 and vr
gr Lr
= 1
Vr =µ r
ρr Lr
which also equals to vr = gr Lr
if gr = 1
Vr = Lr
So we have
Vr =µ r
ρr Lr
= Lr
i.e. υ r
Lr
= Lr
or υr = Lr3/2
If the fluid is same for both model and prototype , υr =1 i.e. Lr =1 i.e. Full Scale model is
to be tested which is meaningless. So the usual practice is to consider only Froude Model
law, by towing the model in water ∴ ρr =1
E310/1 251 Then the resistance scale ratio Rr = ρr Lr vr
2 Q(RN) & (FN) are same for
both model & prototype
= Lr2 vr
2 Qρr =1
But vr = Lr
∴Rr = Lr2 Lr = Lr
3
Also skin friction (Rf) = fL2v2
where ‘f’ is the frictional resistance per unit surface area per unit velocity which is same
for both model and prototype.
∴ (Rf)r = Lr2 vr
2
but Vr = Lr
∴ (Rf)r = Lr3
Then Rw = (R-Rf) = ρ L2 v2φ(FN)
i.e. (Rw)r = ρr Lr2vr
2
and vr = Lr
( )∴ =R Lw r r rρ 3
Thus (Rw), Rf, and so R = Rw+Rf can be determined.
3. Pressure conduit model
In this case the pressure drop is given ∆p = ρv2φ(RN , ∑ /D )
where ∑ /D indicates the roughness ht of surface, which is neglisible in case of laminar
flow and is to be considered only in turbulent flow.
∴ Pressure drop scale ratio = (∆p)r = ρrvr2 and for dynamic similarity (RN)m = (RN)p
i.e. ρr vr Lr
µ r
= 1
4. River Model
a. Rigid river Model is governed by Froude Model Law which is usually distorted model
to increase the depth of flow in the model , so that it can be measured accurately.
In this case the velocity scale ratio is
Vr =1
Nr
Rr( )2 3Sr( )1 2
And for dynamic similarity,
(FN)m = (FN)p
E310/1 252
i.e. vr
gr Lr
= 1 and if gr =1
Then vr = Lr = Dr
vr =1
Nr
Rr( )2 3Sr( )1 2
= Dr
i.e. 1
Nr
Dr2 3 Dr
1 2
Lr1 2 = D r
1Nr
Dr7 6
Lr1 2 = D r
and slope scale ratio Sr =Dr
Lr
It is also equal to Sr =Dr Nr
2
(Rr )4 3
i.e. Dr
Lr
=Dr Nr
2
(Rr )4 3
But for wide river R = D
∴Nr =(Dr )2 3
Lr
b. Movable bed models
In this case also a distorted model is used to get required tractive force in model.
∴ Tractive force τ = γ RS
and the attractive force scale ratio τr =
τr = γr Rr sr
Further this tractive force can be reduced by using light material like coal dust, saw dust,
pumice stone powder etc. Then extreem distortion is not necessary.
Since flow in river is mostly turbulent which can be ensured if Reynolds no is high.
i.e. when V × R >0.007 for turbulent flow
<0.002 for laminar flow
where v is in ‘m/s’
and R = hydraulic mean radius in ‘m’ Another parameter to find turbulent
flow in water is Karman number
i.e. RK =vkυ
=k grs
υ
E310/1 253 where V = shear velocity = gRs
and Rk > 50 or 60 for turbulent flow for sand
> 100 for any other roughness
12.7 Scale effect
Complete similitude can be obtained if all the model laws are satisfied. But in many cases
this is not possible. For example if both Froude model law and Reynolds model law are to
be satisfied as in the case of ship model, we get Lr3 2 = υ r . Thus if some fluid is selected,
υr = 1 and so Lr =1 , that means, full scale model is to be followed which is meaningless.
And also it is not possible to have the liquid of desired viscosity. So under these
conditions the secondary important variables are neglected which leads to some
discrepancy in the prediction of behaviour of Prototype. This discrepancy is called as
Scale effects. So the model results are to be corrected when they are to be transferred to
prototype. Similarly in models surface tension also influence the phenomenon which is
not so in case of prototype. This also leads to scale effect as surface tension is not
considered in model design.
In order to arrive at these scale effects, several models with different scales are tried and
the results are compared to judge the resulting scale effects. In this way an empirical
relationship between scale effects and model size may be arrived which can be used to
correct the model results.
Identify the following as True or False
SAQ 17 A model is said to be distorted where one or more no of characteristics of
model will not be identical to the corresponding to characteristics of prototype.
SAQ 18 Distortion is necessary to have meaningful dimensions of model and to
develop required tractive force in model and to obtain turbulent flow.
SAQ19 In distorted model velocity distribution and kinetic energy are not going to be
reproduced
SAQ20 Secondary forces are not considered in the model design. This leads to scale
effects
SAQ 21 By considering both Reynolds model law and Froude model law we get
υr =Lr3/2 which leads to impracticable model.
SAQ 22 In the above problem if some fluid is used in both model and prototype the
scale of model is Full size
E310/1 254 SAQ 23 The resistance of ship is equal to Lr
3 if same fluid is used for both model and
prototype.
SAQ 24 In river model Vr = Dr
SAQ 25 In model to maintain turbulent VR > 0.007 or VKυ
> 60 for sand
SAQ 26 If some model laws are neglected leads to scale effect
Worked Example (6)
150m long ship with a wetted area of 3000 m2 . A 1:30 scale ship model is towed in
water at 2m / s to produce a resistance of 40N. Given the ratio of skin frictional
resistance per unit area per unit velocity of model and prototype as 1/3 and for prototype it
is 6.0. Calculate the speed of ship and power required in sea water
(γ =10300N/m3 ) Take propeller efficiency as 75%.
Solution:
Vr = Lr =130
=1
5.477
Vp =vm
Lr
=2Lr
= 2 × 5.477 = 10.94m / s
Surface area of model = 3000
30 × 30=
103
m 2
fp = 6.0 fm =2.0
Rf of Prototype = fpAp vp2
= 6 × 3000 × (10.94)2
= 2154304.8 N = 2154.3KN
Rf for model = 2 ×103
× 22 = 26.7N
Rw for model = 40 -26.7 = 13.33N
(Rw)r = ρ Lr3
=10001030
×1
303
∴ (Rw) of prototype = (Rw)model ×10001030
×1
303
=13.33 ×10301000
× 303 = 370707.3N=370.7KN
E310/1 255 ∴ R = Rw+Rf = 370.7+2154.3 = 2525KN
power = R × Vp
0.75
2525 × 10.94 = 36831.3 K W.
Worked Example (7)
A river model with rigid bed has horizontal scale ratio of 1:1000 and vertical scale ratio of
1:100. The discharge in river is 5000 m3/s at roughness coefficient n = 0.03 Find these in
model. Also if the fluid particle takes 1 hr to travel 100 m in the model Determine the
same in river.
Solution:
Vr = Dr =1100
=1
10
Qr = Ar Vr =LrDr Vr =1
100 ×1000×
110
=1
106
∴Qm = Qr × Qp =1
106 × 5000 =1
200m 3 / s
Nr =Dr
2 3
Lr1 2 =
( 1100 )
23
( 11000) 12
= 1.47
Nm = Np × 1.47 = 0.03 × 1.47 = 0.0441
Lr
Vr
=Lr
Dr
=1
1000× 100 =
1100
Tp =Tm
Tr
=1.0hr1/100
= 100hr
Worked Example (8)
A river model with movable bed has horizontal scale of 1 in 500 and vertical scale of 1 in
125. Tractive force in river is 3N/m2 at a slope of 1in 500 and in the model is 0.4N/m2.
calculate the slope of model river if river slope is 1
10000
Sol: τ r =τ m
τ p
=0.43
=1
7.5
Rr = Dr
τ r = γ r Rr sr =γr DrSr
17.5
= 1 ×1
125× Sr
E310/1 256 Sr =
1257.5
= 16.65
Model river slope = Sm = sp × sr
=1
10000×16.65 = 1 in 600
Summary
1. Similitude is defined as the similarity between model and prototype. Complete
similitude between model and prototype can be obtained if both are geometrically
kinematically and dynamically similar.
2. If the ratios of corresponding linear dimensions of model and prototype are constant
then both model and prototype are geometrically similar.
Lr =Lm
Lp
=bm
bp
=dm
dp
3. When the ratio of velocity and acceleration of the homologus points at the
corresponding points are the same between model and prototype, then they are
kinematically similar. This means the flow nets are similar.
vr =vm
vp
=Lr
Tr
, Qr =Qm
Qp
, Qr =Qm
Qp
=Lr
3
Tr
4. The ratio of all similar forces acting on fluid particles at homologus points of model
and prototype which geometrically and kinematically similar are the same then they
are dynamically similar. The dimensionless quantities are constant in both model
and prototype.
( Fi /Fr)n = ( Fi / Fr )P
( Fi /Fg)m = ( Fi / Fg )p
( Fi /Fp)m = ( Fi / Fp)p
( Fi /Fe)m = ( Fi /Fe )p
( Fi /Fs)m = ( Fi / Fr )p
5. Forces ratios (Fi /Fr) = Reynolds no = ρvLµ
and Fi
Fv
= Froude no =vgL
Mach no = Fi Fe =vk c
E310/1 257 Eulers no = Fi Fp =
vp e
Weber no = WN = Fi Fs =v
σ ρL
6. Model laws
In any model law, the governing nondimensional quantity is same for both model
and prototype and from this velocity scale ratio Vr is obtained and from this other scale
ratios are developed viz
Froude Model law
vr
gr Lr
= 1 ; Vr = Lr if gr = 1
Qr = Lr3 / Tr Qr =
Vr
Tr
=Lr
Tr
Force Scale Ratio = F r =ρr Lr
7 2
Tr
etc.
Similarly for other model scales, Vr is obtained and from this other scale ratios can
be developed.
7. Models broadly can be classified as undistorted and distorted models . If one or
many characteristics of model are not identical with the corresponding
characteristics of prototype, then these are called as distorted.
Distorted Models are necessary to get meaningful measurements in model.
8. If both Reynolds Model and FroudeModel laws are considered we get
υr = Lr3/2 But it is not possible to have fluids of required viscosity or if same
fluid is taken, then the model is of full size which is impracticable. So the
secondary forces are neglected leading to scale effects. These scale effects are
going to be determined by considering different models and by comparing
those test results.
9. For ship resistance
Rr = ρr Lr2vr
2 = Lr3
if Froude Model law is considered in the same fluid ie ρr =1
and Vr = Lr
E310/1 258 10. For river model resistance co efficient
Nr =Dr
2 3
Lr
where Vr = Dr
and Sr =Dr Nr
2
(Rr )4 3 =Dr
Lr
called Rr = D r
11. In moving bed river model s
τ r = γr RrSr to maintain turbulent flow VR > 0.007
and V.K
V> 60 for sand
Answers for SAQ
(1) to (15) --- True (16) Velocity model = 1m /s
(17) to (26) --- True
EXERCISE
12.1 Aspillway model is to be built to a geometrically similar scale of 1
40cross a flume ,
of 50 cm width. The prototype is 20 m high and maximum head on it is expected to
be 2m (1) what height of model and what head on the model should be used? (ii) If
the flow over the model at a particular head is 10Lps, what flow per m length of
prototype is expected ?
(iii) If the -ve pressure in the model is 15 cm, what is the -ve pressure in the
prototype ? Is it practicable?
[(I) 0.5 m, 0.05m (ii) 5059.64 lps (iii) 6m , yes]
12. 2 A pipe of diameter 1.8 m is required to transport an oil of sp. gr 0.8 and
viscosity 0.04 poise at the rate of 4 m3/s. Tests were conducted on a 20 cm
diameter pipe using water at 200 c. Find the velocity and rate of flow in the
model . Viscosity of water at 200 c = 0.01 poise
[ 2.829 m/s, 888lps ]
12.3 A model of a submarine of scale 1 /40 is tested in a wind tunnel. Find the speed
of air in wind tunnel if the speed of sub-marine in sea-water is 15 m/s. Also find
the ratio of the resistance between the model and its prototype. Take the values of
Kinematic viscosities for sea-water and air as 0.012 stokes and 0.016 stokes
respectively. The weight density of sea-water and air are given as 1030 kg/m3
and 1.24 kg (f) / m3 respectively.
E310/1 259 [800 m/s. Fm / Fp = 0.00214 ]
12.4 A 1:20 model of a flying boat is towed through water. The prototype is moving
in sea-water of sp weight 1024 kg(f) / m3 at a velocity of 15 m/s. Find the
corresponding speed of the model . Also determine the resistance due to waves
on model if the resistance due to waves of prototype is 500 kg(f) .
[3.354 m/s , 0.061 kg (f) ]
12.5 A ship 250 m long moves in sea-water whose specific weight is 1030 kg(f) /m3 .A
1:125 model of this ship is to be tested in a wind tunnel. The velocity of air in
the wind tunnel around the model is 20 m/s and the resistance of the model is 50
kg(f). Determine the velocity of ship in sea-water and also the resistance of the
ship in sea water. The sp. weight of air is given as 1.24 kg(f) / m3 . Take the
kinematic viscosity of sea water and air as 0.012 stokes and 0.018 stokes
respectively
[0.106 m/s, 18228.7 kg(f) ]
12.6 In a 1:30 model of a stilling basin the height of the hydraulic jump in the model
is found to be 0.25 m. What is the height of the hydraulic jump in the prototype?
If the energy dissipated in the model is 0.002 h.p, Find the corresponding h.p
lost in prototype?
[ 7.5, 296 hp]
12.7 A 1:6 scale model of a canal is made to study wave motion . If in the
prototype canal a waves describes a certain distance in 20 sec, what would be the
time taken by the wave to travel a corresponding distance in the model
[ 8.16 sec]
12.8 A 1:25 model of a navel ship having a submerged area of 4 m2 and length 6 m
has a total drag of 1.50 kg(f) when towed through water at a velocity of 1.15
m/s . Find the total drag on the prototype when moving at the corresponding
speed. Use the relation Ff =12
cf ρAV2 for calculating the skin friction
resistance. The value of cf is given by
cf = 0.0735Re( )1 5
Take Kinematic viscosity for sea water as 0.01 stokes and sp weight for sea
water as 1000kg(f) /m3 .
E310/1 260 [15008 kg (f) ]
12.9 A 1: 25 scale model was tested to estimate the resistance of the prototype ship.
The model has a wetted surface area of 2 m2 . The prototype velocity is 3 m/s
in sea water of density 1.04 g (f) /cm3 . When the model was towed at the
corresponding speed, the total resistance was found to be 2.65 kg(f). The skin
resistance of the model in fresh water is 0.0094 ϑ 1.9 kg(f) / m2 and that of
the ship in sea water is 0.009 ϑ 1.95 kg(f) /m2 .
Calculate (i) The corresponding speed of model and
(ii) The total resistance of ship
[ 0.6 m/s, 42990 kg(f) ]
12.10 A concrete open channel has Manning’s n = 0.014. A 1:64 model of this channel
is needed. Find the value of ‘n’ for model
[nn =0.007 ]
12.11 The mean velocities in the river and its model are respectively 3m/s & 1m./s .
If the slopes in the river and model are 1in 2500 and 1 in 200 respectively,
calculate the length ratio
[ 1in 112.5]
12.12 A river model constructed with a horizontal scale of 1in 4900 and a vertical scale
of 1 in 225 has a discharge ratio of 1 in 1.8 × 107 . Obtain the scale ratios for
velocity and rugosity coefficient [ 0.0612, 2.06]
12.13. A rectangular channel 50 cm wide and 60 cm deep with water supply of 50 lps is
available in the Laboratory for making a model of spillway 50 cm crest length
with a height of 17 m discharging 1000 on crimes under a head of 1.10 m .
Design a suitable model. Will it be distorted.
[ Lr =1
100, Dr =
1342
, yes ]
***