fluid mechanics - i experiments index expt. name...
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FLUID MECHANICS - I EXPERIMENTS
INDEX
Expt. No. Name of The Experiment
1 Determination Of Coefficient Of Discharge For Venturimeter.
2 Determination Of Coefficient Of Discharge For Orificemeter.
3 Determination Of Friction Factor Of a Given Pipe Of Circular Cross Section.
4 Determination Of Loss Of Head Due To Sudden Enlargement Of Pipe.
5 Determination Of Loss Of Head Due To Sudden Contraction Of Pipe.
6 Determination Of Type Of Flow By Reynold’s Number.
7 Verification Of Bernoulli’s Equation For An Incompressible Fluid Flow.
8 Determination Of Coefficient Of Discharge For Rectangular Notch.
9 Determination Of Coefficient Of Discharge For Triangular Notch.
10
Determination Of Coefficient Of Discharge Cd , Coefficient Of Contraction Cc & Coefficient Of Velocity Cv For An Orifice.
11 Determination Of Force Exerted On Stationary Plate By Impact Of Jet.
Date :______________
EXPERIMENT NO :
Name of the Experiment : To determine the coefficient of discharge ( Cd ) for Venturimeter.
Apparatus : Venturimeter fitted across a pipeline leading to a collecting tank,
Stop Watch, U-Tube manometer connected across entry and throat sections etc.
Formula : Theoretical discharge through Venturimeter Q th = [A1.A2(2g.H)1/2] / [A1
2 – A22]1/2
Actual discharge through Venturimeter Q ac = V / t = (A.∆H) / t Where: A1 : Cross section area of Venturimeter at entry section. A2 : Cross section area of Venturimeter at throat section.
H : Pressure head difference in terms of fluid flowing through pipeline system. V : (A.∆H) i.e. Volume of water collected in collecting tank A : Cross section area of collecting tank. ∆H : (H2 – H1) i.e. Depth of water collected in collecting tank. t : Time required to collect the water up to a height ∆H in the collecting tank.
Theory : Venturimeter is a device consisting of a short length of gradual convergence and a long length of gradual divergence. Pressure tapping is provided at the location before the convergence commences and another pressure tapping is provided at the throat section of a Venturimeter. The Difference in pressure head between the two tapping is measured by means of a U-tube manometer. On applying the continuity equation & Bernoulli’s equation between the two sections, the following relationship is obtained in terms of governing variables.
Q th = [A1.A2(2g.H)1/2] / [A1
2 – A22]1/2 -------------------------------------------------------- 1.
Where, H = H m [(ρm /ρw) – 1] ρm & ρw be the densities of manometric liquid & fluid (water) flowing through pipeline system.
In order to take real flow effect into account, coefficient of discharge (Cd ) must be introduced in equation 1 then,
Q ac = Cd.A.(2g.H)1/2 Therefore, Cd = Q ac / Q th
Theoretical discharge is calculated by using equation 1. Actual discharge is calculated by collecting water in collecting tank & noting the time for collection.
Q ac = A.(H2 – H1) / t = V / t = (A.∆H) / t
Procedure : * Note the pipe diameter (d1) and throat diameter (d2) of Venturimeter.
* Note the density of manometric liquid i.e. mercury (ρm) and that of fluid flowing
through pipeline i.e. water (ρw ).
* Start the flow and adjust the control valve in pipeline for maximum discharge.
* Measure the pressure difference (Hm) across the Venturimeter by using U – tube
manometer.
* Measure flow rate i.e. actual discharge (Qac) through Venturimeter by means of
collecting tank.
* Calculate the theoretical discharge (Qth) through Venturimeter by using the formula.
* Decrease the flow rate by adjusting the control valve and repeat the process for at least
five times.
* Determine the coefficient of discharge (Cd) for each flow rate and find the mean value
of coefficient of discharge (Cd) mean.
* Plot a graph of (Qac) on y-axis versus (Qth) on x- axis.
* Calculate the slope of graph of (Qac) versus (Qth), it gives the mean value of
coefficient of discharge (Cd) mean graphically.
Observation :
Diameter of pipe, d1 = ______ m
Diameter of throat, d2 = ______ m
Area of collecting tank, A = ______x______ = ________ m2
Area of pipe at entry, A1 = [(л/4) d12] = [(л/4) ( )2] = ________ m2.
Area of pipe at throat, A2 = [(л/4) d22] = [(л/4) ( )2] = ________ m2.
Density of mercury, ρm =13600 kg / m3.
Density of water, ρw =1000 kg / m3
Observation Table :
Manometric Reading
Pressure
Head Diff.
Tank Reading
Actual Dischar
ge
Left Lim
b
Right
Limb
Diff.
h2 - h1
H =
Hm[(ρm/ρw) -1]
Initial
FinalDiff. H2 - H1
Time t Q ac =
(A.∆H) / t
Q th =
A1A2(2gH)1/2
[A1
2 – A2
2]1/2
Sr. No.
h1 m
h2 m
Hm m m H1
m H2 m
∆H m sec m3 / sec m3 / sec
Cd =
Q ac
Q th
1
2
3
4
5
Sample Calculation : For Observation No. ___.
* Pressure head difference,
H = Hm [(ρm /ρw) – 1]
= ______ [(13600 /1000) – 1]
= ______ [12.6]
= ______ m.
* Actual discharge,
Qac = (A.∆H) / t
= (______ x ______) / ______
= _______ m3 / sec.
* Theoretical discharge,
Q th = [A1.A2(2g.H)1/2] / [A12 – A2
2]1/2
= [ ______ x ______(2 x 9.81 x ____ )1/2] / [( _____ )2 – ( _____ )2]1/2
= _______ m3 / sec.
* Coefficient of discharge
Cd = Q ac / Q th
Cd = _____ / ______
Cd = ______.
* Mean coefficient of discharge,
(Cd) mean= ( ____ +_____+_____+_____+______ ) / 5.
= _______.
* From graph,
(Cd) mean = (∆Y) / (∆X)
= ______ / ______
= _______.
Result : Coefficient of discharge ( Cd ) for Venturimeter is found to be ________ experimentally & _________ graphically.
Experimental Setup : From Storage Flow Control Valve Convergent Cone Throat Divergent Cone Pipeline d1 d2 Hm h2 h1 ∆H U -Tube Manometer H2 H1 Collecting Tank
Experimental Setup To Determine
Coefficient Of Discharge ( Cd ) For Venturimeter
Date :______________
EXPERIMENT NO :
Name of the Experiment : To determine the coefficient of discharge ( Cd ) for an Orificemeter.
Apparatus : An Orificemeter fitted across a pipeline leading to a collecting tank,
Stop Watch, U-Tube manometer etc.
Formula : Actual discharge through Orificemeter Q ac = C.a1.a0(2g.h)1/2 / [a1
2 – a02]1/2
Where: C : Constant i.e. Coefficient of Orificemeter. C = Cd .{1 – (a0
2 / a12)}1/2 / {1 – Cd
2(a02 / a1
2)}1/2 Cd : Coefficient of discharge for Orificemeter. a1 : Cross section area of pipe at inlet i.e. entry section. a0 : Cross section area of Orifice.
h : Pressure head difference in terms of fluid flowing through pipeline system.
Again, Actual discharge through Orificemeter Q ac = V / t = (A.∆H) / t
V : (A.∆H) i.e. Volume of water collected in collecting tank A : Cross section area of collecting tank. ∆H : (H2 – H1) i.e. Depth of water collected in collecting tank. t : Time required to collect the water up to a height ∆H in the collecting tank.
Theory : It works on Bernoulli’s principle and device use for measuring the rate of fluid flowing through a pipe. It is a cheaper device as compared to venturimeter. It consists of flat circular plate which has a circular sharp edge hole called as orifice called as which is concentric with pipe. The orifice diameter is generally kept ½ lines the diameter of pipe.
An Orificemeter is used to measure the discharge in a pipe. An Orificemeter in it’s simplest form consists of a plate having a sharp edged circular hole known as an orifice. The plate is fixed inside the pipe.
A mercury U-tube manometer is inserted to know the difference of pressure head between the two tapping. Orificemeter works on the same principle as that of Venturimeter i.e. by reducing the area of flow passage a pressure difference is developed between the two section and the measurement of pressure difference is used to find the discharge. By applying Bernoulli’s equation between inlet of pipe & throat i.e. orifice section.
(p1 / w) + (v12 / 2g) + z1 = (p2 / w) + (v2
2 / 2g) + z2
When Orificemeter is connected in horizontal pipe, then z1 = z2 Therefore (p1 - p2) / w = (v2
2 / 2g) - (v12 / 2g)
h = (v22 / 2g) - (v1
2 / 2g) --------------------------------------------- 1.
Further if a1 & a2 be the cross section area of Pipe at inlet & that of jet respectively, then by continuity equation
Q = a1v1 = a2v2 a2 = a1v1 / v2 ------------------------------------------------------- a If Cc = Coefficient of contraction = a2 / a0
Cc = Area of jet at vena contracta / Area of orifice a2 = Cc a0 ------------------------------------------------------- b v1 = Cc v2 (a0 / a1) From equation 1; v2 = ( 2gh + v1
2 )1/2 in this equation losses has not been considered and gives theoretical velocity. v2 = ( 2gh + v1
2 )1/2
If Cv= Coefficient of velocity = Actual velocity / Theorotical velocity ∴Actual velocity of jet at vena contracta i.e. at section 2 v2 = Cv ( 2gh + [Cc v2 (a0 / a1)] 2 )1/2
v2 = Cv {(2gh )1/2 /(1- [Cc Cv (a0 / a1)] 2 )1/2 } But Coefficient of discharge Cd = Cc Cv By continuity equation Q = a2v2
Q = Cc a0 v2 Q = Cc Cv a0 {(2gh )1/2 /(1- [Cc Cv (a0 / a1)] 2 )1/2 } Q = Cd a0 {(2gh )1/2 /(1- [Cd (a0 / a1)] 2 )1/2 } If C= Constant of orificemeter, then C = Cd {1 – (a0
2 / a12)}1/2 / {1 – Cd
2(a02 / a1
2)}1/2
Q ac = C.a0(2g.h)1/2 / {1 – (a0
2 / a12)}1/2
Q ac = C.a1.a0(2g.h)1/2 / (a12 – a0
2)1/2
Procedure : * Note the diameter at the inlet of pipe (d1) and the diameter of an orifice (do).
* Note the density of manometric liquid i.e. mercury (ρm) and that of fluid flowing
through pipeline i.e. water (ρw).
* Connect the U-tube manometer to the pressure toppings of orificemeter, one end at the
inlet section and the other end at the section where jet of water leaves from orifice
forming a vena contracta.
* Start the flow and adjust the control valve in pipeline to get the required discharge.
* Measure the pressure difference (Hm) between two sections of orificemeter by using U
- tube mercury manometer.
* Convert the pressure head difference in meters of fluid flowing through pipeline ( i.e.
water ) by using the equation h = Hm [(ρm / ρw) -1]
* Measure flow rate i.e. actual discharge (Qac) through Venturimeter by means
collecting the water in collecting tank for a specified period of time.
Q ac = V / t = (A.∆H) / t
* Change the flow rate by adjusting the control valve and repeat the process or at least
five times.
* Determine the constant (C) of orificemeter and then calculate coefficient of discharge
(Cd) for each flow rate and find the mean value of coefficient of discharge (Cd) mean.
Observation :
Diameter of pipe, d1 = ______ m
Diameter of orifice, do = ______ m
Area of collecting tank, A = ______x______ = ________ m2
Area of pipe at entry, a1 = [(л/4) d12] = [(л/4) ( )2] = ________ m2.
Area of orifice, ao = [(л/4) do2] = [(л/4) ( )2] = ________ m2.
Density of mercury, ρm =13600 kg / m3.
Density of water, ρw =1000 kg / m3
Observation Table :
Manometric Reading
Pressure
Head Diff.
Tank Reading
Actual Discha
rge
Left Limb
Right
Limb
Diff.
h2 - h1
h =
Hm[(ρm/ρw) -1]
Initial
Final
Diff. H2 - H1
Time t Q ac =
(A.∆H) / t
Sr. No.
h1 m
h2 m
Hm m m H1
m H2 m
∆H m sec m3 /
sec
Constant of Orificemet
er
C = Qac [a1
2 – a0
2]1/2
[a1.a0( 2g.h )1/2]
Coefficient
of Discharge
Cd
1
2
3
4
5
Sample Calculation : For Observation No. ___.
* Pressure head difference,
h = Hm [(ρm /ρw) – 1]
= ______ [(13600 /1000) – 1]
= ______ [12.6]
= ______ m.
* Actual discharge,
Qac = (A.∆H) / t
= (______ x ______) / ______
= _______ m3 / sec.
* Constant of Orificemeter, C = Qac [a1
2 – a02]1/2 / [a1.a0( 2g.h )1/2]
= ______ [ _____2 – _____2 ]1/2 / [ _____ x _____( 2 x 9.81 x ______ )1/2]
= ___________ / __________
= __________
* To Find Coefficient of Discharge (Cd),
By Using Relation
C = Cd .{1 – (a02 / a1
2)}1/2 / {1 – Cd2(a0
2 / a12)}1/2
_______ = Cd .{1 – ( _____2 / _____2)}1/2 / {1 – Cd2(______2 / _______2)}1/2
∴
Cd = _________
* Mean Constant of Orificemeter,
(C) mean = ( ____ +_____+_____+_____+______ ) / 5.
= _______.
* Mean Coefficient of Discharge for Orificemeter,
(Cd) mean = ( ____ +_____+_____+_____+______ ) / 5.
= _______.
Result : * Constant of orificemeter ( C ) is found to be ________ * Coefficient of discharge for orificemeter (Cd) is found to be ________
Date : ____________
EXPERIMENT NO :
Name of the Experiment : To determine the Friction Factor ‘ F ’ for a pipe.
Apparatus : U – tube manometer connected across a pipe line, Stop Watch,
Collecting tank etc.
Formula : Head loss due to friction in pipe
hf = dgVlf..2..4 2
OR hf = dg
VlF..2
.. 2
Where : F = friction factor = (4f) l = length of pipe V = Velocity of flow through pipe. d = Diameter of pipe. g = Acceleration due to gravity. f = Coeff. of friction
Theory : The experimental set up consists of a large number of pipes of different diameters. The pipes have tapping at certain distance so that a U – Tube manometer is connected in between them. The flow of water through a pipeline is regulated by operating a control valve which is provided in main supply line, for measuring the head loss. The length of the pipe is considered as a distance between the two pressure tapping, to which a U – Tube mercury manometer is fitted. Actual discharge through pipeline is calculated by collecting the water in collecting tank and by noting the time for collection.
∴ Velocity of flow = V = aQ
atHA /).(
=
Where : A = Area of tank. H = Depth of water collected in tank. t = Time required to collect the water up to a height “H” in the tank. a = Area of pipe. Q = Discharge through pipe. Now applying Bernoulli’s equation between two pressure tapping, we have
w
ghhz
wPz
wP mm
mBA ..
)(ρ
+−+=+ ∵ =W Weight of water
gW w .ρ=
mw
mm
w
B
w
A hg
ghg
Pg
P−=−
...
.. ρρ
ρρ
ggZ
Zw
w
...
ρρ
=
−=
−1
. w
mm
w
BA hgPP
ρρ
ρ
gghZ
hZw
wmm .
.).(ρ
ρ−=−
−= 1
w
mmf hh
ρρ
gPPh
w
BAf .ρ
−=
Procedure :
* Note down the diameter of pipe (d). * Note the density of manometric liquid ( mρ ) and that of fluid (water) flowing through a pipe i.e. ( wρ ). * Connect the U – tube manometer to the pipe in between two pressure tappings. * Start the flow and adjust the control valve in pipe line for required discharge. * Measure the pressure difference at two points A & B of a pipe by means of a U– tube manometer. * By collecting the water in collecting tank for a particular period of time. * Determine the velocity of flow (V) and frictional head loss (hf) by using appropriate equations. * Determine the friction factor (f) in pipe by using Darcy – Weisbach formula. * Change the flow rate by adjusting the control valve & repeat the process for at least five times. * Find out the mean friction factor (f) mean of the pipe. * Plot a graph of velocity of flow (V) on y – axis verses frictional head loss (hf) on x – axis which shows a straight line.
Observation : l = Length of Pipe = ______ m
d = Dia of Pipe = cm
A = Area of collecting tank = _______ x _______ = _______ m2
mρ = Density of mercury = 13600 kg / m3 wρ = Density of water = 1000 kg / m3
Observation Table :
Sr. No.
Manometic Reading Tank Reading
Left Lim
b
Right
Limb
Diff. (hB - hA)
Initial
height
Final height
Diff H2-H1
HA m
HB m
hm m
Frictional Head Loss
−= 1
w
mmf hh
ρρ
gPPh
w
BAf .ρ
−=
Meter
H1 m
H2 m
H m
Time
t
Sec
Actual Discharge
tHAQac.
=
m3/sec
Velocity of flow
aQ
V ac=
m/sec
frict-ion
factor F
fmean
01. 02. 03. 04. 05.
Sample Calculation : 1)
* a = c/s area of pipe = ( )==
4.
4. 22 ππ d m2
* A = Area of tank = _______ x ________ = _________ m2 For Reading No. 1
* Frictional head loss =
−= 1
w
mmf hh
ρρ
−−−−−−= 1
100013600
fh = __________ m
* Actual Discharge = ∗==
tHAQac.
acQ = ______ m3 / Sec
* Velocity of flow = a
QV ac= =
=V _____ m / Sec
* Friction factor = == 2...2
Vldgh
F f =
OR
* Coeff. of friction = f = =2..4....2
Vldgh f
* Mean friction factor = fmean = =++++
Result : The friction factor “ F ” for the pipe is found to be ________.
Date : ____________
EXPERIMENT NO :
Name of the Experiment : To determine loss of head & power Loss due to Sudden Expansion.
Apparatus : Pipe of smaller diameter connected to larger diameter, inlet,
outlet valves, collecting tank, stop watch etc.
Formula : Losses due to Sudden Expansion :
gVVhe .2
)( 221 −=
Where : he = Loss of head due to sudden expansion. V1 = Velocity of flow at smaller section. V2 = Velocity of flow at larger Section.
Theory : Loss of energy duet to change of velocity of the flowing fluid in magnitude or direction is called as minor loss of energy. Consider a fluid flowing through a pipe line which has sudden enlargement. Consider two section 1 – 1 and 2 – 2 before and after enlargement. Let, P1 = Pressure intensity at section 1 – 1. V1 = Velocity of flow at section 1 – 1. A1 = Area of pipe at section 1 – 1. P2, V2 and A2 = Corresponding values of pressure, velocity & area at section 2 - 2. Due to sudden change of diameter, the liquid flowing from smaller pipe is not able to fallow abrupt change of boundary and turbulent eddies are formed, since the flow separates from the boundary. Let, P1 = Pressure intensity of the liquid eddies on Area A2 – A1, he = Loss of head due to expansion. Applying Bernoulli’s equation at section 1 – 1 and 2 – 2.
=++ 1
211
.2Z
gV
wP
ehZg
VwP
+++ 2
222
.2
But Z1 = Z2
−+
−=
gV
gV
wP
wP
he .2.2
22
2121 ------------------------------------------------------------------- 1.
Consider the control volume of liquid between 2 sections. Fx = P1 A1 + P1 (A2 - A1) P2 A2 = ( P1 - P2 ) A2 ----------------------------------------- 2. Momentum of liquid / sec at section 1 –1 = Mass x Velocity = ϱ A1 V1·V1 = ϱ A1 V1
2 Similarly Momentum of liquid / sec at section 2 – 2 = ϱ A2 V2
2 ∴ Change of momentum / Sec = ϱ A2 V2
2 – ϱ A2 V2 x V1 = ϱ A2 (V2
2 – V1 V2) ----------------------------------- 3. Net force acting on the control vol. in the direction of flow must be equal to the rate of change of momentum per second. Hence equating equation 2 and 3.. ( P1 - P2 ) A2 = ϱ A2 (V2
2 – V1 V2)
∴ρ
21 PP − = V2
2 – V1 V2
Dividing throughout by “g”
g
VVVgPP 21
2221 .
.−
=−ρ
OR g
VVVwP
wP 21
2221 .−
=−
Substituting in equation 1.
−+
−=
gV
gV
gVVVhe .2.2. 2
22
1212
2
On solving
gVVhe .2
)( 221 −=
Procedure :
* Arrange and check the apparatus as shown in fig.
* Measure diameter of pipe and dimensions of measuring tank and record.
* Open the inlet valve, keeping the outlet valve opened.
* Connect the manometer to sudden fallings & to one of the pipes / pipe fittings
and check that there is no air bubble entrapped.
* Open partially the outlet valve, keeping the common inlet valve fully open.
* Let the flow become constant and take the readings.
* Open both the valves, slightly about 2 minutes. Open the pressure tapping and
wait till mercury surface in both limbs of the manometer becomes constant or
still. Take readings of each limb, record and check.
* Collect the discharge and measure the time require to fill up to 5 cm.
* Simultaneously take manometer reading. Repeat procedure up to 6 to7 times.
Observation :
A1 = Area of flow at section 1 -1 = ____________ cm2 V1 = Velocity of flow at section 1 - 1 = ________ cm / sec. A2 = Area of flow at section 2 - 2 = __________ cm2 V2 = Velocity of flow at section 2 - 2 = ________ cm / sec. he = Loss of head due to sudden expansion.
Observation Table :
Manometer Reading
(cm) Sr. No. h
1 h2
hm= h2 -
h1
hw=12.6 x hm
∆V cm3
∆t (Sec
)
Q cm3/s
V1 cm/
s
V2 cm/
s gVVhe .2
)( 221 −=
cm
Power Lost
P=ρ.Q.g.hw
Watt
Mean PowerWatt
01. 02. 03. 04. 05.
Sample Calculation : * hm = h2 – h1 = _______-_______ = _______ cm * hw = 12.6 x hm = _______ x ________ = _______ cm
* Actual Discharge = ∗==
∆=
tHA
tVQ .
Q = ______ cm3 / Sec
* Inlet Velocity = V1 = ( )21
.4dQ
π
1V = = cm / sec
* Outlet Velocity = V2 = 122
21 V
dd
× = = cm / sec
* Loss of head due to Sudden Expansion = gVVhe .2
)( 221 −=
eh = = cm
* Power Lost = P = ρ .Q . g .hw =________________ = ____________ Watt
Result : Head loss due to sudden expansion was found to be ________ and corresponding power loss is found to be __________.
Date : ____________
EXPERIMENT NO :
Name of the Experiment : To determine loss of head & power Loss due to Sudden Contraction.
Apparatus : Pipe of larger diameter connected to smaller diameter, inlet,
outlet valves, collecting tank, stop watch etc.
Formula : Losses due to Sudden Contraction : :
gVhc .2.5.0 2
2=
Where : hc = Loss of head due to sudden contraction.
Theory : Loss of energy duet to change of velocity of the flowing fluid in magnitude or direction is called as minor loss of energy. Consider a liquid flowing in a pipe which has a sudden contraction in area. Consider tow section 1 – 1 and 2 – 2, before and after contraction. As the fluid flows from larger pipe to smaller pipe, the area of flow goes on decreasing and becomes minimum at section C – C. This section is called vena- contracta. After section C – C sudden enlargement takes place. The loss of head duet to sudden enlargement from Vena-contract to smaller pipe. Let; Ac = area of flow at Vena-contracta Vc = Velocity of flow at Vena-contracta A2 = area of flow at section 2 – 2 V2 = Velocity of flow at section 2 – 2 hc = Loss of head due to sudden contraction Now, hc = actually loss of head due to enlargement from Vena - contracta to section 2 - 2 and is given by
hc = (Vc – V2 )2
−= 1
.2 2
2
VV
gVh c
c
From continuity equation; AC VC = A2 V2 i .e.
cc
cc
CA
AAA
AV 11
222
===
Substituting in equation 1
−= 11
.2
22
cc Cg
Vh
If valve of CC is not given, then the head loss due to contraction is given as
gVhc .2.5.0 2
2=
Procedure :
* Arrange and check the apparatus as shown in fig.
* Measure diameter of pipe and dimensions of measuring tank and record.
* Open the inlet valve, keeping the outlet valve opened.
* Connect the manometer to sudden fallings & to one of the pipes / pipe fittings
and check that there is no air bubble entrapped.
* Open partially the outlet valve, keeping the common inlet valve fully open.
* Let the flow become constant and take the readings.
* Open both the valves, slightly about 2 minutes. Open the pressure tapping and
wait till mercury surface in both limbs of the manometer becomes constant or
still. Take readings of each limb, record and check.
* Collect the discharge and measure the time require to fill up to 5 cm.
* Simultaneously take manometer reading. Repeat procedure up to 6 to7 times.
Observation :
A2 = Area of flow at section 2 - 2 = __________ cm2
V2 = Velocity of flow at section 2 - 2 = ________ cm / sec.
hc = Loss of head due to sudden contraction.
Observation Table :
Manometer Reading
(cm) Sr. No. h
1 h2
hm= h2 -
h1
hw=12.6 x hm
∆V cm3
∆t (Sec
)
Q cm3/s
V2 cm/
s gVhc .2.5.0 2
2=
cm
Power Lost
P=ρ.Q.g.hw
Watt
Mean PowerWatt
01. 02. 03. 04. 05.
Sample Calculation :
* hm = h2 – h1 = _______-_______ = _______ cm * hw = 12.6 x hm = _______ x ________ = _______ cm
* Actual Discharge = ∗==
∆=
tHA
tVQ .
Q = ______ cm3 / Sec
* Outlet Velocity = V2 = ( )2
2
.4dQ
π = = cm / sec
* Loss of head due to Sudden Contraction = gVhc .2.5.0 2
2=
ch = = cm
* Power Lost = P = ρ .Q . g .hw =________________ = ____________ Watt
Result : Head loss due to sudden contraction was found to be ________ and corresponding power loss is found to be __________.
Date : ____________
EXPERIMENT NO :
Name of the Experiment : To determine the type of flow by using Reynold’s Number.
Apparatus : Reynold’s experimental arrangement, Collecting tank, Stop
watch, Scale, colour dye ( Potassium Permagnet ) etc.
Formula : µ
ρ DVRe..
=
Where : Re = Reynold’s number ( Dimensionless Parameter ). V = Average velocity in cm / sec D = Diameter of pipe in cm. ρ = Mass density of fluid ( Kg / m3 ) µ = Dynamic viscosity ( N - s / m2 or Kg / m. sec)
Theory : The classification of flow is based mainly on viscosity of a fluid or liquid. The viscosity that is seen earlier depends upon velocity gradient (dx, dg) is considered through Reynolds Number defined as below.
µ
ρ DVRe..
=
Reynolds carried out experiments to decide limiting values of Reynolds number to quantifiably decide wheeler the flow is laminar, turbulent or transition. The flows con visualize by passing a streak of dye and observing its motion. Laminar Flow : A flow is said to be laminar when the various fluid particles moves in layer with one layer of fluid living smoothly over on adjacent layer.A laminar flow is one in which the fluid particles moves in layers or laminar with one layer sliding over the other. Therefore there is no exchange of fluid particles from one layer to the other and hence no transfer of later of momentum to be adjacent layers. The particles, in the layer having lower velocity, obstruct the fluid particles in the layer with higher velocity. This obstruction force is called viscous resistance or viscosity. The laminar flow is one in which fluid layers glide over each another. It has low velocity and high viscous resistance. Turbulent Flow : There is a continuous transfer of momentum to adjacent layers. Fluid particles occupy different relative position at different places. It is one in which, the particles get thoroughly mixed on (called turbulence). The turbulent flow has higher velocity. The flow in canals, pipes and rivers is usually turbulent flow.
Transition Flow : The transition flow has intermediate properties between the laminar and turbulent flow. In laminar the forces should be considered to calculate the friction loss and in the turbulent flow only the internal forces are considered because the effect of viscous force is negligible as compared to internal forces. Reynolds carried out experiments to decide limiting values of Reynolds number to quantifiably decide whether the flow is laminar, turbulent or transition. These limits are as below.
Sr. No. Type of Flow Reynolds Number 01. Laminar Flow < 2100 02. Transition Flow 2100 – 3000 03. Turbulent > 3000
The flow can be visualized by passing a streak of dye and observing its motion. In the laminar, low velocity flow the streak line is only slightly zig – zag. In the turbulent flow, the dye thoroughly mixes up in the flow. Thus passing through a glass pipe and observing the velocity at different mixing stages of the dye is the principle on which Reynolds apparatus is based.
Procedure : * Diameter of a pipe, size of measuring tank at room temperature was noted
down.
* The tank was filled to some height by opening inlet valve and closing control
valve.
* Colour dye was filled in dye tank.
* Control valve was open slightly and also the inlet valve such a way that the
water level in the tank remains constant. This happens when in coming
discharge is equal to the out going discharge.
* The discharge was measured.
* The whole procedure is repeated for 3 times.
Observation :
Diameter of pipe = D = _________ cm
Area of pipe = a = ( )==
4.
4. 22 ππ D cm2
Area of collecting Tank = A = ______ x ______ = ______ cm2
Dynamic viscosity µ = 10 – 6 N - s / m2 or Kg / m. sec
Observation Table :
Sr. No.
H1 (cm)
H2 (cm)
∆H (cm)
Time
t (Sec)
Discharge
=QtHA
∆∆.
Velocity
aQV = µ
ρ DVRe..
= Type of Flow
01. 02. 03. 04.
Sample Calculation : * Depth of water collected in cillecteing tank = ∆H = H2 - H1
∆H = _______ - _______ ∆H = _______ cm
* Discharge = tHAQ
∆∆
=.
== cm3 / sec
* Velocity of flow in pipe = aQV = == cm / sec
* Reynolds number = µ
ρ DVRe..
= ==
Result : For the first discharge the Reynold’s number is found to be
______ therefore the flow will be ____________________ For second discharge the Reynold’s number found to be _______ therefore the flow will be ___________________ For third discharge the Reynold’s number found to be _______ therefore the flow will be ___________________
Date : ____________
EXPERIMENT NO :
Name of the Experiment : To verify ‘ Bernoulli’s Theorem ’.
Apparatus : Bernoulli’s apparatus, Controlling valve at inlet and outlet, Discharge Measuring Tank, Scale, Stopwatch etc.
Formula : Total Energy = =++ Zg
VwP
.2
2
Constant
Where, P / w = Pressure energy V2 / 2g = Kinetic energy Z = Potential energy
Theory : The Bernoulli’s theorem states that the total energy of non Viscous in compressible fluid in a steady state of flow, remains constant along a stream line Daniel Bernoulli’s enunciated in 1738 that is “ In any stream flowing steadily without friction, the total energy contained in a given mass is some at energy contained in a given mass is some at energy point in its path of flow.” This statement is called Bernoulli’s theorem with reference to section 1 – 1 and 2 – 2 along the length of steady flow in the stream tube shown in fig. The total energy at section 1 – 1 is equal to the total energy - at section 2 – 2 as stated in Bernoulli’s theorem. With usual notations, the expression for total energy contained in a unit wt of fluid at section 1 – 1 and 2 – 2 is given by Total energy at Section 1 – 1 = P1 / W + V1
2 / 2g +Z1 Total energy at section 2 – 2 = P2 / W + V2
2 / 2g +Z2 Where, P1 / W = pressure energy at section 1 – 1 V1
2 / 2g = Kinetic energy at section 1 – 1 Z1 = Potential energy at section 1 – 1 P2 / W = Pressure energy at section 2 – 2 V2
2 / 2g = Kinetic energy at section 2 – 2 Z2 = Potential energy at section 2 – 2 Thus applying Bernoulli’s theorem between section 1 – 1 and 2 – 2 we find P1 / W + V1
2 / 2g + Z1 = P2 / W + V22 / 2g + Z2
In MKS system the pressure energy, kinetic energy and potential energy measured in meter of fluid column per unit wt of fluid equation is modified by taking into loss of energy due to friction between section 1 – 1 and 2 – 2 is written as
P1 / W + V1
2 / 2g + Z1 = P2 / W + V22 / 2g + Z2 + ( ∆H )1 / 2
Where ( ∆H ) 1 / 2 represents the loss of energy between section 1 – 1 and 2 – 2
Procedure :
* Open the measuring tank valve fully, to keep the tank empty. Close the outlet valve. * Open the inlet valve and let water rise to some height ‘h’ in the inlet tank. Measure this height on the piezometer. Now open the outlet valve slightly. If water level in the tank falls, close the outlet valve slightly and vice-versa. * Thus adjust the outlet valve fill the water level remains constant at ‘h’, and also readings on each of the piezometer. * Check if reading is correctly written. Close the measuring tank valve. Measure the discharge, i.e. note rise in water level in 5 or 10 sec., write these and also measure and note length and breath of the tank. This completes on run. Take at least three runs by changing the discharge. * Note down the area of the conduit at various gauge points. * Open the supply valve and adjust the flow so that the water level in the inlet tanks remains constant. * Measure the height of water level (above an arbitrarily selected suitable plane) in different remains constant. * Measure the discharge of the conduit with the help of measuring tank. * Repeat steps 2 to 4 for two more discharges. * Plot graph between total energy and distance of gauge points starting from u/s side of conduit.
Observation : Area of collecting tank = A = L x B = ______ x _______ = ________ cm2 Difference in water level in collecting tank = ∆h = ________ cm Time required for rise of water level by 10 cm = ∆t = ________ sec.
Discharge = Qac = =∆∆
=tHA
TimeWaterOfVolume . = m3/sec
Observation Table :
Sr. No.
Piezometric head
ZwP+
cm
Duct area (a)
cm2.
Velocity
aQV =
cm/sec
Velocity Head
gV
.2
2
cm
Total Energy
Zg
VwP
++.2
2
cm 01. 02. 03. 04. 05.
Sample Calculation :
* Discharge = Qact = =∆∆
=tHA. cm3/sec
* Duct area = a = 4 x L = _____ x _____ = _________ cm2
* Velocity = aQV = = = cm/sec
* Velocity head = g
V.2
2
= = cm
* Total head = Zg
VwP
++.2
2
= ( ) =++ cm
* Draw the graph : a) No. of tubes to - ZwP+ cm
b) No. of tubes to - g
V.2
2
cm
c) No. of tubes to - Zg
VwP
++.2
2
cm
Result : The total energy of a streamline, while the particle moves from one
point to another. Bernoulli’s theorem for an incompressible fluid flow is verified.
Date : ____________
EXPERIMENT NO :
Name of the Experiment : To determine coefficient of discharge for a Rectangular Notch.
Apparatus : Channel with rectangular notch, Point gauge, Collecting
tank, Stop watch, Scale etc.
Formula : 1. tHA
tVQac ∆
∆=
∆=
.
2. ( ) 23
.232 HgbQth =
3. th
acd Q
QC =
Where : V = Volume of water collected in tank. b = Width of Notch. H = Head of water over still.
Theory : A Notch is a device used for measuring the rate of flow of a liquid through a small channel or a tank. It may be defined as an opening in the side of a tank or a small channel in such a way treat the liquid surface edge of the opening. Consider a rectangular notch provided in channel or tank carrying water. Let, H = Head of water of still or crest. b = width of notch. For finding the discharge of water flowing over notch, consider an elementary horizontal strip of water of thickness ‘dh’ and length surface of water. The area of strip = b x dh Theoretical velocity = Vth = Hg..2 Discharge through strip aQ = Cd x Area of strip x Vth dQ = Cd x L x dh x Hg..2 Where Cd = Coefficient of discharge. ∴ The total discharge Q :
dhhgbCQH
d ...2..0∫=
∫=H
d dhhgbCQ0
...2..
( )
23
.2..23HgbCQ d=
( ) 23.232 HgbQth =
Procedure :
* The tank dimensions were measured.
* The flow in the channel having rectangular notch was started.
* The flow was kept constant.
* The head of water in piezometer of constant time interval for collecting tank
was noted.
* Open slightly the valve without increase the rotation suddenly after fixed time
interval.
* Also note the head over the still after each interval.
Observation :
Area of tank = A = ______ x _______ = cm2
Width of rectangular Notch = b = _______ cm
Time required to collect water to a depth ∆H = ∆t = constant = _________ sec.
Observation Table :
Point Gauge
Reading Discharge Measurement Sr. No. Initial
(cm) Final (cm)
Diff. (H) (cm)
∆H (cm)
∆t (sec) t
HAt
VQac ∆∆
=∆
=.
(cm3 / s)
( ) 23
.232 HgbQth =
(cm3 / s)
th
acd Q
QC =
01. 02. 03. 04. 05
Sample Calculation :
* Volume of water collected in tank.= V = A. =∆H _____ x _____ = _____ cm3
* Actual Discharge = tHA
tVQac ∆
∆=
∆=
. = = ________ cm3 / s
* Theoretical Discharge = ( ) 23
.232 HgbQth = = ( ) ( ) 2
39812
32
×
__________=thQ cm3 / s
* Coefficient of Discharge = th
acd Q
QC = = = _______
* Mean Coefficient of Discharge = meandC = 5
++++ = _____
Result : Coefficient of discharge (Cd ) for rectangular notch was found to
be _________
Date : ____________
EXPERIMENT NO :
Name of the Experiment : To determine coefficient of discharge ( Cd )
for a triangular Notch.
Apparatus : Channel with triangular Notch, Point gauge and Collecting tank, Stop watch, Scale etc.
Formula : 1. tHA
tVQac ∆
∆=
∆=
.
2. ( ) 25
.22
tan158 HgQth
=θ
3. th
acd Q
QC =
Where : V = Volume of water collected in tank. θ = Angle of Notch. H = Head of water over the notch.
Theory : In hydraulics engineering, notches are commonly used to regulate flow in rivers and other open channels. The relation between water level up stream of the notch and the discharge over it is generally known as θ that the discharge at any time may be found by observing the up stream water liquid. Notches usually have sharp edges so that the water springs clear of the plate as it passes through the notch. It is provided in the side walls of a tank, near top edge. These have small dimensions. Notches are use for emptying tank or for discharge measurement. The discharge equation depends upon the shape and thickness of wall. A triangular weir is on ordinary weir is having a triangular or ‘V’ shaped opening or notch provided in its body so that water is discharged through this opening only. Let ‘H’ be the head above the crest of the weir. Consider a horizontally elementary strip of thickness ‘∆h’ at a depth ‘h’ below the water surface. It ‘X’ is width of strip then, X = 2 ( H +h ) tan(θ /2) The area of strip is ( X ∆h ) or { 2 ( H +h ) tan(θ /2) } and the technical velocity of the water flowing through the strip will be Hg..2 . Thus if dθ is the discharge through the strip then, dθ = Cd x 2 ( H +h ) tan(θ /2 )dh Hg..2 The total discharge ‘q’ for the entire triangular notch may be integration above expression within limit O to H. Then,
( ) ..2.2tan)(2.0
ghdhHHhCQH
d θ+×= ∫
Assuming coefficient Cd to be constant for entire notch.
( ) ( ).2tan)(20
dhhhHCQH
d ∫+×= θ
( )H
d hhHhHCQ0
2523
52.
322tan)(2
−+×= θ
( ) 25
.22
tan158 HgCQ d
=θ
If the vector angle θ equal to 900 then for (θ/2) = 450 and tan (θ/2) = 1
( ) 25
.2158 HgCQ d=
For Cd assumed to be 0.6 then, Q = 1.418 (H) 5/2 For discharge it is simplified as Q = K(H) 5/2 Where K is constant for Notch
gCK d .22
tan158
=θ
Procedure :
* Length and breadth of measuring tank is measured, also angle of triangular
Notch is measured.
* Waste valve of the opening is open, then the inlet valve is slightly open, were
the flow over the still just starts, the inflow is stop. When this overflow stops
fully, the initial gauge reading is measured.
* The inlet valve is slightly open with the jerk. When the constant level is a
acquired final gauge reading is recorded.
* The discharge is then measured in the collecting tank.
* The same procedure was repeated for at least 5 times.
Observation :
Area of tank = A = ______ x _______ = cm2
Angle of notch = θ = _______
Time required to collect water to a depth ∆H = ∆t = constant = _________ sec.
Observation Table :
Point Gauge
Reading Discharge Measurement Sr. No. Initial
(cm) Final (cm)
Diff. (H) (cm)
∆H (cm)
∆t (sec) t
HAt
VQac ∆∆
=∆
=.
(cm3 / s)
( ) 25
.22
tan158 HgQth
=θ
(cm3 / s)
th
acd Q
QC =
01. 02. 03. 04. 05
Sample Calculation : * Volume of water collected in tank.= V = A. =∆H _____ x _____ = _____ cm3
* Actual Discharge = tHA
tVQac ∆
∆=
∆=
. = = ________ cm3 / s
* Theoretical Discharge = ( ) 25
.22
tan158 HgQth
=θ
( ) 25
981.22
tan158
×
=thQ
__________=thQ cm3 / s
* Coefficient of Discharge = th
acd Q
QC = = = _______
* Mean Coefficient of Discharge = meandC = 5
++++ = _____
Result : Coefficient of discharge (Cd ) for triangular notch was found to be _________
Date : ____________
EXPERIMENT NO :
Name of the Experiment : Determination of coefficient of discharge, coefficient of contraction, coefficient of velocity of orifice.
Apparatus : Inlet tank which is fed from on overhead tank through a pipe
network sharp edge orifice, hook gauge attached to the inlet tank, Stop watch, Scale etc.
Formula : HgaQth ..2×=
tHA
tVQac ∆
∆=
∆=
.
Where : a = Area of orifice. Q = Constant head of water in inlet tank. V = Volume of water collected in tank.
A = C / S area of tank. ∆H = Depth of water collected in tank.
∆t = Time require to collect the water in collecting tank.
Theory : Orifice is a small opening of any cross section such as circular,
triangular, rectangular, on a side or on the bottom of the tank, through which a fluid flows. Orifices are used for measuring he rate of flowing fluid. The water is allowed to flow through an orifice fitted to tank and a constant head ‘H’. The water is collected in measuring tank for known time ‘ t ‘. The height of water in the measuring tank is noted. Then the actual discharge through the orifice.
tHA
tVQac ∆
∆=
∆=
.
Coefficient of discharge = th
acd Q
QC =
Coefficient of velocity = VelocitylTheorotica
VelocityActual
Consider a liquid particle which is at vena contract at any time status the position along the jet at P. Jet x = Horizontal distance traveled by particles y = Vertical distance traveled by particle. v = actual velocity of jet. ∴ Horizontal distance, x = v. t ---------------------------------------------------------- 1.
Vertical distance y = 2.21 tg --------------------------------------------------------------- 2.
∴ From equation 1 and 2.
)(2
22 a
vxt −−−−−−−−= & )(.22 b
gyt −−−−−−−−−=
Equate (a) & (b)
yxgv.2. 2
2 =
yxgv.2. 2
=
But, Theoretical velocity, HgVth ..2=
∴Coefficient of velocity = th
v VvC = =
yxg.2. 2
xHg..2
1
Hy
xCv ..4
2
=
Coefficient of contraction = v
dc C
CC =
Procedure :
* The diameter of the orifice, dimension of measuring tank, and diameter of
pipeline were noted.
* The x and y movement of the pointer was checked to be jerk free and
smooth.
* The inlet controlled valve was opened. The inlet tank was allowed to fill the
over flow started. The inlet valve was from adjusted till the water level in the tank
becomes constant as checked by piezometer reading.
* After 3 to 5 min, when steady of flow acquired, it and valve of x and y
were measured open pointer was kept at center of jet.
* The discharge was measured and head ‘h’ was calculated again. The
procedure was repeated for a set of 4 different reading.
Observation :
Diameter of orifice, d = ______ cm
Cross sectional area of an orifice = a = ( )4
. 2dπ = _______ = _______ cm2
Area of collecting tank = A = ______ x ______ = _______ cm2
Observation Table :
Sr. No.
∆H (cm)
∆t (cm) t
HAQac ∆∆
=.
(cm3/s)
H
(cm)HgaQth ..2×=
(cm3/s) X
(cm) Y
(cm) th
acd Q
QC =
Mean
Cd
01. 02. 03. 04.
Sr. No. Hy
xCv ..4
2
= Mean Cv v
dc C
CC =
Mean
Cc 01. 02. 03. 04.
Sample Calculation :
* Actual discharge = tHA
tVQac ∆
∆=
∆=
. = ________= cm3 /sec
* Theoretical discharge = HgaQth ..2×= = _________ = _______cm3 /sec
* Coefficient of discharge = th
acd Q
QC = = _________=
* Coefficient of velocity = th
v VvC = =
HyxCv ..4
2
= = = _______
* Coefficient of contraction = v
dc C
CC = = __________=
Result : The mean values of hydraulic coefficients are as follows :
a) Coefficient of discharge, Cd = _____
b) Coefficient of velocity, Cv = _______
c) Coefficient of contraction, Cc = _______
Date : ____________
EXPERIMENT NO :
Name of the Experiment : To study the Impact of Jet i.e. to verify the momentum equation.
Apparatus : Scale, Measuring weights, collecting tank, impact of jet
apparatus, (i.e. nozzle vane, transparent cylinder etc.)
Formula : Force Exerted = F = aQ 2.ρ
Where : ρ = Mass density of water. Q = Discharge in tank. a = Area of nozzle.
Theory : Momentum equation is based one Newton second law of motion which states that the algebraic sum of external force applied to central volume of fluid in any distance. The external forces included the component of weights. of the fluid and of forces exerted externally open the boundary surface of the control volume. If a vertical water jet moving with velocity ‘V’ is made to strike a target which is free to move in vertical direction then the force will be exerted on the target by the impact of jet. According to momentum equation this force (which is also equal to the force required to bring back the target in its original position) must be equal to the rate of change of momentum of the jet flow in the direction. Applying momentum equation in x – direction.
( )inout xxx VVQF −=− ..ρ
( )VCosVQFx −=− βρ ....
( )βρ .1... CosVQFx −= For flat plate β = 90°
VQFx ..ρ= For hemispherical cup, β = 180°
VQFx ...2 ρ= Here ρ is the mass density, Q is the discharge through nozzle, V is the velocity at exists of nozzle and ‘a’ is area of nozzle.
∴ Fx = aQ 2.ρ
While for curved hemispherical Vane the force, Fx = aQ 2..2 ρ
Procedure :
* Note down the dimension as area of collecting tank, mass density of water and
diameter of nozzle.
* The flat plate is inserted.
* When the jet is not running, note down the reading of upper disc.
* The water supply is admitted to the nozzle and the flow rate adjusted to its
max valve.
* As the jet strikes the vane, position of upper disc is changed. Now place the
weights to bring back the upper disc to its original position.
* The procedure is repeated for each valve of flow by reducing water supply.
* The procedure is repeated for 4 to 5 reading.
Observation :
Diameter of nozzle = d = 10 mm = 0.01m
Mass density of water = ρ = 1000 kg / m3.
Area of collecting tank = A = _______ x _______ = ______ cm2
Area of nozzle = Cross sectional area of pipe = a = ( )4
. 2dπ
a = _______ = _______ cm2
Observation Table : * For Horizontal Flat Vane : * When jet is not running, position of upper disc = ________ cm
Discharge Measurement
Balancing
Sr. No.
Intial
H1 cm
Final
H2 cm
Time ∆t Sec
tHAQ
∆∆
=.
cm3/s
W
(gm)
F
(N)
Practical
Force 'F =
aQ 2.ρ
(N)
% error
= 100'
×−F
FF
01. 02. 03. 04. 05.
* For Curved Vane : * When jet is not running, position of upper disc = ________ cm
Discharge Measurement
Balancing
Sr. No.
Intial
H1 cm
Final
H2 cm
Time ∆t Sec
tHAQ
∆∆
=.
cm3/s
W
(gm)
F
(N)
Practical
Force 'F =
aQ 2..2 ρ (N)
% error
= 100'
×−F
FF
01. 02. 03. 04. 05.
Sample Calculation :
* Area of nozzle = Cross sectional area of pipe = a = ( )4
. 2dπ
a = _______ = _______ cm2 * Depth of water collected in tank = ∆H = H2 – H1 = ___________= ______ cm
* Actual discharge = tHA
tVQ
∆∆
=∆
=. = ________= cm3 /sec
* Balancing weight = W = __________ gm.
* Force = F = NW _________81.981.91000
=×=×
* Practical Force ( For Horizontal Flat Vane) = 'F = aQ 2..2 ρ
'F = = ________ N
* Practical Force ( For Curved Vane) = 'F = aQ 2..2 ρ
'F = = ________ N
Result : While verifying the law of momentum it is observed that
analytical and experimental values of force are approximately equal.