prof.balasubramanya compiled fm notes

7/29/2019 Prof.balasubramanya Compiled FM notes

1/97

FLOW THROUGH PIPESDefinition of flow through pipesA pipe is a closed conduit carrying a fluid under pressure. Fluid motion in a pipe

is subjected to a certain resistance. Such a resistance is assumed to be due to

Friction. In reality this is mainly due to the viscous property of the fluid.

Reynolds Number (Re)

It is defined as the ratio of Inertia force of a flowing fluid and the Viscous force.

Re=(Inertia force/Viscous force) =( V D/ )Classification of pipe flow:

Based on the values of Reynolds number (Re), flow is classified as Follows:

Laminar flow or Viscous FlowIn such a flow the viscous forces are more predominent compared to inertia Forces.

Stream lines are practically parallel to each other or flow takes place In the form of

telescopic tubes. This type of flow occurs when Reynolds number Re< 2000. In laminarflow velocity increases gradually from zero at the boundary to Maximum at the center.

Laminar flow is regular and smooth and velocity at any point practically remains constant

in magnitude & direction. Therefore, the flow is also known as stream Line flow.There will be no exchange of fluid particles from one layer to another. Thus there

will be no momentum transmission from one layer to another. Ex: Flow of thick oil

in narrow tubes, flow of Ground Water, Flow of Blood in blood vessels.

Transition flow:

In such a type of flow the stream lines get disturbed a little. This type of flowoccurs when 2000< Re < 4000.

Laminar flow Transition flow Turbulent flow

Water

Dye

Glass tube


2/97

Hydraulic Grade Line & Energy Grade Line

A Line joining the peizometric heads at various points in a flow is known as Hydraulic

Grade Line (HGL)Energy Grade Line (EGL)

It is a line joining the elevation of total energy of a flow measured above a datum, i.e.

EGL Line lies above HGL by an amount V2/2g.

Losses in Pipe Flow

Losses in pipe flow can be two types viz:-a)Major Loss

b)Minor Loss

a)Major Loss: As the name itself indicates, this is the largest of the losses in a pipe. Thisloss occurs due to friction only. Hence, it is known as head loss due to friction (hf)

b)Minor Loss: Minor losses in a pipe occurs due to change in magnitude or direction offlow.

Minor losses are classified as (i) Entry Loss, (ii) Exit loss, (iii) Sudden expansion

loss (iv) Sudden contraction loss (v) Losses due to bends & pipe fittings.

Head Loss due to Friction

Consider the flow through a straight horizontal pipe of diameter D, Length L, between two sections (1) & (2)as shown. Let P1 & P2 be the pressures at these sections. To is the shear stress acting along the pipe

boundary.

(1)

(1) (2)

(2)

L

Dp1p2

Flow (V)


3/97

From II Law of NewtonForce = Mass x accn. But acceleration = 0, as there is no change in velocity, however the reasonthat pipe diameter is uniform or same throughout.

Applying Bernoullis equation between (1) & (2) with the centre line of the pipe asdatum & considering head loss due to friction hf,.

Substituting eq (2) in eq.(1)

From Experiments, Darcy Found that

( )

( ) )1(4

4

44..

0

021

0

2

21

0

2

2

2

1

=

=

+

=

D

LPPor

DLD

PP

DLxD

PD

Pei

forces

fhg

VpZ

g

VpZ +++=++

22

2

222

2

111

21 ZZ =

21 VV =

Pipe is horizontal

Pipe diameter is

same throughout

)2(21 =

fhPP

)4(8

2

0 = Vf


4/97

f=Darcys friction factor (property of the pipe materials Mass density of the liquid.V = velocityEquations (3) & (4)

But,

from Continuity equation

& (5) & (6) are known as DARCY WEISBACH Equation

Pipes in Series or Compound PipeD1, D2, D3, D4 are diameters.

L1, L2,L3, L4 are lengths of a number of Pipes connected in series

(hf)1, (hf)2, (hf)3 & (hf)4 are the head loss due to friction for each pipe.

The total head loss due to friction hffor the entire pipe system is given by

orD

VLfhf

84 2=

)5(2

2

=

=

gD

fLVh

g

f

)6(8

52

2

=

Dgh

fLQhf

D2

D3

D4D1Q

L2 L4L1

4321 hfhfhfhfhf +++=

5

3

2

2

3

5

2

2

2

2

5

1

2

2

1 888

Dg

QfL

Dg

QfL

Dg

QfLhf

+++=


5/97

D1, D2 and D3 are the pipe diameters. Length of each pipe is same, that is, L1=L2=L3For pipes in parallel hf1=hf2=hf3 i.e

Equivalent pipe

In practice adopting pipes in series may not be feasible due to the fact that they may be of

unistandard size (ie. May not be comemercially available) and they experience otherminor losses. Hence, the entire system will be replaced by a single pipe of uniform

diameter D, but of the same length L=L1+ L2+ L3 such that the head loss due to friction

for both the pipes, viz equivalent pipe & the compound pipe are the same.For a compound pipe or pipes in series.

for an equivalent pipe

D1

D2

D3

Q Q

Q1

Q2

Q3

L = L1 = L2 = L3

Pipes in Parallel

D2D1 D3 == Q D

L= L1+L2+L3L

1L

3

Q

321 hfhfhfhf ++=

)1(888

5

3

2

2

3

5

2

2

2

2

5

1

2

2

1 ++=Dg

QfL

Dg

QfL

Dg

QfLhf

)2(8

5

1

2

2

=Dg

fLQhf


6/97

Equating (1) & (2) and simplifying

Or

5

3

35

2

25

1

15 D

LDL

DL

DL ++=

5

1

5

3

3

5

2

2

5

1

1

++=

D

L

D

L

D

L

LD


7/97

Problems

1) Find the diameter of a Galvanized iron pipe required to carry a flow of 40lps of water, if the loss of head

is not to exceed 5m per 1km. Length of pipe, Assume f=0.02.

Solution:-

D=?, Q=40lps = 40x10-3 m3/shf=5m, L=1km = 1000m. f=0.02

Darcys equation is

2) Two tanks are connected by a 500mm diameter 2500mm long pipe. Find the rate of

flow if the difference in water levels between the tanks is 20m. Take f=0.016. Neglect

minor losses.Solution:-

Applying Bernoullis equation between (1) & (2) with (2) as datum & considering head

loss due to friction hfonly,

Z1 = 20m, Z2 = 0 (Datum); V1=V2 = 0 (tanks are very large)

p1=p2=0 (atmospheric pressure)

Therefore From (1)

20+0+0=0+0+0+hf

Or, hf= 20m. But

mmmD 22022.0 =

5

28

Dg

fLQhf

=

2

152

2500016.08

5.081.920

=xx

xxxQ

lpsmQ 8.434sec/4348.0 3 ==


8/97

3) Water is supplied to a town of 0.5million inhabitants from a reservoir 25km away and

the loss of head due to friction in the pipe line is measured as 25m. Calculate the size of

the supply main, if each inhabitant uses 200 litres of water per day and 65% of the dailysupply is pumped in 8 hours. Take f=0.0195.

Solution:-

Number of inhabitants = 5million = 5,00,000Length of pipe = 25km = 25,000m.

Hf= 25m, D=?

Per capita daily demand = 200litres.

Total daily demand = 5,00,000x200= 100x106 litres.

Daily supply = 65/100 x 100x106 = 65,000m3.

Supply rate

4) An existing pipe line 800m long consists of four sizes namely, 30cm for 175m, 25cm

dia for the next 200m, 20cm dia for the next 250m and 15cm for the remaining length.

Neglecting minor losses, find the diameter of the uniform pipe of 800m. Length toreplace the compound pipe.

Solution:-

L=800mL1=175m D1=0.3m

L2=200m D2=0.25m

L3=250m D3=0.20m

L4=175m D4=0.15m

For an equivalent pipe

D = Diameter of equivalent pipe = 0.189m less than or equal to 19cm.

=52

28

Dg

fLQhf

5

1

2

2

2581.9

)1248.2(000,25195.08

=xx

xxxD

mD 487.1

=

+++=5

4

4

5

3

3

5

2

2

5

1

1

5 D

L

D

L

D

L

D

L

D

L


9/97

5) Two reservoirs are connected by four pipes laid in parallel, their respective diametersbeing d, 1.5d, 2.5d and 3.4d respectively. They are all of same length L & have the same

friction factors f. Find the discharge through the larger pipes, if the smallest one carries

45lps.Solution:-

D1=d, D2 =1.5d, D3=2.5d, D4=3.4d

L1=L2=L3=L4= L.

f1=f2=f3=f4=f.

Q1=45x10-3m3/sec, Q2=? Q3=? Q4=?For pipes in parallel hf1=hf2=hf3=hf4 ,i.e.

5

1

5555 15.0

175

2.0

250

25.0

200

3.0

175

800

+++

=D

5

4

2

4

5

3

2

3

5

2

2

2

5

1

2

1

D

Q

D

Q

D

Q

D

Q===

( ) sec/124.010455.1 32

1

23

5

2 mxxd

dQ =

=

( ) sec/4446.010455.2 32

1

23

5

2 mxxd

dQ =

=

( ) sec/9592.010454.3 32

1

23

5

2 mxxd

dQ =

=


10/97

6) Two pipe lines of same length but with different diameters 50cm and 75cm are made

to carry the same quantity of flow at the same Reynolds number. What is the ratio of

head loss due to friction in the two pipes?Solution:-

D1=0.5m, D2 =0.75m

L1=L2Q1=Q2

(Re)1 = (Re)2,

Reynolds number Re=

7) A 30cm diameter main is required for a town water supply. As pipes over 27.5cmdiameter are not readily available, it was decided to lay two parallel pipes of same

diameter. Find the diameter of the parallel pipes which will have the combined discharge

equal to the single pipe. Adopt same friction factor for all the pipes.Solution:-

222 VD

2

222

1

111

VDVD=

2211 DVDV =

21 75.05.0 VV =

( )21 = ( )21 =

21 5.1 VV =

gD

fLVhf

2

2

=

2

2

2

1

1

2

2

1

V

Vx

D

D

hf

hf=

375.35.1

5.0

75.02

2

2 =

=

V

Vx

From Darcys equation

)1(8

52

2

=Dg

fLQhf

)2(2

82

52

=Dg

QfL

hf


11/97

Equating

8) Two reservoirs are connected by two parallel pipes. Their diameter are 300mm &350mm and lengths are 3.15km and 3.5km respectively of the respective values of

coefficient of friction are 0.0216 and 0.0325. What will be the discharge from the larger

pipe, if the smaller one carries 285lps?

Solution:-

D1=300mm=0.3m, D2=-.350m

L1=3150m L2=3500m

F1=0.0216 f 2=0.0325

Q1=0.285m3/sec Q2=?

For parallel pipes

5

15

4

275.0

=D

mmD 275.0205.0 =

or

=

=5

2

2

2

222

5

1

2

2

111 88

Dg

QLf

Dg

QLfhf

2

1

5

122

5

2

2

1112

=DLf

DQLfQ

21

5

52

23.035000325.0

35.0285.031500216.0

=xx

xxxQ

sec/324.0 32 mQ =


12/97

9) Consider two pipes of same lengths and having same roughness coefficient, but with

the diameter of one pipe being twice the other. Determine (I) the ratio of discharges

through these pipes, if the head loss due to friction for both the pipes is the same. (ii) theratio of the head loss due to friction, when both the pipes carry the same discharge.

Solution:-

f1=f2 D1=2D2 L1=L2

(i)Given hf1=hf2 Q1/Q2=?


(ii) Given Q1/Q2, hf1/hf2=?

10) Two sharp ended pipes are 50mm & 105mm diameters and 200m length are

connected in parallel between two reservoirs which have a water level difference of 15m.If the coefficient of friction for each pipes of 0.0215. Calculate the rate of flow in each

pipe and also diameter of a single pipe 200m long which would give the same discharge,if it were substituted for the Original two pipes.Solution

D1=0.015m, D2=0.105m, L1=L2=200m

H=15m, f1=f2=0.0215,

a) Q1=?, Q2=?

(b) D=?, when Q=Q1+Q2

a) For parallel pipes

656.52 2

5

2

22

5

2

1

2

1 =

=

=

D

D

D

D

Q

Q

03125.02

85

2

2

5

1

2

5

1

2

2

211

2

1 =

=

==

D

D

D

D

Dg

QLf

hf

hf

=

=5

2

2

2

222

5

1

2

2

111 88

Dg

QLf

Dg

QLfhf

sec/1063.32000215.08

05.081.915 3321

52

1 mxxx

xxhxQ =

=

sec/023.02000215.08

105.081.915 3221

52

1 mxx

xxhxQ =

= ( ) sec/02684.00232.01063.3 2321 mxQQQ =+=+= b

52

28

Dg

fLQhf

=

( ) 51

2

2

1581.902684.02000215.08

=

xxxxxD

cmmD 12.111112.0 ==


13/97

11) Two pipes with diameters 2D and D are first connected in parallel and when a

discharge Q passes the head loss 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 the pipes are of same length and have the same frictionfactors.

SolutionH1 = head loss due to friction = hf= hf2

i.e.

Case(iii)

12) Two reservoirs are connected by a 3km long 250mm diameter. The difference in

water levels being 10m. Calculate the discharge in lpm, if f=0.03. Also find the

percentage increase in discharge if for the last 600m a second Pipe of the same diameteris laid parallel to the first.

Solution

Applying Bernoullis equation between (1) & (2) with (2) as datum and

considering head loss due to friction hf

52

2

52

2

)2(88

DgfLQ

DgfLQhf

+

=

+=

552

2

22

1

1

18

Dg

fLQH

)4(80312.1

52

2

2 =Dg

xflQxH

2

52

52

2

2

1

80312.1

802256.0

flQx

Dgx

Dg

xflQx

H

H

=

021876.00312.1

02256.0

2

1 ==H

H71.45

1

2 =H

H

Or

fhg

VpZ

g

VpZ +++=++

22

2

122

2

111

mhh ff 100000010 =+++=++

52

28

Dg

fLQhf

=


14/97

Case (ii)

Change in discharge =

( ) 21

52

300003.08

1025.081.9

=xx

xxxQ

sec/03624.0 3mQ =

321 orhfhfhfhf +=

( )

+=5

2

1

5

2

1

2 25.0

2/600

25.0

2400

81.9

03.0810

QQ

x

x

2

166.647210 Q=

sec/0393.0 31 mQ =

( )QQQ = 1

( )03624.00393.0 =

sec/10066.3 33 mxQ =

1001

xQ

Q% increase in discharge =

%46.810003624.0

10066.3 3

==

x

x


15/97

MINOR LOSSES IN PIPES

Minor losses in a pipe flow can be either due to change in magnitude or direction of flow.

They can be due to one or more of the following reasons.i)Entry loss

ii)Exit loss

iii)Sudden expansion lossiv)Sudden contraction loss

v)Losses due to pipe bends and fittings

vi)Losses due to obstruction in pipe.

Equation for head loss due to sudden enlargement or expansion of a pipe

Consider the sudden expansion of flow between the two section (1) (1)& (2) (2) as

shown.P1 & P2 are the pressure acting at (1) (1) and (2) (2), while V1 and V2 are the velocities.

From experiments, it is proved that pressure P1 acts on the area (a2 a1) i.e. at the point

of sudden expansion.

From II Law of Newton Force = Mass x Acceleration.

Consider LHS of eq(1)

Consider RHS of eq(1)

Mass x acceleration = x vol x change in velocity /time

=volume/time x change in velocity

Substitution (ii) & (iii) in eq(i)

Both sides by (sp.weight)

Applying Bernoullis equation between (1) and (2) with the centre line of the pipe as

datum and considering head loss due to sudden expansion hLonly.

( ) )(21 iiiVVxQx

( ) ( )21212 VVpQppa =

( ) ( )21221 VVVpp = or

zontalpipeishoriCZZ 21 =


16/97

In Eq(V) hL is expressed in meters similarly, power (P) lost due to sudden expansion is

Equations for other minor losses

Loss due to entrance and exit

Loss due to bends & fittings

Problems

g

VpZ

g

VpZ

22

2

222

2

111 ++=++

( ) ( )g

VVVVVhL

2

2 222

1212 +=

g

VVVVVhL

2

22 222

1

2

221 +=

g

VVVVVhL

2

22 222

121

2

2 +=

g

VVVVhL

2

2 212

1

2

2 +=

( )

g

VVhL

2

2

21 =

g

VhL

25.0

2

2=Sudden contraction loss

gVh

exitL2

2

=

g

KVhL

2

2

=

K=coefficient


17/97

1) A 25cm diameter, 2km long horizontal pipe is connected to a water tank. The pipe

discharges freely into atmosphere on the downstream side. The head over the centre line

of the pipe is 32.5m, f=0.0185. Considering the discharge through the pipe

Applying Bernoullis equation between (A) and (B) with (B) as datum & considering all

losses.

2) The discharge through a pipe is 225lps. Find the loss of head when the pipe issuddenly enlarged from 150mm to 250mm diameter.

Solution :

D1=0.15m, D2 = 0.25m Q=225lps = 225m3/secHead loss due to sudden expansion is

3) The rate of flow of water through a horizontal pipe is 350lps. The diameter of the pipe

is suddenly enlarge from 200mm to 500mm. The pressure intensity in the smaller pipe is

15N/cm2. Determine (i) loss of head due to sudden enlargement. (ii) pressure intensity inthe larger pipe (iii) power lost due to enlargement.

SolutionQ=350lps=0.35m3/s

D1=0.2m, D2=0.5m, P1=15N/cm2

hL=?, p2=?, P=?

From continuity equation

( )

g

VVhL

2

12

=

gX

D

Q

D

Q

2

1442

2

2

1

=

2

2

2

2

1

2

2 11

2

16

=

DDg

Q

mhL 385.3=

2

222

2

25.0

1

15.0

1

81.92

225.016

=

xx

x


18/97

Applying Bernoullis equation between (1) (1) and (2) (2) with the central line ofthe pipe as datum and considering head loss due to sudden expansion h L only.

4) At a sudden enlargement of an horizontal pipe from 100 to 150mm, diameter, the

hydraulic grade line raises by 8mm. Calculate the discharge through the pipe system.

Solution

Applying Bernoullis equation between (1) & (2) with the central line of the pipe asdatum and neglecting minor losses (hL) due to sudden expansion.


Lhg

VpZ

g

VpZ +++=++

22

2

222

2

111

( )ntalpipehorizoZZ 021 ==

463.462.19

78.1

81.90

62.19

14.11

81.9

1500

2

2

2

+++=++p

22

2 /67.16/68.166 cmNmkNp ==

LQhP =

463.435.081.9 xx=

kWP 32.15=

( ))1(

2

2

21

=g

VVhL

)2(108,31

12

2 =

+

+ mx

pZ

pZGiven

Lhg

VpZ

g

VpZ +++=++

22

222

2

211

1


19/97

Discharge

5) Two reservoirs are connected by a pipe line which is 125mm diameter for the first 10m

and 200mm in diameter for the remaining 25m. The entrance and exit are sharp and thechange of section is sudden. The water surface in the upper reservoir is 7.5m above that

in the lower reservoir. Determine the rate of flow, assuming f=0.001 for each of the

types.

Solution


Applying Bernoullis equation between (1) & (2) in both the reservoirs with the water in

the lower reservoir as datum and considering all losses

2

2

1

2

4

15.0

4

1.0xV

xV

x =

21 25.2 VV =

smx

V /25.01274.0

108 21

3

2 =

=

25.04

15.0

4

2

2

2

2 xx

VD

Q

==

2

2

1

2

4

2.0

4

125.0V

xV

x =

21 56.2 VV =

{ }1434.2243.52768.362.19

5.72

2 +++=V

( ) ( ) ( )

+++=++ g

Vg

VVg

Vxxg

V22

56.22

56.21001.02

5.25.0}005.7

2

2

2

2

2

2

2

1


20/97

FLOW MEASUREMENTS

Flow Through Orifices

An orifice is an opening of any cross section, at the bottom or on the side walls of a

container or vessel, through which the fluid is discharged. If the geometric characteristicsof the orifice plus the properties of the fluid are known, then the orifice can be used to

measure the flow rates.

Classification of orifices

Flow through an orifice

As the fluid passes through the orifice under a head H, the stream lines converge andtherefore the jet contracts. The stream lines which converge are mostly those from near

the walls and they do so because stream lines cannot make right angled bend in motion.

This phenomenon occurs just down stream of the orifice, and such a section where thearea of cross section of the jet is minimum is know as VENA CONTRACTA.

The pressure at Vena Contracta is assumed to be atmospheric and the velocity is assumed

to be the same across the section since the stream lines will be parallel and equallyspaced. Downstream of Vena contracta the jet expands and bends down. Figure shows

the details of free flow through a vertical orifice.

Applying Bernoulli's equation between (B) & (C) with the horizontal through BC as

datum and neglecting losses (hL)

( ) smV /6.416.21 21

2 ==

sec/1445.06.4

4

2.0 22

mxx

Q =

=

Based on shape circulartriangular rectangular

Based on sizeSmall orifice

(when the headover the orifice is

more than fivetimes its size I.e.

H>5d, Largeorifice

Based on shape ofthe u/s edgeSharp edgeBell mouth

Based onflow FreeSubmerged

VVV == 21 ,0


21/97

Velocity V in Eq(1) is known as TORRICELLIS VELOCITY.

Hydraulic Coefficients of an orifice

i)Coefficient of discharge (Cd): It is defined as the ratio of actual discharge (Qact) to thetheoretical discharge (Qth)

Value of Cd varies in the range of 0.61 to 0.65

ii) Coefficient of Velocity (Cv): It is defined as the ratio of actual velocity (Vact) to the

theoretical velocity (Vth).

Value of Cv varies in the range of 0.95 to 0.99

Coefficient of Contraction (Cc): It is defined as the ratio of the area of cross section of

the jet at Vena of cross section of the jet at Vena Contracta (a c) to the area of the orifice

(a).

Value of Cc will be generally more than 0.62.

Relationship between the Hydraulic Coefficients of an orifice


Actual discharge Qact = ac x Vact

Theoretical discharge Qth = a x Vth

Equation for energy loss through an orifice

Applying Bernoullis equation between the liquid surface (A) and the centre of jet andVena Contracta (C) and considering losses (hL).

=

th

actV

V

VC

=

a

aC cC


22/97

Torricellis equation

Equation for Coefficient of Velocity (CV) (Trajectory method)

Consider a point P on the centre line of the jet, such that its horizontal and vertical

coordinates are x and y respectively.By definition, velocity

Since, the jet falls through a vertical distance y under the action of gravity during this

time (t)

Equating equations (1) & (2)

,HZA =

,0=AV)(0 cityactualvelopp BA ==

Lhg

VaH +++=++

20000

2

)2

(2

g

VaHhL =

gHCButV Va 2=

)(2

VL HxCHh =

)1( 2VL CHh =

t

xVa =

aV

xt=

Or

Or


23/97

But

2

1

2

=

g

y

V

x

a

gHCV Va 2=

2

1

2

2

=

g

y

gHC

x

V

2

1

2

1

2

1

2

1

2

1

2

1

22 y

gx

Hg

xCV =

Hy

xCV

2=

=

yH

xCV

4

2

Or


24/97

Problems

1. The head of water over the centre of an orifice 30mm diameter is 1.5m. If thecoefficient of discharge for the orifice is 0.613, Calculate the actual discharge.

Solution:

d=30mm = 3x10-3H=1.5m

Cd=0.613

2. Compensation water is to be discharge by two circular orifices under a constant head

of 1.0m, measured from the centre of the orifices. What diameter will be required to give

a discharge of 20x103 m3 per day? Assume Cd for each notch as 0.615.

Solution: d=? H=1m. Qtotal = 20x103 m3/day Cd=0.615.

we know

3. A jet of water issuing from an orifice 25mm diameter under a constant head of 1.5m

falls 0.915m vertically before it strikes the ground at a distance of 2.288m measured

;th

actd

Q

QC =

gHxaCd 2=

smxQact /1035.233=

lpsQact 35.2=

sm /1157.03

=

gHaCQ dact 2=

181.924

615.01157.02

xxxxdx =

mmmd 5.2322325.0 ==


25/97

horizontally from the Vena Contracta. The discharge was found to be 102lpm. Determine

the hydraulics coefficients of the orifice and the head due to resistance.

Solution: d=25mm=25x10-3

H=1.5m, y=0.915m, x=2.288m

Qact=102lpm = 102/60 = 1.7lps = 1.7x10-3m3/sec, Cd=?, Cc=?, hL=?

4. The head of water over a 100mm diameter orifice is 5m. The water coming out of the

orifice is collected in a circular tank 2m diameter. The time taken to collect 45cm ofwater is measured as 30secs. Also the coordinates of the jet at a point from Vena Contract

are 100cm horizontal and 5.2cm vertical. Calculate the hydraulic coefficients of theorifice.Solution:

D=100mm=0.1m, H=5m

Qact = Area of collecting tankxheight of water collected / time

X=100cm = 1m, y=5.2cm = 0.052m

Cd=?, Cv=?, Cc=?

5. The coordinates of a point on the jet issuing from a vertical orifice are 0.4m & 0.003m.

Neglecting air resistance, determine the velocity of the jet and the height of water above

the orifice in the tank.Solution.

smxx

/0471.030

45.0

4

2 32

==

98.0

5052.04

1

4

22

=

=

=

xxyH

xCv

605.0581.921.0

40471.02

=

==xxx

x

Q

QC

th

actd

618.098.0

605.0===

V

dC

C

CC


26/97

X=0.4m, y=0.3m, V=? H=?

Assume

We know

6. A vertical orifice is fitted 0.2m above the bottom of a tank containing water to a depthof 2m. If G=0.98. What is the vertical distance from the orifice of a point on the jet 0.6m

away from the Vena Contracta?

SolutionHead over the orifice H=(2-0.2)=1.8m

CV=0.98, y=?, x=0.6m

7. A closed tank contains water to a height of 2m above a sharp edged orifice 1.5cm

diameter, made in the bottom of the tank. If the discharge through the orifice is to be 4lps.

1=VC

yH

xCV

4

2

=

==

=

2

2

2

2

22

103.04

4.0

4

4

xxyxG

xH

xyHxG

H=1.33m

mmmxx

y

xyxor

yH

xCV

52052.098.08.14

6.0

8.14

6.0)98.0(,

4

2

2

22

2

==

=

=

=


27/97

Workout the pressure at which air should be pumped into the tank above water. Take

Cd=0.6.

Solution

Q=4lps = 4x10-3m3/s

D=1.5x10-2m, Cd=0.6

PA=?

Total head over the orifice

8. A closed tank contains 3m depth of water and an air space at 15kpa pressure. A 5cm

diameter orifice at the bottom of the tank discharge water to the tank B containingpressurized air at 25kpa. If Cd = 0.61 for the orifice. Calculate the discharge of water

from tank A.Solution

d=5cm = 5x10-2m Cd=0.61.


H=1.9806m

9. A tank has two identical orifices in one of its vertical sides. The upper orifice is 4m

below the water surface and the lower one 6m below the water surface. If the value of Cvfor each orifice is 0.98, find the point of intersection of the two jets.Solution.

333 /10772.11/772.11 mkNxmNair==

+=

AphH

gHaCQ dact 2=

( )

+=

3

223

10772.11281.92

4

105.16.0104

x

Pxxx

xxxx A

)(/83.02

GaugemkNPA =

( )

+=

+=81.9

25153

BA pphH


28/97

Given Cv is same for both the orifices

from figure

Substituting eq(1) in eq(2) and simplifying

Again

10. Two orifices have been provided in the side of the tank, one near the bottom and the

other near the top. Show that the jets from these two orifices will intersect a planethrough the base at the same distance from the tank if the head on the upper orifice is

equal to the height of the lower orifice above the base. Assume Cv to be the same for

both the orifices.

22

2

2

11

2

1

44 Hy

x

Hy

x=

)(44

21

22

2

11

2

1 xxHy

x

Hy

x=

)1(5.164 2121 == yoryyy

( )

)2(2

46

21

21

+=

+=

yy

yy

my

y

yy

4

25.0

25.1

2

2

22

=

=

+=

givesHy

xCV

22

2

2

4=

mx

xx

x

6.9

64498.0

2

2

2

=

=

(points of intersection of the jets from theVena contracts)


29/97

Solution.

To show that x1=x2 when H1=y2

from figure y1=[y2+(H2-H1)---(1)

Problems on Orifices

A 4cm dia orifice in the vertical side of a tank discharges water. The water surface in the

tank is at a constant level of 2m above the centre of the orifice. If the head loss in the

orifice is 0.2m and coefficient of contraction can be assumed to be 0.63. Calculate (I) thevalues of coefficient of velocity & coefficient of discharge, (ii) Discharge through the

orifice and (iii) Location of the point of impact of the jet on the horizontal plane located

0.5m below the centre of the orifice.

Solution

Head loss

22

2

2

11

2

1

1 44, 2 Hy

x

Hy

x

CCGiven VV ==

Or

Or

22

2

12112 HyHHHHy =+

0)( 122212

1 =+ HHyHHH

00

0)(; 222222

221

=

=+= yHyyHyyH

substituting

=

g

VaHhL

2

2

=

81.9222.0

2

x

VaOr


30/97

Coefficient of Velocity

Coefficient of discharge

(ii) Discharge through the orifice

(iii) Coefficient of velocity

An orifice has to be placed in the side of a tank so that the jet will be at a maximumhorizontal distance at the level of its base. If the depth of the liquid int the tank is D, what

is the position of the orifice? Show that the jets from the two orifices in the side of the

tank will intersect at the level of the base if the head on the on the upper orifice is equal

to the height of the orifice above the base.Solution:

943.0246.6

943.5===

V

VC av

Cvd xCCC = 63.0949.0 x=

598.0=dC

gHaCQ dact 2=

281.9204.04

598.0 2 xxxxx

=

yH

xCv

4

2

=

24 xyHCV =

x

Jet

Orifice

D

h

y


31/97

By definition, Velocity V=x/t

But

and

For x to be maximum

Vt =

gHV 2=

2

2

1gty =

2

22

1

)(

= gHx

gHD

( )HDHx = 42

)(4 HDHx =

Or

0=dH

dx

0)2(4 = HD

2/DH =


32/97

We know, x=Vt,

12gHV =

2

22

1gtHy =+

2

122

1

=

gH

xg


33/97

FLOW THROUGH PIPESDefinition of flow through pipes

A pipe is a closed conduit carrying a fluid under pressure. Fluid motion in a pipeis subjected to a certain resistance. Such a resistance is assumed to be due to

Friction. In reality this is mainly due to the viscous property of the fluid.

Reynolds Number (Re)It is defined as the ratio of Inertia force of a flowing fluid and the Viscous force.

Re=(Inertia force/Viscous force) =( V D/ )Classification of pipe flow:

Based on the values of Reynolds number (Re), flow is classified asFollows:Laminar flow or Viscous Flow

In such a flow the viscous forces are more predominent compared to inertiaForces. Stream lines are practically parallel to each other or flow takes placeIn the form of telescopic tubes.This type of flow occurs when Reynolds number Re< 2000.In laminar flow velocity increases gradually from zero at the boundary toMaximum at the center.Laminar flow is regular and smooth and velocity at any point practically remainsconstant in magnitude & direction. Therefore, the flow is also known as stream

Line flow.There will be no exchange of fluid particles from one layer to another.

Thus there will be no momentum transmission from one layer to another.Ex: Flow of thick oil in narrow tubes, flow of Ground Water, Flow of

Blood in blood vessels.

Transition flow:In such a type of flow the stream lines get disturbed a little.

This type of flow occurs when 2000< Re < 4000.

Turbulent Flow: This is the most common type of flow that occurs in nature( flow in rivers, pipes).This flow will be random,erratic,unpredictable. Thus motion of fluid particles result in eddy currents

& they mix up. Streamlines are totally disturbed or cross each other.

The velocity changes in direction and magnitude from point to point.There will be transfer of momentum between the particles as they are continuously colliding witheach other.

There will be considerable loss of energy in this type of flow.This type of flow cannot be truly mathematically analysed and any analysis is possible by stastical

evaluation.For this type of flow in a pipe Re> 4000.

(REYNOLDS EXPERIMENT:Refer Fig.(1)

Hydraulic Grade Line & Energy Grade LineA Line joining the peizometric heads at various points in a flow is known as HydraulicGrade Line (HGL)Energy Grade Line (EGL)

It is a line joining the elevation of total energy of a flow measured above a datum, i.e.

.2

2

g

VpZ ++

EGL Line lies above HGL by an amount V2/2g.(Refer Fig.(2))

Losses in Pipe FlowLosses in pipe flow can be two types viz:-a)Major Loss


34/97

b)Minor Lossa)Major Loss: As the name itself indicates, this is the largest of the losses in a pipe. This lossoccurs due to friction only. Hence, it is known as head loss due to friction (hf)

b)Minor Loss: Minor losses in a pipe occurs due to change in magnitude or direction of flow.Minor losses are classified as (i) Entry Loss, (ii) Exit loss, (iii) Sudden expansion loss (iv)

Sudden contraction loss (v) Losses due to bends & pipe fittings.

Head Loss due to Friction (DARCY-WEISBACH Equation)Consider the flow through a straight horizontal pipe of diameter D, Length L, between two sections (1) & (2)

as shown in fig.(3). Let P1 & P2 be the pressures at these sections. o is the shear stress acting along thepipe boundary.

From II Law of NewtonForce = Mass x accn. But acceleration = 0, as there is no change in velocity,however the reason that pipe diameter is uniform or same throughout.

( )

( ) )1(4

4

44..

0

021

0

2

21

0

2

2

2

1

==

+

=

D

LPPor

DLD

PP

DLxD

PD

Pei

forces

Applying Bernoullis equation between (1) & (2) with the centre line of the pipeas datum & considering head loss due to friction hf,.

fhg

VpZ

g

VpZ +++=++

22

2

222

2

111

21 ZZ =Pipe is horizontal

21 VV =Pipe diameter is same throughout

)2(21 =

fhPP

Substituting eq (2) in eq.(1)

)3(4

40

0 ==l

Dhor

D

Lxh

f

f

From Experiments, Darcy Found that

)4(8

2

0 = Vf

f=Darcys friction factor (property of the pipe materialsMass density of the liquid.

V = velocityEquations (3) & (4)

L

DhV

f f

48

2

=

or ,D

VLfhf

8

4 2=


35/97

But,

)5(2

2

=

=

gD

fLVh

g

f

from Continuity equation2

4D

QV

=

)6(8

52

2

=

Dgh

fLQhf

& (5) & (6) are known as DARCY WEISBACH Equation

Pipes in Series or Compound PipeD1, D2, D3, D4 are diameters.(fig.4)

L1, L2,L3, L4 are lengths of a number of Pipes connected in series

(hf)1, (hf)2, (hf)3 & (hf)4 are the head loss due to friction for each pipe.

The total head loss due to friction hf for the entire pipe system is given by

4321 hfhfhfhfhf +++=

5

4

2

2

4

5

3

2

2

3

5

2

2

2

2

5

1

2

2

1 8888

Dg

QfL

Dg

QfL

Dg

QfL

Dg

QfLhf

+++=

Pipes in parallel

D1, D2 and D3 are the pipe diameters.(Fig.5)

Length of each pipe is same, that is, L1=L2=L3For pipes in parallel hf1=hf2=hf3

i.e

)1(

888

5

3

2

5

2

2

5

1

2

5

3

2

2

3

5

2

2

2

2

5

1

2

2

1

321

321

==

==

D

Q

D

Q

D

Q

orDg

QfL

Dg

QfL

Dg

QfL

From continuity equation Q= Q1+Q2+Q3--------(2)

Equivalent pipeIn practice adopting pipes in series may not be feasible due to the fact thatthey may be of unistandard size (ie. May not be comemercially available)and they experience other minor losses. Hence, the entire system will be

replaced by a single pipe of uniform diameter D, but of the same lengthL=L1+ L2+ L3 such that the head loss due to friction for both the pipes, viz

equivalent pipe & the compound pipe are the same (Fig.6).For a compound pipe or pipes in series

321 hfhfhfhf ++=

)1(888

5

3

2

2

3

5

2

2

2

2

5

1

2

2

1 ++=Dg

QfL

Dg

QfL

Dg

QfLhf


36/97

for an equivalent pipe )2(8

5

1

2

2

=Dg

fLQhf

Equating (1) & (2) and simplifying 53

3

5

2

2

5

1

1

5 D

L

D

L

D

L

D

L++=

or

5

1

5

3

3

5

2

2

5

1

1

++=

D

L

D

L

D

L

LD

PROBLEMS

1) Find the diameter of a Galvanized iron pipe required to carry a flow of 40lps of water, if the lossof head is not to exceed 5m per 1km. Length of pipe, Assume f=0.02.

Solution:-D=?, Q=40lps = 40x10-3 m3/shf=5m, L=1km = 1000m. f=0.02

Darcys equation is5

28

Dg

fLQhf

=

=fhg

fLQD

2

28

5

1

2

23

581.9

)1040(100002.08

=

xx

xxxxD

mmmD 22022.0 ==

2) Two tanks are connected by a 500mm diameter 2500mm long pipe. Find the rate of flow if thedifference in water levels between the tanks is 20m. Take f=0.016. Neglect minor losses.Solution:-

Applying Bernoullis equation between (1) & (2) with (2) as datum & considering head loss due tofriction hfonly, (Fig.7).

)1(22

2

222

2

111 +++=++ fh

g

VpZ

g

VpZ

Z1 = 20m, Z2 = 0 (Datum); V1=V2 = 0 (tanks are very large)

p1=p2=0 (atmospheric pressure)

Therefore From (1)20+0+0=0+0+0+hfOr, hf= 20m.

But5

28

Dg

fLQhf

=


37/97

2

152

2500016.08

5.081.920

=xx

xxxQ

lpsmQ 8.434sec/4348.0 3 ==

3) Water is supplied to a town of 0.5million inhabitants from a reservoir 25km away and the lossof head due to friction in the pipe line is measured as 25m. Calculate the size of the supply main,if each inhabitant uses 200 litres of water per day and 65% of the daily supply is pumped in 8 hours. Take f=0.0195.Solution:-Number of inhabitants = 5million = 5,00,000Length of pipe = 25km = 25,000m.Hf= 25m, D=?

Per capita daily demand = 200litres.Total daily demand = 5,00,000x200= 100x106 litres.Daily supply = 65/100 x 100x106 = 65,000m3.

Supply rate sec/1248.260605.8

000,65 3

mxxQ =

=

=52

28

Dg

fLQhf

5

1

2

2

2581.9

)1248.2(000,25195.08

=xx

xxxD

mD 487.1=

4) An existing pipe line 800m long consists of four sizes namely, 30cm for 175m, 25cm dia for thenext 200m, 20cm dia for the next 250m and 15cm for the remaining length. Neglecting minor

losses, find the diameter of the uniform pipe of 800m. Length to replace the compound pipe.Solution:-L=800mL1=175m D1=0.3m

L2=200m D2=0.25m

L3=250m D3=0.20m

L4=175m D4=0.15m

For an equivalent pipe

+++=5

4

4

5

3

3

5

2

2

5

1

1

5D

L

D

L

D

L

D

L

D

L

5

1

555515.0

175

2.0

250

25.0

200

3.0

175

800

+++

=D

D = Diameter of equivalent pipe = 0.189m less than or equal to 19cm.

5) Two reservoirs are connected by four pipes laid in parallel, their respective diameters being d,1.5d, 2.5d and 3.4d respectively. They are all of same length L & have the same friction factors f.Find the discharge through the larger pipes, if the smallest one carries 45lps.


38/97

Solution:-D1=d, D2 =1.5d, D3=2.5d, D4=3.4d

L1=L2=L3=L4= L.

f1=f2=f3=f4=f.

Q1=45x10-3m3/sec, Q2=? Q3=? Q4=?

For pipes in parallel hf1=hf2=hf3=hf4

i.e.

5

4

2

4

5

3

2

3

5

2

2

2

5

1

2

1

D

Q

D

Q

D

Q

D

Q===

( ) sec/124.010455.1 32

1

23

5

2 mxxd

dQ =

=

( ) sec/4446.010455.2 32

1

23

5

2 mxxd

dQ =

=

( ) sec/9592.010454.3 321

23

5

3 mxxd

dQ =

=

6) Two pipe lines of same length but with different diameters 50cm and 75cm are made to carrythe same quantity of flow at the same Reynolds number. What is the ratio of head loss due tofriction in the two pipes?Solution:-D1=0.5m, D2 =0.75m

L1=L2Q1=Q2

(Re)1 = (Re)2, ?2

1 =hf

hf

Reynolds number Re=

VD

2

222

1

111

VDVD=

2211 DVDV = ( )21 = ( )21 =

21 75.05.0 VV = 21 5.1 VV =

From Darcys equationgD

fLVhf

2

2

=

2

2

21

1

2

2

1

V

Vx

D

D

hf

hf =

375.35.1

5.0

75.02

2

2 =

=

V

Vx

7) A 30cm diameter main is required for a town water supply. As pipes over 27.5cm diameter arenot readily available, it was decided to lay two parallel pipes of same diameter. Find the diameter


39/97

of the parallel pipes which will have the combined discharge equal to the single pipe. Adoptsame friction factor for all the pipes.

Fig.(8)

Solution:- )1(8 52

2

= DgfLQhf

)2(2

8

2

52

= Dg

QfL

hf

Equating

=

52

2

52

22

88

Dg

QfL

Dg

fLQ

55

1 4

11

DD=

or51

5

4

275.0

=D

mmD 275.0205.0 =

8) Two reservoirs are connected by two parallel pipes. Their diameter are 300mm & 350mm andlengths are 3.15km and 3.5km respectively of the respective values of coefficient of friction are0.0216 and 0.0325. What will be the discharge from the larger pipe, if the smaller one carries285lps?Solution:-D1=300mm=0.3m, D2=-.350m

L1=3150m L2=3500m

F1=0.0216 f 2=0.0325

Q1=0.285m3/sec Q2=?

For parallel pipes

=

=5

2

2

2

222

5

1

2

2

111 88

Dg

QLf

Dg

QLfhf

2

1

5

122

5

2

2

1112

=DLf

DQLfQ

2

1

5

52

2 3.035000325.0

35.0285.031500216.0

= xx

xxxQ

sec/324.03

2 mQ =

9) Consider two pipes of same lengths and having same roughness coefficient, but with thediameter of one pipe being twice the other. Determine (I) the ratio of discharges through thesepipes, if the head loss due to friction for both the pipes is the same. (ii) the ratio of the head lossdue to friction, when both the pipes carry the same discharge.


40/97

Solution:-f1=f2 D1=2D2 L1=L2(i)Given hf1=hf2 Q1/Q2=?


=52

28

Dg

fLQhf

5

2

2

2222

5

1

2

2111 88

Dg

QLf

Dg

QLf

=

656.52 2

5

2

22

5

2

1

2

1 =

=

=

D

D

D

D

Q

Q

(ii) Given Q1/Q2, hf1/hf2=?

03125.02

85

2

2

5

1

2

5

1

2

2

211

2

1 =

=

==

D

D

D

D

Dg

QLf

hf

hf

10) Two sharp ended pipes are 50mm & 105mm diameters and 200m lengthare connected in parallel between two reservoirs which have a water leveldifference of 15m. If the coefficient of friction for each pipes of 0.0215.Calculate the rate of flow in each pipe and also diameter of a single pipe200m long which would give the same discharge, if it were substituted for theOriginal two pipes.SolutionD1=0.015m, D2=0.105m, L1=L2=200m

H=15m, f1=f2=0.0215, a) Q1=?, Q2=? (b) D=?, when Q=Q1+Q2a) For parallel pipes

=

=5

2

2

2

222

5

1

2

2

111 88

Dg

QLf

Dg

QLfhf

sec/1063.32000215.08

05.081.915 332

152

1 mxxx

xxhxQ =

=

sec/023.02000215.08

105.081.915 3221

52

2 mxx

xxhxQ =

=

b)

( ) sec/02684.00232.01063.3 2321 mxQQQ =+=+=

52

2

8Dg

fLQhf

= ( ) 51

2

2

1581.902684.02000215.08

=

xxxxxD

cmmD 12.111112.0 ==

11) Two pipes with diameters 2D and D are first connected in parallel and


41/97

when a discharge Q passes the head loss 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 the pipes are of

same length and have the same friction factors.

Solution (Fig.9)

H1 = head loss due to friction = hf= hf2

i.e.

=52

2

1

)2(

8

Dg

fLQhf

)1(

)(

852

2

2

2

=Dg

fLQhf

)2(21 =+ QQQ

QQQ =+ 2266.5 QQ66.6

12 =

)3(02256.0866.6

18

52

2

52

2

1 =

=Dg

QfLx

Dg

QfL

H

Case(iii) 212 hfhfH +=

52

2

52

2

)2(

88

Dg

fLQ

Dg

fLQhf

+

=

+=

552

2

22

1

1

18

Dg

fLQH

)4(80312.1

52

2

2 =Dg

xflQxH

2

52

52

2

2

1

80312.1

802256.0

flQx

Dgx

Dg

xflQx

H

H

=

021876.00312.1

02256.0

2

1 ==H

H

or 71.451

2 =H

H

12) Two reservoirs are connected by a 3km long 250mm diameter. Thedifference in water levels being 10m. Calculate the discharge in lpm, if f=0.03.

Also find the percentage increase in discharge if for the last 600m a secondPipe of the same diameter is laid parallel to the first.

SolutionApplying Bernoullis equation between (1) & (2) with (2) as datum andconsidering head loss due to friction hf (Fig.11)

fhg

VpZ

g

VpZ +++=++

22

2

122

2

111

mhh ff 100000010 =+++=++


42/97

52

28

Dg

fLQhf

= ( )

2

152

300003.08

1025.081.9

=xx

xxxQ

sec/03624.0 3mQ =

Case (ii) 321 orhfhfhfhf += 321 orhfhfhfhf += (Fig.12)

( )

+=5

2

1

5

2

1

2 25.0

2/600

25.0

2400

81.9

03.0810

QQ

x

x

2

166.647210 Q= sec/0393.03

1 mQ =Change in discharge = ( )QQQ = 1

( )03624.00393.0 = sec/10066.3 33 mxQ =

% increase in discharge = 1001

xQ

Q

%46.810003624.0

10066.3 3==

xx

MINOR LOSSES IN PIPESMinor losses in a pipe flow can be either due to change in magnitude or directionof flow. They can be due to one or more of the following reasons.i)Entry lossii)Exit lossiii)Sudden expansion lossiv)Sudden contraction lossv)Losses due to pipe bends and fittingsvi)Losses due to obstruction in pipe.

Equation for head loss due to sudden enlargement or expansion of a pipe

Consider the sudden expansion of flow between the two section (1) (1)& (2) (2)as shown in Fig.13P1 & P2 are the pressure acting at (1) (1) and (2) (2), while V1 and V2 are the

velocities.From experiments, it is proved that pressure P1 acts on the area (a2 a1) i.e. at

the point of sudden expansion.From II Law of Newton Force = Mass x Acceleration. ---------------(1)


43/97

Consider LHS of eq(1)

( ) )(1212211 iaapapapforces ++=( ) )(, 212 iippaforcesor =

Consider RHS of eq(1)

Mass x acceleration =

x vol x change in velocity /time =volume/time x change in velocitySubstitution (ii) & (iii)

( ) ( )21212 VVQppa = or, ( ) ( )21221 VVVpp = Both sides by (sp.weight)

( ))(21221 iv

g

VVVpp

=

Applying Bernoullis equation between (1) and (2) with the centre line of the pipeas datum and considering head loss due to sudden expansion hLonly

gVpZ

gVpZ

22

2

222

2

111 ++=++

zontalpipeishoriCZZ 21 =

Lhg

VVpp=

=

2

2

2

2

121

( ) ( )g

VVVVVhL

2

2 222

1212 +=

g

VVVVVhL

2

22 222

1

2

221 +=

g

VVVVVhL

2

22 22212122 +=

g

VVVVhL

2

2 212

1

2

2 += OR( )

g

VVhL

2

2

21 =

In Eq(V) hL is expressed in meters similarly, power (P) lost due to sudden

expansion is )(viQhP f =

Equations for other minor losses (Fig.14 a,b,c)

Sudden contraction lossg

VhL

2

5.02

2=

Loss due to entrance and exitg

Vh

entryL2

5.0 2=

g

Vh

exitL2

2

=


44/97

Loss due to bends & fittingsg

KVhL

2

2

=

K=coefficient

Problems1) A 25cm diameter, 2km long horizontal pipe is connected to a water tank. Thepipe discharges freely into atmosphere on the downstream side. The head overthe centre line of the pipe is 32.5m, f=0.0185. Considering the discharge throughthe pipe

Applying Bernoullis equation between (A) and (B) with (B) as datum &considering all losses.(Fig.15)

exitlossssfrictionloentryloss

g

vpZ

g

VPZ BBB

AAA +++++=++

22

22

g

V

gD

fLV

g

V

g

V

222

5.0

200005.32

2222

+++++=++

+++= 1

25.0

20000185.05.01

25.32

2 X

g

V

267.75.32 V=

smV /06.2= 4

2DQ

= sec/101.006.24

25.0 32 mxx =

lpsQ 101=

2) The discharge through a pipe is 225lps. Find the loss of head when the pipe issuddenly enlarged from 150mm to 250mm diameter.Solution: D1=0.15m, D2 = 0.25m Q=225lps = 225m3/secHead loss due to sudden expansion is

gX

D

Q

D

Q

2

1442

2

2

1

=

2

2

2

2

1

2

2 11

2

16

=

DDg

Q

2

222

2

25.0

1

15.0

1

81.92

225.016

=

xx

x

mhL 385.3=

3) The rate of flow of water through a horizontal pipe is 350lps. The diameter ofthe pipe is suddenly enlarge from 200mm to 500mm. The pressure intensity inthe smaller pipe is 15N/cm2. Determine (i) loss of head due to suddenenlargement. (ii) pressure intensity in the larger pipe (iii) power lost due toenlargement.


45/97

Solution : (Fig.16)Q=350lps=0.35m3/sD1=0.2m, D2=0.5m, P1=15N/cm2

hL=?, p2=?, P=?

From continuity equationsm

x

x

D

QV /14.11

2.0

35.044

221

1

===

smx

x

D

QV /78.1

5.0

35.04422

2

2 ===

( ) ( )mofwater

xg

VVhL 463.4

81.92

78.114.11

2

2

21 =

=

=

Applying Bernoullis equation between (1) (1) and (2) (2) with the central line ofthe pipe as datum and considering head loss due to sudden expansion h L only.

Power lost

Lhg

VpZg

VpZ +++=++ 22

2

222

2

111

( )ntalpipehorizoZZ 021 ==

463.462.19

78.1

81.90

62.19

14.11

81.9

1500

2

2

2

+++=++p

22

2 /67.16/68.166 cmNmkNp ==Power lost LQhP = 463.435.081.9 xx=

kWP 32.15=

4) At a sudden enlargement of an horizontal pipe from 100 to 150mm, diameter,

the hydraulic grade line raises by 8mm. Calculate the discharge through the pipesystem.

Solution( )

)1(2

2

21

=g

VVhL (Fig.17)

)2(108, 3112

2 =

+

+ mx

pZ

pZGiven

Applying Bernoullis equation between (1) & (2) with the central line of the pipe asdatum and neglecting minor losses (hL) due to sudden expansion.

Lhg

VpZg

VpZ +++=++ 22

2

222

2

111

02

2

1

2

211

22 =

+

+

+

+ Lh

g

VVpZ

pZ


2

2

1

2

4

15.0

4

1.0xV

xV

x =


46/97

21 25.2 VV =

( ) ( )0

81.92

25.2

81.92

25.2108

2

22

2

2

2

23 =

+

+ x

VV

x

VVx

01274.0108 223 = Vx

smx

V /25.01274.0

108 21

3

2 =

=

Discharge 25.04

15.0

4

2

2

2

2 xx

VD

Q

==

smx /10428.4 33= or lpsQ 425.4=

5) Two reservoirs are connected by a pipe line which is 125mm diameter for thefirst 10m and 200mm in diameter for the remaining 25m. The entrance and exitare sharp and the change of section is sudden. The water surface in the upperreservoir is 7.5m above that in the lower reservoir. Determine the rate of flow,assuming f=0.001 for each of the types.

Solution

From continuity equation 2

2

1

2

4

2.0

4

125.0V

xV

x =

21 56.2 VV =

Applying Bernoullis equation between (1) & (2) in both the reservoirs with thewater in the lower reservoir as datum and considering all losses

ansionlosssuddenssfrictionloentrylossg

VpZg

VpZ BBBAA

A exp22

22

+++++=++

( )

+++++=++g

VV

g

VfL

g

V

222

5.0000}005.7

2

21

2

11

2

1

( ) ( ) ( )

+

++=++g

V

g

VV

g

Vxx

g

V

22

56.2

2

56.21001.0

2

5.25.0}005.7

2

2

2

2

2

2

2

1

{ }1434.2243.52768.362.19

5.72

2 +++=V

( ) smV /6.416.21 21

2 ==sec/1445.06.4

4

2.0 22

mxx

Q =

=

Additional ProblemsFlow through Pipes

1) Water flows upwards through a vertical pipeline. A mercury manometerconnected between two points 10m apart shows a reading of 40cm of mercury


47/97

when discharge is 450lpm. If the friction factor is 0.02. Determine the size of thepipe.

2) A town having a population of 1.2lakhs is to be supplied with water from areservoir 4km away, and it is stipulated that half the daily supply at the rate of

140lpcd should be delivered in 8 hours. Determine the size of the concrete pipesto be laid, if the available head is 12m K for concrete pipes = 0.3mm.

3) Two reservoirs are connected by three pipes of same length laid in parallel,and the diameters are D, 2D & 3D respectively. If the coefficients of friction of allthe three pipes is same, and the discharge in the smallest pipe is 30lps,determine the flow rates in the other two pipes.

4) Two reservoirs are connected by a long pipes 300mm, diameter carrying150lps. If another pipe of the same material is to be laid in parallel to carry twicethis discharge, what should be its diameter? Neglect minor losses.

5) Three pipes are connected in parallel between two points and the totaldischarge is 3 cumecs. If the pipes are of length 1200m, 1400m & 1600mdiameter 1m, 0.8m & 1.2m respectively, and friction factor is the same for all thepipes, determine the discharge in each pipe and the pressure difference requiredto maintain the flow, assuming f=0.02.

6) A 450mm concrete pipe 1800m long connects two reservoirs whose differencein water level is 15m. What is the discharge? If another concrete pipe line300mm diameter is introduced in parallel what would be the percentage increasein discharge and the discharge in each pipe. If the parallel pipe is introduced.

a)In the first half of the length. b)In the second half of the lengthc)In the middle one third of the length. Assume f=0.03 for all pipes and samedifference in the reservoir levels.

7) A 450mm, concrete pipeline 200m long connects two reservoirs whosedifference in water levels is 15m. What is the discharge?a)What is the percentage increase in discharge if another pipe line of the samediameter is introduced is parallel for the second half of the length?b)If a 30% increase in discharge is desired, what diameter pipe should beintroduced in parallel for the second half of the length? Assume f=0.03 for all thepipes and the difference in reservoir levels same in the both the cases. Neglectminor losses.

8) Two pipes of 5cm diameter and 10cm diameter are connected in series. Theyhave the same length and friction factor. If the head loss in the 10cm pipes is 1m,what is the head loss in the 5cm pipe? If the discharge through the 10cm pipe is10lps, what is the discharge through the 5cm pipe?


48/97

P1V1P2

V2

flow

Area = a1Area = a2

9) A pipe has D=40cm, L=10m, f=0.02. What is the length of an equivalent pipewhich has D=20cm and f=0.02.

10) An 8cm diameter pipe carrying water has an abrupt expansion 12cmdiameter at a section. If a differential mercury manometer connected to upstream

and downstream sections of the expansion indicate a gauge reading of 2cm.Estimate the discharge in the pipe.

11) When a sudden contraction form 50cm diameter to 25cm is introduced in ahorizontal pipe line the pressure changes from 105kps to 69kps. Assuming acoefficient of contraction of 0.65, calculate the flow rate. Following thiscontraction if there is a sudden enlargement to 50cm and if the pressure in the25cm diameter section is 69kps, what is the pressure in the 50cm section?

12) Three pipes A, B & C with details as given in the following are connected inseries.

Calculate a) the size of a pipe of length 125m and f=0.020, equivalent to the pipeline ABC b) the length of an 8cm diameter (f=0.015) pipe equivalent to the pipeline ABC.

13) A horizontal pipe line carrying water at 0.03 m3/s reduces abruptly from 15cmto 10cm diameter. Taking contraction coefficient CC=0.60. Determine the

pressure loss across the contraction. How this pressure loss compares with theloss that would result if the flow direction is reversed?

14) Two pipes of diameter 40cm and 20cm are 300m each in length. When thepipes are connected in series and the discharge through the pipe line is 0.1m3/s ,

find the loss of head incurred. What would be the loss of head incurred. Whatwould be the loss of head in the system to pass the same total discharge whenthe pipes are connected in parallel, take f=0.03 for each pipe.

(P.S: FOR ANSWERS TO THE ABOVE PROBLEMS, DOWNLOAD THECONTENTS OF SESSION-9, VTU.AC.IN (E-LEARNING).

1 2


49/97

V1

V

entry

V

Fitting collar

V

entry

Fig.13

Fig 14.(a)

Fig 14.(b)

Fig 14. (c)

Fig 15

V2

exit

exit


50/97

V1

V2

flow

V1 V2

flow

Z1+p1/ Z2+p2/

Fig 17

Fig 16

1

1

2

2

1

1

2

2

Q

P2


51/97

FLOW MEASUREMENTSFlow Through Orifices

An orifice is an opening of any cross section, at the bottom or on the side walls ofa container or vessel, through which the fluid is discharged.If the geometric characteristics of the orifice plus the properties of the fluid areknown, then the orifice can be used to measure the flow rates.

Classification of orificesBased on shape: circular, triangular, rectangular

Based on size :Small orifice (when the head over the orifice is more than five times its size I.e. H>5d,Large orificeBased on shape of the u/s edge :Sharp edge, Bell mouth

Based on flow: Free, Submerged

Flow through an orificeAs the fluid passes through the orifice under a head H, the stream lines convergeand therefore the jet contracts. The stream lines which converge are mostlythose from near the walls and they do so because stream lines cannot make rightangled bend in motion. This phenomenon occurs just down stream of the orifice,and such a section where the area of cross section of the jet is minimum is knowas VENA CONTRACTA.

The pressure at Vena Contracta is assumed to be atmospheric and the velocityis assumed to be the same across the section since the stream lines will beparallel and equally spaced.

Downstream of Vena contracta the jet expands and bends down.Figure(18) shows the details of free flow through a vertical orifice.Applying Bernoulli's equation between (B) & (C) with the horizontal through BCas datum and neglecting losses (hL)

Lhg

VpZ

g

VpZ +++=++

22

2

222

2

111

;21 ZZ = ,1 H

p=

VVV == 21 ,0

02

00002

+++=++g

VH

)1(2 = gHorV Theoretical velocityVelocity V in Eq(1) is known as TORRICELLIS VELOCITY.

Hydraulic Coefficients of an orifice


52/97

i)Coefficient of discharge (Cd): It is defined as the ratio of actual discharge (Q act)

to the theoretical discharge (Qth).

=

th

actd

Q

QC .

Value of Cd varies in the range of 0.61 to 0.65

ii) Coefficient of Velocity (Cv): It is defined as the ratio of actual velocity (Vact) to

the theoretical velocity (Vth).

=

th

actV

V

VC Value of Cv varies in the range of

0.95 to 0.99Coefficient of Contraction (Cc): It is defined as the ratio of the area of cross

section of the jet at Vena of cross section of the jet at Vena Contracta (a c) to the

area of the orifice (a).

=

a

aC cC

Value of Cc will be generally more than 0.62.Relationship between the Hydraulic Coefficients of an orificeFrom continuity equation

Actual discharge Qact = ac x VactTheoretical discharge Qth = a x Vth

th

actc

th

act

V

Vx

a

a

Q

Q=

or

ccd xCCC =Equation for energy loss through an orifice

Applying Bernoullis equation between the liquid surface (A) and the centre of jetand Vena Contracta (C) and considering losses (hL).

LCC

CAA

A hg

VpZ

g

VpZ +++=++

22

22

,HZA = )(0 atmospherepp BA ==,0=AV )(0 cityactualvelopp BA ==

Lhg

VaH +++=++

20000

2

)2

(2

g

VaHhL =

gHCButV Va 2=Torricellis equation

)(2

VL HxCHh =

)1(2

VL CHh =Equation for Coefficient of Velocity (CV) (Trajectory method)


53/97

Consider a point P on the centre line of the jet, such that its horizontal andvertical coordinates are x and y respectively.By definition, velocity

t

xVa = or,

aV

xt=

Since, the jet falls through a vertical distance y under the action of gravity duringthis time (t)

2

2gty = or

)2(2 2

1

=

g

yt

Equating equations (1) & (2)

2

1

2

=

g

y

V

x

a

But ,

gHCV Va 2=

2

1

2

2

=

g

y

gHC

x

V

2

1

2

1

2

1

2

1

2

1

2

1

22 y

gx

Hg

xCV =

Hy

xCV

2=

or

=

yH

xCV

4

2

Problems1. The head of water over the centre of an orifice 30mm diameter is 1.5m. If thecoefficient of discharge for the orifice is 0.613, Calculate the actual discharge.Solution: d=30mm = 3x10-3 H=1.5m

Cd=0.613

;th

actd

QQC = thdact xQCQ = gHxaCd 2=

( ) 213

5.181.924

)1030(613.0 xxx

xxx

=

smxQact /1035.233= or lpsQact 35.2=


54/97

2. Compensation water is to be discharge by two circular orifices under aconstant head of 1.0m, measured from the centre of the orifices. What diameterwill be required to give a discharge of 20x103 m3 per day? Assume Cd for each

notch as 0.615.Solution: d=?H=1m. Qtotal = 20x103 m3/day Cd=0.615.

60602411020

21 3

xxxxxQact = sm /1157.0 3=

we know gHaCQ dact 2=

181.924

615.01157.02

xxxxd

x

=

mmmd 5.2322325.0 ==

3. A jet of water issuing from an orifice 25mm diameter under a constant head of1.5m falls 0.915m vertically before it strikes the ground at a distance of 2.288m

measured horizontally from the Vena Contracta. The discharge was found to be102lpm. Determine the hydraulics coefficients of the orifice and the head due toresistance.Solution: d=25mm=25x10-3

H=1.5m, y=0.915m, x=2.288mQact=102lpm = 102/60 = 1.7lps = 1.7x10-3m3/sec, Cd=?, Cc=?, hL=?

976.05.1915.04

288.2

4

22

===xxyH

xCV

( )638.0

5.181.921025

4107.123

3

=

==

xxxxx

xx

Q

QC

th

actd

999.0976.0

638.0=

===

v

dCVCd

C

CCxCCC

( 21 vL CHheadlossh =( ) 2976.015.1 = mmmhL 2.710712.0 ==

4. The head of water over a 100mm diameter orifice is 5m. The water coming outof the orifice is collected in a circular tank 2m diameter. The time taken to collect45cm of water is measured as 30secs. Also the coordinates of the jet at a pointfrom Vena Contract are 100cm horizontal and 5.2cm vertical. Calculate the

hydraulic coefficients of the orifice.Solution:D=100mm=0.1m, H=5mQact = Area of collecting tankxheight of water collected / time

smxx

/0471.030

45.0

4

2 32

==


55/97

X=100cm = 1m, y=5.2cm = 0.052mCd=?, Cv=?, Cc=?

98.05052.04

1

4

22

=

=

=

xxyH

xCv

605.0581.921.0

40471.02

=

==

xxxx

QQC

th

actd

618.098.0

605.0===

V

dC

C

CC

5. The coordinates of a point on the jet issuing from a vertical orifice are 0.4m &0.003m. Neglecting air resistance, determine the velocity of the jet and the heightof water above the orifice in the tank.Solution.X=0.4m, y=0.3m, V=? H=?

Assume

We knowyH

xCV

4

2

=

==

=

2

2

2

2

22

103.04

4.0

4

4

xxyxG

xH

xyHxG

H=1.33m

smxxxgHGV /115.533.181.9212 ===

6. A vertical orifice is fitted 0.2m above the bottom of a tank containing water to a

depth of 2m. If G=0.98. What is the vertical distance from the orifice of a point onthe jet 0.6m away from the Vena Contracta?SolutionHead over the orifice H=(2-0.2)=1.8mCV=0.98, y=?, x=0.6m

mmmxx

y

xyx

or

yH

xCV

52052.098.08.14

6.08.14

6.0)98.0(,

4

2

2

22

2

==

=

=

=

7. A closed tank contains water to a height of 2m above a sharp edged orifice1.5cm diameter, made in the bottom of the tank. If the discharge through the


56/97

orifice is to be 4lps. Workout the pressure at which air should be pumped into thetank above water. Take Cd=0.6.

Solution: (fig.19)Q=4lps = 4x10-3m3/sD=1.5x10-2m, Cd=0.6

PA=? 333 /10772.11/772.11 mkNxmNair ==


+=

AphH

gHaCQ dact 2=

( )

+=

3

223

10772.11281.92

4

105.16.0104

x

Pxxx

xxxx A

)(/83.0 2 GaugemkNPA =

8. A closed tank contains 3m depth of water and an air space at 15kpa pressure.

A 5cm diameter orifice at the bottom of the tank discharge water to the tank Bcontaining pressurized air at 25kpa. If Cd = 0.61 for the orifice. Calculate the

discharge of water from tank A.Solution: fig(20)d=5cm = 5x10-2m Cd=0.61.


( )

+=

+=81.9

25153

BA pphH

H=1.9806m

9806.181.924

05.0

61.02

2

xxx

x

xgHaCQ dact

==lpssmxQact 47.7/1047.7

33 ==

9. A tank has two identical orifices in one of its vertical sides. The upper orifice is4m below the water surface and the lower one 6m below the water surface. If thevalue of Cv for each orifice is 0.98, find the point of intersection of the two jets.

Solution.

yH

xCV

4

2

=

Given Cv is same for both the orifices

22

2

2

11

2

1

44 Hy

x

Hy

x=

)(44

21

22

2

11

2

1 xxHy

x

Hy

x=


57/97

)1(5.164 2121 == yoryyy

from figure(21)( )

)2(2

46

21

21

+=+=

yy

yy

Substituting eq(1) in eq(2) and simplifying

my

yyy

4

25.025.1

2

2

22

== +=

Again givesHy

xCV

22

2

2

4=

mx

xx

x

6.9

64498.0

2

2

2

=

=

(points of intersection of the jets from the Vena contracts)

10. Two orifices have been provided in the side of the tank, one near the bottomand the other near the top. Show that the jets from these two orifices willintersect a plane through the base at the same distance from the tank if the headon the upper orifice is equal to the height of the lower orifice above the base.

Assume Cv to be the same for both the orifices.

Solution.To show that x1=x2 when H1=y2from figure(22)y1=[y2+(H2-H1)---(1)

22

2

2

11

2

11 44

,2 Hy

xHy

xCCGiven VV ==

or 2211 44 HyHy =])([ 221122 HyHHHy =+

22

2

12112 HyHHHHy =+0)( 12221

2

1 =+ HHyHHH substituting

00

0)(; 222222

221

==+= yHyyHyyH

11. A 4cm dia orifice in the vertical side of a tank discharges water. The watersurface in the tank is at a constant level of 2m above the centre of the orifice. Ifthe head loss in the orifice is 0.2m and coefficient of contraction can be assumedto be 0.63. Calculate (I) the values of coefficient of velocity & coefficient ofdischarge, (ii) Discharge through the orifice and (iii) Location of the point ofimpact of the jet on the horizontal plane located 0.5m below the centre of theorifice.


58/97

Solution

2 2 9.81 2V gH x x= =6.264 /V m s= Head loss

2

2

L

Vah H

g

=

20.2 2

2 9.81

Va

x

=

5.943 /aV m s=

Coefficient of Velocity5.943

0.9436.246

av

VC

V= = =


d v CC C xC = 0.949 0.63x= 0.598dC =(ii) Discharge through the orifice

2act d Q C a gH =

20.598 0.04 2 9.81 24

x x x x x

= 34.707 10 / 4.71x m s lps= =

(iii) Coefficient of velocity2

4v

xC

yH= or

24 VyHC x= 24 0.5 2(0.949)x x x = 1.898x m=

12. An orifice has to be placed in the side of a tank so that the jet will be at amaximum horizontal distance at the level of its base. If the depth of the liquid intthe tank is D, what is the position of the orifice? Show that the jets from the twoorifices in the side of the tank will intersect at the level of the base if the head onthe on the upper orifice is equal to the height of the orifice above the base.Solution: fig(23)By definition, Velocity V=x/t

Vt=But

2V gH= and 21

2y gt=

2

1( )2 2

xD H ggH

=

or

( )2 4x H D H= or

4 ( )x H D H= For x to be maximum


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0dx

dH=

4( 2 ) 0D H =/ 2H D =

We know, x=Vt, 12V gH=

2

2

1

2y H gt+ =

2

1

1

2 2

xg

gH

=

( )2 1 24 (1)x H y H= +

( )2 14 / 2 (2)x H y L= + Equating (1) & (2)

1 1 2 1 2 24 4 4 4H y H H H H yH+ = +

1 2 1 24 4H y yH H H= =

13. Two tanks with orifices in the same vertical plane are shown in figure.What should be the spacing x for the jets to intersect in the plane of the base?

Assume CV=0.98 for each orifices. x=?

Solution: fig(25)2 2

1 2

1 1 2 2

(1)4 4

x x

y H y H=

Assuming the coefficient of velocity CV to be the same for both the orifices, we

have (CV1) = (CV2)

where

1 12 2 0.4 1.6y H m= = =

2 22 2 1.6 0.4y H m= = =

1 20.4 , 1.6H m H m= =2 2

1 2

4 1.6 0.4 4 0.4 1.6

x x

x x x x =

2

1

11 14

V

xC

y H=

2

10.98

4 1.6 0.4

x

x x

= 1 21.568x m x= =

[ ]1 2 2 1.568 3.136x x x x m = + = =

14. A large tank has a circular sharp edged orifice 25mm diameter in thevertical side. The water level in the tank is 0.6m above the centre of the orifice.The diameter of the jet at Vena contracta is measured as 20mm. The water of


60/97

the jet is collected in a tank 1.2m long x 0.6m wide and the water level rised from0.15 to 0.75 in 7 minutes. Calculate the orifice coefficients.Solution: Dia of jet at vena contracta dc=20mm

Dia of orifice = d = 25mmHead over the orifice H = 0.6m

Depth of water collected in the measuring tank h=(0.75-0.15)=0.6mDepth of water collected in the measuring tank A = 1.2x0.6 = 0.72m 2

Time taken for collecting 0.6m of water t=7min=7x60=420secTherefore actual discharge Qact = Area of measuring tank x depth of water

collected / time taken

0.72 0.6. .

420act

Ah xi e Q

t

= =

3 31.0286 10 /x m s=

2thQ a gH =( )

23

25 102 9.81 0.6

4

xx x x x

=

3 21.684 10 /thQ x m s=


actd

th

QC

Q=

3

3

1.0286 10

1.684 10

x

x

=

0.61dC =

Coefficient of contractionCc = area of jet at vena contracta / area of orifice

2

2

4

4C

c

d

C x d

=

2 220

0.6425c

d

d

= = =

We know, Cd=CcxCV

Therefore, Coefficient of velocity0.61

0.64

dV

c

CC

C= =

0.953VC = Coefficient of resistance 21

C

V

CC

=

2

11

0.953

=

0.1008rC =

15. A jet of water issuing from a vertical orifice in a tank under a constanthead of 4m. If the depth of water in the tank is 12m, at what depth another orificeto be mounted vertically below the former one, so that both the jets meet at acommon point on the horizontal at the bottom of the tank? Assume C vto be the

same for both the orifices = 0.98.

Solution: fig(26)


61/97

2 2

1 2

1 21 1 2 2

,4 4

V V

x xC C

y H y H= =

From figure, (y1-y2)=(H2-H1)

( ) ( )1 1 12 4 8y y H m= = =

( ) ( )2 2 212y y H H m= = Equating the values of CV

2 2

1 2

1 1 2 24 4

x x

y H y H=

1 1 2 24 4y H y H=

1 2x x=Q

1 1 2 2y H y H=

( )2 2

8 4 12x H H= 2

2 212 32 0H H + =2

2

12 12 4 1 32

2 1

x xH

x

+ =

8 4mor m=H2 = 8m is the correct answer.

Hence, the second orifice should be 4m below the first orifice.16.

Water is to be discharged by two circular orifices under a constant head of 1mabove their centres. What should be the diameter of the orifices to give a

discharge of 20Mlpd? Assume a coefficient of discharge of 0.62.Solution.Total discharge=20Mlpd(million litres per day)

620 10231.48

24 60 60

xlps

x x= =

Therefore Discharge per orifices231.48

115.742

Q lps= =

20.11574 /orQ m s=

But,2dQ C a gH =

0.11574

0.62 2 9.81 1a

x x x

=

20.04214m=


62/97

12

24 0.04214

4

d xa d

= =

=0.2316mTherefore, Diameter of each orifice d = 231.6mm

17. What is the discharge through the 60mm diameter orifice shown in figure,assuming the oil level remains constantSolution. Fig(27)

Head of the orifice H100

20.9 9.81x

= +

13.326H mofoil = 2dQ C a gH =

( )30.65 60 10 2 9.81 13.3264

x x x x x x =

30.02972 /m s= 29.72lps=

18. What is the discharge through a sharp edged slot 0.2 long x 10mm wideat the bottom of a tank 0.5m diameter with 3m depth of water constant?Solution.Head over the orifice H=(2-0.2)=1.8m

0.98, ?, 0.6VC y x m= = =2

4V

xC

yH=

22 0.6

(0.98) 4 1.8xyx

= 0.052 52y m mm = =

19. A vertical orifice is fitted 0.2m above the bottom of a tank containingwater to a depth of 2m. If CV=0.98. What is the vertical distance from the orifice

of a point on the jet 0.6m away from the Vena contracta?Solution.

100.61 0.2 2 9.81 3

100x x x x x

=

3 29.36 10 /x m s=9.36Q lps =

20. The coordinates of a point on the jet issuing from a vertical orifice are0.4m & 0.3m. Neglecting air resistances, determine the velocity of the jet and theheight of water above the orifice in the tank. Assume CV=0.98.

Solution.X=0.4m, y=0.3m, V=?, H=?


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2

4V

xC

yH=

20.40.98

4 0.3x xH=

0.1388H m =

2 2 9.81 0.1388 1.65 /V gH x x m s= = =

Mouth PiecesA mouth piece is a short tube or pipe connected in extension with an orifice

Classification of Mouth PiecesDepending on the position with respect to the tank: External, Internal

Depending on shape :Cylindrical,Convergent, Divergent

Nature of flow: Running Full,Running Free

External Cylindrical Mouthpiece fig(28)

It is a short pipe whose length is two or three times the diameter.H=Head over the centre of the mouth pieceVO=Velocity of the liquid at Vena Contracta

ac=Area of flow at Vena Contracta

V1=Velocity of liquid at outlet

a1=Area of mouth piece at outlet.

Cc=coefficient of contraction

Applying continuity equation between & (1) &(1)accc=a1v1

1

1c

c

a

V Va =

1

0.62cac

c coefficientofcontractiona

= = =

1

1(1)

0.62cV V =

As the jet flows from to (1) (1) there will be loss of head due to suddenenlargement of flow, and this value can be calculated from the relation.

( )

2

12 1

1 0.62

2 2

C

L

VV

V Vh

g g

= =

2

1

0.375(2)

2Lh V

g=

Applying Bernoullis equation between (A) and (1) (1) with the centre line of themouth piece as datum and considering head loss hL due to sudden expansion.

2 2

1 11

2 2

A AA L

p V p VZ Z h

g g + + = + + +


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11 0, , 0( )A

ppAZ Z H atmosphericpressure

= = = =

0( )AV Negligible=2 2

1 10.3750 0 0 0

2 2

V VH

g g

+ + = + + +

2

1

1.375

2H V

g=

or

1

2

1.375

gHV =

1 0.853 2 (3)V gH = By definition, Coefficient of velocityCV=Actual velocity/Theoretical velocity

0.853 2. ,2

V gHi e CgH=

0.853VC =At the exit of the mouth piece CC=1

1 0.853 0.853d c vC C xC x = = =Hence, for an external cylindrical mouth piece Cd=(=0.853) is more than that of

an orifice.Pressure head at Vena contracta

Applying Bernoullis equation between (A) & with the centre line of the mouthpiece as datum & neglecting losses.

22

2 2

c cA AA c L

p Vp VZ Z hg g

+ + = + + +

, 0, 0, 0, ?cA A C LPpA

H V Z Z h

= = = = = =

2

0 0 0 02

c cp VHg

+ + = + + +

2

2

c cp VHg

=

But,2

1 11.375 ,2 0.62

CV VH V

g= =

1

20.853 2

1.375

gHV gH = =

0.853 2&

0.62C

gHV =


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2

0.853 2 1

0.62 2

CgHp

H xg

=

1.893Cp

H H

=

0.893Cp

H

=

Negative sign indicates that the pressure at the Vena contracta is less th

prof.balasubramanya compiled fm notes

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