FLOW THROUGH PIPES - · PDF fileIn 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)

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FLOW THROUGH PIPES Definition of flow through pipes A 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. Reynold’s Number (Re) It is defined as the ratio of Inertia force of a flowing fluid and the Viscous force. Re=(Inertia force/Viscous force) =( ρ V D/µ ) Classification of pipe flow: Based on the values of Reynold’s number (Re), flow is classified as Follows: Laminar flow or Viscous Flow In 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 Reynold’s number Re< 2000. In laminar flow 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 flow occurs when 2000< Re < 4000.

Laminar flow Transition flow Turbulent flow

Water

Dye

Glass tube



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. This loss 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 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 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.

.2

2

gVpZ ++

γ



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

Applying Bernoulli’s equation between (1) & (2) with the centre line of the pipe as datum & considering head loss due to friction hf,.

( )

( ) )1(44

44..

0

021

0

2

21

0

2

2

2

1

−−−=−

=−

−−+

=Σ∴

DLPPor

DLDPP

DLxDPDPei

forces

τ

πτπ

πτππ

fhg

VpZg

VpZ +++=++22

222

2

211

1 γγ

21 ZZ =

21 VV =

Pipe is horizontal

Pipe diameter is same throughout

)2(21 −−−=−

∴ fhPPγ

(1)

(1) (2)

(2)L

Dp1p2

Flow (V)



Substituting eq (2) in eq.(1)

From Experiments, Darcy Found that

f=Darcy’s friction factor (property of the pipe materials Mass density of the liquid. V = velocity Equations (3) & (4)

But,

from Continuity equation

& (5) & (6) are known as DARCY – WEISBACH Equation Pipes in Series or Compound Pipe D1, 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.

)3(4

40

0 −−−==lDh

orD

Lxh ff

γττγ

)4(8

20 −−−= Vf ρτ

LDh

Vf f

482 γ

ρ = or DVLfhf γρ

84 2

=

)5(2

2

−−−

=∴

=

gDfLVh

g

f

ργ

24DQV

π=

)6(852

2

−−−

=∴

DghfLQhf



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

D1, D2 and D3 are the pipe diameters. Length of each pipe is same, that is, L1=L2=L3 For pipes in parallel hf1=hf2=hf3 i.e

D2D3

D4D1Q

L2 L4L1

4321 hfhfhfhfhf +++=

54

2

24

53

2

23

52

2

22

51

2

21 8888

DgQfL

DgQfL

DgQfL

DgQfLhf ππππ

+++=

D1

D2

D3

Q Q

Q1

Q2

Q3

L = L1 = L2 = L3

Pipes in Parallel

)1(

888

53

2

52

2

51

2

53

2

23

52

2

22

51

2

21

321

321

−−−−==

==

DQ

DQ

DQor

DgQfL

DgQfL

DgQfL

πππ

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



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 other minor 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

Equating (1) & (2) and simplifying

Or

D2D1 D3 == Q DL= L1+L2+L3

L1 L3

Q

321 hfhfhfhf ++=

)1(88853

2

23

52

2

22

51

2

21 −−−++=

DgQfL

DgQfL

DgQfLhf πππ

)2(851

2

2

−−−=Dg

fLQhf π

53

352

251

15 D

LDL

DL

DL

++=

51

53

352

251

1

++=

DL

DL

DL

LD



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/s hf=5m, L=1km = 1000m. f=0.02 Darcy’s 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 Bernoulli’s equation between (1) & (2) with (2) as datum & considering head loss due to friction hf only,

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. But

=∴fhg

fLQD 2

28π

51

2

23

581.9)1040(100002.08

=∴−

xxxxxxD

π

mmmD 22022.0 =−

)1(22

222

2

211

1 −−−+++=++ fhg

VpZg

VpZγγ

5

28Dg

fLQhf π=

21

52

2500016.085.081.920

=xx

xxxQ π

lpsmQ 8.434sec/4348.0 3 ==



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 daily supply is pumped in 8 ½ hours. Take f=0.0195. Solution:- Number of inhabitants = 5million = 5,00,000 Length 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 to replace the compound pipe. Solution:- L=800m L1=175m D1=0.3m L2=200m D2=0.25m L3=250m D3=0.20m L4=175m D4=0.15m For an equivalent pipe

sec/1248.260605.8

000,65 3mxx

Q =

=

= 52

28Dg

fLQhf π

51

2

2

2581.9)1248.2(000,25195.08

=xxxxxD

π

mD 487.1=

+++= 54

453

352

251

15 D

LDL

DL

DL

DL



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. 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.

51

5555 15.0175

2.0250

25.0200

3.0175

800

+++

=∴D

54

24

53

23

52

22

51

21

DQ

DQ

DQ

DQ

===

( ) sec/124.010455.1 321

235

2 mxxd

dQ =

= −

( ) sec/4446.010455.2 321

235

2 mxxd

dQ =

= −

( ) sec/9592.010454.3 321

235

2 mxxd

dQ =

= −



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 Reynold’s 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, Reynold’s number Re=

7) A 30cm diameter main is required for a town water supply. As pipes over 27.5cm diameter 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:-

µρ 222 VD

2

222

1

111

µρ

µρ VDVD

=∴

2211 DVDV =

21 75.05.0 VV =

( )21 ρρ =Θ ( )21 µµ =

21 5.1 VV =

gDfLVhf 2

2

=

22

21

1

2

2

1

VVx

DD

hfhf

=∴

375.35.15.075.0

2

2

2 =

=

VVx

From Darcy’s equation

)1(852

2

−−−

=Dg

fLQhf π

)2(28

2

52 −−−

=Dg

QfLhf π



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 f2=0.0325 Q1=0.285m3/sec Q2=? For parallel pipes

=

52

2

52

2 28

8Dg

QfL

DgfLQ

ππ

551 4

11DD

=∴

51

5

4275.0

=D

mmD 275.0205.0 ≥=

or

=

= 52

2

2222

51

2

2111 88

DgQLf

DgQLfhf ππ

21

5122

52

2111

2

=∴DLfDQLfQ

21

5

52

2 3.035000325.035.0285.031500216.0

=∴xx

xxxQ

sec/324.0 32 mQ =



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) the ratio 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=? From Darcy’s equation

(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

= 52

28Dg

fLQhf π

52

2

2222

51

2

2111 88

DgQLf

DgQLf

ππ=∴

656.52 25

2

225

2

1

2

1 =

=

=

DD

DD

QQ

03125.02

85

2

2

5

1

251

2

2211

2

1 =

=

==

DD

DD

DgQLf

hfhf

π

=

= 52

2

2222

51

2

2111 88

DgQLf

DgQLfhf ππ



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 friction factors. Solution H1 = head loss due to friction = hf = hf2i.e.

sec/1063.32000215.08

05.081.915 3321

52

1 mxxxxxhxQ −=

=

sec/023.02000215.08

105.081.915 3221

52

1 mxxxxhxQ =

=

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

52

28Dg

fLQhf π=

( ) 51

2

2

1581.902684.02000215.08

=∴xx

xxxDπ

cmmD 12.111112.0 ==∴

= 52

21

)2(8

DgfLQhf π

)1()(

852

22

2−−−

=Dg

fLQhf π

)2(21 −−−=+ QQQ

QQQ =+ 2266.5 QQ66.61

2 =∴

)3(02256.0866.618

52

2

52

2

1 −−−=

=∴Dg

QfLxDg

QfLH

ππ



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 diameter is laid parallel to the first. Solution Applying Bernoulli’s equation between (1) & (2) with (2) as datum and considering head loss due to friction hf

52

2

52

2

)2(88

DgfLQ

DgfLQ

hf ππ+

=

+=∴ 552

2

2 21

118

DgfLQHπ

)4(80312.152

2

2 −−−=DgxflQxH

π

2

52

52

2

2

1

80312.1802256.0

flQxDgx

DgxflQx

HH π

π=∴

021876.00312.102256.0

2

1 ==HH

71.451

2 =HH

Or

fhg

VpZg

VpZ +++=++22

212

2

211

1 γγ

mhh ff 100000010 =∴+++=++

52

28Dg

fLQhf π=



Case (ii)

Change in discharge =

( ) 21

52

300003.081025.081.9

=∴xx

xxxQ π

sec/03624.0 3mQ =∴

321 orhfhfhfhf +=

( )

+= 5

21

5

21

2 25.02/600

25.02400

81.903.0810 QQ

xx

π

2166.647210 Q=

sec/0393.0 31 mQ =

( )QQQ −=∆ 1

( )03624.00393.0 −=

sec/10066.3 33 mxQ −=∆

1001

xQQ∆

% increase in discharge =

%46.810003624.0

10066.3 3

==−

xx



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 loss iv)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 ρ =volume/time x change in velocity

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

Both sides by (sp.weight)

( ) )(1212211 iaapapapforces −−−−+−+=∑

( ) )(, 212 iippaforcesor −−−−=∑

( ) )(21 iiiVVxQx −−−−ρ

( ) ( )21212 VVpQppa −=−

( ) ( )21221 VVVpp −=− ρor

( ) )(21221 ivg

VVVpp−−−

−=

−∴

γ



Applying Bernoulli’s equation between (1) and (2) with the centre line of the pipe as datum and considering head loss due to sudden expansion hLonly.

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

Equations for other minor losses

gVpZ

gVpZ

22

222

2

211

1 ++=++γγ

zontalpipeishoriCZZ 21 =

LhgVVpp

=−

=

−∴

2

22

2121

γ

( ) ( )g

VVVVVhL 22 2

22

1212 −+−=

gVVVVVhL 2

22 22

21

2221 −+−

=

gVVVVVhL 2

22 22

2121

22 −+−

=

gVVVVhL 2

2 212

12

2 −+=

( )gVVhL 2

221 −=

)(viQhP f −−−= γ

gVhL 2

5.02

2=Sudden contraction loss



Loss due to entrance and exit

Loss due to bends & fittings

Problems

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 Bernoulli’s equation between (A) and (B) with (B) as datum & considering all losses.

gVh entryL 2

5.0 2

=

gVh exitL 2

2

=

gKVhL 2

2

=

K=coefficient

exitlossssfrictionloentrylossg

vpZg

VPZ BBB

AAA +++++=++

22

22

γγ

gV

gDfLV

gV

gV

2225.0

200005.32

2222

+++++=++

+++= 1

25.020000185.05.01

25.32

2 Xg

V

267.75.32 V=

smV /06.2=

4

2DQ π= sec/101.006.24

25.0 32mxx =π

lpsQ 101=



2) The discharge through a pipe is 225lps. Find the loss of head when the pipe is suddenly enlarged from 150mm to 250mm diameter. Solution : D1=0.15m, D2 = 0.25m Q=225lps = 225m3/sec Head 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 in the larger pipe (iii) power lost due to enlargement. Solution Q=350lps=0.35m3/s D1=0.2m, D2=0.5m, P1=15N/cm2hL=?, p2=?, P=? From continuity equation

Applying Bernoulli’s equation between (1) (1) and (2) (2) with the central line of the pipe as datum and considering head loss due to sudden expansion hL only.

( )gVV

hL 212−

=

gX

DQ

DQ

2144

22

21

−=ππ

2

22

21

2

2 11216

−=

DDgQπ

mhL 385.3=

2

222

2

25.01

15.01

81.92225.016

−=

πxxx

smxx

DQV /78.1

5.035.044

222

2 ===ππ

( ) ( ) mofwaterxg

VVhL 463.481.92

78.114.112

221 =

−=

−=



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 Bernoulli’s equation between (1) & (2) with the central line of the pipe as datum and neglecting minor losses (hL) due to sudden expansion.

From continuity equation

Lhg

VpZg

VpZ +++=++22

222

2

211

1 γγ

( )ntalpipehorizoZZ 021 ==

463.462.19

78.181.9

062.1914.11

81.91500

22

2

+++=++p

222 /67.16/68.166 cmNmkNp ==

LQhP γ=

463.435.081.9 xx=

kWP 32.15=

( ) )1(2

221 −−−

−=

gVVhL

)2(108, 311

22 −−−=

+−

+ − mxpZpZGiven

γγ

Lhg

VpZg

VpZ +++=++22

222

2

211

1 γγ

02

21

221

12

2 =

+−

+

+−

+ Lh

gVVpZpZ

γγ



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 the change 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 From continuity equation

Applying Bernoulli’s 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

415.0

41.0 xVxVx ππ

=

21 25.2 VV =

( ) ( ) 081.92

25.281.9225.2108

222

22

223 =

−

+−

+∴ −

xVV

xVVx

01274.0108 22

3 =−− Vx

smxV /25.01274.0108 2

13

2 =

=

−

25.0415.0

4

2

2

22 xxVDQ ππ

==

smx /10428.4 33−=

lpsQ 425.4=

2

2

1

2

42.0

4125.0 VxVx ππ

=

21 56.2 VV =∴



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 characteristics of the orifice plus the properties of the fluid are known, then the orifice can be used to measure the flow rates.

Classification of orifices

ansionlosssuddenssfrictionloentrylossg

VpZg

VpZ BBB

AAA exp

22

22

+++++=++γγ

( )

−

+++++=++gVV

gVfL

gV

2225.0000}005.7

221

211

21

{ }1434.2243.52768.362.19

5.72

2 +++=V

( ) smV /6.416.21 21

2 ==∴

( ) ( ) ( )

+−

++=++g

Vg

VVg

Vxxg

V22

56.22

56.21001.02

5.25.0}005.72

22

22

22

1

sec/1445.06.44

2.0 22

mxxQ =

=π

Based 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 orifice

Based on shape of the u/s edge

Sharp edge Bell mouth

Based on flow Free Subme



Flow through an orifice As the fluid passes through the orifice under a head H, the stream lines converge and therefore 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 the area 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 equally spaced. 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)

Velocity V in Eq(1) is known as TORRICELLI’S VELOCITY. Hydraulic Coefficients of an orifice i)Coefficient of discharge (Cd): It is defined as the ratio of actual discharge (Qact) to the theoretical 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).

Lhg

VpZg

VpZ +++=++22

222

2

211

1 γγ

;21 ZZ = VVV == 21 ,0

02

00002

+++=++∴g

VH

)1(2 −−−= gHorV

,1 Hp=

γ

Theoretical velocity

=∴

th

actd Q

QC



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 (ac) 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 From continuity equation Actual discharge Qact = ac x Vact Theoretical discharge Qth = a x Vth

Equation for energy loss through an orifice Applying Bernoulli’s equation between the liquid surface (A) and the centre of jet and Vena Contracta (C) and considering losses (hL).

=∴

th

actV V

VC

=∴

aaC c

C

th

actc

th

act

VVx

aa

QQ =∴

ccd xCCC =Or

LCC

CAA

A hg

VpZg

VpZ +++=++22

22

γγ

)(0 atmospherepp BA ==



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)

,HZ A =

,0=AV)(0 cityactualvelopp BA ==

Lhg

VaH +++=++∴2

00002

)2

(2

gVaHhL −=

gHCButV Va 2=

)( 2VL HxCHh −=

)1( 2VL CHh −=

txVa =

aVxt =Or

2

2gty = )2(2 21

−−−

=

gyt

Or



But

21

2

=

gy

Vx

a

gHCV Va 2=

21

22

=

gy

gHCx

V

21

21

21

21

21

21

22 y

gxHg

xCV =

HyxCV 2

=

=

yHxCV 4

2

Or



Problems 1. The head of water over the centre of an orifice 30mm diameter is 1.5m. If the coefficient of discharge for the orifice is 0.613, Calculate the actual discharge. Solution: d=30mm = 3x10-3 H=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

;th

actd Q

QC =

thdact xQCQ = gHxaCd 2=

( )213

5.181.924

)1030(613.0 xxxxxx−

= π

smxQact /1035.2 33−=

lpsQact 35.2=

60602411020

21 3

xxxxxQact =

sm /1157.0 3=

gHaCQ dact 2=



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 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=25x10-3H=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 of water 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 the orifice. Solution: D=100mm=0.1m, H=5m Qact = Area of collecting tankxheight of water collected / time

181.924

615.01157.02

xxxxdxπ=

mmmd 5.2322325.0 ==

976.05.1915.04

288.24

22

===xxyH

xCV

( ) 638.05.181.921025

4107.123

3

=

==−

−

xxxxxxx

QQC

th

actd

π

999.0976.0638.0

=

==∴=

v

dCVCd C

CCxCCC

( )21 vL CHheadlossh −=

( )2976.015.1 −=

mmmhL 2.710712.0 ==

smxx /0471.03045.0

42 3

2

==π



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

We know

98.05052.04

14

22

=

=

=

xxyHxCv

605.0581.921.0

40471.02

=

==xxx

xQQC

th

actd π

618.098.0605.0

===V

dC C

CC

1=VC

yHxCV 4

2

=

==∴

=

2

2

2

2

22

103.044.0

4

4

xxyxGxH

xyHxG

smxxxgHGV /115.533.181.9212 ===

H=1.33m



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 on the jet 0.6m away from the Vena Contracta? Solution Head 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. 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

mmmxx

y

xyxor

yHxCV

52052.098.08.14

6.0

8.146.0)98.0(,

4

2

2

22

2

==

=

=

=

333 /10772.11/772.11 mkNxmNair−==γ

+=γ

AphH

gHaCQ dact 2=

( )

+= −

−−

3

223

10772.11281.92

4105.16.0104

xPxxxxxxx Aπ

)(/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 B containing pressurized 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. Total head over the orifice

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 Cv for each orifice is 0.98, find the point of intersection of the two jets. Solution.

Given Cv is same for both the orifices

( )

−+=

−

+=81.925153

γBA pphH

9806.181.92405.061.02

2

xxxxxgHaCQ dactπ

==

lpssmxQact 47.7/1047.7 33 == −

yHxCV 4

2

=

22

22

11

21

44 Hyx

Hyx

=

)(44 21

22

2

11

21 xx

Hyx

Hyx

−=

)1(5.164 2121 −−−==∴ yoryyy



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

( )

)2(2

46

21

21

−−−+=

−+=

yy

yy

my

y

yy

4

25.0

25.1

2

2

22

=∴

=

+=

givesHy

xCV22

22

4=

mx

xxx

6.9

64498.0

2

22

=∴

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

22

22

11

21

1 44,

2 Hyx

HyxCCGiven VV

=∴=

2211 44 HyHy =

Or

])([ 221122 HyHHHy =−+

Or



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) the values 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

222

12112 HyHHHHy =−+

0)( 122212

1 =−+− HHyHHH

00

0)(; 222222221

=∴

=−+−= yHyyHyyH

substituting

281.922 xxgHV ==

smV /264.6=

−=

gVaHhL 2

2

smVa /943.5=

−=

81.9222.0

2

xVa

Or



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 maximum horizontal 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:

943.0246.6943.5

===VVC a

v

Cvd xCCC = 63.0949.0 x=

598.0=dC

gHaCQ dact 2=

281.9204.04

598.0 2 xxxxxπ=

lpssmx 71.4/10707.4 3 == −

yHxCv 4

2

=

24 xyHCV =

2)949.0(25.04 xxx =∴

mx 898.1=



By definition, Velocity V=x/t

But

and

For x to be maximum

- - - - - - - - - - - - - - - - - - - - - - -

x

Jet Orifice

D

h

y

Vt∴=

gHV 2=

2

21 gty =

2

221)(

=−∴

gHxgHD

( )HDHx −= 42

)(4 HDHx −=

Or

0=dHdx



We know, x=Vt,

0)2(4 =− HD

2/DH =∴

12gHV =

22 2

1 gtHy =+

2

1221

=

gHxg



FLOW THROUGH PIPES Definition of flow through pipes A 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. Reynold’s Number (Re) It is defined as the ratio of Inertia force of a flowing fluid and the Viscous force. Re=(Inertia force/Viscous force) =( ρ V D/µ ) Classification of pipe flow: Based on the values of Reynold’s number (Re), flow is classified as Follows: Laminar flow or Viscous Flow In 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 Reynold’s number Re< 2000. In laminar flow 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 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 with each 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. (REYNOLD’S EXPERIMENT:Refer Fig.(1) 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.

.2

2

gVpZ ++

γ

EGL Line lies above HGL by an amount V2/2g.(Refer Fig.(2)) 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. This loss 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 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. τo is the shear stress acting along the pipe boundary. From II Law of Newton Force = 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(44

44..

0

021

0

2

21

0

2

2

2

1

−−−=−

=−

−−+

=Σ∴

DLPPor

DLDPP

DLxDPDPei

forces

τ

πτπ

πτππ

Applying Bernoulli’s equation between (1) & (2) with the centre line of the pipe as datum & considering head loss due to friction hf,.

fhg

VpZg

VpZ +++=++22

222

2

211

1 γγ

21 ZZ = Pipe is horizontal 21 VV = Pipe diameter is same throughout

)2(21 −−−=−

∴ fhPPγ

Substituting eq (2) in eq.(1)

)3(4

40

0 −−−==lDh

orD

Lxh ff

γττγ

From Experiments, Darcy Found that

)4(8

20 −−−= Vf ρτ

f=Darcy’s friction factor (property of the pipe materials Mass density of the liquid. V = velocity Equations (3) & (4)

LDh

Vf f

482 γ

ρ =

or , DVLfhf γρ

84 2

=



But,

)5(2

2

−−−

=∴

=

gDfLVh

g

f

ργ

from Continuity equation 2

4DQV

π=

)6(852

2

−−−

=∴

DghfLQhf

& (5) & (6) are known as DARCY – WEISBACH Equation Pipes in Series or Compound Pipe D1, 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 +++=

54

2

24

53

2

23

52

2

22

51

2

21 8888

DgQfL

DgQfL

DgQfL

DgQfLhf ππππ

+++=

Pipes in parallel D1, D2 and D3 are the pipe diameters.(Fig.5) Length of each pipe is same, that is, L1=L2=L3 For pipes in parallel hf1=hf2=hf3 i.e

)1(

888

53

2

52

2

51

2

53

2

23

52

2

22

51

2

21

321

321

−−−−==

==

DQ

DQ

DQor

DgQfL

DgQfL

DgQfL

πππ

From continuity equation Q= Q1+Q2+Q3--------(2) 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 other minor 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 (Fig.6). For a compound pipe or pipes in series

321 hfhfhfhf ++=

)1(88853

2

23

52

2

22

51

2

21 −−−++=

DgQfL

DgQfL

DgQfLhf πππ



for an equivalent pipe )2(851

2

2

−−−=Dg

fLQhf π

Equating (1) & (2) and simplifying 53

352

251

15 D

LDL

DL

DL

++=

or

51

53

352

251

1

++=

DL

DL

DL

LD

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/s hf=5m, L=1km = 1000m. f=0.02

Darcy’s equation is 5

28Dg

fLQhf π=

=∴fhg

fLQD 2

28π

51

2

23

581.9)1040(100002.08

=∴−

xxxxxxD

π

mmmD 22022.0 == 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 Bernoulli’s equation between (1) & (2) with (2) as datum & considering head loss due to friction hf only, (Fig.7).

)1(22

222

2

211

1 −−−+++=++ fhg

VpZg

VpZγγ

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.

But 5

28Dg

fLQhf π=



21

52

2500016.085.081.920

=xx

xxxQ π

lpsmQ 8.434sec/4348.0 3 ==

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 daily supply is pumped in 8 ½ hours. Take f=0.0195. Solution:- Number of inhabitants = 5million = 5,00,000 Length 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 3mxx

Q =

=

= 52

28Dg

fLQhf π

51

2

2

2581.9)1248.2(000,25195.08

=xxxxxD

π

mD 487.1= 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 to replace the compound pipe. Solution:- L=800m L1=175m D1=0.3m L2=200m D2=0.25m L3=250m D3=0.20m L4=175m D4=0.15m

For an equivalent pipe

+++= 54

453

352

251

15 D

LDL

DL

DL

DL

51

5555 15.0175

2.0250

25.0200

3.0175

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. 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.

54

24

53

23

52

22

51

21

DQ

DQ

DQ

DQ

===

( ) sec/124.010455.1 321

235

2 mxxd

dQ =

= −

( ) sec/4446.010455.2 321

235

2 mxxd

dQ =

= −

( ) sec/9592.010454.3 321

235

3 mxxd

dQ =

= −

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 Reynold’s 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, ?2

1 =hfhf

Reynold’s number Re= µ

ρ VD

2

222

1

111

µρ

µρ VDVD

=∴

2211 DVDV = ( )21 ρρ = ( )21 µµ =

21 75.05.0 VV = 21 5.1 VV =

From Darcy’s equation gD

fLVhf 2

2

=

22

21

1

2

2

1

VVx

DD

hfhf

=∴

375.35.15.075.0

2

2

2 =

=

VVx



7) A 30cm diameter main is required for a town water supply. As pipes over 27.5cm diameter 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. Fig.(8)

Solution:- )1(852

2

−−−

=Dg

fLQhf π )2(2

82

52 −−−

=Dg

QfLhf π

Equating

=

52

2

52

2 28

8Dg

QfL

DgfLQ

ππ

551 4

11DD

=∴

or 51

5

4275.0

=D

mmD 275.0205.0 ≥= 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 f2=0.0325 Q1=0.285m3/sec Q2=?

For parallel pipes

=

= 52

2

2222

51

2

2111 88

DgQLf

DgQLfhf ππ

21

5122

52

2111

2

=∴DLfDQLfQ

21

5

52

2 3.035000325.035.0285.031500216.0

=∴xx

xxxQ

sec/324.0 32 mQ =

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) the ratio 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=?

From Darcy’s equation

= 52

28Dg

fLQhf π

52

2

2222

51

2

2111 88

DgQLf

DgQLf

ππ=∴

656.52 25

2

225

2

1

2

1 =

=

=

DD

DD

QQ

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

03125.02

85

2

2

5

1

251

2

2211

2

1 =

=

==

DD

DD

DgQLf

hfhf

π

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

=

= 52

2

2222

51

2

2111 88

DgQLf

DgQLfhf ππ

sec/1063.32000215.08

05.081.915 3321

52

1 mxxxxxhxQ −=

=

sec/023.02000215.08

105.081.915 3221

52

2 mxxxxhxQ =

=

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

52

28Dg

fLQhf π= ( ) 5

1

2

2

1581.902684.02000215.08

=∴xx

xxxDπ

cmmD 12.111112.0 ==∴

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 friction factors. Solution (Fig.9)

H1 = head loss due to friction = hf = hf2 i.e.

= 52

21

)2(8

DgfLQhf π

)1()(

852

22

2−−−

=Dg

fLQhf π

)2(21 −−−=+ QQQ

QQQ =+ 2266.5 QQ66.61

2 =∴

)3(02256.0866.618

52

2

52

2

1 −−−=

=∴Dg

QfLxDg

QfLH

ππ

Case(iii) 212 hfhfH +=

52

2

52

2

)2(88

DgfLQ

DgfLQ

hf ππ+

=

+=∴ 552

2

2 21

118

DgfLQHπ

)4(80312.152

2

2 −−−=Dg

xflQxHπ

2

52

52

2

2

1

80312.1802256.0

flQxDgx

DgxflQx

HH π

π=∴

021876.00312.102256.0

2

1 ==HH

or 71.451

2 =HH

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 diameter is laid parallel to the first. Solution Applying Bernoulli’s equation between (1) & (2) with (2) as datum and considering head loss due to friction hf (Fig.11)

fhg

VpZg

VpZ +++=++22

212

2

211

1 γγ

mhh ff 100000010 =∴+++=++

52

28Dg

fLQhf π= ( ) 2

152

300003.081025.081.9

=∴xx

xxxQ π



sec/03624.0 3mQ =∴ Case (ii) 321 orhfhfhfhf += 321 orhfhfhfhf += (Fig.12)

( )

+= 5

21

5

21

2 25.02/600

25.02400

81.903.0810 QQ

xx

π

2

166.647210 Q= sec/0393.0 31 mQ =

Change in discharge = ( )QQQ −=∆ 1 ( )03624.00393.0 −= sec/10066.3 33 mxQ −=∆

% increase in discharge = 1001

xQQ∆

%46.810003624.0

10066.3 3

==−

xx

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 loss iv)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 in Fig.13 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. ---------------(1) Consider LHS of eq(1)

( ) )(1212211 iaapapapforces −−−−+−+=∑

( ) )(, 212 iippaforcesor −−−−=∑



Consider RHS of eq(1) Mass x acceleration = ρ x vol x change in velocity /time ρ =volume/time x change in velocity Substitution (ii) & (iii) ( ) ( )21212 VVQppa −=− ρ

or, ( ) ( )21221 VVVpp −=− ρ Both sides by (sp.weight)

( ) )(21221 ivg

VVVpp−−−

−=

−∴

γ

Applying Bernoulli’s equation between (1) and (2) with the centre line of the pipe as datum and considering head loss due to sudden expansion hLonly

gVpZ

gVpZ

22

222

2

211

1 ++=++γγ

zontalpipeishoriCZZ 21 =

LhgVVpp

=−

=

−∴

2

22

2121

γ

( ) ( )g

VVVVVhL 22 2

22

1212 −+−=

gVVVVVhL 2

22 22

21

2221 −+−

=

gVVVVVhL 2

22 22

2121

22 −+−

=

gVVVVhL 2

2 212

12

2 −+= OR ( )

gVVhL 2

221 −=

In Eq(V) hL is expressed in meters similarly, power (P) lost due to sudden expansion is )(viQhP f −−−= γ Equations for other minor losses (Fig.14 a,b,c)

Sudden contraction loss g

VhL 25.0

22=

Loss due to entrance and exit gVh entryL 2

5.0 2

=

g

Vh exitL 2

2

=

Loss due to bends & fittings g

KVhL 2

2

=

K=coefficient



Problems 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 Bernoulli’s equation between (A) and (B) with (B) as datum & considering all losses.(Fig.15)

exitlossssfrictionloentrylossg

vpZg

VPZ BBB

AAA +++++=++

22

22

γγ

gV

gDfLV

gV

gV

2225.0

200005.32

2222

+++++=++

+++= 1

25.020000185.05.01

25.32

2 Xg

V

267.75.32 V=

smV /06.2= 4

2DQ π= sec/101.006.24

25.0 32mxx =π

lpsQ 101=

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

gX

DQ

DQ

2144

22

21

−=ππ

2

22

21

2

2 11216

−=

DDgQπ

2

222

2

25.01

15.01

81.92225.016

−=

πxxx

mhL 385.3=

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 in the larger pipe (iii) power lost due to enlargement. Solution : (Fig.16) Q=350lps=0.35m3/s D1=0.2m, D2=0.5m, P1=15N/cm2hL=?, p2=?, P=?



From continuity equation smxx

DQV /14.11

2.035.044

221

1 ===ππ

smxx

DQV /78.1

5.035.044

222

2 ===ππ

( ) ( ) mofwaterxg

VVhL 463.481.92

78.114.112

221 =

−=

−=

Applying Bernoulli’s equation between (1) (1) and (2) (2) with the central line of the pipe as datum and considering head loss due to sudden expansion hL only. Power lost

Lhg

VpZg

VpZ +++=++22

222

2

211

1 γγ

( )ntalpipehorizoZZ 021 ==

463.462.19

78.181.9

062.1914.11

81.91500

22

2

+++=++p

222 /67.16/68.166 cmNmkNp ==

Power lost LQhP γ= 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 pipe system.

Solution ( ) )1(2

221 −−−

−=

gVVhL (Fig.17)

)2(108, 311

22 −−−=

+−

+ − mxpZpZGiven

γγ

Applying Bernoulli’s equation between (1) & (2) with the central line of the pipe as datum and neglecting minor losses (hL) due to sudden expansion.

Lhg

VpZg

VpZ +++=++22

222

2

211

1 γγ

02

21

221

12

2 =

+−

+

+−

+ Lh

gVVpZpZ

γγ

From continuity equation

2

2

1

2

415.0

41.0 xVxVx ππ

=

21 25.2 VV =

( ) ( ) 081.92

25.281.9225.2108

222

22

223 =

−

+−

+∴ −

xVV

xVVx

01274.0108 22

3 =−− Vx



smxV /25.01274.0108 2

13

2 =

=

−

Discharge 25.0415.0

4

2

2

22 xxVDQ ππ

==

smx /10428.4 33−= or lpsQ 425.4= 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 the change 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

From continuity equation 2

2

1

2

42.0

4125.0 VxVx ππ

=

21 56.2 VV =∴ Applying Bernoulli’s equation between (1) & (2) in both the reservoirs with the water in the lower reservoir as datum and considering all losses

ansionlosssuddenssfrictionloentrylossg

VpZg

VpZ BBB

AAA exp

22

22

+++++=++γγ

( )

−

+++++=++gVV

gVfL

gV

2225.0000}005.7

221

211

21

( ) ( ) ( )

+−

++=++g

Vg

VVg

Vxxg

V22

56.22

56.21001.02

5.25.0}005.72

22

22

22

1

{ }1434.2243.52768.362.19

5.72

2 +++=V

( ) smV /6.416.21 21

2 ==∴

sec/1445.06.44

2.0 22

mxxQ =

=π

Additional Problems Flow through Pipes

1) Water flows upwards through a vertical pipeline. A mercury manometer connected between two points 10m apart shows a reading of 40cm of mercury when discharge is 450lpm. If the friction factor is 0.02. Determine the size of the pipe. 2) A town having a population of 1.2lakhs is to be supplied with water from a reservoir 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 pipes to 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 all the 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 carrying 150lps. If another pipe of the same material is to be laid in parallel to carry twice this discharge, what should be its diameter? Neglect minor losses. 5) Three pipes are connected in parallel between two points and the total discharge is 3 cumecs. If the pipes are of length 1200m, 1400m & 1600m diameter 1m, 0.8m & 1.2m respectively, and friction factor is the same for all the pipes, determine the discharge in each pipe and the pressure difference required to maintain the flow, assuming f=0.02. 6) A 450mm concrete pipe 1800m long connects two reservoirs whose difference in water level is 15m. What is the discharge? If another concrete pipe line 300mm diameter is introduced in parallel what would be the percentage increase in 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 length c)In the middle one – third of the length. Assume f=0.03 for all pipes and same difference in the reservoir levels. 7) A 450mm, concrete pipeline 200m long connects two reservoirs whose difference in water levels is 15m. What is the discharge? a)What is the percentage increase in discharge if another pipe line of the same diameter is introduced is parallel for the second half of the length? b)If a 30% increase in discharge is desired, what diameter pipe should be introduced in parallel for the second half of the length? Assume f=0.03 for all the pipes and the difference in reservoir levels same in the both the cases. Neglect minor losses. 8) Two pipes of 5cm diameter and 10cm diameter are connected in series. They have 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 is 10lps, what is the discharge through the 5cm pipe? 9) A pipe has D=40cm, L=10m, f=0.02. What is the length of an equivalent pipe which has D=20cm and f=0.02. 10) An 8cm diameter pipe carrying water has an abrupt expansion 12cm diameter 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.



P1 V1 P2

V2

flow

Area = a1 Area = a2

11) When a sudden contraction form 50cm diameter to 25cm is introduced in a horizontal pipe line the pressure changes from 105kps to 69kps. Assuming a coefficient of contraction of 0.65, calculate the flow rate. Following this contraction if there is a sudden enlargement to 50cm and if the pressure in the 25cm 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 in series. Calculate a) the size of a pipe of length 125m and f=0.020, equivalent to the pipe line ABC b) the length of an 8cm diameter (f=0.015) pipe equivalent to the pipe line ABC. 13) A horizontal pipe line carrying water at 0.03 m3/s reduces abruptly from 15cm to 10cm diameter. Taking contraction coefficient CC=0.60. Determine the pressure loss across the contraction. How this pressure loss compares with the loss that would result if the flow direction is reversed? 14) Two pipes of diameter 40cm and 20cm are 300m each in length. When the pipes 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. What would be the loss of head in the system to pass the same total discharge when the pipes are connected in parallel, take f=0.03 for each pipe. (P.S: FOR ANSWERS TO THE ABOVE PROBLEMS, DOWNLOAD THE CONTENTS OF SESSION-9, VTU.AC.IN (E-LEARNING). Fig.13

1 2



V1

V

entry

V

Fitting collar

V

entry

Fig 14.(a) Fig 14.(b) Fig 14. (c) Fig 15

V2

exit

exit

Q



V1 V2

flow

V1 V2

flow

Z1+p1/ Z2+p2/

Fig 17 Fig 16

1

1

2

2

1

1

2

2

P2



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 characteristics of the orifice plus the properties of the fluid are known, then the orifice can be used to measure the flow rates.

Classification of orifices Based 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 orifice Based on shape of the u/s edge :Sharp edge, Bell mouth Based on flow: Free, Submerged Flow through an orifice As the fluid passes through the orifice under a head H, the stream lines converge and therefore 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 the area 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 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 BC as datum and neglecting losses (hL)

Lhg

VpZg

VpZ +++=++22

222

2

211

1 γγ

;21 ZZ = ,1 Hp=

γ

VVV == 21 ,0

02

00002

+++=++∴g

VH

)1(2 −−−= gHorV Theoretical velocity Velocity V in Eq(1) is known as TORRICELLI’S VELOCITY. Hydraulic Coefficients of an orifice



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

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 0.99 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 (ac) to the area of the orifice (a).

=∴

aaC c

C

Value of Cc will be generally more than 0.62. Relationship between the Hydraulic Coefficients of an orifice From continuity equation Actual discharge Qact = ac x Vact Theoretical discharge Qth = a x Vth

th

actc

th

act

VVx

aa

QQ

=∴

or ccd xCCC = Equation for energy loss through an orifice Applying Bernoulli’s equation between the liquid surface (A) and the centre of jet and Vena Contracta (C) and considering losses (hL).

LCC

CAA

A hg

VpZg

VpZ +++=++22

22

γγ

,HZ A = )(0 atmospherepp BA == ,0=AV )(0 cityactualvelopp BA ==

Lhg

VaH +++=++∴2

00002

)2

(2

gVaHhL −=

gHCButV Va 2= Torricellis equation

)( 2VL HxCHh −=

)1( 2VL CHh −=

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



txVa = or,

aVxt =

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

2

2gty = or )2(2 21

−−−

=

gyt

Equating equations (1) & (2)

21

2

=

gy

Vx

a

But , gHCV Va 2=

21

22

=

gy

gHCx

V

21

21

21

21

21

21

22 y

gxHg

xCV =

HyxCV 2

=

or

=

yHxCV 4

2

Problems 1. The head of water over the centre of an orifice 30mm diameter is 1.5m. If the coefficient 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 Q

QC = thdact xQCQ = gHxaCd 2=

( )213

5.181.924

)1030(613.0 xxxxxx−

= π

smxQact /1035.2 33−= or lpsQact 35.2= 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.

60602411020

21 3

xxxxxQact = sm /1157.0 3=

we know gHaCQ dact 2=



181.924

615.01157.02

xxxxdx π=

mmmd 5.2322325.0 == 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 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=25x10-3H=1.5m, y=0.915m, x=2.288m Qact=102lpm = 102/60 = 1.7lps = 1.7x10-3m3/sec, Cd=?, Cc=?, hL=?

976.05.1915.04

288.24

22

===xxyH

xCV

( ) 638.05.181.921025

4107.123

3

=

==−

−

xxxxxxx

QQC

th

actd

π

999.0976.0638.0

=

==∴=

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 out of the orifice is collected in a circular tank 2m diameter. The time taken to collect 45cm of water 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 the orifice. Solution: D=100mm=0.1m, H=5m Qact = Area of collecting tankxheight of water collected / time

smxx /0471.03045.0

42 3

2

==π

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

98.05052.04

14

22

=

=

=

xxyHxCv

605.0581.921.0

40471.02

=

==xxx

xQQC

th

actd π

618.098.0605.0

===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 height of water above the orifice in the tank. Solution. X=0.4m, y=0.3m, V=? H=? Assume

We know yHxCV 4

2

=

==∴

=

2

2

2

2

22

103.044.0

4

4

xxyxGxH

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 on the jet 0.6m away from the Vena Contracta? Solution Head over the orifice H=(2-0.2)=1.8m CV=0.98, y=?, x=0.6m

mmmxx

y

xyxor

yHxCV

52052.098.08.14

6.0

8.146.0)98.0(,

4

2

2

22

2

==

=

=

=

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. Workout the pressure at which air should be pumped into the tank above water. Take Cd=0.6. Solution: (fig.19) Q=4lps = 4x10-3m3/s D=1.5x10-2m, Cd=0.6 PA=? 333 /10772.11/772.11 mkNxmNair

−==γ

Total head over the orifice

+=γ

AphH

gHaCQ dact 2=



( )

+= −

−−

3

223

10772.11281.92

4105.16.0104

xPxxxxxxx Aπ

)(/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 B containing 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. Total head over the orifice

( )

−+=

−

+=81.925153

γBA pphH

H=1.9806m

9806.181.92405.061.02

2

xxxxxgHaCQ dactπ

==

lpssmxQact 47.7/1047.7 33 == − 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 Cv for each orifice is 0.98, find the point of intersection of the two jets. Solution.

yHxCV 4

2

=

Given Cv is same for both the orifices

22

22

11

21

44 Hyx

Hyx

=

)(44 21

22

2

11

21 xx

Hyx

Hyx

−=

)1(5.164 2121 −−−==∴ yoryyy

from figure(21) ( )

)2(246

21

21

−−−+=−+=

yyyy

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

yyy

425.0

25.1

2

2

22

=∴=

+=



Again givesHy

xCV22

22

4=

mxxx

x

6.9644

98.0

2

22

=∴

=

(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 bottom and the other near the top. Show that the jets from these two orifices will intersect a plane through 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. Solution. To show that x1=x2 when H1=y2from figure(22) y1=[y2+(H2-H1)---(1)

22

22

11

21

1 44,

2 Hyx

HyxCCGiven VV

=∴=

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

222

12112 HyHHHHy =−+ 0)( 12221

21 =−+− HHyHHH

substituting

00

0)(; 222222221

=∴=−+−= yHyyHyyH

11. 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) the values 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

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

2

2LVah H

g

= −

2

0.2 22 9.81

Vax

= −

5.943 /aV m s= Coefficient of Velocity



5.943 0.9436.246

av

VCV

= = =

Coefficient of discharge d v CC C xC= 0.949 0.63x= 0.598dC =

(ii) Discharge through the orifice 2act dQ C a gH=

20.598 0.04 2 9.81 24

x x x x xπ= 34.707 10 / 4.71x m s lps−= =

(iii) Coefficient of velocity 2

4vxCyH

= 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 a maximum horizontal 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: fig(23) By definition, Velocity V=x/t

Vt∴= But

2V gH= and 212

y gt=

21( )2 2

xD H ggH

∴ − =

or

( )2 4x H D H= − or

4 ( )x H D H= − For x to be maximum

0dxdH

=

4( 2 ) 0D H− = / 2H D∴ =

We know, x=Vt, 12V gH=

22

12

y H gt+ =

2

1

12 2

xggH

=

( )21 24 (1)x H y H= + − − −

( )214 / 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 21 2

1 1 2 2

(1)4 4

x xy 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 21 2

4 1.6 0.4 4 0.4 1.6x x

x x x x∴ =

21

11 14VxCy H

=

210.98

4 1.6 0.4x

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 the vertical 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 the jet is collected in a tank 1.2m long x 0.6m wide and the water level rised from 0.15 to 0.75 in 7 minutes. Calculate the orifice coefficients. Solution: Dia of jet at vena contracta dc=20mm Dia of orifice = d = 25mm Head over the orifice H = 0.6m Depth of water collected in the measuring tank h=(0.75-0.15)=0.6m Depth of water collected in the measuring tank A = 1.2x0.6 = 0.72m2Time taken for collecting 0.6m of water t=7min=7x60=420sec Therefore actual discharge Qact = Area of measuring tank x depth of water collected / time taken

0.72 0.6. .420act

Ah xi e Qt

= =

3 31.0286 10 /x m s−=



2thQ a gH=( )2325 10

2 9.81 0.64

xx x x xπ

−

= 3 21.684 10 /thQ x m s−=

Coefficient of discharge act

dth

QCQ

=

3

3

1.0286 101.684 10

xx

−

−

=

0.61dC∴ =

Coefficient of contraction Cc = area of jet at vena contracta / area of orifice

2

2

44C

c

dC xd

ππ

=2 220 0.64

25c

dd

= = =

We know, Cd=CcxCV Therefore, Coefficient of velocity 0.610.64

dV

c

CCC

= =

0.953VC∴ = Coefficient of resistance

2

1C

V

CC

=

2

1 10.953

= −

0.1008rC =

15. A jet of water issuing from a vertical orifice in a tank under a constant head of 4m. If the depth of water in the tank is 12m, at what depth another orifice to be mounted vertically below the former one, so that both the jets meet at a common point on the horizontal at the bottom of the tank? Assume Cvto be the same for both the orifices = 0.98. Solution: fig(26)

2 21 2

1 21 1 2 2

,4 4V V

x xC Cy 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 21 2

1 1 2 24 4x xy H y H

=

1 1 2 24 4y H y H=

1 2x x=

1 1 2 2y H y H=

( )2 28 4 12x H H= −



22 212 32 0H H∴ − + =

2

212 12 4 1 32

2 1x 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 1m above 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 10 231.4824 60 60

x lpsx x

= =

Therefore Discharge per orifices 231.48 115.74

2Q lps= =

20.11574 /orQ m s= But,

2dQ C a gH= 0.11574

0.62 2 9.81 1a

x x x

=

20.04214m= 1

2 24 0.042144d xa dπ

π = ∴ =

=0.2316m Therefore, 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 constant Solution. Fig(27)

Head of the orifice H 10020.9 9.81x

= +

13.326H mofoil∴ = 2dQ C a gH=

( )30.65 60 10 2 9.81 13.3264

x x x x x xπ −= 30.02972 /m s= 29.72lps=



18. What is the discharge through a sharp edged slot 0.2 long x 10mm wide at 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

4VxCyH

=

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 containing water 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 3100

x 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 are 0.4m & 0.3m. Neglecting air resistances, determine the velocity of the jet and the height of water above the orifice in the tank. Assume CV=0.98. Solution. X=0.4m, y=0.3m, V=?, H=?

2

4VxCyH

= 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 Pieces A mouth piece is a short tube or pipe connected in extension with an orifice Classification of Mouth Pieces Depending 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 piece VO=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

11c

c

aV Va

∴ =

1

0.62cac c coefficientofcontractiona

= = =

11 (1)

0.62cV V∴ = −−−

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

( )2

12 11 0.62

2 2C

L

V VV Vh

g g

− − = =

21

0.375 (2)2Lh V

g= − − −

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

2 21 1

12 2A A

A Lp V p VZ Z h

g gγ γ+ + = + + +

11 0, , 0( )A

ppAZ Z H atmosphericpressureγ γ

= = = =

0( )AV Negligible= 2 2

1 10.3750 0 0 02 2V VH

g g∴ + + = + + +

21

1.3752

H Vg

=

or

121.375

gHV =

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

0.853 2. ,

2VgH

i 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 Bernoulli’s equation between (A) & © © with the centre line of the mouth piece as datum & neglecting losses.

22

2 2c cA A

A c Lp Vp VZ Z h

g gγ γ+ + = + + +

, 0, 0, 0, ?cA A C L

PpA H V Z Z hγ γ

= = = = = =

2

0 0 0 02

c cp VHgγ

∴ + + = + + +

2

2c cp VH

gγ∴ = −

But, 2

1 11.375 ,2 0.62CV VH V

g= =

12

0.853 21.375

gHV gH∴ = =

0.853 2&

0.62CgH

V =

20.853 2 1

0.62 2C gHp H x

gγ

= −

1.893Cp H Hγ

= −

0.893Cp Hγ

∴ = −

Negative sign indicates that the pressure at the Vena contracta is less than atmospheric pressure or the pressure is negative Problems 1. Find the discharge from a 80mm diameter external mouthpiece, fitted to a side of a large vessel if the head over the mouth piece is 6m. Solution.

( )2380 80 104

d mm a x xπ −= ∴ =

3 25.026 10x m−= 2dQ C a gH= For a cylindrical mouth piece Cd=0.853

30.853 5.026 10 2 9.81 6Q x x x x x−∴ =



30.04652 /m s= 46.52Q lps=

2. An external cylindrical mouthpiece of 100mm diameter is discharging water under a constant head of 8m. Determine the discharge and absolute pressure head of water at Vena contracta. Take Cd=0.855 and CC for Vena contracta =0.62. Take atmospheric pressure head =10.3m of water Solution. H=8m, Q=?, Cd=0.855, Cc=0.62

23 20.1100 0.1 7.854 10

4xd mm m x mπ −= = ∴ =

10.3ap mofwaterγ

=

32 0.855 7.854 10 2 9.81 8dQ C a gH x x x x x−= = 30.08413 / sec 84.13Q m lps= =

We know

0.893 ( 0.62)cc

p H whenCγ

= − =

0.893 8 7.144 ( )cp x mofwater Gaugeγ

= − = −

c c cp p pabsolute gaugeγ γ γ

∴ = −

(10.3 7.144)= − 3.156 ( )m Abs= 3. An external cylindrical mouth piece 60mm diameter fitted in the side of a tank discharges under a constant head of 3m, for which CV=0.82 Determine i) the discharge in lps ii) absolute pressure at Vena contracta iii) Maximum head for steady. Flow assuming that separation occurs at 2.5m of water absolute. Local barometer reads 760mm Hg. Solution. (i)Discharge(Q) At the exit of the mouth piece CC=1

1 0.82d c vC C xC x∴ = =

2dQ C a gH=

( )2360 100.82 2 9.81 3

4

x xx x x xπ −

=



30.0178 / 17.8m s lps= = (ii) Absolute Pressure head at Vena contracta Applying Bernoulli’s equation between (A) & © © with the centre line of the mouth piece as datum and neglecting losses hL

22

12 2C CA A

A Lp Vp VZ Z h

g gγ γ+ + = + + +

2

0 0} {0 02

c Cp VHgγ

+ + = + + +

2

(1)2

C Cp VHgγ

= + − − −

From Continuity equation Q=aV QVa

=

( )23

0.0178

0.62 60 104

CC

QVC xa x x xπ −

= =

10.154 /cV m s∴ = (1)FromEq∴ 2

2c cp VH

gγ∴ = −

210.15432 9.81x

= −

=2.255m of water (Gauge)

c atim c

Gauge

p p pγ γ γ

∴ = −

760atmp mmofmercuryγ

= 1 1 2 210.336 ( )mofwater S H S H= =

( )10.336 2.555 7.781( )c

abs

p Absoluteγ

∴ = − =

(iii) Hmax for steady flow Applying Bernoulli’s equation between Vena contracta and exit of the mouth piece with the centre line of the mouth piece as datum & considering head loss hL due to sudden expansion of flow.

2 21 1

12 2c C

C Lp V p VZ Z h

g gγ γ+ + = + + +

2

min 0.62, 0.3752C LVAssu gC h

g= =

( )2 2 2/ 0.62 0.3750 7.836 0 019 62 2 2

V V Vg g

− + = + + +−



2

7.836 1.22652V

g− = −

2 7.836 6.3892 1.226V mofwater

g = =

2VweknowV C gH= 2

2

12 V

VH xg C

=

max 2

16.3890.82

H x∴ = 9.5mofwater=

Alternatively, 0.82c ap p Hγ γ

= −

-7.781=0-0.82xHmax Hmax=9.49m of water

Submerged Orifice A fully submerged orifice is one in which the entire outlet side or the downstream side is completely under the liquid. It is also known as a drowned orifice. Consider points (1) and (2) situated upstream of orifice and at the Vena contracta respectively. H1=Height of water above the top of the orifice on the upstream side. H2=Height of water above the bottom level of the orifice. H=Difference in water level b=width of orifice Ca=Coefficient of discharge. Height of water above the centre of orifice on upstream side

2 1 1 21 2 2

H H H HH − − = + =

Height of water above the centre of the orifice on the downstream side

1 2

2H H H− = −

Applying Bernoulli’s equation between (1) and (2) with the horizontal passing through (A) & (B) as datum and neglecting losses. (hL)

{2 2

1 1 2 21 22 2 L

p V p VZ Z hg gγ γ

+ + = + + +

1 1 2 2 1 21 2 0, ;

2 2p H H p H HZ Z Hγ γ

+ + = = = = −

1 0V = (negligible)



21 2 1 2 20 0 0 0

2 2 2H H H H VH

g+ +

∴ + + = + − + +

22

2Vor H

g= 2 2V gH∴ =

From continuity equation Discharge Qact=Cdxarea of orifice x velocity

( )2 1 2act dQ C xb H H x gH∴ = − Large Rectangular Orifice

An orifice is said to be large when the head acting on it is five times the depth of the orifice. Unlike in the case of a small orifice, the discharge cannot be calculated from the equation for the reason that the velocity is not constant over the entire cross section of the jet. Consider an elementary horizontal strip of depth “dh” at a depth h below the free surface of the liquid as shown. Area of the strip Theoretical velocity through the strip 2gh=

Discharge through the elementary strip 2dQ bxdhx gh= Therefore through the entire orifice is obtained by integrating the above equation between the limits H1= and H2.

2

1

. , 2H

dH

i e Q C xbx ghdh= ∫

122

1

2H

dH

C b g h dh= ∫

3 32 2

2 1223dQ C b g x H H

= −

Problems 1. Find the discharge through a fully submerged orifice of width 2m if the difference of water levels on both the sides of the orifice be 800mm. The height of water from the top and bottom of the orifice are 2.5m and 3m respectively. Take Cd=0.6 Solution. For a submerged orifice.

( )2 1 2dQ C b H H x gH= − Where, Cd=0.6, b=2m, H2=3m, H1=2.5m, H=800mm = 0.8m

( )0.6 2 3 2.5 2 9.81 0.8Q x x x x x∴ = − 32.377 /Q m s= 2. Find the discharge through a totally drowned orifice 1.5m wide and 1m deep, if the difference of water levels on both the sides of the orifice is 2.5m, Take Cd=0.62.



Solution. b=1.5m, d=1m, H=2.5m, Cd=0.62

2dQ C a gH=

0.62 1.5 1 2 9.81 2.5x x x x x= 36.513 /Q m s=

3. Find the discharge through a rectangular orifice 3m wide and 2m deep fitted to a water tank. The water level in the tank is 4m above the top edge of the orifice. Take Cd=0.62. Solution.

For a rectangular orifice 3 32 2

2 12 23 dQ C b g H H

= −

where B=3m, Cd=0.62, H2=(4+2)=6m, H1=4m.

3 32 22 0.62 3 2 9.81 6 4

3Q x x x x x

= −

4. A rectangular orifice 1m wide and 1.5m deep is discharging water from a vessel. The top edge of the orifice is 0.8m below the water surface in the vessel. Calculate the discharge through the orifice if Cd=0.6. Also calculate the percentage error if the orifice is treated as small. Solution.

For a rectangular orifice. 3 32 2

2 12 23 dQ C b g H H

= −

3 32 22 0.6 1 2 9.81 2.3 0.8

3x x x x x

= −

3arg 4.912 /l eQ m s=

2small dQ C a gH=

1.50.6 1 1.5 2 9.81 0.82

x x x x x = +

34.963 / secsmallQ m=

arg

arg

% 100small l e

l e

Q QError x

Q−

=

4.963 4912 1004.912

x− =

% error=1.04 %



Flow Over Notches & Weirs A notch is an opening made in the side wall of a tank such that the liquid surface in the tank is below the upper edge of the opening. Generally notches are made of metallic plates and their use is limited to laboratory channels. A weir is a masonry/concrete structures built across an open channel so as to rise the water level on the upstream jside and to allow the excess water to flow over the entire length onto the downstream side. Classification a) Depending on shape: i) Rectangular ii) Triangular iii) Trapezoidal b) Depending on the shape of the crest i) Sharp crested ii) Broad crested. c) Depending on flow i) Free ii) Submerged d) Depending on Ventilation i) Fully aerated ii) Depressed iii) Clinging or Drowned. Definition sketch of a Notch.(fig.31) a)Flow over a Triangular Notch. Figure(32) shows the details of flow over a V – notch. 2θ=Central angle H=Head over the notch Consider an elemental strip of thickness “dh” at a depth “h” below the free surface as shown. Discharge through the strip dq=area x velocity

2 2dq xdhx gh= Discharge over the entire notch

0 0

2 2Q H

dq xdhx gh=∫ ∫

12

0

2 2H

Q g xh dh= ∫



In the above equation x can be eliminated.Interms of θ & H. In otherwords

tan( )

xH h

θ =−

or

( ) tanx H h θ= −

( ) ( )12

0

2 2 tanH

Q g H h h dhθ∴ = −∫

1 32 2

0

2 2 tanH

g Hh h dhθ

= − ∫

integrating 3 52 2

0

2 22 2 tan3 5

H

Q g Hh hθ

= −

∫

52 2 22 2 tan

3 5g Hθ = −

( ) 52

5 34 2 tan

15g Hθ

−=

528 2 tan (1)

15Q g H θ= − − −

528 2 tan

15act dQ C g H θ∴ =

dC =Coefficient of discharge In deriving Eq(1), velocity of approach Va is neglected. If the head due to this velocity of approach is considered,

then, ( )5522

8 2 tan (3)15act d aQ C g H h ha θ

= + − − − −

Problems 1. A triangular notch discharges 0.0110m3/s under a head of 0.2m. Find the angle of the notch, if Cd=0.626.

Solution. 30.0113 / , 0.2

0.626d

Q m s H mC= ==

528 2 tan

15 dQ C g H θ=

( )52

15 0.0110tan8 0.626 2 9.81 0.2

xx x x

θ =

022.6θ = Therefore angle of the notch 02 45.2θ= = 2. A right angled triangular notch discharges 0.143m3/s. Find the head over the notch if Cd=0.6.



Solution.

0 0

3

2 90 , 450.143 / , ?0.6d

Q m s HC

θ θ= =

= ==

528 2 tan

15 dQ C g H θ=

0

15 0.1438 0.6 19.62 tan 45

H xx x

=

0.3995 399.5H mm∴ = = 3. 150lpm of water is expected to flow down an irrigation furrow. Design the weir, if a minimum head of 100mm is desired. Assume Cd=0.61.

Solution: 3 3150150 2.5 2.5 10 /60

Q lpm lps x m s−= = = =

100 0.1 , 0.61dH mm m C= = = 528 2 tan

15 dQ C g H θ=

( )

3

52

15 2.5 10tan8 0.61 19.62 0.1

xxx x

θ−

=

00.5486 28.75θ= ∴ = 02 57.5 60centralangleθ∴ = = ≈

Hence, it is suggested to use a 600 V – notch. 4. Calculate the top width and depth of a triangular notch capable of discharging 700lps. The weir discharges 5.7 lps when the head over the crest is 7.5cm. Take Cd=0.62. Solution. 3700 0.7 / , ?Q lps m s H= = =

3 31 15.7 10 / , 0.075 , 0.62dQ x m s H m C−= = =

528 2 tan

15 dQ C g H θ= 5 52 2

1

8 152 tan 2 tan15 8d d

Q C g H x C g HQ

θ θ∴ =

52

1 0.075Q HQ

=

52

3

0.75.7 10 0.075

Hx −

=

0.514H m∴ =

Also,



528 2 tan

15 dQ C g H θ=

or

( ) ( )1 52 2

15 0.7tan8 0.62 2 9.81 0.514

xx x x

θ =

0 0

2.52368.38 2 136.76orθ θ

=

∴ = =

Top width of the notch = 02 0.514 tan 68.38

2.594x x

metres==

Flow over a Rectangular Notch Figure(33) shows the details of the flow over a rectangular notch. L=length of the notch H=head over the notch Consider a small strip of thickness dh at a depth h below the liquid surface Discharge through the strip dq=areaxvelocity

2dq Lxdhx gh=

Total discharge 12

0 0

2Q H

dq L g h dh=∫ ∫

32

0

223

H

Q L g h= ∫

322 2

3Q L g H∴ =

322 2 (1)

3act dQ C g LH= − − −

Cd=Coefficient of discharge, its average value is about 0.62. End Contraction When the length of the weir(L) is less than the width of the channel (B), the nappe contracts at the sides, and this is knows as end contractions.(fig34) According to Francis, the effective length of flow over the notch is given by Substituting this value in EQ(1) and simplifying

( )2 32 0.1 (2)3 2dQ C g L nH H= − − − −

A notch without end contraction is known as a suppressed notch. Velocity of approach (Va)



The total head over the weir will be the sum of static head (H) and velocity head

(ha), velocity head 2

2a

aVh

a= is due to the velocity of the liquid approaching the

notch. On similar lines, considering a strip of uniform thickness dh at a depth h below the liquid surface. Discharge through the strip dq=area x velocity. ( )2 adQ Lxdhx g H h= + Therefore Total discharge is given by

( )12

0 0

2Q H

adq L g H h dh= +∫ ∫

( )3322

223

L g H ha ha

= + −

( )3322

2 23 d aQ C g L H h ha

= + −

Empirical Formula

(i)Francis Formula ( ) ( )33221.84 0.1Q L nH H ha ha

= − + −

(ii)(ii)(ii)Bazin’s formula 320.0030.405 2Q L g H

H = +

(iii)(iii)(iii)Rehbock formula ( ) ( )32

0.053 0.0110.403 2 0.0011

HQ L g H

Z +

= + +

Considering velocity of approach and End contraction, we have

( ) ( )3322

2 2 0.13 d aQ C g L nH H h ha

= − + −

Considering velocity of approach and End contraction, we have

( ) ( )3322

2 2 0.13 d aQ C g L nH H h ha

= − + −

5.Find the discharge over a rectangular notch of crest length 400mm. When the head of water over the crest is 50mm. Take Cd=0.6.

Solution. 322 2

3 dQ C g LH=

( )32

2 0.6 2 9.81 0.4 0.053

x x x x x= 3 37.92 10 / 7.92Q x m s lps−= =



6. A rectangular weir 9m long is divided into 3 bays by two vertical post each 300mm wide. If the head of water over the weir is 500mm, Calculate the discharge, given Cd=0.62.

Solution. ( )322 2 0.1

3 dQ C g L nH H= −

n=number of end contractions=6. L=clear length of weir=9-3x0.3=8.1

( )322 0.62 2 9.81 8.1 0.1 6 0.5 0.5

3Q x x x x x x∴ = −

35.05 /Q m s= 7. The discharge over a rectangular weir is 0.4m3/s when the head of water is 0.20m. What would be the discharge if the head is increase to 0.3m? Solution. Q1=0.4m3/s H1=0.20m Q2=? H2=0.3m

32

2 2

1 1

Q HQ H

=

32 3

20.3, 0.4 0.735 /

0.20or Q x m s = =

8. A rectangular channel 6m wide carries a flow of 1.5m3/s. A rectangular sharp crested weir is to be installed near the end of the channel to create a depth of 1m upstream of the weir. Calculate the necessary height . Assume Cd=0.62. Solution: fig(35) L=6m, Q=1.5m3/s, Cd=0.62. Y=(Z+H)=1m Velocity of approach Va= Discharge/area of flow in the channel

1.5 0.25 /1 6aV m sx

∴ = =

Head due to velocity of approach 2

2aV hag=

230.25 3.185 10

2 9.81ah x mx

−∴ = =

But,

( )3322

2 2 ( )3 d aQ C g L H h ha

= + −



( ) ( )3 3

3 32 221.5 0.62 2 9.81 6 3.185 10 3.185 103

x x x x x H x x− − = + −

0.266H m∴ = , And height of the weir

( )1 0.2660.734Z m

= −

=

9. A rectangular sharp crested weir is required to discharge 2.645m3/s of wate under a head of 1.2m. If the coefficient of discharge is 0.6 and the velocity of approach near the weir is 1m/s. Find the length of the weir. Solution. Q=2.645m3/s H=1.2m Cd=0.6 Va=1m/s L=?


2a

aVh

g=

And height of the weir ( )Z y H= −

( )1 0.2660.734Z m

= −

=

10. A rectangular sharp crested weir is required to discharge 2.645m3/s of water under a head of 1.2m. If the coefficient of discharge is 0.6 and the velocity of approach near the weir is 1m/s. Find the length of the weir. Solution. Q=2.645m3/s H=1.2m Cd=0.6 Va=1m/s L=?


2a

aVh

g=

21 0.050972 9.81ah m

x∴ = =

21 0.05097

2 9.81ah mx

∴ = =

( )3322

2 2 ( )3 d aQ C g L H h ha

= + −

( ) ( )

−+= 2

323

05097.005097.02.181.926.032645.2 xLxxxx



( )3322

2 2 ( )3 d aQ C g L H h ha

= + −

( ) ( )3 32 2

22.645 0.6 2 9.81 1.2 0.05097 0.050973

x x x xLx = + −

1.076 107.6L m cm∴ = = 11. Water passing over a rectangular notch flow subsequently over a right angled triangular notch. The length of the rectangular notch is 600mm and Cd=0.62. if the Cd value for the V-notch is 0.60, what will be its washing head, when the head on the rectangular notch is 20cm. Solution. Rectangular Notch. L=600mm=0.6m Cd=0.62, H=0.2m

322 2

3 dQ C g LH=

( )32

2 0.62 2 9.81 0.6 0.23

x x x x x= 30.09825 /Q m s= 0 02 90 , 45 , ?, 0.6dH Cθ θ= = = =

Since the same discharge of 0.09825m3/s is passing over the V-notch, we have

528 2 tan

15 dQ C g H θ=

( )52

80.09825 0.6 2 9.81 tan15

x x x x H θ=

52 0.0693.

0.3438H

H m=

∴ =

Types of Nappe



The equations derived for the discharge over notches were under the assumption that pressure under the nappe is atmosphere. However, when the liquid if flowing over the notch (suppressed), it touches the walls of the channel and the air gets dissolved or entrained in water, continuation of this process results in a negative pressure i.e. partial vaccum under the nappe. Finally the nappe gets deflected closer to the weir wall. The pressure on the inner side of the the nappe decides its type in the following ways. a)Free Nappe In this type, the stream of water passing over the weir the springs clear of the weir. (fig.36a) b) Depressed NappeIn this type, a partial vaccum is created between the nappe and the weir. Discharge for such a flow situation is 8 to 10% greater than that with a free nappe.(fig.36b) c) Clinging NappeIn this type the nappe totally adheres to the face of the weir. The discharge in this case would be 20 to 30% more than that in a fully aerated nappe).(fig36c) Ventilation of Weirs The nappe emerging from a weir should be of a correct form, so that the discharge equations derived for them are valid. For accurate gauging of flow the nappe should spring clear or it should be free. In other words the space between the weir and nappe should be maintained under atmospheric conditions, particularly when the weir is suppressed. In practice ventilation holes are made on the weir walls so that air circulates freely between the weir and the nappe. This is known as ventilation or ventilation of weirs.(fig36d). Submerged Weir A weir is said to be submerged when water level on both the upstream and downstream sides are above the crest level of the weir as shown in figure (37). H1 and H2 are the heads over the weir on the upstream and downstream sides. In the case of submerged weir, it is necessary to derive the discharge equation considering that the flow over the weir is a combination of a free weir and a submerged orifice. In other words, the flow Q1 between H1 and H2 is considered as a free weir and Q2 between H2 and the weir crest as a submerged orifice.

For a free weir )1()(232 2

3

2111 −−−−= HHgLCdQ

For a submerged orifice. )2()(2 21221 −−−−= HHgLHCdQ )3(21 −−−+=∴ QQQ

Cd1=0.58 and Cd2 = 0.80 are usually considered for the weir and the orifice. As in the earlier cases the head due to velocity of approach ha=Va2/2g can also be considered.



In such a case )4()(232 2

323

2111 −−−−+−= hahaHHgLCdQ

)5()(2 21221 −−−+−= haHHgLHCdQ In all the above equation L=length of the notch or weir. Problems 1) A submerged weir 1m high spans the entire width of a rectangular channel 7m wide. Estimate the discharge when the depth of water is 1.8m on the upstream side and 1.25m on the downstream side of the weir. Assume Cd=0.62 for the weir. Solution (fig.38). Qsubmerged weir= Qweir+Qsubmerged orifice

( ) ( )212211 2232

32 HHgLHCHHxgLCd d −+−=

( ) ( )25.08.062.1925.0762.02325.08.062.19762.0

32

−+= xxxxxxxxx

smQ /25.6 3= 2) The upstream and downstream water surfaces are 150mm and 75m above the crest of a drowned weir. If the length of the weir is 2.5m, find the discharge, the coefficients of discharge for the free and drowned portions may be taken as 0.58 and 0.8 respectively. Allow for velocity of approach. Solution. H1=1500mm=15m, H2=75mm=0.075m L=2.5m, Cd1=0.58, Cd2=0.8

( ) ( )

−+−= 21222

3

211 2232 HHgLHCdHHgxLxxCdQ

( )

−+− )075.015.0(62.19075.05.28.0075.015.062.195.258.0

32

23

xxxxx

smQ /2698.0 3= Velocity of approach

softheweiruareaofflowQVa /

=

smx

/719.015.05.2

2698.0=

=

mxg

Vh aa 00264.0

81.92719.0

2

22

==

( )

+−+

+−=

0264.0)075.015.0(62.19075.05.28.0

0264.0075.015.062.195.258.032

23

xxx

xxQ

smQ /271.0 3=



OGEE WEIR When the weir is suppressed and its height is large, the nappe emerging out may be subjected to the problems of ventilation. Hence, in such cases the weir profile downstream is constructed conforming to the shape of the lower side of the nappe. Such a weir is known as a spillway or ogee weir. The cross section of an ogee weir will be shown(39). The coordinates of the spillway profile can be worked out for the head H using the equation.

yHx 85.085.1 2= The u/s face of the spillway is generally kept vertical.

The discharge equation for an ogee weir will be 23

232 LHgCQ d=

Same as that for a suppressed rectangular notch. Problem 1. Calculate the discharge over an ogee weir of 8.5m length, when the head over the crest is 2.15m and Cd=0.61. Solution. L=8.5m, H=2.15m, Cd0.61

23

232 LHgCQ d=

23

)15.2(5.881.9261.032 xxxxx= smQ /27.48 3=

Broad Crested Weir A weir is said to broad crested when its width (parallel to flow) b is greater than 0.5xmaximum head acting on it (fig.40).

xHbei 5.0,. > Let L=length of the weir H=Head of water u/s of the weir w.r.t. the crest h=Depth of water over the weir crest V=Vel. Of flow over the weir Applying Bernoulli’s equation between (1) and (2) with the crest of the weir as datum & neglecting losses (hL)

gVhH

gVpZ

gVpZ

2000

2{}

22

222

2

211

1

++=++

++=++γγ

)1()(2 −−−−=∴ hHgV Discharge over the weir Q=area of flow over the weir x vel. of flow over the weir. i.e. )(2)( hHgxhxLQ −=

Actual discharge ( ) )2(2 −−−−= hHgLhCQ dact From Equation (2), we see that Qact is a function of h for a given value of H.



Q∴ is maximum when 0=∂dhQ

( ) 021

=

−

∂∂

∴ hHhh

( ) ( ) 021

21

=

∂∂

−+−∂

hhhHhH

dhhx

( ) ( ) ( ) 0121 1

211

21

=

−+−− − xhHRHhx

0)()(2

21

21 =

−+−

− hHhH

h

( ) HhhHh 232 =∴−=

max)(32 orQconditionfHorh −=

Substituting the value of h in eq(2) and simplifying.

( )

−

= HHgxHLCQ dact 3

2232

max

32

32 HgLHxCd=

( ) 23

max 233

2 HgLCQ dact =

Problems 1. Determine the discharge over a broad crested weir 26m long, the upstream level of water is measured as 0.5m above the crest level. The height of the weir is 0.6m and the width of the approach channel is 36m. Take Cd=0.9. Solution.

For a broad crested weir. ( ) 23

max 233

2 HgLCQ dact =

23

)5.0(62.19269.033

2 xxxx=

( ) smQact /105.14 3max =

Since, the width of the channel, we have to consider the velocity of approach Va

=

hechannelAreaofflowQVei a int

,.

smx

/356.0)5.06.0(36

105.14=

+



322

10466.681.92

356.02

−=== xxg

Vh aa

( ) ( )23

max 233

2 haHgLCQ dact +=

( )23

310466.65.062.19269.033

2 −+= xxxxx

( ) smQact /379.14 3max =

2. A reservoir discharges water at 60,000 m3/day over a broad crested weir, the head of length of the weir, if Cd=0.65. Solution. Q=60,000m3/day=60,000/24x60x60=0.694m3/s H=500mm=0.5m Cd=?, L=?

( ) 23

max 233

2 HgLCQ dact =

=∴23

)5.0(62.1965.02

694.033

xxx

xL

metresL 77.1= 3. A channel of 45m2 cross sectional area, discharging 50 cumecs of water is to be provided with a broad crested weir. If the crest of the weir is 1.6m below the upstream water surface, find the length of the weir, if Cd=0.85. Solution.

( ) ( )23

max 233

2 haHgLCQ dact +=

gx

AQ

gVwhereha a

21

2

22

==

mx 629.062.19

14550 2

=

=

( )23

0629.06.162.1985.033

250 +=∴ xLxx

L1076.350 = or mmL 1.16089.16 ≈=



B

A

Air

y

C

R x

Fig 18 Fig 19.

H

h=2m

Vena contracta

Centreline of orifice



Y2

H1=4m O1

O2

H1

Y1

X1=X2

Y2

Fig 21 Fig 22

X1 = X2

Y 1

H2=6m

H2 Point of intersection of the jet



H

X

Y

V

H3 y

H1

H2

X1=X2=X3

Y2

Y1

Fig 23 Fig 24

D



H1=0.6m

2m Y1

H2=1.6m

Y2

X1 X2

Y=12m H2=?

H1=4m

Y1

Y2

X1=X2

Fig 25 Fig 26

X=?



100 KN/m2

Air

2m

Oil s=0.9

Fig 27 Fig 28

H

VENA CONTRACTA

MOUTH PIECE



H2 H1

H

1 2

H2 H H1

h

b

d dh

H Flow Q

Velocity of approach Va

Channel Bed

Dead water Z

Notch

Apex or crest

Nappe

Fig 29 Fig 30 Fig 31



H

ha

dh

h x

θ θ

h

dh H

L

Fig 32 Fig 33

Elemental Strip



B L Flow

Notch Side wall

y

H

Z

Nappe

Nappe

Fig 34 Fig 35 Fig 36. (a)



Fig 36. (b) Fig 36 (c)

flow

Partially Ventilated



Nappe

Ventilation holes

Q1

Q2 H2

H1

Weir

1.8m

H1=0.8m

1m

H2=0.25m

1.25m

Fig 36.(d) Fig 37

Fig 38



H

v h flow

Channel bed

Fig 39 Fig 40

Crest of ogee weir

Top surface of nappe

Concrete or masonry

Solid boundary coinciding with the bottom surface of the nappe

weir



(Additional Problems) 1.Find the diameter of a sharp edged orifice sufficient to discharge 3x106 litres of water per day. Under a constant head of 12m. Take Cd=0.62 2.2.A closed tank partially with water upto a height of 2m has an orifice of 2.75cm diameter at its bottom. Determine the air pressure over the water surface for a discharge of 5 lps, through the orifice, Cd=0.62 3. Water is discharged vertically upwards through a 5cm diameter orifice under a pressure of 180N/cm2. CV for the orifice is 0.92. Determine the height to which the jet will rise, when i) there is no air friction ii) when the friction produces a retardation of 2m/s2. 4. In performing an experiment to determine the different coefficients of a sharp edged orifice, a jet of water issuing horizontally from the orifice of 2.5cm diameter under a constant head of 1.5m fell through 0.9m vertically and struck the ground at 2.3m horizontally from Vena contracta. The time required to discharge 91litres of water was found to be 53 seconds. Find the values of Cd, CV and CC for the orifice. 5. A Jet of water issuing from a vertical orifice 0.025m diameter under a constant head of 1.5m falls 0.915m vertically, before it strikes the ground at a distance of 2.288m, measured from Vena contracta horizontally. The discharge was found to be 0.102 m3/min. Determine Cd, CV, CC, & Cr for the orifice. 6. The centre of an orifice is situated 60cm above the bottom of a vessel containing water to a depth of 2.4m. Assuming CV=0.98. Estimate how far the water in the jet issuing from the orifice in the side of the vessel will fall in moving horizontally a distance of 1m from the Vena contracta. 7. A large tank has a circular sharp edged orifice 2.5cm diameter in a vertical side at a depth 0.6m below the constant water level. The jet diameter at Vena contract is found to be 2cm. The water is found to be 2cm . The water discharged is collected in a vessel having the inside dimensions of 1.2m x 0.6m x 0.6m. The time required to fill the vessel is 7minutes. Calculate Cd, CV & CC. 8. Oil of S=0.85 issues from a 5cm diameter orifice under a pressure of 12N/cm2. The diameter of jet at Vena contracta is 4cm and the discharge is 20 lps. What is the coefficient of velocity? 9. A rectangular orifice 0.9m wide, 1.2m deep is discharging water from a vessel. The top edge of the orifice is 0.6m below the water surface in the vessel. Calculate the discharge through the orifice if Cd=0.6 and the percentage error if the orifice is treated as a small orifice. 10. A closed tank contains kerosene S=0.8 to a depth of 3m. The top portion of the tank contains air under a pressure of 27.5kpa. If a sharp edged circular orifice of 3cm diameter Cd=0.62 is provided at the bottom of the tank, estimate the discharge through the orifice. 11. A standard mouthpiece 5cm in diameter discharges water. Determine the maximum head under which the flow would take place without the occurrence of separation at Vena contracta. The coefficient of discharge for the mouth piece is 0.8, separation occurs when the absolute pressure is 2.44m of water.



12. Find the discharge through an external cylindrical mouth piece of 10cm in diameter flowing under a head of 3.5m. What is the pressure at Vena contracta? Take CC=0.62. 13. Calculate the value of V - notch angel to discharge 400lps under a head of 0.6m, assuming Cd=0.60. 14. During a test in a laboratory the water passing through a venturimeter is made to flow over a 900V notch. Diameter at inlet is 25cm an that at the throat is 10cm. The pressure difference is 0.34m when the head over the V-notch is steady at 18.2cm. If Cd for the venturimeter is 0.97, what is the coefficient of discharge of the V-Notch? 15. Discharge varies in a rectangular channel from 0.0057 to 0.142 m3/s. At what height should a 900 V-notch be placed above the bed of the channel, if the depth should not exceed 1.2m, in the channel? What will be its depth for minimum discharge? 16. A 900 V-notch is provided in a rectangular channel 1.5m wide in order to measure the channel discharge the channel is designed to carry a maximum discharge of 0.4m3/s. With a depth not exceeding 1.8m. Find the position of the apex of the notch from the bed of the channel. 17. A sharp crested rectangular weir 1.5m long and 90cm high is installed in a rectangular channel 1.5m wide. If the head on the weir is 30cm, find the discharge a)Neglecting velocity of approach b)Considering velocity of approach 18.A cipolletti weir of 40cm bottom width is Installed in a channel 75cm wide & 45cm deep. If the head over the weir crest is 25cm, find the discharge over the weir. a)Neglecting vel. of approach b)Taking vel. of approach. 19. A rectangular channel is 4.5m wide. Water flows at a depth of 1.2m at a velocity of 90cm/s. A sharp crested weir is constructed across the channel and the depth in the channel rises upto 1.75m. What should be the height of the weir? 20.Two 900 V – notches & one cipolletti weir are to be used side by side to measure a discharge of 0.85m3/s through a channel. If the head should not exceed 30cm, what should be the dimensions of the weirs? 21. Find the discharge through a trapezoidal notch which is 1.2m wide at the top and 0.5m at the bottom and is 40cm in height. The head of water on the notch is 30cm. Cd for the rectangular portion is 0.62 while for triangular portion is 0.60. 22. A discharge of 1500m3/s is to pass over a rectangular weir. The weir is divided into a number of openings each of span 7.5m. If the velocity of approach is 3m/s, find the number of openings needed in order the head of water over the crest is not to exceed 1.8m.



23. A rectangular channel 1.5m wide has a discharge of 200lps, which is measured by a right angled V – notch. Find the position of the apex of the notch from the bed of the channel if the maximum depth of water is not to exceed 1m. Take Cd=0.62. 24. During an experiment in a laboratory 0.9 litres of water is flowing per second over a V-notch when the head is 5cm and 5 litres/sec, when the head is 10cm. Determine the values of k & n in the equation Q=KHn. 25. A triangular weir has one side sloping at 450 and the other side Z(H):1V. Calculate the values of Z which gives a discharge of 0.12m3/s under a head of 0.3m. Cd=0.6 26. Water flows through a 900 V-notch with a free surface of water at a height of 10cm from the crest. The corresponding flow rate is 1000lpm. If the height above the crest is increased by 50%, what would be the new flow rate? 27. A rectangular channel 6m wide carries a flow of 1.5m3/s. A rectangular sharp crested weir is to be installed near the end of the channel to create a depth of 1m. Upstream of the weir, calculate the necessary weir height. Assume Cd=0.62. 28. A suppressed weir having a crest length of 4m, discharges under a head of 40cm. The height of the weir crest above the channel bed is 1m. Find the rate of discharge. 29. A rectangular sharp crested weir is required to discharge 645 m3/s of water under a head of 1.2m. If the coefficient of discharge is 0.6 and the velocity of approach near the weir is 1 m/s, find the length of the weir required. 30. A discharge of 1800m3/s is to pass over a rectangular weir which is divided into a number of spans of 10m each. If the velocity of approach is 4m/s, find the number of opening required such that the head over the crest does not exceed 2m. 31. A waste weir has to pass a flood discharge of 3m3/s. find the length of the broad crested weir to suit the discharge such that the depth of water over the weir crest will not be more than 42cm. Take Cd=0.97. 32. A 3m wide rectangular irrigation canal carries water with a discharge of 6m3/s. what height of rectangular weir installed across the canal will raise the water surface to a level of 2m above the floor? 33. A rectangular weir 0.75m high & 1.5m long is to be used for discharging water from a tank under a head of 0.5m. Estimate the discharge a) when it is used as suppressed weir. b) when it is used as a contracted weir. Take Cd=0.61. 34. A trapezoidal sharp crested weir has a base width of 1.2m & side slopes of 1.5H to 1V. Calculate the discharge over the weir for a head of 35cm. Take Cd=0.62. 28. Calculate the bottom width and side slopes of a trapezoidal notch to discharge 4m3/s at a head of 1.5m and 1.2m3/s at a head of 0.75. Assume Cd=0.63.35. A Cipolletti weir of crest length 0.6m discharges water under of 0.36m. Find the discharge over the weir if the channel is 0.8m wide & 0.5m deep. Take Cd=0.6.



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FLOW THROUGH PIPES - · PDF fileIn 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)