NPTEL PROBLEMS

NPTEL

PROBLEMS

Basic equations

1. The following are the data given of a change in diameter, effected in laying a water supply pipe line. The change in diameter is gradual from 200 mm at A to 500 mm at B. Pressure at A and B are 78.5 kN/m2 and 58.9 kn/m2 respectively with the end B being 3 m higher than A. If the glow in the pipe line is 200 lps, find:

a) direction of flow and b) the head lost in friction between A and B. (10)

Solution:

Q = 0.2 m3/s;D A = 0.2 m; D B = 0.5 m; g = 10 m/s2 (assumed)

p A = 78.5 kN/m2; p B = 58.9 kN/m2; Z B – Z A = Z = 3 m; h f = ?

22 m0314.04

== AA DA π and 22 m1965.04

== BB DA π

From discharge continuity equation, we get

m/s3662.6vA ==AA

Q and m/s0186.1vB ==BA

Q

From modified Bernoulli’s equation applied between A and B, we have

LBBB

AAA hZ

gVpZ

gVp

+++=++22

22

γγ

Lh++×

+××

=×

+×× 3

1020186.1

1010109.58

1023662.6

1010105.78 2

3

32

3

3

h L = 0.9345 m

The flow is always from higher pressure to lower pressure and hence from A to B.

2. Water flows up a conical pipe 450 mm diameter at the lower end and 250 mm diameter at 2.3 m above the lower end. If the pressure and velocity at the lower end are 63 kN/m2 (gauge) and 4.1 m/s,

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assuming a head loss in the pipe to be 10% of the pressure head at the lower end, calculate the discharge through the pipe. Also calculate the pressure and velocity at the upper end (8)

Solution

D A = 0.45 m; D B = 0.25 m; Z B – Z A = Z = 2.3 m; V A = 4.1 m/s;

p A = 63 kN/m2 (gauge); h L = 10%(p A /γ); Q = ?; p B = ?; V B = ?

ρ = 1000 kg/m3, g = 10 m/s2 (assumed)

From modified Bernoulli’s equation applied between A and B, we have

LBBB

AAA hZ

gVpZ

gVp

+++=++22

22

γγ

22 m1590.04

== AA DA π and 22 m0491.04

== BB DA π

From discharge continuity equation, we have Q = A A V A = A B V B

m/s284.13vv AB ==B

B

AA

m63.01010

10631.0 3

3

=×

××=Lh

Substituting in the modified Bernoulli’s equation, we get

63.03.2102

284.131010102

1.410101063 2

3

2

3

3

++×

+×

=×

+×× Bp

Simplifying, we get p B = -102.83 kN/m2 (gauge)

3. A pipe 400 mm diameter carries water at a velocity of 2.5 m/s. The pressure head at pints A and B are given as 30 m and 23 m respectively, while the datum head at A and B are 28 m and 30 m respectively. Find the loss of head between A and B.

Solution:

D = 0.4 m; V = 2.5 m/s; =g

pA

ρh A = 30 m; =

gpB

ρh B = 23 m; Z A = 28 m;

Z B = 30 m; h L = ?

As the pipe is of uniform diameter V A = V B = 2.5 m/s

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1

2

( ) ( ) mgg

hL 16.02

2535.02

vv35.0 2221 =

−=

−=

Applying modified Bernoulli’s equation between A and B

LBBB

AAA hZ

gVpZ

gVp

+++=++22

22

γγ

Lhg

Vg

V+++=++ 30

22328

230

22

h L = 5 m

4. A conical tube of length 2 m is fixed vertically with its smaller end upwards. The velocity of flow at the smaller end is 5 m/s while at the lower end it is 2 m/s. The pressure head at the smaller end is 2.5

m/s of liquid. The loss of head in the tube is ( )g2vv35.0

221 − , where v 1 is the velocity at the smaller end

and v 2 is the velocity at the lower end respectively. Determine the pressure head at the lower end. Flow takes place in the downward direction.

Solution:

L = 2 m, v 1 = 5 m/s, v 2 = 2 m/s, g = 10 m/s2 =g

pρ

1 2.5 m, ?2 =g

pρ

Let the smaller end be represented as 1 and lower end as 2 as shown in Fig.

Applying modified Bernoulli’s equation between 1 and 2

=g

pρ

2 5.39 m

A horizontal venturimeter with inlet diameter 200 mm diameter and throat diameter 100 mm is used to measure the flow of oil of specific gravity 0.8. The discharge of oil through venturimeter is 60 lps. Find the reading of the oil-mercury differential manometer. Take C d = 0.98. (08)

Solution:

d 1 = 0.2 m; d 2 = 0.1 m; Q = 0.06 m3/s; C d = 0.98; S m = 13.6 (assumed)

LhZg

VpZg

Vp+++=++ 2

222

1

211

22 γγ

16.00222

255.2

22

2

+++=++g

pg γ

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x

2

1

x

x = ?

22211 m 03142.02.0

44=×==

ππ da

22222 m 007854.01.0

44=×==

ππ da

ghaa

aaCQ d 222

21

21 ×−

=

h×××−

×= 102

007854.003142.0007854.003142.098.006.0

22

h = 2.849 m of oil

But

mm81.17

18.06.13849.21

=∴

−=⇒

−=

x

xSS

xho

m

6. A venturimeter is to be fit in a 200 mm diameter horizontal pipe line. The inlet pressure is 100 kPa. If the maximum flow of oil (s=0.85) is 200 Lps, calculate the least diameter of the throat, so that the pressure does not fall below 250 mm mercury (vacuum). Assume that 3% of the differential head is lost between the inlet and the throat. (08)

Solution: D 1 = 0.2 m; p 1 = 100 kPa; Q = 0.2 m3/s; s = 0.85; p 2 ≥ - 0.25 m Hg;

h L = 3% h;Assume ρ = 1000 kg/m3; g = 10 m/s2.

Pressure head at inlet = p 1 / ρg = 100 x 103/(0.85 x 1000 x 10) = 11.765 m of liquid

Pressure head at outlet = p 2 /ρg = - 0.25 m of Hg = - 0.25 x 13.6/0.85 = - 4 m of liquid

Differential head = 11.765 – (-4) = 15.765 m of liquid, h L = 3% of 15.765 = 0.473 m of liquid.

985.0765.15

473.0765.15=

−=

−

=hhhC L

d

The discharge through the venturimeter is given by

ghaa

aaCQ d 222

21

21 ×−

=

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22211 m 03142.02.0

44=×==

ππ da

765.1510203142.003142.0985.02.0

22

22 ×××

−=

aa

Solving the above equation, we get =d 2 = 12.13 mm.

Note: Due to mistake in DATA, the result is not compatible (may be in discharge).

7. A vertical venturimeter carries a liquid of relative density 0.8 and has inlet and throat diameters of 150 mm and 75 mm respectively. The pressure connection at the throat is 150 mm above that at the inlet. If the actual rate of flow is 40 Lps and the C d = 0.96, calculate the pressure difference between the inlet and throat in kN/m2.

Solution:

S o = 0.8; Q = 40 x 10-3 m3/s; d 1 = 0.15 m; d 2 = 0.075 m; Z 2 –Z 1 = 0.15 m

C d = 0.96; p 1 – p 2 = ?

22211 m 01767.015.0

44=×==

ππ da

22222 m 004418.0075.0

44=×==

ππ da

The discharge through the venturimeter is given by

ghaa

aaCQ d 222

21

21 ×−

=

h×××−

×= 102

004418.001767.0004418.001767.096.004.0

22

h = 4.17 m

But for a vertical venturimeter,

+−

+= 2

21

1 zgS

pzgS

phoo ρρ

x

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( )

15.017.4 21

1221

−

−=

−−

−=

gSpp

zzgSpph

o

o

ρ

ρ

( )15.017.421 +=− gSpp o ρ

p 1 - p 2 = 34.56 kN/m2 (Ans).

The differential manometer reading x

m26.0

18.06.1317.41

=∴

−=⇒

−=

x

xSS

xho

m

8. Water flows upward a vertical 300 mm x 500 mm venturimeter with a C d = 0.98. The deflection of manometer, filled with a liquid of S = 1.25 is 1.18 m. Determine the discharge if the distance between the two pressure tapings is 457 mm. Work the problem from the first principles.

Solution:

d 1 = 0.5 m; d 2 = 0.3 m; C d = 0.98;

S m = 1.25; x = 1.18 m;

Z 2 – Z 1 = Z = 0.457 m; Q = ?

98.0=

−

=hhhC L

d

Squaring both sides, we get

C d2 h = h – h L and

h L = h (1 – C d2) = 0.0396 h

where

( )2121 ZZ

gpph −+

−=

ρ

Also

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m295.0

10.125.118.11

=∴

−⇒

−=

hSS

xho

m

From Modified Bernoulli’s equation between sections 1 and 2, we get

LhZg

pZg

p+++=++ 2

222

1

211

2gv

2gv

ρρ

( ) LhZZgpp

+−+−

=−

2121

21

22

2gvv

ρ … (01)

But from discharge continuity equation, we have

Q = a 1 v 1 = a 2 v 2 or d 12v 1 = d 2

2v 2

Substituting in Eq. 01, we get hhdd 0396.01

2gv

4

1

222 +=

−

295.00396.15.03.01

2gv 42

2 ×=

− or v 2 = 2.655 m/s

Q = a 2 v 2 = (π/4)d 22 v 2 = 0.1876 m3/s (Ans)

A 45o degree bend is connected in a pipe line, the diameters at the inlet and outlet of the bend being 600 mm and 300 mm respectively. Find the force exerted by water on the bend if intensity of pressure at inlet to bend is 88.29 kPa and rate of flow of water is 600 lps. (Jan/Feb 2006)

Solution: θ = 450, D 1 = 0.6 m, D 2 = 0.3 m p 1 = 88.29 kPa, Q = 0.6 m3/s Assume g = 10 m/s2 and ρ = 1000 kg/m3

2827.04

6.04

221

1 =×

==ππ DA m2

07068.04

3.04

222

2 =×

==ππ DA m2

From discharge continuity equation, we have Q = A V

122.21

1 ==AQV m/s and 488.8

22 ==

AQV m/s

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Applying Bernoulli’s equation between Sections 1 and 2, we get

2

222

1

211

2gv

2gv Z

gpZ

gp

++=++ρρ

But Z 1 = Z 2 Substituting and solving for p 2 = 54.5 kPa F x = ρ Q [V 1 – V 2 cos θ ] + p 1 A 1 – p 2 A 2 cos θ = 19911.4 N F y = ρ Q [ – V 2 sin θ ] – p 2 A 2 sin θ = - 6322.2 N

208912.63224.19911 2222 =+=+= yx FFF N

Acting at o

x

y

FF

616.174.199112.6322tantan 11 =

=

= −−α with horizontal.

July/Aug 2005

12.Water flows up a reducing bend of weight 80 kN place in a vertical plane. For the bend, the inlet diameter is 2 m, outlet diameter is 1.3 m, angle of deflection is 120

o and vertical height (distance between the inlet and the outlet) is 3 m. If the discharge is 8.5 m3/s, pressure at the inlet is 280 kPa and the head loss is half the kinetic head at the exit, determine the force on the bend. (12)

Solution:

W = Weight of the reducing bend acting downwards = 80 kN (↓), d 1 = 2 m,

d 2 = 1.3 m, θ = 120o, Z = 3 m, Q = 2.5 m3/s, p 1 = 280 kPa. h L = 0.5g

V2

22

Assume g = 10 m/s2, ρ = 1000 kg/m3, F x = ? and F y = ?

142.342

4

221

1 =×

==ππ dA m2 and 327.1

43.1

4

222

2 =×

==ππ dA m2/s

Applying discharge continuity equation we have Q = A 1 V 1 = A 2 V 2

Applying Modified Bernoulli’s equation between the two sections of the bend shown in Fig. we get

1

2

45o

V2

p1A1

p2A2

α

Fx

Fy

F

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LhZg

pZg

p+++=++ 2

222

1

211

2gv

2gv

ρρ

Substituting the values, we get

1028835.15.03

1028835.1

1010000

1020.7958

10100010280 22

223

××++

×+

×=+

×+

×× p

p 2 = 247.656 kPa

Forces acting on the bend in x and y direction respectively are F x = ρQ[V 1 –V 2 cos θ ]+p 1 A 1 – p 2 A 2 cos θ = 1,048,423.63N F y =–W+ρQ[–V 2 sin θ] –p 2 A 2 sin θ = – 288,688.06+80,000= – 368,688.06N

871,111,360.368688.0663.1048423 2222 =+=+= yx FFF N

Acting at o

x

y

FF

375.1963.104842306.288688tantan 11 =

=

= −−α with horizontal.

July 2006

In a 45o bend, a rectangular air duct of 1.0 m2 cross-sectional area is gradually reduced to 0.5 m2. Find the magnitude of the force required to hold the duct in position, if the velocity of flow is 20.0 m/s at 1 m2 cross-section and the pressure at both sections is 40 kN/m2. Specific weight of air is 11.0 N/m3. (10)

Solution:

θ = 450, A 1 = 1.0 m2, A 2 = 0.5m2, γ = 11.0 N/m3 p 1 = p 2 = 40.0 kPa, v 1 = 20.0 m/s, Assume g = 10 m/s2

ρ = γ/g = 1.10 kg/m3 From discharge continuity equation, we have Q = A 1 V 1 = A 2 V 2 20 = 0.5V 2

α

Fx

Fy

F

1

2

45o

V2

p1A1

p2A2

V1

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Hence V 2 = 40 m/s and Q = 20 m3/s Forces acting on the bend in x and y direction respectively are F x = ρ Q [V 1 – V 2 cos θ ] + p 1 A 1 – p 2 A 2 cos θ = 25,675.61N F y = ρ Q [ – V 2 sin θ ] – p 2 A 2 sin θ = − 14,764.39 N

22yx FFF += = 29617.97 N

Acting at o

x

y

FF

9.29tan 1 =

= −α with horizontal

) 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 = 0.0046. Considering the discharge through the pipe

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

2 2

entry loss friction loss exit loss2 2

A A B BA B

P V p VZ Zg gγ γ

+ + = + + + + +

gV

gDfLV

gV

gV

2225.0

200005.32

2222

+++++=++

2 4 0.0046 200032.5 1 0.5 12 0.25V

g× × = + + +

2.06m/sV =

4

2DQ π=

230.25 2.06 0.101m / s

4π × × =

lps101=Q

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

exit entry

Q

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Solution: D 1 =0.15m, D 2 = 0.25m, Q = 225 lps = 225 m

3/s

Head loss due to sudden expansion is

gDQ

DQhL 2

1442

22

21

×

−=

ππ

2

22

21

2

2 11216

−=

DDgQπ

2

222

2

25.01

15.01

81.92225.016

−

×××

=π

3.385mLh =

3) The rate of flow of water through a horizontal pipe is 350lps. The diameter of the pipe 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 = 350 lps = 0.35 m3/s, D 1 = 0.2 m, D 2 = 0.5 m, P 1 = 15 N/cm2 h L = ?, p 2 = ?, P = ?

From continuity equation m/s14.112.035.044

221

1 =××

==ππD

QV and m/s78.15.035.044

222

2 =××

==ππD

QV

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( ) ( ) waterofm463.481.92

78.114.112

221 =

×−

=−

=gVVhL

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 h L only.

( )horizontal pipe021 == ZZ

463.462.19

78.181.9

062.1914.11

81.91500

22

2

+++=++p

222 N/cm67.16kN/m68.166 ==p

Power lost LhQP γ= 463.435.081.9 ××=

kW32.15=P

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

)2(m108,Given 311

22 −−−×=

+−

+ −

γγpZpZ

1

1

2

2

V1 V2

flow P2 P1

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

Lhg

VpZg

VpZ +++=++22

222

2

211

1 γγ0

2

21

221

12

2 =

+−

+

+−

+ Lh

gVVpZpZ

γγ

From continuity equation 2

2

1

2

415.0

41.0 VV ×

×=×

× ππ

21 25.2 VV =

( ) ( ) 081.92

25.281.92

25.21082

222

22

23 =

×−

+×

−+×∴ − VVVV

01274.0108 22

3 =−× − V or m/s25.01274.0108 2

13

2 =

×=

−V

Discharge 25.04

15.04

2

2

22 ×

×==

ππ VDQ /sm10428.4 33−×= or lps425.4=Q

5) Two reservoirs are connected by a pipe line which is 125 mm diameter for the first 10 m and 200 mm in diameter for the remaining 25 m. The entrance and exit are sharp and the change of section is sudden. The water surface in the upper reservoir is 7.5 m 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 VV ×

=× ππ or 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

1

1

2

V1

V2 flow

Z1+p1/γ Z2+p2/γ

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2 2 entry friction sudden exit

2 2 loss loss expansion loss lossA A B B

A Bp V p VZ Z

g gγ γ

+ + = + + + + + +

( )22 2 2 21 21 1 1 2 2 2

1 2

0.5 47.5 0 0 0 0 02 2 2 2

V VV L V L V Vfg g d d g g

− + + = + + + + + + +

( ) ( ) ( ) ( )2 2 2 2 22 2 2 2 2 20.5 2.56 10 2.56 25 2.564 0.0017.5

2 2 0.125 0.2 2 2V V V V V V

g g g g

× × −× = + + + +

( )2

2

2

7.5 3.2768 2.597 2.4336 119.62

3.976m/s

V

V

= + + +

=

230.2 4.6 0.125 m /s

4Q π × = × =

. 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

m, H=1.5m, C d =0.613

;th

actd Q

QC =

thdact QCQ ×=

gHaCd 2×=

( )123(30 10 )0.613 2 9.81 1.5

4π

−×= × × × × ×

3 32.35 10 m /sactQ −= ×

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. Q total = 20x103 m3/day C d =0.615.

sm /1157.0 3=

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3 3act

1 120 10 0.1157m /s2 24 60 60

Q = × × × =× ×

act 2dQ C a gH=

20.1157 0.615 2 9.81 1

4

0.2326m=232.6mm

d

d

π= × × × ×

=

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=25x10-3 H=1.5m, y=0.915m, x=2.288m

Q act =102 lpm = 102/60 = 1.7 lps = 1.7x10-3 m

3/s, C d =?, C c =?, h L =?

976.05.1915.04

288.24

22

=××

==yHxCV

( ) 638.05.181.921025

4107.123

3

=

×××

××==

−

−

πth

actd Q

QC

or

=

yHxCV 4

2

999.0976.0638.0

=

==∴×=

v

dCVCd C

CCCCC

( )2head loss 1L vh H C= −

( )2976.015.1 −=

0.0712m 71.2mmLh = =

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 30

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seconds. 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, x=100cm = 1m, y=5.2cm = 0.052m,

Cd=?, Cv=?, Cc=?

Area of collecting tank height of watercollected timeactQ ×

=

605.0581.921.0

40471.02

=

×××

×==

πth

actd Q

QC

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=? C v =1 (assume)

We know 2 1 2 9.81 1.33 5.115m/svV C gH= = × × × =

yHxCV 4

2

=

H = 1.33m

98.05052.04

14

22

=

××

=

=

yHxCv

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

××

=×

=∴

=×

2

2

2

2

22

103.044.0

4

4

v

v

CyxH

xCyH

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C V =0.98, y=?, x=0.6m

2

22

2

2

4

0.6or, (0.98)4 1.8

0.6 0.052m 52mm4 1.8 0.98

VxCyH

y

y

=

=× ×

= = = × ×

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 4 lps. Workout the pressure at which air should be pumped into the tank above water. Take C d =0.6.

Solution

Q=4 lps = 4x10-3 m3/s, D=1.5x10-2 m, C d =0.6, P A =?

3 3 311.772 N/m 11.772 10 kN/mairγ −= = ×

Total head over the orifice=

+=

γAphH

gHaCQ dact 2=

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 = 5x10-2 m C d =0.61.

Total head over the orifice ( )15 25

3 1.9806m9.81

A Bp pH hγ

− − = + = + =

( )223

3

1.5 104 10 0.6 2 9.81 2 or

4 11.772 10APx

π−

−−

× × = × +

20.83 kN/m (Gauge)AP =

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20.052 0.61 2 9.81 1.9806 7.47 lps4act dQ C a gH π ×

= = × × =

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 2 1 24 6 1.5y y or y y∴ = = 01

from figure ( )1 2 26 4 2y y y= + − = + 02

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

2 2

2

1.5 24m

y yy

= +∴ =

Again gives4 22

22

HyxCV =

22

2

0.984 4 6

9.6m

x

x

=× ×

∴ =

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.

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

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Solution.

To show that x 1 =x 2 when H 1 =y 2 f rom figure

y 1 =[y 2 +(H 2 -H 1 )] ---(1)

Given 2 2 9.81 2 6.264 m/sV gH= = × × =

( )

( )

( ) ( )

( )

1 1 2 2

2 2 1 1 2 2

22 1 1 2 1 2 2

1 2

22 2 2 2 2 2

2 2 2 2 2 2

2 2

2 2

4 4

;

0

0

0

y H y H

y H H H y H

y H H H H y H

H y

y H y y H y

y y H y y H

y H

y H

=

+ − =

+ − =

=

− + − =

− − − =

− =

=

11. A 40 mm dia orifice in the vertical side of a tank discharges water. The water surface in the tank is at a constant level of 2 m above the centre of the orifice. If the head loss in the orifice is 0.2 m 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

−=

gVaHhL 2

2

2

0.2 22 9.81

Va = − ×

2 2 9.81 2 6.264 m/sV gH= = × × =

Coefficient of Velocity yHxCv 4

2

= or 5.943m/saV =

943.0246.6943.5

===VVC a

v

Coefficient of discharge

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(ii) Discharge through the orifice gHaCQ dact 2=

(iii) Coefficient of velocity

224 xyHCV =

24 0.5 2 (0.949) 1.898mx∴ = × × =

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:

By definition, Velocity xVt

=

tVx =∴

But gHV 2=

and 2

21 gty =

Cvd CCC ×= 63.0949.0 ×= 0.598=

x

Jet

Orifice

H

h

y

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( )HDHx −= 42

or 4 ( )x H D H= −

For x to be maximum ( )24 0d DH HdH

− =

22 2

1 gtHy =+

We know, x=V t, 12gHV =

2

1221

=

gHxg

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 C V = 0.98 for each orifices. x = ?

Solution:

Assuming the coefficient of velocity CV to be the same for both the orifices, we have (C V1 ) = (C V2 )

2

221)(

=−∴

gHxgHD

2 21 2

1 1 2 2

(1)4 4

x xy H y H

= − − −

4( 2 ) 0D H− = 2/DH =∴

X=?

H1=0.6m

2m Y1

H2=1.6

Y2

X1 X2

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where

2 22 2 1.6 0.4my H= − = − =

1 20.4m, 1.6mH H= =

2 21 2

4 1.6 0.4 4 0.4 1.6x x

∴ =× × × ×

210.98

4 1.6 0.4x

∴ =× ×

or

[ ]1 2 2 1.568 3.136mx x x∴ = + = × =

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 = 0.72m2

Time taken for collecting 0.6m of water t = 7min = 7x60=420 s

Therefore actual discharge

Area of collecting tank height of watercollected timeactQ ×

=

0.72 0.6. .420act

Ahi e Qt

× = =

3 31.0286 10 m /s−= ×

( )2325 102 9.81 0.6

4π

−×= × × × × 3 31.684 10 m /s−= ×

1 12 2 0.4 1.6y H m= − = − =

21

11 14VxCy H

=

1 21.568x m x= =

2thQ a gH=

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Coefficient of discharge

3

31.0286 101.684 10

−

−

×= ×

Coefficient of contraction

Area of jet at veena contractaarea of orificecC =

2

24

4Cc

dCd

ππ

= ×

We know, Cd=Cc x C V Therefore, Coefficient of velocity

Coefficient of resistance 2

1 1CV

CC

= −

or

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:

actd

th

QCQ

=

0.61dC∴ =

2 220 0.6425c

dd

= = =

0.610.64

dV

c

CCC

= =

0.953VC∴ =

21 1

0.953 = −

0.1008rC =

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y=12 m

H2=?

H1=4m

y1

y2

x1=x2

From figure, (y 1 -y 2 )=(H 2 -H 1 )

Equating the values of CV

8m or 4m=

H 2 = 8m is the correct answer.

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

2 21 2

1 21 1 2 2

,4 4V V

x xC Cy H y H

= =

( ) ( )1 1 12 4 8y y H m= − = − =

( ) ( )2 2 212y y H H m= − = −

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+ ± −

∴ =

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100 KN/m2

Air

2m

Oil s=0.9

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 10 231.48lps24 60 60

×= =

× ×

Therefore Discharge per orifices 231.48 115.74 lps2

Q = = 2or 0.11574m /sQ =

But, 0.115740.62 2 9.81 1

a =

× × ×

20.04214 m=

12 24 0.04214

4da 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.

Head of the orifice

H 10020.9 9.81

= + ×

13.326m of oilH∴ =

( )30.65 60 10 2 9.81 13.3264π −= × × × × × × 30.02972 m /s= 29.72 lps=

2dQ C a gH=

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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

22 0.6(0.98)

4 1.8y

= × ×

0.052m 52 mmy∴ = =

19. A vertical orifice is fitted 0.2m above the bottom of a tank containing water to a depth of 2m. If C V = 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

Q = × × × × ×

3 39.36 10 m /s−= ×

9.36 lpsQ∴ =

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 C V = 0.98.

Solution.

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

20.40.98

4 0.3 H=

× ×

2 2 9.81 0.1388 1.65 m/sV gH= = × × =

Mouth Pieces

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

2

4VxCyH

=

2

4VxCyH

=

0.1388H m∴ =

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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

It is a short pipe whose length is two or three times the diameter.

H=Head over the centre of the mouth piece

V o =Velocity of the liquid at Vena Contracta © ©

a c =Area of flow at Vena Contracta

V 1 =Velocity of liquid at outlet

a 1 =Area of mouth piece at outlet.

C c =coefficient of contraction

Applying continuity equation between CC & (1) &(1)

a c c c =a 1 v 1

1

coefficient of contraction 0.62cc

a ca

= = =

11c

c

aV Va

∴ =

11 (1)

0.62cV V∴ = − − −

H

Vena Contracta

Mouth Piece

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

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.

11 0, , 0 (atmospheric pressure)A

Ap pZ Z Hγ γ

= = = =

0(Negligible)AV =

or

By definition, Coefficient of velocity

C V =Actual velocity/Theoretical velocity

At the exit of the mouth piece C C = 1

1 0.853 0.853d c vC C C∴ = × = × =

( )2

12 11 0.62

2 2C

L

V VV Vh

g g

− − = =

21

0.375 (2)2Lh V

g= − − −

2 21 1

12 2A A

A Lp V p VZ Z h

g gγ γ+ + = + + +

2 21 10.3750 0 0 0

2 2V VH

g g∴ + + = + + +

21

1.3752

H Vg

=

121.375

gHV =

1 0.853 2 (3)V gH∴ = − − −

0.853 2. ,

2VgH

i e CgH

=

0.853VC∴ =

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Hence, for an external cylindrical mouth piece C d (=0.853) is more than that of an orifice.

Pressure head at Vena contracta

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

But,

20.853 2 1

0.62 2C gHp H

gγ

= − ×

Negative sign indicates that the pressure at the Vena contracta is less than atmospheric pressure or the pressure is negative

Flow measurements

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∴ = −

21 11.375 ,

2 0.62CV VH V

g= =

12

0.853 21.375

gHV gH∴ = =

0.853 2&

0.62CgH

V =

1.893Cp H Hγ

= −

0.893Cp Hγ

∴ = −

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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.

( )2380mm 80 104

d a π −= ∴ = × × 3 25.026 10 m−= ×

For a cylindrical mouth piece C d =0.853

30.853 5.026 10 2 9.81 6Q −∴ = × × × × × 30.04652m /s= 46.52lps=

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 C d =

0.855 and C C for Vena contracta =0.62. Take atmospheric pressure head =10.3m of water

Solution.

H=8m, Q=?, C d =0.855, C c =0.62

23 20.1100mm 0.1m 7.854 10 m

4d π −×

= = ∴ = ×

10.3m of waterapγ

=

32 0.855 7.854 10 2 9.81 8dQ C a gH −= = × × × × ×

30.08413m /s 84.13 lpsQ = =

We know

0.893 (when 0.62)cc

p H Cγ

= − =

0.893 8 7.144m of water (Gauge)cpγ

= − × = −

absolute gaugec c cp p pγ γ γ

∴ = −

3.156m(Abs)=

2dQ C a gH=

(10.3 7.144)= −

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3. An external cylindrical mouth piece 60mm diameter fitted in the side of a tank discharges under a constant head of 3m, for which C V =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 C C =1

1 0.82d c vC C C∴ = × = ×

( )2360 100.82 2 9.81 3

4π − × × = × × × ×

30.0178 m /s 17.8lps= =

(ii) Absolute Pressure head at Vena contracta

Applying Bernoulli’s equation between (A) & CC with the centre line of the mouth piece as datum and neglecting losses h L

From Continuity equation Q = aV

( )23

0.0178

0.62 60 104

CC

QVC a π −

= = × × × ×

10.154m/scV∴ = From Eq (1)∴

2dQ C a gH=

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

= + − − −

QVa

=

2

2c cp VH

gγ∴ = −

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210.15432 9.81

= − ×

=2.255m of water (Gauge)

760mm of mercuryatmpγ

= 1 1 2 210.336m of water ( )S H S H= =

( )10.336 2.555 7.781(Absolute)c

abs

pγ

∴ = − =

(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 7.836 6.389m of water2 1.226V

g = =

we know 2VV C gH=

2

21

2 V

VHg C

= ×

max 216.389

0.82H∴ = × 9.5m of water=

Alternatively,

-7.781=0-0.82xHmax

c atim c

Gauge

p p pγ γ γ

∴ = −

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− = −

0.82c ap p Hγ γ

= −

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Hmax=9.49m of water

4. 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 gH= − ×

Where, Cd=0.6, b=2m, H 2 =3m, H 1 =2.5m, H=800mm = 0.8m

( )0.6 2 3 2.5 2 9.81 0.8Q∴ = × × − × × × 32.377 m /sQ =

5. 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, C d =0.62

0.62 1.5 1 2 9.81 2.5= × × × × ×

36.513m /sQ =

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

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

3 32 22 0.62 3 2 9.81 6 4

3Q

= × × × × × −

6. 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.

2dQ C a gH=

3 32 2

2 12 23 dQ C b g H H

= −

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Solution.

For a rectangular orifice.

3 32 22 0.6 1 2 9.81 2.3 0.8

3

= × × × × × −

3arg 4.912m /sl eQ =

1.50.6 1 1.5 2 9.81 0.82

= × × × × × +

or 34.963m /ssmallQ =

arg

arg

%Error 100small l e

l e

Q QQ

−= ×

4.963 4912 1004.912

− = ×

% error=1.04 %

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

Solution.

( )52

15 0.0110tan8 0.626 2 9.81 0.2

θ = × × × ×

Therefore angle of the notch

8. A right angled triangular notch discharges 0.143m3/s. Find the head over the notch if Cd=0.6.

3 32 2

2 12 23 dQ C b g H H

= −

2small dQ C a gH=

30.0113 / , 0.20.626d

Q m s H mC

= ==

528 2 tan

15 dQ C g H θ=

022.6θ =

02 45.2θ= =

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Solution.

0 0

3

2 90 , 450.143 m /s, ?0.6d

Q HC

θ θ= =

= ==

0.3995 399.5mmH∴ = =

9. 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 3150150lpm 2.5lps 2.5 10 m /s60

Q x −= = = =

100mm 0.1m, 0.61dH C= = =

( )

3

52

15 2.5 10tan8 0.61 19.62 0.1

θ−

× = × × ×

02 central angle 57.5 60θ∴ = = ≈

Hence, it is suggested to use a 600 V – notch.

10 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.

5 52 2

1

8 152 tan 2 tan15 8d d

Q C g H C g HQ

θ θ∴ = ×

528 2 tan

15 dQ C g H θ=

0

15 0.1438 0.6 19.62 tan 45

H xx x

=

528 2 tan

15 dQ C g H θ=

00.5486 28.75θ= ∴ =

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=

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0.514mH∴ =

Also,

or

( ) ( )1 52 2

15 0.7tan 2.5238 0.62 2 9.81 0.514

θ = × = × × ×

0 068.38 2 136.76orθ θ∴ = =

Top width of the notch 02 0.514 tan 68.38 2.594 m= × × =

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.

( )32

2 0.6 2 9.81 0.4 0.053

= × × × × ×

3 37.92 10 m /s 7.92lpsQ −= × =

11. 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.

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∴ = × × × − × × × or 35.05m /sQ =

52

1 0.075Q HQ

=

52

30.7

5.7 10 0.075H

x − =

528 2 tan

15 dQ C g H θ=

322 2

3 dQ C g LH=

( )322 2 0.1

3 dQ C g L nH H= −

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12. 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 3

20.3or, 0.4 0.735 m /s

0.20Q = × =

13. 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:

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 m/s1 6aVx

∴ = =

32

2 2

1 1

Q HQ H

=

y

H

Z

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Head due to velocity of approach

230.25 3.185 10 m

2 9.81ah −∴ = = ××

But,

( ) ( )3 3

3 32 221.5 0.62 2 9.81 6 3.185 10 3.185 103

H − − = × × × × × + × − ×

0.266 mH∴ = , And height of the weir 0.734Z m=

14. A rectangular sharp crested weir is required to discharge 2.645 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 1m/s. Find the length of the weir.

Solution.

Q=2.645m3/s H=1.2m

Cd=0.6 Va=1m/s L=?


And height of the weir ( ) ( )1 0.266Z y H= − = −

0.734 mZ =

2

2aV hag

=

( )3322

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

= + −

2

2a

aVh

g=

21 0.05097m2 9.81ah∴ = =

×

( )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 0.05097 1 0.2663

x L = × × × × + − = −

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15. 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=?


21 0.05097 m2 9.81ah∴ = =

×

( ) ( )3 32 2

22.645 0.6 2 9.81 1.2 0.05097 0.050973

x L = × × × × + −

1.076m 107.6 cmL∴ = =

16. 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

( )32

2 0.62 2 9.81 0.6 0.23

= × × × × × or

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

2

2a

aVh

g=

( )3322

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

= + −

30.09825 /Q m s=

0 02 90 , 45 , ?, 0.6dH Cθ θ= = = =

322 2

3 dQ C g LH=

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( )52

80.09825 0.6 2 9.81 tan15

H θ= × × × ×

17. 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.

Qsubmerged weir= Qweir+Qsubmerged orifice

( ) ( )212211 2232

32 HHgLHCHHxgLCd d −+−=

( ) ( )2 30.62 7 19.62 0.8 0.25 0.62 7 0.25 19.62 0.8 0.253 2

= × × × × × + × × × × −

smQ /25.6 3=

18. 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.

H 1 =1500mm=15m, H 2 =75mm=0.075m

L=2.5m, C d 1=0.58, C d 2=0.8

( ) ( )32

1 1 2 2 2 1 22 2 23 d dQ C L g H H C LH g H H

= × × × − + −

( )32

2 0.58 2.5 19.62 0.15 0.075 0.8 2.5 0.075 19.62 (0.15 0.075)3

× × − + × × × −

528 2 tan

15 dQ C g H θ=

52 0.0693.

0.3438H

H m=

∴ =

1.8m

H1=0.8m

1m

H2=0.25m

1.25m

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smQ /2698.0 3=

Velocity of approach area of flow/s of the weira

QV =

0.2698 0.719 m/s2.5x0.15

= =

2 20.719 0.00264 m2 2 9.81

aa

Vhg

= =×

( )32

2 0.58 2.5 19.62 0.15 0.075 0.026430.8 2.5 0.075 19.62 (0.15 0.075) 0.0264

Q × × − + = + × × × − +

smQ /271.0 3=

19. 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=

322 0.61 2 9.81 8.5 (2.15)

3= × × × × ×

348.27 m /sQ =

20. 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 =

322 0.9 26 19.62 (0.5)

3 3= × × × ×

( ) 3act max 14.105m /sQ =

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

i.e,Area of flow in the channela

QV =

14.105 0.356m/s36 (0.6 0.5)

= × +

322

10466.681.92

356.02

−=== xxg

Vh aa

( ) ( )23

max 233

2 haHgLCQ dact +=

( )32

32 0.9 26 19.62 0.5 6.466 103 3

−= × × × × + ×

( ) 3max 14.379m /sactQ =

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

1.77mL =

21. 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 +=

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22 1where 2 2

aa

V Qhg A g

= = ×

250 1 0.629m45 19.62

= × =

( )32250 0.85 19.62 1.6 0.0629

3 3L∴ = × × × +

L1076.350 =

or 16.089m 16.1mL = ≈

Water Hammer

22. A hydraulic pipeline 3 km long, 500 mm diameter is used is used to convey water with a velocity of 1.5 m/s. Determine the pressure growth of the valve provided at the outflow end is closed in (i) 20 s (ii) 3.5 s. Consider pipe to be rigid and take bulk modulous of elasticity of water as K water = 20 x 108 N/m2 (10)

Solution:

L = 3000 m; d=0.5m;V=1.5m/s; t 1 = 20 s; t 2 = 3.5 s; K=20x108 N/m2; ρ = 1000 kg/m3 (Assumed)

Critical time 2 LTC

= , where Celerity 92 10 1414.2m/s

1000KCρ

×= = =

Hence 2 3000 4.24 s1414.2

T ×= =

Case (i)

t 1 > T, Hence the valve closure is gradual.

Instantaneous rise in pressure is given by

1000 3000 1.5 225 kPa20i

LVpt

ρ × ×= = = (Ans)

Case (ii)

t 2 < T, Hence the valve closure is Sudden with pipe rigid.


1000 1.5 1414.2 2.1213 MPaip VCρ= = × × = (Ans)

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23 Water flowing with a velocity of 1.5 m/s in a rigid pipe of diameter 500 mm is suddenly brought to rest. Find the instantaneous rise in pressure if bulk modulous of water is 1.962 GPa. (04)

Solution:

V = 1.5 m/s; K = 1.962 GPa; ρ = 1000 kg/m3 (Assumed)

Celerity 91.962 10 1400.7m/s

1000KCρ

×= = =


1000 1.5 1400.7 2.1 MPaip VCρ= = × × = (Ans)

3. A steel penstock of 1000 mm diameter has a thickness of 20 mm. Water is flowing initially with a velocity of 2.0 m/s. Flow velocity is brought to rest by closing a valve at the end of the pipeline. Bulk modulous of water is 2 x 109 N/m2 and elastic modulous of pipe material is 2 x 1011 N/m2. If the length of the pipe is 2000 m, find the pressure rise in terms of head of water when:

(i) Water is compressible and pipe is elastic (08)

(ii) Water is compressible and pipe is rigid (02)

Solution:

V = 2.0m/s; d=1m; t=20x10-3m; K=2.0GPa; E=200GPa; 1 14m

= and ρ = 1000 kg/m3 (Assumed)

Case (i) The valve closure is Sudden with pipe Elastic.

3 9

10002.0 2.309 MPa1 1 1 1

2 20 10 200 10

ip VD

K t E

ρ

−

= = = + + × ×

(Ans)

Instantaneous rise in Pressure head =

62.309 10 230.9 m of water1000 10

ipgρ

×= =

× (Ans)

Case (ii) The valve closure is Sudden with pipe rigid.

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Celerity 92 10 1414.2m/s

1000KCρ

×= = =


1000 2.0 1414.2 2.828 MPaip VCρ= = × × =

Instantaneous rise in Pressure head=H i

=62.828 10 282.8 m of water

1000 10ipgρ

×= =

× (Ans)

24. What is the maximum pressure rise due to sudden closure of a valve in a pipe of 300 mm diameter conveying water with a velocity of 1.8 m/s? The pipe wall is 18 mm thick. The E pipe = 210 GPa and K water = 2.1 GPa. Also find the hoop stress developed. (06)

Solution:

V = 1.8 m/s; d=0.3m; t =18x10-3 m; K = 2.1 GPa; E = 210 GPa;

1 14m

= and ρ = 1000 kg/m3 (Assumed)

The valve closure is Sudden with pipe Elastic.

3 9

10001.8 2.091 MPa1 1 1 1

2.1 18 10 210 10

ip VD

K t E

ρ

−

= = = + + × ×

(Ans)

Hoop stress developed is given by

6

1 32.091 10 0.3 17.425 MPa

2 2 18 10ip dft −

× ×= = =

× ×(Ans)

Open channel flow – Flow meassurements

Establish a relation between Chezy’s C and Manning’s n

Soln: Chezy’s equation is 0RSCV =

Manning’s equation is 21

0321 SR

nV =

Equating the two equations 21

21

032

021 1 SR

nSCR =

611 R

nC =

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Manning’s n has dimensions. The dimensions of n being

13TL−

2. A rectangular channel 1.5 m wide with a bed slope of 0.0001 carries water to a depth of 1.2m. The channel has Manning’s n = 0.025. Calculate the rate of uniform flow in the channel.

Soln: B = 1.5m, y = 1.2m, n = 0.025,

21

32

oSRnAQ =

2m8.12.15.1where =×== ByA

( ) m9.32.125.12 =×+=+= yBP

m4615.09.38.1

===PAR

( ) ( ) /sm43.00001.04615.0025.08.1 32

132

=×=∴Q

3. Calculate the uniform depth of flow in a rectangular channel of 3m width designed to carry 10 cumecs of water. Given Chezy’s C = 65 and channel bed slope= 0.025 %.

Ans. B = 3m, y = ?, Q = 10 cumecs, C = 65, S o = 0.025%

Chezy’s eqn is Q = AC(RS o )0.5

A = By = 3y m2

( ) m232 yyBP +=+= and

+

==y

yPAR

233

Substituting all values in Chezy’s eq

+

××=100025.0

23365310

yyy

21

238726.1

+

×=y

yy ( ) 23

21

238726.1 yy =+

( ) 3235065.3 yy =+ 0519.10013.73 =−−∴ yy

Solving by trial and error y = 3.21m

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4. Find the rate of flow of water through a triangular channel having the total angle between the sides as 60. Take the value of n = 0.015 and the slope bed as 1m in 1km. The depth of flow is 1.6m.

21

032

equationsManning'm6.1

10001km1inm1,015.0,30?,

SRnAQy

SnQ oo

==

===== θ

22 m478.130tan6.1tan212 ==

××= oyyA θ

m695.330tan16.12tan12 22 =+××=+= oyP θ

m40.0695.3478.1

===PAR

/sm789.01000

14.0015.0478.1 32

1

23

=

××=∴Q

5. Water flows at a velocity of 1 m/s in a rectangular channel 1m wide. The bed slope is 2x10-3 and n = 0.015. Find the depth of flow under uniform flow conditions.

Soln: V = 1 m/s, B = 1m, S = 2x10-3, n = 0.015, y = ?

From Manning’s equation 21

321

oSRn

V =

A = By = y m2

P = B+2y = 1+2y m

m21 y

yPAR

+==

Substituting, we get, ( )213

2

310221015.0

11 −×

+

=y

y

19425.03354.021

or3354.021

23

32

==

+

=

+ y

yy

y

Solving for y we get y = 0.3176 m.

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6. In a rectangular channel, the bed width is 2.5 m and the bed slope is 1 in 500. If the depth of flow is constant at 1.7m calculate (a) the hydraulic mean depth (b) the velocity of flow (c) the volume rate of flow. Assume that the value of coefficient C in the Chezy’s formula is 50.

Soln: ??,,7.1,500

1,m5.2 ===== VRmySB o Q = ?, C = 50

2m25.47.15.2 =×== ByA

( ) m9.57.125.22 =×+=+= yBP

m72.09.525.4

===PAR

m/s9.1500

172.050 =××== oRSCV

/sm066.89.1325.4 3=×== AVQ

7. An open channel of trapezoidal section base width 1.5m and side slopes 60˚ to the horizontal is used to convey water at a constant depth of 1m. If the channel bed slope is 1: 400. Compute the

discharge in cumecs. The Chezy’s constant may be evaluated using the relation ( )RC

/2.0187

+= Where

R is the hydraulic radius (VTU, Aug 2005)

( ) ( ) 2022 m077.230tan115.1tan =×+×=+ θyBy

P = ( ) ( ) 81.330tan1125.1tan12 022 =+××+=++ θyB m

m545.081.3

077.2radiusHyd ===PAR

oRSACQ =eqnschezy'From /sm25.54001545.046.68077.2 3=×××=

8. A channel 5m wide at the top and 2m deep has sides sloping 2v:1H. Find the volume rate of flow when the depth of water is constant at 1m. Take C = 53. What would be the depth of water, if the flow were to be doubled.

Soln: From fig T = B+2yz

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m322125 =∴××+= BB

Given the depth of flow y = 1m

222 m5.312113 =×+×=+= zyByA

m24.521112312

22

=

+××+=++= zyBP

m669.024.55.3

===PAR

oRSACQ =eqnsChezy'From /sm8.41000

1669.0535.3 3=

×××=

?,/sm6.98.42,Now 13 ==×= yQ

oRSACQ =eqnsChezy'From

2

1

2112

11

21123

213

532136.9

+×+

+××

×+×=

y

yyyy

Solving by trial and error 1y = 1.6m

9. A trapezoidal channel 1.8 m wide at the bottom and having sides of slope 1:1 is laid on a slope of 0.0016. If the depth of the water is 1.5m, find the rate of uniform flow. Assume n = 0.014.

Soln: B = 1.8m, n = 1, S o = 0.0016, y = 1.5m, Q = ?, n = 0.014.

222 m95.4)5.115.18.1( =×+×=+= zyByA

m21.7)5.115.128.1(12( 22 =+××+=++= zyBP

m687.021.795.4

===PAR

21

32

eqnsManning'From oSRnAQ = /sm11)0016.0()687.0(

014.095.4 32

132

=×=

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10. A concrete lined trapezoidal channel with side slope 2H:IV has a base width of 3m and carries 5.5 m3/s of water on a slope of 1m 10000. Find the depth of flow. Assume n = 0.011

Soln: z = 2, B = 3m, Q = 5.5 m3/s, y = ?, n = 0.011, 10000

10 =S

21

32

oSRnAQ =

22 23 yyzyByA +=+=

yyBzyBP 47.4321212 22 +=++=++=

)47.4323(

2

yyy

PAR

++

==

21

32

22

000,101

47.4323

011.0235.5

+

+

+=∴

yyyyy

Solving by trial and error y = 1.32m

11. A trapezoidal channel is designed to convey 1.5 m3/s of water at a depth of 1m if the mean velocity of flow is 0.5 m/s and side slopes are1:1, find the base width and the bed slope. Take C = 60

Soln: Q = 1.5 m3/s y = 1m, V = 0.5m/s, z = 1, B = ?, C = 60

2m35.05.1eqncontinuityFrom ===

VQA

zyByA 2+=

m21113 2 =∴×+×= BB

212 zyBP ++= ( ) m828.411122 2 =+××+=

m621.0828.43

===PAR

oRSCV =eqnschezy'fromNow

oS××= 621.0605.0 or 8947

110118.1 4 =×= −oS

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12. Water flows through a channel of circular section of 600mm diameter at the rate of 200 lps the slope of the channel is 1m in 2.5 km and the depth of flow is 0.45m. Calculate the mean velocity and the value of Chezy’s coefficient

Soln: Q = 200lps = 0.20 m3/s

AC = y = 0.45m

OC = r = 600/2 = 300 mm = 0.3m

OA = (AC-OC) = (0.45-0.3) = 0.15m

From triangle OAB

αcosOBOA

= 5.03.0

15.0cos ==α

( ) 094.2120180

120180and60 000 =×==−==παθα

( ) 22 m2275.02

094.22sin094.23.0 =

×

−=A

Wetted perimeter P = 2rθ = 2x0.3x2.094 = 1.2564m

m181.02564.12275.0radiusHydraulic ===

PAR

oRSACQ =eqnsChery'From

×

=

25001181.02275.0

2.0C = 103.3

m/s88.02275.0

2.0eqncontinuityFrom ===AQV

13. An open channel has a cross section semicircular at the bottom with vertical sides and is 1.2m wide. It is laid at a bed slope of 0.375m per km. Calculate the values of chezy’s C and Manning’s n, if the depth of flow is 1.2m while the discharge is 0.85 m3/s

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Soln: C = ? n = ? y = 1.2m, Q = 0.85 m3/s,

1000375.0

=oS

Area of flow A = (Area of rectangle 1A + Area of semicircle 2A )

220.60.6 1.2 1.2856 m

2A π ×

= × + =

Wetted perimeter P = [2 x 0.6 + 0.6π] = 3.085 m

Hydraulic radius m417.0085.3

2856.1===

PAR

oRSACQ =eqnsChezy'From

87.52

1000375.0417.02856.1

85.0=

×=∴C

Relation between Manning’s n and Chezy’s C is 611 R

nC =

( ) 0163.09.52

417.061

61

=

=

=∴

CRn

14. Water is conveyed in a channel of semicircular cross section with a stage of 1 in 2500. The Chezy’s coefficient C has a value of 56. If the radius of the channel is 0.55 m, what will be the volume rate of flow in m3/s flowing when the depth is equal to the radius?

If the channel had been rectangular of the same width, what would be the discharge for the same slope and value of C ?.

Soln: Case (i) Semicircular channel

C = 56, r = 0.55m, Q = ?, y = r = 0.55m i.e the channel is flowing full. 2500

10 =S

222

m475.02

55.02

=×

==ππ rA

m728.155.0 =×== ππ rP

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mPAR 275.0

728.1475.0

===

/sm279.02500

1275.056475.0 3=

×××== oRSACQ

Case (ii) Rectangular channel

B = 1.1m, y = 0.55m, C = 56, Q = ? 2500

10 =S

A = By = 1.1 x 0.55 = 0.605 m2

P = B + 2y =1.1 + 2x0.55 = 2.2m

m275.02.2

605.0===

PAR

oRSACQ = /sm3553.02500

1275.056605.0 3=

×××=

15. A rectangular channel conveys a discharge of 9.6 cumecs. If the width of channel is 6m, find the depth of flow. Take C = 55 and bed slope = 2x10-4

Soln:

oRSACQ = ;m6 2yByA == yBP 2+= m)26( y+=

yy

PAR

266+

== , 455, 2 10oC S −= = ×

Substituting all values in Eq. (i),

410226

65566.9 −××

+

××=y

yy

705.026

3

=+ yy

Solving by trial and error my 92.1=

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16. A flow of 100 lps flow down in a rectangular flume of 60cm width and having adjustable bottom slope if Chezy’s constant C is 56, find the bottom slope necessary for uniform flow with a depth of flow of 30cm. Also calculate the conveyance K for the flume.

Soln:

Area of flow A = By = 0.6x0.3 = 0.18 m2

Wetted perimeter P = B + 2y = 0.6 + 2x0.3 = 1.2 m

Hydraulic radius R=A/P = 0.18/1.2 = 0.15m

From Chezy’s formula oRSACQ =

152411056.6

18.015.0561.0 4

22

2

22

2

0 =×=××

==∴ −

RACQS

Conveyance RACK = 15.05618.0 ×= /sm904.3 3=

17. A channel of trapezoidal section has a bottom width of 5m, one side is sloping at 400 with the vertical and the other has a slope of 1V to 2H. If the depth of flow is 1.5m, find the bed slope required to discharge 35 cumecs. Taking Manning’s n = 0.017.

××+××+×= 5.13

215.126.1

215.15A 2m69.10=

( ) m313.105.126.15.135 2222 =++++=P

m0365.1313.1069.10

===PAR and oSR

nAQ 2

132

=

( ) ( )

6.3381

0365.1017.069.1035

21

32

0

=

×=

S

S

18. An earthen canal in good condition is 16.8 wide at the bottom and has side slopes of 2H to 1V. One side slope extends to a height of 2.52m above the bottom level and the other side extends flat to a distance of 150m and rises vertically. If the slope of the canal is 69cm per 1584m, estimate the discharge when the depth of water is 2.52m. Assume C = 35

Sol:

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Wetted perimeter P = length AB+BC+CD+DE+EF.

P 72.01506.38.180.1604.552.2 2222 +++++= =177.2 m

Cross section area

A=A 1 +A 2 +A 3 +A 4 ( ) 72.01506.372.052.22152.280.1604.552.2

21

×+×++×+××= =160.5m2

Hydraulic radius R = A/P = 160.5/177.2 = 0.906m

From Chezy’s equation oRSACQ =

/sm6.11184.1569.0906.0355.160 3=

×××=

19. A circular sewer of 500mm internal dia, has a slope of 1in 144. Find the depth when the discharge is 0.3 m3/s. Take Chezy’s C = 50.

Soln:

Let 2α be the angle subtended by the free surface at the centre.

Area of flow

−= αα 2

212 SinrA

Wetted perimeter dP α=

Now, oRSACQ = 21

21

23

P

SCA o=

Squaring both the sides P

SACQ 032

2 =

Substituting all values; 144

12

2150

3.0

322

2 ×

−×

=d

Sinr

α

αα

Substituting d = 2r = 0.5m

( ) ααα 8522 3 =− Sin

Solving by trial & error 0143radians5.2 ==α

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The corresponding depth of flow

( )0143cos12

−=dy ( )0143cos1

25.0

−= m45.0=∴ y

20. A trapezoidal channel having a cross sectional area A 1 , wetted perimeter P 1 , Manning’s n is laid to a slope of S, carries a certain discharge Q 1 , at a depth of flow equal to y. To increase the discharge, the base width of the channel is widened by x, keeping all other parameters same. Prove that

2

1

5

1

3

1

2 11

+÷

+=

Px

Ayx

QQ

213

2

1

111

1:ln SPA

nAQSo

×=

In the second case,

213

2

2

222

1 SPA

nAQ

×=

23

2 2 1 22

1 2 2 1

Q A P AQ P P A

∴ = ×

22

21

51

52

3

1

2

PP

AA

QQ

=

Now, xPP += 12 and yxAA += 12

Substituting these values & simplifying

2

1

1

5

1

1

1

2

+

+=∴

xPP

AyxA

QQ

or 2

1

5

1

3

1

2 11

+÷

+=

Px

Ayx

QQ

21. Water is flowing through a circular open channel at the rate of 400lps. When the channel is having a bed slope of 1in 9000, find the diameter of the channel, if the depth of flow is 1.25 times the radius of the channel. Take n = 0.015.

b b

y

of 83

Solution:

21

32

oSRnAQ =

From Figure 25.025.0cos ==r

rα

radians8235.148.104,52.75 00 === θα

( ) 222 0655.22

8235.12sin8235.122 rrSinrA =

×

−=

−=

θθ

rrrP 647.38235.122 =×== θ

rrr

PAR 566.0

647.30655.2 2

===

( ) 21

2

90001566.0

015.00655.24.0

32

×=∴ rr

38

9936.04.0 r= m71.09936.0

4.0 83

=

=r

m422.12channeltheofDiameter ===∴ rD

A rectangular channel carries water at the rate of 2.25 m3/s when the slope of the channel is 0.025 % find the most economical dimensions of the channel if the manning’s N=0.020

Soln: For a rectangular channel

A = By (i)

P = B+2y (ii)

Condition for most economic channel is B = 2y and R = y/2 (iii)

From Manning’s equation 2 13 2

0AQ R Sn

=

2 13 22 0.0252.25

0.020 2 100y y y × = ×

212or, 62

QA mV

= = =

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But, A = By or 6 = 2y+y or y=1.732m

And hydraulic radius R = y/2=0.833m

From chezy’s equation 0RSCV =

24

0 22 19.4264 10

70 0.866 1061S −

= = × = ×

2. A rectangular channel is designed for maximum efficiency, if the wetted perimeter is 8m and the bed slope is 1 in 100 calculate the discharge, given manning’s n = 0.025

Soln: for a rectangular channel P = (B+2y) & for it to be most efficient B = 2y & R = y/2

P = (2y+2y) = 4y

i.e, 8 = 4y or y = 2m ; B = 4m and R = 1m

Area of cross section A = By = 4x2 = 8m

From Manning’s equation,

2 13 2

0AQ R Sn

= 1

2 238 1(1)

0.025 1000Q = × ×

310.12m /s=

3. A rectangular channel 4.5m wide 1.2 m deep is laid on a slope of 0.0009 and is laid with rubber masonry n = 0.017, what saving in excavation and lining can be had by using the best hydraulic dimensions, but at the same time keeping the same shape, discharge and slope.

Soln: From Manning’s equation 2 13 2

0AQ R Sn

=

Where, A = By = 4.5x1.2 = 5.40m

P = B + 2y = 4.5 + 2x1.2 = 6.9m

5.4 0.7826m6.9

ARP

= = = 0009.00 =S 38.09 m /sQ =

Now, considering the channel to be most economical or best hydraulic section B = 2y; R=y/2 and A=2y

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Again, from Manning’s equation ( ) ( ) ( )12

2232

8.09 0.00090.017 2

y y = × ×

227.5,246.3,81.0,62.1 mAmBmRmy ====

% Saving in excavation in terms of area=2.5%.

4. Find the maximum discharge for least excavation of a rectangular channel 3m wide, when C = 65 and bed slope 1m in 1.25km

Soln: for a best hydraulic rectangular channel B = 2y; y = B/2 and R = y/2

From Chezy’s equation 0RSACQ =

3 3 13 652 2 2 1250

Q = × × × × sm /165.7 3=

5. It is proposed to provide a rectangular channel of best section of area 12.5m2, find the breadth and depth. If the bed slope is 1 in 2000, find the discharge, take C = 45.

Soln: for a best rectangular channel

B = 2y and R = y/2

A = By i.e 12.5 = (2y)(y) 1212.5 2.5 (Depth)

2y m ∴ = =

and, 2 and 5.0 (breadth)B y m= =


( ) 112.5 45 1.252000

Q = ×

sm /06.14 3=

6. A triangular channel section 20m, what is the apex angle and depth for the condition of maximum discharge.

Soln: When the channel carries the maximum discharge it will be most economical or best hydraulic

section. For such a channel 22

yR =

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And side slopes are 045=θ with the vertical.

( ) 2 2 01 2 tan tan tan 452

A y y y yθ θ= × = =

2 or 20 4.47mA y y A∴ = = = =

7. An open channel is to be excavated in trapezoidal section with side slopes of 1:1 find the proportions for minimum excavation.

Soln: for a trapezoidal channel 2 given 1(side slope)A By zy z= + =

2A By y∴ = + (i)

AB yy

= −

(ii)

22 1P B y z= + + 2112 ++= yB yBP 22+= (iii)

Substituting eq(ii) in eq(iii)

yyyAP 22+−= 1.82AP y

y∴ = + (iv)

For minimum excavation the channel has to be most economical ie 0=∂∂

yP (z the side slope is

constant)

From eq (iv) 0=∂∂

yP , gives 082.12 =

+

−y

A 282.1 yA =∴

Equating eq(iv) and eq(v)

By + y = 1.82y or By = 0.82y or B = 0.82y or 82.0=yB

8. A discharge of 170 cubic meters per minute of water is to be carried in a trapezoidal channel of best hydraulic Efficiency. The bed slope is 1 in 5000 and side slopes is 1:1 compute the bottom width and depth of flow Chezy’s C = 50

Soln: for a trapezoidal channel

A = By + yz (i) and B = A/y - yz (ii)

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22 1P B y z= + +

Substituting eq (ii) in eq(iii)

22 1AP yz y zy

= − + +

For best hydraulic efficiency wetted perimeter P should be minimum i.e

(n the side slope is constant) 0=∂∂

yP

22 1 0A zy y zy y

∂∴ − + + = ∂

22given, 2 1 0A z z

y−

− + + = ( )2 22 1A z z y∴ = + −

Substituting eq (i) in eq(v)

2 2 2 22 1By y z y z y z+ = + −

22 2 1B yz y z+ = +

Substituting n=1 21122 +=+ yyB

( )2 2 2B y∴ = −

B = 2x0.41y or B = 0.82y

2

2Hydraulic radius

22 1A By y z yRP B y z

+= = =

+ +


3 3170where, 170m / min 2.83 m /s60

Q = = =

( )2 12.83 0.82 502 5000yy y ∴ = + × ×

Solving for y = 1.572m (depth)

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And B = 0.82x1.572 = 1.302 m (bottom width)

9. A trapezoidal channel of best section carries a discharge of 13.7 cumecs at velocity of 0.9m/s the side slopes are2H:IV, find the bed width and depth of flow. Find also the bed slope if the value of Manning’s n = 0.025.

Soln: From continuity equation Q = AV

Area of cross section 213.7 15.22m0.9

QAV

= = =

2A By y z= + (i)

AB yzy

= − (ii)

22 1P B y z= + + (iii)

Substituting eq(ii) in eq(iii)

22 1AP yz y zy

= − + + (iv)

For best section, 0=∂∂

yP 22 1 0A yz y z

y y ∂

− + + = ∂

22 2 1 0A z z

y −

∴ − + + =

( )2 22 1A z z y∴ = + −

Substituting z = 2 the given value of the side slopes. ( )2 22 1 2 2A y= + −

( ) 22 5 1 A y= − ( ) 2or, 15.22 2 5 1 y= −

y = 2.48m (Depth of flow)

15.22and, 2 2.48 1.174 m (Bottom Width)2.48

B = − × =

2.48also, 1.24m2 2yR = = =

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From Manning’s equation

2 13 2

0AQ R Sn

=

( ) ( )2 13 2

010.9 1.24

0.025S= ×

40

13.8 102631.41

S − = × =

11. A trapezoidal channel of best form has a cross sectional area of 37.2m and side slopes of 0.5H:IV, if the bed slope is 1 in 2000 and Chezy’s C = 65. Compute the total flow in the channel.

Soln: For a best trapezoidal section ( )2 22 1 A z z y= + − or ( )2 237.2 2 1 .05 .05 y= + −

or y = 4.629 m

Hydraulic radius 4.629 2.315m2 2yR = = =

From Chezy’s equation

0RSACQ = 3137.2 65 2.315 82.26 m /s2000

= × × =

12. A power canal of trapezoidal section has to be excavated through hard clay at the least cost. Determine the dimensions of the channel given discharge equal to 14 m3/s bed slope 1 in 2500 and Manning’s n = 0.020. Assume side slopes at 60˚ with the horizontal.

Soln: The cost of the channel will be least when it is economical section.

We know 2A By y z= + (i)

AB yzy

= − (ii)

22 1P B y z= + + (iii)

Substituting (ii) in (iii)

22 1AP yz y zy

= − + +

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For most economical channel 0=∂∂

yP

22 2 1 0A z z

y−

∴ − + + = ( )2 22 1 A z z y= + − (iv)

Given side slopes as 60˚ with the horizontal 13

z =

From eq (iv) 22

31

3112 yA

−

+= 2or, (1.7321)A y=

For a most economical trapezoidal channel R = y/2

Now from Manning’s equation 2 13 2

0AQ R Sn

=

2 12 3 2(1.732) 114

0.02 2 2500y y = ×

Simplifying, 2.60m (Depth)y =

From Eq( ), 3.00m (Bed Width)ii B =

13. A trapezoidal channel with side slopes of 1:1 has to be designed to carry at a velocity of so that amount of concrete lining for the bed and sides is minimum. Calculate the area of lining required for 1m length of the canal.

Soln: from continuity equation Q = AV 252

10 mVQA ===

For a most economical trapezoidal channel ( )2 22 1 A z z y= + −

Substituting z = 1 (side slope) and A = 5m solving for y we have

2 25 (2 1 1 1) y= + − 1.66m (Depth)y∴ =

Also, AB yzy

= −( )

5 1 (1.66) 1.364 m (bed width)1.66

= − × =

Wetted perimeter 22 1 6.06mP B y z= + + =

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Area of lining for 1 m length of the canal Px1 = 6.06m2

14. Determine the bed width and discharge of the most economical trapezoidal channel with side slopes of IV:2H and bed slope of 1m per km and depth of flow equal to 1.25m. Roughness coefficient of channel is 0.024.

Soln: For a most economical trapezoidal channel ( )2 22 1 A z z y= + −

Substituting z = 2 and y = 1.25m

( )2 22 1 2 2 1.25A = + − × 23.863 m (Area of cross section)=

But, Bed Width AB yzy

= −

( )3.8632 1.25 0.59m

1.25B∴ = − × =

1.25Also, Hydraulic radius 0.625m2 2yR = = =


0AQ R Sn

=( ) ( )

23

123.863 106.25

0.024 1000 = × ×

33.721 m /s (discharge)Q∴ =

15. Find the maximum velocity and maximum discharge through a circular sewer 0.75m radius given n = 0.016 channel bed slope = 0.1 percent.

Soln: Case (a) Maximum velocity

For a maximum velocity 0128 45 2.247 radiansθ ′= =

2 sin 2Area2

A r θθ = −

2 2sin 2 2.2470.75 2.247 1.538m2

× = − =

wetted Perimeter 2 2 0.75 2.247 3.37mP rθ= = × × =

1.538Hydraulic Radius, 0.456m3.37

ARP

= = =

Now from Manning’s equation 2 13 2

0AQ R Sn

=

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23

121 0.1(0.456)

0.016 100 = ×

1.17 m/sV∴ =

Case (b) Maximum discharge

For maximum discharge 154 2.688 radiansθ = =

21.733 , 4.032 , 0.43A m P m R m= = =

Now from Manning’s equation

12

23

0AQ R Sn

=( )( ) ( )

23

121.733 0.10.43

0.016 100 = ×

31.961m /s=

16. Find the depth of flow for maximum velocity in a circular sewer 1.50m diameter

Soln: for maximum velocity ( )0 0 0180 128 45 51 15α ′ ′∴ = − =

The corresponding depth of flow y is given by

( )αcosrry += ( )αcos1+= r

( )01.50 1 cos51 15 1.215 m2

y ′= + =

17. Find the depth of flow for maximum discharge in a circular sewer 1.25m diameter.

Soln: Condition for maximum discharge is

( )0 0154 and 180 154 26θ α= = − =

S

S-y√2

y

S

y√2

E F

B

A

C

D

S/√2

(S/√2)-y

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The corresponding depth of flow is ( )αcos1+= ry ( )026cos1

225.1

+= , 1.1875 my =

8. The cross section of an open channel is a square with diagonal vertical S is the side of the square and y is the portion of the water line below the apex, show that for maximum discharge, ratio is 0.127, while it is 0.414 for maximum velocity.

Soln: From figure 2ySBE −= yEF 2=

( )Wetted Perimeter, 2 2 2P S S y= + − 2 1Area of Flow 22

A S y y= − × × 22 yS −=

For maximum discharge, 0=∂∂

yQ

3

0Ay P

∂= ∂

2 32

1 3 0 or 3 0A P A PA P A P AP y y y y

∂ ∂ ∂ ∂− = − = ∂ ∂ ∂ ∂

Substituting for A & P solving 0.127yS

= (Proved)

For maximum velocity 0=∂∂

yV or 0=

∂∂

PA

y

Substituting for A & P solving 2 1 0.414 (proved)yS

= − =

19. A rectangular channel 5.5 m wide and 1.25m depth has a slope of 1 in 900 determine the discharge when Manning’s n = 0.015 if it is desired to increase the discharge to a maximum. By changing the size of the channel but keeping the same quantity of lining determine the new dimensions and percentage increase in discharge.

Soln: 25.5 1.25 6.875mA = × = 5.5 2 1.25 8 mP = + × =

6.875 0.859 m8

ARP

= = =

Now from Manning’s equation

2 13 2

0AQ R Sn

= ( )23

12 36.875 10.859 13.805 m /s

0.015 900 = × × =

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Let B and y be the new width and depth. In order to have the same amount of lining the wetted perimeter should be the same or unchanged.

i.e., P = B + 2y = 8.0

For maximum discharge i.e. the channel to be best efficient B = 2y

2y + 2y = 8 or y = 2m (depth)

B = 2x2 = 4m (bed width)

24 2 8A m= × = mPAR 1

88

===

And discharge 2 13 2

0AQ R Sn

=

12 2 338 1(1) 17.778 m /s

0.015 900 = × =

Increase in discharge 17.778 13.805 100 28.78%13.805

−= × =

20. A trapezoidal channel carries a discharge of 28.5 m3/s, with a mean velocity of 1.5 m/s when lined with rubble masonry n = 0.017 one side is vertical and the other has a slope of 2H:IV. Determine the minimum slope and dimensions of the channel.

Sol: From continuity equation Q = AV

2195.15.28 m

VQA ===

But from figure 212

A By y z= + (i)

Wetted perimeter 2 2 2P B y z y y= + + + 21B y y z= + + +

( )21 1P B y z= + + + (ii)

( )22

1 1 1 02

A z zy−

− + + + =

Substituting A = 19m, z = 2 & solving y = 2.92m (depth)

19 1 2 2.92 3.60m (Bed width)2.92 2

B = − × × =

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Hydraulic radius 1.456 mAR

P= =

Now from manning’s equation 2 13 2

0AQ R Sn

=

Substituting for V, n & R solving for S

40

13.94 10 or (bed slope)2537.9

S − = ×

Specific Energy

. Establish a relation between the alternate depths for a horizontal rectangular open channel.

Soln: We know that the alternate depths y 1 & y 2 represents the same specific energy E

2

1 1 21

(for super critical flow)2

QE ygA

= + (i)

2

2 2 22

(for sub critical flow)2

QE ygA

= + (ii)

Equating the two equations in order to satisfy the definition of alternate depths we have 2 2

1 22 21 22 2

Q Qy ygA gA

+ = +

(iii)

2

2 1 2 21 2

1 1rewriting, ( )2Qy y

g A A

− = −

Substituting 2

31 1, 2 2 , and c

Q qA By A By q yB g

= = = =

( )

−=− 22

221

2

2

1211

2 yByBgQyy

−= 22

21

2

2 112 yygBQ 2 22

2 12 21 22

y yqg y y

−=

( ) ( )( )

+−

=− 22

21

12123

12 2 yyyyyyyyy c

2 231 2

1 2

2c

y y yy y

∴ = +

(iv)

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2. Water flows in a channel at a velocity 2 m/s and at a depth 2.5 m calculate the specific energy.

Soln: we know specific energy g

VyE2

2

+=

2substituting, 2.5m, 2m/s, 9.81m/sy V g= = =

222.5 2.71m2 9.81

E

= + = ×

3. Water flows at 12.5 cumecs in an horizontal rectangular channel 2 m wide, a velocity of 1.25 m/s. Calculate the specific energy critical depth, critical velocity and the minimum specific energy.

Soln.

From continuity equation Q = AV 21025.1

5.12 mVQA ===∴

But, A = By mBAy 5

210

===∴ , 2

specific energy2VE y

g= +

21.255 5.08m2 9.81

E∴ = + =×

Discharge per unit width 312.5 6.25 m /s/m width2

QqB

= = =

Critical depth 31

2

=

gqyc

m585.181.925.6 3

12

=

=

Area corresponding to critical depth 22 1.585 3.17 mc cA By= = × =

Critical velocity 12.5 3.943 m/s3.17c

c

QVA

= = =

Minimum specific energy min32 cE y=

( )min3 1.585 2.3775m2

E∴ = × =

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4. A discharge of 18m3/s flow through a horizontal rectangular channel 6m wide at a depth of 1.6m. Find, (a) the specific energy, (b) the critical depth (c) minimum specific energy, (d) alternate depth corresponding to the given depth of 1.6m, (e) state of flow.

Soln: area of cross section A = By

26 1.6 9.6mA = × =

(a) Specific energy 2

2

2gAQyE +=

2181.62 9.81 9.6

= + × ×

(b) Discharge per unit width 318 3m /s6

QqB

= = =

Critical depth 31

2

=

gqyc

1

2 33 0.972 m9.81

= =

(c) Minimum Specific energy cyE

23min = 3 (0.972) 1.457m

2= × =

(d) Alternate depth y can be calculated from the equation

+=

21

22

213 2

yyyyyc

Where, 1 1.6 m, 0.972mcy y= =

Substituting these values of 1 and cy y

In the above equation we get a quadratic equation in 2y , which has two roots

2 20.45m 0.632 my y= − =

Considering only the +ve root we have my 632.02 =

(e) Now that y 1 < y c and referring to the specific energy diagram, we can conclude that the flow is in the sub critical state.

5. 11.32 cumecs of water flows through a rectangular channel 3m wide. At what depth will the specific energy be 2.25m?. Also calculate the corresponding Froude number

Soln: Discharge per unit width 311.32 3.773 m /s/m3

q∴ = =

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Critical depth 1 1

2 3 33.773 1.132 m9.81c

qyg

= = =

From Continuity equation ,AVQ = 11.32 m/s3

Vy

∴ =×

Substituting these values of V in the specific energy equation g

VyE2

2

+=

211.32 12.25

3 2 9.81y

y

= + × × or 3 22.25 0.726 0y y− + =

Solving by trial and error my 66.0=

We know that the relation between alternate depths is given by 21

22

213 2

yyyyyc +

=

Substituting the values of y 1 and y c and solving we have mymy 506.0,17.2 22 −==

Considering only the positive value of y 2 we have y 2 = 2.17m (alternate depth)

Froude number

=

gyVF

11

1 1 1

11.32 2.253 0.66 9.81 0.66

V QFgy By gy

∴ = = = =

× × ×

22

2 2 2

11.32 0.3763 2.17 9.81 2.17

V QFgy By gy

= = = =

× × ×

(6) The specific energy in a rectangular channel is 5 N-m/N. Calculate the critical depth if the width of the channel is 10m. Calculate the maximum discharge.

Soln: We know E min = 3/2 y c

Considering the given value of specific energy of 5 N-m/N as the minimum specific energy, we have:

Critical depth 2 2 5 3.333 3cy E m= = × =

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From the relation 32

cygq

=

( ) ( )121

3 2 322 3.33 9.81 19.032 m /s/mcq y g= = × =

Therefore, total discharge Q q B= × ( )19.03 10= × 3190.32m /s=

(7) A rectangular channel is to carry a discharge of 25 cumecs at a slope of 0.006. Determine the width of the channel for the critical flow. Take n = 0.016.

Soln: If B is the bed width of the channel, then by definition B

Q 25=

ByByQ

AQV 25

===

For critical flow, 32

cygq

= 31

2

=

gqyor c

12 325 1

9.81B = ×

2

3

4cy

B∴ =


01V R Sn

=

( )213

2

006.02016.0

125

+=

c

c

c yBBy

By

Substituting 2

3

4cy

B= , in the above equation and simplifying

( )12

13

23

25 1 4 0.077580.0164

B

B BB

= +

( )2

5 33

, 0.512

8

Bor

B

=

+

Solving by trial and error mB 3=

Hydraulic jump

Derive an equation for the loss of energy due to a hydraulic jump in a horizontal rectangular open channel.

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Soln: applying Bernoulli’s equation between 1,1 and 2,2 with the channel bed as datum and considering head loss due to the jump

Eg

VyZg

VyZ ∆+++=++22

22

22

21

11

Since the channel is horizontal 21 ZZ =

Eg

Vyg

Vy ∆++=+∴22

22

2

21

1 or 2 2

1 22 1( )

2V VE y y

g−

∆ = − − + (1)

From continuity equation 2211 VAVAQ ==

111 y

qByQV == ,

22 y

qV =

Substituting these values of in eq (1)

)(112 122

221

2

yyyyg

qE −−

−=∆

2 222 1

2 12 21 2

( )2

y yq y yg y y

−= − −

(2)

But for a rectangular channel the general equation of hydraulic jump is

)(22121

2

yyyygq

+=

21 2

1 2( )2

y yq y yg

= + (3)

Substituting eq 3 in eq 2

( )2 2

1 2 2 11 2 2 12 2

1 2

( )2 2

y y y yE y y y yy y−

∆ = + × − − ( )( ) )(4 12

21

21212 yy

yyyyyy

−−+−

=( ) ( )2

2 1 2 1 1 2

1 2

4

4

y y y y y y

y y

− + − =

( )32 1

1 24y y

Ey y−

∆ = (4)

In eq 4 E is expressed in meters. If E is to be expressed as an energy loss in terms of KW or as power lost. P Q Eγ= ∆ (5)

Where P will be in KW, γ specific weight of the liquid KN/m

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Q discharge m3/s and ΔE the head lost in m

2. A rectangular channel 3m wide carrying 5.65 cumecs of water at a velocity of 6m/s discharges into a channel where a hydraulic jump is obtained what is the height of the jump? Calculate the critical depth also

Soln: From the continuity eqn Q = AV

15.65 3 6y= × × 1 0.314m (initial depth)y∴ =

35.65 1.883 m /s/m3

QqB

= = =

Now from the equation of hydraulic jump in a rectangular channel.

2 21

2 3 31

8 0.314 8 1.8831 1 1 1 1.3686 m2 2 9.81 0.314y qy

gy

× = − + + = − + + = ×

Height of the jump h j =(1.3686-0.314)=1.0547m

Critical depth y c is calculated from the equation. gqyc

23 =

1 1

2 23 31.883 0.712 m9.81c

qyg

= = =

3. In a rectangular channel 2.4 m wide the discharge is 9.1m3/s. If a hydraulic jump occurs and the depth before the jump is 0.75 m. find the height of the jump energy head loss and power lost by energy dissipation.

Soln: From the relation BQq = , we have 39.1 3.792 m /s/m

2.4q = =

discharge per unit width.

From the equation of hydraulic jump in a rectangular channel.

++−= 31

21

2811

2 gyqyy

Substituting all values and solving for 2y

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2

2 30.75 8 3.7921 1 1.637 m

2 9.81 0.75y

× = − + + = ×

Height of the jump ( )2 1 (1.637 0.75) 0.887jh y y m= − = − =

Energy head loss( )3 3

2 1

1 2

0.887 0.1422m4 4 0.75 1.637

y yE

y y−

∆ = = = × ×

Power lost P Q Eγ∆ = ∆ = 9.81x9.1x(0.1422) = 12.69 kW

4. Flow over a spill way is 3 cumces/meter width, the supercritical velocity down the spillway is 12.15m/s. What must be the depth of the tail water to cause a hydraulic jump at the apron? What is the energy lost per unit width? What is the total head of flow before and after the jump.

Soln: from continuity eq Q = AV 11VByQ =∴

1 1 11

3or 0.247m12.15

qq y V yV

= ∴ = = =

For a rectangular channel A = By

y 1 is the initial depth of flow from the equation of hydraulic jump in a rectangular channel.

++−= 31

21

2811

2 gyqyy

Substituting all values and solving

2

2 30.247 8 31 1 2.606 m

2 9.81 0.247y

× = − + + = ×

2y is the depth of tail water required for the formation of hydraulic jump

Energy lost ( ) ( )606.2247.04

247.0606.24

3

21

312

xxyyyyE −

=−

=∆ =5.10m

Total head of flow before the jump

g

VyH2

21

11 +=212.15(0.247 ) 7.771 m

2 9.81x= + =

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From continuity eq

22

3 1.151 m/s2.606

qVy

= = =

Total head of flow after the jump

2 22

2 21.1512.606 2.673m

2 2 9.81VH y

g= + = + =

×

5. The stream issuing from beneath a vertical sluice gate is 0.3m deep at vena contracta. Its mean velocity is 6 m/s a standing wave is created on the level bed below the sluice gate. Find the height of the jump the loss of head and the power dissipated per unit width of sluice.

Soln; from the continuity eq Q = AV = ByV

31 1or, 0.3 6 1.8m /s/mQ q y V

B= = = × = or 1 0.3 my =

Conjugate depth or depth after the jump 2y is given by

2 21

2 3 31

8 0.3 8 1.81 1 1 1 1.341 m2 2 9.80 0.3y qy

gy

× = − + + = − + + = ×

Height of the jump 2 1( ) (1.341 0.3) 1.0414mjh y y= − = − =

( )3 32 1

1 2

1.0414 0.702m4 4 0.3 1.341

y yE

y y−

∆ = = = × ×

Power dissipated / unit width = 9.81x1.8x(0.702) = 12.394 kW/m width

6. If the velocity when the water enters the channel is 4 m/s and Froude number is 1.4 obtain a) The depth of flow after the jump b) the loss of specific energy due to the formation of the jump.

Soln: From the definition of Froude number we have

11

1 1

4.01.4 0.832m9.81

VFgy y

= ⇒ = =×

The depth of flow after the jump is given by

{ }21

12 811

2Fyy ++−= , { }2

20.832 1 1 8 1.4 1.283m

2y = − + + × =

Loss of specific energy 2 22 1

2 12 2V VE y y

g g ∆ = + +

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From continuity equation

2211 VyVy =

1 12

2

0.832 4 2.593 m/s1.283

y VVy

× ∴ = = =

Loss of specific energy

2 22.593 41.283 0.832 0.0217 m19.62 19.62

E ∆ = + ≈ + =

7. In a rectangular channel 0.6m wide a jump occurs where the Froude number is 3. the depth after the jump is 0.6m estimate the total loss of head and the power dissipated by the jump.

Soln: from the eq of Hydraulic jump { }21

1

2 81121 F

yy

++−=

3,6.0 12 == Fmy { } 16.03811216.0 2

1

=++−= xy

Head loss due to the jump 21

312

4)(

yyyyE −

=∆

3(0.6 0.16) 0.2232 m4 0.16 0.6

E −∆ = =

× ×

8. The depth and velocity of water downstream of a sluice gate in a horizontal rectangular channel is 0.4m and 6m/s respectively. Examine whether a hydraulic jump can possibly occur in the channel. If so find the depth after the jump and head loss due to the jump.

Soln: the value of initial Froude number is calculated from the relation

11

1

6 3.0299.81 0.4

VFgy

= = =×

Since, F1 (=3.029)>1 i.e, the flow is supercritical, a Hydraulic jump will occur. Now, from the relation.

{ }21

12 811

2Fyy ++−=

Substituting all values

( ){ }22

0.4 1 1 8 3.029 1.525 m (Downstream depth)2

y = − + + × =

Loss of head due to the jump

( )21

312

4 yyyyE −

=∆( )31.525 0.4

0.584m4 0.4 1.525

−= =

× ×

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9. A rectangular channel 5m wide carries a discharge of 6 cumecs. If the depth on the downstream of the hydraulic jump is 1.5m, determine the depth upstream of the jump. What is the energy dissipated?

Soln: Discharge per unit width BQq =

32

6 1.2 m /s/m width, given 1.5m5

q y∴ = = =

Therefore depth before or upstream of the jump

++−= 3

2

22

1811

2 gyqyy

Substituting 22

1 38 1.21 1

2 9.814 1.5yy

× = − + + ×

1 0.1208m (Depth upstream of the jump)y∴ =

Energy dissipated ( )2

1

312

4 yyyyE −

=∆( )

( ) ( )

31.5 0.12083.62 m

4 0.1208 1.5−

= =× ×

10. Determine the flow rate in a horizontal rectangular channel 1.5m wide in which the depths before and after the hydraulic jumps are 0.25m and 1.0m.

Soln: From the equation of hydraulic jump. 22

31 1

1 81 12

y qy gy

= − + +

Substituting y 1 = 0.25m, y 2 = 1m, g = 9.81 m/s2 and solving for q

2

31 1 81 1

0.25 2 9.81 0.25q = − + + ×

Flow rate in the rectangular channel

( ) ( ) 31.238 1.5 1.587 m /sQ q B= × = × =

11. Water flows at the rate of 1.25 cumecs in a channel of rectangular section 1.5m wide. Calculate the critical depth, if a hydraulic jump occurs at a point where the upstream depth is 0.30m, what would be the rise of water level produced and the power lost in the jump?

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Soln: Critical depth cy is given by

gBQ

gqyc 2

223 ==

12 31.25 1 0.414m

1.5 9.81cy ∴ = × =

From the equation of Hydraulic jump

++−= 31

31

2811

2 yyyy c

3

2 30.3 8 (0.414)1 1 0.55 m (Conjugate Depth)2 0.3

y × = − + + =

Energy loss due to the jump 21

312

4)(

yyyyE −

=∆

3(0.55 0.3) 0.0237 m4 0.3 0.55

E −∆ = =

× ×

Power lost due to the jump

9.81 1.25 0.0237 0.29kWP Q Eγ= ∆ = × × =

12. A sluice spans a channel of rectangular section 15m wide having an opening of 0.6m depth discharges water at the rate of 40 cumecs. If a hydraulic jump is formed on the downstream side of the sluice, determine the probable height of crest above the upper edge of the sluice

Soln.

From the equation of hydraulic jump

+−= 31

21

281

2 gyqyy

2

2 30.6 8 2.671 1.285 m (conjugate depth)2 9.81 0.6

y × = − + = ×

myy 685.0)60.0285.1()( 12 =−=−

Loss of energy due to the jump

( ) ( )3 32 1

1 2

1.285 0.60.104m

4 4 0.6 1.285y y

Ey y x x− −

∆ = = =

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13.In the case of a hydraulic jump in a rectangular channel, prove that ( )

+−

=)1(32

14

93

rrra ,

2

1

where, ,c

yEa ry y

∆= =

Soln: We know ( )

33 2

3 12 1 1

1 2 1 2

1

4 4

yyy y y

Ey y y y

− − ∆ = =

Substituting ryy

=1

2 , ( )2

321

21

331

4)1(

41

yry

yyryE −

=−

=∆

Raising both sides to the power 3

( )9613

3 32

14

y rE

y−

∆ =

Dividing both the sides by 3cy

( )9631

3 3 32

164c c

y rEy y y

−∆=

× ( ) 39

3113

2

164 c

r y yy y

− = × ×

33 91

3 3( 1)64c c

yE ry r y

∆ −∴ =

(1)

++−== 31

3

1

2 81121

yyr

yy c

+=+

3

1

2 81)12(yyr c

32 2

1

4 1 4 1or,8

cyr ry

+ + −=

( )8

143

1

+=

∴

rryyc ( )3

1

1or,

2c r ry

y+

=

(2)

From Eq (1) and (2)

( )( )

3 9

3

1 264 1c

rEy r r r

− ∆= × +

( )( )

+−

=132

14

93

rrra (Proved)

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14. In a rectangular channel, the discharge per unit width is 2.5 m3/s/m, a hydraulic jump occurs and the loss of energy is 2.68 N.m/N. Determine the depths before and after the jump.

Soln: Equation for energy loss is ( )32 1

1 24y y

Ey y−

∆ = (1) 21

2 31

81 12y qy

gy

= − + +

22

31 2

1 8 2.5, 1 12 9.81

yy y

× ∴ = − + + × 31

1 5.0971 12 y

= − + +

(2)

From Eq (1) 21

3

1

231

4

1

yyyyy

E

−

=∆

68.24

1097.51121

2

3

31

21

=

−

++−

=∆y

yy

E

3

21 3 3

1 1 1

1 5.097 1 5.0971 1.5 2.68 4 2.68 4 1 12 2

yyy y y

+ − = × × = × × − + +

3

1 3 31 1

1 5.097 5.0971 1.5 5.36 1 12

yy y

+ − = − + +

(3)

Solving Eq (3) by trial and error y 1 =0.2644 m.

2 30.2644 5.0971 1 2.067 m (Conjugate depth)

2 0.2644y

= − + + =

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