1

MEMORIAL UNIVERSITY OF NEWFOUNDLAND FACULTY OF ENGINEERING AND APPLIED SCIENCE

Hydraulics Engineering 6713

Problems & Solutions

Dr. Leonard Lye Professor of Civil Engineering

2

EL = 0 m

PA=?

A

B

TUTORIAL 1 TURBULENT PIPE FLOW

1. In a chemical processing plant it is desired to deliver benzene at 50°C (rel. density 0.86) to point B with a pressure of 550 KN/m2. A pump is located at point A 21 m below point B, and the 2 points are connected by 240 m of plastic pipe having an inside diameter of 50 mm. If the volume flow rate is 110 liters/min, calculate the required pressure at the outlet of the pump. How much would the required pressure change if welded steel pipes are used instead of plastic pipes? (µ = 0.00042 N-s/m2).

001833.0min/110 LQ sm /3

934.0

4

05.0

001833.02

A

Qv sm /

41056.99562400042.0

05.0934.0860Re

vD

From Moody diagram for smooth pipes, ƒ = 0.018

Therefore 84.32

934.0

05.0

240018.0

2

g

h f m

Applying Bernoulli between A & B:

fBA hg

P

g

P 21

03.9084.32181.9860

10550 3

g

PA

m

03.9081.9860AP 6.759 2/ mKN If welded steel pipes are used, ε = 0.000046 m

Therefore 00091.005.0

000046.0

D

From Moody diagram, with Re = 41056.9 and 00091.0D

,

ƒ = 0.022 and 70.42

934.0

05.0

240022.0

2

g

h f m

89.9081.9860AP 8.766 2/ mKN

EL = 21 m PB = 550 KN/m2

L = 240 m

D = 50 mm

µ = 0.00042 Ns/m2

ρ = 0.86 * 1000 = 800 kg/m3

3

2. What diameter of cast iron pipe would be required to ensure that a discharge of 0.20 m3/s would not cause a head loss in excess of 0.01 m/ 100 m of pipe length? Assume water temperature of 20°C.

2.0Q sm /3 ,100

01.0

L

h f, 000244.0 m

From the Darcy-Weisbach equation,

g

v

D

Lfh f

2

2

1026.12

4

24

2.010001.0

522

2

D

f

gDD

f

53305.001.0

D

f

fD 05.335 --------- (1)

D

D

D

vD 252878

10007.1

4

2.0Re

62

--------- (2)

DD

000244.0

--------- (3)

Assume ƒ = 0.02 (arbitrary value)

From (1) 92.0D m

From (2) 51074.2Re

From (3) 00027.0D

From Moody diagram, new ƒ = 0.017

From (1) 89.0D m

From (2) 51084.2Re

From (3) 000274.0D

From Moody diagram, new ƒ = 0.017 (closest one can read)

Therefore 89.0D m

4

hf = 50m

L = 2000 m

(1)

(2)

3. A 2000 m long commercial steel pipeline of 200 mm diameter conveys water at 20°C between two reservoirs, as shown in the Figure below. The difference in water level between the reservoirs is maintained at 50 m. Determine the discharge through the pipeline. Neglect the minor losses.

mm045.0 mmD 200

Applying Bernoulli between (1) and (2):

fh 0005000

Therefore,

g

vf

g

v

D

Lf

22.0

2000

250

22

20981.0 fv

fv

0981.02 --- (1)

vvvD

19861010007.1

2.0Re

6

--- (2)

000225.0200

045.0

D

Assume flow is in fully rough zone, with 000225.0D

, 015.0f

From (1), 557.2v m/s

From (2), 51008.5Re

From Moody diagram, new 016.0f

From (1), 476.2v m/s

From (2), 51092.4Re

From Moody diagram, new 016.0f

Therefore 4

2.0476.2

2

Q 0.078 m3/s

4. Compare answers for Q1, Q2, and Q3 using explicit equations.

5. Use the Hazen-Williams and Manning’s equations to solve Q1, Q2, and Q3.

6. Use Flowmaster to solve Q1, Q2, and Q3.

5

TUTORIAL 2 SIMPLE PIPE PROBLEMS AND MINOR LOSSES

1. A 6-km-long, new cast-iron pipeline carries 320 litres/s of water at 30°C. The pipe diameter is 30

cm. Compare the head loss calculated from (a) the Hazen-Williams formula, (b) the Manning formula, and (c) the Darcy-Weisbach formula.

6000L m, 32.0Q m3/s, 300D mm, 610804.0 m2/s, 000244.0 m, 130HWC ,

Manning’s 011.0n a) Hazen-Williams formula

54.063.2278.0 SDCQ HW 54.063.23.0130278.032.0 S

21.054.0 S

60000556.0

fhS

Therefore 6.333fh m

b) Manning formula

n

SD

n

SRv

21

32

21

32

4

527.4

4

3.0

32.02

A

Qv m/s

011.0

1778.0527.4

21

S

60000784.0 fh

S

Therefore 4.470fh m

c) Darcy-Weisbach formula

fg

fg

v

D

Lfh f 65.20890

2

527.4

3.0

6000

2

22

00081.0D

, 6

6107.1

10804.0

3.0527.4Re

From Moody diagram, 019.0f

Therefore 9.39665.20890019.0 fh m

6

2. Two reservoirs 1200 m apart are connected by a 50 cm smooth concrete pipe. If the two reservoirs have an elevation difference of 5 m, determine the discharge in the pipe by (a) the Hazen-Williams formula, (b) the Manning formula, and (c) the Darcy-Weisbach formula.

a. Hazen-Williams formula:

Q 54.063.2278.0 SDCHW

Q 54.063.2 )1200/5(5.0140278.0

Q = 0.326 m3/s

b. Manning formula:

4

422

13

2

D

n

SD

VAQ

4

5.0

011.0

12005125.0 2

21

32

Q

Q = 0.288 m3/s c. Darcy-Weisbach formula:

0018.0500

9.0

mm

D

5fh , 610007.1 m2/s (assume 20°C water)

Using Swamee and Jain’s explicit equation for Q,

Q =

1200

55.081.95.0

10007.1784.1

7.3

0018.0ln

1200

55.081.95.0965.0

62

Q = 000025133.00004865.0ln034489.0

Q = 0.261 m3/s

5 m L = 1200 m

D = 0.5 m

7

3. An old pipe 2 m in diameter has a roughness of ε=30mm. A 12-mm-thick lining would reduce the roughness to ε=1mm. How much in annual pumping costs would be saved per kilometer of pipe for water at 20°C with discharge of 6 m3/s? The pumps and motors are 80% efficient, and power costs 4 cents per kilowatt-hour.

Old pipe: D = 2 m, ε=30mm, 91.1

4

2

62

A

Qv m/s

Head loss of old pipe = fg

fh f 96.922

91.1

2

1000 2

m/km

015.02000

30

D

,

6

6108.3

10007.1

291.1Re

vD

From Moody diagram, 044.0f

Therefore, 09.4044.096.92 fh m/km

New pipe D = 2000 – 24 = 1.976 m, ε = 1 mm, 957.1

4

976.1

62

A

Qv m/s

fg

fh f 739.982

957.1

976.1

1000 2

m/km

000506.01976

1

D

,

6

61084.3

10007.1

976.1957.1Re

vD

From Moody diagram, 017.0f

Therefore, 679.10165.0739.98 fh m/km Saving in head = 4.09 – 1.679 = 2.411 m/km

Energy = Power x Time = 212.1554243658.0

411.2681.924365

e

Qh KW

Therefore annual savings = 1554.2 x $ 0.04 = $ 62168/km

8

4. What size commercial steel pipe is needed to convey 200 L/s of water at 20°C 5 km with a head drop of 4 m? The line connects two reservoirs, has a reentrant entrance, a submerged outlet, four standard elbows, and a globe valve.

Applying Bernoulli between 1 and 2:

foGVelbowse hhhhh 4000400

g

v

D

Lf

g

vk

g

vk

g

vk

g

vk oGVelbowse

222224

22222

0.1ek , 9.0elbowsk , 0.10GVk , 0.1ok Therefore,

Df

g

v 50000.10.109.040.1

24

2

D

f

g

v50006.15

24

2

222

2546.042.0

4

DDD

Qv

Therefore,

D

f

D50006.15

2546.048.78

4

2

---------(1)

DD

000046.0

---------(2)

DD

vD 256830

10007.1

2546.0Re

6

--------- (3)

Assume 02.0f , from (1) 54

4821.601121.148.78

DD

By trial and error 62.0D m

From (2), 000074.0D

,

51014.4Re 0145.0f

From (1), 54

69953.401121.148.78

DD 583.0D m

From (2), 000079.0D

,

5104.4Re

4 m

Submerged

outlet

Reentrant

entrance

(2)

(1)

Q = 0.2 m3/s

Standard elbows

Globe valve

9

New 0145.0f , therefore 583.0D m

Note: Equivalent length method would have been easier

5. What is the equivalent length of 50 mm diameter pipe, f=0.022, for (a) a re-entrant pipe entrance, (b) a sudden expansion from 50 mm to 100 mm diameter, (c) a globe valve and a standard tee?

022.0f , 50D mm

f

kDLe

a. 0.1k for re-entrant pipe entrance.

Therefore, 27.2022.0

05.01 eL m

b. Sudden expansion from 50 mm to 100 mm.

5625.0100

5011

22

22

2

1

D

Dk

Therefore 278.1022.0

05.05625.0

eL m

c. 0.10k for globe valve, 8.1k for standard tee.

Therefore 8.11K

82.26022.0

05.08.11 eL m

6. Solve Q1 and Q1 using Flowmaster.

10

g

v

2

2

1g

v

2

2

2

TUTORIAL 3 EGL, HGL, AND PIPES IN SERIES

1. Sketch the energy grade line and the hydraulic grade line for the compound pipe shown below.

Consider all the losses and the change in velocity and pressure heads.

2. Two sections of cast-iron pipe connected in series bring water from a reservoir and discharge it into air at a location 100 m below the water surface elevation in the reservoir through a globe valve. The first pipe section is 400 mm diameter and is 1000 m long, and the second pipe section is 200 mm diameter and 1200 m long. If the water temperature is 10°C, and square connections are used, determine the discharge. Sketch the EGL and HGL.

Energy between (A) and (B):

1002

2

221

g

vhhhhh vfcfe

g

vhe

25.0

2

1 , g

vfh f

24.0

10002

111

g

vhc

233.0

2

2 33.0ck (assumption)

g

vfh f

22.0

12002

222

g

v

g

vkh vv

210

2

2

2

2

2

EGL

HGL

100 m

(A)

(B) D1 = 0.4 m D2 = 0.2 m L1 = 1000 m L2 = 1200 m

he

hc

11

Therefore,

g

vf

g

vf

25.0

4.0

1000

2)33.0

2.0

1200101(100

2

11

2

22

From continuity,

2211 VAVA

2

2

1

22.0

44.0

4vv

Substituting for 1v , we get:

21

2

2600025.15636.11

1962

ffv

2

4

6

1

1

111 1063.7

1031.1

4.0Re v

vDv

2

5

6

2

2

222 1053.1

1031.1

2.0Re v

vDv

00065.01

1 D

0178.01 f

0013.02

2 D

0205.02 f

Solving for 2v ,

)0205.0(6000)0178.0(25.15636.11

19622

2

v

78.32 v m/s

94.078.325.01 v m/s 54

1 1088.278.31063.7Re 019.01 f 55

2 1078.578.31053.1Re 021.02 f

Use new values of 1f and 2f to calculate 2v ,

)021.0(6000)019.0(25.15636.11

19622

2

v

739.32 v m/s

935.0739.325.01 v m/s 54

1 1085.2739.31063.7Re 019.01 f 55

2 1072.5739.31053.1Re 021.02 f

Therefore, 117.04

2.0739.3

2

2211

AvAvQ m3/s 0.12m3/s

12

3. Two new cast-iron pipes in series connect two reservoirs. Both pipe are 300 m long and have diameters of 0.6 m and 0.4 m, respectively. The elevation of water surface in reservoir A is 80 m. The discharge of 10°C water from reservoir A to reservoir B is 0.5 m3/s. Find the elevation of the surface of reservoir B. Assume a sudden contraction at the junction and a square-edge entrance.

77.1

4

6.0

5.02

1

1

A

Qv m/s

98.3

4

4.0

5.022

v m/s

5

6

111 1008.8

1031.1

6.077.1Re

Dv

6

6

222 1022.1

1031.1

4.098.3Re

Dv

00043.01

1 D

, 00065.0

2

2 D

017.01 f , 018.02 f , from Moody diagram

Bernoulli between two reservoirs,

g

v

g

v

g

v

g

v

g

vH

22018.0

224.0

26.0

300017.0

25.0

2

2

2

2

2

2

2

1

2

1

gggggH

2

98.3

2

98.3

4.0

300018.0

2

98.324.0

2

77.1

6.0

300017.0

2

77.15.0

22222

807.0899.101938.0357.10798.0 H

337.13H m Therefore, the surface elevation of reservoir B = 80-13.337 = 66.66 m

80 m

H

Q = 0.5 m3/s

L1 = 300 m L2 = 300 m D1 = 0.6 m D2 = 0.4 m

13

4. Pipeline AB connects two reservoirs. The difference in elevation between the two reservoirs is 10 m. The pipeline consists of an upstream section, D1 = 0.75 m and L1 = 1500 m, and a downstream section, D2 = 0.5 m and L2 = 1000 m. The pipes are cast-iron and are connected end-to-end with a sudden reduction of area. Assume the water temperature at 10°C. Compute the discharge capacity using the graphical approach.

See figure of previous problem except different lengths and diameters. Bernoulli between two reservoirs,

g

v

g

vf

g

v

g

vf

g

vH

225.0

1000

224.0

275.0

1500

25.0

2

2

2

22

2

2

2

11

2

1 --- (1)

From continuity,

2211 AvAv

4

5.0

4

75.0 2

2

2

1 vv

12 25.2 vv

Substituting for 2v in (1)

2

121 05.51694.10135.0 vffH

22

22

21

75.0

)4(05.51694.10135.0

QffH

21

2 05.51694.10135.01234.5 ffQH

00035.01

1 D

, 00052.0

2

2 D

Assume Q = 0.5 m3/s:

Therefore smv /132.1

4

75.0

5.021

smv /546.22

5

6

111 1048.6

1031.1

75.0132.1Re

Dv

015.01 f

5

6

222 1071.9

1031.1

5.0546.2Re

Dv

018.02 f

Therefore H = 14.304 m too high

Assume Q = 0.4 m3/s

smv /9054.01

smv /037.22

5

1 1018.5Re

016.01 f

5

2 1078.7Re

0175.02 f

Therefore H = 9.03 m too low

14

H (m) vs. Q (m3/s)

From the graph, Q = 0.42 m3/s for H = 10 m

15

TUTORIAL 4 HYDRAULICS 6713

BRANCHING PIPES AND PIPE NETWORKS

1. A two-loop pipe network has node designations as shown below. Inflows of 0.4 m3/s and 0.45 m3/s enter points A and B, respectively. Equal withdrawals are made at points C, D, and F. The pipe characteristics are as follows:

Pipe Length (m) Diameter (m) Friction factor ƒ

AB 500 0.4 0.017

BC 400 0.5 0.016

AF 650 0.5 0.014

BE 750 0.35 0.015

CD 700 0.4 0.013

DE 550 0.5 0.016

EF 900 0.6 0.015

A B

F

C

E D

1 + 2 +

0.4 m3/s 0.45 m3/s 0.28333 m3/s

0.28333 m3/s

0.28333 m3/s

L = 500, D = 0.4, f = 0.017 L = 400, D = 0.5, f = 0.016

L = 900, D = 0.6, f = 0.015 L = 550, D = 0.5, f = 0.016

L = 700,

D = 0.4,

f = 0.013

L = 650,

D = 0.5,

f = 0.014

L = 750,

D = 0.35,

f = 0.015

A B C

F E D

Loop 1 Loop 2

16

Trial 1 Loop Pipe R Q hL 2hL/Q new Q

1

AB 68.619 0.4 10.979 54.895 0.202

BE 177.067 0.4 28.331 141.655 0.202

EF 14.352 0.28333 1.152 8.13186 0.08533

AF 24.072 0 0 0 -0.198

40.462 204.6819

ΔQ = 0.198 0.198

2

BC 16.93 0.45 3.428 15.237 0.4623

CD 73.462 0.16667 2.041 24.492 0.17897

DE 23.279 -0.1166 -0.317 5.435 -0.10436

EB 177.067 -0.202 -7.225 71.535 -0.1897

-1.439 116.699

ΔQ = -0.0123 -0.0123 Trial 2

Loop Pipe R Q hL 2hL/Q new Q

1

AB 68.619 0.202 2.7999 27.72178 0.124039

BE 177.067 0.1897 6.3719 67.1787 0.111739

EF 14.352 0.08533 0.1045 2.449314 0.007369

AF 24.072 -0.198 -0.9437 9.532323 -0.27596

8.3326 106.8821

ΔQ = 0.07796 0.077961

2

BC 16.93 0.4623 3.6183 15.65347 0.421699

CD 73.462 0.17897 2.353 26.29491 0.138369

DE 23.279 -0.1043 -0.2535 4.858183 -0.14496

EB 177.067 -0.1117 -2.2108 39.57043 -0.15234

3.507 86.377

ΔQ = 0.0406 0.040601 Trial 3

Loop Pipe R Q hL 2hL/Q new Q

1

AB 68.619 0.12404 1.05577 17.02306 0.084586

BE 177.067 0.15234 4.10928 53.9488 0.112886

EF 14.352 0.00737 0.00078 0.211669 -0.03208

AF 24.072 -0.2759 -1.8331 13.28584 -0.31541

3.33265 84.46936

ΔQ = 0.03945 0.039454

2

BC 16.93 0.4217 3.01068 14.27878 0.40115

CD 73.462 0.13837 1.40652 20.32984 0.11782

DE 23.279 -0.1449 -0.4891 6.749034 -0.16551

EB 177.067 -0.1128 -2.2565 39.97821 -0.13344

1.67146 81.33586

ΔQ = 0.02055 0.02055

17

Trial 4 Loop Pipe R Q hL 2hL/Q new Q

1

AB 68.619 0.08459 0.491 11.60894 0.068126

BE 177.067 0.13344 3.1529 47.2557 0.116976

EF 14.352 -0.0320 -0.0147 0.920823 -0.04854

AF 24.072 -0.3154 -2.3948 15.18531 -0.33187

1.23433 74.97077

ΔQ = 0.01646 0.016464

2

BC 16.93 0.40115 2.7244 13.58295 0.39261

CD 73.462 0.11782 1.01977 17.31064 0.10928

DE 23.279 -0.1655 -0.6376 7.705758 -0.17405

EB 177.067 -0.1169 -2.4230 41.42657 -0.12552

0.68344 80.02592

ΔQ = 0.00854 0.00854 Trial 5

Loop Pipe R Q hL 2hL/Q new Q

1

AB 68.619 0.06813 0.31851 9.350066 0.062184

BE 177.067 0.12552 2.78974 44.45092 0.119574

EF 14.352 -0.0485 -0.0338 1.39349 -0.05449

AF 24.072 -0.3318 -2.6512 15.97752 -0.33782

0.4232 71.172

ΔQ = 0.00595 0.005946

2

BC 16.93 0.39261 2.60963 13.29375 0.389475

CD 73.462 0.10928 0.87729 16.05582 0.106145

DE 23.279 -0.1740 -0.7052 8.103419 -0.17719

EB 177.067 -0.1195 -2.5315 42.34373 -0.12271

0.2502 79.79672

ΔQ = 0.003135 0.003135

To three decimal places, the final flow values are: AB = 0.062 m3/s BC = 0.390 m3/s CD = 0.106 m3/s DE = -0.177 m3/s EF = -0.054 m3/s AF = -0.338 m3/s EB = -0.123 m3/s

18

2. Determine the flow into and out of each reservoir in the Figure below if the connecting pipes are made of the same material with ε = 0.05 mm and water temperature at 20°C. The pipe characteristics are as follows:

Reservoir Elevation (m) Pipe Length (m) Diameter (m)

A 100 a 3000 0.8

B 80 b 4000 1.2

C 70 c 5000 0.6 Try both the iterative method and the graphical method. Plotting first few points from table:

P≈ 81.4 m (Assume fully turbulent flows, correct for ƒ when near correct answer)

70

75

80

85

90

-2 -1 0 1 2

P

ΣQ

J

A

a

c b

B

C

19

Assume fully turbulent flow Initial estimates: 𝜀𝑎

𝐷𝑎=

0.05

800= 6.3 × 10−5, 𝑓𝑎 = 0.011, 𝑅𝑎 = 8.3212

𝜀𝑏

𝐷𝑏=

0.05

1200= 4.2 × 10−5, 𝑓𝑏 = 0.010 , 𝑅𝑏 = 1.3282

𝜀𝑐

𝐷𝑐=

0.05

600= 8.3 × 10−5, 𝑓𝑐 = 0.0115, 𝑅𝑐 = 61.099

Trial 1 (assume P = 90 m) 𝑧𝐴 − 𝑃 = 𝑅𝑎𝑄𝑎

2 = 100− 90 = 10 𝑃 − 𝑧𝐵 = 𝑅𝑏𝑄𝑏

2 = 90− 80 = 10 𝑃 − 𝑧𝐶 = 𝑅𝑐𝑄𝑐

2 = 90− 70 = 20 So,

𝑄𝑎 = 10

8.3212

12

= 1.096

𝑄𝑏 = 10

1.3283

12

= 2.744 (−)

𝑄𝑐 = 20

61.099

12

= 0.572 (−)

𝑄 = −2.219

Continue trial with a smaller P. e.g. Trial 4, P = 81.5 m 𝑄𝑎 = 1.491 𝑄𝑏 = 1.063 (−) 𝑄𝑐 = 0.434 (−)

J

A

a

c b

B

C

100

70

80

P

20

𝑄 = −0.0055

Now check Re and get new 𝑓s. 𝑅𝑒𝑎 = 2.25 × 106 new 𝑓𝑎 = 0.0121 𝑅𝑎 = 9.1533

𝑅𝑒𝑏 = 9.97 × 105 𝑓𝑏 = 0.0125 𝑅𝑏 = 1.6603 𝑅𝑒𝑐 = 8.56 × 105 𝑓𝑐 = 0.0131 𝑅𝑐 = 69.60 With P = 81.5 m,

𝑄 = 0.0646

Try P = 81.7 m

𝑄 = −0.0079

Try P = 81.68 m 𝑄𝑎 = 1.4147 𝑄𝑏 = 1.0059 (−) 𝑄𝑐 = 0.4097 (−)

𝑄 = −0.00084 (𝑔𝑜𝑜𝑑 𝑒𝑛𝑜𝑢𝑔𝑕)

Should check Re and get new 𝑓s. But it should be very similar to last values.

3. Four pipes are connected in parallel. Their characteristics are as follows:

Pipe No. Diameter (m) Length (m) Roughness (mm)

1 0.15 3000 0.06

2 0.30 3000 0.06

3 0.45 3000 0.09

4 0.60 3000 0.09 Determine the discharge through each pipe if the total flow is 1.4 m3/s. Assume that the pipe flow is fully turbulent. Total flow = 1.4 m3/s

21

Q = 1.4 m3/s Q = 1.4 m3/s B A

[1]

[2]

[3]

[4]

Four pipes in parallel. Assume fully turbulent flows, 𝜀1

𝐷1= 0.0004, 𝑓1 = 0.016, 𝑅1 = 52228.42

𝜀2

𝐷2= 0.0002, 𝑓2 = 0.014 , 𝑅2 = 1428.12

𝜀3

𝐷3= 0.0002, 𝑓3 = 0.016, 𝑅3 = 214.93

𝜀4

𝐷4= 0.00015, 𝑓4 = 0.016, 𝑅4 = 51.00

𝑄 = 𝑄𝑖 = 1.4 𝑚3/𝑠

𝑄1 = 𝑕𝐴𝐵𝑅1

12

;𝑄2 = 𝑕𝐴𝐵𝑅2

12

;𝑄3 = 𝑕𝐴𝐵𝑅3

12

;𝑄4 = 𝑕𝐴𝐵𝑅4

12

Or

𝑕𝐴𝐵 =𝑄2

1

𝑅1

12

+1

𝑅2

12

+1

𝑅3

12

+1

𝑅4

12

2

=1.42

1

52228.42 1

2 +

1

1428.12 1

2 +

1

214.93 1

2 +

1

51.00 1

2

2

= 34.29 𝑚

Therefore,

𝑄1 = 0.026𝑚3

𝑠,𝑄2 = 0.155

𝑚3

𝑠,𝑄3 = 0.399

𝑚3

𝑠,𝑄4 = 0.82 𝑚3/𝑠

22

4. Two reservoirs have a difference in elevation of 6 m and are connected by a pipeline which consists of a single 600 mm diameter pipe 3000 m long, feeding a junction from which 2 pipes, each 300 mm diameter and 3000 m long, lead in parallel to the lower reservoir. If ƒ = 0.04, calculate the flow rate between reservoirs.

321 QQQ ------(1)

688

5

2

2

2

222

5

1

2

2

111 gD

QLf

gD

QLf

------(2)

5

3

2

2

333

5

2

2

2

222 88

gD

QLf

gD

QLf

------(3a)

Since 32 ff , 32 LL , 32 DD , so 32 QQ ------(3b)

From (2)

634.408051.1272

2

2

1 QQ ------(4)

From (1) and (3b), 21 2QQ

Substitute into (4):

634.4080451.1272

2

2

2 QQ

Therefore 0013071.02

2 Q , 0362.032 QQ m3/s, and 0723.02 21 QQ m3/s

5. Solve Q1, Q2, Q3, and Q4 using WaterCad.

6 m

Q3

Q2 Q1

D2 = 300 mm L2 = 3000m

D1 = 600 mm L1 = 3000m f = 0.04

23

TUTORIAL 5 HYDRAULICS 6713

UNIFORM FLOWS IN OPEN CHANNELS

1. The cross-section of a canal is as shown below. The canal slope is 1/4000.

a. Determine the discharge if Chezy’s C is 60 m½/s.

RSACQ

2182322332

1mA

mP 65.1536423

mP

AR 15.1

Therefore,

s

mQ

3

31.1800025.015.16018

b. Determine the discharge if Manning’s n is 0.025.

s

m

n

SARQ

321

32

21

32

50.12025.0

00025.015.118

c. What value of C corresponds to n=0.025?

s

m

n

RC

21

61

61

9.40025.0

15.1

d. What value of n corresponds to C = 60 m½/s?

017.060

15.1

60

61

61

R

n

2. A trapezoidal canal has a bottom width of 5 m, side slopes of 1:2 and a slope of 0.0004. Manning’s n is 0.014. The depth is 2 m. Determine the discharge.

n

SARQ

21

32

3 m

2 m 1:3 1:3

24

21822255 mA

mP 944.1316425

mR 29.1

Therefore,

s

mQ

321

32

49.30014.0

0004.029.118

3. Calculate for the same canal as in Problem 2 the water depth when the discharge is 75 m3/s. Answer must be accurate to the nearest cm.

n

SARQ

21

32

Since A and R are functions of depth, rearranging Mannings eqn.

32

21

ARS

Qn , where the LHS is known = 52.50

2254102

1455 yyyyyA

525425 22 yyyP

525

410

y

yR

Since Q > 30.49s

m3

(from Q2), y must be > 2 m.

y A P R R⅔ A R⅔

3 33 18.42 1.792 1.475 48.68

4 52 22.89 2.272 1.728 89.87

3.1 34.72 18.86 1.841 1.502 52.15

3.2 36.48 19.31 1.889 1.528 55.75

3.11 34.89 18.91 1.846 1.505 52.50

Ans: y = 3.11 m

4. A reinforced concrete aqueduct of rectangular cross-section is to be designed to carry 10 m3/s with a velocity of 2 m/s. Determine the water depth and the width of the cross-section so that the required slope of the aqueduct is minimized.

y

b

25

VAQ

A210 , 25mA For minimum slope, P must be minimized for a given cross section.

yy

P 25

0252

ydy

dP

52 2 y

Therefore,

my 582.1

mb 162.3

5. Design a trapezoidal cross-section canal with an area of 60 m2, a hydraulic radius of 2 m, and side slopes of 1:3.

260mA

mR 2 Side slope = 1:3

P

AR , therefore

1029230 22 yByyBmP ------------------- (1)

23)32(2

160 yByyyBBA ------------------- (2)

From (1), multiplying by y, we get:

yyBy 30325.6 2 ------------------- (3)

(3) – (2) gives:

06030325.3 2 yy

650.6

1.1030

325.32

60325.3490030

y

my 053.6 (not feasible) or m992.2

Therefore, mB 08.11992.2325.630

6. Solve Q1, Q2, and Q3 using Flowmaster.

1:3 1:3 y

B

26

TUTORIAL 6

HYDRAULICS 6713

ENERGY CONCEPTS IN OPEN CHANNEL FLOW

1. Water is flowing in a rectangular channel at a velocity of 3 m/s and a depth of 2.5 m. Determine

the changes in water surface elevation for the following alterations in the channel bottom:

a. An increase (upward step) of 20 cm, neglecting losses.

s

mv 3

my 5.21

Check whether flow is sub or supercritical.

61.05.281.9

31

gy

vFr , subcritical (water level will drop or encounter a hump)

21 EhE

2

2

2

22

1

2

12

2.02 gy

qy

gy

qy

2

2

2

22

2

81.92

)5.23(2.0

5.281.92

)5.23(5.2

yy

2

2

22

734.52.0959.2

yy

2

2

2

867.2759.2

yy or 0867.2759.2)(

2

2

3

22 yyyf

Using Newton’s Method:

)('

)(

2

20,2,2

yf

yfyy n , where 2

2

22 518.53)(' yyyf

y2 y1

h = 0.2 m

TEL (2) (1)

27

y2,0 f(y2) f’(y2) y2,n

2.0 -0.169 0.964 2.175

2.175 0.104 2.190 2.128

2.128 0.0096 1.843 2.123

2.123 0.0005 1.807 2.123

Therefore, y2 = 2.123 m

b. The maximum increase allowable for the specified upstream flow conditions to

remain unchanged, neglecting losses.

mg

qyc 79.13

2

, mEc 684.279.12

3

Therefore,

mEEh c 274.0684.2959.2max 1

c. A “well-designed” decrease (downward step) of 20 cm.

21 2.0 EE

2

2

2

22

5.72.0959.2

gyy

2

2

2

867.2159.3

yy , or 0867.2159.3)(

2

2

3

22 yyyf

2

2

22 318.63)(' yyyf

y2,0 f(y2) f’(y2) y2,n

2.7 -0.479 4.811 2.8

2.8 0.052 5.830 2.79

2.79 -0.0053 5.725 2.791

2.791 0.0004 5.735 2.791

Therefore, y2 = 2.791 m

28

2. Water is flowing in a rectangular channel whose width is 5 m. The depth of flow is 2 m and the

discharge is 25 m3 /s. Determine the changes in depth for the following alterations in the

channel width:

a. An increase of 50 cm, neglecting losses.

s

mQ

3

25 , mb 51

Therefore,

msmq //5 3

1 , my 21

Check:

56.03

gy

qFr , subcritical therefore with decrease in q, depth increases

Neglecting energy losses, 21 EE

2

2

2222

1

2

112

319.222

252

2 gy

qym

ggy

qyE

msmq //545.45.5

25 3

2

Therefore,

2

2

2

22

545.4319.2

gyy , or 0053.1319.2)(

2

2

3

22 yyyf

2

2

22 638.43)(' yyyf

y2,0 f(y2) f’(y2) y2,n

2.0 -0.223 2.724 2.08

2.08 0.019 3.332 2.074

2.074 -0.0009 3.2852 2.074

my 074.22

q2 q1 5 m 5.5 m

(2) (1)

29

b. A decrease of 25 cm, assuming a “well-designed” transition.

A decrease in width means an increase in depth.

21 EE

2

2

2

22

75.4/25319.2

gyy

419.1319.2)(2

2

3

22 yyyf

2

2

22 638.43)(' yyyf

y2,0 f(y2) f’(y2) y2,n

2.0 0.143 2.724 1.948

1.948 0.0112 2.3493 1.943

1.943 -0.00049 2.3141 1.943

my 943.12

c. The maximum decrease allowable for the specified upstream flow conditions to

remain unchanged, neglecting losses.

For no change in upstream flow condition, mEyc 546.13

2

Maximum q at section 2 should be,

msmgqm //02.6)773.0(546.12 32 Therefore,

mq

Qbc 152.4

02.6

25

Therefore,

Maximum decrease

m848.0152.45

30

3. A lake discharges into a steep channel. At the channel entrance the lake level is 2.5 m above the

channel bottom. Neglecting losses, find the discharge for the following geometries:

a. Rectangular section, b = 4 m.

mb 4 Rectangular channel.

At the channel entrance, depth =yc

Assuming no losses , cEE 1 (since v1=0)

Therefore, mEy cc 667.15.23

2

3

2

At critical flow, msmgyq c //74.6667.181.9 333

Therefore, smQ /96.26474.6 3

b. Trapezoidal section, b = 3 m, side slopes = 1:2.5.

2

22

22 gA

Qy

g

vyE c

ccc

At critical depth, 13

2

gA

BQ

Therefore,

c

cc

cccy

yy

yB

AyE

5.25.232

5.25.22

13

2

2

c

ccc

y

yyy

106

5.235.2

2

225.231062515 ccccc yyyyy

E1 2.5 m yc

(1)

(2)

1: 2.5

B

3 m

yc

31

015165.122

cc yy

Therefore, myc 909.125

72.3116

25

5.1215425616

22 83.14909.15.2909.13 mA

mB 09.25909.1106

s

m

B

gAQ

333

71.3509.25

83.1481.9

32

TUTORIAL 7

HYDRAULICS 6713

MOMENTUM CONCEPTS IN OPEN CHANNEL FLOW

1. A 3 m wide rectangular channel carries 15 m3/s of water at 0.6 m depth before entering a

hydraulic jump. Compute the downstream water depth and the critical depth.

msmq //53

15 3

Critical depth, mg

qyc 60.1

81.9

53

2

3

2

smy

qv /33.8

6.0

5

1

1

43.36.0

33.81

gFr , supercritical

mFry

y 38.4143.3812

6.0181

2

22

11

2

2. A long rectangular channel 3 m wide carries a discharge of 15 m3 /s. The channel slope is 0.004

and the Manning’s roughness coefficient is 0.01. At a certain point in the channel where the

flow reaches the normal depth,

a. Determine the state of the flow. Is it supercritical or subcritical?

msmq //53

15 3

From question 1, myc 60.1

From Manning’s eqn, n

SARQ

21

32

byA 11 , by

by

P

AR

1

1

1

11

2 , mb 3

Therefore, 21

32

1

11 004.0

32

33

01.0

115

y

yy

Solving for y1, my 08.11 , 42.13

1 gy

qFr

Since cyy 1 , flow is supercritical.

33

b. If a hydraulic jump takes place at this depth, what is the sequent depth at the jump?

mFry

y 698.1142.1812

08.1181

2

22

11

2

c. Estimate the energy head loss through the jump.

Head loss

myy

yyE 032.0

08.17.14

62.03

21

3

12

Or WEQP 4709032.0159810

3. A spillway, as shown, has a flow of 3 m3 /s per meter of width occurring over it. What depth y2

will exist downstream of the hydraulic jump? Assume there is no energy loss over the spillway.

For no losses:

2

1

2

12

0

2

022 gy

qy

gy

qy

2

1

2

12

2

2

3

52

35

gyy

g

my 312.01

49.5312.0

3

33

1

1

ggy

qFr

Therefore, my

y 27.2149.5812

212

(0) (1) (2)

5 m

y1

y2

34

TUTORIAL 8

HYDRAULICS 6713

DESIGN AND ANALYSIS OF CULVERTS

1. A rectangular concrete conduit is to be used as a culvert on a slope of 0.02. The culvert is 15 m

long and has a cross-section of 2.13 m x 2.13 m. If the tail water elevation is 1.8 m above the

crown at the outlet, determine the head water elevation necessary to pass a 10 m3/s discharge.

Assume a square-edged entrance (Ke = 0.5).

“outlet control”

mL 15

013.0n

5.0eK

mR 5325.013.24

13.2 2

22 5369.413.2 mA

HWLSTWh oL

LSTWHWh oL

1502.093.3 HWhL

63.3 HWhL

HW

SoL

S = 0.02

Q = 10 m3/s

TW

hL

1.8 m

2.13 m

2.13 m

2.13 m

35

2

2

34

2

212

gA

Qg

R

LnKh eL

2

2

34

2

5369.42

1012

5325.0

15013.05.0

gghL

24762.011152.05.0 Lh

mhL 39995.0

Therefore,

mHW 03.439995.063.3

mHWelevation 33.43.003.4

[Using Culvert Master, HW elevation = 4.327m]

(see attached printout)

2. A culvert is 11 m long and has upstream and downstream inverts of 263.4 and 263.1 meters,

respectively. The downstream tailwater is below the downstream pipe invert.

a. For a square-edged entrance and Manning’s n of 0.013, what is the minimum diameter

for a concrete circular culvert (in mm) required to pass 1.4 m3/s under a roadway with a

maximum allowable headwater elevation of 265.2 m?

b. What is the headwater elevation for the selected culvert?

For the given condition Inlet control

Use orifice equation,

ghACQ d 2

265.2 m

263.1 m

h

265.4 m D Q = 1.4 m3/s TW

36

28.1

24.2632.265

DDh

Assume 62.0dC

28.12

462.04.1

2 Dg

D

Squaring both sides and simplifying,

28.16532.496.1 4 D

D -----------(1)

By trial, mD 736.0

Closest size available is probably 0.75m,

Therefore use mD 75.0

With mD 75.0 , from (1)

2

75.075.06532.496.1 4 HW

mHW 7062.1

Or mHWelevation 106.265

Using Culvert Master, mD 75.0 , mHWelvation 131.265

Reason for the difference is that Culvert Master uses a slightly different form of equation for inlet

control (if Cd = 0.61, we get the same answer as Culvert Master).

3. Twin 1220 by 910 mm box culverts (n = 0.013, 90° and 15° wingwall flares entrance) carry 8.5

m3/s along a 31 m length of pipe constructed at a 1.0 percent slope. The tailwater depth is 0.61

m.

a. What is the headwater depth?

b. Are the culverts under inlet or outlet control conditions?

Best to use Culvert Master:

37

a. Headwater depth = 2.634 m (remember to subtract the upstream invert elevation from the

headwater elevation).

b. Culvert under inlet control.

[See attached printout from Culvert Master]

4. A 12.2 m long 920 by 570 mm concrete arch pipe (n =0.013, groove-end with headwall entrance)

constructed at a 0.8 percent slope carries 1.84 m3/s.

a. If there is a constant tailwater depth of 0.3 m, what is the headwater depth for both

inlet and outlet control conditions?

b. Is the culvert flowing under inlet or outlet control conditions?

c. What would be the result if the tailwater was 0.5 m deeper?

Use Culvert Master:

a. Inlet control headwater depth is 2.34 m, outlet control headwater depth is 2.13 m.

b. Culvert is flowing under inlet control.

c. If TW is 0.5 m deeper, we get outlet control and headwater depth is 2.36 m.

[See printouts from Culvert Master]

5. Twin culverts are proposed to discharge 6.5 m3/s. The culverts will be 36.6 m long and have

inverts of 20.1 and 19.8 m. The design engineer analyzed the following three culvert systems.

Which of the following proposed culverts will result in the highest headwater elevation? The

lowest? Tailwater elevation is below the downstream invert. [Hint: Use Culvert-Master to either

solve the problems or use it to check your solutions].

a. 1200 mm circular concrete pipes (n = 0.013, square-edged entrance);

b. 1200 x 910 mm concrete box culverts (n = 0.013, 90° and 15° wingwall flares entrance);

c. 1630 x 1120 mm steel and aluminum arches (n = 0.025 and Ke= 0.5).

Use Culvert Master:

a. 1200 mm circular concrete pipes (n = 0.013, Ke = 0.5), HW elevation = 21.92 m

b. 1220 x 910 mm concrete box culverts (n = 0.013, 90° and 15° wingwall flares entrance), HW

elevation = 21.94 m.

c. 1630 x 1120 mm steel and aluminum arches (n= 0.025, Ke = 0.5), HW elevation = 21.70 m.

d. Therefore, box culverts have the highest headwater; the arches have the lowest.

(See attached printouts from Culvert Master).

38

TUTORIAL 9

HYDRAULICS 6713

PUMPS

1. From the manufacturer’s data, a pump of 254 mm impeller diameter has a capacity of 76 L/s at

a head of 18.6 m when operating at a speed of 900 rpm. It is desired that the capacity be about

95 L/s at the same efficiency. Determine the adjusted speed of the pump and the corresponding

head.

mmD 2541 , slQ /761 , mH 6.181 , rpmN 9001 , slQ /952

Therefore,

2

1

1

2

1

2

1

2 25.1

H

H

N

N

Q

Q

rpmN 112590025.12

mH 06.296.1825.1 2

2

2. The following performance curves were obtained from a test on a 216 mm double entry

centrifugal pump moving water at a constant speed of 1350 rpm:

Q (m3/min) 0 0.454 0.905 1.36 1.81 2.27 2.72 3.8

H (m) 12.2 12.8 13.1 13.4 13.4 13.1 12.2 9.0

η 0 0.26 0.46 0.59 0.70 0.78 0.78 0.74

Plot H vs. Q and η vs. Q. If the pump operates in a system whose demand curve is given by H =

5+ Q2, find the operating point of the pump and the power required. In the demand curve, Q is

given in m3/min.

39

At operating point, min

72.23m

Q , mH 2.12 , 78.0

Input power = KWQH

96.678.0

12.12

60

72.29810

0

2

4

6

8

10

12

14

16

18

20

22

0 0.5 1 1.5 2 2.5 3 3.5 4

H (

m)

Q (m3/s)

Pump Curve

System Curve

operating point

0

20

40

60

80

100

0 0.5 1 1.5 2 2.5 3 3.5 4

η%

Q (m3/s)

Efficiency

40

3. With reference to the pump data in Problem 2, if the pump is run at 1200 rpm, find the

discharge, head, and required power.

rpmN 13501 , rpmN 12002

89.01350

1200 3

1

1

22

1

1

2

1

2

1

2

P

P

H

H

Q

Q

N

N

Therefore,

min42.272.289.0

3

2

mQ

mH 66.92.1289.0 2

2

KWP 91.496.689.0 3

2

4. Water is pumped between two reservoirs in a pipeline with the following characteristics: D =

300 mm, L = 70 m, ƒ = 0.025, ΣK = 2.5. The radial flow pump characteristic curve is approximated

by the formula:

Hp = 22.9 + 10.7Q – 111Q2

Where Hp is in meters and Q is in m3/s. Determine the discharge Q and pump head H for the

following situations:

a. Total static head = 15 m, one pump placed in operation;

b. Total static head = 15 m, with two identical pumps operating in parallel;

c. Total static head = 25 m.

System Curve:

4

2

5

281.0

D

QK

D

fLQ

gHH sp

4

2

5

2

3.0

5.2

3.0

70025.0

81.9

81.0 QQHH sp

295.84 QHH sp

Pump Curve: 21117.109.22 QQH p

a. mH s 15 (one pump)

41

Operating point when: 22 1117.109.2295.8415 QQQ

Or,

09.77.1095.195 2 QQ

s

mQD

32

23.09.391

41.797.10

95.1952

9.795.19547.107.10

Therefore, operating head,

mHo 49.1923.095.8415 2

b. For two pumps in parallel:

2

2

75.2735.59.222

1112

7.109.22 QQQQ

H p

Equating this to the system curve, 2

0

275.2735.59.228515 QQQ oo

09.735.58.1122

oo QQ

s

mQo

3

29.0

Therefore,

mHo 2.2229.08515 2

c. Since the static head is greater than the single pump shut off head (ie. 25 > 22.9), it is

necessary to operate with two pumps in series. The combined pump curve is:

22 2224.218.451117.109.222 QQQQH

The system demand curve is changed since Hs = 25m. It becomes: 28525 QHo

Equating the pump curve and system curve, we get: 22

2224.218.458525 ooo QQQ

08.204.213072

oo QQ

s

mQo

3

3.0

Therefore,

mHo 7.323.08525 2

42

5. A pumping system is to deliver 28.3 L/s of water at 15°C. The suction line is 152 mm in diameter

in a 91 m long cast iron pipe. The suction inlet is 6 m above the reservoir level. The atmospheric

pressure of 101 kPa exists over the reservoir. The required NPSH of the pump is 2 m. Determine

whether the system will have a cavitation problem. (Vapour pressure at 15°C is 16.8 kPa,

kinematic viscosity of water is 1.14 x 10-6 m2/s).

Lsvatm hH

PPNPSH

s

m

A

Qv 56.1

152.04

0283.0

2

5

61008.2

1014.1

152.056.1Re

vD

0016.0152.0

00024.0

D

From Moody Diagram, 023.0f

mgg

v

D

fLh f 708.1

2

56.1

152.0

91023.0

2

22

Minor losses, (assume 1 exit and 1 bend)

4.19.05.0 K

Therefore,

mgg

vhm 174.0

2

56.14.1

24.1

22

And,

mNPSH 24.2174.0708.1681.9

68.1101

Since NPSH > 2m (required), there is no cavitation problem.