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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 50C (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? ( = 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, = 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 6.7592/mKN
If welded steel pipes are used, = 0.000046 m
Therefore 00091.005.0
000046.0
D
From Moody diagram, with Re = 41056.9 and 00091.0D
,
= 0.022 and 70.42
934.0
05.0
240022.0
2
g
h f m
89.9081.9860AP 8.7662/mKN
EL = 21 m PB = 550 KN/m
2 L = 240 m
D = 50 mm
= 0.00042 Ns/m2
= 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 20C.
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 = 0.02 (arbitrary value)
From (1) 92.0D m
From (2) 51074.2Re
From (3) 00027.0D
From Moody diagram, new = 0.017
From (1) 89.0D m
From (2) 51084.2Re
From (3) 000274.0D
From Moody diagram, new = 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 20C 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 Mannings 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 30C. 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 ,
Mannings 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 f
hS
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 20C water)
Using Swamee and Jains 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 =30mm. A 12-mm-thick lining would reduce the roughness to =1mm. How much in annual pumping costs would be saved per kilometer of pipe for water at 20C 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, =30mm, 91.1
4
2
62
A
Qv m/s
Head loss of old pipe = fgfh f 96.92
2
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, = 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 20C 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) 544821.601121.1
48.78DD
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 10C, 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
22
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
196222
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
196222
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 10C 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 10C. 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)
2121 05.51694.10135.0 vffH
22
22
21
75.0
)4(05.51694.10135.0
QffH
212 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
51 1018.5Re
016.01 f
52 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
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
Q = 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
Q = -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
Q = 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
Q = 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
Q = 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
Q = 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
Q = 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
Q = 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
Q = 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
Q = 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 = 0.05 mm and water temperature at 20C. 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 when near correct answer)
70
75
80
85
90
-2 -1 0 1 2
P
Q
J
A
a
c b
B
C
19
Assume fully turbulent flow Initial estimates:
=
0.05
800= 6.3 105, = 0.011, = 8.3212
=0.05
1200= 4.2 105, = 0.010 , = 1.3282
=0.05
600= 8.3 105, = 0.0115, = 61.099
Trial 1 (assume P = 90 m) =
2 = 100 90 = 10 =
2 = 90 80 = 10 =
2 = 90 70 = 20 So,
= 10
8.3212
12
= 1.096
= 10
1.3283
12
= 2.744 ()
= 20
61.099
12
= 0.572 ()
= 2.219
Continue trial with a smaller P. e.g. Trial 4, P = 81.5 m = 1.491 = 1.063 () = 0.434 ()
J
A
a
c b
B
C
100
70
80
P
20
= 0.0055
Now check Re and get new s. = 2.25 10
6 new = 0.0121 = 9.1533 = 9.97 10
5 = 0.0125 = 1.6603 = 8.56 10
5 = 0.0131 = 69.60 With P = 81.5 m,
= 0.0646
Try P = 81.7 m
= 0.0079
Try P = 81.68 m = 1.4147 = 1.0059 () = 0.4097 ()
= 0.00084 ( )
Should check Re and get new s. 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, 11
= 0.0004, 1 = 0.016, 1 = 52228.42
22
= 0.0002, 2 = 0.014 , 2 = 1428.12
33
= 0.0002, 3 = 0.016, 3 = 214.93
44
= 0.00015, 4 = 0.016, 4 = 51.00
= = 1.4 3/
1 = 1
12
;2 = 2
12
;3 = 3
12
;4 = 4
12
Or
=2
1
11
2 +
1
21
2 +
1
31
2 +
1
41
2
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
Therefore,
1 = 0.0263
,2 = 0.155
3
,3 = 0.399
3
,4 = 0.82
3/
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 = 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 m
3/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 Chezys C is 60 m/s.
RSACQ
2182322332
1mA
mP 65.1536423
mP
AR 15.1
Therefore,
s
mQ
3
31.1800025.015.16018
b. Determine the discharge if Mannings 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 m/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. Mannings 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 R A R
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 Newtons 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 31 , 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 32
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(546.1232
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 //74.6667.181.9333
Therefore, smQ /96.26474.63
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
mFryy 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 Mannings 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 Mannings 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 , 42.131
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?
mFryy 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 Mannings 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
DD -----------(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 manufacturers 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 22
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
0 0.26 0.46 0.59 0.70 0.78 0.78 0.74
Plot H vs. Q and 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
%
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 31
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 22
KWP 91.496.689.0 32
4. Water is pumped between two reservoirs in a pipeline with the following characteristics: D =
300 mm, L = 70 m, = 0.025, K = 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 49.1923.095.84152
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 2.2229.085152
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 7.323.085252
42
5. A pumping system is to deliver 28.3 L/s of water at 15C. 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 15C is 16.8 kPa,
kinematic viscosity of water is 1.14 x 10-6 m2/s).
Lsvatm hHPP
NPSH
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.