CHAPTER 12
ENERGY PRINCIPLES IN OPEN CHANNELS
12.1. A trapezoidal channel a bed width of 6-m and a side slope of 1:1 is discharging
water at a rate of 8.0 m3/s. Calculate the specific energy of water if the depth of f
...
CHAPTER 12
ENERGY PRINCIPLES IN OPEN CHANNELS
12.1. A trapezoidal channel a bed width of 6-m and a side slope of 1:1 is discharging
water at a rate of 8.0 m3/s. Calculate the specific energy of water if the depth of flow is
2.0-m. What is the alternate depth to the observed depth?
Solution: E = y + 2
2
2gA
Q
A = (6.0 + 1 x 2.0) x 2.0 = 16.0 m2
E = 2.0 + 2.013 m
2 x 9.81 x (16.0)
(8.0)
2
2
=
Then 2.013 = y2 +
2 2
19.62 x (6 x y 2 y 2 )
64
+
= y2 +
2 2
(6 x y 2 y 2 )
3.26
+
Whence y2 = 0.215 m
12.2. For a trapezoidal channel with a bed width, b = 6.0 ft and a side slope, t = 2, find
the alternate depth if: a) Q = 150 ft3/s, b) Q = 75 ft3/s, and c) Q = 300 ft3/s
Solution: where
AB
A
Q g
= ,
A = by + ty2 = 6y + 2y2 = 2y (3+y)
B = b + 2ty = 6 + 4y = 2(3+2y)
a)
c c
c
c c
yy
y
y y
3 2
3
2 (3 )
32.2
150 2
+
+
= +
13.217 =
c c
c
c c
yy
y
y y
3 2
3
(3 )
2
2
+
+
+
Solving by trial and error we get yc = 2.115 ft.
b)
c c
c
c c
yy
y
y y
3 2
3
2 (3 )
32.2
75 2
+
+
= +
c c
c
c c
yy
y
y y
3 2
3
6.608 (3 )
2
2
+
+
= +
Solving by trial and error, we get yc = 1.43 ft.
c)
c c
c
c c
yy
y
y y
3 2
3
2 (3 )
32.2
300 2
+
+
= +Elementary Hydraulics
202
c c
c
c c
yy
y
y y
3 2
3
26.434 (3 )
2
2
+
+
= +
Solving by trial and error, we get yc = 3.06 ft.
12.3. A cement lined rectangular channel 6.0 m wide carries water at a rate of 11.0 m3/s.
Determine the critical depth and the critical velocity.
Solution: q = Q/b = 11/6 = 1.833 m2/s
yc = 0.67 m
9..81
(1.67)
1/ 3
2
3
2
⎟ ⎟ =
⎞ ⎠
qg = ⎜ ⎜ ⎝ ⎛
Vc = q/yc = 1.67/0.657 = 2.62 m/s
12.4. A broad-crested weir of a height h is placed in a channel of width b. If the upstream
depth is y1 and the upstream velocity head and frictional losses are neglected, develop a
theoretical equation for the discharge in terms of the upstream depth of flow. Assume the
flow over the broad-crested weir to be critical.
Solution:
Applying the Bernoulli equation,
y1 +
g
V
h y
g
V
c
c
2 2
2 2
1
= + +
Neglecting the velocity head at section 1,
y1 = h + 3/2 yc or yc = 2/3(y1-h)
3
1
yc3 = q 2 / g = [2 / 3(y − h)] , or
3
1
2 3
( )
23
y h
qg
⎟ −
⎞ ⎠
⎛⎜⎝
= , which can be written as
q = 3 / 2
1
3 / 2
( )
23
g ⎟ y − h
⎞ ⎠
⎛⎜⎝
Therefore, Q = b 3 / 2
1
3 / 2
( )
23
g ⎟ y − h
⎞ ⎠
⎛⎜⎝
And for g = 9.81 m2/s, Q = 1.706 b (y1-h)3/2
12.5. A trapezoidal channel having a bed width of 7.0 ft and a side slope of 3:2 is
discharging 300 ft3/s of water. If the specific energy is equal 5.7 ft, calculate the alternate
depths and their corresponding slopes to satisfy these conditions. Take n = 0.025.
Solution: A = 7.0 y + 1.5 y2
2 2 2 2
2
2
2
(7y 1.5y )
1397.5
y
(7y 1.5y )
(500)
2
Q
+
= +
+
= + = +
g
y
gA
E y = 5.7 ft.
By trail and error, y1 = 5.5 ft, and y2 = 1.95 ft.Chapter 12. Energy Principles in Open Channels
203
Then, A1 = 83.875, P1 = 26.83, R1 = 3.126, V1 = 3.57 ft/s
S1 = 2
2/3 )
1.49 x (3.126)
0.025 x 3.57
( = 0.000785 ft/ft
A2 = 19.35, P2 = 14.02, R2 = 1.38, V1 = 15.5 ft/s
S2 = 2
2/3 )
1.49 x (1.
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