5. 14
Problem 4:
Rectangular channel q= 100/5= 20 cfs/ft
y
(m)
E
(m)
0.50 25.34
1.00 7.21
1.50 4.26
2.00 3.55
2.50 3.49
3.00 3.69
3.50 4.01
4.00 4.39
5.00 5.25
6.00 6.17
7.00 7.13
8.00 8.10
9.00 9.08
y= 2.0 ft is on the supercritical limb.
Trapezoidal channel
Q= 200 cfs, b= 6.0 ft, m=2
y
(ft)
A
(ft2
)
V
(fps)
E
(ft)
0.50 3.50 57.14 51.20
1.00 8.00 25.00 10.70
1.50 13.50 14.81 4.91
2.00 20.00 10.00 3.55
2.50 27.50 7.27 3.32
3.00 36.00 5.56 3.48
3.50 45.50 4.40 3.80
4.00 56.00 3.57 4.20
5.00 80.00 2.50 5.10
6.00 108.00 1.85 6.05
7.00 140.00 1.43 7.03
8.00 176.00 1.14 8.02
9.00 216.00 0.93 9.01
10.00 260.00 0.77 10.01
y= 2.8 is on the subcritical limb.
0.0
2.0
4.0
6.0
8.0
10.0
0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0
E (ft)
y
(ft)
0.0
2.0
4.0
6.0
8.0
10.0
0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0
E (ft)
y
(ft)
6. 15
Triangular channel
Q= 45 cfs, m= 1.5
y
(ft)
A
(ft2
)
V
(fps)
E
(ft)
0.50 0.38 120.00 224.10
1.00 1.50 30.00 14.98
1.50 3.38 13.33 4.26
2.00 6.00 7.50 2.87
2.25 7.59 5.93 2.80
2.50 9.38 4.80 2.86
3.00 13.50 3.33 3.17
3.50 18.38 2.45 3.59
4.00 24.00 1.88 4.05
5.00 37.50 1.20 5.02
6.00 54.00 0.83 6.01
7.00 73.50 0.61 7.01
8.00 96.00 0.47 8.00
9.00 121.50 0.37 9.00
10.00 150.00 0.30 10.00
y= 2.0 m is on the supercritical limb.
Circular Channel
d0= 5 ft and Q= 100 cfs
y
(ft) θ
A
(ft2
)
E
(ft)
1.5 1.16 4.95 7.83
1.8 1.29 6.36 5.63
2.1 1.41 7.83 4.64
2.4 1.53 9.32 4.19
2.7 1.65 10.82 4.03
3.0 1.77 12.30 4.03
3.3 1.90 13.75 4.12
3.6 2.03 15.13 4.28
3.9 2.17 16.43 4.48
4.2 2.32 17.61 4.70
4.5 2.50 18.61 4.95
4.8 2.74 19.37 5.21
y= 3.2 ft is on the subcritical limb.
0
1
2
3
4
5
0 1 2 3 4 5 6 7 8 9 10
E (ft)
y
(ft)
0.0
2.0
4.0
6.0
8.0
10.0
0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0
E (ft)
y
(ft)
7. 16
Problem 5:
No. The streamlines curve up. Pressure distribution can not be hydrostatic if the flow has
a vertical velocity component.
Problem 6:
Equation 2.12 is
2
2
2gA
Q
y
E +
=
Taking the derivative of both sides with respect to x
dx
dy
dy
dA
gA
Q
dx
dy
dx
dA
gA
Q
dx
dy
A
dx
d
g
Q
dx
dy
gA
Q
dx
d
dx
dy
dx
dE
3
2
3
2
2
2
2
2
2
2
)
1
(
2
)
2
( −
=
−
=
+
=
+
=
)
1
( 2
2
2
r
r F
dx
dy
dx
dy
F
dx
dy
dx
dy
T
gA
V
dx
dy
dx
dE
−
=
−
=
−
=
)
1
(
/
2
r
F
dx
dE
dx
dy
−
=
Problem 7:
2
2
2gA
Q
y
E +
=
As y approaches infinity so does A, and the second term in the expression for E
approaches zero. Then, as y approaches infinity, the upper limb becomes asymptotical to
the straight line E=y. This straight line makes a 45º angle with each of the E and y axes.
As y approaches zero so does A, and the second term approaches infinity. Therefore, the
lower limb is asymptotical to the E axis, which can also be defined by y= 0.
Problem 8:
From Figure 2.11, the minimum specific energy, EBmin, at section B required to pass 290
cfs is 4.0 ft. The corresponding minimum specific energy at A is EAmin= 4.0 +1.0= 5.0 ft.
The flow has a specific energy of 5.5 ft at A under the given flow conditions. This is
larger than the required specific energy.
9. 18
Problem 10:
(a) Recall from Equations 2.15, 2.3 and 2.17 that for rectangular channels
2
2
2gy
q
y
E +
=
3
2
g
q
yc =
c
c E
y
3
2
=
In this problem ∆z= 0.5 ft, yA= 3.0 ft, qA= 60/12= 5 cfs/ft and qB= 60/6= 10 cfs/ft.
At section A
ft
EA 04
.
3
)
0
.
3
)(
2
.
32
(
2
)
0
.
5
(
0
.
3 2
2
=
+
=
At section B
ft
1.46
2
.
32
)
10
(
3
2
=
=
cB
y
ft
y
E c
B 19
.
2
)
46
.
1
)(
2
/
3
(
)
2
/
3
(
min =
=
=
The corresponding specific energy required at A is
ft
z
E
E B
A 69
.
2
5
.
0
19
.
2
min
min =
+
=
∆
+
=
Because EA>EAmin, choking will not occur. Then EB=(EA – ∆z) and
2
2
2
2
2
55
.
1
)
2
.
32
(
2
)
10
(
2
54
.
2
50
.
0
04
.
3
B
B
B
B
B
B
B
y
y
y
y
gy
q
y +
=
+
=
+
=
=
−
By trial and error yB= 2.23 ft.
(b) Givens are ∆z= 1.0 ft, yA= 3.0 ft, qA= 60/12= 5 cfs/ft and qB= 60/6= 10 cfs/ft.
As in part (a) we can determine EA= 3.04 ft and EBmin= 2.19 ft. Also,
ft
z
E
E B
A 19
.
3
0
.
1
19
.
2
min
min =
+
=
∆
+
=
Because EA>EAmin, choking will occur, the depth at B will be equal to the critical depth
1.46 ft, and flow at A will adjust to acquire the minimum required specific energy.
10. 19
2
2
2
2
2
39
.
0
)
2
.
32
(
2
)
5
(
2
19
.
3
Aadj
Aadj
Aadj
Aadj
Aadj
A
Aadj
y
y
y
y
gy
q
y +
=
+
=
+
=
By trial and error we find yAadj= 3.15 ft.
Problem 11:
Q
(cfs)
y
(ft)
A
(ft2
)
YCA
(ft3
)
V
(fps)
M
(ft3
)
145 1.00 8.00 3.67 18.13 85.29
145 2.00 20.00 17.33 7.25 49.98
145 3.00 36.00 45.00 4.03 63.14
145 4.00 56.00 90.67 2.59 102.33
435 1.00 8.00 3.67 54.38 738.24
435 2.00 20.00 17.33 21.75 311.16
435 3.00 36.00 45.00 12.08 208.24
435 4.00 56.00 90.67 7.77 195.61
Problem 12:
Q
(cfs)
y
(ft)
θ
(rad.)
A
(ft2
)
AYC
(ft3
)
M
(ft3
)
10.00 0.50 0.84 0.77 0.16 4.17
10.00 1.00 1.23 2.06 0.85 2.36
10.00 1.50 1.57 3.53 2.25 3.13
10.00 2.00 1.91 5.01 4.39 5.01
20.00 0.50 0.84 0.77 0.16 16.20
20.00 1.00 1.23 2.06 0.85 6.88
20.00 1.50 1.57 3.53 2.25 5.76
20.00 2.00 1.91 5.01 4.39 6.87
Problem 13:
Givens: d0= 36 inches= 3.0 ft. Q= 20 cfs, yJ1= 1.0 ft.
Q/(g0.5
d0
2.5
)= 20/[(32.2)0.5
(3.0)2.5
]= 0.226
yJ1/d0= 1.0/3.0= 0.33
From Figure 2.29, yJ2/d0= 0.67. Therefore, yJ2= 0.67 (3.0)= 2.0 ft
Let us now verify mathematically if the hydraulic jump equation (Equation 2.21) is
satisfied. For d0= 36 inches= 3.0 ft. Q= 20 cfs, yJ1= 1.0 ft, by using the expressions from
Table 2.1, we obtain θ= 1.23 rad., A= 2.06 ft2
, and AYC= 0.85 ft3
. Then, by using
11. 20
Equation 2.19, we find MJ1= 6.88 ft3
. Likewise for d0= 36 inches= 3.0 ft. Q= 20 cfs, yJ2=
2.0 ft, by using the expressions from Table 2.1, we obtain θ= 1.91 rad., A= 5.01 ft2
, and
AYC= 4.39ft3
. Then, by using Equation 2.19, we find MJ2= 6.88 ft3
. Because MJ1= MJ2 the
jump equation is satisfied with yJ2= 2.0 ft.
Problem 14:
Givens: d0= 1.0m, Q= 0.75 m3
/s, yJ1= 0.30m.
Q/(g0.5
d0
2.5
)= 0.75/[(9.81)0.5
(1.0)2.5
]= 0.24
yJ1/d0= 0.30/1.0= 0.3
From Figure 2.29, we find yJ2/d0= 0.76. Therefore, yJ2= 0.76 (1.0)= 0.76 m.
Let us now verify mathematically if the hydraulic jump equation (Equation 2.21) is
satisfied. For d0= 1.0m, Q= 0.75 m3
/s, yJ1= 0.30m, by using the expressions from Table
2.1, we obtain θ= 1.16 rad., A= 0.20 m2
, and AYC= 0.02 m3
. Then, by using Equation
2.19, we find MJ1= 0.31 m3
. Likewise for d d0= 1.0m, Q= 0.75 m3
/s, yJ2= 0.76m, by using
the expressions from Table 2.1, we obtain θ= 2.12 rad., A= 0.64 m2
, and AYC= 0.31m3
.
Then, by using Equation 2.19, we find MJ2= 0.31 m3
. Because MJ1= MJ2 the jump
equation is satisfied with yJ2= 0.76 m.
Problem 15:
(a) Givens: Trapezoidal channel, Q= 200 cfs, b= 6.0 ft, m= 2, yJ1= 1.0 ft.
To use Figure 2.28 let us evaluate the dimensionless parameters
13
.
1
)
0
.
6
(
)
2
.
32
(
)
0
.
2
)(
200
(
5
.
2
5
.
0
5
.
1
5
.
2
5
.
0
5
.
1
=
=
b
g
Qm
and
33
.
0
0
.
6
)
0
.
1
)(
0
.
2
(
1
=
=
b
myJ
From Figure 2.28, we obtain myJ2/b= 1.6. Then yJ2= 1.6(6.0)/2.0= 4.80 ft. Let us now
determine if this depth satisfy the jump equation, MJ1= MJ2
By using the expressions given in Table 2.1
3
2
2
1
1
2
1
1
1
2
1
159
)]
0
.
6
(
3
)
0
.
1
)(
0
.
2
(
2
[
6
)
0
.
1
(
)
0
.
1
)](
0
.
1
)(
0
.
2
(
0
.
6
[
2
.
32
)
200
(
)
3
2
(
6
)
(
ft
M
b
my
y
y
my
b
g
Q
M
J
J
J
J
J
J
=
+
+
+
=
+
+
+
=
Likewise for the trial value of yJ2= 4.8 ft
12. 21
3
2
2
2
2
2
2
2
2
2
2
159
)]
0
.
6
(
3
)
8
.
4
)(
0
.
2
(
2
[
6
)
8
.
4
(
)
8
.
4
)](
8
.
4
)(
0
.
2
(
0
.
6
[
2
.
32
)
200
(
)
3
2
(
6
)
(
ft
M
b
my
y
y
my
b
g
Q
M
J
J
J
J
J
J
=
+
+
+
=
+
+
+
=
Because MJ1= MJ2 we conclude that yJ2= 4.80 ft.
(b) The head loss due the hydraulic jump is calculated using hLj= EJ1-EJ2 or
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
= 2
2
2
2
2
2
1
1
2
1
]
)
[(
2
]
)
[(
2 J
J
J
J
J
J
Lj
y
my
b
g
Q
y
y
my
b
g
Q
y
h
ft
hLj 79
.
5
]
80
.
4
))
80
.
4
(
2
0
.
6
)[(
2
.
32
(
2
200
80
.
4
]
0
.
1
))
0
.
1
(
2
0
.
6
)[(
2
.
32
(
2
200
0
.
1 2
2
2
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
=
Problem 16:
(a) Givens: Trapezoidal channel, Q= 1320 m3
/s, b= 10.0m, m= 2, yJ2= 13.3m.
To use Figure 2.28 let us evaluate the dimensionless parameters
77
.
3
)
10
(
)
81
.
9
(
)
0
.
2
)(
1320
(
5
.
2
5
.
0
5
.
1
5
.
2
5
.
0
5
.
1
=
=
b
g
Qm
and
66
.
2
0
.
10
)
3
.
13
)(
0
.
2
(
2
=
=
b
myJ
From Figure 2.28, we obtain myJ1/b= 0.75. Then yJ2= 0.75(10)/2.0= 3.75 m. Let us now
determine if this depth satisfy the jump equation, MJ1= MJ2
By using the expressions given in Table 2.1
3
2
2
2
2
2
2
2
2
2
2
2812
)]
0
.
10
(
3
)
75
.
3
)(
0
.
2
(
2
[
6
)
75
.
3
(
)
75
.
3
)](
75
.
3
)(
0
.
2
(
0
.
10
[
81
.
9
)
1320
(
)
3
2
(
6
)
(
m
M
b
my
y
y
my
b
g
Q
M
J
J
J
J
J
J
=
+
+
+
=
+
+
+
=
Also for the section before the jump
13. 22
3
2
2
1
1
2
1
1
1
2
1
2818
)]
0
.
10
(
3
)
3
.
13
)(
0
.
2
(
2
[
6
)
3
.
13
(
)
3
.
13
)](
3
.
13
)(
0
.
2
(
0
.
10
[
81
.
9
)
1320
(
)
3
2
(
6
)
(
m
M
b
my
y
y
my
b
g
Q
M
J
J
J
J
J
J
=
+
+
+
=
+
+
+
=
We will accept yJ1= 3.75 ft since MJ1 is nearly equal to MJ2.
(b) The head loss due the hydraulic jump is calculated using hLj= EJ1-EJ2 or
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
= 2
2
2
2
2
2
1
1
2
1
]
)
[(
2
]
)
[(
2 J
J
J
J
J
J
Lj
y
my
b
g
Q
y
y
my
b
g
Q
y
h
m
hLj 7
.
10
]
3
.
13
))
3
.
13
(
2
0
.
10
)[(
81
.
9
(
2
1320
3
.
13
]
75
.
3
))
75
.
3
(
2
0
.
10
)[(
81
.
9
(
2
1320
75
.
3 2
2
2
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
=
Problem 17:
2
2
2
2
2
2
2
1
1
2
y
gy
q
y
gy
q
+
=
+
( )
2
1
2
2
2
1
2
2
1
1
1
y
y
y
y
g
q
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
( )( )
1
2
1
2
2
1
1
2
2
2 y
y
y
y
y
y
y
y
g
q
+
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
( ) ( ) 0
1
2
1
2
2
1
2
1
2 =
⎥
⎦
⎤
⎢
⎣
⎡
+
−
− y
y
y
y
g
q
y
y
( ) 0
1
2
1
2
2
1
2
=
+
− y
y
y
y
g
q
( ) 0
2
1
2
2
2
1
2
=
+
− y
y
y
gy
q
Divide both sides by y1
2
14. 23
0
2
1
2
2
1
2
2
3
1
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
y
y
y
y
gy
q
Noting that
3
1
2
1
gy
q
Fr = we obtain
0
2 2
1
1
2
2
1
2
=
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
r
F
y
y
y
y
This quadratic equation has two roots. One of the roots will yield a negative value for the
ratio y2/y1 and has no physical meaning. The second root will yield
( )
1
8
1
2
1 2
1
1
2
−
+
= r
F
y
y
Problem 18:
Equation 2.16 can be written as
1
1
2
8
2
1
2
2
1 −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
y
y
Fr
or
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
1
2
2
1
2
2
1
2
1
2
1
y
y
y
y
Fr (*)
Equation 2.28
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
−
=
−
−
+
= 2
2
2
1
2
2
1
2
2
2
2
2
1
2
1
1
1
2
)
(
2
2 y
y
g
q
y
y
gy
q
y
gy
q
y
hLJ
( ) ( )
2
2
2
1
2
2
2
1
1
2
1
2
2
2
1
2
2
2
1
2
2
1
2
)
(
2
)
( y
y
y
F
y
y
y
y
y
y
gy
q
y
y
h r
LJ −
+
−
=
−
+
−
=
Substituting Equation * into this expression we have
( ) ( )
1
2
2
3
1
3
2
2
1
2
2
1
2
1
2
2
2
1
2
1
2
1
2
1
4
1
)
(
4
1
)
( y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
hLJ −
+
−
−
−
=
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
+
−
=
15. 24
( )
2
1
3
1
2
1
2
2
3
1
3
2
2
1
2
1
2
2
2
2
1
2
1 4
)
(
4
4
4
1
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
hLJ
−
=
+
−
+
−
−
=
Problem 19:
The hydraulic jump equation on a sloped channel would include the component of the
weight of water in the flow direction. With reference to Equation 2.18, the jump equation
becomes
2
0
1
2
J
U
D
J M
A
A
xS
M =
+
∆
+
Therefore for the same yJ1, the magnitude of MJ2 would be larger. An inspection of any
specific momentum would reveal that, on the subcritical limb, a larger MJ2 corresponds to
a larger yJ2.
Problem 20:
If the friction forces are not negligible, the hydraulic jump equation includes a friction
force term opposing the flow. With reference to Equation 2.18, the hydraulic jump
equation becomes
2
1 J
f
J M
F
M =
−
γ
Therefore for the same yJ1, the magnitude of MJ2 would be smaller. An inspection of a
specific momentum diagram would reveal that, on the subcritical limb, a smaller MJ2
corresponds to a smaller yJ2.
