Chapter 2
2.3
Applying Newton second law
γQ
g
(V2 − V1 ) = P1 − P2 + W sin θ − FR
Assume θ=0 and no friction FR =0
P1 = γ z1 A1
P2 = γ z2 A2
Q2 Q2
−
= z1 A1 − z2 A2
gA2 gA1
Q2
Q2
+ z1 A1 =
+ z2 A2
gA1
gA2
Q2
+ zA
gA
Force / Unit weight
Fs =
2.4
Q=4 x 5 x 5 = 100 m3/s
V
4
F=
=
= 0.57 Sub critical
gy
9.81 × 5
Energy equation neglecting losses
H1 = E1 = E2 − ∆z
2
y1 +
5+
V1
= E2 + 0.2
2g
Q2
42
= y2 +
+ 0.2
2 × 9.81
2 gA2
5.61 = y2 +
1002
2 × 9.81 × (5 × y2 ) 2
y2 − 5.61 × y2 + 20.387 = 0
3
2
Trial and error will give y2 = 2.6 m.
14
Chapter 2
2
H = y1 +
5+
2
V1
V
+ 0.2 = y2 + 2
2g
2g
42
1002
+ 0.2 = y2 +
2 × 9.81
2 × 9.81 × (5 × y2 ) 2
6.015 = y2 +
20.387
2
y2
y2 − 6.015 × y2 + 20.387 = 0
3
2
y2 = 5.285 m rise with water level.
2.5
F1 =
V1
Q A1
4
=
=
= 0.571
gy1
gy1
9.81 × 5
Energy equation neglecting losses
H1 = E1 = E2 + ∆z
q = Q/B = 20 m3/m
2
V
42
y1 + 1 = 5 +
= 5.8155 = E2 + 0.3
2g
2 × 9.81
5.8155 − 0.3 = y2 +
q2
2
2 gy2
y2 − 5.5155 y2 + 20.387 = 0
3
2
Using trial and error y2 = 4.52 m
(ii)
F1 = 0.571
2
H 1 = y1 +
2
V1
V
+ 0.2 = y 2 + 2 = E 2
2g
2g
q2
42
+ 0.2 = y 2 +
5+
2
2 × 9.81
2 gy 2
15
Chapter 2
y 2 − 6.015 y 2 + 20.387 = 0
3
2
Using trial and error, y2 = 5.285 m
2.6
Q = 400 m3 / s
V1 = 400 / 5 x 10 = 8 m / sec
V
8
F1 = 1 =
= 1.14 > 1
gy1
9.81 × 5
Super critical
If the step is a rise water surface rises
If the step is a drop water surface drops
2.7
Q = 96 m3 / s
Section1
B1 = 8 m Y1 = 4 m
V1 = 3 m / sec
H1= H2= H3= H4= H5
Hi= Ei
2
V
H1 = y1 + 1 = 4.459m
2g
F1 =
3
= 0.479 < 1 Sub critical
9.81 × 4
q1 = Q/B1= 96/8 = 12 m3/s / m
Section 2
B2 = 7.5 m
q1 = Q/B1= 96/7.5 = 12.8 m3/s / m
Q2
E2 = E1 = y2 +
= 4.459
( B2 y2 ) 2 × 2 × 9.81
y2 − 4.459 y2 + 8.35 = 0
y2 = 3.91 m
3
2
16
Chapter 2
Section 3
B3 = 7 m
q1 = Q/B1= 96/7 = 13.71 m3/s / m
q32
E1 = E3 = y3 + 2
= 4.459
y3 × 2 × 9.81
y2 − 4.459 y2 + 9.58 = 0
y2 = 3.79 m
3
2
Section 4
B4 = 6.5 m
q1 = Q/B1= 96/ 6.5= 14.77 m3/s / m
q42
E1 = E4 = y4 + 2
= 4.459
y4 × 2 × 9.81
y2 − 4.459 y2 + 11.12 = 0
y2 = 3.6 m
3
2
Section 5
B4 = 6 m
q1 = Q/B1= 96/ 6= 16 m3/s / m
q52
E1 = E5 = y5 + 2
= 4.459
y5 × 2 × 9.81
y 2 − 4.459 y 2 + 13.048 = 0
y2 = 3.11 m
3
2
2.8
(i)
h =20 m
Q= 40 m3/s b= 4 m
E1= E2 , V1 = 0, V2 =Q/By2 = q/y2, q= 10 m3/s / m
h+
V1
q2
= y2 +
2× g
2 × g × y22
y2 − 20 y2 + 5.097 = 0
y2= 0.511 m
V2=10/0.511=19.57 m/s
V
F2 =
= 8.47 > 1 Supercritical
g×y
3
2
17
Chapter 2
[
]
y2
1 + 8 F22 − 1
2
0.511
y3 =
1 + 8 × 8.742 − 1 = 6.066m
2
y3 =
[
]
(ii)
Thrust on the gate
Fth = Fs1 − Fs 2
Fs1 =
Q2
402
+ z1 A1 =
+ 10 × (20 × 4) = 802.038m3
gA1
9.8 × (20 × 4)
Fs 2 =
402
0.511
Q2
+ z2 A2 =
+
× (0.511 × 4) = 80.31m3
9.8 × (0.511 × 4)
2
gA2
Fth = 802.038 − 80.31 = 721.72m3
Thrust force = γFth=9.81 x 721.72 = 7080.07 kN
(iii)
Energy losses
E2 = E3 + H L
H L = E2 − E3
H L = y2 +
q2
q2
−
−
y
3
2 gy22
2 gy32
H L = 0.511 +
10 2
102
−
−
= 13.85m
6
.
066
2 × 9.8 × 0.5112
2 × 9.8 × 0.662
2.9
E= y+
q2
2 gy 2
q 2 = 2 gy 2 ( E − y )
1
q = (2 gy 2 E − 2 gy 3 ) 2
−1
dq 1
= (2 gy 2 E − 2 gy 3 ) 2 4 gyE − 6 gy 2 = 0
dy 2
[
]
Maximum q for critical depth
18
Chapter 2
4 gyE − 6 gy 2 = 0
2
E
3
y
2
tan θ = c =
E 3
yc =
2.10
(i) Channel width remains constant
Use a step and determine if it is a rise or down
V
4
F1 = 1 =
= 0.639 < 1 Sub critical flow
gy1
9.81
Δz is positive (step rise)
E1 > E2
E1 = E2 + ∆z
q=
Q
= V y = 16m 3 / sec/ m
b
and
y1 = y2 + ∆z + (704 − 703.54)
∆z = 3.54 − y2
2
V
q2
y1 + 1 = y2 +
+ ∆z
2g
2 gy22
4+
16
16 2
= y2 +
+ (3.54 − y2 )
2 × 9.81
2 × 9.81 × y22
y2 = 3.198m
∆z = 3.54 − 3.198 = 0.34m
Bottom elevation = 700+0.34=700.34m
(ii)
Channel bottom level 0.5 of transition at elevation 700.2 m
b1= 8m
y1 = 4 m; y2 = 3.34 m
Q= 32 x 4 =128 m3/s
E1 = E 2 + ∆z
2
2
V
V
y1 + 1 = y 2 + 2 + ∆z
2g
2g
19
Chapter 2
4+
16
128 2
= 3.34 +
+ 0.2
2 × 9.81
2 × 9.81× (b2 × 3.34) 2
4.815 = 3.54 +
74.856
b22
b2 = 7.66m
2.11
(i)
V1 A1 = V2 A2
V2 =
0.8 × 2 × b
y2 × b
V2 =
1.6
y2
0.8
= 0.18 < 1
2 × 9.81
E1 = E2 + ∆z
F1 =
Sub critical
2
2
V
V
y1 + 1 = y 2 + 2 + ∆z
2g
2g
(1.6 y2 ) 2
0.8 2
+ 0.15
2+
= y2 +
2 × 9.81
2 × 9.81
3
2
y 2 − 1.882 y 2 + 0.1305 = 0
y 2 = 1.84m
(ii)
y2 = 2 m; V1 = 0.8 m; q = 1.6 m3/sec/m
2
V
E1 = y1 + 1 = 2.033
2g
Maximum Δz occurs at minimum E and Fr=1
y2 = 3
q 2 3 1.62
=
= 0.64
g
9.81
2
V2
1.62
= 0.64 +
= 0.96
2g
2 × 9.81 × 0.642
= E1 − E2 = 2.033 − 0.96 = 1.073 m
E2 = y 2 +
∆zmax
20
Chapter 2
2.12
Assume b1=10 m
Q1= Q2= 10 x 2 x 0.8 = 16 m3/s
E1 = y 2 +
q 22
2 gy 22
E1 y 22 = y 23 +
q 22
2g
q 2 = (2 gE1 y 22 − 2 gy 23 )1 2
Q
b2
Q is constant. Therefore, maximum q2 corresponds to minimum b2.
q2 =
[
]
−1
dq2 1
= (2 gy22 E − 2 gy23 ) 2 4 gyE − 6 gy 2 = 0
dy2 2
y2 =
2
E1
3
From problem 2.11, E1 = 2.033 m
2
× 2.033 = 1.355 m
3
q2 = (2 × 9.81 × 2.033 × 1.3552 − 2 × 9.81 × 1.3553 )1 2
y2 =
= 4.941 m3 /sec/m
b2 = 16
= 3.238 m
4.941
b2 3.238
=
= 0.32
b1
10
2.13
(i)
Q = 80 ft3 / sec
E1 = E 2
2
2
V
V
y1 + 1 = y 2 + 2
2g
2g
y1 +
Q2
Q2
=
+
y
2
2 g (b1 y1 ) 2
2 g (b2 y 2 ) 2
21
Chapter 2
80 2
80 2
8+
= y2 +
2 × 32.2 × (5 × 8) 2
2 × 32.2 × (4 × y 2 ) 2
8 + 0.0621 = y 2 +
6.211
y2
2
y 2 − 8.0621 y 2 + 6.211 = 0
3
2
y 2 = 7.964 ft
(ii)
80
= 2 ft/sec
5×8
2
V
=
= 0.125 < 1 subcritical
Fr =
gy
9.81 × 8
V1 =
E1 = E2 + ∆z
y1 +
Q2
Q2
y
=
+
+ ∆z
2
2 g (b1 y1 ) 2
2 g (b2 y2 ) 2
802
802
+ ∆z
7
.
964
=
+
2 × 32.2 × (5 × 8) 2
2 × 32.2 × (5 × 7.964) 2
8.0621 = 8.0266 + + ∆z
8+
y2 − 8.0621 y2 + 6.211 = 0
3
2
∆z = 0.035 ft
2.14
80
= 8 m/sec
0.5 × 20
8
Fr =
= 3.612
9.81 × 0.5
V1 =
[
[
]
y1
1 + 8 F12 − 1
2
0.5
y2 =
1 + 8 × (3.162) 2 − 1 = 2.316 m
2
Losses H L = E1 − E 2
y2 =
]
2
H L = y1 +
2
V1
V
− y2 − 2
2g
2g
22
Chapter 2
82
1  80 
− 2.316 −
H L = 0.5 +


2 × 9.81
2 × 9.81  20 × 23.6 
= 1.294 m
2
2.15
Q = 80 m3 / s
54
V1 =
= 11.25 m/s
8 × 0.6
11.25
Fr =
= 4.63
9.81 × 0.6
y
y2 = 1 1 + 8 F12 − 1
2
0.6
1 + 8 × (4.63) 2 − 1 = 3.646 m
y2 =
2
[
]
[
]
2.16
Losses at junction =
0.2V22
2g
10
= 0.67 m/s
10 ×1.5
0.67
= 0.173 subcritical
Fr =
9.81×1.5
E1 = E2 + ∆z + Losses
V1 =
2
y1 +
2
V1
V
0.2V22
= y2 + 2 + 0.1 +
2g
2g
2g
1.5 +
0.67 2
10 2
= y 2 + 1.2 ×
+ 0.1
2 × 9.81
2 × 9.81 × (8 y 2 ) 2
y 2 − 1.432 y 2 + 0.0956 = 0
3
2
y 2 = 0.29 m
2.17
B =8m
Q = 8 × 2 × 3 = 48 m 3 / s
E1 + ∆z = E 2
32
48 2
2+
+ 0.15 = y 2 +
2 × 9.81
2 × 9.81× (8 y 2 ) 2
23
Chapter 2
2.609 = y 2 +
1.835
y 22
y 2 − 2.609 y 2 + 1.835 = 0
3
2
y 2 = 2.245 m
2.18
F=
V
3
=
= 0.533
g× y
9.81 × 3
V1 y1 = V2 y2
3 × 3 = V2 y2
V2 = 9 / y2
E1 = E2 + ∆z
2
32
1  9
  + 0.3
3+
= y2 +
2 × 9.81
2 × 9.81  y2 
y2 − 3.16 y2 + 4.128 = 0
3
2
y2 = 1.66 m
2.19
V1 = 2 g (160 − 120)
= 28 m/sec
Q = BV1 y1
y1 =
1200
= 2.14 m
20 × 28
V2
28 2
=
= 37.35
F =
gy1 9.81 × 2.14
2
r1
[
]
y1
1 + 8F22 − 1
2
2.14
y2 =
1 + 8 × 37.35 − 1 = 17.46 m
2
1200
= 3.44 m/s
V2 =
20 × 17.46
Energy losses = E 1 -E 2
y2 =
[
]
28 2
3.44 2
= 2.14 +
− 17.46 −
= 24 m
2 × 9.81
2 × 9.81
24
Chapter 2
2.20
Q= 18 m3/sec; B= 6 m; q= 18/6 = 3 m3 / s / m
18
= 10 m/s
6 × 0.3
10 2
V2
Fr12 =
=
= 33.97
gy1 9.81 × 0.3
V1 =
[
]
0.3
1 + 8 × 33.97 − 1 = 2.328 m sequential depth
2
q2
E= y+
2 gy 2
at y = 0.3
y2 =
32
= 5.397
2 × 9.81 × 0.32
0.4587
5.397 = y +
y2
E = 0.3 +
y2 − 5.397 y2 + 0.4587 = 0
3
2
y2 = 5.38 m alternate depth
Head loss = E1 -E 2
= y1 +
q12
q22
−
−
y
2
2 gy12
2 gy22
32
32
−
2
.
328
−
2 × 9.81× 0.32
2 × 9.81× 2.3282
= 5.397 − 2.413 = 2.98 m
= 0.3 +
2.21
(i)
q= 15 ft3/sec/ft
Q= 6 x 5 x 3 = 90 ft3/sec
25
Chapter 2
Fr1 =
V
3
=
= 0.2364 subcritical
32.2 × 5
g×y
2
32
V1
E1 = y1 +
=5+
= 5.14
2g
2 × 32.2
E1 = E2 = E3 = E4 = E5 = E6
Section 2 :
E1 = E2
5.14 = y2 +
90 2
2 × 32.2 × 5.82 × y22
y2 − 5.14 y2 + 3.74 = 0
3
2
y2 = 4.99 ft
Section 3 :
E1 = E3
5.14 = y3 +
90 2
2 × 32.2 × 5.6 2 × y32
y3 − 5.14 y3 + 4.01 = 0
3
2
y3 = 4.977 ft
Section 4 :
E1 = E4
5.14 = y4 +
902
2 × 32.2 × 5.42 × y42
y4 − 5.14 y4 + 4.31 = 0
3
2
y4 = 4.965 ft
Section 5 :
E1 = E 5
5.14 = y 5 +
90 2
2 × 32.2 × 5.2 2 × y 52
y 5 − 5.14 y 5 + 4.65 = 0
3
2
y 5 = 4.95 ft
Section 6 :
E1 = E 6
26
Chapter 2
90 2
5.14 = y 6 +
2 × 32.2 × 5 2 × y 62
y 6 − 5.14 y 6 + 5.031 = 0
3
2
y 6 = 4.933 ft
(ii)
H1 = H 2
E1 + ∆z = E2
2
2
V1
V
+ ∆z = y2 + 2
2g
2g
for the water surface to be horizontal
y1 + ∆z = y2
y1 +
2
2
9=
902
52 y22
V1
V
= 2
2g 2g
y2 = 6 ft
∆z = y2 − y1 = 6 − 5 = 1 ft
2.22
Q= 50 ft3/sec
50
= 3.33 ft/sec
V1 =
3× 5
50
= 5 ft/sec
V2 =
2.5 × 4
H1 = H 2
E1 = E 2 + ∆z
2
2
V
V
y1 + 1 = y 2 + 2 + ∆z
2g
2g
3.33 2
52
+ ∆z
= 2.5 +
2 × 32.2
2 × 32.2
∆z = 0.28 ft
3+
27
Chapter 2
2.23
[
]
1.5
1 + 8 × 25 − 1 = 9.88 ft
2
V1
Fr = 5 =
32.2 ×1.5
V1 = 34.75 ft/sec
Y2 =
34.75 ×1.5
= 5.275 ft/sec
9.88
head loss = E1 − E2
V2 =
34.752
5.2752
− 9.88 −
= 9.939 ft
1.5 +
2 × 32.2
2 × 32.2
2.24
Assume rectangular cross section and horizontal channel bed
V2
4.3
Fr2 =
=
= 0.3323
gy 2
32.2 × 5.2
[
[
]
y2
1 + 8 F22 − 1
2
5.2
y2 =
1 + 8 × (0.3323) 2 − 1 = 0.968 ft
2
q = 5.2 × 4.3 = 22.36 ft 3 / sec/ ft
y1 =
]
22.36
= 23.099 ft/sec
0.968
E1 = E3 neglecting losses through the gate
V1 =
2
2
V
V
y1 + 1 = y 3 + 3
2g
2g
23.099 2
y3 = 0.968 +
= 9.25 ft
2 × 32.2
22.36
V3 =
= 2.41 ft/sec
∴ y 3 = 9.16 ft
9.25
22.36
V3 =
= 2.44 ft/sec
∴ y 3 = 9.157 ft
9.16
2.25
A = (8 + 2 y ) y
Q2
Q2
E= y+
= y+
2 gA2
2 × 9.8 × (8 + 2 y ) 2 y 2
28
Chapter 2
Y
0.3
0.5
0.7
0.8
0.9
1
1.1
1.2
1.4
1.5
1.6
2
4
E (Q=10)
1.066
0.7515
0.8177
0.886
0.9655
1.051
1.14
1.233
E (Q=20)
3.363
1.506
1.171
1.145
1.162
1.204
1.262
1.33
E (Q=40)
12.55
4.527
2.58
2.183
1.948
1.815
1.7477
1.723
1.756
1.8
1.85
2.142
4.02
2.26
D = Hydraulic depth
α = Velocity Coefficient
d = ycosθ
E=H =Z+
p
γ
+
αQ 2
2 gA 2
γd cos θ αQ 2
αQ 2
2
= Z+
+
= Z + y cos θ +
γ
2 gA 2
2 gA 2
αQ 2
dE
dA
2
= 0 = cos θ +
(−2 A −3 )
2g
dy
dy
cos θ 2 =
V2 =
αQ 2 B
g
gD cos θ
A
3
at critical depth
where
dA
=B
dy
2
α
V2
= 1 = Fr2
gD cos θ 2
α
Fr =
V
gD cos θ 2
α
29
E (Q=50)
19.44
6.79
3.64
2.96
2.54
2.27
2.11
2.018
1.957
1.968
1.997
2.22
4.03
Chapter 2
2.28
V1 = 2 × 32.2(400 − z )
Q = 80000 = 200 × y1 2 × 32.2(400 − z )
y1 =
80000
49.844
=
200 2 × 32.2(400 − z )
400 − z
3
V 2 2 × 32.2 × (400 − z )
= 0.04 × (400 − z ) 2
F =
=
49.844
gy1
32.2 ×
400 − z
y2 1
=
1 + 8 F12 − 1
y1 2
2
r1
[
]
3
220 − z
1

=  1 + 8 × 0.04 × (400 − z ) 2 − 1
49.844
2

400 − z
3


(220 − z ) × 400 − z = 24.922  1 + 0.32(400 − z ) 2 − 1


By trial and error, z = 166.5 ft.
30