OPEN CHANNEL FLOW
251
CHAPTER FOURTEEN
OPEN CHANNEL FLOW
OPEN CHANNEL FLOW
253
4
See Pr
o
b. 14.8
fo
r
d
= 0
.
50
m
;
Prob. 14.9
f
or
d
= 2.50
m
a. d = 0.50 m; A = 0.50 m2; R = 0.25 m
14.15 a. Depth = 3.0 ft:
1
b. Depth = 6.0 ft:
A = (4)(12) + 1
2 (4)(4)
2
14.16 AR2/3 = 2/12/1
(0.015)(150)
1.49 (1.49)(0.001)
nQ
S
= 47.75
OPEN CHANNEL FLOW
255
OPEN CHANNEL FLOW
257
OPEN CHANNEL FLOW
259
Trapezoid: y = 0.4545
1.73 1.73
A = 0.5126 ft; R = y/2 = 0.2563 ft
Semicircle: A =
22 2(0.4545)
;
2
A
y y
 
  = 0.5379 ft
14.39 a. When y = yc, NF = 1.0 = 3/ 2
( )
h
Q Q Q
A gy by gy b g y
gy
 
b. Minimum E occurs when y = yc: From Eq. 14.18:
e.
= 5.5 ;
2.0 F
Q Q
N
A by y gy
 
OPEN CHANNEL FLOW
261
0
C
i
rcu
l
ar
C
h
ann
e
l
Q = 1.45 m3/s n = 0.015 Finished concrete
y(m)
(rad) A(m2) T(m) yh(m) NF E(m) Velocity
y less than D (m/s)
0.10 1.171 0.0450 0.6633 0.068 39.478 52.982 32.211
0.25 1.896 0.1707 0.9747 0.175 6.481 3.928 8.494
0.40 2.462 0.3300 1.1314 0.292 2.597 1.384 4.394
0.60 3.142 0.5655 1.2000 0.471 1.193 0.935 2.564
y greater than D
0.70 3.476 0.6849 1.1832 0.579 0.888 0.928 2.117
0.90 4.189 0.9099 1.0392 0.876 0.544 1.029 1.594
1.00 4.601 1.0071 0.8944 1.126 0.433 1.106 1.440
Part f of problem: Slopes for given y and alternate depth
y(m) R(m) S
0.50 0.2649 0.0140 S for given y
Problem 14.40 Procedure: Refer to Table 14.2 for geometry of a partially full circular pipe.
a) For given Q, D, and y: Compute
, A, T using equations in Table 14.2.
d) Specific energy for y = 0.50 m:
e) Velocity = v = Q/A, NF =
 
/ /
h h
v gy Q A gy. See spreadsheet
Chapter 14
262
f
)
Co
m
pute
W
P
=
D
/2 (See
T
a
b
l
e
14.2
)
. Co
m
pute
R
=
A
/
W
P
.
14.41 Triangular channel
z = 1.5 n = 0.022
Q = 0.68 ft3/s
y(ft) A(ft2) V(ft/s) T(ft) yh(ft) NF E(ft)
0.20 0.060 0.60 0.100 6.316 2.194
0.30 0.135 0.90 0.150 2.292 0.694
0.418 0.262 2.594 1.254 0.209 1.000 0.523 Critical depth
0.60 0.540 1.80 0.300 0.405 0.625
0.80 0.960 2.40 0.400 0.197 0.808
1.00 1.500 3.00 0.500 0.113 1.003
1.10 1.815 3.30 0.550 0.089 1.102
1.30 2.535 3.90 0.650 0.059 1.301
Slopes at given depth and alternate depth
y(ft) R(ft) S
Problem 14.41 Procedure: Refer to Table 14.2 for geometry of a triangular channel.
a) For given Q, z, and y: Compute A, T using equations in Table 14.2.
b) Minimum specific energy:
c) Specific energy versus y: See spreadsheet using equation in b).
OPEN CHANNEL FLOW
263
e) Velocity = v = Q/A, NF =
/ /
h h
v gy Q A gy. See spreadsheet
14.42 Trapezoidal channel
z = 0.75 n = 0.013
Q = 0.80 ft3/s b = 3.000 ft
y(ft) A(ft2) V(ft/s) T(ft) Yh(ft) NF E(ft)
0.1 0.308 2.602 3.15 0.098 1.467 0.205
0.20 0.630 1.270 3.30 0.191 0.512 0.225
0.30 0.968 0.827 3.45 0.280 0.275 0.311
0.4770 1.602 0.499 3.72 0.431 0.134 0.481 Alternate de
pth
0.60 2.070 0.386 3.90 0.531 0.093 0.602
0.80 2.880 0.278 4.20 0.686 0.059 0.801
1.00 3.750 0.213 4.50 0.833 0.041 1.001
1.10 4.208 0.190 4.65 0.905 0.035 1.101
1.30 5.168 0.155 4.95 1.044 0.027 1.300
1.50 6.188 0.129 5.25 1.179 0.021 1.500
Slopes at given depth and alternate depth
y(ft) R(ft) S
0.05 0.0486 0.264 Slope for given depth
Problem 14.42 Procedure: Refer to Table 14.2 for geometry of a trapezoidal channel.
OPEN CHANNEL FLOW
265
14.48 Q = 2.48H
5/
2
H(in) H(ft) Q(ft3/sec)
0 0 0
4 .333 .159
6 .500 .439
14.49 Q = 3.07H1.53
H1.53 = Q/3.07 H = (Q/3.07)1/1.53
14.50 L = 8.0 ft; Qmin = 3.5 ft3/s; Qmax = 139.5 ft3/s
H(ft) Q(ft3/sec)
0.25 3.434
1.50 61.469
2.25 118.077
14.51 a) Q = 50 ft3/s; L = 4.0 ft; Q = 4.00 LHn; n = 1.58
14.52 Trapezoidal channel Long-throated flume Design C: H = 0.84 ft; Q = K1(H + K2)n
n
14.54 Rectangular chann Long-throated flume Design A: H = 0.35 ft; Q = bcK1(H + K2)n
14.55 Rectangular chann Long-throated flume Design C: H = 0.40 ft; Q = bcK1(H + K2)n
14.56 Circular channe g-throated flume Design B: H = 0.25 ft; Q = D2.5 K1 (H/D + K2)
n
14.57 Circular channe g-throated flume Design A: H = 0.09 ft; Q = D2.5 K1 (H/D + K2)
n
14.58 Rectangular chann Long-throated flume Design B: Q = 1.25 ft3/s; Find H.
14.59 Circular channe g-throated flume Design C: Q = 6.80 ft3/s; Find H.
14.60 Select a long-throated flume for 30 gpm < Q < 500 gpm. Using 449 gpm = 1.0 ft3/s,
H Q(ft3/s) Q(gpm)
0.20 0.271 121.7
OPEN CHANNEL FLOW
267
14.61 Given 50 m
3
/h < Q , 180 m
3
/h; Convert to ft
3
/s; 0.4907 ft
3
/h < Q < 1.766 ft
3
/h
Find H for each limiting flow rate.
H(m) H(ft) Q(ft3/s) Q(m3/h)
0.100 0.328 0.622 63.38
0.150 0.492 1.190 121.3