WORKED EXAMPLES 235
Solution
Initially, assume a spillway length of 200 m and a constant discharge coeffi-
cient of 0.75. Thus, the spillway capacity is given by (equation (4.19))
Q
2
3
Cd 2
g1/2bH3/22
3
0.754.43^200 H3/2^443 H3/2(m^3 s^1 ).Note that the maximum inflow of 3300 m^3 s^1 has to be reduced by flood
routing to 443 3 3/22300 m^3 s^1 or less.
Taking the time step in the computation (routing period) as 10 h
36 000 s, compute the following:
H (m) V ( 106 m^3 ) O (m^3 s^1 )
2
∆V
t
O (m^3 s^1 )0.5 045 0156 02 656
1.0 090 0443 05 443
1.5 138 0814 08 480
2.0 188 1253 11 697
2.5 243 1751 15 251
3.0 300 2302 18 969
Using the above values and the given inflow hydrograph, now compute the
following table (interpolating values from above as necessary) – equation (4.8):
T (h) I (m^3 s^1 )I 1 I 2 O (m^3 s^1 )
2
∆V
t O (m^3 s^1 ) 2
∆V
tO (m^3 s^1 )0 0200 0000 00 000 00 000
10 0960 1160 0068 01 024 01 160
20 1720 2680 0263 03 176 03 704
30 2480 4200 0679 06 018 07 376
40 3240 5720 1259 09 220 11 738
50 2860 6100 1761 11 798 15 320
60 2480 5340 2031 13 076 17 138
70 2100 4580 2107 13 441 17 656
80 1720 3820 2049 13 163 17 261
90 1340 3060 1895 12 433 16 223
100 0960 2300 1678 14 733
The maximum outflow is 2107 m^3 s^2 (2300 m^3 s^1 ) and the maximum
head on the spillway is H(2107/443)2/32.83 m. For a preliminary design
this is a satisfactory result. Note that the maximum outflow is equal to the
inflow at that time. Should a more accurate result be required, a slightly
shorter spillway could be considered, as well as the variation of the coeffi-
cient of discharge (see worked example 4.3).