For comparison the hydrostatic force on the gate is
P(3 0.826)^2 g/223.15 kN m^1.
(ii)First trial
Referring to Fig. 6.9 assume gate opening a1.0 m
y 2 0.6051.00.605 m
V 2 3.547/0.6055.86 m s^1 ;Fr 2 V 2 /(gy 2 )1/22.40
conjugate depth to y 2
y^13
y
2
2
( 1 1
8 F r
22 ) ( 1 1
8
2 .4
2
)1.77 m.
Asy 3 2.1y^13 (1.77) submerged flow conditions exist at the gate.
Assuming no energy less between sections 1 and 2 from the energy
equation:
y 1 V^21 /(2g)yV^22 /(2g)
3 3.547^2 /(19.6 32 )y5.86^2 /19.6 y1.32 m.
From the momentum equation between sections 2 and 3:
y^2 /2qV 2 /gy^23 /2qV 3 /g
1.32^2 /23.5475.86/9.812.1^2 /23.547^2 /9.812.1
2.992.816
Second trial
Assume gate opening a0.92 m
y 2 0.6050.920.557 m; V 2 6.37 m s; Fr 2 2.73;y^13 1.89 m (2.1).
The same procedure as in the first trial results in y1.0 m and
y^2 /2qV 2 /g2.803 2.816.
Therefore the gate opening is a0.925 m.
ComputingCdfromqfor this gate opening:
Cd 0.5.
3.547
0.925(19.63)1/2
q
a 2
g y
1
0.605
2
286 GATES AND VALVES
