and from Equation 2.144
3
3
Hmin = 2 yc = 2 1.46 ft = 2.19 ft
Assuming no energy loss, the specific head upstream of the dam is
H = Hmin + ∆Z = 2.19 + 4.0 = 6.19 ft
To determine the depth of flow upstream of the dam (y) solve the specific head equation (Equation
2.137)
2
q
2
10 + y = 6.19ft
H =
+ y =
2
2
2g y
64.4 y
or
y3 - 6.19 y2 + 102/64.4 = 0
The solution is
y = 6.14 ft
As the normal depth is only 5 ft, the backwater is
∆y = 6.14 - 5.00 = 1.14 ft
That is, the depth upstream of the dam is increased 1.14 ft by the 4.0 ft high dam when the flow is
1,000 cfs.
2.89