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

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