Manning's equation (7-2) to determine depth of

entrance. Therefore, a width of chute can he se-

water as in the first solution, find dw=0.493 foot.

lected and the appropriate length and depth of drop

From figure 7-4, with q equals 12.5, sine of angle

determined from the curves in figure 7-3. For this

of slope equals 0.243 and dw equals 0.493 foot,

design select a width of 2 feet. Then H/W = =

determine the depth of air to be 0.311 foot. Thus,

0.5 and Q/W5/2 = 25/(2)5/2 = 4.42**. **From figure 7-3

total depth is 0.804 foot. Use 0.80 foot. Wall

find a curve that matches these values. This is

height is 1.5 times 0.80 foot, or 1.20 feet. This

found on the curve for D/w 1.0, on the chart for

design is shown in figure 7-6.

B/W=4. Therefore, B=8 feet and D=2.0 feet. Using

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