In a conduit with a diameter of 4.5 ft, the depth of flow is 4.0 ft. (a) Determine the hydraulic radius, hydraulic depth, and section factors for critical and normal flows. (b) Determine the alternate depth of flow that will carry the same discharge.

(a) 1.125 ft, Section factor = 22.78

(b) 42.75 ft

Explanation:

Hydraulic radius is given by Where

A = Cross sectional area of flow and

P = Perimeter  h

Since the cross section is a circle  then at depth 4 of 4.5 the perimeter

 where r = 2.25 and θ = 102.1 °

perimeter = 10.1 ft and the  area = =  11.39 ft²

Therefore

Section factor is given by for critical flow = Z = A×√D

= 11.39 ft² × √(4 ft) = 22.78

for normal flow Z = = 22.78

(b) The alternate depth of flow is given by

for a given flow rate, we have from chart for flow in circular pipes

Alternative depth = 0.9×45 = 42.75 ft

energy can cross the boundaries of a closed system in the form of heat and mass

explanation:

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