Question

A 36" RCP (n=0.015) conveys 9.7 ft^3/s at a slope of 1.5%.
a. Calculate the depth of flow in the pipeline
b- Calculate the velocity of flow in the pipeline

Answer 1

Given,

Diameter of pipe (D) = 36 inches = 36/12 = 3 ft

Discharge (Q) = 9.7 ft³/s

Slope of the channel (S) = 1.5% = 0.015

Let the depth of the flow in the channel = y

Velocity of flow in the channel = V

Let the pipe is partially filled making an angle of at the center as shown in figure.

Area of flow for the assumed condition (A) = ( * D² / 4) * (/360 - sin/(2*)) ................(1)

Wetted perimeter (P) = * D * (/360) ...............................................................................(2)

Therefore, Hydraulic mean depth (R) = A/P .........................................................................(3)

From the continuity equation, Q = A * V ...............................................................................(4)

From Manning's equation, V = (1/n) * R^(2/3) * S^(1/2)

Therefore, Q = [( * D² / 4) * {/360 - sin/(2*)}] * [(1/n) * R^(2/3) * S^(1/2) ]........................(5)

Replacing R with A/P in the above equation we get,

Adopting trial and error method,

Put, = 134.35 degrees in the above equation

Therefore,

9.7 = 2411 * [(134.35/360) - {sin(134.35) / (2 * )}]^(5/3)

9.7 = 9.6999

Therefore, LHS = RHS (condition satisfied)

Therefore assumed value of is correct.

Therefore, = 134.35

[ Note: Here only one trial is shown. But it may not be the case that in first trial itself we will get the result. So until the condition is satisfied do the trial and error method with changed value of . This is the easiest method to solve this type of problems]

From equation (1),

A = ( * 3² / 4) * (134.35/360 - sin134.35 / (2*))

A = 1.833 ft²

Therefore from continuity equation,

V = Q / A = 9.7 / 1.833 = 5.29 ft / s

From the figure,

Depth of flow (y) = (D/2) - OB

From triangle OAB,

OA = Radius of pipe = 3/2 = 1.5 ft

OB = OA * cos (/2)

= 1.5 * cos(134.35/2)

OB = 0.583 ft

Therefore, Depth of flow (y) = 3 / 2 - 0.583 = 0.917 ft

Summery:

Depth of flow = 0.917 ft

Velocity of flow = 5.29 ft / s