Laminar flow of oil in a 2-in Schedule 40 steel pipe has an average velocity of 10.72 ft/s. Find the velocity at (a) the center of the pipe, (b) at the wall of the pipe, and (c) at a distance of 0.6 inches from the centerline.

The answers to the questions are;

(a) The velocity at the center of the pipe is 21.44 ft/s

(b) The velocity at the wall of the pipe is 0 ft/s

(c) The velocity at a distance of 0.6 inches from the center-line is 19.63 ft/s.

Explanation:

To solve the question, we note that

The velocity profile in the cross section of a circular pipe with laminar flow is given by

U = 2×v×[1 - (r/r₀)²]

Where

U = The sought velocity at a point

r = Pipe radius where velocity is sought

r₀ = Internal radius of pipe = for  2-in Schedule 40 steel pipe =  2.067 in 52.6 mm

v = Average velocity of flow = 10.72 ft/s = 3.2675 m/s

Therefore we have

(a) The velocity at the center of the pipe

At the center r = 0 so we have

U = 2×v×[1 - (r/r₀)²]

At center U = 2×10.72 ft/s×[1 - (0/2.067 in)²] =  2×10.72 ft/s = 21.44 ft/s

(b) The velocity at the wall of the pipe is given by

r = r₀ ⇒ U = 2×v×[1 - (r/r₀)²] ⇒ U = 2×v×[1 - (r₀/r₀)²]

= U = 2×v×[1 - (1)²] =  2×v×0 = 0

The velocity at the wall of the pipe is 0 ft/s

(c) The velocity at a distance of 0.6 inches from the center-line is given by

U = 2×v×[1 - (r/r₀)²] = 2×10.72 ft/s×[1 - (0.6/2.067)²]  =  19.63 ft/s.

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