Question

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.

Answer 1

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.

Step-by-step 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.

Answer 2

(a) 21.44 ft/s

(b) 0 ft/s

(c) 19.51 ft/s

Step-by-step explanation:

2 in = 2/12 ft = 0.167 ft

For steady laminar flow, the function of the fluid velocity in term of distance from center is modeled as the following equation:

`v(r) = v_c\left[1 - (r^2)/(R^2)\right]`

where R = 0.167 ft is the pipe radius and
`v_c` is the constant fluid velocity at the center of the pipe.

We can integrate this over the cross-section area of the in order to find the volume flow

`\dot{V} = \int\limits {v(r)} \, dA \\= \int\limits^R_0 {v_c\left[1 - (r^2)/(R^2)\right]2\pi r} \, dr\\ = 2\pi v_c\int\limits^R_0 {r - (r^3)/(R^2)} \, dr\\ = 2\pi v_c \left[(r^2)/(2) - (r^4)/(4R^2)\right]^R_0\\= 2\pi v_c \left((R^2)/(2) - (R^4)/(4R^2)\right)\\= 2\pi v_c \left((R^2)/(2) - (R^2)/(4)\right)\\= 2\pi v_c R^2/4\\=\pi v_c R^2/2\\A = \pi R^2\\\dot{V} = Av_c/2\\`

So the average velocity

`v = \dot{V} / A = v_c/2 = 10.72`

`v_c = 10.72*2 = 21.44 ft/s`

b) At the wall of the pipe, r = R so
`v(R) = v_c(1 - 1) = 0 ft/s`

c) At a distance of 0.6 in = 0.6/12 = 0.05 ft

`v(0.05) = v_c(1 - 0.05^2/0.167^2) = 0.91v_c = 0.91*21.44 = 19.51 ft/s`