Question

A fluid, n-Butyl alcohol is flowing at 60℉ with a velocity of 80 ft/s through a tube with a diameter of 0.20 inches. Assuming that the velocity distribution is parabolic, determine the magnitude of the shear stress at the wall of the tube.

Answer 1

Assumption:

1. n - Butyl alcohol can be assumed to be a newtonian fluid.

2. Viscocity of n - Butyl alcohol is assumed to be 2.5 cP (from any standard reference) = 0.0025 Pa-s

3. Maximum local fluid velocity at the center of the pipe is 80 ft/s.

Solution

Since the velocity distribution within the cross sectional area of the pipe is parabolic, the local fluid velocity (u ) in m/s is zero at the wall of the pipe and maximum at the center of the pipe.

Let "y" be the distance from the wall and "u" be the local fluid velocity as defined above.

The shear stress is then defined as

$\tau$ = $\mu$ ( du / dy) ............. (1) where

$\tau$ = Shear stress in N/m²

$\mu$ = Absolute fluid viscocity in Pa-s

du / dy = shear rate or velocity gradient in m/s / m

Since the velocity distribution within the cross sectional area of the pipe is parabolic, du / dy = 0 at the center of the pipe and du / dy = maximum at the wall of the pipe

du / dy = (u_center - u_wall ) / (0.5*D - 0) .............. (2) ; Refer assumption 1,3

u_center = 80 ft/s = 80 ft/s * 0.3048 m/ft = 24.4 m/s

0.5*D = 0.5 * 0.2 inches * 0.0254 m / inches = 0.00254 m

From (2)

du / dy = (24.4-0) / (0.00254 - 0) = 9606.3 /s

From (1)

$\tau$ ( N/ m²) = 0.0025 * 9606.3 = 24 N/m²