The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the pressure gradient, ∂p/∂x, where x is in the flow direction, if the pipe is: a. Horizontal b. Vertical with flow up. c. Vertical with flow down.

a) - 8 lb / ft^3

b) - 70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

(dP - y*L*sin (Q)) / L = 2*t / r

dP / L - y*sin (Q) = 2*t / r

Where dP / L = - dP/dx,

dP / dx = - 2*t / r - y*sin (Q)

Where r = D / 2,

dP / dx = - 4*t / D - y*sin (Q)

a) Horizontal Pipe Q = 0

Hence, dP / dx = - 4*2 / 1 - 62.4*sin (0)

dP / dx = - 8 + 0

dP/dx = - 8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence, dP / dx = - 4*2 / 1 - 62.4*sin (pi/2)

dP / dx = 8 - 62.4

dP/dx = - 70.4 lb / ft^3

c) Vertical flow down Q = - pi/2

Hence, dP / dx = - 4*2 / 1 - 62.4*sin (-pi/2)

dP / dx = - 8 + 62.4

dP/dx = 54.4 lb / ft^3