Question

The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe is 1.85 lb/ft^2.

Determine the pressure gradient ∂p/∂x, wherexis the flow direction when (a) the pipe is horizontal, (b) the pipe is vertical with the flow up, and (c) the pipe is vertical with the flowdown.

Answer 1

a)
`-7.4(lb)/(ft^3)`

b)
`-69.8(lb)/(ft^3)`

c)
`55 (lb)/(ft^3)`

Step-by-step explanation:

1) Notation

`\tau=1.85(lb)/(ft^2)` represent the shear stress defined as "the external force acting on an object or surface parallel to the slope or plane in which it lies"

R represent the radial distance

L the longitude

`\theta=0\degree` since at the begin we have a horizontal pipe, but for parts b and c the angle would change.

D represent the diameter for the pipe

`\gamma=62.4(lb)/(ft^3)` is the specific weight for the water

2) Part a

For this case we can use the shear stress and the radial distance to find the pressure difference per unit of lenght, with the following formula

`(2\tau)/(r)=(\Delta p -\gamma Lsin\theta)/(L)`

`(2\tau)/(r)=(\Delta p)/(L)-\gamma sin\theta`

If we convert the difference's into differentials we have this:

`-(dp)/(dx)=(2\tau)/(r)+\gamma sin\theta`

We can replace
`r=(D)/(2)` and we have this:

`(dp)/(dx)=-[(4\tau)/(D)+\gamma sin\theta]`

Replacing the values given we have:

`(dp)/(dx)=-[(4x1.85(lb)/(ft^2))/(1ft)+62.4(lb)/(ft^3) sin0]=-7.4(lb)/(ft^3)`

3) Part b

When the pipe is on vertical upward position the new angle would be
`\theta=\pi/2`, and replacing into the formula we got this:

`(dp)/(dx)=-[(4x1.85(lb)/(ft^2))/(1ft)+62.4(lb)/(ft^3) sin90]=-69.8(lb)/(ft^3)`

4) Part c

When the pipe is on vertical downward position the new angle would be
`\theta=-\pi/2`, and replacing into the formula we got this:

`(dp)/(dx)=-[(4x1.85(lb)/(ft^2))/(1ft)+62.4(lb)/(ft^3) sin(-90)]=55 (lb)/(ft^3)`