Question

A 2 nominal schedule 40 PVC pipe is 75 ft long. It is to convey water. The available pump can provide a pressure drop of 1.84 psi. Determine the expected flow rate of water in the pipe.

Answer 1

Answer:

10.22 m^3/s

Explanation:

To estimate the pressure drop in a pipe we use the Darcy-Weisbach equation

$\frac{(deltaP)}{L} =\alpha *(\frac{density}{2})*(\frac{v^{2}}{D})$ (equation 1)

With:

$\alpha$ = Darcy-Weisbach friction coefficient

L = length of duct or pipe

v = velocity of fluid

D= hydraulic diameter

Also flow rate is:

$Q=v*A$

Where v is:

$v=\frac{Q}{A}$

Area as a function of the diameter is:

$A=\pi *\frac{D^{2}}{4}$

So

$v=\frac{4*Q}{\pi*D^{2}}$ (equation 2)

For a laminar regime the the Darcy-Weisbach friction coefficient is function of the Reynolds number (Re) as:

$\alpha=\frac{64}{Re}$

$Re=\frac{density*v*D}{u}$

With: v =velocity, D= diameter if the pipe and u= viscosity.  

With this information alpha would be:

$\alpha=\frac{64*u}{density*v*D}$ (equation 3)

Replacing equation 3 in equation 1 we have:

$\frac{(deltaP)}{L} =\frac{32*u*v}{D^{2}}$

And finally replacing the value for v in this equation we have:

$\frac{(deltaP)}{L} =\frac{128}{\pi}*\frac{u*Q}{D^{4}}$

Clearing for Q we get an expression to estimate the expected flow rate in the pipe.

$Q=\frac{deltaP}{L}*\frac{\pi}{128}*\frac{D^{4}}{u}$

We know  

Delta P = 1.84 psi or $\frac{lb}{in^{2}}$

L= 75 ft or 900 in

D for a 2 nominal schedule 40 PVC is 2.047 in. In tables you find External diameter and internal diameter. For calculations you use internal diameter (ID)

U for water at 20°C is $2.034*10^{-5}\frac{lb*s}{ft^{2}}$ or $1.4113*10^{-7}\frac{lb*s}{in^{2}}$

$Q=\frac{1.84\frac{lb}{in^{2}}}{900in}*\frac{\pi}{128}*\frac{(2.047in)^{4}}{1.4113*10^{-7}\frac{lb*s}{in^{2}}} = 6.24*10^{3}\frac{in^{3}}{s}$

So the flow expected for this pipe is $6.24*10^{3}\frac{in^{3}}{s}$ or $10.22 \frac{m^{3}}{s}$