Question

For Laminar flow conditions, what size pipe will deliver 90 gpm of medium oil at 40 °F (v = 6.55 * 10^-5)?

Answer 1

1.693 feet

Step-by-step explanation:

We have given the pipe deliver 90 gpm

We know that 1 cubic feet per second =448.833 gpm

So 90 gpm will be equal to
`(90)/(448.833)=0.2(feet^3)/(sec)`

If d is the diameter of the pipe then
`V_(avg)=(Q)/(A)=(0.2)/((\pi )/(4)d^2)=(0.255)/(d^2)`

For the pipe flow critical Reynolds number =2300

We have given
`\\u =6.55* 10^(-5)ft^2/sec`

So
`(V_(avg)* d)/(\\u )=2300`

`(0.255* d)/(d^2* 6.55* 10^(-5))=2300`

d=1.693 feet