Question

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

Answer 1

1.693 feet

Step-by-step explanation:

We have given Q = 90 gpm

We know that 1 cubic feet per second = 443.833 gpm

So 90 gpm will be equal to
`(90)/(443.833)=0.202(ft^3)/(sec)`

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

We know that for pipe flow critical Reynolds number =2300

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

Value of
`\\u` is
`6.55* 10^(-5)` given in question

So
`(0.255* d)/(d^2* 6.5* 10^(-5))=2300`

d=1.693 feet