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

Answer:

1.693 feet

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 $\frac{90}{443.833}=0.202\frac{ft^3}{sec}$

Let d is the diameter of the pipe then $V_{avg}=\frac{Q}{A}=\frac{0.2}{\frac{\pi }{4}d^2}=\frac{0.255}{d^2}$

We know that for pipe flow critical Reynolds number =2300

So $\frac{V_{avg}d}{\nu }=2300$

Value of $\nu$ is $6.55\times 10^{-5}$ given in question

So $\frac{0.255\times d}{d^2\times 6.5\times 10^{-5}}=2300$

d=1.693 feet