Question

For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is Δp1. If the length of the pipe is then doubled, what is the relation of the new pressure drop Δp2 to the original pressure drop Δp1 at the original mass flow rate?

delta P2=?

Answer 1

Answer:

$\Delta p_{2} = 2\cdot \Delta p_{1}$

Explanation:

The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.

$\Delta p_{2} = 2\cdot \Delta p_{1}$