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 p1. if the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?

∆p2 = 2* ∆p1

Explanation:

Given that all other factors remain constant. The pressure drop across the pipeline is directly proportional to the length.

i.e ∆p ~ L

Therefore,

∆p2/L2 = ∆p1/L1

Since L2 = 2 * L1

∆p2/2*L1 = ∆p1/L1

Eliminating L1 we have,

∆p2/2 = ∆p1

Multiplying both sides by 2

∆p2 = 2 * ∆p1

radial fractures in a piece of glass look most like which of the following?

a. spokes of a wheel

b. a circle

c. a zig zag

d. none of the above

the correct answer is a. spokes of a wheel

hope this (: