Question

A liquid flows with same velocity through pipes 1 & 2 having same diameter. If the length of the second pipe is twice that of first pipe, what should be the ratio of head loss in two pipes?

• A. 1:2
• B. 2:1
• C. 1:4
• D. 4:1

Hint:

When all other factors remain constant, the head loss in a pipe system is linearly proportional to the length. Doubling the length always doubles the frictional loss.

Answer 1

The Correct Option is A

Step 1: Recall what controls friction head loss.
The Darcy-Weisbach relation gives \( h_f = \dfrac{f L v^2}{2gD} \). Since velocity \( v \), diameter \( D \) and the friction factor \( f \) are the same for both pipes, \( h_f \) is directly proportional to the length \( L \) alone.

Step 2: Write the ratio using the given lengths.
Let pipe 1 have length \( L \) and pipe 2 have length \( 2L \), since pipe 2 is twice as long. \[ \frac{h_{f1}}{h_{f2}} = \frac{L}{2L} = \frac{1}{2} \]

Step 3: State the answer as a ratio.
\[ \boxed{h_{f1} : h_{f2} = 1:2} \]