Question

A liquid flows through two similar pipes 1 and 2. If the ratio of their flow velocities V$_1$:V$_2$ be 2:3, what will be the ratio of the head loss in the two pipes?

• A. 8:27
• B. $\sqrt{2}$:$\sqrt{3}$
• C. 2:3
• D. 4:9

Hint:

For similar pipes, the frictional head loss is proportional to the square of the velocity (\(h_f \propto V^2\)).
Simply square the velocity ratio to find the head loss ratio:
\[ (2:3)^2 \rightarrow 4:9 \] This relationship is highly common in pipeline design problems.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When fluid flows through a pipe, it experiences a loss of energy (head loss) due to friction against the pipe wall.
This frictional head loss can be calculated using the Darcy-Weisbach equation.
Key Formula or Approach:
The Darcy-Weisbach equation for frictional head loss (\(h_f\)) is:
\[ h_f = \frac{f L V^2}{2gD} \] Where:
- \(f\) is the friction factor
- \(L\) is the length of the pipe
- \(D\) is the diameter of the pipe
- \(V\) is the flow velocity
- \(g\) is the acceleration due to gravity

Step 2: Detailed Explanation:
The problem states that the two pipes are "similar", which means they have identical physical properties:
- Same length (\(L_1 = L_2\))
- Same diameter (\(D_1 = D_2\))
- Same friction factor (\(f_1 = f_2\))
Since \(f\), \(L\), \(D\), and \(g\) are constant for both pipes, the head loss (\(h_f\)) is directly proportional to the square of the flow velocity (\(V^2\)):
\[ h_f \propto V^2 \] We can write the ratio of the head losses as:
\[ \frac{h_{f1}}{h_{f2}} = \left( \frac{V_1}{V_2} \right)^2 \] Given the velocity ratio \(V_1 : V_2 = 2 : 3\):
\[ \frac{h_{f1}}{h_{f2}} = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \] Therefore, the ratio of head loss in the two pipes is \(4:9\).

Step 3: Final Answer:
The ratio of the head loss in the two pipes is 4:9.