Question

For flowing through the same discharge, the diameters of a pipe at section 1 and section 2 are 50 mm and 100 mm respectively. If the velocity of fluid at section 1 is 8 m/s, then the velocity at section 2 is

• A. 8 m/s
• B. 6 m/s
• C. 4 m/s
• D. 2 m/s

Hint:

From the continuity equation, velocity is inversely proportional to the square of the diameter ($v \propto 1/D^2$).
If you double the diameter, you increase the area by a factor of 4, so the velocity must decrease by a factor of 4.
In this problem, $D_2 = 2D_1$, so $v_2 = v_1/4 = 8/4 = 2$ m/s.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question describes fluid flowing through a pipe that changes diameter. Given the velocity at the smaller section, we need to find the velocity at the larger section for the same flow rate.

Step 2: Key Formula or Approach:
This problem is an application of the principle of conservation of mass, expressed through the

Continuity Equation. For an incompressible fluid, the volume flow rate ($Q$) is constant.
\[ Q = A_1 v_1 = A_2 v_2 \] where:
$A_1, v_1$ are the area and velocity at section 1.
$A_2, v_2$ are the area and velocity at section 2.
The area of a circular pipe is $A = \frac{\pi}{4} D^2$.

Step 3: Detailed Explanation:
Let's plug the area formula into the continuity equation:
\[ \frac{\pi}{4} D_1^2 v_1 = \frac{\pi}{4} D_2^2 v_2 \] The term $\pi/4$ cancels out, leaving a direct relationship between velocity and diameter:
\[ D_1^2 v_1 = D_2^2 v_2 \] Rearrange to solve for the unknown velocity, $v_2$:
\[ v_2 = v_1 \left( \frac{D_1}{D_2} \right)^2 \] We are given:
- $D_1 = 50$ mm
- $D_2 = 100$ mm
- $v_1 = 8$ m/s
Substitute the values:
\[ v_2 = 8 \text{ m/s} \times \left( \frac{50 \text{ mm}}{100 \text{ mm}} \right)^2 \] \[ v_2 = 8 \times \left( \frac{1}{2} \right)^2 \] \[ v_2 = 8 \times \frac{1}{4} \] \[ v_2 = 2 \text{ m/s} \]

Step 4: Final Answer:
The velocity at section 2 is 2 m/s.