Question

A fluid flows through a pipe with a varying cross-sectional area. If the velocity of the fluid is \( v_1 = 4 \, \text{m/s} \) at a point where the cross-sectional area is \( A_1 = 2 \, \text{m}^2 \), and the velocity at another point where the cross-sectional area is \( A_2 = 1 \, \text{m}^2 \) is \( v_2 \), what is the velocity \( v_2 \)?

• A. \( 8 \, \text{m/s} \)
• B. \( 4 \, \text{m/s} \)
• C. \( 2 \, \text{m/s} \)
• D. \( 1 \, \text{m/s} \)

Hint:

For fluids flowing through a pipe with varying cross-section, use the principle of continuity \( A_1 v_1 = A_2 v_2 \) to find the velocity at different points in the pipe.

Answer 1

The Correct Option is A

A fluid flows through a pipe with variable cross-sectional areas. At a cross-section with area \( A_1 = 2 \, \text{m}^2 \), the fluid velocity is \( v_1 = 4 \, \text{m/s} \). At another cross-section with area \( A_2 = 1 \, \text{m}^2 \), the velocity is \( v_2 \). Determine \( v_2 \). Step 1: Apply the continuity principle The continuity principle for incompressible fluid flow asserts that the volumetric flow rate is constant. This is expressed as: \[ A_1 v_1 = A_2 v_2 \] Step 2: Input known values Substitute the given values \( A_1 = 2 \, \text{m}^2 \), \( v_1 = 4 \, \text{m/s} \), and \( A_2 = 1 \, \text{m}^2 \) into the continuity equation: \[ 2 \times 4 = 1 \times v_2 \] \[ 8 = v_2 \] Answer: The velocity \( v_2 \) is \( 8 \, \text{m/s} \).