Question

A cylindrical pipe has a radius of \( 0.1 \, \text{m} \). If the speed of water flowing through the pipe is \( 2 \, \text{m/s} \), calculate the volume flow rate of water through the pipe.

• A. \( 0.0628 \, \text{m}^3/\text{s} \)
• B. \( 0.0314 \, \text{m}^3/\text{s} \)
• C. \( 0.1256 \, \text{m}^3/\text{s} \)
• D. \( 0.02 \, \text{m}^3/\text{s} \)

Hint:

The volume flow rate is directly proportional to the cross-sectional area and the velocity of the fluid. Larger pipes or faster flowing fluids result in higher flow rates.

Answer 1

The Correct Option is A

The volume flow rate \( Q \) through a pipe is given by the formula: \[ Q = A \times v \] Where: - \( Q \) is the volume flow rate, - \( A \) is the cross-sectional area of the pipe, - \( v = 2 \, \text{m/s} \) is the speed of water. For a cylindrical pipe, the cross-sectional area \( A \) is given by: \[ A = \pi r^2 \] Where \( r = 0.1 \, \text{m} \) is the radius of the pipe. Substituting the value of \( r \): \[ A = \pi \times (0.1)^2 = \pi \times 0.01 = 0.0314 \, \text{m}^2 \] Now, calculating the volume flow rate: \[ Q = 0.0314 \times 2 = 0.0628 \, \text{m}^3/\text{s} \] Thus, the volume flow rate of water through the pipe is \( 0.0628 \, \text{m}^3/\text{s} \).