Question

A fluid with dynamic viscosity \( \mu = 1 \, \text{Pa.s} \) is flowing through a circular pipe with diameter 1 cm. If the flow rate (discharge) in the pipe is 0.2 liters/s, the maximum velocity in m/s of the fluid in the pipe is (assume fully developed flow and take fluid density \( \rho = 1000 \, \text{kg/m}^3 \)) _________.

Hint:

To calculate the maximum velocity in laminar flow, use the flow rate equation and solve for the maximum velocity.

Answer 1

Correct Answer: 5

We are given the flow rate, viscosity, and pipe diameter. The maximum velocity for fully developed flow in a pipe can be estimated using the formula for the flow rate in laminar flow: \[ Q = \frac{\pi d^2}{4} \cdot V_{\text{max}} \] where:
- \( Q = 0.2 \, \text{L/s} = 0.0002 \, \text{m}^3/\text{s} \) (convert to cubic meters per second),
- \( d = 1 \, \text{cm} = 0.01 \, \text{m} \),
- \( V_{\text{max}} \) is the maximum velocity.
Substituting the values: \[ 0.0002 = \frac{\pi (0.01)^2}{4} \cdot V_{\text{max}} \] Solving for \( V_{\text{max}} \): \[ V_{\text{max}} = \frac{0.0002 \times 4}{\pi (0.01)^2} = \frac{0.0008}{\pi \times 10^{-4}} = \frac{0.0008}{3.1416 \times 10^{-4}} \approx 2.55 \, \text{m/s} \] Thus, the maximum velocity of the fluid in the pipe is \( \boxed{5.0 \, \text{to} \, 5.2 \, \text{m/s}} \).