Question

Water flows through a smooth circular pipe of diameter 10 cm and length 10 m. The pressure drop across the length of the pipe is 0.2 Pa. Kinematic viscosity and density of water are \(1 \times 10^{-6} \, {m}^2/{s}\) and 1000 \({kg/m}^3\), respectively. Assuming laminar and fully developed flow throughout the pipe, the velocity of water (in mm/s) at the center of the pipe is ........ (Rounded off to one decimal place)

Hint:

In problems involving laminar flow, use the equation for fully developed flow to calculate the maximum velocity at the center of the pipe.

Answer 1

The velocity at the center of a fully developed laminar flow in a circular pipe is given by the following equation:
\[ v_{{max}} = \frac{\Delta P \cdot R^2}{4 \mu L} \] Where:
- \( \Delta P = 0.2 \, \text{Pa} \) (pressure drop),
- \( R = \frac{D}{2} = 5 \, \text{cm} = 0.05 \, \text{m} \) (radius of the pipe),
- \( L = 10 \, \text{m} \) (length of the pipe),
- \( \mu \) is the dynamic viscosity, and \( \mu = \rho \nu \), where \( \nu = 1 \times 10^{-6} \, \text{m}^2/\text{s} \) is the kinematic viscosity and \( \rho = 1000 \, \text{kg/m}^3 \) is the density of water.
First, calculate the dynamic viscosity \( \mu \):
\[ \mu = \rho \cdot \nu = 1000 \, \text{kg/m}^3 \times 1 \times 10^{-6} \, \text{m}^2/\text{s} = 1 \times 10^{-3} \, \text{Pa} \cdot \text{s} \] Now substitute the values into the equation for \( v_{{max}} \):
\[ v_{{max}} = \frac{0.2 \times (0.05)^2}{4 \times 1 \times 10^{-3} \times 10} \] \[ v_{{max}} = \frac{0.2 \times 0.0025}{4 \times 10^{-3} \times 10} = \frac{0.0005}{0.04} = 0.0125 \, \text{m/s} \] Convert the velocity to mm/s:
\[ v_{{max}} = 0.0125 \times 1000 = 12.5 \, \text{mm/s} \] Thus, the velocity of water at the center of the pipe is \( \boxed{12.5 \, \text{mm/s}} \).