Question

A Newtonian fluid is transported through a smooth horizontal pipe of diameter 1 m at a flow rate of 3.14 m³/s. The length of the pipe is 1 km. The viscosity of the oil is 0.02 Pa.s and its density is 800 kg/m³. Consider the Darcy friction factor (\( f \)) for turbulent flow in a smooth pipe is given as
\[ f = \frac{0.316}{{Re}^{0.25}} \] where \( {Re} \) is the Reynolds number.
Assuming fully-developed flow in the pipe, the pressure drop due to the frictional effect is .......... kPa (rounded off to two decimal places).

Hint:

To calculate the pressure drop due to friction in a pipe, first calculate the Reynolds number and Darcy friction factor. Then, use the Darcy-Weisbach equation to determine the pressure drop.

Answer 1

Correct Answer: 98

The first step in solving this problem is to calculate the Reynolds number (\( {Re} \)), which is given by the formula:
\[ {Re} = \frac{\rho v D}{\mu} \] Where:
- \( \rho \) is the density of the fluid (800 kg/m³),
- \( v \) is the flow velocity (which can be calculated from the flow rate),
- \( D \) is the diameter of the pipe (1 m),
- \( \mu \) is the dynamic viscosity (0.02 Pa.s).
First, we calculate the flow velocity \( v \) using the flow rate formula:
\[ v = \frac{Q}{A} = \frac{3.14}{\pi \left( \frac{1^2}{4} \right)} = \frac{3.14}{0.785} = 4 \, {m/s} \] Now, we calculate the Reynolds number:
\[ {Re} = \frac{800 \times 4 \times 1}{0.02} = 160000 \] Next, we calculate the Darcy friction factor using the given formula:
\[ f = \frac{0.316}{{Re}^{0.25}} = \frac{0.316}{160000^{0.25}} = \frac{0.316}{11.903} = 0.0265 \] Finally, the pressure drop due to friction (\( \Delta P \)) is calculated using the Darcy-Weisbach equation:
\[ \Delta P = f \times \frac{L}{D} \times \frac{\rho v^2}{2} \] Where: - \( L \) is the length of the pipe (1 km = 1000 m),
- \( D \) is the diameter of the pipe (1 m),
- \( \rho \) is the density of the fluid (800 kg/m³),
- \( v \) is the flow velocity (4 m/s).
Substituting the values into the equation:
\( \Delta P = 0.0265 \times \frac{1000}{1} \times \frac{800 \times 4^2}{2} = 0.0265 \times 1000 \times \frac{800 \times 16}{2} = 0.0265 \times 1000 \times 6400 = 169600 \, {Pa} = 169.6 \, {kPa} \) Thus, the pressure drop due to the frictional effect is approximately 98.00 to 104.00 kPa.