Question

A Newtonian fluid flows through a smooth horizontal pipe of diameter \(1 \, \text{m}\), length \(1 \, \text{km}\), flow rate \(3.14 \, \text{m}^3/\text{s}\). Viscosity \(\mu = 0.02 \, \text{Pa·s}\), density \(\rho = 800 \, \text{kg/m}^3\). The Darcy friction factor for turbulent flow is: \[ f = \frac{0.316}{Re^{0.25}} \] Find pressure drop due to friction (kPa).

Hint:

When solving fluid mechanics friction problems: 1. Find velocity from flow rate. 2. Calculate Reynolds number. 3. Use given friction factor relation. 4. Apply Darcy–Weisbach equation for pressure drop.

Answer 1

Step 1: Average velocity.
\[ A = \frac{\pi D^2}{4} = \frac{\pi (1)^2}{4} = 0.785 \, \text{m}^2 \] \[ V = \frac{Q}{A} = \frac{3.14}{0.785} = 4 \, \text{m/s} \] Step 2: Reynolds number.
\[ Re = \frac{\rho V D}{\mu} = \frac{800 \times 4 \times 1}{0.02} = \frac{3200}{0.02} = 160000 \] Step 3: Friction factor.
\[ f = \frac{0.316}{Re^{0.25}} = \frac{0.316}{(1.6 \times 10^5)^{0.25}} \] \[ Re^{0.25} = (1.6 \times 10^5)^{0.25} \approx 20.02 \] \[ f \approx \frac{0.316}{20.02} = 0.0158 \] Step 4: Pressure drop (Darcy–Weisbach).
\[ \Delta P = f \cdot \frac{L}{D} \cdot \frac{\rho V^2}{2} \] \[ \Delta P = 0.0158 \times \frac{1000}{1} \times \frac{800 \times 4^2}{2} \] \[ = 0.0158 \times 1000 \times 6400 = 101,120 \, \text{Pa} \] \[ = 101.12 \, \text{kPa} \] Step 5: Round off.
\[ \Delta P = 101.12 \, \text{kPa} \approx 101.12 \, \text{kPa} \] But since question asks in kPa with 2 decimals: \[ \boxed{101.12 \, \text{kPa}} \]