Question

Water of density \( 1000 \, \text{kg/m}^3 \) is pumped at a volumetric flow rate of \( 3.14 \times 10^{-2} \, \text{m}^3/\text{s} \), through a pipe of inner diameter \( 10 \, \text{cm} \) and length \( 100 \, \text{m} \), from a large Reservoir 1 to another large Reservoir 2 at a height \( 50 \, \text{m} \) above Reservoir 1, as shown in the figure. The flow in the pipe is in the turbulent regime with a Darcy friction factor \( f = 0.06 \), and a kinetic energy correction factor \( \alpha = 1 \). Take \( g = 9.8 \, \text{m/s}^2 \). If all minor losses are negligible, and the pump efficiency is \( 100\% \), the pump power, in \( \text{kW} \), rounded off to 2 decimal places, is: \includegraphics[width=0.5\linewidth]{q59 CE.PNG}

Hint:

For pipe flow problems, always calculate head loss using the Darcy-Weisbach equation and include elevation differences for pump head calculations.

Answer 1

Step 1: Bernoulli's equation with pump head. The energy equation between Reservoir 1 and Reservoir 2 with a pump is: \[ h_p = h_L + z_2 - z_1, \] where: - \( h_p \) is the pump head, - \( h_L \) is the head loss in the pipe, - \( z_2 - z_1 = 50 \, \text{m} \) is the elevation difference. Step 2: Calculate head loss \( h_L \). The head loss in the pipe is given by: \[ h_L = \frac{f L v^2}{D 2g}, \] where: - \( f = 0.06 \) (Darcy friction factor), - \( L = 100 \, \text{m} \) (pipe length), - \( D = 0.1 \, \text{m} \) (pipe diameter), - \( v = \frac{Q}{A} = \frac{3.14 \times 10^{-2}}{\pi (0.1/2)^2} \approx 4 \, \text{m/s} \) (flow velocity). Substitute values: \[ h_L = \frac{0.06 \cdot 100 \cdot 4^2}{0.1 \cdot 2 \cdot 9.8} = \frac{96}{1.96} \approx 49 \, \text{m}. \] Step 3: Total pump head \( h_p \). \[ h_p = h_L + z_2 - z_1 = 49 + 50 = 99 \, \text{m}. \] Step 4: Pump power. The pump power is: \[ P = \rho g Q h_p, \] where: - \( \rho = 1000 \, \text{kg/m}^3 \), - \( g = 9.8 \, \text{m/s}^2 \), - \( Q = 3.14 \times 10^{-2} \, \text{m}^3/\text{s} \), - \( h_p = 99 \, \text{m} \). Substitute values: \[ P = 1000 \cdot 9.8 \cdot 3.14 \times 10^{-2} \cdot 99 = 30506 \, \text{W} = 30.50 \, \text{kW}. \] Step 5: Conclusion. The pump power is \( 30.50 \, \text{kW} \).