Find the power required to drive a centrifugal pump which delivers 0.02 m³/s of water to a height of 30 m through a 10 cm diameter pipe and 90 m long. The overall efficiency of the pump is 70% and f = 0.009 in the formula of head loss.

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Pump Power Requirement Calculation

Problem Statement

Find the power required to drive a centrifugal pump which delivers 0.02 m³/s of water to a height of 30 m through a 10 cm diameter pipe and 90 m long. The overall efficiency of the pump is 70% and f = 0.009 in the formula \(h_f = \frac{4fLV^2}{2gD}\).

Given Data & Constants

  • Discharge, \(Q = 0.02 \, \text{m}^3/\text{s}\)
  • Static head (height), \(h_s = 30 \, \text{m}\)
  • Pipe diameter, \(D = 10 \, \text{cm} = 0.1 \, \text{m}\)
  • Pipe length, \(L = 90 \, \text{m}\)
  • Overall efficiency, \(\eta_o = 70\% = 0.70\)
  • Friction factor, \(f = 0.009\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

1. Calculate Velocity of Water in Pipe (V)

The velocity is the discharge divided by the cross-sectional area of the pipe.

$$ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.1)^2 \approx 0.007854 \, \text{m}^2 $$ $$ V = \frac{Q}{A} = \frac{0.02}{0.007854} \approx 2.546 \, \text{m/s} $$

2. Calculate Head Loss due to Friction (\(h_f\))

We use the given Darcy-Weisbach formula to find the head lost to friction in the pipe.

$$ h_f = \frac{4 \cdot f \cdot L \cdot V^2}{2 \cdot g \cdot D} $$ $$ h_f = \frac{4 \times 0.009 \times 90 \times (2.546)^2}{2 \times 9.81 \times 0.1} $$ $$ h_f = \frac{4 \times 0.009 \times 90 \times 6.482}{1.962} $$ $$ h_f = \frac{21.00}{1.962} \approx 10.70 \, \text{m} $$

3. Calculate Total Manometric Head (\(H_m\))

The total head the pump must work against is the sum of the static lift height and the frictional head loss.

$$ H_m = \text{Static Head} + \text{Frictional Head} = h_s + h_f $$ $$ H_m = 30 \, \text{m} + 10.70 \, \text{m} = 40.70 \, \text{m} $$

4. Calculate the Power Required (Shaft Power)

First, we find the power delivered to the water (Water Power). Then, we use the overall efficiency to find the required input power (Shaft Power).

$$ \text{Water Power, } P_w = \rho \cdot g \cdot Q \cdot H_m $$ $$ P_w = 1000 \times 9.81 \times 0.02 \times 40.70 $$ $$ P_w \approx 7985.34 \, \text{W} $$ $$ \text{Shaft Power, } P_s = \frac{\text{Water Power}}{\eta_o} = \frac{P_w}{0.70} $$ $$ P_s = \frac{7985.34}{0.70} \approx 11407.6 \, \text{W} $$
Final Result:

The power required to drive the pump is approximately \(11.41 \, \text{kW}\).

Explanation of Terms

Static Head (\(h_s\)): This is the vertical distance the water needs to be lifted (30 m). It represents the change in potential energy of the water.

Frictional Head Loss (\(h_f\)): As water flows through the 90 m long pipe, energy is lost due to friction between the water and the pipe walls. This energy loss is equivalent to needing to pump the water an extra vertical distance, which we calculated as 10.70 m.

Manometric Head (\(H_m\)): This is the total equivalent height the pump must overcome. It's the sum of the static head and all losses in the system. Here, \(H_m = 30 + 10.70 = 40.70\) m.

Overall Efficiency (\(\eta_o\)): This accounts for all losses within the pump and its motor (mechanical friction, hydraulic losses, electrical losses). It's the ratio of the useful power delivered to the water to the actual electrical power consumed by the motor.

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