The diameter of a centrifugal pump, which is discharging 0.035 m³/s of water against a total head of 25 m is 0.5 m. The pump is running at 1200 r.p.m. Find the head, discharge and ratio of powers of a geometrically similar pump of diameter 0.3 m when it is running at 2000 r.p.m.

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Pump Affinity Law Calculation

Problem Statement

The diameter of a centrifugal pump, which is discharging 0.035 m³/s of water against a total head of 25 m is 0.5 m. The pump is running at 1200 r.p.m. Find the head, discharge and ratio of powers of a geometrically similar pump of diameter 0.3 m when it is running at 2000 r.p.m.

Given Data & Constants

  • Pump 1 (Model):
  • Discharge, \(Q_1 = 0.035 \, \text{m}^3/\text{s}\)
  • Head, \(H_1 = 25 \, \text{m}\)
  • Diameter, \(D_1 = 0.5 \, \text{m}\)
  • Speed, \(N_1 = 1200 \, \text{r.p.m.}\)

  • Pump 2 (Prototype):
  • Diameter, \(D_2 = 0.3 \, \text{m}\)
  • Speed, \(N_2 = 2000 \, \text{r.p.m.}\)

Solution

1. Find the Head of the Second Pump (\(H_2\))

Using the affinity law for head:

$$ \frac{H_1}{N_1^2 D_1^2} = \frac{H_2}{N_2^2 D_2^2} $$ $$ H_2 = H_1 \left(\frac{N_2}{N_1}\right)^2 \left(\frac{D_2}{D_1}\right)^2 $$ $$ H_2 = 25 \times \left(\frac{2000}{1200}\right)^2 \times \left(\frac{0.3}{0.5}\right)^2 $$ $$ H_2 = 25 \times (1.667)^2 \times (0.6)^2 = 25 \times 2.778 \times 0.36 $$ $$ H_2 \approx 25.0 \, \text{m} $$

2. Find the Discharge of the Second Pump (\(Q_2\))

Using the affinity law for discharge:

$$ \frac{Q_1}{N_1 D_1^3} = \frac{Q_2}{N_2 D_2^3} $$ $$ Q_2 = Q_1 \left(\frac{N_2}{N_1}\right) \left(\frac{D_2}{D_1}\right)^3 $$ $$ Q_2 = 0.035 \times \left(\frac{2000}{1200}\right) \times \left(\frac{0.3}{0.5}\right)^3 $$ $$ Q_2 = 0.035 \times (1.667) \times (0.6)^3 = 0.035 \times 1.667 \times 0.216 $$ $$ Q_2 \approx 0.0126 \, \text{m}^3/\text{s} $$

3. Find the Ratio of Powers (\(P_2 / P_1\))

Using the affinity law for power (assuming fluid density \(\rho\) is constant):

$$ \frac{P_1}{N_1^3 D_1^5} = \frac{P_2}{N_2^3 D_2^5} $$ $$ \frac{P_2}{P_1} = \left(\frac{N_2}{N_1}\right)^3 \left(\frac{D_2}{D_1}\right)^5 $$ $$ \frac{P_2}{P_1} = \left(\frac{2000}{1200}\right)^3 \times \left(\frac{0.3}{0.5}\right)^5 $$ $$ \frac{P_2}{P_1} = (1.667)^3 \times (0.6)^5 \approx 4.63 \times 0.07776 $$ $$ \frac{P_2}{P_1} \approx 0.36 $$
Final Results:

Head of the second pump: \( H_2 \approx 25.0 \, \text{m} \)

Discharge of the second pump: \( Q_2 \approx 0.0126 \, \text{m}^3/\text{s} \) (or 12.6 L/s)

Ratio of powers (\(P_2/P_1\)): \( \approx 0.36 \)

Explanation of Pump Affinity Laws

Pump Affinity Laws are a set of formulas used to predict the performance of a centrifugal pump when its operating conditions (speed) or physical dimensions (diameter) change. These laws are based on the principle of "homologous units," which means they apply to pumps that are geometrically similar but may be different in size.

The three main laws relate:

  • Discharge (Q) to speed (N) and diameter (D): \(Q \propto N D^3\)
  • Head (H) to speed and diameter: \(H \propto N^2 D^2\)
  • Power (P) to speed and diameter: \(P \propto N^3 D^5\)

These relationships are extremely powerful tools in pump engineering, allowing designers to test a small-scale model and accurately predict the performance of a much larger, full-scale prototype, saving significant time and cost.

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