⚙️ Pump Affinity Laws
Predict performance changes in centrifugal pumps due to rotation speed variations or impeller trim diameter modifications. Overlays original and scaled Q-H curves.
⚡ Fortran 90 Engine
Double Precision (IEEE 754)
✓ ISO / ASME Validated
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📦 Fortran Code
4.4 KB
📅 Released
Jun 2026
⏱️ Latency
< 1 ms
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📋 Sizing Inputs
Units:
SI Metric (m³/h, m, kW)
📊 Scaling Performance Results
Original (Before)
Flow Rate ($Q_1$):
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Pump Head ($H_1$):
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Shaft Power ($P_1$):
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NPSH Required:
—
Efficiency ($\eta_1$):
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Predicted (After)
Flow Rate ($Q_2$):
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Pump Head ($H_2$):
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Shaft Power ($P_2$):
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NPSH Required:
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Efficiency ($\eta_2$):
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Affinity Laws Active
Calculations are within standard valid regimes.
📈 Operating Point & Curve Overlay
📘 Calculation Methodology: Centrifugal Pump Affinity Laws
Mathematical Model & Theory
The affinity laws govern centrifugal pump performance scaling when modifying rotational speed $N$ or trimming impeller diameter $D$:
$$\frac{Q_1}{Q_2} = \left(\frac{N_1}{N_2}\right)\left(\frac{D_1}{D_2}\right), \quad \frac{H_1}{H_2} = \left(\frac{N_1}{N_2}\right)^2\left(\frac{D_1}{D_2}\right)^2, \quad \frac{P_1}{P_2} = \left(\frac{N_1}{N_2}\right)^3\left(\frac{D_1}{D_2}\right)^3$$
Assumptions
- Geometric and kinematic similarity maintained across operating speeds.
- Constant hydraulic efficiency across modest speed/diameter trimming adjustments ($\pm 20\%$).
Academic References
- Karassik, I. J. et al.: Pump Handbook, McGraw-Hill.
- Stepanoff, A. J.: Centrifugal and Axial Flow Pumps, Krieger.
Worked Engineering Example
Problem Statement:
A pump operating at $N_1 = 1750\text{ RPM}$ delivers $Q_1 = 100\text{ m}^3/\text{h}$ at head $H_1 = 40\text{ m}$ consuming $P_1 = 15\text{ kW}$. Find performance if driven by a VFD at $N_2 = 1400\text{ RPM}$.
Step-by-step Solution:
1. Speed ratio: $\alpha = 1400 / 1750 = 0.80$.
2. $Q_2 = 100 \times 0.80 = 80.0\text{ m}^3/\text{h}$.
3. $H_2 = 40 \times (0.80)^2 = 40 \times 0.64 = 25.6\text{ m}$.
4. $P_2 = 15 \times (0.80)^3 = 15 \times 0.512 = 7.68\text{ kW}$.
Final Result:
At 1400 RPM: $Q = \mathbf{80\text{ m}^3/h}$, Head = $\mathbf{25.6\text{ m}}$, Power = $\mathbf{7.68\text{ kW}}$ (48.8% power savings).
A pump operating at $N_1 = 1750\text{ RPM}$ delivers $Q_1 = 100\text{ m}^3/\text{h}$ at head $H_1 = 40\text{ m}$ consuming $P_1 = 15\text{ kW}$. Find performance if driven by a VFD at $N_2 = 1400\text{ RPM}$.
Step-by-step Solution:
1. Speed ratio: $\alpha = 1400 / 1750 = 0.80$.
2. $Q_2 = 100 \times 0.80 = 80.0\text{ m}^3/\text{h}$.
3. $H_2 = 40 \times (0.80)^2 = 40 \times 0.64 = 25.6\text{ m}$.
4. $P_2 = 15 \times (0.80)^3 = 15 \times 0.512 = 7.68\text{ kW}$.
Final Result:
At 1400 RPM: $Q = \mathbf{80\text{ m}^3/h}$, Head = $\mathbf{25.6\text{ m}}$, Power = $\mathbf{7.68\text{ kW}}$ (48.8% power savings).