Electric Pump Shaft Power Calculator

Shaft Power Equation:

\[ P_{\text{shaft}} = \frac{\rho \times g \times Q \times H}{\eta} \]

Density (ρ):

kg/m³

Flow Rate (Q):

m³/s

Head (H):

Efficiency (η):

(0-1)

Shaft Power (P_shaft):

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1. What is Pump Shaft Power?

Shaft power is the mechanical power required at the pump shaft to deliver the hydraulic power to the fluid. It accounts for the pump's efficiency and represents the actual power needed to drive the pump.

2. How Does the Calculator Work?

The calculator uses the shaft power equation:

\[ P_{\text{shaft}} = \frac{\rho \times g \times Q \times H}{\eta} \]

Where:

\( P_{\text{shaft}} \) — Shaft power (W)
\( \rho \) — Fluid density (kg/m³)
\( g \) — Gravitational acceleration (9.81 m/s²)
\( Q \) — Volumetric flow rate (m³/s)
\( H \) — Total dynamic head (m)
\( \eta \) — Pump efficiency (0-1)

Explanation: The equation calculates the power required to move a fluid with given properties at a specified flow rate against a certain head, adjusted for the pump's efficiency.

3. Importance of Shaft Power Calculation

Details: Accurate shaft power calculation is essential for proper pump selection, motor sizing, and energy consumption estimation. It helps prevent underpowering or overpowering the pump system.

4. Using the Calculator

Tips:

For water at standard conditions, density is approximately 1000 kg/m³
Flow rate should be in cubic meters per second (m³/s)
Head is the total dynamic head in meters
Efficiency is a decimal value between 0 and 1 (e.g., 0.8 for 80% efficiency)

5. Frequently Asked Questions (FAQ)

Q1: What's the difference between shaft power and hydraulic power?
A: Hydraulic power is the theoretical power needed to move the fluid, while shaft power includes the pump's inefficiency and is the actual power required at the pump shaft.

Q2: How do I convert the result to horsepower?
A: 1 kW ≈ 1.341 horsepower. Multiply the kW result by 1.341 to get horsepower.

Q3: What's a typical pump efficiency range?
A: Centrifugal pumps typically range from 40% to 85% efficiency, with larger pumps generally being more efficient.

Q4: Why is density important in the calculation?
A: Denser fluids require more power to pump at the same flow rate and head because they have more mass per unit volume.

Q5: How does head affect power requirements?
A: Power is directly proportional to head - doubling the head doubles the power requirement (all else being equal).