What Does This Boiler Feed Pump Power Calculator Do?
This calculator estimates the power associated with one boiler feed pump operating point. It starts with feedwater flow and total pump head, uses feedwater temperature to estimate water density, and then calculates hydraulic power delivered to the liquid.
Pump efficiency converts hydraulic power into required shaft power. Motor efficiency is then used to estimate the electrical input needed to provide that shaft output. A separate margin can be applied to shaft power to produce a preliminary motor-output requirement.
The tool focuses only on power. It does not calculate feedwater flow, total dynamic head, NPSH, pipe friction, control-valve loss, electrical cable size or starting current.
How Boiler Feed Pump Power Is Calculated
Hydraulic power depends on liquid density, gravitational acceleration, volumetric flow and pump head. The required pump shaft power is higher because no pump transfers all shaft energy to the liquid. Electrical input is higher again because the motor also has losses.
Hydraulic power (kW) = ρ × g × Q × H ÷ 1000Pump shaft power = Hydraulic power ÷ Pump efficiencyEstimated electrical input = Pump shaft power ÷ Motor efficiencyDesign motor output = Pump shaft power × (1 + Margin % ÷ 100)
Feedwater density in kg/m³9.80665 m/s²Volumetric flow in m³/sTotal pump head in metresHow to Use the Calculator
- Select Metric or US Units. Metric mode accepts m³/h and metres. US mode accepts US GPM and feet.
- Enter design feedwater flow. Use the required pump flow at the operating condition being checked.
- Enter total pump head. Flow and head must represent the same duty point.
- Enter pump efficiency. Use the efficiency shown near the duty point on a reliable pump curve.
- Enter motor efficiency. This estimates electrical input power from shaft output.
- Enter feedwater temperature. The calculator uses temperature to estimate density.
- Apply a visible motor-output margin. Use the allowance required by the project basis rather than hiding it inside other inputs.
Understanding the Power Results
Hydraulic power
Hydraulic power is the useful rate of energy transferred to the feedwater. It is determined by flow, head and density and does not include pump or motor losses.
Pump shaft power
Shaft power is the mechanical power the pump must receive at its shaft to produce the required hydraulic output. It is calculated by dividing hydraulic power by pump efficiency.
Estimated electrical input
Electrical input is the approximate power drawn by the motor at the calculated shaft load. It is found by dividing shaft power by motor efficiency. Actual measured input can differ because motor efficiency varies with load and because drives or transmission components may add further losses.
Design motor output
This is the calculated shaft requirement after applying the selected margin. It is a preliminary value to compare with available motor-output ratings. It is not an instruction to select the closest smaller motor.
Combined efficiency
The combined pump-and-motor efficiency is the product of the two entered efficiencies. It shows the approximate fraction of electrical input converted into useful hydraulic power.
Worked Example
Consider a feed pump operating at 12.27 m³/h and 164.16 m total head. Feedwater temperature is 105°C, pump efficiency is 70%, motor efficiency is 90%, and the selected motor-output margin is 10%.
| Stage | Calculation | Approximate result |
|---|---|---|
| Feedwater density | Temperature interpolation at 105°C | 954.7 kg/m³ |
| Hydraulic power | 954.7 × 9.80665 × (12.27 ÷ 3600) × 164.16 | 5.24 kW |
| Pump shaft power | 5.24 ÷ 0.70 | 7.48 kW |
| Electrical input | 7.48 ÷ 0.90 | 8.31 kW |
| Design motor output | 7.48 × 1.10 | 8.23 kW |
The preliminary design motor output is therefore about 8.23 kW, or 11.04 hp. The final selected rating should be rounded up according to available motor sizes and checked against the pump manufacturer's maximum absorbed-power requirement over the intended operating range.
What Values Should You Enter?
The result is only as reliable as the operating-point data and efficiency assumptions entered. Keep every value tied to the same design condition.
Design flow
Enter the volumetric flow required from the pump. Do not mix a maximum flow with head from a different point.
Total pump head
Enter the required head at the selected flow. The power calculation assumes this head has already been established.
Pump efficiency
Use efficiency at the duty point. Efficiency near shut-off or far from the best-efficiency region may be substantially lower.
Motor efficiency
Use a realistic efficiency for the expected motor size and load. Nameplate or manufacturer data is preferable.
Feedwater temperature
Temperature changes water density and therefore changes hydraulic power for a fixed volumetric flow and head.
Motor-output margin
Apply only the explicit allowance required by the design basis. Avoid duplicating margins already present in flow or head.
Hydraulic Power, Shaft Power and Motor Power
These values describe different points in the energy path and should not be used interchangeably.
| Power value | Meaning | Typical use |
|---|---|---|
| Hydraulic power | Useful power transferred to the feedwater | Checks the physical duty created by flow and head |
| Pump shaft power | Mechanical power required at the pump shaft | Compared with the pump absorbed-power curve |
| Electrical input power | Approximate electrical demand at the motor terminals | Preliminary energy and loading estimate |
| Design motor output | Shaft requirement after the selected margin | Starting point for choosing an available motor rating |
Why Efficiency Changes the Required Power
Flow and head establish the hydraulic duty, but efficiency determines how much input power is needed to create it. A lower pump efficiency increases shaft power even when flow and head remain unchanged. A lower motor efficiency increases electrical input without changing the shaft power required by the pump.
Efficiency should therefore be taken from the expected operating point, not copied from a pump's maximum published efficiency. The same pump can operate at different efficiencies as flow changes.
Common Boiler Feed Pump Power Calculation Mistakes
- Using mismatched flow and head: power must be calculated from values that occur at the same operating point.
- Entering efficiency as a whole number in a manual formula: 70% must be used as 0.70 when dividing outside this calculator.
- Confusing hydraulic power with motor rating: hydraulic power excludes both pump and motor losses.
- Using maximum efficiency for every duty: actual duty-point efficiency may be lower than the best published value.
- Ignoring feedwater density: hot feedwater has a lower density than cold water, which affects hydraulic power.
- Applying the same margin twice: check whether flow, head or manufacturer absorbed power already includes an allowance.
- Selecting the next smaller motor: preliminary calculated output should be rounded up and verified, not rounded down.
- Using electrical input as motor rated output: motor nameplate output and electrical input are different quantities.
Using the Result for Preliminary Motor Selection
Use the design motor output as a screening value, then compare it with the available motor-output ratings used on the project. The chosen motor should provide adequate output at the required operating condition without relying on continuous overload.
Manufacturer data remains essential. Review the pump's certified absorbed-power curve across the expected operating range, not only at one calculated point. The maximum expected shaft demand may occur away from the nominal duty, depending on the pump curve and operating limits.
Electrical input from this calculator is an estimate based on the entered motor efficiency. Actual power consumption may also be influenced by motor loading, supply conditions, variable-frequency-drive losses and the final selected equipment.
Frequently Asked Questions
What is the basic boiler feed pump power formula?
Hydraulic power is density multiplied by gravity, volumetric flow and pump head. Pump shaft power is hydraulic power divided by pump efficiency.
Is pump shaft power the same as motor input power?
No. Pump shaft power is mechanical output delivered by the motor to the pump. Electrical input is higher because the motor is not 100% efficient.
Should I use pump efficiency or motor efficiency?
Use both when both shaft power and electrical input are required. Pump efficiency converts hydraulic power to shaft power, while motor efficiency converts shaft output to electrical input.
Why does feedwater temperature affect pump power?
Temperature changes water density. For fixed volumetric flow and head, a lower density produces slightly lower hydraulic power.
What motor power margin should I enter?
Use the margin specified by the project or equipment-selection basis. Avoid adding another margin when the pump absorbed-power data or duty inputs already include one.
Can this calculator provide the final motor size?
No. It provides a preliminary power requirement. Final selection requires available motor ratings and a check of the pump manufacturer's certified absorbed-power curve over the full intended operating range.