
ENGINEERING TOOL
Pump power
The hydraulic power of a pump is P = ρ·g·Q·H: density × gravity × flow × total head. With flow in m³/h and head in meters, P [kW] = Q × H × specific gravity ÷ 367. Shaft power (BHP) is the hydraulic power divided by the pump efficiency at the duty point.
At a glance
- Hydraulic (useful) power is P = ρ·g·Q·H. In practical form, kW = flow[m³/h] × head[m] × relative density / 367.
- Shaft power (BHP) is always higher than hydraulic: BHP = hydraulic ÷ efficiency. No pump is 100% efficient.
- BHP to CV: multiply by 1.0139, because 1 hp = 745.6999 W and 1 CV = 735.4988 W (NIST SP 811).
- Motor selection considers the absorbed-power curve, starting torque, duty and manufacturer criteria.
- Use efficiency from the model curve at duty flow. Maximum efficiency at BEP does not automatically represent another operating point.
Updated
Answers to guide your selection
What is the formula for pump power?
Hydraulic power is P_h = ρ·g·Q·H, in watts, with ρ in kg/m³, g = 9.80665 m/s², Q in m³/s and H in meters. In practice, P_h [kW] = Q [m³/h] × H [m] × specific gravity ÷ 367. Shaft power, or BHP, is P_h divided by the pump efficiency at the duty point, read from the model curve.
How do I convert BHP to CV?
Multiply by 1.0139. BHP is in mechanical horsepower (745.6999 W) and CV is metric horsepower (735.4988 W), per NIST SP 811. Example: 10 BHP × 1.0139 = 10.14 CV. The other way round, CV × 0.98632 = hp. For kW, BHP × 0.7457.
Why is shaft power higher than hydraulic power?
Efficiency relates power transferred to the liquid to power received at the shaft; the difference is the pump's mechanical, volumetric and hydraulic losses. According to the Hydraulic Institute, at 50% efficiency the motor must deliver twice the hydraulic power. Compare models at the same duty point and with the same fluid; peak efficiency alone does not determine installation consumption.
Data for assessing pump power
Use the information below to start the assessment with the technical team.
Discuss your duty point with engineering
Send the flow, total head, fluid and the pump curve or model. FB Bombas engineering checks hydraulic and shaft power at the duty point and indicates the pump and drive.
Selection depends on process conditions and confirmed data. The assessment and supply scope will be defined with the technical team.
Information for the assessment
- Operating flow (m³/h)
- Total dynamic head (m)
- Fluid and density, if not water
- Model curve with efficiency and power at the duty point; if unavailable, identify model and speed
- Operating regime (continuous or intermittent) — for the motor margin
The pump power formulas
Hydraulic power is transferred to the liquid; shaft power includes pump losses. Motor selection requires the power curve, starting torque and service conditions.
P_h = ρ · g · Q · HPractical: kW = Q[m³/h]·H[m]·relative density / 367. It is the energy delivered to the fluid.
P_shaft = P_h / ηη = efficiency at the duty point, expressed as a fraction. Consult the model curve.
P_h → P_shaft → P_motorConfirm motor and margin for the curve and duty. Service factor is not a generic percentage.
How to convert BHP to CV and kW
BHP (brake horsepower) is shaft power in mechanical horsepower. Per NIST SP 811, 1 hp = 745.6999 W and 1 CV (metric horsepower) = 735.4988 W; so 1 hp = 1.0139 CV, and to go from BHP to CV you multiply by 1.0139. NIST also lists the electric horsepower, of 746 W: check which unit the nameplate or proposal uses. The factors in the table are rounded.
| From | To | Multiply by |
|---|---|---|
| kW | CV | × 1.35962 |
| kW | HP | × 1.34102 |
| CV | kW | × 0.735499 |
| HP | kW | × 0.745700 |
| CV | HP | × 0.986320 |
| HP (BHP) | CV | × 1.01387 |
Hydraulic power formula in practical units
All forms come from P_h = ρ·g·Q·H, or from P_h = Q·Δp when the known value is pressure, with g = 9.80665 m/s² (standard gravity), 1 bar = 100 kPa and 1 kgf/cm² = 98.0665 kPa, per NIST SP 811. SG is the fluid specific gravity (water ≈ 1) at the pumping temperature.
| Inputs | Hydraulic power | Where the constant comes from |
|---|---|---|
| Q in m³/h, H in m | P_h [kW] = Q × H × SG ÷ 367 | 3,600 ÷ 9.80665 = 367.1 |
| Q in L/s, H in m | P_h [kW] = Q × H × SG ÷ 102 | 1,000 ÷ 9.80665 = 102.0 |
| Q in m³/h, Δp in bar | P_h [kW] = Q × Δp ÷ 36 | 3,600 ÷ 100 = 36 |
| Q in m³/h, Δp in kgf/cm² | P_h [kW] = Q × Δp ÷ 36.71 | 3,600 ÷ 98.0665 = 36.71 |
Worked power calculation
Water at 20 °C (specific gravity 0.998, NIST), 50 m³/h and 40 m of total head, with an assumed efficiency of 70% at the duty point. In a real selection, efficiency comes from the model curve at that flow.
| Step | Calculation | Result |
|---|---|---|
| Hydraulic power | 50 × 40 × 0.998 ÷ 367 | 5.44 kW |
| Shaft power (BHP in kW) | 5.44 ÷ 0.70 | 7.77 kW |
| Shaft power in CV | 7.77 × 1.35962 | 10.6 CV |
| Shaft power in hp (BHP) | 7.77 × 1.34102 | 10.4 hp |
The motor does not come from this point alone: selection considers the absorbed-power curve over the whole expected flow range, starting torque and duty.
The head that feeds this calculation includes the piping head loss; and poorly sized suction compromises the NPSH. Request all three calculations in the same conversation.
Frequently Asked Questions
How do I calculate the power of a pump?
Hydraulic (useful) power is P_h = ρ·g·Q·H, where ρ is the fluid density, g gravity, Q the flow, and H the total head. In practical form, P_h [kW] = Q[m³/h]·H[m]·relative density / 367. Shaft power (what the motor must deliver) is higher: divide the hydraulic power by the pump efficiency. FB engineering checks both against the model curve at the duty point.
What is the difference between hydraulic, shaft (BHP) and motor power?
Hydraulic power is transferred to the liquid. Shaft power also accounts for pump losses and depends on efficiency at the duty point. The motor must meet required power and torque across the intended range. Motor rated power, efficiency and service factor are distinct information.
What is pump efficiency and why does it matter?
Efficiency is the ratio of hydraulic power to shaft power. Read the value on the model curve at the intended flow and speed. BEP identifies the maximum on that curve; selection must consider duty point and operating range, without assigning one percentage to the whole FBCN series.
What margin should I use to size the motor?
Select the motor using the absorbed-power curve, starting torque, duty and installation conditions. Confirm margin and rated power against pump and motor manufacturer criteria. Do not treat service factor as a universal 10%, 15% or 25% addition to a calculation.
How do I convert kW to CV and HP in pump power?
Check whether the document uses kW, metric CV or mechanical HP. Rounded NIST factors are 1 CV ≈ 0.735499 kW and 1 HP ≈ 0.745700 kW. To convert kW, use approximately 1.35962 CV or 1.34102 HP per kW. Record the unit alongside the value and verify power on the nameplate and curve.
Does fluid density change the pump power?
Yes, directly. Hydraulic power is proportional to density: pumping a fluid denser than water (a brine, for example) requires more power for the same flow and head; a lighter oil requires less. That is why engineering asks for the water temperature (which sets the density) or the informed density for other fluids. Note: head in meters of column is already independent of density — what density scales is the energy per meter. According to the Hydraulic Institute, a specific gravity of 1.1 requires 10% more power than 1.0.
What is BHP in a pump?
BHP (brake horsepower) is the power the pump shaft receives, expressed in hp. It adds the pump losses to the hydraulic power: BHP = hydraulic power ÷ efficiency. On the model curve, absorbed power varies with flow, so a BHP value applies to one duty point. In CV, BHP × 1.0139; in kW, BHP × 0.7457.
How do I calculate power from pressure, as in gear pumps?
When the known value is pressure, as with positive displacement pumps, use P_h = Q × Δp. With Q in m³/h and Δp in kgf/cm², P_h [kW] = Q × Δp ÷ 36.71; with Δp in bar, ÷ 36. Example: 10 m³/h and 7 kgf/cm² give 10 × 7 ÷ 36.71 = 1.91 kW of hydraulic power. Shaft power is that value divided by the efficiency at the duty point, which engineering checks for the fluid and speed.
NIST: power conversion factors
Hydraulic Institute: power and efficiency curves
Hydraulic Institute: why pump power is higher than expected
Assessment depends on installation data, the curve and requested scope. Confirm conditions in the technical proposal. For fire systems, documentation must meet the project and standards and requirements applicable to the location.
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