Formula used
- 3-phase electrical input power:
P(kW)=√3·V·I·PF/1000 - 1-phase electrical input power:
P(kW)=V·I·PF/1000 - Shaft output power:
Pout(kW)=Pin(kW)·η - Torque-based shaft power:
P(kW)=2π·N·T/(60·1000) - Hydraulic power:
Ph(kW)=SG·9.81·Q(m³/s)·H(m) - Pump shaft power:
Pshaft=Ph/ηpumpand motor input:Pin=Pshaft/ηmotor - Operating and sizing:
Poper=P·LoadFactor,Prated=Poper·(1+ServiceMargin) - Power conversion:
HP=kW×1.3410
How to use this calculator
- Select the calculation basis that matches your known data.
- Enter realistic efficiency and power factor values.
- Set load factor for duty cycle or partial loading.
- Add a service margin to handle site variability and aging.
- Click Calculate to see results above the form.
- Use the export buttons to save a CSV or PDF record.
Note: Always verify against manufacturer curves and local electrical standards.
Example data table
| Scenario | Inputs | Typical use | Expected outputs |
|---|---|---|---|
| Electrical | 3-phase, 400 V, 30 A, PF 0.85, η 90%, load 100%, margin 10% | Dewatering pump | ~15.9 kW shaft, ~21–22 HP, rating ~17.5 kW |
| Mechanical | Torque 180 N·m, 1450 RPM, η 92%, load 80%, margin 15% | Concrete mixer drive | ~21.9 kW shaft (at 100%), sized by load and margin |
| Hydraulic | Flow 120 m³/h, head 35 m, SG 1.00, pump η 75%, motor η 90% | Site water transfer | Hydraulic ~11.4 kW, shaft ~15.1 kW, input ~16.8 kW |
These examples are approximate; real systems vary by friction, temperature, and controls.
1) Electrical input power for site motor loads
Construction equipment is commonly checked using measured voltage and current. Real power depends on power factor (PF), so the same current can represent different kW on different loads. Using PF-based kW improves generator sizing, feeder selection, and cost forecasting for temporary power distribution.
2) Efficiency chain from supply to usable shaft output
Electrical input becomes mechanical output after copper, iron, and ventilation losses. Typical motor efficiency ranges are 85–95% depending on size and class. If you start from required shaft power, divide by motor efficiency to estimate supply kW, then check that protective devices and starters match the expected current.
3) Torque and speed method for rotating machinery
For conveyors, mixers, and hoists, torque at operating speed directly defines shaft demand. Power rises linearly with torque and speed, so worst-case points (starts, jams, and peak throughput) should be evaluated. Add a service margin when duty is cyclic, shock-prone, or temperature is high.
4) Hydraulic power for pumps, dewatering, and slurry transfer
Pump duties convert flow and head into hydraulic power. Consistent units are critical: head in meters and flow in m³/s. The calculator links hydraulic power to pump shaft power using pump efficiency, and then to electrical input using motor efficiency. This supports practical selection when pipeline losses and elevation vary.
5) Example data for quick validation and reporting
| Case | Key inputs | Check value |
|---|---|---|
| 3‑phase electrical | 400 V, 50 A, PF 0.85, η 90% | Pin ≈ 29.4 kW, Pout ≈ 26.5 kW |
| Torque method | 1450 rpm, 120 N·m, η 92% | Pshaft ≈ 18.2 kW, Pin ≈ 19.8 kW |
| Dewatering pump | Q 0.03 m³/s, H 35 m, SG 1.0, ηpump 75%, ηmotor 90% | Prated often 20–25 kW with margin |
Use these cases to sanity-check field readings and document assumptions in your calculation reports.
FAQs
1) Which basis should I choose?
Choose electrical when you know V, I, and PF; mechanical when you know torque and speed; hydraulic when you know pump flow, head, and efficiencies.
2) What is the difference between operating and rated power?
Operating power is what the motor normally delivers. Rated power includes margin for starts, peaks, and environmental conditions, so the motor can run reliably without overheating.
3) Why does power factor matter?
PF adjusts apparent power to real kW. Lower PF means higher current for the same kW, affecting cable sizing, generator loading, and protective device settings.
4) How should I enter efficiencies?
Use realistic field values. Motor efficiency is typically 85–95%. Pump efficiency varies widely (50–85%). If unknown, use conservative values and validate with manufacturer curves.
5) What units are assumed for hydraulic inputs?
Flow is entered in m³/s (the calculator also accepts L/s and m³/h conversions). Head is in meters. Specific gravity is dimensionless, with water close to 1.0.
6) Is the current estimate accurate for all motors?
It is an engineering estimate using rated voltage, PF, and efficiency. Actual current varies with load, VFD settings, harmonics, and supply voltage deviations.
7) What service margin should I apply?
Common margins are 10–25% depending on duty. Use higher margins for frequent starts, variable loads, high ambient temperature, or dusty sites.