Welcome to “Back of the Envelope” — a column where each article gives you one engineering calculation you can do on a napkin, without opening software. Because the best engineers don’t reach for the calculator first. They reach for intuition.
The Rule
Pump shaft power (kW) ≈ Flow (m³/h) × Head (m) ÷ 270
For imperial: Pump shaft power (hp) ≈ Flow (gpm) × Head (ft) ÷ 3960
Then multiply by about 1.15 for motor power (to account for pump efficiency and service factor).
When to Use It
You’re in a meeting. Someone says “we need to pump 100 m³/h up 30 meters.” The vendor will email a quote next week. But you need a ballpark number RIGHT NOW — to check if the budget is even close, or to catch a gross error in someone’s datasheet.
This formula gives you pump hydraulic power. Add 15–30% for pump + motor inefficiency and you have motor power. It takes 30 seconds. It will be within ±15% of the final number. That’s enough for 90% of early-stage decisions.
The Math
The hydraulic power equation:
P = ρ × g × Q × H
Where:
- ρ = density of water = 1000 kg/m³
- g = 9.81 m/s²
- Q = flow rate (m³/s)
- H = head (m)
Convert Q from m³/h to m³/s: Q(m³/s) = Q(m³/h) ÷ 3600
So: P(W) = 1000 × 9.81 × (Q/3600) × H = 2.725 × Q × H
In kW: P(kW) = 2.725 × Q × H ÷ 1000 = Q × H ÷ 367
Then divide by pump efficiency (typically 0.70–0.80): Q × H ÷ (367 × 0.75) ≈ Q × H ÷ 275
I round to 270 (easier mental math). The error is <2%.
Examples
Example 1 — Cooling water pump:
- Flow: 200 m³/h
- Head: 25 m
- Power ≈ 200 × 25 ÷ 270 = 18.5 kW hydraulic
- Motor ≈ 18.5 ÷ 0.75 × 1.15 = 28.3 kW → pick 30 kW motor
- Reality check: A 200 m³/h, 25m pump typically ships with a 30 kW motor. ✓
Example 2 — Boiler feed pump:
- Flow: 10 m³/h
- Head: 150 m
- Power ≈ 10 × 150 ÷ 270 = 5.6 kW hydraulic
- Motor ≈ 5.6 ÷ 0.70 × 1.15 = 9.2 kW → pick 11 kW motor
- (Pump efficiency lower for small high-head pumps, use 0.70)
Example 3 — Sanity check someone else’s number:
- A vendor quotes 75 kW motor for 100 m³/h at 20 m.
- Hydraulic power = 100 × 20 ÷ 270 = 7.4 kW
- Even with inefficiency: 7.4 ÷ 0.70 × 1.15 = 12.2 kW
- The 75 kW number is off by 6×. Something is wrong — wrong flow, wrong head, or wrong pump.
When NOT to Use It
- High-viscosity fluids (>100 cP): Viscous losses dominate, use dedicated correlation
- Slurries (>10% solids): Density correction needed, and derating for solids
- Multiphase flow: Gas-liquid mixtures require completely different approach
- Very high head, multistage pumps (>500m): Interstage losses add up
- Final detailed engineering: Use manufacturer’s pump curve. Always.
The Takeaway
P(kW) ≈ Q(m³/h) × H(m) ÷ 270
Memorize this. It will save you from looking foolish in meetings more times than you can count.
Next time on Back of the Envelope: Quick Pipe Size Estimation from Flow Rate.