Pump Power & Energy Calculator

Flow and total head must describe the same operating point.

This selection changes review context only. Formulas and example inputs stay unchanged. Check the model assumptions and applicability for the actual process.

Identify the actual asset, process, work boundary, source documents, units and assumptions. Confirm that the selected model or template applies before using its results.

1 Inputs

Minimum: 0

Minimum: 0

Minimum: 0.1

Minimum: 1 · Maximum: 100

Minimum: 1 · Maximum: 100

Minimum: 0

Minimum: 0

2 Results

Annual energy cost
$4,566
Hydraulic power
4.08 kW
Shaft power
5.83 kW
Motor input power
6.34 kW
Overall efficiency
64.4%
Pₕ = QHSG/367.3; Pₑ = Pₕ/(ηₚηₘ)
Flow and total head must describe the same operating point. Efficiencies are fractions of transmitted power. Compare the calculated electrical input with measurements and pump curves; this does not assess cavitation or mechanical suitability.

Worked example

Example values are illustrative. Replace them with your measured data.

InputValue
Flow rate Q50 m³/h
Total head H30 m
Fluid specific gravity1
Pump efficiency70 %
Motor efficiency92 %
Operating hours / year6000 h
Electricity rate0.12 $/kWh

Annual energy cost: $4,566

Related tools

About this calculator

How it works

Hydraulic power (kW) = flow (m³/h) × head (m) × specific gravity ÷ 367.3. Shaft power = hydraulic power ÷ pump efficiency. Motor input = shaft power ÷ motor efficiency. Annual energy = motor input × hours, and cost = energy × electricity rate. Overall efficiency = pump efficiency × motor efficiency.

Worked example (illustrative inputs)

50 m³/h, 30 m head, water (SG 1.0), 70% pump and 92% motor efficiency, 6,000 hours a year at $0.12 per kWh. Hydraulic power = 1,500 ÷ 367.3 ≈ 4.08 kW. Shaft ≈ 5.83 kW. Motor input ≈ 6.34 kW. Energy ≈ 38,048 kWh, costing about $4,566 a year. Overall efficiency = 64.4%.

How to read the result

Overall efficiency is the share of the electricity you pay for that actually moves fluid. The tool shows 60% or more as good and 45–60% as a warning. Because pump power varies with roughly the cube of speed, an oversized or throttled pump wastes energy. A trimmed impeller or a variable-speed drive is the usual remedy.

When to use it

Use it to estimate running cost, compare duty points or check how much a lower head or a better efficiency would save.

À propos de ce calculateur

Traduction (brouillon) — à relire.

Fonctionnement

Puissance hydraulique (kW) = débit (m³/h) × hauteur (m) × densité ÷ 367,3. Puissance sur l'arbre = puissance hydraulique ÷ rendement de la pompe. Puissance absorbée = puissance sur l'arbre ÷ rendement du moteur. Énergie annuelle = puissance absorbée × heures, et coût = énergie × prix de l'électricité. Rendement global = rendement pompe × rendement moteur.

Exemple chiffré (valeurs d'illustration)

50 m³/h, 30 m de hauteur, eau (densité 1,0), rendements de 70 % (pompe) et 92 % (moteur), 6 000 h/an à 0,12 $/kWh. Puissance hydraulique = 1 500 ÷ 367,3 ≈ 4,08 kW. Arbre ≈ 5,83 kW. Absorbée ≈ 6,34 kW. Énergie ≈ 38 048 kWh, soit environ 4 566 $/an. Rendement global = 64,4 %.

Lire le résultat

Le rendement global est la part de l'électricité payée qui déplace réellement le fluide. L'outil le juge bon à 60 % ou plus et en alerte entre 45 et 60 %. Comme la puissance d'une pompe varie à peu près avec le cube de la vitesse, une pompe surdimensionnée ou étranglée gaspille de l'énergie. Une roue rognée ou un variateur de vitesse est la solution habituelle.

Quand l'utiliser

À utiliser pour estimer un coût de fonctionnement, comparer des points de fonctionnement ou mesurer le gain d'une hauteur plus faible ou d'un meilleur rendement.

حول هذه الحاسبة

ترجمة (مسودة) — للمراجعة.

طريقة الحساب

القدرة الهيدروليكية (كيلوواط) = التدفق (م³/ساعة) × الضاغط (م) × الثقل النوعي ÷ 367.3. قدرة العمود = القدرة الهيدروليكية ÷ كفاءة المضخة. قدرة دخل المحرك = قدرة العمود ÷ كفاءة المحرك. الطاقة السنوية = قدرة الدخل × الساعات، والتكلفة = الطاقة × سعر الكهرباء. الكفاءة الكلية = كفاءة المضخة × كفاءة المحرك.

مثال محسوب (قيم توضيحية)

50 م³/ساعة، ضاغط 30 م، ماء (ثقل نوعي 1.0)، كفاءة مضخة 70% ومحرك 92%، 6000 ساعة سنوياً بسعر 0.12 دولار لكل كيلوواط ساعة. القدرة الهيدروليكية = 1500 ÷ 367.3 ≈ 4.08 كيلوواط. العمود ≈ 5.83 كيلوواط. دخل المحرك ≈ 6.34 كيلوواط. الطاقة ≈ 38,048 كيلوواط ساعة، بتكلفة نحو 4,566 دولاراً سنوياً. الكفاءة الكلية = 64.4%.

قراءة النتيجة

الكفاءة الكلية هي حصة الكهرباء المدفوعة التي تحرّك السائل فعلاً. تعدّها الأداة جيدة عند 60% فأكثر وتحذّر بين 45% و60%. ولأن قدرة المضخة تتغير تقريباً مع مكعب السرعة، فإن المضخة كبيرة الحجم أو المخنوقة بصمام تهدر الطاقة، والحل المعتاد هو تقليم الدفّاعة أو استخدام محرك متغير السرعة.

متى تستخدمه

استخدمها لتقدير تكلفة التشغيل أو مقارنة نقاط التشغيل أو معرفة وفر ضاغط أقل أو كفاءة أفضل.

In one line

Flow and total head must describe the same operating point. Efficiencies are fractions of transmitted power. Compare the calculated electrical input with measurements and pump curves; this does not assess cavitation or mechanical suitability.

Worked example

EXAMPLE

Illustrative values (the calculator defaults), not data from a real site.

Inputs
Flow rate Q50 m³/h
Total head H30 m
Fluid specific gravity1
Pump efficiency70 %
Motor efficiency92 %
Operating hours / year6000 h
Electricity rate0.12 $/kWh
Results
Annual energy cost$4,566
Hydraulic power4.08 kW
Shaft power5.83 kW
Motor input power6.34 kW
Overall efficiency64.4%

Frequently asked questions

What's the difference between hydraulic, shaft and motor power?

Hydraulic power is the useful power added to the fluid (Q·H·SG/367.3). Divide by pump efficiency for the shaft power the motor must deliver, then by motor efficiency for the electrical input power you actually pay for.

Where does 367.3 come from?

It bundles fluid density and gravity for SI units: P(kW) = ρ·g·Q·H. With Q in m³/h and H in m, the constant works out to approximately 367.3 (3.6×10⁶ ÷ 9.81).

What if I don't know my pump efficiency?

Check the manufacturer's pump curve or nameplate for efficiency at your actual operating point, not just the best-efficiency point — running far from BEP means real efficiency can be well below the curve's peak. With no data at all, 60–75% is a reasonable placeholder for a mid-sized centrifugal pump in decent condition, but treat the result as approximate.

How is this different from the Motor Efficiency Upgrade Savings tool?

This calculator sizes the whole hydraulic system — flow, head, and both pump and motor efficiency — to find your actual energy cost today. The Motor Efficiency tool instead compares an existing motor against a premium-efficiency replacement, holding the mechanical load constant, to estimate the payback of swapping motors specifically.

Why does overall efficiency matter if the pump is already installed?

A low overall efficiency (pump × motor) tells you how much of your electricity bill is genuinely useful work versus losses, which helps justify impeller trimming, VFDs or right-sizing. Pumps run roughly a fifth of all industrial electricity use, so even modest efficiency gains on a continuously-running pump add up over a year.

What effect does a variable-speed drive have here?

A VFD lets the pump match flow to actual demand instead of running at full speed and throttling a valve, and because power scales roughly with the cube of speed, cutting speed by 20% can cut power by close to 50%. Re-run this calculator at the reduced flow and head your VFD would actually deliver to estimate the savings.

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