VFD Energy Savings (Affinity Laws)

Ideal cube-law scenario for a centrifugal load with negligible static head and similar efficiency.

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1 Inputs

Minimum: 0

Minimum: 0 · Maximum: 100

Minimum: 0

Minimum: 0

Minimum: 0

2 Results

Annual energy savings
$17,568
Power at 80% speed
25.6 kW
Energy saved / year
146,400 kWh
Energy reduction
49%
CO₂ avoided / year
58.6 t
P₂ = P₁ × (n₂/n₁)³
Ideal cube-law scenario for a centrifugal load with negligible static head and similar efficiency. It excludes drive losses and minimum-flow constraints. Compare measured system curves and duty profiles before specifying a drive.

Worked example

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

InputValue
Power at full speed50 kW
Average operating speed80 % of full
Operating hours / year6000 h
Electricity rate0.12 $/kWh
Grid emission factor0.4 kg/kWh

Annual energy savings: $17,568

Related tools

About this calculator

How it works

You enter full-speed power (kW), average operating speed (% of full), hours, electricity rate and grid emission factor. By the affinity laws power scales with speed cubed: P = P_full × (speed%)³. Saving = (P_full − P) × hours × rate.

Worked example (illustrative inputs)

50 kW at 80% speed: 50 × 0.8³ = 25.6 kW. Over 6,000 h: (50 − 25.6) × 6,000 = 146,400 kWh saved, 49% less, ≈ $17,568 at $0.12 and ≈ 58.6 t CO₂.

How to read the result

The cube law is the ideal case for a system with friction only. Systems with static head save less, so treat the figure as an upper bound. It compares running at full speed with running slower; VFD losses are not included.

When to use it

Use it on throttled pumps or dampered fans that run below full flow most of the time.

À propos de ce calculateur

Traduction (brouillon) — à relire.

Fonctionnement

Vous saisissez la puissance à pleine vitesse (kW), la vitesse moyenne (% du maximum), les heures, le tarif et le facteur d'émission. Selon les lois de similitude, la puissance varie avec le cube de la vitesse : P = P_pleine × (vitesse %)³. Économie = (P_pleine − P) × heures × tarif.

Exemple chiffré (valeurs d'illustration)

50 kW à 80 % de vitesse : 50 × 0,8³ = 25,6 kW. Sur 6 000 h : (50 − 25,6) × 6 000 = 146 400 kWh économisés, 49 % de moins, ≈ 17 568 $ à 0,12 $ et ≈ 58,6 t de CO₂.

Lire le résultat

La loi du cube est le cas idéal d'un circuit sans hauteur statique. Les circuits avec hauteur statique économisent moins : considérez le chiffre comme un maximum. Il compare la marche à pleine vitesse et à vitesse réduite ; les pertes du variateur ne sont pas incluses.

Quand l'utiliser

Utilisez-le pour des pompes vannées ou des ventilateurs à registre qui tournent la plupart du temps sous leur débit maximal.

حول هذه الحاسبة

ترجمة (مسودة) — للمراجعة.

طريقة الحساب

تُدخل القدرة عند السرعة الكاملة (kW) ومتوسط سرعة التشغيل (% من الكاملة) والساعات والتعرفة ومعامل الانبعاث. وفق قوانين التشابه تتناسب القدرة مع مكعب السرعة: ‏P = P_كاملة × (السرعة%)³. التوفير = (P_كاملة − P) × الساعات × التعرفة.

مثال محسوب (قيم توضيحية)

50 kW عند سرعة 80%: ‏50 × 0.8³ = 25.6 kW. وعلى 6000 ساعة: ‏(50 − 25.6) × 6000 = 146,400 kWh موفّرة، أي أقل بنسبة 49%، ≈ 17,568 $ عند 0.12 $ و≈ 58.6 طن من CO₂.

قراءة النتيجة

قانون المكعب هو الحالة المثالية لنظام فيه احتكاك فقط. الأنظمة ذات الرفع الساكن توفّر أقل، فاعتبر الرقم حدًا أعلى. وهو يقارن التشغيل بالسرعة الكاملة بالتشغيل الأبطأ، دون احتساب فقد المغيّر.

متى تستخدمه

استخدمها على المضخات المخنوقة بصمامات أو المراوح ذات الخوانق التي تعمل معظم الوقت دون التدفق الكامل.

In one line

Ideal cube-law scenario for a centrifugal load with negligible static head and similar efficiency. It excludes drive losses and minimum-flow constraints. Compare measured system curves and duty profiles before specifying a drive.

Worked example

EXAMPLE

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

Inputs
Power at full speed50 kW
Average operating speed80 % of full
Operating hours / year6000 h
Electricity rate0.12 $/kWh
Grid emission factor0.4 kg/kWh
Results
Annual energy savings$17,568
Power at 80% speed25.6 kW
Energy saved / year146,400 kWh
Energy reduction49%
CO₂ avoided / year58.6 t

Frequently asked questions

Why do VFDs save so much energy on pumps and fans?

Because power on centrifugal pumps and fans varies with the cube of speed. Cutting speed to 80% drops power to roughly half, and cutting to 50% speed uses only around an eighth of full power — throttling a valve or damper wastes that potential, while a VFD captures it by actually slowing the motor.

Is the cube law always accurate?

It's the ideal case for a system that's all friction loss with no static head. Real systems with static lift or a fixed pressure requirement save somewhat less than the cube law predicts, because the portion of head that doesn't scale with flow limits how much speed reduction actually helps.

Does this apply to positive-displacement pumps or compressors?

No — the cube-law affinity relationship is specific to centrifugal machines, meaning pumps and fans. Positive-displacement equipment behaves very differently and needs a different savings analysis.

How do I estimate my average operating speed percentage?

Pull it from your VFD or building-management-system trend data if available, using a true average rather than just the setpoint, or estimate from how the process typically runs across a day or season. Using a rough guess will proportionally skew the savings estimate, since it's cubed.

What's a typical payback for a VFD retrofit?

It varies by application, but on throttled pumps and dampered fans with meaningful part-load operating hours, paybacks in roughly the 1–3 year range are common. Systems that run mostly at full speed, or have little static head, will see smaller savings and longer paybacks.

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