Physics · Energy and energy balance

Efficiency of an energy conversion

Efficiency compares intended energy output with complete input for the same system boundary and time interval.

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Formula

η = \frac{E _{nutz}}{E _{zu}} = \frac{P _{nutz}}{P _{zu}}

η is dimensionless; percent is 100η %. Input and output energies use the same unit; P̄ in W denotes each average power over the same interval.

Variables & units

Variables & units
Symbol	Meaning	Unit
$η$	Fraction of input energy delivered as defined useful output	dimensionless
$E_{\mathrm{nutz}}$	Useful output energy over the chosen interval	J
$E_{\mathrm{zu}}$	Fully accounted input energy over that same interval	J
$P_{\mathrm{nutz}}$	Average useful output power over the same interval, written P̄useful	W
$P_{\mathrm{zu}}$	Average input power over the same interval, written P̄in	W

LaTeX

\eta=\frac{E_\mathrm{nutz}}{E_\mathrm{zu}}=\frac{\overline P_\mathrm{nutz}}{\overline P_\mathrm{zu}}

QUANTA STUDYquanta-study.de

An energy converter receives E_in. Useful energy E_use leaves to the right, while loss energy E_loss exits downward. Below is η = E_use/E_in. — Efficiency η compares useful output energy with input energy; losses remain part of the balance.

Efficiency calculator

Find the efficiency, useful energy or energy input, optionally with powers or for two converters in a row.

Calculation model

Find

Energy input E_in

All the energy put in over the same period, e.g. 360 kJ, i.e. 0.1 kWh from the socket.

Useful energy Eᵤ

The part that is put to use, e.g. 324 kJ that actually heat the water.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for η, substitute, evaluate

1. Rearrange for η
$\eta=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}$
2. Substitute values (in SI units)
- E_in = 360 kJ = 360 000 J
- Eᵤ = 324 kJ = 324 000 J
$\frac{324000}{360000}$
3. Result
η = 90 %

Result: Efficiency η

η = 90 %

With E_in = 360 kJ and Eᵤ = 324 kJ, η = 90 %.

Quick tips

Formula triangle: useful energy on top, η and energy input below. Cover the quantity you need, e.g. energy input = useful energy ÷ η.
Check: η lies between 0 and 100 %. If more comes out than goes in, input and output are swapped or an input is missing. Stages in a row multiply: 90 % · 80 % = 72 %.

Assumptions and limits

Fully accounted positive input, nonnegative useful output and unchanged storage within the same boundary and interval. Efficiency lies between zero and one. No storage discharge, COP or omitted auxiliary inputs. All energies refer to the same interval.

Derivation & background

Complete accounting gives $E_{\mathrm{in}}=E_{\mathrm{use}}+E_{\mathrm{other,out}}+\Delta E_{\mathrm{stored}}$ . Energy excluded from useful output does not disappear. Equal endpoint storage and complete inputs imply η at most one. The energy ratio equals the ratio of matching average powers; constant operation permits constant power readings. For varying loads, total efficiency is not an arbitrary mean of individual efficiencies. Heat-pump COP and the Carnot bound have different definitions and assumptions.

Exam blueprint

Validity range

Input and output refer to the same declared boundary, useful-output definition and time interval; input energy must be positive. The energy ratio equals the ratio of powers averaged over that same interval. Individual instantaneous readings are not interchangeable for varying operation. The bound $\eta\le1$ assumes complete input accounting and no net discharge of stored energy. A heat pump’s COP counts only drive work in the denominator and is a different measure. Carnot limits cyclic heat engines between two reservoirs, not every converter.

Derivation steps

Define useful output energy as a fraction of fully accounted input energy for the same system boundary and time interval.

1Draw the system boundary, choose interval endpoints, and specify useful output, such as lifting energy or delivered heat.
2Account for all relevant energy inputs over the interval as Ein. Auxiliary drives also belong in the input when they cross the boundary.
3The balance is $E_{\mathrm{in}}=E_{\mathrm{use}}+E_{\mathrm{other}}+\Delta E_{\mathrm{stored}}$ . The difference between input and useful output is therefore not automatically heat alone.
4For $E_{\mathrm{in}}>0$ define $\eta=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}$ . With complete input accounting and no net discharge of stored energy, useful output cannot exceed input.
5With changing power, determine energies through $\int_{t_0}^{t_1}P(t)\,\mathrm dt$ . Both average powers use the same interval Δt: $\overline P=\frac E{\Delta t}$ .
6The common duration cancels to give $\eta=\frac{\overline P_{\mathrm{use}}}{\overline P_{\mathrm{in}}}$ . Constant operation permits constant power values; convert the decimal to percent only at the end.

Rearrangements

Find useful energy

E_\mathrm{nutz}=\eta E_\mathrm{zu}

Use η as a decimal, for example 0.8 for 80%.

Task variant

Over one common interval, a device receives 750 J and delivers 600 J usefully. All inputs are included and stored energy is unchanged. Find efficiency and other output.

Approach: $\eta=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}=\frac{600}{750}=0.80=80\%$ .
With unchanged storage, $E_{\mathrm{other}}=750-600=150\,\mathrm J$ .
This energy is conserved but is not part of the specified useful output.

Common mistakes

Interpreting losses as energy that disappeared or as negative total energy.

Total energy is conserved; some is converted into forms that are not the intended output.

Exam context

Energy conversion, electrical devices, motors, and quantitative energy balances.

Formula cluster

Energy conversion

The general ratio applies to devices; Carnot additionally assumes a heat engine.

η = 1 − T_k/T_w

Carnot Efficiency

η = Pout/Pin

Machine efficiency

Eel = P·t = U·I·t

Electrical Energy at Constant Voltage and Current

Worked example

For $E_{\mathrm{in}}=750\,\mathrm J$ and $E_{\mathrm{use}}=600\,\mathrm J$ over one interval, $\eta=\frac{600}{750}=0.80=80\%$ . The difference $150\,\mathrm J$ must be accounted for as other output or storage change. With unchanged storage it is entirely other output, but not automatically heat alone.

Applications

Evaluate motors, lifting devices, conversion chains and energy systems with an explicit boundary; include startup, auxiliary consumption and storage changes in accounting.

How do you calculate with Efficiency of an Energy Conversion?

Here is how to work through a typical Efficiency of an Energy Conversion (η = Enutz/Ezu = P̄nutz/P̄zu) task step by step:

1
Task
Over one common interval, a device receives 750 J and delivers 600 J usefully. All inputs are included and stored energy is unchanged. Find efficiency and other output.
Solution path
1. Approach: $\eta=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}=\frac{600}{750}=0.80=80\%$ .
2. With unchanged storage, $E_{\mathrm{other}}=750-600=150\,\mathrm J$ .
3. This energy is conserved but is not part of the specified useful output.

Understand and apply the equation

How much input energy serves the intended purpose? Efficiency answers this only after the device boundary, interval, and useful output have been defined. Learn the balance through a lifting drive, changing operating phases, and an installation with auxiliary controls.

How much input serves the intended purpose?

A lifting drive consumes electricity, but your task initially concerns work delivered to the lifted body. Before dividing numbers, imagine a boundary around the device: what enters, what should leave, and does the device store energy during the interval? These three questions determine what an efficiency value means.[1][2]

You need energy in joules, power as energy per time, and conservation of energy. Eta denotes a ratio. Inputs and outputs in the following balance are positive energy amounts; only the storage change may be positive or negative. Heat is not automatically a loss: its usefulness depends on the chosen purpose.[1][2]

First specify useful output and system boundary

\eta_E=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}},\qquad E_{\mathrm{in}}>0

The ratio is a definition; specify what counts as useful output before calculation.[1]

A motor can receive electrical energy and deliver mechanical work. A heater instead aims to supply heat at a chosen location. The same heat can therefore be unwanted output in one application and useful output in another. Efficiencies are comparable only with compatible purpose, input, output and operating conditions. The ratio itself does not decide which output is useful.[1]

System boundary determines whether controls, power supply, bearings and auxiliary equipment are included. Shaft output divided by motor electrical input is not automatically whole-system efficiency including controls. Fuel-based efficiency also needs its heating-value convention; different reference inputs must not be silently mixed.

Conservation explains the upper bound

E_{\mathrm{in}}=E_{\mathrm{use}}+E_{\mathrm{other,out}}+\Delta E_{\mathrm{stored}}

All inputs and outputs use the same interval; other, out includes outputs not counted as useful.[2]

If initial and final storage are equal and no input is missing, useful output is part of total input: $0\le\eta\le1$ . The rest has not disappeared but leaves through other channels. 750 J input and 600 J useful output give 80%;150 J must be accounted for. Without storage information, those 150 J are not necessarily waste heat.

A discharging store can temporarily deliver more output power than contemporaneous input. That does not mean conversion efficiency above one: stored energy decreases. During charging, some input remains inside. A complete charge-discharge cycle needs compatible endpoint storage or explicit accounting for its change.

Connect energy and power through matching averages

\begin{aligned}E_{\mathrm{in}}&=\int_{t_0}^{t_1}P_{\mathrm{in}}(t)\,\mathrm dt\\E_{\mathrm{use}}&=\int_{t_0}^{t_1}P_{\mathrm{use}}(t)\,\mathrm dt\\\eta_E&=\frac{\overline P_{\mathrm{use}}}{\overline P_{\mathrm{in}}}\end{aligned}

Both average powers use the same interval length.

For constant powers the common duration cancels in $E=P\Delta t$ . For changing powers the ratio of their averages also works. Averaging individual instantaneous efficiencies is generally incorrect: phases receiving more input energy must count more heavily. With positive input power throughout, precisely this weighting gives interval efficiency. Idle operation also consumes energy and can lower overall efficiency.

\begin{aligned}\eta_E&=\frac{\int\eta(t)P_{\mathrm{in}}(t)dt}{\int P_{\mathrm{in}}(t)dt}\\{} \eta(t)&=\frac{P_{\mathrm{use}}(t)}{P_{\mathrm{in}}(t)}\end{aligned}

This identity requires positive or otherwise defined input power in the instantaneous ratio.

Energy accumulates over both operating phases

QUANTA STUDYquanta-study.de

For the second example, the operating phase changes after

20\,\mathrm s

. Finally

18.0=10.2+7.8

, in kilojoules. Final heights determine interval efficiency; slopes represent powers.[1][2]

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Check percentages and rearrangements

\begin{aligned}E_{\mathrm{use}}&=\eta E_{\mathrm{in}}\\{} E_{\mathrm{in}}&=\frac{E_{\mathrm{use}}}{\eta}\ (\eta>0)\\{} \eta_{\%}&=100\eta\end{aligned}

$\eta=0.80$ is written 80%; use 0.80 in products.

Numerator and denominator are both energies or both compatible powers. Match joules and kilojoules; watts are not joules. 600 J useful output from 0.750 kJ input gives $\frac{600}{750}=0.80$ rather than 800. A dimensionless ratio is necessary but does not prove that accounting boundaries match. A percentage is also dimensionless.

Uncertainty in either input or output affects efficiency. A very small denominator can make the ratio especially sensitive. Instruments should cover the same interval; startup consumption, run-down and auxiliary power must not be assigned arbitrarily to another period. Meaningful comparisons state operating point and task as well as η.

Example: electrical lifting drive over a complete interval

A load gains potential energy

Given: The accounting boundary contains the drive, not the lifted load. The drive receives average electrical power 120 W for 180 s and delivers lifting work to a 45.0 kg load raised through 18.0 m; g=9.81 m/s². The load has the same kinetic energy at both endpoints, so delivered lifting work is mgh. Stored energy inside the drive is unchanged in this model. The potential-energy increase of the external load is not counted again as storage inside the drive.
Find: Find efficiency for useful lifting work and other output.

Approach: Find input and lifting work over the same interval.
$\begin{aligned}E_{\mathrm{in}}&=P_{\mathrm{in}}\Delta t\\E_{\mathrm{use}}&=mg\Delta h\\\eta&=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}\end{aligned}$
Substitute the given values with matching units.
$E_{\mathrm{in}}=120\cdot180=21600\,\mathrm J$
Calculate the chosen useful output as increased potential energy.
$E_{\mathrm{use}}=mgh=45.0\cdot9.81\cdot18.0=7946.1\,\mathrm J$
Form efficiency and remaining accounted output.
$\begin{aligned}\eta&=\frac{7946.1}{21600}=0.367875\\{} E_{\mathrm{other,out}}&=13653.9\,\mathrm J\end{aligned}$

Result: Useful lifting efficiency is 36.7875%, about 36.8%. Other outputs total 13.6539 kJ within the specified boundary.

Check: Average useful power is $\frac{7946.1\,\mathrm J}{180\,\mathrm s}=44.145\,\mathrm W$ . Its input-power ratio also gives $\frac{44.145}{120}=0.367875$ . The balance $7946.1+13653.9=21600$ , in joules, closes. A different storage change would alter the other output.

Mechanical power here describes lifting work delivered to the external load per time, $\overline P_{\mathrm{use}}=W_{\mathrm{lift}}/\Delta t$ . This puts the already calculated useful output on the same interval length as electrical input. The load’s potential energy is still not additionally counted as storage inside the drive.[6][2]

Example: combine two operating phases correctly

A device changes operating point

Given: Phase A:20 s at 500 W input and 350 W useful output. Phase B:40 s at 200 W input and 80 W useful output. Constant powers within each phase, one accounting boundary and no storage change.
Find: Find full-cycle efficiency and compare with averaging ratios.

Approach: Find phase energies separately and then add them.
$\begin{aligned}E_{\mathrm{in}}&=\sum_iP_{\mathrm{in},i}\Delta t_i\\E_{\mathrm{use}}&=\sum_iP_{\mathrm{use},i}\Delta t_i\\\eta&=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}\end{aligned}$
Substitute the given values with matching units.
$E_{\mathrm{in}}=500\cdot20+200\cdot40=18000\,\mathrm J$
Separately sum useful output energies.
$E_{\mathrm{use}}=350\cdot20+80\cdot40=10200\,\mathrm J$
Divide total energies.
$\eta_E=\frac{10200}{18000}=0.5666667$

Result: Overall efficiency 56.6667%, not 55% from the simple mean of phase efficiencies.

Check: Average powers are $300\,\mathrm W$ input and $170\,\mathrm W$ useful; $\frac{170}{300}=\frac{10200}{18000}$ . Phase efficiencies $0.70$ and $0.40$ have input-energy weights $\frac{10}{18}$ and $\frac8{18}$ . A time-only mean would give $0.50$ . Other outputs are $18000-10200=7800\,\mathrm J$ .

Conversion chains and additional inputs

\eta_{\mathrm{chain}}=\prod_i\eta_i

The product rule requires each defined useful output to be the next stage’s complete input.

Stages at 90% and 80% jointly give 72%, not 170% or 85%. From 1000 J initial input, stage 1 delivers 900 J to stage 2, which produces 720 J. Other outputs 100 J and 180 J sum to 280 J. Additional cooling pumps, parallel paths or stored energy may invalidate the simple product rule and require whole-system accounting.

Combined heat and power may define both heat and electrical work as useful outputs. Their energy sum can define overall energy efficiency, but does not imply equal capacity to perform work. State purpose and boundary; a large combined value does not prove large electrical efficiency.

Explore further: COP and the Carnot limit

\begin{aligned}\mathrm{COP}_{HP}&=\frac{Q_H}{W_{\mathrm{in}}}\\{} Q_H&=Q_C+W_{\mathrm{in}}\end{aligned}

Environmental heat QC is an additional heat-pump energy input.[3]

The model heat pump delivers $Q_H=1800\,\mathrm J$ of heating from $W_{\mathrm{in}}=600\,\mathrm J$ of work and $Q_C=1200\,\mathrm J$ from the cold reservoir. Hence $\mathrm{COP}=\frac{Q_H}{W_{\mathrm{in}}}=3$ . This measure deliberately omits one energy input from its denominator. A heat engine instead has $\eta=\frac{W_{\mathrm{net}}}{Q_H}$ ; its reversible limit between constant-temperature reservoirs is $1-\frac{T_C}{T_H}$ , with temperatures in kelvin.[3][4]

Carnot is not a general limit for electric-motor calculations. Efficiency 80% alone specifies neither output power nor energy consumed for a task. Small and large devices with the same η may have very different consumption. Evaluate efficiency together with useful energy, duration and operating point.[5]

Carnot efficiency adds a temperature bound to the general energy ratio for an appropriate heat engine. It replaces neither complete input accounting nor the useful-output definition; this specific reservoir bound is not a general motor limit for the electrical lifting drive.[4]

Motor alone or the complete installation?

Additional inputs count when the system boundary includes them. This comparison therefore keeps motor output unchanged and changes only the accounting scope.[1][2]

Motor alone or the complete installation?

Given: A motor draws 800 W and delivers 640 W of shaft power. Separately supplied controls draw another 40 W. Both operate steadily for 120 s. All shaft work counts as useful output; there is no storage change. All values are supplied model inputs.
Find: Find motor efficiency, complete-installation efficiency, and its energy balance.

Approach: the installation receives both inputs. Compare the same useful output against two different boundaries.
$\begin{aligned}\eta_m&=\frac{P_{\mathrm{shaft}}}{P_m}\\\eta_{\mathrm{system}}&=\frac{P_{\mathrm{shaft}}}{P_m+P_{\mathrm{aux}}}\\E_{\mathrm{other}}&=(P_m+P_{\mathrm{aux}}-P_{\mathrm{shaft}})\Delta t\end{aligned}$
Substitute powers and form energies over the common interval.
$\begin{aligned}\eta_m&=\frac{640}{800}=0.80\\\eta_{\mathrm{system}}&=\frac{640}{840}\approx0.761905\\E_{\mathrm{in}}&=840\cdot120\,\mathrm J=100800\,\mathrm J\\E_{\mathrm{use}}&=640\cdot120\,\mathrm J=76800\,\mathrm J\\E_{\mathrm{other}}&=200\cdot120\,\mathrm J=24000\,\mathrm J\end{aligned}$

Result: The model motor has 80% efficiency; the installation including controls has approximately 76.19%. The controls change the accounting boundary, not the specified motor output.

Check: The balance closes: $100800=76800+24000$ , in joules. Motor losses are $160\,\mathrm W$ , and the controls contribute $40\,\mathrm W$ ; together they equal the calculated $200\,\mathrm W$ of other output.

Five solved conversion balances

Calculate · Foundation

A device receives 750 J and delivers 600 J usefully with zero storage change. Find η and other output.

Show hint

Assume matching boundary and time.

Learning goal: Find useful fraction and other energy outputs.

Show solution and calculation

Approach: divide useful energy by complete input.
Thus $\eta=0.80=80\%$ .
With no storage change, other output is $150\,\mathrm J$ .

Check: both outputs total $750\,\mathrm J$ .

\begin{aligned}\eta&=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}=\frac{600}{750}=0.80\\E_{\mathrm{other}}&=E_{\mathrm{in}}-E_{\mathrm{use}}=150\,\mathrm J\end{aligned}

Calculate · Application

A device receives constant 200 W, delivers useful output at $\eta=0.80$ and runs 30 s. Find useful energy.

Show hint

First integrate power into energy.

Learning goal: Find useful energy from constant power operation.

Show solution and calculation

Approach: obtain input energy from power and duration, then multiply by efficiency.
Results are $6000\,\mathrm J$ input, $4800\,\mathrm J$ useful output, and $160\,\mathrm W$ useful power.
With unchanged storage, other outputs total $1200\,\mathrm J$ .

Check: $160\,\mathrm W\cdot30\,\mathrm s=4800\,\mathrm J$ .

\begin{aligned}E_{\mathrm{in}}&=P_{\mathrm{in}}\Delta t=200\cdot30\,\mathrm J=6000\,\mathrm J\\E_{\mathrm{use}}&=\eta E_{\mathrm{in}}=0.80\cdot6000\,\mathrm J=4800\,\mathrm J\\P_{\mathrm{use}}&=\eta P_{\mathrm{in}}=160\,\mathrm W\end{aligned}

Calculate · Application

Two conversion stages have η1=0.90,η2=0.75 and initial input 2.40 kW. No auxiliary input or storage change. Find final useful power.

Show hint

The complete intermediate output becomes the next input.

Learning goal: Combine two conversion-stage efficiencies.

Show solution and calculation

Approach: all first-stage output becomes second-stage input.
Their product is $\eta_{\mathrm{tot}}=0.675$ .
Stage one delivers $2.16\,\mathrm{kW}$ , and stage two $1.62\,\mathrm{kW}$ .
Other outputs are $0.24+0.54=0.78\,\mathrm{kW}$ , closing $2.40=1.62+0.78$ .

\begin{aligned}\eta_{\mathrm{tot}}&=\eta_1\eta_2=0.675\\P_1&=0.90\cdot2.40\,\mathrm{kW}=2.16\,\mathrm{kW}\\P_{\mathrm{use}}&=\eta_2P_1=1.62\,\mathrm{kW}\\P_{\mathrm{other}}&=2.40-1.62=0.78\,\mathrm{kW}\end{aligned}

Interpret · Application

During 10 s a device receives 100 J electrically, supplies 130 J useful energy and outputs another 20 J. Its initial store has enough energy. Find storage change. Can 130% be claimed as conversion efficiency without qualification?

Show hint

The complete balance includes storage change.

Learning goal: Distinguish storage discharge from apparent excess efficiency.

Show solution and calculation

Approach: Input minus all outputs gives storage change, here −50 J.
Storage discharge funds part of useful output.
The ratio 130 to 100 describes only this selected interval boundary, not steady conversion efficiency without storage change.
A complete boundary must explicitly include storage flow.

\Delta E_{\mathrm{store}}=E_{\mathrm{in}}-E_{\mathrm{use}}-E_{\mathrm{other}}=100-130-20=-50\,\mathrm J

[1]

Error analysis · Application

A device runs first for 1 s at input 100 W and η=0.9, then for 9 s at input 10 W and η=0.5. No storage change and same useful-output definition. Someone reports 70% by averaging both efficiencies. Find correct interval efficiency.

Show hint

Weight by input energies rather than just duration or state count.

Learning goal: Average efficiency using energy weights rather than arithmetic means.

Show solution and calculation

Approach: Input energies are 100 J and 90 J, useful energies 90 J and 45 J.
Total useful output is 135 J from 190 J input, about 71.0526%.
Neither a simple mean nor duration-only weighting preserves this energy balance when input powers differ.
Other outputs total 55 J.

\eta_{\mathrm{interval}}=\frac{0.9\cdot100\cdot1+0.5\cdot10\cdot9}{100\cdot1+10\cdot9}=\frac{135}{190}\approx0.710526

[1]

In brief

Specify useful output, every input, interval, and storage change. Then form the energy ratio; changing power requires adding energies. Auxiliary inputs change the system balance. Heat-pump COP and discharging storage do not violate energy conservation.[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant scientific connections; preserved existing field and energy balances, derivations, examples, exercises, figures and model limits.
October 1, 2026Clarified system boundaries and energy weighting; added an auxiliary-controls installation example and an original energy–time graph. Preserved existing examples and all five exercises; clarified solution approaches.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Efficiency of an Energy Conversion. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/wirkungsgrad

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Frequently asked questions about Efficiency of an Energy Conversion

How do I calculate efficiency?+

First define the system boundary, time interval and intended useful energy output. Then use $\eta=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}$ with positive input. For 600 J useful output from 750 J complete input, the result is 0.80 or 80%. Both energies need matching units, and auxiliary consumption must not silently be omitted. A power ratio gives the same result when both powers are averaged over that exact interval or remain constant. Use the decimal value when rearranging: $E_{\mathrm{use}}=\eta E_{\mathrm{in}}$ . Checking $E_{\mathrm{use}}+E_{\mathrm{other}}+\Delta E_{\mathrm{stored}}=E_{\mathrm{in}}$ additionally establishes complete accounting. The ratio alone does not identify the physical form of remaining energy.

Where does the energy “missing” from efficiency go?+

Energy is missing only from intended usefulness, not from total accounting. With 750 J input and 600 J useful output, the difference is 150 J. It can leave as heat, sound or another output, while some may remain stored inside the system. A motor warming during startup illustrates why not every difference has already crossed the boundary. Only unchanged endpoint storage makes the entire 150 J other output. Whether heat counts as useful depends on purpose: it is intended output for a heater but usually not for a lifting device. Identify energy forms, boundary and endpoint states instead of treating “loss” as destruction.

Why can efficiency not exceed 100%?+

With complete input accounting and no net discharge of stored energy, intended output cannot exceed total supply. These assumptions give $\eta\le1$ . A larger measured ratio calls for checking units, intervals, boundary and possible storage discharge. A discharging store can briefly deliver more output power than concurrent external input without creating energy. A heat pump with $\mathrm{COP}=3$ , for example, delivers 1800 J heating from 600 J drive work and 1200 J environmental heat. Environmental heat is an additional energy input; COP counts only work in its denominator. It therefore is not conversion efficiency using complete energy input.

When can I use energy or power to calculate η?+

With constant powers, the common duration cancels from the useful-to-input energy ratio. With varying powers, this remains true for averages over exactly the same start and end times: $\eta=\frac{\overline P_{\mathrm{use}}}{\overline P_{\mathrm{in}}}$ . Individual instantaneous readings cannot replace those averages. The article’s cycle has 20 s at 500 W input and 350 W useful output, followed by 40 s at 200 W input and 80 W useful output. Total input is 18000 J and useful output 10200 J, giving 56.67% overall efficiency. Neither the simple mean of 55% nor the time-weighted mean of 50% is correct; phase efficiencies require input-energy weighting.

How does general efficiency differ from Carnot efficiency?+

General efficiency compares defined useful output with complete input within a chosen boundary. The Carnot bound $\eta_C=1-\frac{T_C}{T_H}$ specifically applies to a cyclic heat engine between two reservoirs at constant absolute temperatures. It limits work obtained from absorbed heat: a reversible ideal cycle attains it, while irreversible operation falls below it. For $T_H=600\,\mathrm K$ and $T_C=300\,\mathrm K$ , this bound is 50%. It is not a ceiling for an electric motor or gearbox with different energy inputs. Use kelvin and compare matching performance definitions. A heat pump’s COP describes heating benefit per drive work and needs its own definition and reservoir balance.

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Variables & units

Symbol

Meaning

Unit

η

Fraction of input energy delivered as defined useful output

dimensionless

E_{\mathrm{nutz}}

Useful output energy over the chosen interval

E_{\mathrm{zu}}

Fully accounted input energy over that same interval

P_{\mathrm{nutz}}

Average useful output power over the same interval, written P̄useful

P_{\mathrm{zu}}

Average input power over the same interval, written P̄in

Derivation & background

Exam blueprint

Validity range

Derivation steps

Define useful output energy as a fraction of fully accounted input energy for the same system boundary and time interval.

1Draw the system boundary, choose interval endpoints, and specify useful output, such as lifting energy or delivered heat.
2Account for all relevant energy inputs over the interval as Ein. Auxiliary drives also belong in the input when they cross the boundary.
3The balance is $E_{\mathrm{in}}=E_{\mathrm{use}}+E_{\mathrm{other}}+\Delta E_{\mathrm{stored}}$ . The difference between input and useful output is therefore not automatically heat alone.
4For $E_{\mathrm{in}}>0$ define $\eta=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}$ . With complete input accounting and no net discharge of stored energy, useful output cannot exceed input.
5With changing power, determine energies through $\int_{t_0}^{t_1}P(t)\,\mathrm dt$ . Both average powers use the same interval Δt: $\overline P=\frac E{\Delta t}$ .
6The common duration cancels to give $\eta=\frac{\overline P_{\mathrm{use}}}{\overline P_{\mathrm{in}}}$ . Constant operation permits constant power values; convert the decimal to percent only at the end.

How do you calculate with Efficiency of an Energy Conversion?

Here is how to work through a typical Efficiency of an Energy Conversion (η = Enutz/Ezu = P̄nutz/P̄zu) task step by step:

Task

Over one common interval, a device receives 750 J and delivers 600 J usefully. All inputs are included and stored energy is unchanged. Find efficiency and other output.

Solution path

Approach: $\eta=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}=\frac{600}{750}=0.80=80\%$ .
With unchanged storage, $E_{\mathrm{other}}=750-600=150\,\mathrm J$ .
This energy is conserved but is not part of the specified useful output.

How much input serves the intended purpose?

First specify useful output and system boundary

\eta_E=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}},\qquad E_{\mathrm{in}}>0

The ratio is a definition; specify what counts as useful output before calculation.[1]

Conservation explains the upper bound

E_{\mathrm{in}}=E_{\mathrm{use}}+E_{\mathrm{other,out}}+\Delta E_{\mathrm{stored}}

All inputs and outputs use the same interval; other, out includes outputs not counted as useful.[2]

Connect energy and power through matching averages

\begin{aligned}E_{\mathrm{in}}&=\int_{t_0}^{t_1}P_{\mathrm{in}}(t)\,\mathrm dt\\E_{\mathrm{use}}&=\int_{t_0}^{t_1}P_{\mathrm{use}}(t)\,\mathrm dt\\\eta_E&=\frac{\overline P_{\mathrm{use}}}{\overline P_{\mathrm{in}}}\end{aligned}

Both average powers use the same interval length.

\begin{aligned}\eta_E&=\frac{\int\eta(t)P_{\mathrm{in}}(t)dt}{\int P_{\mathrm{in}}(t)dt}\\{} \eta(t)&=\frac{P_{\mathrm{use}}(t)}{P_{\mathrm{in}}(t)}\end{aligned}

This identity requires positive or otherwise defined input power in the instantaneous ratio.

Energy accumulates over both operating phases

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For the second example, the operating phase changes after

20\,\mathrm s

. Finally

18.0=10.2+7.8

, in kilojoules. Final heights determine interval efficiency; slopes represent powers.[1][2]

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Check percentages and rearrangements

\begin{aligned}E_{\mathrm{use}}&=\eta E_{\mathrm{in}}\\{} E_{\mathrm{in}}&=\frac{E_{\mathrm{use}}}{\eta}\ (\eta>0)\\{} \eta_{\%}&=100\eta\end{aligned}

$\eta=0.80$ is written 80%; use 0.80 in products.

Example: electrical lifting drive over a complete interval

A load gains potential energy

Given: The accounting boundary contains the drive, not the lifted load. The drive receives average electrical power 120 W for 180 s and delivers lifting work to a 45.0 kg load raised through 18.0 m; g=9.81 m/s². The load has the same kinetic energy at both endpoints, so delivered lifting work is mgh. Stored energy inside the drive is unchanged in this model. The potential-energy increase of the external load is not counted again as storage inside the drive.
Find: Find efficiency for useful lifting work and other output.

Approach: Find input and lifting work over the same interval.
$\begin{aligned}E_{\mathrm{in}}&=P_{\mathrm{in}}\Delta t\\E_{\mathrm{use}}&=mg\Delta h\\\eta&=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}\end{aligned}$
Substitute the given values with matching units.
$E_{\mathrm{in}}=120\cdot180=21600\,\mathrm J$
Calculate the chosen useful output as increased potential energy.
$E_{\mathrm{use}}=mgh=45.0\cdot9.81\cdot18.0=7946.1\,\mathrm J$
Form efficiency and remaining accounted output.
$\begin{aligned}\eta&=\frac{7946.1}{21600}=0.367875\\{} E_{\mathrm{other,out}}&=13653.9\,\mathrm J\end{aligned}$

Result: Useful lifting efficiency is 36.7875%, about 36.8%. Other outputs total 13.6539 kJ within the specified boundary.

Example: combine two operating phases correctly

A device changes operating point

Given: Phase A:20 s at 500 W input and 350 W useful output. Phase B:40 s at 200 W input and 80 W useful output. Constant powers within each phase, one accounting boundary and no storage change.
Find: Find full-cycle efficiency and compare with averaging ratios.

Approach: Find phase energies separately and then add them.
$\begin{aligned}E_{\mathrm{in}}&=\sum_iP_{\mathrm{in},i}\Delta t_i\\E_{\mathrm{use}}&=\sum_iP_{\mathrm{use},i}\Delta t_i\\\eta&=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}\end{aligned}$
Substitute the given values with matching units.
$E_{\mathrm{in}}=500\cdot20+200\cdot40=18000\,\mathrm J$
Separately sum useful output energies.
$E_{\mathrm{use}}=350\cdot20+80\cdot40=10200\,\mathrm J$
Divide total energies.
$\eta_E=\frac{10200}{18000}=0.5666667$

Result: Overall efficiency 56.6667%, not 55% from the simple mean of phase efficiencies.

Conversion chains and additional inputs

\eta_{\mathrm{chain}}=\prod_i\eta_i

The product rule requires each defined useful output to be the next stage’s complete input.

Explore further: COP and the Carnot limit

\begin{aligned}\mathrm{COP}_{HP}&=\frac{Q_H}{W_{\mathrm{in}}}\\{} Q_H&=Q_C+W_{\mathrm{in}}\end{aligned}

Environmental heat QC is an additional heat-pump energy input.[3]

Motor alone or the complete installation?

Additional inputs count when the system boundary includes them. This comparison therefore keeps motor output unchanged and changes only the accounting scope.[1][2]

Motor alone or the complete installation?

Given: A motor draws 800 W and delivers 640 W of shaft power. Separately supplied controls draw another 40 W. Both operate steadily for 120 s. All shaft work counts as useful output; there is no storage change. All values are supplied model inputs.
Find: Find motor efficiency, complete-installation efficiency, and its energy balance.

Approach: the installation receives both inputs. Compare the same useful output against two different boundaries.
$\begin{aligned}\eta_m&=\frac{P_{\mathrm{shaft}}}{P_m}\\\eta_{\mathrm{system}}&=\frac{P_{\mathrm{shaft}}}{P_m+P_{\mathrm{aux}}}\\E_{\mathrm{other}}&=(P_m+P_{\mathrm{aux}}-P_{\mathrm{shaft}})\Delta t\end{aligned}$
Substitute powers and form energies over the common interval.
$\begin{aligned}\eta_m&=\frac{640}{800}=0.80\\\eta_{\mathrm{system}}&=\frac{640}{840}\approx0.761905\\E_{\mathrm{in}}&=840\cdot120\,\mathrm J=100800\,\mathrm J\\E_{\mathrm{use}}&=640\cdot120\,\mathrm J=76800\,\mathrm J\\E_{\mathrm{other}}&=200\cdot120\,\mathrm J=24000\,\mathrm J\end{aligned}$

Result: The model motor has 80% efficiency; the installation including controls has approximately 76.19%. The controls change the accounting boundary, not the specified motor output.

Five solved conversion balances

Calculate · Foundation

A device receives 750 J and delivers 600 J usefully with zero storage change. Find η and other output.

Show hint

Assume matching boundary and time.

Learning goal: Find useful fraction and other energy outputs.

Show solution and calculation

Approach: divide useful energy by complete input.
Thus $\eta=0.80=80\%$ .
With no storage change, other output is $150\,\mathrm J$ .

Check: both outputs total $750\,\mathrm J$ .

\begin{aligned}\eta&=\frac{E_{\mathrm{use}}}{E_{\mathrm{in}}}=\frac{600}{750}=0.80\\E_{\mathrm{other}}&=E_{\mathrm{in}}-E_{\mathrm{use}}=150\,\mathrm J\end{aligned}

Calculate · Application

A device receives constant 200 W, delivers useful output at $\eta=0.80$ and runs 30 s. Find useful energy.

Show hint

First integrate power into energy.

Learning goal: Find useful energy from constant power operation.

Show solution and calculation

Approach: obtain input energy from power and duration, then multiply by efficiency.
Results are $6000\,\mathrm J$ input, $4800\,\mathrm J$ useful output, and $160\,\mathrm W$ useful power.
With unchanged storage, other outputs total $1200\,\mathrm J$ .

Check: $160\,\mathrm W\cdot30\,\mathrm s=4800\,\mathrm J$ .

\begin{aligned}E_{\mathrm{in}}&=P_{\mathrm{in}}\Delta t=200\cdot30\,\mathrm J=6000\,\mathrm J\\E_{\mathrm{use}}&=\eta E_{\mathrm{in}}=0.80\cdot6000\,\mathrm J=4800\,\mathrm J\\P_{\mathrm{use}}&=\eta P_{\mathrm{in}}=160\,\mathrm W\end{aligned}

Calculate · Application

Two conversion stages have η1=0.90,η2=0.75 and initial input 2.40 kW. No auxiliary input or storage change. Find final useful power.

Show hint

The complete intermediate output becomes the next input.

Learning goal: Combine two conversion-stage efficiencies.

Show solution and calculation

Approach: all first-stage output becomes second-stage input.
Their product is $\eta_{\mathrm{tot}}=0.675$ .
Stage one delivers $2.16\,\mathrm{kW}$ , and stage two $1.62\,\mathrm{kW}$ .
Other outputs are $0.24+0.54=0.78\,\mathrm{kW}$ , closing $2.40=1.62+0.78$ .

\begin{aligned}\eta_{\mathrm{tot}}&=\eta_1\eta_2=0.675\\P_1&=0.90\cdot2.40\,\mathrm{kW}=2.16\,\mathrm{kW}\\P_{\mathrm{use}}&=\eta_2P_1=1.62\,\mathrm{kW}\\P_{\mathrm{other}}&=2.40-1.62=0.78\,\mathrm{kW}\end{aligned}

Interpret · Application

During 10 s a device receives 100 J electrically, supplies 130 J useful energy and outputs another 20 J. Its initial store has enough energy. Find storage change. Can 130% be claimed as conversion efficiency without qualification?

Show hint

The complete balance includes storage change.

Learning goal: Distinguish storage discharge from apparent excess efficiency.

Show solution and calculation

Approach: Input minus all outputs gives storage change, here −50 J.
Storage discharge funds part of useful output.
The ratio 130 to 100 describes only this selected interval boundary, not steady conversion efficiency without storage change.
A complete boundary must explicitly include storage flow.

\Delta E_{\mathrm{store}}=E_{\mathrm{in}}-E_{\mathrm{use}}-E_{\mathrm{other}}=100-130-20=-50\,\mathrm J

[1]

Error analysis · Application

A device runs first for 1 s at input 100 W and η=0.9, then for 9 s at input 10 W and η=0.5. No storage change and same useful-output definition. Someone reports 70% by averaging both efficiencies. Find correct interval efficiency.

Show hint

Weight by input energies rather than just duration or state count.

Learning goal: Average efficiency using energy weights rather than arithmetic means.

Show solution and calculation

Approach: Input energies are 100 J and 90 J, useful energies 90 J and 45 J.
Total useful output is 135 J from 190 J input, about 71.0526%.
Neither a simple mean nor duration-only weighting preserves this energy balance when input powers differ.
Other outputs total 55 J.

\eta_{\mathrm{interval}}=\frac{0.9\cdot100\cdot1+0.5\cdot10\cdot9}{100\cdot1+10\cdot9}=\frac{135}{190}\approx0.710526

[1]

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant scientific connections; preserved existing field and energy balances, derivations, examples, exercises, figures and model limits.
October 1, 2026Clarified system boundaries and energy weighting; added an auxiliary-controls installation example and an original energy–time graph. Preserved existing examples and all five exercises; clarified solution approaches.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Efficiency of an Energy Conversion. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/wirkungsgrad

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Frequently asked questions about Efficiency of an Energy Conversion

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Efficiency of an energy conversion

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Applications

How do you calculate with Efficiency of an Energy Conversion?

How much input serves the intended purpose?

First specify useful output and system boundary

Conservation explains the upper bound

Connect energy and power through matching averages

Energy accumulates over both operating phases

Check percentages and rearrangements

Example: electrical lifting drive over a complete interval

A load gains potential energy

Example: combine two operating phases correctly

A device changes operating point

Conversion chains and additional inputs

Explore further: COP and the Carnot limit

Motor alone or the complete installation?

Motor alone or the complete installation?

Five solved conversion balances

A device receives 750 J and delivers 600 J usefully with zero storage change. Find η and other output.

A device receives constant 200 W, delivers useful output at η=0.80\eta=0.80η=0.80 and runs 30 s. Find useful energy.

Two conversion stages have η1=0.90,η2=0.75 and initial input 2.40 kW. No auxiliary input or storage change. Find final useful power.

During 10 s a device receives 100 J electrically, supplies 130 J useful energy and outputs another 20 J. Its initial store has enough energy. Find storage change. Can 130% be claimed as conversion efficiency without qualification?

A device runs first for 1 s at input 100 W and η=0.9, then for 9 s at input 10 W and η=0.5. No storage change and same useful-output definition. Someone reports 70% by averaging both efficiencies. Find correct interval efficiency.

In brief

Sources for this explanation

Frequently asked questions about Efficiency of an Energy Conversion

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More formulas: Heat and thermodynamics

Retain Efficiency of an Energy Conversion for exams

Efficiency of an energy conversion

Variables & units

Efficiency calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Energy conversion

Worked example

Applications

How do you calculate with Efficiency of an Energy Conversion?

How much input serves the intended purpose?

First specify useful output and system boundary

Conservation explains the upper bound

Connect energy and power through matching averages

Energy accumulates over both operating phases

Check percentages and rearrangements

Example: electrical lifting drive over a complete interval

A load gains potential energy

Example: combine two operating phases correctly

A device changes operating point

Conversion chains and additional inputs

Explore further: COP and the Carnot limit

Motor alone or the complete installation?

Motor alone or the complete installation?

Five solved conversion balances

A device receives 750 J and delivers 600 J usefully with zero storage change. Find η and other output.

A device receives constant 200 W, delivers useful output at η=0.80\eta=0.80η=0.80 and runs 30 s. Find useful energy.

Two conversion stages have η1=0.90,η2=0.75 and initial input 2.40 kW. No auxiliary input or storage change. Find final useful power.

During 10 s a device receives 100 J electrically, supplies 130 J useful energy and outputs another 20 J. Its initial store has enough energy. Find storage change. Can 130% be claimed as conversion efficiency without qualification?

A device runs first for 1 s at input 100 W and η=0.9, then for 9 s at input 10 W and η=0.5. No storage change and same useful-output definition. Someone reports 70% by averaging both efficiencies. Find correct interval efficiency.

In brief

A device receives constant 200 W, delivers useful output at $\eta=0.80$ and runs 30 s. Find useful energy.

A device receives constant 200 W, delivers useful output at $\eta=0.80$ and runs 30 s. Find useful energy.