Electrical Energy and Power
4. Efficiency and Energy Use
Learning outcomes
- I can calculate efficiency.
- I can explain energy losses.
- I can compare efficient technologies.
- I can evaluate electrical energy consumption.
- I can suggest methods for conserving energy.
Efficiency and Energy Use
What Is Efficiency?
Efficiency describes how much of the energy supplied to a device is transferred in a useful way.
No real electrical device transfers all of its input energy into the desired output. Some energy is usually transferred to the surroundings in unwanted forms, often as thermal energy or sound.
Efficiency can be calculated using:
efficiency = useful energy output / total energy input
As a percentage:
efficiency = useful energy output / total energy input × 100%
Efficiency can also be calculated using power:
efficiency = useful power output / total power input × 100%
Understanding Efficiency
Suppose an electrical device receives:
100 J
of energy.
If:
80 J
is transferred usefully, its efficiency is:
efficiency = 80/100 × 100%
efficiency = 80%
The remaining:
100 − 80 = 20 J
is transferred in other ways.
The energy has not disappeared.
Conservation of Energy
The law of conservation of energy states that energy cannot be created or destroyed.
Therefore:
total energy input = total energy output
For a device:
input energy = useful energy output + unwanted energy transfers
For example:
500 J input = 400 J useful + 100 J unwanted
Energy is conserved even if the device is inefficient.
Energy Losses
We often describe unwanted energy transfers as energy losses.
However, the energy is not actually destroyed.
It has usually been transferred to the surroundings in forms that are less useful for the intended purpose.
Common unwanted transfers include:
- thermal energy
- sound
- vibration
- frictional heating
- electrical resistance
Example: Electric Motor
An electric motor receives:
1000 J
of electrical energy.
It produces:
850 J
of useful mechanical energy.
The remaining energy is:
1000 − 850
= 150 J
This may be transferred as:
- thermal energy
- sound
- vibration
Efficiency:
850/1000 × 100%
= 85%
Efficiency Using Power
Because power is the rate of energy transfer, efficiency can also be calculated using:
efficiency = useful power output / input power × 100%
Suppose a motor receives:
P_input = 800 W
and produces:
P_useful = 680 W
Then:
efficiency = 680/800 × 100%
efficiency = 85%
The unwanted power transfer is:
800 − 680
= 120 W
Calculating Useful Energy
The efficiency equation can be rearranged.
Using efficiency as a decimal:
useful energy = efficiency × input energy
Suppose:
input energy = 2000 J
efficiency = 75%
Convert:
75% = 0.75
Then:
useful energy = 0.75 × 2000
useful energy = 1500 J
Calculating Input Energy
The equation can also be rearranged:
input energy = useful energy / efficiency
Suppose a machine must provide:
900 J
of useful energy and is:
60% efficient.
Convert:
60% = 0.60
Then:
input energy = 900/0.60
input energy = 1500 J
The unwanted transfer is:
1500 − 900
= 600 J
Efficiency Cannot Normally Exceed 100%
Efficiency cannot normally be greater than:
100%
A claimed efficiency of 120%, for example, would mean the useful energy output is greater than the total energy input being counted.
That would violate conservation of energy if all relevant energy inputs and outputs were properly included.
An answer above 100% usually indicates:
- calculation error
- incorrect measurements
- missing energy input
- incorrect interpretation of the system
Useful Energy Depends on Purpose
Whether an energy transfer is useful depends on what the device is designed to do.
Electric heater
Useful output:
thermal energy
Electric motor
Useful output:
mechanical energy
Lamp
Useful output:
light
Speaker
Useful output:
sound
Thermal energy from a heater is useful.
Thermal energy from a motor is usually unwanted.
Energy Flow
Consider a lamp receiving:
100 J
of electrical energy.
Suppose:
25 J → useful light
75 J → thermal energy
Efficiency:
25/100 × 100%
= 25%
Energy conservation:
100 J = 25 J + 75 J
The lamp does not lose 75 J from existence. That energy is transferred into the surroundings.
Sankey Diagrams
A Sankey diagram represents energy transfers using arrows.
The width of each arrow represents the amount of energy transferred.
For example:
100 J input
→ 80 J useful output
→ 20 J unwanted output
A more efficient device has a greater proportion of its input arrow continuing into the useful output.
Electrical Resistance and Energy Loss
When current flows through resistance, electrical energy is transferred into thermal energy.
The heating power is:
P = I²R
Therefore, unwanted heating increases with:
- greater resistance
- greater current
Because current is squared, increasing current can dramatically increase heating losses.
Effect of Current on Energy Loss
Suppose a wire has:
R = 2 Ω
and carries:
I = 3 A
Heating power:
P = I²R
P = 3² × 2
P = 18 W
If current doubles to:
6 A
then:
P = 6² × 2
P = 72 W
Doubling the current produces:
four times the heating power
Reducing Electrical Transmission Losses
Electrical power transmission provides an important application of efficiency.
Power transmitted is approximately:
P = VI
For the same transmitted power, increasing voltage allows the current to decrease.
Heating losses in transmission wires are:
P_loss = I²R
Therefore:
higher transmission voltage → lower current → lower I²R losses
This is one major reason electricity is transmitted at very high voltages.
Example: Transmission Loss
Suppose a transmission line has:
R = 5 Ω
and carries:
I = 100 A
Power lost:
P_loss = I²R
P_loss = 100² × 5
P_loss = 50,000 W
or:
50 kW
If the current were reduced to:
20 A
then:
P_loss = 20² × 5
P_loss = 2000 W
or:
2 kW
The current has fallen by a factor of 5, but the heating loss has fallen by a factor of 25.
Comparing Efficient Technologies
Two devices designed for the same purpose may use very different amounts of electrical energy.
A fair comparison should consider:
- useful output
- input power
- operating time
- efficiency
- energy consumption
- expected lifetime
- required performance
Simply choosing the device with the lowest power rating does not always produce the best comparison.
Efficient Lighting
Lighting is a useful example.
An inefficient lamp may transfer a large proportion of its electrical input into unwanted thermal energy.
A more efficient lamp produces the required illumination while using less electrical power.
Modern LED lighting can provide high luminous output while requiring much less electrical power than many older lighting technologies.
Comparing Two Lamps
Suppose two lamps provide similar useful illumination.
Lamp A:
P = 60 W
Lamp B:
P = 10 W
Both operate for:
5 h
Lamp A
60 W = 0.060 kW
E = Pt
E = 0.060 × 5
E = 0.30 kWh
Lamp B
10 W = 0.010 kW
E = 0.010 × 5
E = 0.05 kWh
For the same operating time:
Lamp B uses:
0.25 kWh less energy
Percentage Energy Saving
Energy saving can be expressed as a percentage.
percentage saving = (original use − new use) / original use × 100%
Using the previous lamps:
percentage saving = (0.30 − 0.05)/0.30 × 100%
percentage saving ≈ 83.3%
This comparison assumes the lamps provide a sufficiently similar useful lighting output.
Efficient Motors
Electric motors are used in:
- fans
- pumps
- refrigerators
- air-conditioning systems
- elevators
- industrial machinery
- electric vehicles
Improving motor efficiency can reduce energy consumption significantly because many motors operate for long periods.
Efficiency can be improved through:
- lower electrical resistance
- reduced friction
- improved magnetic design
- appropriate motor sizing
- efficient control systems
Heating Devices and Efficiency
Electrical resistance heaters can convert a very large proportion of their electrical input into thermal energy at the point of use.
If the goal is simply to produce heat, that thermal output is useful.
However, this does not automatically mean every electric heating system is the most energy-efficient way to heat a building.
The performance of the entire system must be considered.
Heat Pumps
A heat pump uses electrical energy to move thermal energy from one place to another.
Examples include:
- refrigerators
- air conditioners
- heat-pump heating systems
Instead of producing all the thermal energy directly from electrical resistance, a heat pump transfers thermal energy between environments.
For heat pumps, engineers often use a measure called coefficient of performance (COP) rather than ordinary energy-conversion efficiency.
This distinction is important because a heat pump can deliver more thermal energy to a room than the electrical energy it consumes by transferring additional energy from the environment.
This does not violate conservation of energy.
Electrical Energy Consumption
Energy consumption depends on both power and time.
E = Pt
For household electricity:
- P is often measured in kW
- t is measured in hours
- E is measured in kWh
For example:
P = 2 kW
t = 3 h
E = 2 × 3
E = 6 kWh
Evaluating Energy Consumption
When evaluating an appliance, ask:
- What is its power?
- How long does it operate?
- How often is it used?
- What useful output does it provide?
- How efficient is it?
- Does it consume standby power?
- Is there a more efficient alternative?
A high-power appliance used briefly may consume less energy than a low-power appliance operating continuously.
Example: Kettle vs Television
Kettle
P = 2.0 kW
t = 6 minutes = 0.10 h
E = 2.0 × 0.10
E = 0.20 kWh
Television
P = 0.10 kW
t = 5 h
E = 0.10 × 5
E = 0.50 kWh
The television has much lower power but consumes more energy in this example because it operates much longer.
Daily Energy Consumption
Suppose a device has:
P = 500 W = 0.50 kW
and operates for:
6 h/day
Daily energy:
E = 0.50 × 6
E = 3 kWh/day
For 30 days:
E = 3 × 30
E = 90 kWh
Long operating times can make even moderate-power devices significant energy consumers.
Calculating Energy Cost
If electricity costs:
$0.30/kWh
and a device uses:
90 kWh
then:
cost = 90 × 0.30
cost = $27
Reducing energy consumption can therefore reduce both:
- energy demand
- operating cost
Comparing Two Appliances
Suppose two appliances perform the same task.
Appliance A
Power = 1200 W
Operating time = 2 h
Appliance B
Power = 800 W
Operating time = 2.5 h
Appliance A:
E = 1.2 × 2
E = 2.4 kWh
Appliance B:
E = 0.8 × 2.5
E = 2.0 kWh
Appliance B has the lower total energy consumption even though it operates longer.
Standby Power
Some devices consume power while waiting for use.
Examples can include:
- televisions
- game consoles
- computers
- printers
- chargers
- network equipment
- smart appliances
Suppose ten devices each average:
3 W
of unnecessary standby power.
Total:
P = 10 × 3
P = 30 W
Over 24 hours:
E = 0.030 × 24
E = 0.72 kWh/day
Over 365 days:
E = 0.72 × 365
E ≈ 263 kWh
Small continuous loads can become significant over time.
Energy-Efficient Appliances
Many modern appliances are designed to reduce energy consumption.
Examples include:
- LED lighting
- efficient refrigerators
- high-efficiency motors
- heat pumps
- variable-speed air conditioners
- efficient power supplies
- smart controls
When comparing devices, energy labels can help estimate their expected consumption under standardized conditions.
Energy Labels
Energy labels may provide information such as:
- efficiency rating
- estimated annual energy consumption
- power consumption
- capacity
- performance
However, actual energy use depends on how the device is used.
A highly efficient appliance that operates unnecessarily for long periods can still consume substantial energy.
Energy Conservation
Energy conservation in everyday language can mean reducing unnecessary energy consumption.
This is different from the law of conservation of energy.
Law of conservation of energy
Energy cannot be created or destroyed.
Conserving energy
Reducing unnecessary energy consumption or unwanted energy transfers.
The two ideas are related but should not be confused.
Reducing Lighting Energy Use
Lighting energy consumption can be reduced by:
- using efficient lamps
- switching lights off when unnecessary
- using daylight where practical
- using occupancy sensors
- using appropriate lighting levels
Both technology and user behavior affect consumption.
Reducing Heating and Cooling Energy
Heating and cooling can require large amounts of energy.
Consumption can be reduced through:
- insulation
- reducing unwanted air leakage
- appropriate thermostat settings
- efficient heating/cooling equipment
- shading
- efficient windows
- regular maintenance
Good insulation reduces unwanted thermal energy transfer between the building and its surroundings.
Building Insulation
Buildings lose or gain thermal energy through:
- walls
- roofs
- windows
- floors
- air leakage
Improved insulation reduces the rate of unwanted thermal energy transfer.
This reduces the energy required to maintain a comfortable indoor temperature.
Reducing Appliance Energy Use
Useful strategies include:
- switching off unnecessary devices
- avoiding unnecessary standby operation
- selecting efficient appliances
- using appropriate operating modes
- maintaining equipment
- using timers or automatic controls
- operating appliances efficiently
The best strategy depends on the device.
Efficient Refrigeration
Refrigerators transfer thermal energy from inside the refrigerator to the surrounding room.
Energy consumption can be reduced by:
- good insulation
- effective door seals
- minimizing unnecessary door opening
- appropriate temperature settings
- keeping heat-transfer surfaces functioning properly
- using efficient compressors
A damaged door seal increases thermal energy entering the refrigerator, causing the cooling system to operate more.
Efficient Air Conditioning
Air-conditioning systems can be major electrical loads.
Efficiency can be improved through:
- suitable temperature settings
- clean filters
- effective building insulation
- closing doors and windows
- shading from direct sunlight
- efficient variable-speed systems
- appropriate equipment sizing
Reducing the cooling load can be just as important as improving the air conditioner's efficiency.
Efficiency and Device Lifetime
Energy efficiency is not the only factor when evaluating technology.
Other considerations include:
- purchase cost
- operating cost
- durability
- maintenance
- useful performance
- materials
- environmental impact
- expected lifetime
A good evaluation considers several factors rather than focusing on a single number.
Energy Payback
Some energy-efficient technologies require more energy or resources to manufacture but reduce energy consumption during use.
A complete evaluation can therefore consider:
energy used during manufacture + energy used during operation + energy involved at end of life
This is part of a broader life-cycle analysis.
Improving an Inefficient System
Suppose a motor system has:
input power = 1000 W
useful output power = 700 W
Efficiency:
700/1000 × 100%
= 70%
An improved motor produces the same:
700 W
useful output at:
800 W input.
New efficiency:
700/800 × 100%
= 87.5%
Power saved:
1000 − 800
= 200 W
If operated for:
10 h
energy saved:
0.200 × 10
= 2 kWh
Energy Saving Over Time
Small improvements can become significant when repeated.
Suppose an efficient device saves:
0.5 kWh/day
Over one year:
0.5 × 365
= 182.5 kWh
If electricity costs:
$0.30/kWh
annual saving:
182.5 × 0.30
= $54.75
The longer a device operates, the more important efficiency can become.
Evaluating Whether an Upgrade Is Worthwhile
Suppose an efficient appliance costs:
$150 more
but saves:
$50 per year
Simple payback time:
payback time = additional cost / annual saving
payback time = 150/50
= 3 years
If the appliance lasts much longer than three years, the upgrade may provide financial savings after the payback period.
However, a complete evaluation might also consider maintenance, performance and changing electricity prices.
Energy Audits
An energy audit examines where energy is being consumed and identifies opportunities to reduce unnecessary consumption.
An audit might examine:
- lighting
- heating
- cooling
- appliances
- standby loads
- insulation
- operating schedules
- equipment efficiency
The largest energy consumers are often the best places to investigate first.
Example Household Energy Audit
Suppose daily consumption is:
air conditioning = 12 kWh
water heating = 5 kWh
lighting = 2 kWh
electronics = 3 kWh
other = 2 kWh
Total:
24 kWh/day
Air conditioning accounts for:
12/24 × 100%
= 50%
Improving cooling efficiency could therefore have a larger impact than focusing only on very small loads.
Reducing Peak Demand
Energy use can also be evaluated in terms of when electricity is required.
If many high-power devices operate simultaneously, the total power demand can become large.
For example:
air conditioner = 2.5 kW
oven = 3.0 kW
kettle = 2.0 kW
water heater = 3.5 kW
Total:
11.0 kW
Reducing simultaneous operation can reduce peak electrical demand even if total daily energy use remains similar.
Efficiency vs Conservation
Efficiency and conservation are related but different.
Efficiency
Obtaining more useful output from a given energy input.
Example:
replacing an inefficient lamp with an efficient LED.
Conservation
Reducing unnecessary energy use.
Example:
switching the lamp off when nobody needs it.
The most effective approach often combines both:
use efficient technology + avoid unnecessary use
Comparing Technologies Fairly
A fair comparison should ensure that devices provide approximately the same useful service.
For example, comparing two lamps only by wattage is incomplete.
Better questions include:
- Do they provide similar illumination?
- How much electrical power does each require?
- How long will each last?
- What is the total energy consumption?
- What is the operating cost?
Similarly, two air conditioners should be compared at similar cooling performance rather than simply comparing their electrical power.
Improving Efficiency
Efficiency can often be improved by reducing unwanted energy transfers.
Examples include:
Motors
Reduce friction and electrical resistance.
Buildings
Improve insulation.
Lighting
Use more efficient light sources.
Power transmission
Reduce current by using higher transmission voltage.
Electronics
Use efficient power supplies.
Heating systems
Reduce unwanted heat loss.
Cooling systems
Reduce heat entering the cooled space.
Common Mistakes
Mistake: Saying inefficient devices destroy energy.
Energy is conserved. It is transferred into less useful forms.
Mistake: Confusing efficiency with energy conservation.
Efficiency describes useful output compared with input. Conservation can also refer to reducing unnecessary use.
Mistake: Writing efficiency as more than 100%.
Ordinary energy-conversion efficiency cannot exceed 100%.
Mistake: Forgetting to multiply by 100% when expressing efficiency as a percentage.
Mistake: Using 80 instead of 0.80 in rearranged efficiency calculations.
Convert percentages to decimals when multiplying or dividing.
Mistake: Assuming lower power always means greater efficiency.
Useful output must also be considered.
Mistake: Assuming high efficiency automatically means low energy consumption.
Operating time and required output matter.
Mistake: Calling all thermal energy wasted.
Thermal energy is useful in heating devices.
Mistake: Ignoring standby consumption.
Small continuous loads can accumulate substantial energy use.
Mistake: Focusing only on small devices while ignoring major energy consumers.
Energy audits should prioritize large or long-running loads.
Key Terms
Efficiency — proportion of input energy or power transferred usefully.
Useful energy — energy transferred in the form required for the intended purpose.
Unwanted energy transfer — energy transferred in a way that does not contribute to the intended function.
Energy loss — common term for energy transferred into unwanted forms rather than destroyed.
Conservation of energy — principle that energy cannot be created or destroyed.
Energy consumption — amount of energy used or transferred over a period of time.
Kilowatt-hour (kWh) — unit of energy equal to 3.6 MJ.
Power — rate of energy transfer.
Standby power — electrical power consumed while a device is waiting or not performing its main function.
Energy-efficient technology — technology designed to provide a required useful output using less input energy.
Energy conservation — reduction of unnecessary energy consumption.
Sankey diagram — diagram representing energy transfers with arrow widths proportional to energy amounts.
Energy audit — systematic investigation of energy consumption and opportunities for improvement.
Payback time — time required for financial savings to recover an additional initial cost.
Life-cycle analysis — evaluation considering impacts across manufacture, operation and end of life.
Peak demand — maximum rate of electrical energy use during a particular period.
Key Takeaways
- Efficiency measures the proportion of input energy transferred usefully.
- efficiency = useful energy output / total energy input × 100%.
- Efficiency can also be calculated using power.
- Energy cannot be created or destroyed.
- Inefficient devices transfer more energy into unwanted forms.
- Thermal energy, sound and friction are common unwanted transfers.
- Whether an energy transfer is useful depends on the device's purpose.
- Sankey diagrams can represent useful and unwanted energy transfers.
- Electrical resistance can produce unwanted heating.
- Transmission losses are proportional to I²R.
- High-voltage transmission reduces current and therefore reduces heating losses.
- Energy consumption depends on both power and operating time.
- E = Pt.
- Household electrical energy is commonly measured in kWh.
- Low-power devices can consume significant energy when operated continuously.
- Efficient technologies should be compared based on similar useful outputs.
- LED lighting, efficient motors, insulation and efficient cooling systems can reduce energy consumption.
- Energy labels can help compare appliances.
- Efficiency and energy conservation are different but complementary approaches.
- Energy audits help identify where the greatest savings can be achieved.
- The most effective energy strategy is usually to combine efficient technology with reduced unnecessary use.
Check Your Understanding
- Define efficiency.
- Write the equation for efficiency using energy.
- Write the equation for efficiency using power.
- A device receives 500 J and produces 400 J of useful output. Calculate its efficiency.
- Where does the remaining 100 J in Question 4 go?
- A motor is 75% efficient and receives 2000 J. Calculate its useful energy output.
- A machine must provide 1200 J of useful energy and is 80% efficient. Calculate the required input energy.
- Explain why ordinary energy-conversion efficiency cannot exceed 100%.
- Explain why the term "energy loss" does not mean that energy has been destroyed.
- A wire carries 5 A and has a resistance of 2 Ω. Calculate the heating power.
- If the current in Question 10 doubles, calculate the new heating power.
- Explain why high-voltage transmission reduces energy losses.
- A 2 kW appliance operates for 4 hours. Calculate its energy consumption in kWh.
- A 50 W device operates continuously for 24 hours. Calculate its daily energy consumption in kWh.
- Explain why a lower-power appliance is not necessarily more efficient.
- Compare energy efficiency and energy conservation.
- Suggest three ways a household could reduce energy used for heating or cooling.
- A new appliance costs $200 more but saves $40 per year in electricity. Calculate its simple payback time.
- Explain why an energy audit should usually focus first on high-power or long-running devices.
- Evaluate the statement: "The best way to reduce electrical energy use is simply to buy lower-power appliances."