8.3.1 Power

Power as a rate of transfer

Definition

Power

Power is the rate at which energy is transferred, which is the same as the rate at which work is done.

Definition

Watt

The watt is the unit of power, equal to an energy transfer of one joule each second.

Power is the rate at which energy is transferred, and because work done and energy transferred are the same quantity it is also the rate of doing work.
Power answers how many joules each second, not how many joules altogether.
Two machines that transfer the same energy have different powers if they take different times, and the quicker one is the more powerful.
The unit of power is the watt, $W\text{W}$ , and $J/s1\ \text{W} = 1\ \text{J/s}$ .
A device rated at $W60\ \text{W}$ transfers $J60\ \text{J}$ every second it is switched on.
Larger powers are quoted in kilowatts and megawatts, where $W1\ \text{kW} = 1000\ \text{W}$ and $W1\ \text{MW} = 1\,000\,000\ \text{W}$ .
A large power does not guarantee a large total energy transfer, because a small power running for a long time can transfer more.

The power equation

Power is calculated from $\dfrac{E}{t}$ .
Here $P$ is the power in watts, $E$ is the work done or energy transferred in joules and $t$ is the time taken in seconds.
Rearranging gives $\times t$ for the energy transferred and $\dfrac{E}{P}$ for the time taken.
Convert every time into seconds first, so $s2.5\ \text{minutes} = 150\ \text{s}$ and $s1\ \text{hour} = 3600\ \text{s}$ .
Convert kilowatts into watts before substituting, so $W2.4\ \text{kW} = 2400\ \text{W}$ .
For a lift or a hoist, work out $\times g \times \Delta h$ first and then divide by the time.
For a machine pulling against a steady force, work out $\times d$ first and then divide by the time.

Example

Power of an electric motor

A motor does $J3600\ \text{J}$ of work in $s30\ \text{s}$ .
The equation is $\dfrac{E}{t}$ .
Substituting gives $\dfrac{3600}{30}$ .
The power is $120\ \text{W}$ , which means the motor transfers $J120\ \text{J}$ every second.

Example

Power of a lift raising a load

A lift raises a load of mass $kg500\ \text{kg}$ through a vertical height of $m12\ \text{m}$ in $s20\ \text{s}$ , with $10\ \text{N/kg}$ .
The useful energy transferred is $\times g \times \Delta h = 500 \times 10 \times 12$ .
This gives $60\,000\ \text{J}$ .
The power is $\dfrac{E}{t} = \dfrac{60\,000}{20}$ .
The useful power output of the lift is $3000\ \text{W}$ , which is $kW3.0\ \text{kW}$ .

Example

Energy transferred by a rated appliance

A $kW2.4\ \text{kW}$ shower runs for $minutes8.0\ \text{minutes}$ .
Converting the values gives $2400\ \text{W}$ and $\times 60 = 480\ \text{s}$ .
Rearranging gives $\times t = 2400 \times 480$ .
The energy transferred is $1\,152\,000\ \text{J}$ , which is $kJ1152\ \text{kJ}$ .

Practical

Measuring your own power output

Aim: to measure the useful power output of a student climbing a flight of stairs.
Apparatus: bathroom scales or a balance, metre rule, stopwatch reading to $s0.01\ \text{s}$ , a staircase with a countable number of steps, and a calculator.
Variables: the time taken is the measured variable, the power output is the calculated variable, and the student, the staircase and the vertical height climbed are kept the same for every run.
Method, set-up:
- Measure the student's mass in kilograms on the scales and record it.
- Measure the vertical height of one step with the metre rule, then multiply by the number of steps to get the vertical height of the whole flight.
- Mark a clear start line at the bottom step and a finish line at the top step so every run covers the same vertical height.
Method, measurements:
- Start the stopwatch as the student's foot leaves the start line and stop it as the foot lands on the finish line.
- Climb at a brisk but safe walking pace, keeping one hand free for the handrail.
- Record the time in seconds.
- Repeat three times with a rest between runs, then calculate the mean time.
- Repeat with a second student so two power outputs can be compared.
Results: the student who climbs the same flight in the shorter time has the greater power output, and a heavier student climbing in the same time has a greater power output still.
Maths: calculate the useful energy transferred from $\times g \times \Delta h$ with $10\ \text{N/kg}$ , then calculate the power from $\dfrac{E}{t}$ ; a student of mass $kg60\ \text{kg}$ climbing $m3.0\ \text{m}$ in $s5.0\ \text{s}$ transfers $J1800\ \text{J}$ and has a power output of $W360\ \text{W}$ .
Watch out: reaction time in starting and stopping the stopwatch is the largest source of uncertainty, so use a longer flight and repeat every run; the height must be the vertical rise of the flight rather than the length of the staircase; the figure found is the useful power only, because the body also dissipates energy by heating.
Safety: keep the staircase clear, do not run, use the handrail, and let anyone who feels unwell sit the activity out.

Comparing power in real machines

Two cranes raising identical loads to the same height do the same work, so the crane that finishes sooner has the greater power.
A $kW2.0\ \text{kW}$ kettle and a $W10\ \text{W}$ phone charger differ by a factor of $200$ in the joules they transfer each second.
The charger left plugged in overnight can still transfer more energy in total than the kettle used for two minutes.
The power rating printed on an appliance gives the energy it transfers each second while it is working normally.
A car engine quoted in kilowatts is being described by how quickly it can do work, not by how far the car can travel.
Comparing powers fairly means comparing the energy transferred over the same time, or the time taken to transfer the same energy.

Exam technique

Answering power questions

Write $\dfrac{E}{t}$ before substituting, and show any conversion of minutes into seconds on its own line.
Work out the energy first when the question gives a mass and a height, or a force and a distance, then divide by the time.
Give the unit as $W\text{W}$ , or as $kW\text{kW}$ only when the question asks for kilowatts.
When comparing two devices, say which quantity is the same and which differs, such as the same energy is transferred in a shorter time, so the power is greater.
Use the word rate in the definition, because an answer that says only that power is energy does not gain the mark.

Common Mistake

Do not give a power in joules, because joules measure the energy transferred while watts measure the rate of transfer.
Do not substitute a time in minutes or hours, because the equation needs seconds.
Do not assume the more powerful device always transfers more energy, because that holds only for equal times.
Do not confuse $kW\text{kW}$ with $kJ\text{kJ}$ , because one is a power and the other is an energy.

Self review

Define power and give its unit.
Write the equation linking power, energy transferred and time.
Complete the statement $J/s1\ \text{W} = \ldots\ \text{J/s}$ .
Describe how to measure your own power output on a flight of stairs.
Explain how a $W10\ \text{W}$ charger can transfer more energy than a $kW2\ \text{kW}$ kettle.

8.3.2 Efficiency

Efficiency of a device

Definition

Efficiency

Efficiency is the ratio of useful energy transferred by a device to the total energy supplied to it.

Efficiency compares the energy a device transfers usefully with the total energy supplied to it.
The useful transfer is the one the device is designed for, so it is the gravitational store for a hoist and the light for a lamp.
Every other transfer is unwanted, and it is dissipated into the thermal energy stores of the device and its surroundings.
Efficiency is a ratio of two energies measured in the same unit, so it has no unit.
It can be written either as a decimal between $0$ and $1$ or as a percentage between $0%0\%$ and $100%100\%$ .
No real device reaches an efficiency of $1$ , because some energy is always dissipated.
A higher efficiency means a larger share of the joules supplied is doing the job you wanted.

The efficiency equation

Efficiency is calculated from $device\text{efficiency} = \dfrac{\text{useful energy transferred by the device}}{\text{total energy supplied to the device}}$ .
Multiply a decimal by $100$ to turn it into a percentage, so $0.75$ becomes $75%75\%$ .
Divide a percentage by $100$ to turn it into a decimal, so $40%40\%$ becomes $0.40$ .
Rearranging gives $supplied\text{useful energy transferred} = \text{efficiency} \times \text{total energy supplied}$ .
Rearranging again gives $transferredefficiency\text{total energy supplied} = \dfrac{\text{useful energy transferred}}{\text{efficiency}}$ .
The total energy supplied is shared between the useful transfer and the unwanted transfers, so the total equals the useful energy plus the wasted energy.
Because both energies are measured over the same time, the same ratio can be found from powers using $input\dfrac{\text{useful power output}}{\text{total power input}}$ .
Put both quantities into the same unit before dividing, so convert kilojoules into joules first.

Efficiency from a Sankey diagram

The arrow entering a Sankey diagram on the left has a width set by the total energy supplied.
The branch drawn straight on is normally the useful transfer, while the branches that turn away are the unwanted ones.
Reading the two widths and dividing the useful one by the total gives the efficiency straight away.
A diagram with $J500\ \text{J}$ entering and a useful branch of width $J125\ \text{J}$ shows an efficiency of $125500=0.25\dfrac{125}{500} = 0.25$ .
Adding the widths of every branch leaving a point checks the reading, because they must equal the width of the arrow entering.

Increasing the efficiency of a device

Efficiency rises when the unwanted transfers are made smaller, because fewer joules are then dissipated.
Lubricating moving parts with oil or grease lowers the friction between surfaces, so less energy is transferred by heating.
Streamlining a shape lowers air resistance, so less energy is dissipated into the air.
Fitting thermal insulation slows the unwanted transfer by heating out of a device, so more of the energy stays where it is useful.
Replacing a filament lamp with a light emitting diode raises efficiency, because a smaller share of the energy supplied warms the surroundings.
Replacing worn bearings and gears lowers the friction inside a machine, so less of the energy supplied is dissipated.

Example

Efficiency of an electric motor

A motor is supplied with $J500\ \text{J}$ and transfers $J125\ \text{J}$ usefully to raise a load.
The equation is $supplied\text{efficiency} = \dfrac{\text{useful energy transferred}}{\text{total energy supplied}}$ .
Substituting gives $efficiency=125500\text{efficiency} = \dfrac{125}{500}$ .
The efficiency is $0.25$ , which is $0.25 \times 100 = 25\%$ .
The remaining $J375\ \text{J}$ is dissipated, mostly into the thermal energy stores of the motor and the air around it.

Example

Finding the wasted energy

A hair dryer has an efficiency of $0.85$ and is supplied with $J60\,000\ \text{J}$ during one drying session.
Rearranging gives $000\text{useful energy transferred} = \text{efficiency} \times \text{total energy supplied} = 0.85 \times 60\,000$ .
The useful transfer is $J51\,000\ \text{J}$ .
The energy dissipated in unwanted ways is $J60\,000 - 51\,000 = 9000\ \text{J}$ .

Example

Efficiency from power ratings

An electric winch is rated at $kW1.5\ \text{kW}$ and its useful power output while lifting is $kW1.2\ \text{kW}$ .
Both values are powers measured over the same time, so $efficiency=1.21.5\text{efficiency} = \dfrac{1.2}{1.5}$ .
The efficiency is $0.80$ , which is $80%80\%$ .
The remaining $kW0.3\ \text{kW}$ is dissipated, mostly by heating the motor windings and the gears.

Practical

Measuring the efficiency of a motor lifting a load

Aim: to measure the efficiency of a small electric motor as it raises a known load through a measured height.
Apparatus: low-voltage electric motor with a cotton reel or pulley on its shaft, low-voltage power supply, joulemeter, light string, slotted masses on a hanger, balance, metre rule, clamp stand and boss, and a soft landing mat.
Variables: the mass raised is the independent variable, the efficiency is the calculated variable, and the motor, the supply setting, the string and the vertical height raised are controlled.
Method, set-up:
- Clamp the motor to the stand so the string winds onto the shaft and the load hangs freely, with the mat underneath it.
- Measure the mass of the hanger and masses in kilograms on the balance.
- Fix a metre rule beside the load and mark a start line and a finish line so every run raises the load through the same vertical height.
- Connect the joulemeter between the supply and the motor so that it reads the total energy supplied to the motor.
Method, measurements:
- Zero the joulemeter, then switch on and raise the load from the start line to the finish line.
- Switch off the moment the load reaches the finish line and record the total energy supplied in joules.
- Record the mass raised in kilograms and the vertical height in metres.
- Repeat each load three times and calculate the mean energy supplied.
- Repeat for at least five different masses, keeping the height and the supply setting unchanged.
Results: the measured efficiency rises as the load increases, because the energy dissipated inside the motor stays roughly the same while the useful transfer grows.
Maths: calculate the useful energy transferred from $\times g \times \Delta h$ with $10\ \text{N/kg}$ , then calculate $supplied\text{efficiency} = \dfrac{m \times g \times \Delta h}{\text{total energy supplied}}$ , and plot efficiency against the mass raised.
Watch out: the string must wind on without slipping and the load must not swing, switching off late adds joules that did no useful lifting, and friction in the shaft together with the mass of the string both lower the measured efficiency.
Safety: keep fingers and long hair clear of the turning shaft, place a soft mat under the load, work at low voltage only, and clamp the stand to the bench so it cannot topple.

Exam technique

Setting out an efficiency answer

Decide which energy is useful by asking what the device is for, then put that value on the top of the fraction.
Put the total energy supplied on the bottom of the fraction, and never the wasted energy.
Check the answer lies between $0$ and $1$ , or between $0%0\%$ and $100%100\%$ , before writing it down.
Write no unit after an efficiency, because it is a ratio of two energies.
When an improvement is asked for, name the change and the transfer it reduces, such as oiling the bearings to reduce the energy dissipated by heating.

Common Mistake

Do not divide the total energy supplied by the useful energy transferred, because that gives a value greater than $1$ .
Do not write $J\text{J}$ or $W\text{W}$ after an efficiency, because a ratio has no unit.
Do not treat the wasted energy as the total supplied, because the total is the useful energy and the wasted energy added together.
Do not claim that a device can be $100%100\%$ efficient, because some energy is always dissipated.

Self review

Define efficiency and state its unit.
Write the efficiency equation and say which energy goes on the top.
Convert an efficiency of $0.35$ into a percentage.
Calculate the efficiency of a lamp supplied with $J200\ \text{J}$ that transfers $J24\ \text{J}$ by light.
Name two changes that raise the efficiency of a machine and state the transfer each one reduces.

8.3.1 Power

Power as a rate of transfer

Definition

Power

Power is the rate at which energy is transferred, which is the same as the rate at which work is done.

Definition

Watt

The watt is the unit of power, equal to an energy transfer of one joule each second.

Power is the rate at which energy is transferred, and because work done and energy transferred are the same quantity it is also the rate of doing work.
Power answers how many joules each second, not how many joules altogether.
Two machines that transfer the same energy have different powers if they take different times, and the quicker one is the more powerful.
The unit of power is the watt, $W\text{W}$ , and $J/s1\ \text{W} = 1\ \text{J/s}$ .
A device rated at $W60\ \text{W}$ transfers $J60\ \text{J}$ every second it is switched on.
Larger powers are quoted in kilowatts and megawatts, where $W1\ \text{kW} = 1000\ \text{W}$ and $W1\ \text{MW} = 1\,000\,000\ \text{W}$ .
A large power does not guarantee a large total energy transfer, because a small power running for a long time can transfer more.

The power equation

Power is calculated from $\dfrac{E}{t}$ .
Here $P$ is the power in watts, $E$ is the work done or energy transferred in joules and $t$ is the time taken in seconds.
Rearranging gives $\times t$ for the energy transferred and $\dfrac{E}{P}$ for the time taken.
Convert every time into seconds first, so $s2.5\ \text{minutes} = 150\ \text{s}$ and $s1\ \text{hour} = 3600\ \text{s}$ .
Convert kilowatts into watts before substituting, so $W2.4\ \text{kW} = 2400\ \text{W}$ .
For a lift or a hoist, work out $\times g \times \Delta h$ first and then divide by the time.
For a machine pulling against a steady force, work out $\times d$ first and then divide by the time.

Example

Power of an electric motor

A motor does $J3600\ \text{J}$ of work in $s30\ \text{s}$ .
The equation is $\dfrac{E}{t}$ .
Substituting gives $\dfrac{3600}{30}$ .
The power is $120\ \text{W}$ , which means the motor transfers $J120\ \text{J}$ every second.

Example

Power of a lift raising a load

A lift raises a load of mass $kg500\ \text{kg}$ through a vertical height of $m12\ \text{m}$ in $s20\ \text{s}$ , with $10\ \text{N/kg}$ .
The useful energy transferred is $\times g \times \Delta h = 500 \times 10 \times 12$ .
This gives $60\,000\ \text{J}$ .
The power is $\dfrac{E}{t} = \dfrac{60\,000}{20}$ .
The useful power output of the lift is $3000\ \text{W}$ , which is $kW3.0\ \text{kW}$ .

Example

Energy transferred by a rated appliance

A $kW2.4\ \text{kW}$ shower runs for $minutes8.0\ \text{minutes}$ .
Converting the values gives $2400\ \text{W}$ and $\times 60 = 480\ \text{s}$ .
Rearranging gives $\times t = 2400 \times 480$ .
The energy transferred is $1\,152\,000\ \text{J}$ , which is $kJ1152\ \text{kJ}$ .

Practical

Measuring your own power output

Aim: to measure the useful power output of a student climbing a flight of stairs.
Apparatus: bathroom scales or a balance, metre rule, stopwatch reading to $s0.01\ \text{s}$ , a staircase with a countable number of steps, and a calculator.
Variables: the time taken is the measured variable, the power output is the calculated variable, and the student, the staircase and the vertical height climbed are kept the same for every run.
Method, set-up:
- Measure the student's mass in kilograms on the scales and record it.
- Measure the vertical height of one step with the metre rule, then multiply by the number of steps to get the vertical height of the whole flight.
- Mark a clear start line at the bottom step and a finish line at the top step so every run covers the same vertical height.
Method, measurements:
- Start the stopwatch as the student's foot leaves the start line and stop it as the foot lands on the finish line.
- Climb at a brisk but safe walking pace, keeping one hand free for the handrail.
- Record the time in seconds.
- Repeat three times with a rest between runs, then calculate the mean time.
- Repeat with a second student so two power outputs can be compared.
Results: the student who climbs the same flight in the shorter time has the greater power output, and a heavier student climbing in the same time has a greater power output still.
Maths: calculate the useful energy transferred from $\times g \times \Delta h$ with $10\ \text{N/kg}$ , then calculate the power from $\dfrac{E}{t}$ ; a student of mass $kg60\ \text{kg}$ climbing $m3.0\ \text{m}$ in $s5.0\ \text{s}$ transfers $J1800\ \text{J}$ and has a power output of $W360\ \text{W}$ .
Watch out: reaction time in starting and stopping the stopwatch is the largest source of uncertainty, so use a longer flight and repeat every run; the height must be the vertical rise of the flight rather than the length of the staircase; the figure found is the useful power only, because the body also dissipates energy by heating.
Safety: keep the staircase clear, do not run, use the handrail, and let anyone who feels unwell sit the activity out.

Comparing power in real machines

Two cranes raising identical loads to the same height do the same work, so the crane that finishes sooner has the greater power.
A $kW2.0\ \text{kW}$ kettle and a $W10\ \text{W}$ phone charger differ by a factor of $200$ in the joules they transfer each second.
The charger left plugged in overnight can still transfer more energy in total than the kettle used for two minutes.
The power rating printed on an appliance gives the energy it transfers each second while it is working normally.
A car engine quoted in kilowatts is being described by how quickly it can do work, not by how far the car can travel.
Comparing powers fairly means comparing the energy transferred over the same time, or the time taken to transfer the same energy.

Exam technique

Answering power questions

Write $\dfrac{E}{t}$ before substituting, and show any conversion of minutes into seconds on its own line.
Work out the energy first when the question gives a mass and a height, or a force and a distance, then divide by the time.
Give the unit as $W\text{W}$ , or as $kW\text{kW}$ only when the question asks for kilowatts.
When comparing two devices, say which quantity is the same and which differs, such as the same energy is transferred in a shorter time, so the power is greater.
Use the word rate in the definition, because an answer that says only that power is energy does not gain the mark.

Common Mistake

Do not give a power in joules, because joules measure the energy transferred while watts measure the rate of transfer.
Do not substitute a time in minutes or hours, because the equation needs seconds.
Do not assume the more powerful device always transfers more energy, because that holds only for equal times.
Do not confuse $kW\text{kW}$ with $kJ\text{kJ}$ , because one is a power and the other is an energy.

Self review

Define power and give its unit.
Write the equation linking power, energy transferred and time.
Complete the statement $J/s1\ \text{W} = \ldots\ \text{J/s}$ .
Describe how to measure your own power output on a flight of stairs.
Explain how a $W10\ \text{W}$ charger can transfer more energy than a $kW2\ \text{kW}$ kettle.

8.3.2 Efficiency

Efficiency of a device

Definition

Efficiency

Efficiency is the ratio of useful energy transferred by a device to the total energy supplied to it.

Efficiency compares the energy a device transfers usefully with the total energy supplied to it.
The useful transfer is the one the device is designed for, so it is the gravitational store for a hoist and the light for a lamp.
Every other transfer is unwanted, and it is dissipated into the thermal energy stores of the device and its surroundings.
Efficiency is a ratio of two energies measured in the same unit, so it has no unit.
It can be written either as a decimal between $0$ and $1$ or as a percentage between $0%0\%$ and $100%100\%$ .
No real device reaches an efficiency of $1$ , because some energy is always dissipated.
A higher efficiency means a larger share of the joules supplied is doing the job you wanted.

The efficiency equation

Efficiency is calculated from $device\text{efficiency} = \dfrac{\text{useful energy transferred by the device}}{\text{total energy supplied to the device}}$ .
Multiply a decimal by $100$ to turn it into a percentage, so $0.75$ becomes $75%75\%$ .
Divide a percentage by $100$ to turn it into a decimal, so $40%40\%$ becomes $0.40$ .
Rearranging gives $supplied\text{useful energy transferred} = \text{efficiency} \times \text{total energy supplied}$ .
Rearranging again gives $transferredefficiency\text{total energy supplied} = \dfrac{\text{useful energy transferred}}{\text{efficiency}}$ .
The total energy supplied is shared between the useful transfer and the unwanted transfers, so the total equals the useful energy plus the wasted energy.
Because both energies are measured over the same time, the same ratio can be found from powers using $input\dfrac{\text{useful power output}}{\text{total power input}}$ .
Put both quantities into the same unit before dividing, so convert kilojoules into joules first.

Efficiency from a Sankey diagram

The arrow entering a Sankey diagram on the left has a width set by the total energy supplied.
The branch drawn straight on is normally the useful transfer, while the branches that turn away are the unwanted ones.
Reading the two widths and dividing the useful one by the total gives the efficiency straight away.
A diagram with $J500\ \text{J}$ entering and a useful branch of width $J125\ \text{J}$ shows an efficiency of $125500=0.25\dfrac{125}{500} = 0.25$ .
Adding the widths of every branch leaving a point checks the reading, because they must equal the width of the arrow entering.

Increasing the efficiency of a device

Efficiency rises when the unwanted transfers are made smaller, because fewer joules are then dissipated.
Lubricating moving parts with oil or grease lowers the friction between surfaces, so less energy is transferred by heating.
Streamlining a shape lowers air resistance, so less energy is dissipated into the air.
Fitting thermal insulation slows the unwanted transfer by heating out of a device, so more of the energy stays where it is useful.
Replacing a filament lamp with a light emitting diode raises efficiency, because a smaller share of the energy supplied warms the surroundings.
Replacing worn bearings and gears lowers the friction inside a machine, so less of the energy supplied is dissipated.

Example

Efficiency of an electric motor

A motor is supplied with $J500\ \text{J}$ and transfers $J125\ \text{J}$ usefully to raise a load.
The equation is $supplied\text{efficiency} = \dfrac{\text{useful energy transferred}}{\text{total energy supplied}}$ .
Substituting gives $efficiency=125500\text{efficiency} = \dfrac{125}{500}$ .
The efficiency is $0.25$ , which is $0.25 \times 100 = 25\%$ .
The remaining $J375\ \text{J}$ is dissipated, mostly into the thermal energy stores of the motor and the air around it.

Example

Finding the wasted energy

A hair dryer has an efficiency of $0.85$ and is supplied with $J60\,000\ \text{J}$ during one drying session.
Rearranging gives $000\text{useful energy transferred} = \text{efficiency} \times \text{total energy supplied} = 0.85 \times 60\,000$ .
The useful transfer is $J51\,000\ \text{J}$ .
The energy dissipated in unwanted ways is $J60\,000 - 51\,000 = 9000\ \text{J}$ .

Example

Efficiency from power ratings

An electric winch is rated at $kW1.5\ \text{kW}$ and its useful power output while lifting is $kW1.2\ \text{kW}$ .
Both values are powers measured over the same time, so $efficiency=1.21.5\text{efficiency} = \dfrac{1.2}{1.5}$ .
The efficiency is $0.80$ , which is $80%80\%$ .
The remaining $kW0.3\ \text{kW}$ is dissipated, mostly by heating the motor windings and the gears.

Practical

Measuring the efficiency of a motor lifting a load

Aim: to measure the efficiency of a small electric motor as it raises a known load through a measured height.
Apparatus: low-voltage electric motor with a cotton reel or pulley on its shaft, low-voltage power supply, joulemeter, light string, slotted masses on a hanger, balance, metre rule, clamp stand and boss, and a soft landing mat.
Variables: the mass raised is the independent variable, the efficiency is the calculated variable, and the motor, the supply setting, the string and the vertical height raised are controlled.
Method, set-up:
- Clamp the motor to the stand so the string winds onto the shaft and the load hangs freely, with the mat underneath it.
- Measure the mass of the hanger and masses in kilograms on the balance.
- Fix a metre rule beside the load and mark a start line and a finish line so every run raises the load through the same vertical height.
- Connect the joulemeter between the supply and the motor so that it reads the total energy supplied to the motor.
Method, measurements:
- Zero the joulemeter, then switch on and raise the load from the start line to the finish line.
- Switch off the moment the load reaches the finish line and record the total energy supplied in joules.
- Record the mass raised in kilograms and the vertical height in metres.
- Repeat each load three times and calculate the mean energy supplied.
- Repeat for at least five different masses, keeping the height and the supply setting unchanged.
Results: the measured efficiency rises as the load increases, because the energy dissipated inside the motor stays roughly the same while the useful transfer grows.
Maths: calculate the useful energy transferred from $\times g \times \Delta h$ with $10\ \text{N/kg}$ , then calculate $supplied\text{efficiency} = \dfrac{m \times g \times \Delta h}{\text{total energy supplied}}$ , and plot efficiency against the mass raised.
Watch out: the string must wind on without slipping and the load must not swing, switching off late adds joules that did no useful lifting, and friction in the shaft together with the mass of the string both lower the measured efficiency.
Safety: keep fingers and long hair clear of the turning shaft, place a soft mat under the load, work at low voltage only, and clamp the stand to the bench so it cannot topple.

Exam technique

Setting out an efficiency answer

Decide which energy is useful by asking what the device is for, then put that value on the top of the fraction.
Put the total energy supplied on the bottom of the fraction, and never the wasted energy.
Check the answer lies between $0$ and $1$ , or between $0%0\%$ and $100%100\%$ , before writing it down.
Write no unit after an efficiency, because it is a ratio of two energies.
When an improvement is asked for, name the change and the transfer it reduces, such as oiling the bearings to reduce the energy dissipated by heating.

Common Mistake

Do not divide the total energy supplied by the useful energy transferred, because that gives a value greater than $1$ .
Do not write $J\text{J}$ or $W\text{W}$ after an efficiency, because a ratio has no unit.
Do not treat the wasted energy as the total supplied, because the total is the useful energy and the wasted energy added together.
Do not claim that a device can be $100%100\%$ efficient, because some energy is always dissipated.

Self review

Define efficiency and state its unit.
Write the efficiency equation and say which energy goes on the top.
Convert an efficiency of $0.35$ into a percentage.
Calculate the efficiency of a lamp supplied with $J200\ \text{J}$ that transfers $J24\ \text{J}$ by light.
Name two changes that raise the efficiency of a machine and state the transfer each one reduces.

Power and efficiency

8.3.1 Power

Power as a rate of transfer

The power equation

Power of an electric motor

Power of a lift raising a load

Energy transferred by a rated appliance

Measuring your own power output

Comparing power in real machines

Answering power questions

8.3.2 Efficiency

Efficiency of a device

The efficiency equation

Efficiency from a Sankey diagram

Increasing the efficiency of a device

Efficiency of an electric motor

Finding the wasted energy

Efficiency from power ratings

Measuring the efficiency of a motor lifting a load

Setting out an efficiency answer

Recap questions

Review 8.3 Power and efficiency by teaching Genie

Recap your knowledge with an interactive lesson

Put it into practice with exam-style questions

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8.3 Power and efficiency Revision Guide

Power and efficiency

8.3.1 Power

Power as a rate of transfer

The power equation

Power of an electric motor

Power of a lift raising a load

Energy transferred by a rated appliance

Measuring your own power output

Comparing power in real machines

Answering power questions

8.3.2 Efficiency

Efficiency of a device

The efficiency equation

Efficiency from a Sankey diagram

Increasing the efficiency of a device

Efficiency of an electric motor

Finding the wasted energy

Efficiency from power ratings

Measuring the efficiency of a motor lifting a load

Setting out an efficiency answer

Recap questions

Review 8.3 Power and efficiency by teaching Genie

Recap your knowledge with an interactive lesson

Put it into practice with exam-style questions

Remember key concepts with flashcards

8.3 Power and efficiency Revision Guide