# Calculate the efficiency of a motor in two ways(Activity)

### Practical measurements and calculations as exemplars to learn from

What the activity is for

In this activity a motor is used to lift a load. The efficiency of the motor is calculated in two ways:

• Considering the energy shifted from the chemical store to the gravity store once the load has been lifted.
• Considering the power in the electrical and mechanical pathways as the load is lifted.

• What the activity is for

• a 12 V power pack
• a small motor with pulley/axle attachment
• 2 G clamps
• a 1 m rule
• some Blu-tack
• a 100 g mass holder
• 3 100 g masses
• an SEP energy meter
• 1 m of light, strong, thin string
• Safety note: Both the falling masses and the strained string line present possible hazards.

What happens during this activity

This account includes sample values to give an indication of the kind of values expected. Your own values may differ and should be used.

### Calculating using stores and energy

Set up the equipment to lift a mass of 400 gram through a distance of 80.0 cm. Blu-tack a 1 metre ruler close to the load to show distance moved clearly. Have the SEP energy meter set to read energy and the power pack set at 5 V DC.

Teacher: So, we're using a motor to lift a load from the floor. We have an energy meter, which is connected directly to the power supply to let us know how much energy is being shifted from the store. The kind of store depends on our local power station. Here we burn gas, so it'll be a chemical store.

Switch on the power supply and the energy meter simultaneously. Allow the load to move through a distance of 80 cm and then switch the supply and energy meter off simultaneously. (Use a student to help because three hands are needed and the mass will start to fall back down immediately.)

Teacher: How much energy is shifted to the motor?

Lola: 21.7 joule.

It's probably worth writing this on the board:

Energy shifted from the chemical store is 21.7 joule

Teacher: Excellent, and you used the correct unit too. The gravity store is filling as the load is lifted. How could we calculate how much energy is shifted to the store?

Bill: Gravity force × vertical distance moved.

Teacher: Good. Now we can calculate the energy shifted as the load is lifted further from the earth.

We'd suggest writing this on the board (don't forget the units to make it more intelligible):

energy shifted = 0.4 kilogram  × 9.8 newton kilogram-1 × 0.8 m

energy is 3.14 J

Teacher: We know that energy is neither created nor destroyed, so why is this change not the same as that shifted from the store?

Julie: Energy must have been shifted to other stores: stores that are not useful.

Explain that there is also energy being shifted to thermal stores. Now calculate the efficiency of the motor on the board using the equation.

A clear written-out calculation can serve as a good model, perhaps written on a board.

efficiency = useful energy shiftedtotal energy input to the system × 100 %.

Put the numbers in to get 3.14 joule21.7 joule  × 100 %, which can be worked out to be an efficiency of 14.5 %.

### Calculating using power in pathways

Change the SEP energy meter to read average power. The average power in the electrical pathway can be read directly from the meter. The motor switches this power more or less effectively to mechanical working. The power in the mechanical pathway can be found from force  × velocity.

If all of the power in the electrical working pathway is switched to mechanical working, then the efficiency is 100 %. So efficiency can also be calculated using power values. After this preamble we suggest you get a new set of measurements to get these data.

Teacher: What do we need to know if we want to find the average power in the mechanical working pathway as the load is lifted?

David: The force required to lift the load and the speed of the load.

Lift the load as before, recording the average power shifted to the motor as read from the SEP meter and the time taken to lift the load.

Here are some sample values, showing the line of argument you might follow.

The average input power to the motor is 4.34 watt

The power needed to lift the load(the output power) is 0.4 kilogram  × 9.8 newton inverse kilogram  × 0.8 metre5 second which is 3.92 N  × 0.16 m s-1, and therefore the output power is 0.63 W.

Now for some efficiencies:

efficiency = power in mechanical pathwaypower in electrical pathway × 100 %

Putting the values in, you get: 0.63 watt4.34 watt  × 100, which evaluates to give an efficiency of 14.5 %

This activity has been written containing only one lifting of the load for the energy and power calculations for ease of explanation. It could be run several times and average values used to calculate the efficiencies.

Some possibilities to extend the experiment:

• Change the load size to see if this affects the efficiency of the motor.
• Change the distances moved by the load to see if this affects the efficiency of the motor.