How to Calculate Total Efficiency for Two Devices in Sequence

An electric motor is 80% efficient. It drives a pump that is 75% efficient. What is the efficiency of the complete system?

You might be tempted to add the percentages and get 155%, or average them to get 77.5%. Neither is correct. The total efficiency is 60%.

The reason is simple: the pump doesn’t receive all the energy supplied to the motor. It only receives the 80% that the motor successfully transfers.

For two devices connected in sequence, multiply their efficiencies:

\[ \boxed{\eta_{\text{total}}=\eta_1\times\eta_2} \]

Use decimal efficiencies in this equation. For example:

\[ \eta_{\text{total}}=0.80\times0.75=0.60 \]

Therefore, the complete system is 60% efficient.

Let’s work through why this multiplication works, when you can use it, and how to handle the calculations that commonly appear in HSC Physics questions.

01Why do we multiply efficiencies?

Imagine that an electrical motor receives 100 J of energy.

It converts 80% of this energy into useful mechanical energy, which is transferred to a pump.

The pump then converts 75% of its mechanical energy input into useful energy that raises water.

Follow the energy through the system:

The first device transfers:

\[ E_1=100\times0.80=80\text{ J} \]

The second device receives those 80 J, not the original 100 J.

Its useful output is:

\[ E_2=80\times0.75=60\text{ J} \]

The complete system therefore receives 100 J and produces 60 J of useful output.

\[ \eta_{\text{total}}=\frac{60}{100} \]

\[ \boxed{\eta_{\text{total}}=60\%} \]

Notice something important: the pump’s 25% energy loss is calculated from its own input of 80 J.

It loses 20 J, not 25 J.

This explains why adding the individual percentage losses doesn’t work. Each percentage is calculated using a different energy input.

The mathematical relationship

Efficiency is the ratio of useful output energy to total input energy:

\[ \eta=\frac{E_{\text{useful output}}}{E_{\text{input}}} \]

For the first device:

\[ \eta_1=\frac{E_{\text{intermediate}}}{E_{\text{input}}} \]

For the second device:

\[ \eta_2=\frac{E_{\text{useful output}}}{E_{\text{intermediate}}} \]

Multiplying these expressions gives:

\[ \eta_1\eta_2= \frac{E_{\text{intermediate}}}{E_{\text{input}}} \times \frac{E_{\text{useful output}}}{E_{\text{intermediate}}} \]

The intermediate energy cancels:

\[ \boxed{ \eta_{\text{total}}= \frac{E_{\text{useful output}}}{E_{\text{input}}} =\eta_1\eta_2 } \]

This cancellation is the reason the multiplication rule works.

The output of the first device becomes the input of the second.

Using percentages instead of decimals

You can use either decimal efficiencies or percentages, but you must be consistent.

A useful general formula when working directly with percentages is:

\[ \boxed{ \eta_{\text{total}}(\%)= \frac{\eta_1(\%)\times\eta_2(\%)}{100} } \]

For three devices in sequence, simply multiply all three decimal efficiencies:

\[ \eta_{\text{total}}=\eta_1\eta_2\eta_3 \]

The same principle applies to any number of devices, provided the useful output of each stage is transferred into the next stage without an additional unaccounted loss.

02Energy efficiency versus power efficiency

One common source of mistakes is confusing energy with power.

They are related, but they measure different things.

• Energy, measured in joules (J), tells you how much energy is transferred.
• Power, measured in watts (W), tells you how quickly energy is transferred.

One watt is equivalent to one joule per second:

\[ 1\text{ W}=1\text{ J/s} \]

Power is calculated using:

\[ P=\frac{E}{t} \]

where:

• \(P\) is power in watts (W),
• \(E\) is energy transferred in joules (J),
• \(t\) is time in seconds (s).

You can calculate efficiency using either energy or power.

Using energy:

\[ \eta=\frac{E_{\text{useful output}}}{E_{\text{input}}} \]

Using power:

\[ \eta=\frac{P_{\text{useful output}}}{P_{\text{input}}} \]

The power equation applies when the input and output powers describe the same operating interval and there is no unaccounted accumulation of energy between the stages.

For two devices operating steadily:

\[ \boxed{ \eta_{\text{total}}= \frac{P_{\text{output,2}}}{P_{\text{input,1}}} =\eta_1\eta_2 } \]

This is particularly useful when an HSC question gives you the input power of the first device rather than an amount of energy.

For example, if a motor receives 500 W and is 80% efficient, its useful mechanical output power is:

\[ P_{\text{motor}}=500\times0.80 \]

\[ P_{\text{motor}}=400\text{ W} \]

If the second device is 75% efficient:

\[ P_{\text{useful}}=400\times0.75 \]

\[ P_{\text{useful}}=300\text{ W} \]

The total efficiency is:

\[ \eta_{\text{total}}=\frac{300}{500}=0.60 \]

The system transfers 300 J of useful energy every second for every second it operates under these conditions.

When power changes during operation

Be careful when a device’s efficiency or input power changes during a question.

You cannot always multiply two ordinary average efficiencies and expect to obtain the correct total efficiency.

For a variable system, calculate the total energy entering and leaving over the complete interval:

\[ \eta_{\text{total}}= \frac{\text{total useful energy output}} {\text{total energy input}} \]

This is an energy-weighted calculation, not necessarily an average of the percentages measured at different times.

We’ll return to this idea in the final practice question.

03Make sure the devices really are in sequence

The multiplication rule has an important condition: the useful output of the first device must become the input of the second.

Consider two different arrangements.

In the first arrangement, the devices are in sequence. Multiply their efficiencies.

In the second arrangement, the devices operate independently. You cannot simply multiply their efficiencies.

For example, suppose:

• Device A receives 100 J and produces 80 J of useful energy.
• Device B independently receives another 100 J and produces 60 J of useful energy.

The combined efficiency is:

\[ \eta_{\text{combined}}= \frac{80+60}{100+100} \]

\[ \boxed{\eta_{\text{combined}}=70\%} \]

Multiplying 80% by 60% would give 48%, which is incorrect for this arrangement.

The distinction is about the energy pathway, not simply the number of devices.

What if there is an energy loss between the devices?

Suppose a motor drives a pump through a belt.

The motor converts electrical energy into mechanical energy. The belt transfers that mechanical energy to the pump, but some energy is dissipated through friction. The pump then converts the remaining energy into useful output.

You now have three efficiency stages:

1. Motor efficiency.
2. Belt transmission efficiency.
3. Pump efficiency.

The overall efficiency becomes:

\[ \boxed{ \eta_{\text{total}}= \eta_{\text{motor}} \times\eta_{\text{belt}} \times\eta_{\text{pump}} } \]

For example, if the efficiencies are 80%, 90%, and 75%:

\[ \eta_{\text{total}}= 0.80\times0.90\times0.75 \]

\[ \boxed{\eta_{\text{total}}=54\%} \]

Don’t ignore an intermediate transfer stage just because the question describes it as a connection rather than a separate device.

04How would you measure the total efficiency experimentally?

Consider a motor driving a water pump.

The motor receives electrical energy and transfers mechanical energy through its rotating shaft. The pump uses that mechanical energy to raise water against gravity.

This provides a practical way to measure the efficiency of a two-device system.

What should the experiment show?

If the motor has efficiency \(\eta_m\) and the pump has efficiency \(\eta_p\), the model predicts:

\[ \eta_{\text{total}}=\eta_m\eta_p \]

This prediction assumes that all the useful mechanical energy leaving the motor reaches the pump.

To test the complete system, we don’t actually need to measure the mechanical energy at the shaft.

Instead, we can measure the electrical energy supplied to the motor and the gravitational potential energy gained by the water.

Measuring the input energy

For a stable DC supply, electrical power is:

\[ P_{\text{input}}=VI \]

where:

• \(V\) is the potential difference across the motor, in volts (V),
• \(I\) is the current through the motor, in amperes (A).

Over a time interval \(t\):

\[ E_{\text{input}}=VIt \]

Measuring the useful output energy

The water is raised through a vertical height \(h\).

Its gain in gravitational potential energy is:

\[ E_{\text{useful}}=mgh \]

where:

• \(m\) is the mass of water raised, in kilograms (kg),
• \(g\) is gravitational field strength, approximately \(9.8\text{ m/s}^2\),
• \(h\) is the vertical height gained, in metres (m).

Therefore, the measured total efficiency is:

\[ \boxed{ \eta_{\text{total}}=\frac{mgh}{VIt} } \]

This gives the efficiency as a decimal. Multiply by 100 to express it as a percentage.

Why would a first attempt at the experiment be unreliable?

Imagine recording only the current supplied to the motor and the volume of water pumped.

That isn’t enough.

Current alone doesn’t tell you the electrical input power. You also need the voltage.

Likewise, the volume of water alone doesn’t tell you its gravitational potential energy gain. You need its mass and the vertical height through which it moves.

A better experimental design measures:

• Voltage and current while the motor is operating.
• The time taken for the pumping interval.
• The mass of water collected.
• The vertical height gained by the water.

Keep the height and operating conditions constant, and take the input and output measurements over the same time interval.

There are also limitations to the model. The calculation above treats the gravitational potential energy gain as the useful output. Changes in the water’s kinetic energy or pressure energy are not included. We assume these are negligible or unchanged between the beginning and end of the measurement.

Worked measurement: finding the efficiency from experimental data

Suppose an illustrative experiment produces the following results:

Step 1: Calculate the electrical input energy.

\[ E_{\text{input}}=VIt \]

\[ =12.0\times0.50\times60 \]

\[ E_{\text{input}}=360\text{ J} \]

Step 2: Calculate the useful gravitational potential energy gained.

\[ E_{\text{useful}}=mgh \]

\[ =14.7\times9.8\times1.50 \]

\[ E_{\text{useful}}=216.09\text{ J} \]

Step 3: Calculate the total efficiency.

\[ \eta_{\text{total}}=\frac{216.09}{360} \]

\[ \eta_{\text{total}}=0.60025 \]

\[ \boxed{\eta_{\text{total}}=60.0\%} \]

The complete motor-pump system converts approximately 60% of its electrical input energy into the gravitational potential energy gained by the water.

Notice that we measured the complete system directly. We didn’t need to know the individual efficiencies of the motor and pump.

05Worked examples: applying the multiplication rule

The following examples demonstrate the two main calculation approaches you’ll encounter: working with energy amounts and working with power rates.

06A reliable method for HSC exam questions

When you see two devices connected in sequence, use this process.

There are four mistakes worth watching for.

Mistake 1: Adding the efficiencies.

An 80%-efficient device followed by a 75%-efficient device does not produce a system with 155% efficiency.

Mistake 2: Averaging the efficiencies.

The pump receives less energy than the motor, so the two percentages cannot simply be averaged.

Mistake 3: Using the original input for both devices.

The second device’s input is the first device’s useful output, not the original system input.

Mistake 4: Forgetting intermediate losses.

If a belt, cable, or other transmission stage loses energy, include its efficiency as another multiplication factor.

One final check: if your total efficiency is greater than either individual efficiency in an ordinary, steady, two-stage system, revisit your calculation.

07Questions and solutions

Try each question before reading its solution. The questions progress from straightforward efficiency calculations to problems involving additional energy transfers, experimental measurements, and changes in stored energy.

Question 1

An electric motor drives a water pump.

The motor receives electrical power of 500 W and has an efficiency of 80%. The pump has an efficiency of 65%.

Assume all the useful mechanical power leaving the motor enters the pump.

Calculate:

(a) The total efficiency of the system.

(b) The useful output power of the pump.

Solution 1

The total efficiency is 52%, and the pump produces 260 W of useful output power.

Step 1: Convert both efficiencies into decimals.

\[ \eta_{\text{motor}}=0.80 \]

\[ \eta_{\text{pump}}=0.65 \]

Step 2: Calculate the total efficiency.

\[ \eta_{\text{total}}=0.80\times0.65 \]

\[ \boxed{\eta_{\text{total}}=52\%} \]

Step 3: Calculate the motor’s useful output power.

\[ P_{\text{motor}}=500\times0.80 \]

\[ P_{\text{motor}}=400\text{ W} \]

Step 4: Calculate the pump’s useful output power.

\[ P_{\text{pump}}=400\times0.65 \]

\[ \boxed{P_{\text{pump}}=260\text{ W}} \]

The pump receives 400 W from the motor and converts 260 W into useful output power.

A useful check is:

\[ \frac{260}{500}\times100=52\% \]

This agrees with the calculated total efficiency.

Question 2

An electric motor drives a water pump continuously for 4.0 minutes.

The motor receives a constant electrical input power of 750 W. Its efficiency is 84%, while the pump has an efficiency of 70%.

Calculate:

(a) The total electrical energy supplied to the system.

(b) The mechanical output power of the motor.

(c) The useful output power of the pump.

(d) The total useful energy transferred to the water.

(e) The total energy dissipated by the two devices.

Solution 2

The system receives 180 kJ of electrical energy, transfers 630 W of mechanical power to the pump, and delivers 441 W of useful output power. The water gains 105.84 kJ, while 74.16 kJ is dissipated.

Step 1: Convert the operating time into seconds.

\[ t=4.0\times60=240\text{ s} \]

Step 2: Calculate the electrical input energy.

\[ E_{\text{input}}=Pt \]

\[ =750\times240 \]

\[ \boxed{E_{\text{input}}=180000\text{ J}=180\text{ kJ}} \]

Step 3: Calculate the motor’s mechanical output power.

\[ P_{\text{motor}}=750\times0.84 \]

\[ \boxed{P_{\text{motor}}=630\text{ W}} \]

Step 4: Calculate the pump’s useful output power.

\[ P_{\text{useful}}=630\times0.70 \]

\[ \boxed{P_{\text{useful}}=441\text{ W}} \]

Step 5: Calculate the useful energy transferred to the water.

\[ E_{\text{useful}}=Pt \]

\[ =441\times240 \]

\[ \boxed{E_{\text{useful}}=105840\text{ J}=105.84\text{ kJ}} \]

Step 6: Calculate the energy dissipated.

The motor dissipates the difference between its input and useful output:

\[ E_{\text{loss,motor}}=(750-630)\times240 \]

\[ =28800\text{ J}=28.8\text{ kJ} \]

The pump dissipates:

\[ E_{\text{loss,pump}}=(630-441)\times240 \]

\[ =45360\text{ J}=45.36\text{ kJ} \]

Therefore:

\[ E_{\text{loss,total}}=28.8+45.36 \]

\[ \boxed{E_{\text{loss,total}}=74.16\text{ kJ}} \]

Check the energy balance:

\[ 180-105.84=74.16\text{ kJ} \]

The energy balance is satisfied.

The trap here is calculating the pump’s loss as 30% of the original 750 W. Its input is only 630 W, so its 30% loss must be calculated from that amount.

Question 3

A 1.5 kW electric motor drives a water pump through a belt.

The motor has an efficiency of 82%. The belt transmits 90% of the mechanical power it receives to the pump. The pump has an efficiency of 70%.

Calculate:

(a) The useful output power of the pump.

(b) The total efficiency of the system.

(c) The power dissipated at each stage.

Solution 3

The pump produces 774.9 W of useful output power. The total system efficiency is 51.66%.

The important detail is that the belt introduces an additional energy loss. It must be included as a separate stage.

Step 1: Convert the input power into watts.

\[ P_{\text{input}}=1.5\text{ kW}=1500\text{ W} \]

Step 2: Calculate the motor’s mechanical output power.

\[ P_{\text{motor}}=1500\times0.82 \]

\[ P_{\text{motor}}=1230\text{ W} \]

Step 3: Calculate the power transmitted by the belt.

\[ P_{\text{belt}}=1230\times0.90 \]

\[ P_{\text{belt}}=1107\text{ W} \]

This is the power received by the pump.

Step 4: Calculate the pump’s useful output power.

\[ P_{\text{useful}}=1107\times0.70 \]

\[ \boxed{P_{\text{useful}}=774.9\text{ W}} \]

Step 5: Calculate the total system efficiency.

\[ \eta_{\text{total}}= 0.82\times0.90\times0.70 \]

\[ \eta_{\text{total}}=0.5166 \]

\[ \boxed{\eta_{\text{total}}=51.66\%} \]

Step 6: Calculate the power dissipated at each stage.

Motor:

\[ P_{\text{loss,motor}}=1500-1230 \]

\[ =270\text{ W} \]

Belt:

\[ P_{\text{loss,belt}}=1230-1107 \]

\[ =123\text{ W} \]

Pump:

\[ P_{\text{loss,pump}}=1107-774.9 \]

\[ =332.1\text{ W} \]

The total dissipated power is:

\[ P_{\text{loss,total}}=270+123+332.1 \]

\[ \boxed{P_{\text{loss,total}}=725.1\text{ W}} \]

Check:

\[ 1500-774.9=725.1\text{ W} \]

The result is consistent.

A common mistake would be to multiply only the motor and pump efficiencies, obtaining 57.4%. That ignores the belt’s 10% loss. Every stage that reduces the energy transferred towards the final useful output must be accounted for.

Question 4

A student conducts an experiment using a DC motor connected to a water pump.

During a 90-second test:

• The potential difference across the motor is 18.0 V.
• The current through the motor is 0.40 A.
• The pump raises 30.0 kg of water through a vertical height of 1.20 m.

Assume the water has negligible net change in kinetic energy and that the useful output is its gain in gravitational potential energy.

Calculate:

(a) The electrical input power.

(b) The electrical input energy.

(c) The useful output energy gained by the water.

(d) The total efficiency of the system.

(e) The useful output power of the pump.

A second student suggests calculating efficiency by comparing the mass of water raised per second directly with the electrical input power. Explain why this approach is incorrect.

Take \(g=9.8\text{ m/s}^2\).

Solution 4

The system has an overall efficiency of 54.4%, and the useful output power is 3.92 W.

The key is to convert both the electrical input and the gravitational potential energy gain into comparable quantities before calculating efficiency.

Step 1: Calculate the electrical input power.

For a DC motor:

\[ P_{\text{input}}=VI \]

\[ =18.0\times0.40 \]

\[ \boxed{P_{\text{input}}=7.20\text{ W}} \]

Step 2: Calculate the electrical input energy.

\[ E_{\text{input}}=Pt \]

\[ =7.20\times90 \]

\[ \boxed{E_{\text{input}}=648\text{ J}} \]

Step 3: Calculate the gravitational potential energy gained by the water.

\[ E_{\text{useful}}=mgh \]

\[ =30.0\times9.8\times1.20 \]

\[ \boxed{E_{\text{useful}}=352.8\text{ J}} \]

Step 4: Calculate the total efficiency.

\[ \eta_{\text{total}}= \frac{352.8}{648} \]

\[ \eta_{\text{total}}=0.5444 \]

\[ \boxed{\eta_{\text{total}}=54.4\%} \]

Step 5: Calculate the useful output power.

\[ P_{\text{useful}}=\frac{E_{\text{useful}}}{t} \]

\[ =\frac{352.8}{90} \]

\[ \boxed{P_{\text{useful}}=3.92\text{ W}} \]

Step 6: Explain why the second student’s method is incorrect.

The mass of water raised per second is:

\[ \frac{m}{t}=\frac{30.0}{90} \]

\[ =0.333\text{ kg/s} \]

This is a mass flow rate, not a power.

The units are different:

• Mass flow rate: kg/s.
• Power: J/s or W.

To convert the mass flow rate into useful gravitational power, multiply by gravitational field strength and height:

\[ P_{\text{useful}}=\frac{m}{t}gh \]

\[ =0.333\times9.8\times1.20 \]

\[ \boxed{P_{\text{useful}}=3.92\text{ W}} \]

The second student confused the rate at which mass is raised with the rate at which energy is transferred.

This distinction is important in questions involving pumps, cranes, lifts, and other systems that move objects against gravity.

For more practice with this relationship, see How to Calculate Lifting Power at Constant Speed.

Question 5

This final problem tests a subtle limitation of the multiplication rule.

Two devices, A and B, are connected through a flywheel that can store mechanical energy.

During a test:

• Device A receives 1000 J of electrical energy.
• Device A has an efficiency of 80%.
• The flywheel initially contains no stored energy.
• Device B has an efficiency of 75%.
• By the end of the test, device B has received 600 J from the flywheel.
• The remaining energy is still stored in the flywheel.

Assume there are no energy losses in the flywheel itself, no external energy inputs, and no other energy transfers.

Answer the following:

(a) Calculate the useful energy delivered by device B during the test.

(b) Calculate the overall efficiency if useful output is defined as the energy delivered by device B during the test only.

(c) Suppose the stored energy in the flywheel is also considered a useful final output. Calculate the overall efficiency using this different definition.

(d) A student argues that the total efficiency must be \(80\%\times75\%=60\%\), regardless of the amount of energy stored in the flywheel. Explain precisely why this argument is not correct for the end-of-test measurements.

(e) If device B is allowed to receive all the remaining energy in the flywheel after the test, and its efficiency stays at 75%, what is the final overall efficiency based on the energy delivered by B?

Solution 5

The useful energy delivered by B during the test is 450 J. The overall efficiency is 45% if only B’s output during the test counts, 65% if the remaining stored energy also counts as useful output, and 60% after the flywheel has completely discharged through B.

The key issue is that the two devices have not transferred all their energy through the complete chain by the end of the test.

Step 1: Calculate the useful mechanical energy transferred by device A.

\[ E_A=1000\times0.80 \]

\[ \boxed{E_A=800\text{ J}} \]

Therefore, the flywheel receives 800 J.

Step 2: Calculate the useful energy delivered by device B during the test.

Device B receives only 600 J during the test.

Its useful output is:

\[ E_B=600\times0.75 \]

\[ \boxed{E_B=450\text{ J}} \]

This answers part (a).

Step 3: Calculate the overall efficiency using only B’s output.

The total input is 1000 J, and the useful output is 450 J.

\[ \eta_{\text{total}}= \frac{450}{1000} \]

\[ \boxed{\eta_{\text{total}}=45\%} \]

This answers part (b).

Step 4: Include the remaining stored energy as useful output.

The flywheel still contains 200 J.

If this stored energy is also considered a useful final output, the total useful energy is:

\[ E_{\text{useful,total}}=450+200 \]

\[ E_{\text{useful,total}}=650\text{ J} \]

Therefore:

\[ \eta_{\text{total}}=\frac{650}{1000} \]

\[ \boxed{\eta_{\text{total}}=65\%} \]

This answers part (c).

Notice that the 200 J in the flywheel is not energy that has disappeared. It is still stored in the system.

Step 5: Explain why multiplying the two efficiencies gives a different answer.

The student’s calculation is:

\[ \eta_{\text{total}}=0.80\times0.75 \]

\[ \eta_{\text{total}}=60\% \]

The multiplication itself is mathematically correct. The problem is the assumption behind it.

Multiplying 80% by 75% assumes that all 800 J transferred by device A eventually enters device B.

However, B has received only 600 J by the end of the test.

The other 200 J remains stored in the flywheel.

The simple multiplication rule therefore does not describe the energy delivered by B during this particular measurement interval.

Check the complete energy balance:

All the original input energy is accounted for.

This is the important distinction: stored energy is not automatically a loss, but it is not automatically useful output from the second device either. That depends on what the question defines as useful.

Step 6: Calculate the efficiency after the flywheel completely discharges through device B.

Device B eventually receives the remaining 200 J.

Its additional useful output is:

\[ E_{\text{additional}}=200\times0.75 \]

\[ E_{\text{additional}}=150\text{ J} \]

The total useful energy delivered by B becomes:

\[ E_{\text{B,total}}=450+150 \]

\[ E_{\text{B,total}}=600\text{ J} \]

The overall efficiency is now:

\[ \eta_{\text{total}}=\frac{600}{1000} \]

\[ \boxed{\eta_{\text{total}}=60\%} \]

This answers part (e).

The original multiplication rule now works because all 800 J transferred by A have eventually passed through B.

The lesson is that total efficiency depends on the complete energy pathway and the time interval being considered. When energy is stored between devices, you must account for the change in stored energy before applying the usual multiplication rule.

Once you can track energy through multiple devices, the next useful skill is calculating how quickly that energy is transferred. This connects directly to mechanical power, lifting objects at constant speed, and analysing systems where energy is continuously converted rather than simply stored.