11.2.10 Heat Engines

Efficiency of a Heat Engine

The goal of a heat engine is to transfer thermal energy into useful mechanical work as efficiently as possible
The efficiency of a heat engine can, therefore, be calculated using

efficiency = $\frac{u s e f u l w o r k o u t p u t}{t o t a l e n e r g y i n p u t}$

efficiency = $\frac{W}{Q_{H}} = \frac{(Q_{H} - Q_{C})}{Q_{H}}$

Where:
- W = useful work output (J)
- Q_H = energy transferred from the source (J)
- Q_C = energy transferred to the sink (J)

11-2-9-heat-engine

Source-sink diagram for a heat engine

Since the efficiency of a heat engine can never be 0 (otherwise there would no work!) this means no heat engine can completely convert heat into work

Worked example

Which is a correct statement about an ideal heat engine?

A The efficiency is 75% when the kelvin temperature of the hot source is four times the kelvin temperature of the cold sink

B The maximum efficiency depends on the p-V cycle fo the engine

C The efficiency is decreased when the kelvin temperature of the hot source and the cold sink are decreased by equal amounts

D The efficiency is 25% when the kelvin temperature of the hot source is four times the kelvin temperature of the cold sink

Answer: A

If the hot source is four times the kelvin temperature of the cold sink, then $Q_{H} = 4 Q_{C}$
In the efficiency equation, this is

$\frac{Q_{H} - Q_{C}}{Q_{H}} = \frac{4 Q_{C} - Q_{C}}{4 Q_{C}} = \frac{3 Q_{C}}{4 Q_{C}} = \frac{3}{4} = 0.75$

This efficiency is 75 %

Exam Tip

This equation is given on your data sheet. Make sure all the variables are in the same units i.e. J or kJ. You must convert these into the same units before you do your calculation.

If the efficiency is asked for as a percentage instead of a decimal, remember to × 100

Maximum Theoretical Efficiency

As the efficiency of a thermodynamic system increases, the difference between the temperatures of the source and sink increases
The maximum theoretical efficiency of a heat engine is:

Maximum theoretical efficiency = $\frac{T_{H} - T_{C}}{T_{H}}$

Where:
- $T_{C}$ = temperature in the sink (cold reservoir) (K)
- $T_{H}$ = temperature in the source (hot reservoir) (K)
This equation can be used if an ideal gas is used as a substance for the engine
Therefore, to make an engine as efficient as possible, the source temperature must be as high as possible, and the sink temperature as low as possible
The maximum theoretical efficiency is 100% only if the sink temperature is at 0 K

Worked example

An engineer designs a heat engine that has an inlet temperature of 500 K and an outlet temperature of 300 K. The engineer claims that 100 kJ of thermal energy flows out of the source and 25 kJ of thermal energy flows into the sink.

Determine, with reference to the second law of thermodynamics, whether this engine is thermodynamically possible.

Answer:

Step 1: Determine the efficiency of the proposed engine

The efficiency of this engine would be

$e f f i c i e n c y = \frac{W}{Q_{H}} = \frac{Q_{H} - Q_{C}}{Q_{H}}$

Where:
- Heat transferred in, $Q_{H}$ = 100 kJ
- Heat transferred out, $Q_{C}$ = 25 kJ

Efficiency = $\frac{100 - 25}{100}$ = 0.75 = 75%

Step 2: Determine the maximum theoretical efficiency of the proposed engine

A Carnot engine operating between the same temperatures would have an efficiency of

Maximum theoretical efficiency = $\frac{T_{H} - T_{C}}{T_{H}}$

Where:
- Inlet temperature, $T_{H}$ = 500 K
- Outlet temperature, $T_{C}$ = 300 K

Maximum theoretical efficiency = $\frac{500 - 300}{500}$ = 0.4 = 40%

Step 3: Discuss the proposed engine in relation to the second law

The second law of thermodynamics states that it is impossible for heat to flow from a cooler body to a hotter body without performing work
This law sets an upper limit on the maximum possible efficiency of the transfer of thermal energy to mechanical energy in a heat engine
The maximum possible efficiency of the proposed engine is 40%, but the engineer is proposing an efficiency of 75% i.e. an efficiency greater than the efficiency of the engine
This violates the second law, hence the proposed engine is impossible

Exam Tip

This equation is valid for all idealised reversible engines, irrespective of the particular cycle and the particular working substance. You may be given unfamiliar cycles in the exam (such as the Carnot cycle) to apply this to.

AQA A Level Physics

Revision Notes

Efficiency of a Heat Engine

Worked example

Exam Tip

Maximum Theoretical Efficiency

Worked example

Exam Tip

You've read 0 of your 0 free revision notes

Get unlimited access

Join the 100,000+ Students that ❤️ Save My Exams

Author: Ashika