Gas power cycles: practice problems with step-by-step solutions

Free Gas power cycles practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: Otto cycle, Diesel cycle, Brayton cycle.

Free practice problemEasy
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An ideal Otto cycle has a compression ratio of 10.5. At the start of compression the air is at 295 K, and the highest temperature in the cycle is 1800 K. Use the cold-air-standard assumptions: air is an ideal gas with constant specific heats, k = 1.4, cp = 1.005 kJ/(kg·K) and cv = 0.718 kJ/(kg·K). Find the net work output w per kilogram of air, in kJ/kg.

r = 10.5T1 = 295 KT3 = 1800 Kk = 1.4cv = 0.718 kJ/(kg·K)
kJ/kg
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What you practise

  1. EasyOtto cycle
  2. MediumDiesel cycle
  3. HardBrayton cycle

Key formulas

Isentropic, volume ratio
T2/T1 = rk−1
Isentropic, pressure ratio
T2/T1 = rp(k−1)/k
Otto efficiency
η = 1 − 1/rk−1
Diesel efficiency
η = 1 − (1/rk−1)·(rck − 1)/(k(rc − 1))
Brayton efficiency
η = 1 − 1/rp(k−1)/k
Heat added
Otto: cv(T3 − T2) Diesel, Brayton: cp(T3 − T2)
Brayton net work
w = cp[(T3 − T4) − (T2 − T1)]
Isentropic efficiencies
wc = wc,s/ηc, wt = ηt·wt,s
Cold air
k = 1.4, cp = 1.005, cv = 0.718 kJ/(kg·K)

Learn Gas power cycles from scratch

Before this: Heat engines & refrigerators, Ideal gas law

1What a gas power cycle is

Car engines and gas turbines are heat engines that use air as their working gas. The air is squeezed, heated by burning fuel, expands while it pushes on a piston or turbine, and is then thrown out.

Real engines are complicated. To calculate them, engineers use the air-standard model: a fixed amount of air goes round a closed cycle of four simple steps, and the burning of fuel is replaced by adding heat.

With the cold-air-standard assumptions the air is an ideal gas with constant properties: k = cp/cv = 1.4, cp = 1.005 kJ/(kg·K) and cv = 0.718 kJ/(kg·K).

You will meet three cycles:

  • Otto: the petrol engine. Heat is added at constant volume, like a spark that burns fuel almost instantly.
  • Diesel: the diesel engine. Heat is added at constant pressure while the piston moves.
  • Brayton: the gas turbine. Air flows through a compressor, a combustion chamber and a turbine.

The most important number of a piston engine is its compression ratio r = v1/v2: the largest volume divided by the smallest.

Self-check A cylinder holds 0.5 L when the piston is at the bottom and 0.05 L when it is at the top. What is the compression ratio?
  1. 0.1
  2. 0.45
  3. 10
  4. 9

Answer: 10. r = largest volume / smallest volume = 0.5/0.05 = 10. Dividing the other way round gives 0.1, and the difference of the volumes (0.45 L) is not a ratio.

2Isentropic steps and the Otto cycle

In the ideal cycle, compression and expansion happen without friction and without heat exchange. Such a step is called isentropic. For an ideal gas the temperature then follows the volume ratio:

T2 / T1 = rk−1 = r0.4

Squeezing air to one eighth of its volume heats it from 300 K to about 689 K. This is why a diesel engine does not need a spark.

The Otto cycle has four steps, shown on a pressure–volume diagram:

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  1. 1 → 2: isentropic compression. T rises by the factor r0.4.
  2. 2 → 3: heat added at constant volume: qin = cv·(T3 − T2).
  3. 3 → 4: isentropic expansion. T falls by the same factor r0.4.
  4. 4 → 1: heat rejected at constant volume: qout = cv·(T4 − T1).

The net work per kilogram is w = qin − qout, and the efficiency depends only on the compression ratio:

ηOtto = 1 − 1 / rk−1
Self-check What is the thermal efficiency of an ideal Otto cycle with a compression ratio of 10? (k = 1.4)
  1. 39.8%
  2. 60.2%
  3. 90%
  4. 96.0%

Answer: 60.2%. 100.4 = 2.512, so η = 1 − 1/2.512 = 60.2%. Leaving out the exponent gives 1 − 1/10 = 90%; 39.8% is 1/r0.4, the part that is lost.

3Diesel and Brayton cycles

The Diesel cycle adds its heat at constant pressure. The volume grows during heating by the cutoff ratio rc = v3/v2. At constant pressure the temperature grows by the same factor:

T3 = rc · T2 qin = cp · (T3 − T2)
ηDiesel = 1 − (1 / rk−1) · (rck − 1) / (k · (rc − 1))

The Brayton cycle of a gas turbine is described by the pressure ratio rp = p2/p1 instead of a volume ratio. The isentropic rule then has a different exponent:

T2 / T1 = rp(k−1)/k = rp0.2857
ηBrayton = 1 − 1 / rp(k−1)/k

The compressor uses work and the turbine delivers work. The net work per kilogram is what is left:

wnet = cp · [(T3 − T4) − (T2 − T1)]
Self-check What is the efficiency of an ideal Brayton cycle with a pressure ratio of 8? (k = 1.4)
  1. 44.8%
  2. 55.2%
  3. 56.5%
  4. 87.5%

Answer: 44.8%. 80.2857 = 1.811, so η = 1 − 1/1.811 = 44.8%. Using the volume exponent 0.4 (the Otto formula) gives 56.5%, which is wrong for a pressure ratio.

4The method step by step

  1. Identify the cycle (Otto, Diesel or Brayton) and number the four states 1 to 4, starting before compression.
  2. Convert every temperature to kelvin.
  3. Compression 1 → 2: T2 = T1·r0.4 for a volume ratio, or T1·rp0.2857 for a pressure ratio.
  4. Heat addition 2 → 3: use cv for Otto (constant volume), cp for Diesel and Brayton (constant pressure).
  5. Expansion 3 → 4: divide T3 by the same kind of factor (for Diesel use v4/v3 = r/rc).
  6. Find qout, then w = qin − qout and η = w/qin. Compare η with the formula for that cycle as a check.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

An ideal Otto cycle has a compression ratio of 8. At the start of compression the air is at 300 K, and the highest temperature in the cycle is 1800 K. Use the cold-air-standard assumptions (k = 1.4, cv = 0.718 kJ/(kg·K)). Find the efficiency and the net work per kilogram of air.

1

Draw the cycle

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State 1 is 300 K. The highest temperature is at state 3, so T3 = 1800 K. We need T2 and T4.

2

Compression 1 → 2

r0.4 = 80.4 = 2.297
T2 = 300 · 2.297 = 689.2 K
3

Heat added 2 → 3

Heat goes in at constant volume, so use cv:

qin = 0.718 · (1800 − 689.2) = 797.5 kJ/kg
4

Expansion and heat rejected

T4 = 1800 / 2.297 = 783.5 K
qout = 0.718 · (783.5 − 300) = 347.1 kJ/kg
5

Net work and efficiency

w = 797.5 − 347.1 = 450.4 kJ/kg
η = 450.4 / 797.5 = 0.565 = 56.5%
6

Check the answer

The Otto formula gives η = 1 − 1/2.297 = 56.5%. That is the same, so the temperatures are right.

Answer

η ≈ 56.5% and w ≈ 450 kJ/kg.

6Common mistakes

  • Using °C in the isentropic relation. T2/T1 = r0.4 only works in kelvin.
  • Using the exponent k − 1 = 0.4 with a pressure ratio. A pressure ratio needs (k − 1)/k = 0.2857.
  • Using cp for the Otto cycle or cv for Diesel and Brayton. Constant volume goes with cv, constant pressure with cp.
  • Forgetting the compressor work in a Brayton cycle: the net work is turbine work minus compressor work.
  • Using the exponent k instead of k − 1.
Self-check Air at 27 °C is compressed isentropically with a compression ratio of 8 (k = 1.4). What is the temperature after compression?
  1. 2401 K
  2. 5514 K
  3. 62.0 K
  4. 689.6 K

Answer: 689.6 K. T1 = 300.15 K, so T2 = 300.15 · 80.4 = 689.6 K. Using 27 gives 62.0, leaving out the exponent gives 2401 K, and using the exponent 1.4 gives 5514 K.

7How exam questions look

  • Easy: the Otto cycle. Find its efficiency from r, the temperature T2 after compression, or the net work per kilogram when the highest temperature is given.
  • Medium: the Diesel cycle with a compression ratio and a cutoff ratio. Find T3 at the end of heat addition, the heat added qin, or the efficiency.
  • Hard: the Brayton cycle of a gas turbine. Find the efficiency from the pressure ratio, or the net work per kilogram from the compressor and turbine inlet temperatures.

Every question restates the cold-air-standard values. You only need k = 1.4 and either cv or cp.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
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An ideal Otto cycle has a compression ratio of 6. At the start of compression the air is at 305 K, and the highest temperature in the cycle is 2000 K. Use the cold-air-standard assumptions: air is an ideal gas with constant specific heats, k = 1.4, cp = 1.005 kJ/(kg·K) and cv = 0.718 kJ/(kg·K). Find the net work output w per kilogram of air, in kJ/kg.

r = 6T1 = 305 KT3 = 2000 Kk = 1.4cv = 0.718 kJ/(kg·K)

Find w (kJ/kg)

Show hints
  1. Find T₂ after the isentropic compression.
  2. qin = cv·(T₃ − T₂), η = 1 − 1/rk−1.
  3. T₂ = 624.5 K.
Show the answer and the step-by-step solution

Otto cycle (spark-ignition engine): 1→2 isentropic compression, 2→3 heat addition at constant volume, 3→4 isentropic expansion, 4→1 heat rejection at constant volume. With compression ratio r = v₁/v₂: T₂ = T₁·rk−1 and η = 1 − 1/rk−1. Heat added: qin = cv·(T₃ − T₂); net work w = η·qin.

End of compression

T2 = 305·60.4 = 624.5 K

Heat added at constant volume

qin = 0.718·(2000 − 624.5) = 987.6 kJ/kg

Net work

η = 1 − 1/60.4 = 0.5116
w = η·qin = 0.5116·987.6 = 505.3 kJ/kg

Answer

w = 505.3 kJ/kg
Example 2Medium
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An ideal Diesel cycle has a compression ratio r = 16 and a cutoff ratio rc = v₃/v₂ = 2.2. At the start of compression the air is at 310 K. Use the cold-air-standard assumptions: air is an ideal gas with constant specific heats, k = 1.4, cp = 1.005 kJ/(kg·K) and cv = 0.718 kJ/(kg·K). Find the heat added qin per kilogram of air, in kJ/kg.

r = 16rc = 2.2T1 = 310 Kk = 1.4cp = 1.005 kJ/(kg·K)

Find qin (kJ/kg)

Show hints
  1. Find T₂ (isentropic compression) and T₃ = rc·T₂.
  2. qin = cp·(T₃ − T₂).
  3. T₂ = 939.7 K, T₃ = 2067 K.
Show the answer and the step-by-step solution

Diesel cycle (compression-ignition engine): 1→2 isentropic compression, 2→3 heat addition at constant pressure, 3→4 isentropic expansion, 4→1 heat rejection at constant volume. With compression ratio r = v₁/v₂ and cutoff ratio rc = v₃/v₂: T₂ = T₁·rk−1, T₃ = rc·T₂, qin = cp·(T₃ − T₂) and η = 1 − (1/rk−1)·(rck − 1)/(k·(rc − 1)).

Temperatures

T2 = 310·160.4 = 939.7 K
T3 = 2.2·939.7 = 2067 K

Heat added at constant pressure

qin = 1.005·(2067 − 939.7) = 1133 kJ/kg

Answer

qin = 1133 kJ/kg
Example 3Hard
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An ideal Brayton cycle operates with a compressor pressure ratio of 20. Use the cold-air-standard assumptions: air is an ideal gas with constant specific heats, k = 1.4, cp = 1.005 kJ/(kg·K) and cv = 0.718 kJ/(kg·K). Find its thermal efficiency η in percent.

rp = 20k = 1.4

Find η (%)

Show hints
  1. The ideal Brayton efficiency depends only on the pressure ratio and k.
  2. η = 1 − 1/rp(k−1)/k.
  3. (k − 1)/k = 0.2857, rp0.2857 = 2.354.
Show the answer and the step-by-step solution

Brayton cycle (gas turbine): 1→2 isentropic compression, 2→3 heat addition at constant pressure, 3→4 isentropic expansion in the turbine, 4→1 heat rejection at constant pressure. With pressure ratio rp = P₂/P₁: T₂ = T₁·rp(k−1)/k, T₄ = T₃/rp(k−1)/k and η = 1 − 1/rp(k−1)/k. Net work w = cp·[(T₃ − T₄) − (T₂ − T₁)]. With isentropic efficiencies, the compressor needs ws/ηc and the turbine gives ηt·ws.

Thermal efficiency

η = 1 − 1/200.2857 = 1 − 1/2.354 = 57.51%

Answer

η = 57.51%

Questions and answers

What is Gas power cycles?

Car engines and gas turbines are heat engines that use air as their working gas. The air is squeezed, heated by burning fuel, expands while it pushes on a piston or turbine, and is then thrown out.

Real engines are complicated. To calculate them, engineers use the air-standard model: a fixed amount of air goes round a closed cycle of four simple steps, and the burning of fuel is replaced by adding heat.

How do I solve Gas power cycles problems step by step?

Otto cycle (spark-ignition engine): 1→2 isentropic compression, 2→3 heat addition at constant volume, 3→4 isentropic expansion, 4→1 heat rejection at constant volume. With compression ratio r = v₁/v₂: T₂ = T₁·rk−1 and η = 1 − 1/rk−1. Heat added: qin = cv·(T₃ − T₂); net work w = η·qin.

What are common mistakes in Gas power cycles?
  • That is the heat added. Only the fraction η of it becomes net work.
  • In the Otto cycle heat is added at constant volume: use cv, not cp.
  • Heat is added from state 2 to state 3. Find T2 after compression first.
What do the Gas power cycles exercises cover?
  • Easy: Otto cycle
  • Medium: Diesel cycle
  • Hard: Brayton cycle

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Heat engines & refrigeratorsIdeal gas law

Calculators

Otto and Brayton cycle efficiency calculator

Courses with this topic

Thermodynamics 1FE Mechanical exam practice

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Printable worksheet

Gas power cycles worksheet with answers
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