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Real engine cycles

The Otto, Diesel, Brayton and Rankine cycles: their efficiencies from the compression or pressure ratio and from steam tables, and why each falls short of Carnot.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to follow the temperatures around an Otto or Brayton cycle, find its heat, work and efficiency, compute a Rankine plant's efficiency from steam-table enthalpies, and explain why each cycle falls short of the Carnot limit.

2. What you already have

You know the Carnot limit $1 - T_c/T_h$, the adiabatic relations $TV^{\gamma - 1}$ and $PV^{\gamma}$ constant, and that at constant pressure the heat is the change in enthalpy. This lesson applies all three to the cycles that actually power cars, jets and the electrical grid.

3. Words for this lesson

4. Real engines follow cycles with their own efficiencies

A gasoline engine runs, ideally, the Otto cycle. Air and fuel are compressed adiabatically from volume $V_1$ to $V_2 = V_1/r$; a spark burns the fuel so fast that heat $Q_h$ enters at constant volume; the hot gas expands adiabatically back to $V_1$; and the exhaust rejects heat $Q_c$ at constant volume.

Both adiabats span the same volume ratio $r$, so each multiplies or divides temperatures by the same factor, $r^{\gamma - 1}$: $T_2 = T_1r^{\gamma - 1}$ and $T_3 = T_4r^{\gamma - 1}$. The heats are $Q_h = C_V(T_3 - T_2)$ and $Q_c = C_V(T_4 - T_1)$, and substituting gives

$$e_{\text{Otto}} = 1 - \frac{1}{r^{\gamma - 1}}.$$

With $r = 8$ and air's $\gamma = 1.4$, $e = 1 - 1/8^{0.4} = 0.56$. The efficiency depends on the compression ratio and the gas only — not on the fuel or the peak temperature.

A gas turbine runs the Brayton cycle: air is compressed by a pressure ratio $r_p$, fuel burns at constant pressure in a combustor, and the hot gas expands through a turbine. The same argument with $T \propto P^{(\gamma - 1)/\gamma}$ along the adiabats gives

$$e_{\text{Brayton}} = 1 - \frac{1}{r_p^{(\gamma - 1)/\gamma}}.$$

A steam power plant runs the Rankine cycle, whose working fluid changes phase, so it is computed not from $\gamma$ but from the enthalpies in steam tables: the turbine's work is the drop in enthalpy across it, the boiler's heat the rise in enthalpy through it.

5. The method, step by step, and how to check it

1. Label the corners of the cycle 1 to 4 and write what each process holds fixed.
2. Follow the temperatures. Across an adiabat of volume ratio $r$, multiply or divide by $r^{\gamma - 1}$; across one of pressure ratio $r_p$, by $r_p^{(\gamma - 1)/\gamma}$.
3. Find the heats from the processes where heat flows: $C_V\Delta T$ at constant volume, $C_P\Delta T$ at constant pressure, $\Delta h$ from steam tables for a Rankine cycle.
4. Find the work as heat in minus heat out, and the efficiency as their ratio. For Otto and Brayton, check the result against the closed formula.

Checks. The efficiency must be less than the Carnot efficiency between the cycle's hottest and coldest temperatures, often much less, because heat enters and leaves over a range of temperatures. It must rise with the compression or pressure ratio. For a Rankine cycle, the pump work should be a percent or less of the turbine work; if it is not, liquid and vapor enthalpies have been mixed up. And real engines reach roughly two thirds to three quarters of their ideal cycle efficiency: a gasoline engine with $r = 10$ has an ideal $0.60$ but manages about $0.35$ on the road.

6. Why gasoline engines stop at a compression ratio of about ten

If efficiency rises with $r$, why not build a gasoline engine with $r = 20$? Because the adiabat heats the fuel-air mixture as it compresses it: at $r = 10$, air at $300$ K reaches $300 \times 10^{0.4} = 754$ K before the spark, and at $r = 20$ it would reach $994$ K. Gasoline mixed with air then ignites on its own before the piston reaches the top, a violent uncontrolled combustion called knock that can destroy an engine.

Octane rating measures resistance to knock, which is why high-compression sports cars require premium fuel. A diesel engine turns the same fact into its design: it compresses air alone, to ratios of $15$ to $22$, and injects the fuel at the top of the stroke, where the air is already hot enough to ignite it. The higher compression ratio is why diesel engines are more efficient, and why long-haul trucks and ships use them. The Diesel cycle's ideal efficiency is a little below the Otto cycle's at the same $r$, because its heat is added while the piston is already moving, but its much higher allowed $r$ more than makes up for that.

7. Steam tables and the Rankine cycle

Water is an unusual working fluid: it boils and condenses within the cycle, so no formula in $\gamma$ describes it. Engineers instead read its enthalpy $h$, the energy per kilogram including the $PV$ work, from steam tables compiled from measurement. Superheated steam at $3.0$ MPa and $400$ °C has $h_1 = 3231.7$ kJ/kg. Expanding ideally through a turbine to the condenser at $10$ kPa, where water boils at $45.8$ °C, the entropy stays fixed, and the tables give $h_2 = 2192.7$ kJ/kg for the resulting mix of steam and droplets. Condensed to liquid it has $h_3 = 191.8$ kJ/kg, and pumping it back to $3.0$ MPa costs only about $3$ kJ/kg.

The turbine's work is $h_1 - h_2 = 1039$ kJ/kg, the net work $1036$ kJ/kg, and the heat added $h_1 - h_4 = 3036.9$ kJ/kg, for an efficiency of $0.341$. The Carnot limit between $673$ K and $319$ K is $0.526$. Real plants raise their efficiency by superheating to higher temperatures, reheating the steam between turbine stages, and preheating the feedwater with steam bled from the turbine — each a way of adding heat at higher average temperature, closer to Carnot's ideal of adding it all at the top.

8. Combining cycles

Because the Brayton cycle rejects heat at a high temperature — the exhaust of a gas turbine leaves at $800$ to $900$ K — its waste heat can drive a second cycle. A combined-cycle plant does exactly that: the turbine's exhaust passes through a heat-recovery steam generator that runs a Rankine cycle. If the gas turbine converts a fraction $e_1$ of the fuel's heat and the steam cycle converts a fraction $e_2$ of what the turbine rejects, the overall efficiency is $e_1 + (1 - e_1)e_2$. With $e_1 = 0.40$ and $e_2 = 0.35$, that is $0.40 + 0.60 \times 0.35 = 0.61$, higher than either cycle alone. The combination adds heat at the gas turbine's high temperature and rejects it at the steam cycle's low one, stretching the effective temperature range toward Carnot's.

9. In the world: why a hybrid car uses a different cycle

The engine in a Toyota Prius does not run a pure Otto cycle. It runs the Atkinson cycle, in which the intake valve stays open for part of the compression stroke, so the effective compression ratio is about $10$ while the expansion ratio is $13$. The power stroke then continues past the point where an Otto engine would open its exhaust valve, extracting more work from the hot gas.

The ideal efficiencies show the gain. An Otto cycle at $r = 10$ has $e = 1 - 1/10^{0.4} = 0.60$. Expanding further, to $13$, lowers the exhaust temperature and the rejected heat; the ideal efficiency rises by a few percent, and in practice the Prius engine reaches a peak thermal efficiency of about $40$ percent, among the highest of any production gasoline engine. The cost is lower power for its size, since less air is trapped in each stroke, which the hybrid's electric motor makes up when accelerating. It is a direct application of the cycle analysis: a longer expansion than compression means heat leaves at a lower average temperature.

10. In the world: the jet engine as a Brayton cycle

A jet engine is a gas turbine whose turbine drives only the compressor and a fan, with the rest of the gas's energy used for thrust. The GE90 engines on a Boeing 777 have an overall pressure ratio of about $40$. The ideal Brayton efficiency is then $1 - 1/40^{0.2857} = 1 - 1/2.87 = 0.65$.

That thermal efficiency is only part of the story: the engine must also turn the gas's energy into thrust efficiently, which is why modern engines move a large mass of air slowly through a huge fan rather than a small mass fast. But the pressure ratio is the heart of it. Engines of the 1960s had pressure ratios near $15$ and ideal efficiencies near $0.54$; the climb to $40$ and beyond, made possible by compressor aerodynamics and by turbine blades that survive gas hotter than their own melting point thanks to internal cooling, is why a modern airliner burns roughly half the fuel per passenger-mile of one from that era.

11. Peak temperature does not set the Otto efficiency

Carnot's formula makes it tempting to think that any engine becomes more efficient when its fuel burns hotter. For the ideal Otto cycle it does not: $e = 1 - 1/r^{\gamma - 1}$ contains no temperature at all. A richer mixture burns hotter and does more work per stroke, but it also rejects proportionally more heat, and the fraction converted is unchanged. What raises the efficiency is the compression ratio, which is why engine designers fight knock so hard.

The Carnot formula applies to heat that enters at a single temperature. In every real cycle heat enters over a range — during combustion in an Otto engine, while water warms and then boils in a Rankine plant — and leaves over a range too. The useful way to think about it is that a cycle's efficiency is set by the average temperature at which heat enters and leaves, which is why reheating and regeneration in power plants, both ways of raising that average, work.

12. The efficiency of an ideal gasoline engine

1. A gasoline engine has compression ratio $8$. Write the ideal Otto efficiency.

   $e = 1 - \dfrac{1}{r^{\gamma - 1}}$

   Both adiabats scale temperatures by $r^{\gamma - 1}$.

2. Find the exponent for air.

   $\gamma - 1 = 1.4 - 1 = 0.4$

   Air is diatomic.

3. Evaluate the power.

   $8^{0.4} = 2.2974$

   With a calculator, or $e^{0.4\ln 8}$.

4. Take the reciprocal and subtract from one.

   $e = 1 - \dfrac{1}{2.2974} = 1 - 0.4353 = 0.5647$

   Just over half the heat becomes work in the ideal cycle.

5. Compare with a real engine.

   $e_{\text{real}} \approx 0.30 \text{ to } 0.35$

   Friction, heat lost through the cylinder walls and incomplete combustion take their share.

13. Following an Otto cycle around its corners

1. One mole of air enters at $300$ K; $r = 8$; combustion raises it to $1800$ K. Find the temperature after compression.

   $T_2 = 300 \times 8^{0.4} = 300 \times 2.2974 = 689\ \text{K}$

   The adiabat heats the air before any fuel burns.

2. Find the temperature after the power stroke.

   $T_4 = \dfrac{1800}{2.2974} = 783\ \text{K}$

   The expansion divides by the same factor.

3. Find the heat added at constant volume.

   $Q_h = C_V(T_3 - T_2) = 20.775 \times (1800 - 689) = 23\,080\ \text{J}$

   Air's $C_V$ is $\tfrac{5}{2}R = 20.775$ J/(mol K).

4. Find the heat rejected.

   $Q_c = C_V(T_4 - T_1) = 20.775 \times (783 - 300) = 10\,040\ \text{J}$

   The exhaust leaves hot.

5. Find the net work.

   $W = Q_h - Q_c = 23\,080 - 10\,040 = 13\,040\ \text{J}$

   The first law for the cycle.

6. Find the efficiency and check it.

   $e = \dfrac{13\,040}{23\,080} = 0.565 = 1 - \dfrac{1}{2.2974}$

   The long way and the formula agree.

7. Compare with Carnot between the same extremes.

   $1 - \dfrac{300}{1800} = 0.833 > 0.565$

   Heat enters between $689$ and $1800$ K and leaves between $783$ and $300$ K, never at the extremes, so the Otto cycle falls well short of Carnot.

14. A steam plant from the steam tables

1. Steam enters a turbine at $3.0$ MPa and $400$ °C with $h_1 = 3231.7$ kJ/kg and leaves at $10$ kPa with $h_2 = 2192.7$ kJ/kg. Find the turbine work.

   $w_t = h_1 - h_2 = 3231.7 - 2192.7 = 1039.0\ \text{kJ/kg}$

   For a steady-flow device with no heat loss, work is the drop in enthalpy.

2. Find the pump work.

   $w_p = v\,\Delta P = 0.00101 \times (3000 - 10) \approx 3.0\ \text{kJ/kg}$

   Liquid water is nearly incompressible, so the pump work is its specific volume times the pressure rise.

3. Find the net work.

   $w_{\text{net}} = 1039.0 - 3.0 = 1036.0\ \text{kJ/kg}$

   The pump takes less than one percent of the turbine's work.

4. Find the enthalpy entering the boiler.

   $h_4 = h_3 + w_p = 191.8 + 3.0 = 194.8\ \text{kJ/kg}$

   Condensed water, then pumped.

5. Find the heat added in the boiler.

   $q_h = h_1 - h_4 = 3231.7 - 194.8 = 3036.9\ \text{kJ/kg}$

   Heating and boiling at constant pressure.

6. Find the efficiency.

   $e = \dfrac{1036.0}{3036.9} \approx 0.341$

   Work out over heat in.

7. Compare with the Carnot limit.

   $1 - \dfrac{45.8 + 273}{400 + 273} = 1 - \dfrac{319}{673} = 0.526$

   Most of the heat enters while the water boils at $234$ °C, well below the peak, which is why the Rankine cycle falls short.

8. Find the steam flow for a $500$ MW plant.

   $\dot{m} = \dfrac{500\,000\ \text{kW}}{1036\ \text{kJ/kg}} \approx 483\ \text{kg/s}$

   Nearly half a ton of steam every second.

15. Your turn: a gas turbine has a pressure ratio of $16$. What is its ideal Brayton efficiency with air?

1. Find the exponent.

   $\dfrac{\gamma - 1}{\gamma} = \dfrac{0.4}{1.4} = 0.2857$

   Along an adiabat $T \propto P^{(\gamma - 1)/\gamma}$.

2. Evaluate the power.

   $16^{0.2857} = 2.208$

   With a calculator.

3. Your turn: work this step out. Its working is at the end of the packet.

   Subtract the reciprocal from one.

16. Guided practice

An ideal Otto engine with compression ratio $4$ burns a richer fuel mixture, so its peak temperature rises. What happens to its efficiency?

• It stays the same, $0.4257$: it depends only on the compression ratio
• It rises, as the Carnot formula suggests for a hotter source
• It falls, because more heat is wasted
• It becomes the Carnot efficiency

17. Guided practice

Complete the worked solution: an ideal Otto cycle has compression ratio $4$, takes in air at $300$ K and peaks at $1900$ K. Take $4^{0.4} = 1.7411$ and $\tfrac{1}{1.7411} = 0.5743$. Find the temperatures after each adiabat and the efficiency.

1. Multiply the intake temperature by the compression factor.

   $T_2 = 300 \times 1.7411 =$ _b K

   Compression without heat loss warms the air.

2. Multiply the peak temperature by the reciprocal of the factor.

   $T_4 = 1900 \times 0.5743 =$ _d K

   The power stroke is an adiabat across the same volume ratio, so it divides by the same factor.

3. Subtract the reciprocal from one for the efficiency.

   $e = 1 - \dfrac{1}{r^{0.4}} =$ _e

   The rejected heat is the added heat divided by $r^{0.4}$.

18. Guided practice

Match each engine cycle to how heat enters it. (A gasoline engine in the list has compression ratio $8$.)

19. Practice

For an ideal Otto cycle with air ($\gamma = 1.4$), fill in $1/r^{0.4}$ and the efficiency for each compression ratio, to four decimal places. The values of $r^{0.4}$ are $1.7411$, $2.2974$, $2.5119$ and $3.0314$.

20. Practice

A gas turbine runs an ideal Brayton cycle with a pressure ratio of $10$. Taking air as an ideal gas with $\gamma = 1.4$, what is its efficiency? Take $10^{0.2857} = 1.9307$.

Answer:

21. Practice

An ideal Otto cycle with compression ratio $10$ takes in air at $300$ K, and combustion raises it to $2100$ K. Treating air as an ideal gas with $C_V = 20.775$ J/(mol K), what is the net work per mole per cycle, in joules? Take $10^{0.4} = 2.5119$.

Answer: J

22. Somewhere new

A steam power plant's boiler delivers steam at 10 MPa and 550 °C, and the condenser runs at $10$ kPa. From steam tables, the enthalpy is $3500.9$ kJ/kg entering the turbine and $2139.3$ kJ/kg leaving it; the pump adds $10.1$ kJ/kg, so water enters the boiler at $201.9$ kJ/kg. What is the ideal Rankine efficiency?

Answer:

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

Complete the worked solution: an ideal Otto cycle has compression ratio $8$, takes in air at $300$ K and peaks at $1500$ K. Take $8^{0.4} = 2.2974$ and $\tfrac{1}{2.2974} = 0.4353$. Find the temperatures after each adiabat and the efficiency.

1. Multiply the intake temperature by the compression factor.

   $T_2 = 300 \times 2.2974 =$ _b K

   Compression without heat loss warms the air.

2. Multiply the peak temperature by the reciprocal of the factor.

   $T_4 = 1500 \times 0.4353 =$ _d K

   The power stroke is an adiabat across the same volume ratio, so it divides by the same factor.

3. Subtract the reciprocal from one for the efficiency.

   $e = 1 - \dfrac{1}{r^{0.4}} =$ _e

   The rejected heat is the added heat divided by $r^{0.4}$.

25. What you can do now

You can find the efficiency of the engine cycles that run cars, jets and power plants. Explain to someone why a gasoline engine cannot simply raise its compression ratio to twenty.

Working for the steps left to you

15. Your turn: a gas turbine has a pressure ratio of $16$. What is its ideal Brayton efficiency with air?, step 3