An engine with no waste but the necessary one
Lesson 2 showed that every heat engine must reject some heat. Real engines also waste heat through friction and through heat flowing across large temperature gaps. In 1824 the French engineer Sadi Carnot asked a sharper question. What is the best an engine can do if it wastes nothing except what the second law demands? This lesson uses \(R = 8.31\,\text{J/(mol K)}\) and \(\gamma = 1.4\) for air.
To waste nothing, the engine must be reversible. So heat may flow only between bodies at the same temperature. The gas must therefore take in heat only while it is at \(T_H\), the temperature of the source. It must give out heat only while it is at \(T_C\). Between these two stages it must change its temperature with no heat flow at all, that is, adiabatically.
These conditions fix the cycle: two isotherms joined by two adiabats. It is the Carnot cycle, and an engine that runs on it is a Carnot engine. Follow one mole of air round it in Fig 3.1, between a source at \(500\,\text{K}\) and a sink at \(300\,\text{K}\).
The four legs
A to B is an isothermal expansion at \(500\,\text{K}\), from \(10\,\text{L}\) to \(20\,\text{L}\). The cylinder stands on the hot reservoir. The gas does work, but its temperature and so its internal energy stay fixed. So it takes in heat equal to its work: \(Q_H = nRT_H\ln(V_B/V_A)\), which is \(8.31 \times 500 \times \ln 2 = 2880\,\text{J}\).
B to C is an adiabatic expansion. The cylinder moves to the insulator, and the gas goes on expanding with no heat flow. It does work out of its own internal energy, so it cools. The leg ends when the gas reaches \(300\,\text{K}\), at \(71.7\,\text{L}\). The work is \(nC_v(T_H - T_C)\), which is \(20.775 \times 200 = 4155\,\text{J}\), with \(C_v = \tfrac52 R = 20.775\,\text{J/(mol K)}\).
C to D is an isothermal compression at \(300\,\text{K}\), from \(71.7\,\text{L}\) to \(35.9\,\text{L}\), half the volume. The cylinder stands on the cold reservoir. Work is done on the gas, and it gives out an equal heat, \(Q_C = nRT_C\ln(V_C/V_D)\), which is \(8.31 \times 300 \times \ln 2 = 1728\,\text{J}\).
D to A is an adiabatic compression back to \(10\,\text{L}\). On the insulator again, the work done on the gas raises its temperature from \(300\,\text{K}\) to \(500\,\text{K}\). This work is \(4155\,\text{J}\), the same as on leg B to C. So the two adiabatic legs cancel exactly, and the gas is back at A.
Add the legs. The gas took in \(2880\,\text{J}\) and gave out \(1728\,\text{J}\), so the net work is \(1152\,\text{J}\), the area inside the loop. The efficiency is \(1152/2880 = 0.40\). Note that the adiabatic legs carry large work but no heat. Only the two isotherms exchange heat with the reservoirs.
In the Carnot cycle of Fig 3.1, which statements are true?
Heats in the ratio of temperatures
In the numbers above, both isotherms had the same volume ratio, \(2\). That is no accident. It forces the heats into the ratio of the temperatures: \(1728/2880 = 0.6 = 300/500\). The derivation shows why, using \(TV^{\gamma - 1} = \text{constant}\) on an adiabat from the chapter on the First Law.
\[ Q_H = nRT_H\ln\dfrac{V_B}{V_A} \]
Why. on the isotherm A to B, \(\Delta U = 0\), so the heat taken in equals the work of an isothermal expansion.
\[ Q_C = nRT_C\ln\dfrac{V_C}{V_D} \]
Why. on the isotherm C to D, the heat given out equals the work done on the gas.
\[ T_H V_B^{\gamma - 1} = T_C V_C^{\gamma - 1} \]
Why. B to C is adiabatic, and \(TV^{\gamma - 1}\) stays constant on an adiabat.
\[ T_H V_A^{\gamma - 1} = T_C V_D^{\gamma - 1} \]
Why. D to A is adiabatic too, between the same two temperatures.
\[ \dfrac{V_B}{V_A} = \dfrac{V_C}{V_D} \]
Why. divide step 3 by step 4; the temperatures cancel, and taking the \((\gamma - 1)\)th root leaves the volume ratios.
\[ \dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H} \]
Why. divide step 2 by step 1; by step 5 the two logarithms are equal and cancel.
\[ \eta = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H} \]
Read it. put step 6 into the definition of efficiency from Lesson 1.
The formula uses the absolute temperatures. A source at \(100\,°\text{C}\) and a sink at \(0\,°\text{C}\) do not give \(1 - 0/100 = 1\). In kelvin they are \(373\,\text{K}\) and \(273\,\text{K}\), so \(\eta = 1 - 273/373 = 0.268\). Even between boiling and freezing water, the best engine turns only about \(27\,\%\) of its heat into work.
Read Fig 3.2 for the \(600\,\text{K}\) source. A sink at \(300\,\text{K}\) gives \(\eta = 0.50\), and a sink at \(450\,\text{K}\) gives \(0.25\). At \(T_C = 600\,\text{K}\) the efficiency is zero: with both reservoirs at one temperature, the engine gives back all it takes and does no work. The line reaches \(1\) only at \(T_C = 0\), and absolute zero cannot be reached. So no engine is ever perfect.
A Carnot engine works between \(400\,\text{K}\) and \(200\,\text{K}\). Find its efficiency. Which raises the efficiency more: raising the source by \(50\,\text{K}\), or lowering the sink by \(50\,\text{K}\)?
\(T_H = 400\,\text{K}\), \(T_C = 200\,\text{K}\). The two changes give \(450\,\text{K}\) with \(200\,\text{K}\), or \(400\,\text{K}\) with \(150\,\text{K}\).
\(\eta = 1 - T_C/T_H\) in each case.
Now: \(1 - 200/400 = 0.50\). Hotter source: \(1 - 200/450 = 0.556\). Colder sink: \(1 - 150/400 = 0.625\).
Lowering the sink by \(50\,\text{K}\) helps more than raising the source by \(50\,\text{K}\). In practice the sink is the air or a river, so engineers raise \(T_H\) instead.
A Carnot engine has a source at \(500\,\text{K}\) and an efficiency of \(40\,\%\). Keeping the sink where it is, to what temperature must the source be raised for an efficiency of \(50\,\%\), in kelvin?
A Carnot engine takes \(1000\,\text{J}\) per cycle from a source at \(400\,\text{K}\) and works with a sink at \(300\,\text{K}\). How much heat does it give to the sink per cycle, in joules?
A Carnot engine has efficiency \(1/6\). When its sink is made \(62\,\text{K}\) colder, its efficiency becomes \(1/3\). What is the temperature of the source, in kelvin?
The ceiling for every engine
The derivation used an ideal gas. Would a Carnot engine running on steam or helium do better? Carnot's answer, proved from the second law, is no.
No engine working between two reservoirs can be more efficient than a reversible engine working between the same two reservoirs. All reversible engines between the same two reservoirs have the same efficiency, \(1 - T_C/T_H\), whatever their working substance.
The proof is a machine that would break the second law. Suppose an engine X had an efficiency of \(50\,\%\) between two reservoirs where a Carnot engine has \(40\,\%\). Let X take \(1000\,\text{J}\) from the source, do \(500\,\text{J}\) of work and reject \(500\,\text{J}\). Now run the Carnot engine backwards, as a refrigerator, driven by \(400\,\text{J}\) of the work of X.
A reversed Carnot engine runs every leg backwards, so its heats and work keep their sizes and swap their directions. With \(400\,\text{J}\) of work put in, it takes \(600\,\text{J}\) from the sink and gives \(1000\,\text{J}\) to the source. Now count. The source gives \(1000\,\text{J}\) and receives \(1000\,\text{J}\): no change. The sink receives \(500\,\text{J}\) and gives \(600\,\text{J}\): it loses \(100\,\text{J}\). And \(100\,\text{J}\) of work is left over.
That pair takes heat from one reservoir and turns it wholly into work, which the Kelvin-Planck statement forbids. So no engine X can exist. Every engine between \(T_H\) and \(T_C\) has \(\eta \le 1 - T_C/T_H\). The same argument shows that two reversible engines must tie. The Carnot efficiency is a limit set by the two temperatures alone.
The Carnot efficiency is a ceiling, not a typical value. A real engine has friction, heat flowing across large temperature gaps and gases rushing through valves. Each of these is irreversible and pulls the efficiency below \(1 - T_C/T_H\).
A coal station makes steam at about \(823\,\text{K}\), \(550\,°\text{C}\), and its condenser runs near the river's temperature, about \(303\,\text{K}\). Its Carnot limit is \(1 - 303/823 = 0.63\). Vindhyachal reaches about \(35\,\%\), a little more than half the limit. Now the cooling towers of Lesson 1 make full sense. Even a perfect station would reject \(37\,\%\) of its heat, and the real one rejects about \(65\,\%\).
An inventor claims an engine that takes heat at \(500\,\text{K}\), rejects heat at \(300\,\text{K}\), and has an efficiency of \(60\,\%\). Is the claim possible?
Two Carnot engines work in series. Engine 1 works between \(800\,\text{K}\) and \(T\). Engine 2 takes all the heat rejected by engine 1 and works between \(T\) and \(200\,\text{K}\). If the two engines have equal efficiencies, what is \(T\), in kelvin?
A Carnot engine works between \(400\,\text{K}\) and \(300\,\text{K}\). In each cycle the gas does \(2301\,\text{J}\) of work during its isothermal expansion. What net work does the engine do per cycle, in joules?
In 1824 the young French engineer Sadi Carnot published a short book, Reflections on the Motive Power of Fire. He wanted to know why British steam engines were better than French ones, and whether any engine had a limit. He saw that work comes from heat falling from a hot body to a cold one, as work comes from water falling in a mill. He died of cholera at 36, and his book was nearly forgotten until Clausius and Kelvin rebuilt thermodynamics on it.
- Putting Celsius temperatures into \(\eta = 1 - T_C/T_H\).Convert both to kelvin first: \(T = t + 273\).
- Thinking the adiabatic legs exchange heat because the temperature changes.An adiabatic leg has \(Q = 0\); the temperature changes because the gas does work or has work done on it.
- Expecting a better working substance to beat the Carnot limit.By Carnot's theorem the limit depends only on \(T_H\) and \(T_C\).