University Physics — Year 1 · Bachelor Year 1

24Heat Engines and Machines

A car engine, a power station, a refrigerator and a heat pump are the same object seen from four sides: a fluid taken round a cycle, in contact by turns with a hot source and a cold one, exchanging heat with both and work with a shaft. The first law bookkeeps the energy; the second law, through the entropy balance over one cycle, sets the limit that no cleverness can beat — Carnot’s theorem, the most consequential inequality of engineering. This chapter derives that limit, computes the ideal cycle and the real ones that approach it (Otto, Diesel, Stirling), turns the machine round to make cold and to heat houses, and shows how the created entropy of a real machine is paid for, joule by joule, in lost work.

24.1 Cyclic machines and the two laws

Definition 24.1 (Thermodynamic machine)

A thermodynamic machine is a closed system (the working fluid) taken round a cycle, so that it returns periodically to the same state, while exchanging work $W$ with the outside and heat $Q_i$ with sources at temperatures $T_i$ — thermostats, whose temperature the exchange does not change. A ditherm machine uses two thermostats, the hot source at $T_h$ and the cold source at $T_c < T_h$ . It is an engine if it delivers work ( $W < 0$ ), a refrigerator if it is driven ( $W > 0$ ) to take heat from the cold source, a heat pump if it is driven to give heat to the hot one. All quantities are reckoned over one cycle, algebraically, as received by the fluid.

Theorem 24.2 (The two laws over a cycle; Clausius inequality)

Over one cycle of a ditherm machine,

W + Q_h + Q_c = 0 , \qquad \frac{Q_h}{T_h} + \frac{Q_c}{T_c} = -S_{\mathrm{created}} \leq 0 ,

with equality in the second relation if and only if the cycle is reversible. More generally, for any number of thermostats, $\sum_i Q_i/T_i \leq 0$ (Clausius inequality).

Proof. $U$ and $S$ are state functions, so $\Delta U = 0$ and $\Delta S = 0$ over a cycle. The first law gives $0 = W + Q_h + Q_c$ ; the entropy balance (Theorem 23.1) gives $0 = S_{\mathrm{exch}} + S_{\mathrm{created}}$ with $S_{\mathrm{exch}} = \sum Q_i/T_i$ because each thermostat exchanges at its own fixed temperature (Proposition 23.5). ∎

Corollary 24.3 (Kelvin again)

A monotherm machine ( $Q_c = 0$ ) has $Q_h \leq 0$ , hence $W \geq 0$ : no cycle can convert the heat of a single source into work. A ship cannot sail by cooling the sea — an engine needs a cold source to reject heat to, and the second law says how much it must reject.

The two ways round a ditherm cycle. Left: the engine takes heat from the hot source, rejects part of it to the cold one and delivers the difference as work. Right: the same machine driven backward pumps heat from cold to hot; it is a refrigerator if one wants Q_c, a heat pump if one wants |Q_h|. — The two ways round a ditherm cycle. Left: the engine takes heat from the hot source, rejects part of it to the cold one and delivers the difference as work. Right: the same machine driven backward pumps heat from cold to hot; it is a refrigerator if one wants $Q_c$ , a heat pump if one wants $|Q_h|$ .

24.2 Carnot’s theorem

Definition 24.4 (Efficiency and coefficients of performance)

The efficiency of an engine is the work delivered per unit heat taken from the hot source, $\eta = -W/Q_h$ ; the coefficient of performance (COP) of a refrigerator is the cold produced per unit work, $e_f = Q_c/W$ ; that of a heat pump is the heat delivered per unit work, $e_p = -Q_h/W$ . Each is the ratio “what one wants” over “what one pays”; an efficiency is at most 1, a COP can exceed 1.

Theorem 24.5 (Carnot)

Between thermostats $T_h > T_c$ ,

\eta \leq \eta_C = 1 - \frac{T_c}{T_h}, \qquad e_f \leq \frac{T_c}{T_h - T_c}, \qquad e_p \leq \frac{T_h}{T_h - T_c} = 1 + \frac{T_c}{T_h - T_c},

with equality if and only if the cycle is reversible. The bounds depend on the temperatures only — not on the fluid, not on the mechanism.

Proof. For an engine ( $Q_h > 0$ ): $\eta = -W/Q_h = (Q_h + Q_c)/Q_h = 1 + Q_c/Q_h$ , and the Clausius inequality gives $Q_c/Q_h \leq -T_c/T_h$ . For a refrigerator ( $Q_c > 0$ , $W > 0$ ): $W = -Q_h - Q_c$ with $-Q_h \geq Q_c T_h/T_c$ , so $W \geq Q_c(T_h/T_c - 1)$ and $e_f = Q_c/W \leq T_c/(T_h - T_c)$ . For the heat pump, $e_p = -Q_h/W = (W + Q_c)/W = 1 + e_f$ . Equality is the reversible case $S_{\mathrm{created}} = 0$ throughout. ∎

Example 24.6 (Three numbers)

A steam turbine between $T_h = 800\,\mathrm{K}$ and a river at $T_c = 300\,\mathrm{K}$ : $\eta_C = 0.625$ ; real plants reach about $0.40$ . A domestic refrigerator between $275\,\mathrm{K}$ inside and a kitchen at $298\,\mathrm{K}$ : $e_f \leq 275/23 = 12$ ; real machines give $2$ to $4$ . A heat pump warming a house at $293\,\mathrm{K}$ from air at $273\,\mathrm{K}$ : $e_p \leq 293/20 = 14.7$ ; real pumps give $3$ to $5$ — still three to five times what the same electricity would give in a resistor (Problem 24.1).

Remark 24.7 (Why the efficiency cannot be 1)

An engine receives entropy $Q_h/T_h$ with the heat it takes in; work carries none away; the only exit is heat rejected to the cold source, which carries $|Q_c|/T_c$ . Entropy being uncreatable, at least $Q_h T_c/T_h$ of heat must be thrown away, and the work is what is left. The lower the cold source and the higher the hot one, the better — hence the cooling towers, and the quest for ever hotter turbine blades.

24.3 The Carnot cycle

Definition 24.8 (Carnot cycle)

The Carnot cycle is the reversible ditherm cycle: two reversible isotherms, at $T_h$ (in contact with the hot source) and at $T_c$ (with the cold one), joined by two reversible adiabatic legs (isentropic, with no source in contact). Run clockwise in the $(V, P)$ plane it is an engine; counterclockwise, a refrigerator or heat pump. In the entropy diagram it is a rectangle.

Proposition 24.9 (Carnot cycle of a perfect gas)

For $n$ moles of perfect gas taken round $A \to B$ (isotherm $T_h$ , expansion), $B \to C$ (adiabat), $C \to D$ (isotherm $T_c$ , compression), $D \to A$ (adiabat):

Q_h = nRT_h\ln\frac{V_B}{V_A}, \qquad Q_c = -nRT_c\ln\frac{V_B}{V_A}, \qquad \eta = 1 - \frac{T_c}{T_h} ,

the Carnot value — as it must be for any reversible ditherm cycle.

Proof. On an isotherm $\Delta U = 0$ , so $Q = -W = nRT\ln(V_{\text{final}}/ V_{\text{initial}})$ . On the adiabats $TV^{\gamma-1}$ is constant: $T_hV_B^{\gamma-1} = T_cV_C^{\gamma-1}$ and $T_hV_A^{\gamma-1} = T_cV_D^{\gamma-1}$ , whence $V_C/V_D = V_B/V_A$ and $Q_c = nRT_c\ln(V_D/V_C) = -nRT_c\ln(V_B/V_A)$ . Then $\eta = 1 + Q_c/Q_h = 1 - T_c/T_h$ . ∎

The Carnot cycle of a perfect gas. Left: in the Clapeyron diagram two isotherms and two (steeper) adiabats, run clockwise; the enclosed area is the work delivered. Right: the same cycle in the entropy diagram is a rectangle of height T_h - T_c and width Q_h/T_h; its area is again -W, and the ratio of that area to the one under the top side, Q_h, is 1 - T_c/T_h by inspection. — The Carnot cycle of a perfect gas. Left: in the Clapeyron diagram two isotherms and two (steeper) adiabats, run clockwise; the enclosed area is the work delivered. Right: the same cycle in the entropy diagram is a rectangle of height $T_h - T_c$ and width $Q_h/T_h$ ; its area is again $-W$ , and the ratio of that area to the one under the top side, $Q_h$ , is $1 - T_c/T_h$ by inspection.

Remark 24.10 (Carnot’s cycle is a theorem, not an engine)

Reversible isotherms require infinitely slow heat transfer across a vanishing temperature gap: a Carnot engine delivers its maximal work at zero power. Real machines trade efficiency for power by accepting finite temperature differences in their exchangers (Exercise 24.10); the Carnot value remains the ceiling every design is measured against.

24.4 Real engine cycles

Proposition 24.11 (Otto cycle)

The idealized petrol engine takes air ( $\gamma$ ) round: $1 \to 2$ adiabatic compression from $V_1$ to $V_2$ (compression ratio $r = V_1/V_2$ ); $2 \to 3$ isochoric heating (the combustion); $3 \to 4$ adiabatic expansion back to $V_1$ ; $4 \to 1$ isochoric cooling (the exhaust). Its efficiency is

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

Proof. Heat is exchanged only on the isochores: $Q_h = nC_{V,m}(T_3 - T_2)$ and $Q_c = nC_{V,m}(T_1 - T_4)$ . On the adiabats $TV^{\gamma-1}$ is constant: $T_2 = T_1r^{\gamma-1}$ and $T_3 = T_4r^{\gamma-1}$ , so $T_3 - T_2 = (T_4 - T_1)r^{\gamma-1}$ and $\eta = 1 + Q_c/Q_h = 1 - (T_4 - T_1)/(T_3 - T_2) = 1 - r^{1-\gamma}$ . ∎

Example 24.12 (Numbers for a petrol engine)

$r = 9$ , $\gamma = 1.4$ : $9^{0.4} = 2.41$ and $\eta = 0.58$ . A real engine gives about $0.30$ : the combustion is not instantaneous, the gas is not perfect at $2500\,\mathrm{K}$ , heat leaks to the walls, friction and pumping take their share. Raising $r$ helps — until the air–fuel mixture ignites by itself at the end of the compression (knock); the Diesel engine turns that vice into its principle.

Proposition 24.13 (Diesel cycle)

Replace the isochoric heating of the Otto cycle by an isobaric one, from $V_2$ to $V_3 = \rho V_2$ (cut-off ratio $\rho > 1$ ): fuel injected into air already hot from a compression ratio $r$ of $15$ to $22$ burns as it arrives. Then

\eta_{\text{Diesel}} = 1 - \frac{1}{r^{\gamma-1}}\,\frac{\rho^{\gamma} - 1}{\gamma(\rho - 1)} ,

smaller than the Otto value for the same $r$ but reached with much larger $r$ , hence larger in practice ( $0.40$ to $0.45$ for big engines).

Proof. $Q_h = nC_{P,m}(T_3 - T_2)$ , $Q_c = nC_{V,m}(T_1 - T_4)$ , so $\eta = 1 - (T_4 - T_1)/[\gamma(T_3 - T_2)]$ . With $T_2 = T_1r^{\gamma-1}$ , $T_3 = \rho T_2$ and, on the adiabat $3 \to 4$ ( $V_4 = V_1 = rV_2$ ), $T_4 = T_3(V_3/V_4)^{\gamma-1} = \rho T_1 r^{\gamma-1}(\rho/r)^{\gamma-1} = T_1\rho^{\gamma}$ : $\eta = 1 - (\rho^\gamma - 1)/[\gamma r^{\gamma-1}(\rho - 1)]$ . ∎

Left: the Otto cycle in the Clapeyron diagram (drawn with r = 4): two adiabats and two isochores; the heat enters at top dead centre and leaves with the exhaust. Right: ideal efficiencies against the compression ratio; the dashed lines mark a petrol engine (r = 9) and a Diesel (r = 18, = 2). — Left: the Otto cycle in the Clapeyron diagram (drawn with $r = 4$ ): two adiabats and two isochores; the heat enters at top dead centre and leaves with the exhaust. Right: ideal efficiencies against the compression ratio; the dashed lines mark a petrol engine ( $r = 9$ ) and a Diesel ( $r = 18$ , $\rho = 2$ ).

Proposition 24.14 (Stirling cycle and the regenerator)

Two isotherms ( $T_c$ , $T_h$ ) joined by two isochores. The heat of the isochoric legs, $\pm nC_{V,m}(T_h - T_c)$ , is equal and opposite; if it is stored on the cooling leg in a regenerator (a porous mass the gas flows through) and returned on the heating leg, only the isotherms exchange with the sources, and $\eta = 1 - T_c/T_h$ : the Stirling engine is a Carnot engine in all but the shape of its cycle. Without the regenerator the isochoric heat must come from the hot source and $\eta = \dfrac{nR(T_h - T_c)\ln(V_1/V_2)}{nRT_h\ln(V_1/V_2) + nC_{V,m}(T_h - T_c)}$ , substantially less (Exercise 24.8).

Proof. On the isotherms $Q_h = nRT_h\ln(V_1/V_2)$ and $Q_c = -nRT_c\ln(V_1/V_2)$ , so $-W = Q_h + Q_c + 0$ once the isochoric heats cancel; divide by the heat actually drawn from the hot source in each case. ∎

Remark 24.15 (Steam and gas turbines)

Power stations do not use a gas in a cylinder but water boiled, expanded through a turbine, condensed and pumped back (the Rankine cycle, with two phase changes — Chapter 25), or air compressed, heated by combustion and expanded in a gas turbine (the Brayton cycle, two adiabats and two isobars). Their analysis needs the energy balance of a flowing fluid, given in the Year 2 volume; the Carnot bound and the logic of this chapter apply unchanged.

24.5 Refrigerators and heat pumps

Definition 24.16 (Vapour-compression cycle)

Almost every refrigerator, freezer, air conditioner and heat pump runs a refrigerant round four components: a compressor (receives the work, raises the pressure of the vapour), a condenser (the hot vapour liquefies at high pressure, releasing $|Q_h|$ to the hot side), an expansion valve (the liquid drops to low pressure without work or heat, and partly flashes to vapour) and an evaporator (the rest boils at low pressure, absorbing $Q_c$ from the cold side). The heat is carried as latent heat of vaporization, and the two pressures set the two boiling temperatures (Chapter 25).

The vapour-compression cycle. The refrigerant boils in the evaporator, colder than the cold side, and condenses in the condenser, hotter than the hot side: heat flows the natural way across each exchanger, and the compressor pays for lifting it from T_ ev to T_ cd. — The vapour-compression cycle. The refrigerant boils in the evaporator, colder than the cold side, and condenses in the condenser, hotter than the hot side: heat flows the natural way across each exchanger, and the compressor pays for lifting it from $T_{\mathrm{ev}}$ to $T_{\mathrm{cd}}$ .

Proposition 24.17 (The price of the temperature gaps)

A reversible machine working between its own internal temperatures $T_{\mathrm{ev}} < T_c$ and $T_{\mathrm{cd}} > T_h$ has $e_p = T_{\mathrm{cd}}/(T_{\mathrm{cd}} - T_{\mathrm{ev}})$ , smaller than the Carnot value between the sources; a real machine reaches a fraction (typically $0.4$ to $0.6$ ) of even that. The performance of a heat pump therefore falls as the outdoor temperature drops and as the temperature demanded by the heating circuit rises.

Proof. Apply Theorem 24.5 to the fluid, whose thermostats are effectively the exchanger temperatures; $T_{\mathrm{cd}} - T_{\mathrm{ev}} > T_h - T_c$ lowers the ratio. ∎

Coefficient of performance of a heat pump against outdoor temperature: the Carnot bound between house (20 C) and outdoor air; the bound once the exchanger temperature gaps are counted (T_ cd = 313\, K, T_ ev = T_ out - 8\, K); and a realistic machine at 0.55 of the latter. Even at -10 C the real pump delivers about three joules of heat per joule of electricity, where a resistor gives one. — Coefficient of performance of a heat pump against outdoor temperature: the Carnot bound between house ( $20{}^{\circ}\mathrm{C}$ ) and outdoor air; the bound once the exchanger temperature gaps are counted ( $T_{\mathrm{cd}} = 313\,\mathrm{K}$ , $T_{\mathrm{ev}} = T_{\mathrm{out}} - 8\,\mathrm{K}$ ); and a realistic machine at $0.55$ of the latter. Even at $-10{}^{\circ}\mathrm{C}$ the real pump delivers about three joules of heat per joule of electricity, where a resistor gives one.

Example 24.18 (Heat pump or resistor?)

A house losing $6\,\mathrm{kW}$ at $-5{}^{\circ}\mathrm{C}$ outdoors needs $6\,\mathrm{kW}$ of electricity with resistors; an ideal heat pump from $268\,\mathrm{K}$ to $293\,\mathrm{K}$ would need $6/11.7 = 0.51\,\mathrm{kW}$ ; a real one with exchanger temperatures $260\,\mathrm{K}$ and $313\,\mathrm{K}$ and a $0.55$ quality factor, $e_p = 0.55 \times 313/53 = 3.2$ , needs $1.9\,\mathrm{kW}$ . The weekend problem works the season through, entropy included.

24.6 Entropy and lost work

Theorem 24.19 (Lost work)

A ditherm engine creating the entropy $S_{\mathrm{created}}$ per cycle delivers

-W = Q_h\Bigl(1 - \frac{T_c}{T_h}\Bigr) - T_c\,S_{\mathrm{created}} ,

and a ditherm heat pump delivering $|Q_h|$ consumes $W = |Q_h|(1 - T_c/T_h) + T_c\,S_{\mathrm{created}}$ : every joule per kelvin created costs $T_c$ joules of work, the temperature of the cold source being the rate of exchange between entropy and work.

Proof. From Theorem 24.2, $Q_c = -T_c(Q_h/T_h + S_{\mathrm{created}})$ ; insert into $-W = Q_h + Q_c$ . For the pump, write the same two relations with $Q_h < 0$ . ∎

Method 24.20 (Auditing a machine)

Measure (or compute) the heats exchanged with each source and the work; check the first law; compute $S_{\mathrm{created}} = -\sum Q_i/T_i$ per cycle or per second; the lost power is $T_c\dot S_{\mathrm{created}}$ ; locate the creation (finite temperature gaps in exchangers, friction, throttling, non-quasi-static compression) by applying the entropy balance to each component in turn. A component that creates no entropy is not worth improving.

Example 24.21 (A power plant’s audit)

Hot source $800\,\mathrm{K}$ , river $300\,\mathrm{K}$ , $1.0\,\mathrm{GW}$ of electricity at $\eta = 0.40$ : $\dot Q_h = 2.5\,\mathrm{GW}$ , $\dot Q_c = 1.5\,\mathrm{GW}$ , so $\dot S_{\mathrm{created}} = 1.5/300 - 2.5/800 = 1.9\,\mathrm{MW}/\mathrm{K}$ and the lost power is $300 \times 1.9 = 0.56\,\mathrm{GW}$ : exactly the gap between the Carnot output $0.625 \times 2.5 = 1.56\,\mathrm{GW}$ and the real $1.0\,\mathrm{GW}$ . The river warms by the $1.5\,\mathrm{GW}$ it must carry away; in summer, when it is warm and low, plants throttle back.

24.7 Exercises

Exercise 24.1 ★

A power station takes heat at $800\,\mathrm{K}$ and rejects it to a river at $300\,\mathrm{K}$ . Carnot efficiency; if the real efficiency is $0.40$ and the electrical output $1.0\,\mathrm{GW}$ , heat taken from the fuel and heat rejected to the river per second.

Solution

Solution of Exercise 24.1.

$\eta_C = 1 - 300/800 = 0.625$ . At $0.40$ : $\dot Q_h = 1.0/0.40 = 2.5\,\mathrm{GW}$ from the fuel, $\dot Q_c = 2.5 - 1.0 = 1.5\,\mathrm{GW}$ into the river — more heat is thrown away than is sold.

Exercise 24.2 ★

A refrigerator keeps its interior at $275\,\mathrm{K}$ in a kitchen at $298\,\mathrm{K}$ . Maximum coefficient of performance; with a real COP of $3$ , electrical power needed to remove the $100\,\mathrm{W}$ that leak in through the walls.

Solution

Solution of Exercise 24.2.

$e_f \leq 275/(298 - 275) = 12$ . With $e_f = 3$ : $P = 100/3 = 33\,\mathrm{W}$ (and $133\,\mathrm{W}$ go into the kitchen).

Exercise 24.3 ★

Ideal efficiency of an Otto cycle with $r = 9$ and $\gamma = 1.4$ ; with $r = 11$ (a modern direct-injection engine). Why not $r = 20$ ?

Solution

Solution of Exercise 24.3.

$r = 9$ : $9^{0.4} = 2.41$ , $\eta = 0.585$ ; $r = 11$ : $11^{0.4} = 2.61$ , $\eta = 0.62$ . At $r = 20$ the petrol–air mixture, compressed to about $900\,\mathrm{K}$ , would ignite before the spark (knock), destroying the engine; only the Diesel, which compresses air alone, can go there.

Exercise 24.4 ★

A heat pump warms a house at $300\,\mathrm{K}$ from outdoor air at $280\,\mathrm{K}$ . Maximum $e_p$ ; electrical power for a $5\,\mathrm{kW}$ heating demand with that ideal pump and with a real one of $e_p = 3.5$ ; compare with a resistor.

Solution

Solution of Exercise 24.4.

$e_p \leq 300/20 = 15$ : ideal power $5/15 = 0.33\,\mathrm{kW}$ ; real, $5/3.5 = 1.43\,\mathrm{kW}$ ; resistor, $5\,\mathrm{kW}$ .

Exercise 24.5 ★★

One mole of perfect gas ( $\gamma = 1.4$ ) runs a Carnot cycle between $T_h = 500\,\mathrm{K}$ and $T_c = 300\,\mathrm{K}$ ; the hot isotherm takes it from $1.0\,\mathrm{L}$ to $3.0\,\mathrm{L}$ . Compute $V_C$ , $V_D$ , $Q_h$ , $Q_c$ , $W$ and check the efficiency.

Solution

Solution of Exercise 24.5.

Adiabats: $V_C = V_B(T_h/T_c)^{1/(\gamma-1)} = 3.0 \times (5/3)^{2.5} = 10.8\,\mathrm{L}$ , $V_D = 1.0 \times 3.59 = 3.59\,\mathrm{L}$ (so $V_C/V_D = 3 = V_B/V_A$ ). $Q_h = RT_h\ln3 = 8.314 \times 500 \times 1.099 = 4.57\,\mathrm{kJ}$ , $Q_c = -RT_c\ln3 = -2.74\,\mathrm{kJ}$ , $W = -(4.57 - 2.74) = -1.83\,\mathrm{kJ}$ , $\eta = 1.83/4.57 = 0.40 = 1 - 300/500$ .

Exercise 24.6 ★★

An inventor claims a cyclic machine that takes $1000\,\mathrm{J}$ per cycle from a source at $400\,\mathrm{K}$ , rejects $800\,\mathrm{J}$ to a source at $300\,\mathrm{K}$ and delivers $200\,\mathrm{J}$ of work. Possible? Entropy created per cycle. A second inventor claims $300\,\mathrm{J}$ of work from the same $1000\,\mathrm{J}$ , rejecting $700\,\mathrm{J}$ . Possible? What is the most work anyone can get from those $1000\,\mathrm{J}$ ?

Solution

Solution of Exercise 24.6.

First claim: energy $200 = 1000 - 800$ balances; Clausius $1000/400 - 800/300 = 2.5 - 2.67 = -0.17\,\mathrm{J}/\mathrm{K}$ , so $S_{\mathrm{created}} = 0.17\,\mathrm{J}/\mathrm{K} \geq 0$ : possible ( $\eta = 0.20$ ). Second: $1000/400 - 700/300 = +0.17\,\mathrm{J}/\mathrm{K} > 0$ : entropy would have to be destroyed — impossible. Maximum: $\eta_C = 1 - 300/400 = 0.25$ , i.e. $250\,\mathrm{J}$ .

Exercise 24.7 ★★

Diesel cycle with $r = 18$ , $\rho = 2$ , $\gamma = 1.4$ : efficiency; compare with an Otto cycle of the same $r$ and with the Otto cycle at $r = 9$ . Why can the Diesel use such a large $r$ ?

Solution

Solution of Exercise 24.7.

$r^{\gamma-1} = 18^{0.4} = 3.18$ ; $\rho^\gamma = 2^{1.4} = 2.64$ : $\eta = 1 - 1.64/(1.4 \times 3.18 \times 1) = 0.63$ . Otto at $r = 18$ : $1 - 1/3.18 = 0.69$ ; at $r = 9$ : $0.585$ . The Diesel compresses pure air: no fuel is present to pre-ignite, and the fuel, injected into air at $900\,\mathrm{K}$ , burns as it enters — the knock limit does not exist.

Exercise 24.8 ★★

Stirling cycle of one mole of diatomic gas ( $C_{V,m} = \tfrac52R$ ) between $600\,\mathrm{K}$ and $300\,\mathrm{K}$ with a volume ratio $V_1/V_2 = 2$ . Work per cycle; efficiency with a perfect regenerator and without one.

Solution

Solution of Exercise 24.8.

$Q_h = RT_h\ln2 = 8.314 \times 600 \times 0.693 = 3.46\,\mathrm{kJ}$ , $Q_c = -8.314 \times 300 \times 0.693 = -1.73\,\mathrm{kJ}$ : $W = -1.73\,\mathrm{kJ}$ per cycle. With the regenerator $\eta = 1.73/3.46 = 0.50 = 1 - 300/600$ . Without it the hot source must also supply $C_{V,m}(T_h - T_c) = 2.5 \times 8.314 \times 300 = 6.24\,\mathrm{kJ}$ : $\eta = 1.73/(3.46 + 6.24) = 0.18$ .

Exercise 24.9 ★★

Two identical blocks of heat capacity $C = 4.0\,\mathrm{kJ}/\mathrm{K}$ at $T_1 = 400\,\mathrm{K}$ and $T_2 = 300\,\mathrm{K}$ are used as the sources of an engine until they reach a common temperature $T_f$ . Show that the maximum work is obtained for a reversible engine, that $T_f = \sqrt{T_1T_2}$ then, and compute $W_{\max}$ . What would $T_f$ be if the blocks were simply put in contact, and what is the entropy then created?

Solution

Solution of Exercise 24.9.

Entropy of the two blocks: $\Delta S = C\ln(T_f/T_1) + C\ln(T_f/T_2) = C\ln\bigl(T_f^2/T_1T_2\bigr) = S_{\mathrm{created}} \geq 0$ (the engine’s own entropy returns to its value each cycle), so $T_f \geq \sqrt{T_1T_2}$ , and $W = C(T_1 + T_2 - 2T_f)$ (first law) is largest for the smallest $T_f$ : the reversible case, $T_f = \sqrt{400 \times 300} = 346.4\,\mathrm{K}$ , $W_{\max} = 4000 \times (700 - 692.8) = 29\,\mathrm{kJ}$ . Direct contact: $T_f = 350\,\mathrm{K}$ , no work, $S_{\mathrm{created}} = 4000\ln(350^2/120000) = 82\,\mathrm{J}/\mathrm{K}$ .

Exercise 24.10 ★★★

Engine at maximal power. A reversible engine works between internal temperatures $x$ and $y$ , and exchanges heat with the sources through exchangers of conductance $K$ : $\dot Q_h = K(T_h - x)$ and $|\dot Q_c| = K(y - T_c)$ . (a) Write the power $P$ and the entropy condition of the reversible core. (b) Show that $y = xT_c/(2x - T_h)$ and that $P = \frac{K}{2}\bigl[(T_h + T_c) - u - T_hT_c/u\bigr]$ with $u = 2x - T_h$ . (c) Maximize over $u$ and show that the efficiency at maximal power is $\eta^\ast = 1 - \sqrt{T_c/T_h}$ . (d) Numbers for $T_h = 800\,\mathrm{K}$ , $T_c = 300\,\mathrm{K}$ ; compare with Exercise 24.1.

Solution

Solution of Exercise 24.10.

(a) $P = \dot Q_h - |\dot Q_c| = K(T_h - x) - K(y - T_c)$ ; the reversible core exchanges no net entropy: $K(T_h - x)/x = K(y - T_c)/y$ . (b) Cross-multiply: $y(T_h - x) = x(y - T_c)$ , so $y(T_h - 2x) = -xT_c$ and $y = xT_c/(2x - T_h)$ . Then $P = K(T_h - x)(1 - y/x) = K(T_h - x)(2x - T_h - T_c)/(2x - T_h)$ ; with $u = 2x - T_h$ , $T_h - x = (T_h - u)/2$ and $P = K(T_h - u)(u - T_c)/(2u) = \tfrac{K}{2}[(T_h + T_c) - u - T_hT_c/u]$ . (c) $\dd P/\dd u = \tfrac K2(-1 + T_hT_c/u^2) = 0$ at $u = \sqrt{T_hT_c}$ , a maximum; then $x = (T_h + \sqrt{T_hT_c})/2$ , $y = xT_c/u = (\sqrt{T_hT_c} + T_c)/2$ , and $\eta^\ast = 1 - y/x = 1 - \sqrt{T_c}(\sqrt{T_h} + \sqrt{T_c})/[\sqrt{T_h}(\sqrt{T_h} + \sqrt{T_c})] = 1 - \sqrt{T_c/T_h}$ ; $P_{\max} = \tfrac K2(\sqrt{T_h} - \sqrt{T_c})^2$ . (d) $\eta^\ast = 1 - \sqrt{0.375} = 0.39$ , against $\eta_C = 0.625$ : the real $0.40$ of Exercise 24.1 is an engine built for power, not for efficiency.

Exercise 24.11 ★★★

Absorption refrigerator. A machine with no moving parts takes the heat $Q_h$ from a burner at $T_h$ , the heat $Q_c$ from a cold chamber at $T_c$ , and rejects heat to the room at $T_0$ ( $T_c < T_0 < T_h$ ), with no work. Show that $Q_c/Q_h \leq \bigl(1 - T_0/T_h\bigr)\,T_c/(T_0 - T_c)$ , interpret the two factors, and compute the bound for $T_h = 400\,\mathrm{K}$ , $T_0 = 300\,\mathrm{K}$ , $T_c = 275\,\mathrm{K}$ .

Solution

Solution of Exercise 24.11.

Cycle: $Q_h + Q_c + Q_0 = 0$ and $Q_h/T_h + Q_c/T_c + Q_0/T_0 \leq 0$ . Eliminate $Q_0 = -(Q_h + Q_c)$ : $Q_h(1/T_h - 1/T_0) + Q_c(1/T_c - 1/T_0) \leq 0$ , i.e. $Q_c(T_0 - T_c)/(T_cT_0) \leq Q_h(T_h - T_0)/(T_hT_0)$ , whence $Q_c/Q_h \leq (1 - T_0/T_h)\,T_c/(T_0 - T_c)$ . The first factor is the efficiency of a Carnot engine between burner and room, the second the COP of a Carnot refrigerator between chamber and room: the bound is what one gets by coupling the two ideally. Numbers: $0.25 \times 11 = 2.75$ (real absorption fridges: about $0.6$ ).

Exercise 24.12 ★★★

Heating with electricity, three ways. Electricity is made in a thermal power station of efficiency $0.40$ . Per joule of fuel heat, how much heat reaches a house (a) through a resistor, (b) through a heat pump of $e_p = 3.5$ , (c) if the fuel is burned directly in a boiler of efficiency $0.90$ ? (d) The ideal: a reversible engine between $800\,\mathrm{K}$ and $293\,\mathrm{K}$ driving a reversible heat pump between $273\,\mathrm{K}$ and $293\,\mathrm{K}$ — how much heat per joule of fuel, and why is the answer greater than one without contradicting anything?

Solution

Solution of Exercise 24.12.

(a) $0.40\,\mathrm{J}$ . (b) $0.40 \times 3.5 = 1.4\,\mathrm{J}$ . (c) $0.90\,\mathrm{J}$ . (d) Engine $\eta = 1 - 293/800 = 0.634$ , pump $e_p = 293/20 = 14.65$ : $0.634 \times 14.65 = 9.3\,\mathrm{J}$ (and the engine’s own $0.37\,\mathrm{J}$ rejected at $293\,\mathrm{K}$ could warm the house too: $9.7\,\mathrm{J}$ ). More than one joule because the rest is lifted from the outdoor air: energy is conserved, and the second law only asks that entropy not be destroyed, which a reversible chain respects exactly.

An air-source heat pump in winter: it takes heat from cold outdoor air and delivers three to four times the electrical energy it consumes to the house — the weekend problem.

24.8 Problem: The house heat pump

Problem 24.1

Weekend problem — heating a house through a winter: the bill with resistors, with gas, with an ideal heat pump and with a real one, and where the real one’s electricity goes

A house loses heat at the rate $K(T_{\mathrm{in}} - T_{\mathrm{out}})$ with $K = 250\,\mathrm{W}/\mathrm{K}$ ; it is kept at $T_{\mathrm{in}} = 20{}^{\circ}\mathrm{C}$ . Design outdoor temperature $-5{}^{\circ}\mathrm{C}$ ; heating season: $200$ days with a mean outdoor temperature of $7{}^{\circ}\mathrm{C}$ . Prices: electricity 0.25 € per $\mathrm{kW}\mathrm{h}$ , gas 0.10 € per $\mathrm{kW}\mathrm{h}$ ; $1\,\mathrm{kW}\mathrm{h} = 3.6\,\mathrm{MJ}$ .

Part I — The house and the old ways.

Heating power needed at the design temperature.
Energy needed over the season, in joules and in $\mathrm{kW}\mathrm{h}$ .
Cost of the season with electric resistors.
Cost with a gas boiler of efficiency $0.90$ .
The electricity comes from a thermal power station of efficiency $0.40$ . Fuel energy consumed, per season, by the resistors and by the boiler; comment on “electric heating is thermodynamically wasteful”.

Part II — The ideal heat pump.

Define $e_p$ and prove, from the two laws over one cycle, that $e_p \leq T_{\mathrm{in}}/(T_{\mathrm{in}} - T_{\mathrm{out}})$ .
Ideal $e_p$ at the design temperature, and the electrical power then needed.
Ideal $e_p$ at the season’s mean temperature, and (taking that value for the whole season) the seasonal electricity.
Explain physically why $e_p$ falls when the outdoor temperature falls, and why this is the worst possible behaviour for a heating device.
At the design point the ideal pump draws $0.53\,\mathrm{kW}$ and delivers $6.25\,\mathrm{kW}$ : where do the other $5.7\,\mathrm{kW}$ come from, and why does cooling the cold outdoor air to heat a warm house not violate the second law?

Part III — The real machine. The pump runs a vapour-compression cycle. The refrigerant evaporates at $T_{\mathrm{ev}} = T_{\mathrm{out}} - 8\,\mathrm{K}$ and condenses at $T_{\mathrm{cd}} = 313\,\mathrm{K}$ (the water of a floor-heating circuit runs at $35{}^{\circ}\mathrm{C}$ ); the machine achieves $0.55$ of the Carnot value computed between $T_{\mathrm{ev}}$ and $T_{\mathrm{cd}}$ . Latent heat of the refrigerant: $L = 200\,\mathrm{kJ}/\mathrm{kg}$ .

Name the four components of the cycle, say in which the fluid receives $Q_c$ , gives $|Q_h|$ and receives $W$ , and why the evaporator must be colder than the outdoor air and the condenser hotter than the house.
At the design temperature: $T_{\mathrm{ev}}$ , and the Carnot $e_p$ between $T_{\mathrm{ev}}$ and $T_{\mathrm{cd}}$ .
Real $e_p$ , electrical power drawn, and heat taken from the outdoor air, at the design temperature.
Mass flow rate of refrigerant through the evaporator.
Same questions (real $e_p$ , electrical power) at the mean temperature of $7{}^{\circ}\mathrm{C}$ , where the house loses $3.25\,\mathrm{kW}$ .
Give $e_p$ as a function of $T_{\mathrm{out}}$ ; seasonal electricity and cost, taking the mean-temperature $e_p$ for the season. Compare with Part I.
Fuel energy consumed at the power station for the heat-pump season; compare with the resistors and the boiler.
A cold snap at $-15{}^{\circ}\mathrm{C}$ : heat demand, real $e_p$ , electrical power if the pump could deliver it. The compressor’s capacity actually falls with $T_{\mathrm{ev}}$ and the pump can only deliver $3.8\,\mathrm{kW}$ there: backup resistor power needed, and the meaning of “sizing a heat pump”.
Had the house old radiators needing water at $60{}^{\circ}\mathrm{C}$ ( $T_{\mathrm{cd}} = 338\,\mathrm{K}$ ): real $e_p$ and electrical power at the design temperature; conclude on the choice of emitters.

Part IV — Where the electricity goes.

Show that for a heat pump delivering $|Q_h|$ to the house at $T_{\mathrm{in}}$ from outdoor air at $T_{\mathrm{out}}$ , $W = |Q_h|(1 - T_{\mathrm{out}}/T_{\mathrm{in}}) + T_{\mathrm{out}}S_{\mathrm{created}}$ .
At the design point, entropy created per second by the heat transfer across the condenser (from $313\,\mathrm{K}$ to the house) and across the evaporator (from the outdoor air to $260\,\mathrm{K}$ ).
Electrical power these two temperature gaps cost; fraction of the $1.92\,\mathrm{kW}$ drawn.
Total entropy created per second by the real machine (compare its $1.92\,\mathrm{kW}$ with the ideal $0.53\,\mathrm{kW}$ ); share of the exchangers; name the other sources.
Halve the two gaps ( $4\,\mathrm{K}$ and $10\,\mathrm{K}$ ): entropy created in the exchangers and electrical power saved; what does halving a gap cost the designer?
Sum up in three numbers: kilowatt-hours of heat per kilowatt-hour of electricity over the season; seasonal cost against gas; and, with electricity at $60\,\mathrm{g}$ of CO $_2$ per $\mathrm{kW}\mathrm{h}$ and gas at $200\,\mathrm{g}$ , the carbon emitted by each route.

Solution

Solution of Problem 24.1.

1. $P = 250 \times 25 = 6.25\,\mathrm{kW}$ .

2. Mean loss $250 \times 13 = 3.25\,\mathrm{kW}$ for $200 \times 86400 = 1.73 \times 10^{7}\,\mathrm{s}$ : $E = 5.6 \times 10^{10}\,\mathrm{J} = 15\,600\,\mathrm{kW}\mathrm{h}$ .

3. $15600 \times 0.25 = 3900$ €.

4. Gas $15600/0.90 = 17\,300\,\mathrm{kW}\mathrm{h}$ : $1730$ €.

5. Resistors: $15600/0.40 = 39\,000\,\mathrm{kW}\mathrm{h}$ of fuel; boiler $17\,300\,\mathrm{kW}\mathrm{h}$ : $2.25$ times less. Electricity is work, the form that can do anything; turning it into $20{}^{\circ}\mathrm{C}$ heat in a resistor discards the $60\%$ of the fuel already lost at the plant and the possibility of pumping heat.

6. Over a cycle $W + Q_h + Q_c = 0$ and $Q_h/T_{\mathrm{in}} + Q_c/T_{\mathrm{out}} \leq 0$ with $Q_h < 0$ , $Q_c > 0$ : $Q_c \leq |Q_h|T_{\mathrm{out}}/T_{\mathrm{in}}$ , so $W = |Q_h| - Q_c \geq |Q_h|(1 - T_{\mathrm{out}}/T_{\mathrm{in}})$ and $e_p = |Q_h|/W \leq T_{\mathrm{in}}/(T_{\mathrm{in}} - T_{\mathrm{out}})$ .

7. $293/25 = 11.7$ ; $6.25/11.7 = 0.53\,\mathrm{kW}$ .

8. $293/13 = 22.5$ ; $15600/22.5 = 690\,\mathrm{kW}\mathrm{h}$ (about $170$ €).

9. The work per joule lifted is $(T_{\mathrm{in}} - T_{\mathrm{out}})/ T_{\mathrm{in}}$ , the “height” of the lift; the colder it is outside, the higher each joule must be pumped — and the more joules are needed, since the loss is also proportional to $T_{\mathrm{in}} - T_{\mathrm{out}}$ : the electrical power grows as the square of the temperature difference, worst exactly when heating matters most.

10. From the outdoor air, which is cooled: $Q_c = 6.25 - 0.53 = 5.7\,\mathrm{kW}$ . The second law forbids heat flowing spontaneously from cold to hot; here work pays the lift. Check: entropy taken from the air $5.72/268 = 21.3\,\mathrm{W}/\mathrm{K}$ , entropy given to the house $6.25/293 = 21.3\,\mathrm{W}/\mathrm{K}$ — nothing destroyed, nothing created: the ideal pump.

11. Compressor (receives $W$ ), condenser (gives $|Q_h|$ ), expansion valve (neither), evaporator (receives $Q_c$ ). Heat only flows down a temperature slope: the refrigerant must be colder than the air to take heat from it and hotter than the house water to give heat to it.

12. $T_{\mathrm{ev}} = 268 - 8 = 260\,\mathrm{K}$ ; $e_{p,C} = 313/(313 - 260) = 5.9$ .

13. $e_p = 0.55 \times 5.9 = 3.25$ ; $W = 6.25/3.25 = 1.92\,\mathrm{kW}$ ; $Q_c = 6.25 - 1.92 = 4.33\,\mathrm{kW}$ from the air.

14. $\dot m = 4330/200000 = 22\,\mathrm{g}/\mathrm{s}$ , $1.3\,\mathrm{kg}$ a minute.

15. $T_{\mathrm{ev}} = 272\,\mathrm{K}$ , $313/41 = 7.6$ , $e_p = 4.2$ ; $W = 3.25/4.2 = 0.77\,\mathrm{kW}$ .

16. $e_p = 0.55 \times 313/(321 - T_{\mathrm{out}}) = 172/(321 - T_{\mathrm{out}})$ ( $T_{\mathrm{out}}$ in kelvin). Season: $15600/4.2 = 3700\,\mathrm{kW}\mathrm{h}$ , $930$ € — against $3900$ with resistors and $1730$ with gas.

17. $3700/0.40 = 9300\,\mathrm{kW}\mathrm{h}$ of fuel: $46\%$ less than the boiler and four times less than the resistors, even through a $40\%$ power station.

18. Demand $250 \times 35 = 8.75\,\mathrm{kW}$ ; $T_{\mathrm{ev}} = 250\,\mathrm{K}$ , $313/63 = 5.0$ , $e_p = 2.7$ , $W = 3.2\,\mathrm{kW}$ if it could. Capacity $3.8\,\mathrm{kW}$ : backup $8.75 - 3.8 \approx 5\,\mathrm{kW}$ of resistors. A pump sized for the coldest hour would be oversized — and cycling inefficiently — all winter; one sizes for the design temperature and lets a cheap resistor cover the rare extreme.

19. $338/(338 - 260) = 4.33$ , $e_p = 2.4$ , $W = 6.25/2.4 = 2.6\,\mathrm{kW}$ : $36\%$ more electricity than with the floor at $35{}^{\circ}\mathrm{C}$ . Low-temperature emitters are half the installation.

20. Entropy over a cycle: $Q_h/T_{\mathrm{in}} + Q_c/T_{\mathrm{out}} + S_{\mathrm{created}} = 0$ gives $Q_c = T_{\mathrm{out}}|Q_h|/T_{\mathrm{in}} - T_{\mathrm{out}}S_{\mathrm{created}}$ ; insert in $W = |Q_h| - Q_c$ .

21. Condenser: $6250\,(1/293 - 1/313) = 1.36\,\mathrm{W}/\mathrm{K}$ ; evaporator: $4330\,(1/260 - 1/268) = 0.50\,\mathrm{W}/\mathrm{K}$ .

22. $268 \times 1.86 = 0.50\,\mathrm{kW}$ : a quarter of the $1.92\,\mathrm{kW}$ .

23. $(1.92 - 0.53)/268 = 5.2\,\mathrm{W}/\mathrm{K}$ ; the exchangers make $36\%$ ; the rest comes from the compressor (friction, non-isentropic compression, motor losses), the throttling in the valve (a free expansion), pressure drops, the fans.

24. $T_{\mathrm{cd}} = 303\,\mathrm{K}$ , $T_{\mathrm{ev}} = 264\,\mathrm{K}$ : condenser $6250\,(1/293 - 1/303) = 0.70\,\mathrm{W}/\mathrm{K}$ , evaporator $4330\,(1/264 - 1/268) = 0.25\,\mathrm{W}/\mathrm{K}$ : $0.95\,\mathrm{W}/\mathrm{K}$ instead of $1.86$ , i.e. $268 \times 0.91 \approx 0.25\,\mathrm{kW}$ saved ( $13\%$ ; the whole machine, rated at $0.55$ of a better Carnot value, would gain more). Halving a gap means doubling the exchange surface: bigger, dearer exchangers and more fan power.

25. $15600/3700 = 4.2$ kilowatt-hours of heat per kilowatt-hour of electricity; $930$ € against $1730$ for gas; carbon: $3700 \times 0.060 = 220\,\mathrm{kg}$ against $17300 \times 0.20 = 3500\,\mathrm{kg}$ — fifteen times less.

Terms defined in this chapter

See all 393 terms in the glossary