Worked examples · Sheet 2

Carnot engines and limits on efficiency

Theory used: Efficiency of the Carnot cycle Carnot's theorem Balance of an engine between two reservoirs Carnot refrigerator and heat pump The TS plane

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A Carnot engine operates between a reservoir at $527^{\circ} C$ and a reservoir at $27^{\circ} C$ and in each cycle absorbs $4.0 kJ$ from the hot reservoir. Compute the efficiency, the work produced in one cycle and the heat released to the cold reservoir.

T_{1} = (527 + 273) K = 800 K, T_{2} = (27 + 273) K = 300 K, Q_{1} = 4.0 kJ

η_{C} = 1 - \frac{T _{2}}{T _{1}}, L = η_{C} Q_{1}, ∣ Q_{2} ∣ = Q_{1} - L = \frac{T _{2}}{T _{1}} Q_{1}

η_{C} = 1 - \frac{300 K}{800 K} = 0.625

L = 0.625 \cdot 4.0 kJ = 2.5 kJ, ∣ Q_{2} ∣ = \frac{300}{800} \cdot 4.0 kJ = 1.5 kJ

The efficiency of the Carnot cycle depends only on the temperatures of the reservoirs, which must be expressed in kelvins, because the formula contains their ratio; with the temperatures in degrees Celsius the ratio would be different and the result wrong. The work is the efficiency multiplied by the heat absorbed. The heat released is obtained from the balance of the cycle $L = Q_{1} - ∣ Q_{2} ∣$ or, independently, from the relation $\frac{∣ Q _{2} ∣}{Q _{1}} = \frac{T _{2}}{T _{1}}$ ; the two computations give the same value, and this provides a check of the result.

Exercise 2

An inventor claims to have built an engine that operates between a reservoir at $600 K$ and one at $350 K$ and that in each cycle absorbs $8.0 kJ$ and produces $3.6 kJ$ of work. Determine whether the engine can exist and compute the maximum work per cycle compatible with the reservoirs and with the heat absorbed.

T_{1} = 600 K, T_{2} = 350 K, Q_{1} = 8.0 kJ, L = 3.6 kJ

η = \frac{L}{Q _{1}}, η_{C} = 1 - \frac{T _{2}}{T _{1}}, \frac{Q _{1}}{T _{1}} + \frac{Q _{2}}{T _{2}} \leq 0

η = \frac{3.6}{8.0} = 0.45, η_{C} = 1 - \frac{350}{600} \approx 0.417

Q_{2} = L - Q_{1} = - 4.4 kJ, \frac{8000 J}{600 K} - \frac{4400 J}{350 K} \approx 13.33 J/K - 12.57 J/K = 0.76 J/K > 0

L_{m a x} = η_{C} Q_{1} = \frac{5}{12} \cdot 8.0 kJ \approx 3.3 kJ

By Carnot’s theorem no engine between two reservoirs can have an efficiency greater than that of a reversible engine, that is, $η_{C} \approx 0.417$ . The claimed efficiency is $0.45$ and exceeds this limit, therefore the engine cannot exist. The same conclusion is reached with the balance of an engine between two reservoirs. The heat released is $4.4 kJ$ , and the sum $\frac{Q _{1}}{T _{1}} + \frac{Q _{2}}{T _{2}}$ turns out to be positive, while the second law requires it to be less than or equal to zero. The maximum work is obtained with a reversible engine that absorbs the same heat, and it is about $3.3 kJ$ .

Exercise 3

A kitchen freezer keeps its interior at $- 18^{\circ} C$ , while the kitchen is at $25^{\circ} C$ . To freeze a certain amount of food, $1.2 \cdot 1 0^{5} J$ must be extracted from it. Compute the minimum work required and the corresponding heat released to the kitchen.

T_{2} = 255 K, T_{1} = 298 K, Q_{2} = 1.2 \cdot 1 0^{5} J

ε_{C} = \frac{T _{2}}{T _{1} - T _{2}}, ∣ L ∣_{m i n} = \frac{Q _{2}}{ε _{C}} = Q_{2} \frac{T _{1} - T _{2}}{T _{2}}, ∣ Q_{1} ∣ = Q_{2} + ∣ L ∣_{m i n}

ε_{C} = \frac{255}{43} \approx 5.93

∣ L ∣_{m i n} = 1.2 \cdot 1 0^{5} J \cdot \frac{43}{255} \approx 2.0 \cdot 1 0^{4} J

∣ Q_{1} ∣ \approx 1.2 \cdot 1 0^{5} J + 0.20 \cdot 1 0^{5} J = 1.4 \cdot 1 0^{5} J

The freezer is a refrigerator between the cold reservoir, the interior at $255 K$ , and the hot reservoir, the kitchen at $298 K$ . By Carnot’s theorem the maximum coefficient of performance is that of the Carnot refrigerator, $\frac{T _{2}}{T _{1} - T _{2}}$ ; for the same heat extracted, the maximum coefficient of performance corresponds to the minimum work. Therefore the work spent is at least $2.0 \cdot 1 0^{4} J$ . The heat released to the kitchen is the sum of the heat extracted from the food and of the work received. A real freezer has a coefficient of performance smaller than $ε_{C}$ , and therefore spends more work.

Exercise 4

Two Carnot engines operate in series: the first between a reservoir at $T_{1} = 900 K$ and an intermediate reservoir at temperature $T$ , the second between the reservoir at temperature $T$ and a reservoir at $T_{3} = 300 K$ , and in each cycle the second absorbs exactly the heat released by the first. The first engine absorbs $Q_{1} = 1800 J$ per cycle. Determine $T$ so that the two engines produce the same work, compute the two works and the overall efficiency.

L_{1} = Q_{1} (1 - \frac{T}{T _{1}}), Q = \frac{T}{T _{1}} Q_{1}, L_{2} = Q (1 - \frac{T _{3}}{T}) = Q_{1} \frac{T - T _{3}}{T _{1}}

L_{1} = L_{2} ⟺ T_{1} - T = T - T_{3} ⟺ T = \frac{T _{1} + T _{3}}{2}

T = \frac{900 K + 300 K}{2} = 600 K

L_{1} = 1800 J \cdot \frac{300}{900} = 600 J, Q = 1200 J, L_{2} = 1200 J \cdot \frac{300}{600} = 600 J

η = \frac{L _{1} + L _{2}}{Q _{1}} = \frac{1200 J}{1800 J} = \frac{2}{3} = 1 - \frac{T _{3}}{T _{1}}

The two engines in the

(S, T)

plane, with

T

in kelvins and

S

J/K

up to a constant: the rectangles have the same base

Δ S = \frac{Q _{1}}{T _{1}} = 2.0 J/K

and the same height

300 K

, therefore equal areas, that is, equal works

Each engine is reversible, therefore for each one $\frac{∣ Q _{out} ∣}{Q _{in}} = \frac{T _{cold}}{T _{hot}}$ holds. The first releases to the intermediate reservoir the heat $Q = \frac{T}{T _{1}} Q_{1}$ , which is the heat absorbed by the second; substituting into the work of the second engine, $L_{2}$ turns out to be proportional to $T - T_{3}$ , while $L_{1}$ is proportional to $T_{1} - T$ . The works are equal when the two temperature drops are equal, that is, for $T = 600 K$ . The intermediate reservoir receives and releases the same heat, therefore the pair of engines is equivalent to a reversible engine between $900 K$ and $300 K$ , and the overall efficiency is the Carnot efficiency between the extreme temperatures, $\frac{2}{3}$ . In the $(S, T)$ plane the result can be read directly, because the two engines exchange heat with the same change in entropy $\frac{1800}{900} = \frac{1200}{600} = 2.0 J/K$ , and their cycles are rectangles with the same base.