21.6.1 Designing a Carnot Engine

The Carnot cycle is an ideal-gas cycle that consists of the two adiabatic processes $(Q = 0)$ and two isothermal processes $(Δ E_{th} = 0)$ shown in FIGURE 21.20. These are the two types of processes allowed in a perfectly reversible gas engine. As a Carnot cycle operates,

The gas is isothermally compressed while in thermal contact with the cold reservoir at temperature $T_{C}$ . Heat energy $Q_{C} = | Q_{12} |$ is removed from the gas as it is compressed in order to keep the temperature constant. The compression must take place extremely slowly because there can be only an infinitesimal temperature difference between the gas and the reservoir.
The gas is adiabatically compressed while thermally isolated from the environment. This compression increases the gas temperature until it matches temperature $T_{H}$ of the hot reservoir. No heat is transferred during this process.
After reaching maximum compression, the gas expands isothermally at temperature $T_{H}$ . Heat $Q_{H} = Q_{34}$ is transferred from the hot reservoir into the gas as it expands in order to keep the temperature constant.
Finally, the gas expands adiabatically, with $Q = 0$ , until the temperature decreases back to $T_{C}$ .

Work is done in all four processes of the Carnot cycle, but heat is transferred only during the two isothermal processes.

The figure illustrates the Pressure versus Volume graph of the Carnot engine. The graph represents a cyclic process.

FIGURE 21.20 Full Alternative Text

The thermal efficiency of any heat engine is

η = \frac{W_{out}}{Q_{H}} = 1 - \frac{Q_{C}}{Q_{H}}

We can determine $η_{Carnot}$ by finding the heat transfer in the two isothermal processes.

Process $1 \to 2 :$ Table 21.1 gives us the heat transfer in an isothermal process at temperature $T_{C} :$

Q_{12} = (W_{s})_{12} = n R T_{C} ln (\frac{V_{2}}{V_{1}}) = - n R T_{C} ln (\frac{V_{1}}{V_{2}})

(21.20)

$V_{1} > V_{2}$ , so the logarithm on the right is positive. $Q_{12}$ is negative because heat is transferred out of the system, but $Q_{C}$ is simply the amount of heat transferred to the cold reservoir:

Q_{C} = | Q_{12} | = n R T_{C} ln (\frac{V_{1}}{V_{2}})

(21.21)

Process $3 \to 4 :$ Similarly, the heat transferred in the isothermal expansion at temperature $T_{H}$ is

Q_{H} = Q_{34} = (W_{s})_{34} = n R T_{H} ln (\frac{V_{4}}{V_{3}})

(21.22)

Thus the thermal efficiency of the Carnot cycle is

η_{Carnot} = 1 - \frac{Q_{C}}{Q_{H}} = 1 - \frac{T_{C} ln (V_{1} / V_{2})}{T_{H} ln (V_{4} / V_{3})}

(21.23)

We can simplify this expression. During the two adiabatic processes,

T_{C} V_{2}^{γ - 1} = T_{H} V_{3}^{γ - 1} and T_{C} V_{1}^{γ - 1} = T_{H} V_{4}^{γ - 1}

(21.24)

An algebraic rearrangement gives

V_{2} = V_{3} (\frac{T_{H}}{T_{C}})^{1 / (γ - 1)} and V_{1} = V_{4} (\frac{T_{H}}{T_{C}})^{1 / (γ - 1)}

(21.25)

from which it follows that

\frac{V_{1}}{V_{2}} = \frac{V_{4}}{V_{3}}

(21.26)

Consequently, the two logarithms in Equation 21.23 cancel and we’re left with the result that the thermal efficiency of a Carnot engine operating between a hot reservoir at temperature $T_{H}$ and a cold reservoir at temperature $T_{C}$ is

η_{Carnot} = 1 - \frac{T_{C}}{T_{H}} (Carnot thermal efficiency)

(21.27)

This remarkably simple result, an efficiency that depends only on the ratio of the temperatures of the hot and cold reservoirs, is Carnot’s legacy to thermodynamics.

NOTE Temperatures $T_{H}$ and $T_{C}$ are absolute temperatures.

EXAMPLE 21.4 A Carnot engine

A Carnot engine is cooled by water at $T_{C} = 10 ° C$ . What temperature must be maintained in the hot reservoir of the engine to have a thermal efficiency of 70%?

MODEL The efficiency of a Carnot engine depends only on the temperatures of the hot and cold reservoirs.

SOLVE The thermal efficiency $η_{Carnot} = 1 - T_{C} / T_{H}$ can be rearranged to give

T_{H} = \frac{T_{C}}{1 - η_{Carnot}} = 943 K = 670 ° C

where we used $T_{C} = 283 K$ .

REVIEW A “real” engine would need a higher temperature than this to provide 70% efficiency because no real engine will match the Carnot efficiency.

EXAMPLE 21.5 A real engine

The heat engine of Example 21.2 had a highest temperature of 2700 K, a lowest temperature of 300 K, and a thermal efficiency of 15%. What is the efficiency of a Carnot engine operating between these two temperatures?

SOLVE The Carnot efficiency is

η_{Carnot} = 1 - \frac{T_{C}}{T_{H}} = 1 - \frac{300 K}{2700 K} = 0.89 = 89 %

REVIEW The thermodynamic cycle used in Example 21.2 doesn’t come anywhere close to the Carnot efficiency.