21.3.2 A Strategy for Heat-Engine Problems

The engine of Example 21.1 was not a realistic heat engine, but it did illustrate the kinds of reasoning and computations involved in the analysis of a heat engine.

PROBLEM-SOLVING STRATEGY 21.1
Heat-engine problems

MODEL Identify each process in the cycle.

VISUALIZE Draw the pV diagram of the cycle.

SOLVE There are several steps in the mathematical analysis.

Use the ideal-gas law to complete your knowledge of n, p, V, and T at one point in the cycle.
Use the ideal-gas law and equations for specific gas processes to determine p, V, and T at the beginning and end of each process.
Calculate Q, $W_{s}$ , and $Δ E_{th}$ for each process.
Find $W_{out}$ by adding $W_{s}$ for each process in the cycle. If the geometry is simple, you can confirm this value by finding the area enclosed within the pV curve.
Add just the positive values of Q to find $Q_{H}$ .
Verify that $(Δ E_{th})_{net} = 0$ . This is a self-consistency check to verify that you haven’t made any mistakes.
Calculate the thermal efficiency $η$ and any other quantities you need to complete the solution.

REVIEW Is $(Δ E_{th})_{net} = 0 ?$ Do all the signs of $W_{s}$ and Q make sense? Does $η$ have a reasonable value? Have you answered the question?

EXAMPLE 21.2 Analyzing a heat engine II

A heat engine with a diatomic gas as the working substance uses the closed cycle shown in FIGURE 21.13. How much work does this engine do per cycle, and what is its thermal efficiency?

The figure illustrates a pressure versus volume graph for a heat engine. The graph represents a cyclic process.

FIGURE 21.13 Full Alternative Text

MODEL Processes $1 \to 2$ and $3 \to 4$ are isobaric. Processes $2 \to 3$ and $4 \to 1$ are isochoric.

VISUALIZE The pV diagram has already been drawn.

SOLVE We know the pressure, volume, and temperature at state 4. The number of moles of gas in the heat engine is

n = \frac{p_{4} V_{4}}{R T_{4}} = \frac{(101, 300 {Pa) (1.0 m}^{3})}{(8.31 J/mol K) (300 K)} = 40.6 mol

$p / T = constant$ during an isochoric process and $V / T = constant$ during an isobaric process. These allow us to find that $T_{1} = T_{3} = 900 K$ and $T_{2} = 2700 K$ . This completes our knowledge of the state variables at all four corners of the diagram.

Process $1 \to 2$ is an isobaric expansion, so

(W_{s})_{12} = p Δ V = (3.0 \times 101, 300 Pa) (2.0 m^{3}) = 6.08 \times 10^{5} J

where we converted the pressure to pascals. The heat transfer during an isobaric expansion is

\begin{matrix} Q_{12} & = & n C_{P} Δ T = (40.6 mol) (29.1 J/mol K) (1800 K) \\ = & 21.27 \times 10^{5} J \end{matrix}

where $C_{P} = \frac{7}{2} R$ for a diatomic gas. Then, using the first law,

Δ E_{12} = Q_{12} - (W_{s})_{12} = 15.19 \times 10^{5} J

Process $2 \to 3$ is an isochoric process, so $(W_{s})_{23} = 0$ and

Δ E_{23} = Q_{23} = n C_{V} Δ T = - 15.19 \times 10^{5} J

Notice that $Δ T$ is negative.

Process $3 \to 4$ is an isobaric compression. Now $Δ V$ is negative, so

(W_{s})_{34} = p Δ V = - 2.03 \times 10^{5} J

and

Q_{34} = n C_{P} Δ T = - 7.09 \times 10^{5} J

Then $Δ E_{th} = Q_{34} - (W_{s})_{34} = - 5.06 \times 10^{5} J$ .

Process $4 \to 1$ is another constant-volume process, so again $(W_{s})_{41} = 0$ and

Δ E_{41} = Q_{41} = n C_{V} Δ T = 5.06 \times 10^{5} J

The results of all four processes are shown in TABLE 21.3. The net results for $W_{out}, Q_{net}$ , and $(Δ E_{th})_{net}$ are found by summing the columns. As expected, $W_{out} = Q_{net}$ and $(Δ E_{th})_{net} = 0$ .

TABLE 21.3 Energy transfers in Example 21.2. All energies $\times 10^{5} J$

Process	$W_{s}$	Q	$Δ E_{t h}$
$1 \to 2$	6.08	21.27	15.19
$2 \to 3$	0	$- 15.19$	$- 15.19$
$3 \to 4$	$- 2.03$	$- 7.09$	$- 5.06$
$4 \to 1$	0	5.06	5.06
Net	4.05	4.05	0

The work done during one cycle is $W_{out} = 4.05 \times 10^{5} J$ . Heat enters the system from the hot reservoir during processes $1 \to 2$ and $4 \to 1$ , where Q is positive. Summing these gives $Q_{H} =_{} 26.33 \times 10^{5} J$ . Thus the thermal efficiency of this engine is

η = \frac{W_{out}}{Q_{H}} = \frac{4.05 \times 10^{5} J}{26.33 \times 10^{5} J} = 0.15 = 15 %

REVIEW The verification that $W_{out} = Q_{net}$ and $(Δ E_{th})_{net} = 0$ gives us great confidence that we didn’t make any calculational errors. This engine may not seem very efficient, but $η$ is quite typical of many real engines.

We noted in Example 21.1 that a heat engine’s hot-reservoir temperature $T_{H}$ must exceed the highest temperature reached by the system and the cold-reservoir temperature $T_{C}$ must be less than the coldest system temperature. Although we don’t know what the reservoirs are in Example 21.2, we can be sure that $T_{H} > 2700 K$ and $T_{C} < 300 K$ .