Composite Cycle Efficiency Analysis
We need to determine the efficiency of the composite cycle $1 \to 2 \to 3 \to 4 \to 1$, given the efficiencies of two component cycles:
- Cycle I ($1 \to 2 \to 3 \to 1$): Efficiency $\eta_1$
- Cycle II ($1 \to 3 \to 4 \to 1$): Efficiency $\eta_2$
Detailed Breakdown
Let’s define the heat and work interactions for the two constituent cycles.
Cycle I ($1 \to 2 \to 3 \to 1$)
- Process $1 \to 2$ (Straight line): Gas expands, pressure rises. Heat is absorbed ($Q_{12} > 0$).
- Process $2 \to 3$ (Adiabatic): $Q = 0$.
- Process $3 \to 1$ (Isotherm): Compression. Heat is rejected ($Q_{31} < 0$).
The efficiency $\eta_1$ is given by:
$$\eta_1 = \frac{W_1}{Q_{in}} = \frac{Q_{12} – |Q_{31}|}{Q_{12}} = 1 – \frac{|Q_{31}|}{Q_{12}}$$From this, we can deduce the relationship between the rejected and absorbed heat:
$$|Q_{31}| = Q_{12}(1 – \eta_1) \quad \dots \text{(Eq. 1)}$$Cycle II ($1 \to 3 \to 4 \to 1$)
Note: The cycle effectively runs as $1 \to 3 \to 4 \to 1$.
- Process $1 \to 3$ (Isotherm): Expansion. This is the reverse of process $3 \to 1$ in Cycle I. Heat is absorbed. Since the path is identical but reversed, the heat absorbed here ($Q_{13}$) is equal in magnitude to the heat rejected in Cycle I. $$Q_{13} = |Q_{31}|$$
- Process $3 \to 4$ (Isobar): Compression. Heat is rejected ($Q_{34} < 0$).
- Process $4 \to 1$ (Adiabatic): $Q = 0$.
The only heat input in Cycle II is during the isothermal expansion $1 \to 3$. Thus, the efficiency $\eta_2$ is:
$$\eta_2 = \frac{W_2}{Q_{13}} \implies W_2 = \eta_2 Q_{13}$$Composite Cycle ($1 \to 2 \to 3 \to 4 \to 1$)
We now combine the cycles. The total work done is the sum of the areas enclosed by both loops:
$$W_{total} = W_1 + W_2$$The only heat supplied to the entire composite engine comes from the external heater during process $1 \to 2$. (Process $2 \to 3$ and $4 \to 1$ are adiabatic, and $3 \to 4$ rejects heat). The internal heat exchange along the isotherm $1 \leftrightarrow 3$ is internal to the system boundary and cancels out.
$$Q_{in, total} = Q_{12}$$The efficiency of the composite cycle is:
$$\eta = \frac{W_{total}}{Q_{in, total}} = \frac{W_1 + W_2}{Q_{12}}$$We know $W_1 = \eta_1 Q_{12}$.
We also know $W_2 = \eta_2 Q_{13}$. Using Eq. 1 ($Q_{13} = |Q_{31}| = Q_{12}(1 – \eta_1)$), we get:
$$W_2 = \eta_2 [Q_{12}(1 – \eta_1)]$$Substituting these into the efficiency equation:
$$\eta = \frac{\eta_1 Q_{12} + \eta_2 Q_{12}(1 – \eta_1)}{Q_{12}}$$Canceling $Q_{12}$ from the numerator and denominator:
$$\eta = \eta_1 + \eta_2(1 – \eta_1)$$ $$\eta = \eta_1 + \eta_2 – \eta_1\eta_2$$