THERMAL CYU 32

Efficiencies of Cycles A and B

Let the cycles operate between pressures $P$ and $cP$.

1. Efficiency of Cycle A ($\eta_A$):

Cycle A operates between volumes $2V$ and $3V$.
Work done $W_A = (cP – P)(3V – 2V) = (c-1)PV$.
Heat Absorbed ($Q_{in,A}$) occurs during isochoric pressure rise and isobaric expansion:

• Isochoric ($2V, P \to cP$): $Q_1 = n C_v \Delta T = \frac{3}{2} (2V) (cP – P) = 3(c-1)PV$.
• Isobaric ($cP, 2V \to 3V$): $Q_2 = n C_p \Delta T = \frac{5}{2} (cP) (3V – 2V) = 2.5 c PV$.

$$Q_{in, A} = [3(c-1) + 2.5c] PV = (5.5c – 3) PV$$ $$\eta_A = \frac{W_A}{Q_{in, A}} = \frac{c-1}{5.5c – 3} = \frac{2(c-1)}{11c – 6}$$

2. Efficiency of Cycle B ($\eta_B$):

Cycle B operates between volumes $4V$ and $5V$.
Work done $W_B = (cP – P)(5V – 4V) = (c-1)PV$. (Same work as A).
Heat Absorbed ($Q_{in,B}$):

• Isochoric ($4V, P \to cP$): $Q_1 = \frac{3}{2} (4V) (cP – P) = 6(c-1)PV$.
• Isobaric ($cP, 4V \to 5V$): $Q_2 = \frac{5}{2} (cP) (5V – 4V) = 2.5 c PV$.

$$Q_{in, B} = [6(c-1) + 2.5c] PV = (8.5c – 6) PV$$ $$\eta_B = \frac{c-1}{8.5c – 6} = \frac{2(c-1)}{17c – 12}$$

3. Using the Ratio $k$:

Given $\frac{\eta_A}{\eta_B} = k$.

$$k = \frac{17c – 12}{11c – 6}$$

Solving for $c$ in terms of $k$: $11kc – 6k = 17c – 12 \Rightarrow c(11k – 17) = 6k – 12$.

$$c = \frac{6(k-2)}{11k – 17}$$

Substituting this back into the expression for $\eta_A$ simplifies significantly (after algebraic manipulation):

$$\eta_A = \frac{k-1}{3}$$

And consequently:

$$\eta_B = \frac{\eta_A}{k} = \frac{k-1}{3k}$$