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A cycle followed by an engine is shown in the figure.

A to B: volume constant

B to C: adiabatic

C to D: volume constant

D to A: adiabatic

$V_C = V_D = 2V_A = 2V_B$

a) in which part of the cycle heat is supplied to the engine from outside?

b) in which part of the cycle heat is being given to the surrounding by the engine?

c) what is the work done by the engine in one cycle in terms of $P_A, P_B, V_A$ ?

d) what is the efficiency of the engine?

Answers (1)

(a) Heat is supplied to the engine in part AB where dV=0

dW=PdV=P(0)

dW=0

dQ=dU+dW(First law of thermodynamics)

dQ=dU

$P=\frac{nRT}{V}$

V=constant

$p\propto T$

(b) Heat is given out by the system in a part CD where dV=0 and pressure and temperature decreases

(c ) WD by system= $\int_{B}^{A}PdV+\int_{B}^{C}PdV+\int_{C}^D{}PdV+\int_{D}^{A}PdV$

$\int_{B}^{A}PdV=0 \text{and}\int_{C}^D{}PdV=0$

For adiabatic change

$PV^{\gamma}=K$

$P=\frac{K}{V^{\gamma}}$

$\int_{B}^{C}PdV=\int_{A}^{B}\frac{K}{V^{\gamma }}dV=K\left [ \frac{V^{-\gamma +1}}{1-\gamma } \right ]^{V_{C}}_{V_{B}}$

$\int_{B}^{C}PdV=\frac{K}{1-\gamma}\left [ V_{C}^{1-\gamma}-V_{B}^{1-\gamma} \right ]$

$\int_{D}^{A}PdV=\frac{K}{1-\gamma}\left [ V_{A}^{1-\gamma}-V_{D}^{1-\gamma} \right ]$

$\int_{D}^{A}PdV=\frac{P_{A}V_{A}^{\gamma}V_{A}^{1-\gamma}-P_{D}V_{D}^{\gamma}V_{D}^{1-\gamma}}{1-\gamma}$

$\int_{D}^{A}PdV=\frac{P_{A}V_{A}-P_{D}V_{D}}{1-\gamma}$

$\int_{D}^{A}PdV=\frac{P_{A}V_{A}-P_{D}V_{D}}{1-\gamma}$

$\int_{B}^{C}PdV=\frac{P_{C}V_{C}-P_{B}V_{B}}{1-\gamma}$

Total $WD=\frac{[P_{C}V_{C}-P_{B}V_{B}+P_{A}V_{A}-P_{D}V_{D}]}{1-\gamma}$

For Adiabatic change= $P_{B}V_{B}^\gamma =P_{C}V_{C}^{\gamma}$

$P_C=\frac{P_BV_B^{\gamma}}{V_C^\gamma}=P_B\left (\frac{V_B}{V_C} \right )^\gamma=P_B\left (\frac{V_B}{2V_B} \right )^\gamma=\frac{P_B}{2^\gamma}$
$P_{D}=\frac{P_{A}}{2^\gamma }$

Net WD= $\frac{1}{1-\gamma} [P_BV_C2-\gamma-P_BV_B+P_AV_A-P_AV_D2-\gamma ]$

$=\frac{1}{1-\gamma}[P_B2V_B2^{-\gamma}-P_BV_B+P_AV_A-P_A2V_A2^{-\gamma}]$

$=\frac{1}{1-\gamma}[-P_{B}V_{B}[-2^{1-\gamma }+1]P_{A}V_{A}[-2^{1-\gamma }+1]]$

$=\frac{1}{1-\gamma}[P_{A}V_{A}-P_{B}V_{B}][1-2^{1-\gamma}]$

$2V_B=2V_A\Rightarrow V_B=V_A=\frac{1}{1-\frac{5}{3}}\left [ -2^{1-\frac{5}{3}}+1 \right ][-P_B+P_A]V_A$

$=\frac{3}{2}\left [ 1-\left ( \frac{1}{2} \right )^{\frac{2}{3}} \right ][P_B-P_A]V_A$

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