Question

Two Carnot engines A and B are operated in series. The first one A receives heat at 800 K and rejects to a reservoir at temperature T. The second engine B receives the heat rejected by the first engine and in turn rejects to a heat reservoir at 300 K. Calculate the temperature T for the following situations:
(a) The outputs of the two engines are equal and
(b) The efficiencies of two engines are equal

Answer 1

Given:

$T_1=800K$

$T_3=400K$

To find:

Temperature T

(i) when output of two engines are equal

(ii) when efficiencies of two engines are equal.

Let the output of both engines be W.

Let the engine A take $Q_1$ heat as input at temperature $T_1$ and gives out heat $Q_2$ at temperature $T$ the second engine B receive $Q_2$ as input and give out $Q_3$ at temperature $T_2$ to the sink.

Work done by engine, A $W=Q_1–Q_2$

Work done by engine, B $W=Q_2–Q_3$

Thus,

$Q_1–Q_2=Q_2–Q_3$

Dividing both sides by $Q_1$,

$1–\frac{Q_2}{Q_1}=\frac{Q_2}{Q_1}–\frac{Q_3}{Q_1}$

$⟹1-\frac{T}{T_1}=\frac{Q_2}{Q_1}(1-\frac{Q_3}{Q_2})$

$⟹1-\frac{T}{T_1}=\frac{Q_2}{Q_1}(1-\frac{T_3}{T})$

$⟹1-\frac{T}{T_1}=\frac{T}{T_1}(1-\frac{T_3}{T})$

$⟹\frac{T_1}{T}–1=1-\frac{T_3}{T}$

$⟹\frac{T_1}{T}+\frac{T_3}{T}=2$

$⟹\frac{1}{T}(T1+T3)=2$

$⟹T=\frac{(T_1+T_3)}{2}$

$⟹T=\frac{(900+400)}{2}=650K$

is the Temperature when output of the engines are equal.

Let the efficiency of both engines be $\eta$. Now considering both engines efficiency are equal. This gives,

$1-\frac{T}{T_1}=1-\frac{T_3}{T}⟹\frac{T}{T_1}=\frac{T_3}{T}⟹T^2=T_1\times T_3⟹T^2=800\times 400=320000$

$T=565.68K$ is the temperature when efficiencies of the both the engines are equal.