Two carnot emgines A and B are operated in series. engine A receiepved heat?from a reservoir at 600 K and rejects it to a reservoir at temp T. B receives heat rejected by A and in turn rejects it to reservoir at 100K. Find (η_B /η


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Question Number 52324 by prakash jain last updated on 06/Jan/19

$Two carnot emgines A and B are operated in$ $series . engine A receiepved heat ? from$ $a reservoir at 600 K and rejects it to$ $a reservoir at temp T . B receives$ $heat rejected by A and in turn rejects$ $it to reservoir at 100 K .$ $Find \frac{η_{B}}{η_{A}}$ $η efficiency$
Commented by Tinkutara last updated on 14/Jan/19

Commented by Tinkutara last updated on 14/Jan/19
In my JEE main it appeared with correction!
	Answered by ajfour last updated on 06/Jan/19

	$η_{A} = 1 - \frac{T}{600}$ $η_{B} = 1 - \frac{100}{T}$ $η_{e q} = 1 - \frac{100}{600} = \frac{5}{6} = (η_{A} + η_{B}) - η_{A} η_{B}$ $l e t \frac{η_{B}}{η_{A}} = r$ $. . . .$
	Commented by prakash jain last updated on 06/Jan/19
	Thanks. I thought there is some info missing. But the question came in JEE main 2018 exam
	Commented by ajfour last updated on 06/Jan/19

	$y e s S i r, i i n q u i r e d a n d g o t t o k n o w$ $t h e s a m e, b u t n e i t h e r c o u l d s o l v e$ $n o r f o u n d s o l u t i o n g i v e n, p l e a s e$ $s h a r e t h e s o l u t i o n i f {y o u}^{'} v e$ $u n d e r s t o o d .$
	Commented by ajfour last updated on 06/Jan/19

	$η_{A} = 1 - \frac{Q_{1}}{Q_{0}}, η_{B} = 1 - \frac{Q_{2}}{Q_{1}}$ $η_{e q} = 1 - \frac{Q_{2}}{Q_{0}}$ $(1 - η_{A}) (1 - η_{B}) = \frac{Q_{2}}{Q_{0}} = (1 - η_{e q})$ $\Rightarrow η_{e q} = (η_{A} + η_{B}) - η_{A} η_{B}$
	Commented by prakash jain last updated on 06/Jan/19
	I don't know the answer I just wanted to confirm the same. If you cannot solve a physics question it then no-one else can. My son is appearing for JEE Main next week and asked me this question.
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