Question 486: Thermodynamics & KTG

Question

Two Carnot engines A and B are operated in series. Engine A receives heat from a reservoir at 600 K and rejects heat to a reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 100 K. If the efficiencies of the two energies A and B are represented by $${\eta _A}$$ and $${\eta _B},$$ respectively, then what is the value of $${{{\eta _B}} \over {{\eta _A}}}$$ ?

Accepted Answer

Given that two Carnot engines A and B are operated in series. Efficiency of a Carnot engine $$\eta  = 1 - {{{T_2}} \over {{T_1}}}$$; where T
1
is temperature of source and T
2
is temperature of sink. Therefore,
$${\eta _A} = 1 - {{{T_{2A}}} \over {{T_{1A}}}} = 1 - {T \over {600}}$$; T is heat rejected by A
and $${\eta _B} = 1 - {{{T_{2B}}} \over {{T_{1B}}}} = 1 - {{100} \over T}$$; T is heat rejected by A and received by B
Now, $${{{\eta _B}} \over {{\eta _A}}}{{1 - {{100} \over T}} \over {1 - {T \over {600}}}} = \left( {{{T - 100} \over T}} ight) 	imes {{600} \over {600 - T}}$$
Let T = 350 K, therefore,
$${{{\eta _B}} \over {{\eta _A}}} = {{350 - 100} \over {350}} 	imes {{600} \over {600 - 350}} = {{250} \over {350}} 	imes {{600} \over {25}} = {{600} \over {350}}$$
$$ \Rightarrow {{{\eta _B}} \over {{\eta _A}}} = {{12} \over 7}$$

Answer

$${{12} \over 7}$$

Answer

$${{7} \over 12}$$

Answer

$${{12} \over 5}$$

Answer

$${{5} \over 12}$$

Thermodynamics & KTG

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