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 engines A and B are represented by ηA and ηB, respectively,then what is the value of \frac{\eta_{B} }{\eta _{A}} ?

  • Option 1)

    \frac{12}{7}

  • Option 2)

    \frac{7}{12}

  • Option 3)

    \frac{12}{5}

  • Option 4)

    \frac{5}{12}

 

Answers (2)
A Anuj Pandey
A Aadil Khan

As we learnt

Efficiency of a carnot cycle -

\eta =\frac{W}{Q_{1}-Q_{2}}=1-\frac{T_{2}}{T_{1}}

T_{1}\, and\, T_{2}  are in kelvin
 

- wherein

T_{1}= Source temperature

T_{2}= Sink Temperature

\left ( T_{1} > T_{2}\right )

 

 

\eta _{A}=1-\frac{T}{600}=\frac{w}{Q_{1}}\rightarrow (1)
\eta _{B}=1-\frac{100}{T}=\frac{w}{Q_{2}}=\frac{w}{Q_{1-w}}

\eta _{B}=1-\frac{100}{T}=\frac{1}{\frac{Q_{1}}{w}-1}

from eqn1:

\eta _{B}=1-\frac{100}{T}=\frac{1}{\frac{1}{1-\frac{T}{600}}-1}=\frac{1-\frac{T}{600}}{\frac{T}{600}}

\eta _{B}=\frac{600}{T}-1= 1-\frac{100}{T}\: \: or\: \: \frac{700}{T}=2\: \: or\: \: T=350K

\frac{\eta _{B}}{\eta _{A}}= \frac{1-\frac{100}{T}}{1-\frac{T}{600}}=\frac{12}{7}


Option 1)

\frac{12}{7}

Option 2)

\frac{7}{12}

Option 3)

\frac{12}{5}

Option 4)

\frac{5}{12}

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