Three reversible engines of Carnot type are operating in series as shown between the limiting temperatures of 1100 K and 300 K. Determine the intermediate temperatures if the work output from engines is in proportion of 3 : 2 : 1.

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[latex] ext{Here, }W_{1} : W_{2} : W_{3} = 3 : 2 : 1\\begin{aligned} ext{Efficiency of engine, HE}_{1}, \qquad \qquad &\\frac{W_{1}}{Q_{1}}&=\left\lgroup 1 - \frac{T_{2}}{1100} ight group \Rightarrow Q_{1} = \frac{1100 . W_{1}}{(1100 - T_{2})}\ ext{for HE}_{2} ext{ engine,} \qquad \qquad \qquad \qquad &\\frac{W_{2}}{Q_{2}}&=\left\lgroup 1 - \frac{T_{3}}{T_{2}} ight group\ ext{for HE}_{3} ext{ engine,} \qquad \qquad \qquad \qquad &\\frac{W_{3}}{Q_{3}}&=\left\lgroup 1 - \frac{300}{T_{3}} ight group\ ext{From energy balance on engine, HE}_{1}&\Q_{1}&= W_{1} + Q_{2} \Rightarrow Q_{2} = Q_{1} - W_{1}\ ext{Above gives,} \qquad \qquad \qquad \qquad \qquad &\Q_{1}&= \left\{\frac{1100W_{1}}{(1100 - T_{2})}-W_{1} ight\} = W_{1}\left\{\frac{T_{2}}{1100 - T_{2}} ight\} \ ext{Substituting Q}_{2} ext{ in efficiency of HE}_{2} &\\frac{W_{2}}{W_{1}\left\lgroup\frac{T_{2}}{1100 - T_{2}} ight group }&= \left\lgroup 1 - \frac{T_{3}}{T_{2}} ight group \ ext{or} \quad\qquad \qquad \qquad \qquad \qquad \qquad \frac{W_{2}}{W_{1}}&= \left\lgroup\frac{T_{2}}{1100 - T_{2}} ight group \left\lgroup\frac{T_{2} - T_{3}}{T_{2}} ight group = \left\lgroup\frac{T_{2} - T_{3}}{1100 - T_{2}} ight group \ ext{or} \qquad\qquad \qquad \qquad \qquad \qquad \qquad &\left\{\frac{2}{3}=\left\lgroup\frac{T_{2}-T_{3}}{1100-T_{2}} ight group ight\} \ ext{or} \qquad \qquad 2200 – 2T_{2} = 3T_{2} – 3T_{3}&\& 5T_{2} − 3T_{3} = 2200\ ext{Energy balance on engine HE}_{2} ext{ gives,}&\Q_{2}&= W_{2} + Q_{3}\ ext{Substituting in efficiency of HE}_{2}, &\\frac{W_{2}}{(W_{2}+ Q_{3})}&= \left\lgroup\frac{T_{2}-T_{3}}{T_{2}} ight group \ ext{or}\qquad \qquad \qquad \qquad \qquad \qquad W_{2}. T_{2}&= (W_{2} + Q_{3}) (T_{2} – T_{3})\ ext{or} \quad\qquad \qquad \qquad \qquad \qquad \qquad Q_{3}&= \frac{W_{2}T_{3}}{(T_{2} - T_{3})}\ ext{Substituting Q}_{3} ext{ efficiency of HE}_{3}, &\\frac{W_{3}}{\left\lgroup\frac{W_{2}T_{3}}{T_{2} - T_{3}} ight group }&= \frac{T_{3} - 300}{T_{3}}\\frac{W_{3}}{W_{2}}&= \left\lgroup\frac{T_{3}}{T_{2} - T_{3}} ight group \left\lgroup\frac{T_{3} - 300}{T_{3}} ight group \\frac{1}{2}&= \frac{T_{3} - 300}{T_{2} - T_{3}}\3T_{3} - T_{2}&= 600\ ext{Solving, equations of T}_{2} ext{ and T}_{3}, T_{3}&= 433.33 K\T_{2}&= 700 K\end{aligned}\ ext{Intermediate temperatures: } extbf{700 K} ext{ and } extbf{433.33 K}[/latex]

Question 4.9: Three reversible engines of Carnot type are operating in ser...

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