Question 204: Thermodynamics & KTG

Question

A Carnot engine $(\mathrm{E})$ is working between two temperatures 473 K and 273 K . In a new system two engines - engine $E_1$ works between 473 K to 373 K and engine $E_2$ works between 373 K to 273 K .
If $\eta_{12}, \eta_1$ and $\eta_2$ are the efficiencies of the engines $E, E_1$ and $E_2$, respectively, then

Accepted Answer

We know, efficiency of a carnot engine, $\eta$ or $$e = 1 - {{{T_C}} \over {{T_H}}}$$
where, T$_C$ = temperature of cold sink
T$_H$ = temperature of hot source
So, $${\eta _{12}} = 1 - {{273} \over {473}} = {{200} \over {473}} = 0.423$$
$${\eta _1} = 1 - {{373} \over {473}} = {{100} \over {473}} = 0.211$$
$${\eta _2} = 1 - {{273} \over {373}} = {{100} \over {373}} = 0.268$$
Here, $${\eta _1} + {\eta _2} = 0.211 + 0.268 = 0.479 > 0.423$$
So, $${\eta _{12}} < {\eta _1} + {\eta _2}$$

Answer

$\eta_{12}=\eta_1 \eta_2$

Answer

$\eta_{12}=\eta_1+\eta_2$

Answer

$\eta_{12} \geq \eta_1+\eta_2$

Answer

$\eta_{12}<\eta_1+\eta_2$

Thermodynamics & KTG

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