Question

It is desired to double the efficiency of a carnot engine from 35% by raising its temperature of heat addition, while keeping the temperature of heat rejection constant. What percentage of increase in high temperature is required?

Answer 1

Answer on #69364, Physics / Molecular Physics | Thermodynamics

Question. It is desired to double the efficiency of a Carnot engine from $35\%$ by raising its temperature of heat addition, while keeping the temperature of heat rejection constant. What percentage of increase in high temperature is required?

Given: $\eta = 35\%$ (0.35);

$T_{2} = \text{const};$

$\eta' = 2 \cdot \eta.$

Find: $T_{1}^{\prime}$

Solution. The efficiency of the Carnot engine is defined to be:

$$\eta = \frac{T_{1} - T_{2}}{T_{1}} = 1 - \frac{T_{2}}{T_{1}},$$

where $-T_{1}$ is the absolute temperature of the hot reservoir; $T_{2}$ is the absolute temperature of the cold reservoir. Thus

$$\eta = 1 - \frac{T_{2}}{T_{1}}; \frac{T_{2}}{T_{1}} = 1 - \eta;$$ $$T_{2} = T_{1}(1 - \eta),$$

and

$$T_{2} = T_{1}'(1 - \eta').$$ $$T_{1}(1 - \eta) = T_{1}'(1 - \eta')$$ $$T_{1}(1 - \eta) = T_{1}'(1 - 2\eta);$$ $$\frac{T_{1}'}{T_{1}} = \frac{1 - \eta}{1 - 2\eta} = \frac{1 - 0.35}{1 - 2 \cdot 0.35} = \frac{0.65}{0.3} \approx 2.17 \text{ or } 217\%.$$

Answer: $T_{1}' = 2.17 \cdot T_{1}$, percentage of increase $-217\%$.

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