Certainly! Let's go through the problem step-by-step in detail.
### Part (a): Efficiency of the Heat Engine
To find the efficiency of the heat engine, we use the definition of thermal efficiency for a heat engine, which is given by:
[tex]\[ \eta_{\text{actual}} = \frac{\text{Work Output}}{\text{Thermal Energy Input}} \][/tex]
In this problem:
- The thermal energy input ([tex]\( Q_{\text{in}} \)[/tex]) is given as 3 MW.
- The work output (electric power produced) is 2 MW.
However, this also implies:
- The thermal energy output or the wasted energy ([tex]\( Q_{\text{out}} \)[/tex]) is 1 MW (as [tex]\( 3 \text{ MW} - 2 \text{ MW} = 1 \text{ MW} \)[/tex]).
Using these values, the efficiency is:
[tex]\[ \eta_{\text{actual}} = \frac{Q_{\text{in}} - Q_{\text{out}}}{Q_{\text{in}}} = \frac{3 \text{ MW} - 1 \text{ MW}}{3 \text{ MW}} = \frac{2 \text{ MW}}{3 \text{ MW}} = \frac{2}{3} \approx 0.6667 \][/tex]
So, the efficiency of the heat engine is approximately [tex]\( 0.6667 \)[/tex] or 66.67%.
### Part (b): Maximum Possible Efficiency (Carnot Efficiency)
To find the maximum possible efficiency of a heat engine operating between two temperatures, we use the Carnot efficiency formula, which is:
[tex]\[ \eta_{\text{max}} = 1 - \frac{T_{\text{C}}}{T_{\text{H}}} \][/tex]
where:
- [tex]\( T_{\text{H}} \)[/tex] is the high temperature reservoir (450 K).
- [tex]\( T_{\text{C}} \)[/tex] is the low temperature reservoir (240 K).
Using these temperatures, the maximum possible efficiency is:
[tex]\[ \eta_{\text{max}} = 1 - \frac{240 \text{ K}}{450 \text{ K}} = 1 - \frac{8}{15} \approx 1 - 0.5333 = 0.4667 \][/tex]
So, the maximum possible efficiency, if the heat engine were ideal, is approximately [tex]\( 0.4667 \)[/tex] or 46.67%.
### Summary
- Actual Efficiency: [tex]\( 66.67\% \)[/tex]
- Maximum Possible Efficiency: [tex]\( 46.67\% \)[/tex]
