Q: Consider a heat engine working with a hot reservoir of the furnace gases at 2100 °C , when the cooling water is available at -35.7 °C. then, choose the correct option(s).
(a) Engine efficiency can be made more than 90% without altering operating temperatures
(b) Engine efficiency always remains less than 90 %
(c) Engine efficiency is much lower practically due to irreversibility in the actual cycle
(d) Engine efficiency is 90% but such temperatures are not available practically
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Sol: Temperature of hot Reservoir , T1 = 2100 + 273 = 2373 k
Temperature of cold Reservoir , T2 = -35.7 + 273 = 237.3 K
Maximum Efficiency , $\displaystyle \eta_{max} = 1 – \frac{T_2}{T_1}$
$\displaystyle \eta_{max} = 1 – \frac{237.3}{2373}$
$\displaystyle \eta_{max} = 1 – \frac{1}{10} = \frac{9}{10}$
Percentage efficiency $\displaystyle = \frac{9}{10} \times 100$
= 90 %