InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A cannot heat engine whose sink is at 200 K, has an efficiency 30%. By how much the temperature of the source be increased to have its efficiency equal to 50%. Keeping sink temperature constant. |
|
Answer» From the relation, η = 1 − \(\frac{T_2}{T_1}\) or, \(\frac{T_2}{T_1}\) = 1 - \(\frac{30}{100}\) = \(\frac{7}{10}\) or, T1 = \(\frac{10T_2}{7}\) = \(\frac{10\times200}{7}\)= 285.71 K New efficiency is now 50%, η' = 1 − \(\frac{T_2}{T_1}\) \(\frac{T_2}{T_1}\) = 1 - η' \(\frac{T_2}{T_1}\) = 1 - \(\frac{50}{100}\) or \(\frac{1}{2}\) 2T2 = T1 or, T1 = 2 × 200 K T1 = 400 K Now increase in temperature of source, = 400 K – 285.71 K = 114.3 K |
|