This question was previously asked in

AAI JE (Technical) Official Paper 2018

Option 1 : 210 kJ

ST 1: DC Circuits

2463

20 Questions
20 Marks
25 Mins

__Concept__:

Unavailable work: The difference between the heat input to an engine and the maximum work done by the engine is known as unavailable work.

Wunavailable = Qinput - Wmax

\({\eta _{rev}} = \frac{{{T_1} - {T_2}}}{{{T_1}}}\)

\({\eta _{rev}} = \frac{W_{max}}{{{Q_{input}}}} \)

__Given:__

T1 = 800 K, T2 = 300 K, Q1 = 560 kJ.

\({\eta _{rev}} = \frac{{{T_1} - {T_2}}}{{{T_1}}}\)

\({\eta _{rev}} = \frac{{{800} - {300}}}{{{800}}}=\frac{5}{8}\)

\({\eta _{rev}} = \frac{W_{max}}{{{Q_{input}}}} \)

\(\frac{5}{8} = \frac{W_{max}}{{{560}}} \)

W_{max} = 350 kJ.

Wunavailable = Qinput - Wmax

W_{unavailable} = 560 - 350 = 210 kJ