Heat flows through a wall at a rate of 3 x 105 KJ/hr. The temperatures of two faces of the wall are 327°C and 207°C. If the surroundings are at 27°C

Given:

$Q = 3 × 10^5 kJ/hr, T_1 = 600K, T_2 = 480K, T_0 = 300K$

Available Energy with source = $Q(1 - \frac{T_o}{T_1})$

= $3 × 10^5(1 - \frac{300}{600})$

= $1.5 × 10^5 kJ/hr$

Available Energy with system = $Q(1 - \frac{T_o}{T_1})$

= $3 × 10^5(1 - \frac{480}{600})$

= …