200 kJ of energy is transferred from a heat reservoir at 1050 K to a heat reservoir at 550 K. The ambient temperature is 310 K. The loss of available energy due to heat transfer process is:

Concept:-

Available energy is the maximum amount of energy that could be converted into useful work by ideal processes that reduce the system to a dead state (a state in equilibrium with the earth and its atmosphere). 

Unavailable energy is the product of the lowest temperature of heat rejection (ambient temperature, T₀) and the change of entropy of the system during the process of supplying heat.

Loss of available energy,

\(L.A.E. = T_oQ\left ( \frac{1}{T_2} -\frac{1}{T_1} \right )\;\)

Where

T_o = Ambient temperature 

Q = Heat transfer in kJ

T₁ = Higher or source temperature 

T₂ = Lower or sink temperature 

Calculation:-

Given:-

Q = 200 kJ, To = 310 K, T1 = 1050 K, T2 = 550 K

Loss of available energy due to the direct heat transfer process is, 

\(L.A.E. = T_oQ\left ( \frac{1}{T_2} -\frac{1}{T_1} \right )\;\)

⇒ \(L.A.E. = 310\times 200\left ( \frac{1}{550} -\frac{1}{1050} \right )=53.68\;kJ\;\)

L.A.E. = 53.68 kJ