Demonstration of quality of energy based on exergy

Consider a hot gas of mass m flowing through a pipeline (Fig. 1.26). Due to the heat loss to the surroundings, the temperature of the gas decreases continuously from the inlet state 'a' to the exit state 'b'. Thus, the process is irreversible. However, for sake of simplicity, let us consider the process to be reversible and isobaric between the inlet and outlet. For an infinitesimal reversible process at constant pressure,

Fig. 1.26 Energy quality degrades along the flow direction

1.215

or,

1.216

Thus the slope depends on temperature T. With increase in T, increases and vice versa.

Let, Q unit of heat lost to the surroundings as temperature of gas decreases from to , T₁ being average of the two temperatures.

Heat loss at section 1-1

1.217

Exergy lost with this heat loss at temperature T₁ is

W1 = Q - T0ΔS1 = T1ΔS1 - T0ΔS1 = (T1 - T0) ΔS1

1.218

At section 2-2, let heat loss be same as in section 1-1, Q(T₂ < T₁)

1.219

Exergy loss due to this heat loss at temperature T₂ is

W2 = Q - T0ΔS2 = (T2 - T0) ΔS2

1.220

Considering the exergy loss at temperatures T₁ and T₂

W1 = Q - T0ΔS1 = T1ΔS1 - T0ΔS1 = (T1 - T0) ΔS1

1.221

W2 = Q - T0ΔS2 = (T2 - T0) ΔS2

1.222

As T₁ > T₂ , ΔS₁ < ΔS₂

∴ W1 > W2

1.223