Question 7.6: In the cogeneration installation whose schematic diagram is ......
In the cogeneration installation whose schematic diagram is presented in Fig. E.7.8, find the following:
(a) Define F, P and L.
(b) Carry out an exergy balance.
(c) Determine the exergy efficiency.
(a) The product of the cogeneration plant is the electricity and thermal energy produced, which is
P={\dot{E}}+{\dot{B}}The plant fuel is the combustible consumed, which is basically that combustible chemical exergy
F={\dot{B}}_{f}Both the combustion gases exergy and that of the heat lost by the alternator and the rest of the plant equipment are part of the loss flows
L={\dot{B}}_{\rm g}+\left(1-{\frac{T_{0}}{T_{a l}}}\right){\dot{Q}}_{a l}+\left(1-{\frac{T_{0}}{T_{e q}}}\right){\dot{Q}}_{e q}(b) Undertaking the exergy balance, we have the equation
{\dot{B}}_{f}=(\dot{E}+\dot{{B}})+\left[{\dot{B}}_{\rm g}+\left(1-\frac{T_{0}}{T_{a l}}\right){\dot{Q}}_{a l}+\left(1-\frac{T_{0}}{T_{e q}}\right){\dot{Q}}_{e q}\right](c) The installation exergy efficiency is
\varphi=\frac{\dot{E}+\dot{B}}{\dot{B}_{f}}=1-\frac{\left[\dot{B}_{\rm g}+\left(1-\frac{T_{0}}{T_{a l}}\right)\dot{Q}_{a l}+\left(1-\frac{T_{0}}{T_{e q}}\right)\dot{Q}_{e q}\right]+\dot{D}}{\dot{B}_{f}}As we can see, this exergy efficiency is the sum of two: the electric exergy efficiency, whose value is similar to the electric efficiency of the plant, since the chem-ical exergy of the combustible has a value close to its heating value and the numerator is the same; and the thermal exergy efficiency, which is significantly lower than ther-mal efficiency.
In all these examples we see that, if the energy conversion under consideration were carried out in a perfect way, that is, without losses or destructions of exergy (without irreversibilities), the exergy efficiency would be the unit. This behaviour is what we expect from a coefficient that measures the degree of perfection of the energy transfor-mation under consideration.