Question 7.4: In the cross-flow heat exchanger in Fig. E.7.6, with 1, 2 be......
In the cross-flow heat exchanger in Fig. E.7.6, with 1, 2 being the cold flow states at the inlet and outlet and 3, 4 the corresponding states of the hot flow, define F and P, as well as the exergy efficiency of the equipment in the following two cases:
(a) When T_{1}\gt T_{0}, and the purpose of the exchanger is to heat the cold flow.
(b) When T_{3}\lt T_{0}, and the purpose of the exchanger is to cool the hot flow.
(a) Since the purpose of the exchanger is to heat the cold flow from state 1 to 2, the exchanger product is
P={\dot{B}}_{2}-{\dot{B}}_{1}=\dot{m}_{c}(b_{2}-b_{1})This useful effect is achieved through the hot flow, whose exergy will decrease from state \underline{3} at the inlet to state \underline{4} at the outlet of the exchanger, so that
F=\dot{B}_{3}-\dot{B}_{4}=\dot{m}_{h}(b_{3}-b_{4})This would be the case, for example, of the heat exchanger of a cogeneration plant, in which a flow of hot water (cold flow) is produced via the engine exhaust gases (hot flow). Consequently, the heat exchanger exergy efficiency is
\varphi={\frac{{\dot{B}}_{2}-{\dot{B}}_{1}}{{\dot{B}}_{3}-{\dot{B}}_{4}}}=1-{\frac{\dot{D}}{{\dot{B}}_{3}-{\dot{B}}_{4}}}(b) If the heat exchanger objective is to cool a flow that is already below the ambient temperature, this will be its product, so that
P={\dot{B}}_{4}-{\dot{B}}_{3}={\dot{m}}_{h}(b_{4}-b_{3})Note that although in state \underline{4} the enthalpy is lower than in state 3, its exergy will be greater, since state 4 is farther from equilibrium conditions with the environment so that the difference (b_{4}-b_{3}) is positive. This cooling of the hot flow is achieved by the cold flow which, when receiving heat from the hot flow, decreases its exergy, since it approaches the environmental conditions, so that
F=\dot{B}_{1}-\dot{B}_{2}=\dot{m}_{c}(b_{1}-b_{2})This type of heat exchange is what occurs, for example, in an air-conditioning unit in which the hot flow, the water flow at 12°C is cooled and leaves the unit at \underline{7}^{\circ}C, with a lower enthalpy but greater exergy. This effect is achieved via the cold flow, the refrigerant, which enters the air-conditioning unit with greater exergy than it has at the outlet. Therefore, exergy efficiency is
\varphi={\frac{{\dot{B}}_{4}-{\dot{B}}_{3}}{{\dot{B}}_{1}-{\dot{B}}_{2}}}=1-{\frac{\dot{D}}{{\dot{B}}_{1}-{\dot{B}}_{2}}}In a heat exchanger in which the hot flow is above the ambient temperature and the cold flow below, there will be a decrease in the exergy of both flows, since both the hot and cold flows approach the environmental conditions. Therefore, it does not make sense to define a product in this exchanger, since the equipment acts as an exergy dissipator, and so from a thermodynamic point of view, this is a situation that should be avoided. If a heat exchanger works in this way in any installation, the dissipation of exergy that takes place in it should be attributed to the productive equipment with which it is associated.