Question 6.7: In a heat exchanger, a vapour flow rate of 20 t/h is condens......
In a heat exchanger, a vapour flow rate of 20 t/h is condensed with cold water, its state at the inlet corresponding to a wet vapour at 0.056 bar and quality 0.92 and at the outlet, saturated liquid. The refrigeration water flow rate is 330 kg/s and is heated in the exchanger to 28°C. Given the limited availability of water, it must be recirculated, so to cool it, it is passed through a cooling tower with a 60 kW fan, whose operating conditions are as follows: (1) In summer, the air enters the tower at 30°C with a relative humidity of 60%, leaving it at 25°C and with relative humidity of 100%; (2) In winter, air enters the tower at 8°C and relative humidity of 30%, leaving at 17°C with 95% relative humidity. The temperature of the replacement water is 20°C in summer and 8°C in winter.
If the temperature of the refrigeration water at the entrance of the exchanger is constant, for both operating conditions, and the ambient pressure is 1 atm, calculate:
(a) The mass airflow rate in the cooling tower and mass flow rate of replacement water in summer and winter.
(b) The rate of exergy destruction in the tower in summer.
(c) The exergy efficiency of the tower in summer.
(d) Th rate of exergy destroyed in the exchanger also in summer.
(e) Does it make sense to define the exergy efficiency of the heat exchanger?
In the following Fig. E.6.2 a schema of the installation is shown.
(a) According to the nomenclature adopted in Fig. E 6.2, carrying out an energy balance in the exchanger, we have
\dot{m}_{\nu}(h_{\nu,1}-h_{\nu,2})=\dot{m}_{w}(h_{I}-h_{III})From the thermodynamic data for vapour we have
h_{\nu,2}=133.9\frac{k J}{k g}\quad h_{\nu,1}=h^{\prime}+x_{1}.l=146.6+0.92\cdot2.418,1=2.371,2\,\frac{\mathrm{kJ}}{\mathrm{kg}}Since \dot{m}_{\nu}=20t/h\ \ \mathrm{and}\ \ \,\dot{m}_{w}(h_{I}-h_{II I})=330\cdot4.18(28-T_{I II}) substituting these values in the balance equation, gives T_{III}=19^{\circ}{\mathrm{C}}. The replacement water is evaporated in the tower so that
\dot{m}_{I V}=\dot{m}_{a}(\omega_{1}-\omega_{0})We do not take into account the kinetic energies of the air and water flows or head losses, so in the energy balance in the tower we do not include the power of the fans. Therefore, from the energy balance in the tower, we have
{\dot{m}}_{w}h_{I}+{\dot{m}}_{a}h_{0}=[{\dot{m}}_{w}-{\dot{m}}_{a}(\omega_{1}-\omega_{0})]h_{I I}+{\dot{m}}_{a}h_{1}From the energy balance in the mixing of the return water of the cooling tower with the replacement water we get
\dot{m}_{a}(\omega_{1}-\omega_{0})h_{I V}+[\dot{m}_{w}-\dot{m}_{a}(\omega_{1}-\omega_{0})]h_{I I}=\dot{m}_{w}h_{III}Eliminating h_{II} between these two equations, we have
\dot{m}_{a}=\frac{\dot{m}_{w}(h_{I}-h_{II I})}{h_{1}-h_{0}-(\omega_{1}-\omega_{0})h_{I V}}In summer, since p_{s}(30^{\circ}{\mathrm{C}})=42.42 mbar and p_{s}(25^{\circ}{\mathrm{C}})=31.66 mbar, we have
\omega_{0}=0,622{\frac{p_{s}(30^{\circ}{\mathrm{C}})}{\frac{p_{0}}{\phi_{0}}-p_{s}(30^{\circ}{\mathrm{C}})}}=16\ \frac{\mathrm{g}}{{\mathrm{kg~d~a}}}with the enthalpy of the atmospheric air being
h_{0}=c_{p,a}T_{0}(^{\circ}{\mathrm{C}})+\omega_{0}{\big(}l(0^{\circ}{\mathrm{C}})+c_{p,\nu}T_{0}(^{\circ}{\mathrm{C}}){\big)}=71.8~{\frac{\mathrm{kJ}}{\mathrm{kg~d~a}}}For the air at the output of the cooling tower
\omega_{1}=0.622{\frac{p_{s}(25^{\circ}{\mathrm{C}})}{\displaystyle{\frac{p_0}{\phi_{1}}}-p_{s}(25^{\circ}{\mathrm{C}})}}=20~{\frac{\mathrm{g}}{\mathrm{kg\,d\,a}}}with its enthalpy being
h_{1}=c_{p,a}T_{1}(^{\circ}{\rm C})+\omega_{1}\big(l(0^{\circ}{\rm C})+c_{p,\nu}T_{1}(^{\circ}{\rm C})\big)=76.2\;{\frac{\mathrm{kJ}}{\mathrm{kg~d~a}}}Substituting these values in the previous equation, we have that
{\dot{m}}_{a}=2.618.6{\frac{\mathrm{kg~d~a}}{\mathrm{s}}}As
\nu_{0}=(0.622+\omega_{0})R_{\nu}\frac{T_{0}}{p}=0.881~\frac{\mathrm{m^{3}}}{\mathrm{kg~d~a}}this means that the volume airflow rate is
{\dot{V}}_{0}=\dot m_{a}\nu_{0}=2,307\ {\frac{\mathrm{m^{3}}}{\mathrm{s}}}with the replacement water being
{\dot{m}}_{I V}={\dot{m}}_{a}(\omega_{1}-\omega_{0})=10.5\ {\frac{\mathrm{kg}}{\mathrm{s}}}Values for winter are solved in a totally analogous way, obtaining the following
{\dot{m}}_{a}=376.5{\frac{\mathrm{kg~d~a}}{\mathrm{s}}}\quad{\dot{V}}_{1}={\dot{m}}_{a}\nu_{1}=300.8\ {\frac{\mathrm{m}^{3}}{\mathrm{s}}}\quad{\dot{m}}_{I V}=3.6\ {\frac{\mathrm{kg}}{\mathrm{s}}}\,Note that the results obtained are very different. In summer, you need a mass flow rate of air that is more than seven times the mass flow rate that is needed in winter and the replacement water needed is almost three times of that needed in winter.
(b) In the cooling tower, the inlet air is atmospheric air, and therefore, its exergy is zero. The air that comes out of the tower has physical and chemical exergy, but that exergy is finally destroyed in the environment so that it forms part of the external irreversibilities (losses). Therefore, undertaking an exergy balance in the tower, we have
Calculating each of the terms on the left of this equation. For this, we need to previously know the temperature T_{II}. Undertaking a balance of energy in the mixture with the return water, we have
10.5\,h_{IV(20^{\circ}{\mathrm{C}})}+319.5\,h_{II(t_{II})}=330\,h_{III(19^{\circ}{\mathrm{C}})}{\rightarrow}T_{II}=18.9^{\circ}{\mathrm{C}}According to the statement, the replacement water and the ambient air are not in thermodynamic equilibrium, since their temperatures are different. For the calculation of the exergy of the water mass flow rates we choose as ambient temperature the one corresponding to the replacement water, that is, 20°C.
We now calculate the specific exergy of the water at the states I and II. As their temperatures are known, through Eq. (3.44) we have
Returning to the equation of exergy balance finally gives
\dot I_{C T W}=205.9\,\mathrm{kW}(c) The objective of the tower is to cool the flow of water that circulates through it, that is, to reduce its exergy, so that the more the water cools (more heat dissipates), the better the operation of the tower. It is, therefore, a dissipative equipment, which only makes sense in that, it is an auxiliary component that serves other productive equipment. Together with the exchanger (condenser) its mission is to condense the vapour mass flow. Therefore, it does not make sense to define an exergy efficiency, as the exergy dissipated in it will be attributed to the productive equipment it serves.
(d) From the exergy balance in the heat exchanger, we have
We calculate the exergy of the saturated liquid-vapour mixture at the entrance of the exchanger
b_{\nu1}=h^{\prime}+x_{\nu1}(h^{\prime\prime}-h^{\prime})-h_{0}-T_{0}\biggl(s^{\prime}+{\frac{x_{\nu1}(h^{\prime\prime}-h^{\prime})}{T}}-s_{0}\biggr)=165.5\ {\frac{\mathrm{kJ}}{\mathrm{kg}}}The exergy of the saturated liquid at the outlet of the exchanger and the exergy change between I and III are obtained by applying Eq. (3.44), giving
b_{\nu2}=1.55\ \frac{\mathrm{kJ}}{\mathrm{kg}}\quad b_{I}-b_{II I}=0.44\ \frac{\mathrm{kJ}}{\mathrm{kg}}Finally, the exergy destruction in the heat exchanger gives
\dot{D}_{E X C}=765.6\ k W(e) The objective of the heat exchanger (together with the tower) is to condense vapour, that is, to reduce its exergy. The heat exchanger will form part of an installation in which it is neces-sary to use the produced condensate, in such a way that the heat exchanger is auxiliary equip-ment for some other productive equipment. It is, therefore, a dissipative equipment, which must be analysed considering the productive equipment it serves so that, considered in isola-tion, it does not make sense to define its exergy efficiency.
