Question 6.5: An airflow of 3800 kg/h at 2°C is taken from the outside wit......
An airflow of 3800 kg/h at 2°C is taken from the outside with a relative humidity of 70%, passing through an electrical resistance where it is heated up to 12°C. Then, that air is mixed with another flow of air at 20°C saturated with humidity, so that the mass flow rate of saturated air is double. With the environmental pressure at 1 bar, determine
(a) The absolute and relative humidity of the air after passing through the electrical resistance.
(b) The heat given in the resistance.
(c) The exergy destruction in this heating process.
(d) The temperature and relative humidity of the air resulting from the mixture.
(e) The exergy destruction in the mixture of the two flows
(a) Calculating the absolute humidity of the outside air, the atmospheric air, which we call air in state 0. Since p_{s}(2\mathrm{C})=7.059 mbar, we have
\omega_{0}=0.622{\frac{p_{s}(2^{\circ}C)}{{\frac{p_{0}}{\phi_{0}}}-p_{s}(2^{\circ}C)}}=3~{\frac{g}{k g\;d\,a}}After passing through the electrical resistance, the air in state 1 has the same abso-lute humidity, \omega_{1}=\omega_{0}=3~\mathrm{g/kg} dry air. The relative humidity is
\phi_{1}=\frac{\omega_{1}}{\omega_{1}+\mathrm{0.622}}\,\frac{p_{0}}{p_{s}(12^{\circ}{\mathrm{C}})}=34.2{\%}Before going further, we calculate the mass flow of dry air.
\dot{m}_{a,0}(1+\omega_{0})=3800\;\frac{\mathrm{kg}}{\mathrm{h}}\to\dot{m}_{a,0}=1.050\;\frac{\mathrm{kg}\;\mathrm{d}\,\mathrm{a}}{\mathrm{s}}(b) Applying the energy balance we have
{\dot{Q}}=\dot m_{a,0}(h_{1}-h_{0})=10.6\,k W(c) Undertaking an exergy balance and with the electricity consumption in the resistance being {\dot{{E}}}={\dot{Q}}, we have
{\dot{E}}-{\dot{m}}_{a,0}(b_{1}-b_{0})={\dot{D}}Since we consider air as a mixture of ideal gases, dry air and water vapour, of approximately constant specific heats c_{p,a}=1.004\:{\rm k J}/({\rm k g}\cdot{\rm k}) and c_{p,v}=1.86\:{\rm k J}/({\rm k g}\cdot{\rm k}) and with the pressure being constant, the change of the physical exergy of the air is
b_{1}-b_{0}=b_{1}=\left(c_{p,a}+\omega_{1}c_{p,\nu}\right)\left(T_{1}-T_{0}-T_{0}\;l n\frac{T_{1}}{T_{0}}\right)=0.18\;\frac{\mathrm{kJ}}{\mathrm{kg}\;\mathrm{d}\mathrm{a}}which coincides with the exergy of 1, since b_{0}=0. Substituting in the exergy balance equation gives
\dot{D}=10.4\,\mathrm{kW}This result shows us the highly irreversible nature of this heating process since approximately 98% of the exergy contributed is destroyed.
(d) First, we calculate the mass flow rate of the saturated air at 20°C, which we call air 2, and for which we first determine the absolute humidity. Since p_{s}(20^{\circ}{\mathrm{C}})=23.39 mbar and when saturated \phi_{2}=1, we have
As the mass flow rate of air in state 2 is double, we have the relationship
{\dot{m}}_{a,2}(1+\omega_{2})=2{\dot{m}}_{a,0}(1+\omega_{0})\to{\dot{m}}_{a,2}=2.075\ {\frac{\mathrm{kg}}{s}}If the air resulting from the mixture is 3, from the mass balances in the mixing process, we have that
\dot{m}_{a,1}+\dot{m}_{a,2}=\dot{m}_{a,3}{\rightarrow}\dot{m}_{a,3}=3.125\ \frac{\mathrm{kg}\ \textrm{d}\ \mathrm{a}}{\mathrm{s}} \\ \dot{m}_{a,1}\omega_{1}+\dot{m}_{a,2}\omega_{2}=\dot{m}_{a,3}\omega_{3}\rightarrow\omega_{3}=\mathrm{~10.9~}\frac{\mathrm{~g}}{\mathrm{kg~d~a}}To find the temperature resulting from the mixture we carry out the energy balance, which is
{\dot{m}}_{a.1}h_{1}+{\dot{m}}_{a.2}h_{2}={\dot{m}}_{a.3}h_{3}\to T_{3}=17.6^{\circ}\mathrm{C}As p_{s}(17.6^{\circ}{\mathrm{C}})=19.4 mbar, the relative humidity is
\omega_{3}=0.622{\frac{p_{s}(17.6^{\circ}{\mathrm{C}})}{{\frac{p_{0}}{\phi_{3}}}-p_{s}(17.6^{\circ}{\mathrm{C}})}}\to\phi_{3}={\frac{p_{0}}{0.622~p_{s}(17.6^{\circ}{\mathrm{C}})+\omega_{3}p_{s}(17.6^{\circ}{\mathrm{C}})}}\omega_{3}\\ = 89\%(e) Performing the exergy balance in the mixing process, we have
\dot{m}_{a,1}\left(b_{1}+b_{1}^{c h}\right)+\dot{m}_{a,2}\bigl(b_{2}+b_{2}^{c h}\bigr)-\dot{m}_{a,3}\bigl(b_{3}+b_{3}^{c h}\bigr)=\dot{D}Calculating the physical and chemical exergy of the air in each state. Using Eq. (3.37) for physical exergy, we have
b_{h a}=\left(c_{p,a}+\omega~c_{p,\nu}\right)\left[\left((T-T_{0})-T_{0}l n\frac{T}{T_{0}}\right)\right]+0.461(\omega+0.622)T_{0}l n\frac{p}{p_{0}}\qquad\qquad(3.37) \\ b_{1}=0.18~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ a}}~~~~b_{2}=0.58~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ a}}~~~~b_{3}=0.44~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ a}}Calculating the chemical exergy for which we use Eq. (3.124). Since state 1 has the same humidity as the ambient air, its chemical exergy is zero.
b_{h a}^{c h}=0.461(\omega+0.622)T_{0}\bigg[\frac{0.622}{\omega+0.622}l n\frac{\omega_{0}+0.622}{\omega+0.622}+\frac{\omega}{\omega+0.622}l n\biggl(\frac{\omega}{\omega_{0}}\,\frac{\omega_{0}+0.622}{\omega+0.622}\biggr)\biggr]\qquad\qquad(3.124) \\ b_{1}^{c h}=0\quad b_{2}^{c h}=1.52\ {\frac{\mathrm{kJ}}{\mathrm{kg\,d\,a}}}\quad b_{3}^{c h}=0.53\ {\frac{\mathrm{kJ}}{\mathrm{kg\,d\,a}}}Returning to the balance equation, we have
\dot{D}=1.51\ k\mathrm{W}