Question 6.9: In the AHU of an air conditioning installation, there is a c......
In the AHU of an air conditioning installation, there is a cold battery and a heat battery. The extraction flow is partially recirculated, with the ratio between the recirculated air mass flow and fresh air being 25%. The outside air is at 34°C and 74% relative humidity, with local air conditions at 25°C and 52% relative humidity. The ratio between the latent load of the conditioned room to the sensitive load (associated with dry air) is 1/3. The air enters the heat battery at 14°C and goes to the room at a temperature of 18°C, with the mass flow rate of dry air being 1.2 kg/s. The water tem-perature at the inlet of the heat battery is 62°C, with a variation of 8°C, while in the cold battery the temperature is 7°C, with a variation of 5°C. Determine
(a) The state of the air after mixing with the recirculated air, at the outlet of the cold battery and the entrance to the premises.
(b) The irreversibilities in the mixing process.
(c) Overall exergy efficiency of the AHU
(a) We shall use 0 for the state of the outside air, 1 for the recirculated air, 2 for the air resulting from the mixture, 3 for the air to the outlet of the cold battery and input to the heat battery and 4 for the heat battery output, and therefore, the entrance to the site to be conditioned. As p_{s}(34^{\circ}{\mathrm{C}})=53.2 mbar and p_{s}(25^{\circ}{\mathrm{C}})=31.7 mbar the absolute humidity in state 0 and 1 are respectively,
\omega_{0}=0.622{\frac{p_{s}(34^{\circ}C)}{\frac{p_0}{\phi_{0}}-p_{s}(34^{\circ}C)}}=25\ {\frac{g}{k g\,d a}} \\ \omega_{1}=0.622{\frac{p_{s}(25^{\circ}C)}{\frac{p_0}{\phi_{1}}-p_{s}(25^{\circ}C)}}=10\ {\frac{g}{k g\,d a}}Next, we calculate the mass flow rate of dry air from the outside and the recir-culation. We have the equation
\frac{\dot{m}_{a,1}(1+\omega_{1})}{\dot{m}_{a,0}(1+\omega_{0})}=0.25\quad\dot{m}_{a,0}+\dot{m}_{a,1}=\dot{m}_{a}=1.2\ \frac{\mathrm{kg}}{s}giving
{\dot{m}}_{a,0}=0.83\ {{\frac{\mathrm{kg~d~a}}{s}}{}}\quad{\dot{m}}_{a,1}=0.37\ \frac{\mathrm{kg~d~a}}{s}The state of the air resulting from the mixture, state 2, is obtained by solving the system of the two following equations
\dot{m}_{a,0}h_{0}+\dot{m}_{a,1}h_{1}=\dot{m}_{a}h_{2}(T_{2}) \\ \dot{m}_{a.0}\omega_{0}+\dot{m}_{a.1}\omega_{1}=\dot{m}_{a}\omega_{2}evidently with \,\dot{m}_{a}=\dot{m}_{a,0}+\dot{m}_{a,1} With the equation of the vapour balance we calculate the humidity in state 2, giving \omega_{2}=22\mathrm{g/kg~d~a}. We take this result to the energy balance equation, obtaining
T_{2}=305.3\mathrm{~K~}(32.2^{\circ}{\mathrm{C}})Therefore, the state of the air after mixing is 2(T_{2}=32.2^{\circ}{\mathrm{C}},\omega_{2}=22\,\mathrm{g}/\mathrm{kg\,d\,a}). We now determine the humidity of the air in state 4, at the entrance of the room to be conditioned. According to the relation between the sensitive load and total load, the latent load is one-third of the sensitive load \dot{Q}_{l}=\dot{Q}_{T O T}-\dot{Q}_{s}=\dot{Q}_{s}/3. Writing the energy balance equation for the sensitive load associated with dry air
\dot{m}_{a}c_{p,a}T_{4}+\dot{Q}_{s}=\dot{m}_{a}c_{p,a}T_{1}\rightarrow\dot{Q}_{s}=8.43\,\mathrm{kW}Using the energy balance equation for the latent load gives
{\dot{m}}_{a}\omega_{4}\bigl(l(0^{\circ}{\bf C})+c_{p,\nu}{ T}_{4}\bigr)+{\frac{\dot{Q}_{s}}{3}}={\dot{m}}_{a}\omega_{1}\bigl(l(0^{\circ}{ C})+c_{p,\nu}{T}_{1}\bigr) \\ \omega_{4}=\frac{1.2\cdot0.01\ (2500+1.86\cdot25)-8.43/3}{1.2\ (2500+1.86\cdot18)}\to\omega_{4}=9\ \frac{\mathrm{g}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}Therefore, the state of the air at the inlet to the room is 4(T_{4}=18^{\circ}{\mathrm{C}},\,\omega_{4}=9\,\mathrm{g/kg~d~a)} and the state of the air at the outlet of the cooling battery is 3(T_{3}\,=\,14^{\circ}{\mathrm C},\omega_{3}=\omega_{4}=9\;\mathrm{g}/\mathrm{kg~d~a}).
(b) Since in state 0 the exergy of the air is zero, carrying out a balance of exergy in the mixture of the two flows we have
{\dot{m}}_{a,1}\left(b_{1}+b_{1}^{ch}\right)-{\dot{m}}_{a}\bigl(b_{2}+b_{2}^{c h}\bigr)={\dot{I}}_{MIX}Calculating each of the terms on the left of the equality. The physical exergy is obtained by applying Eq. (3.37), giving
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}=138\ {\frac{\rm{J}}{\rm{kg}\ \mathrm{d}\mathrm{~a}}}\quad b_{2}=5\ \frac{\rm{J}}{\rm{kg}\ \mathrm{d}\mathrm{~a}}The temperature of state 2 is very close to the ambient temperature, and hence, its physical exergy is practically negligible. The chemical exergy is calculated by applying Eq. (3.124), giving
b_{h a}^{c h}=0.461(\omega+0.622)T_{0}\biggl[\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}=801~{\frac{\mathrm{J}}{\mathrm{kg\,d\,a}}}~~~~b_{2}^{c h}=26~{\frac{\mathrm{J}}{\mathrm{kg\,d\,a}}}Returning to the equation of exergy balance, we get
\dot{I}_{MIX}=188\ \mathrm{W}(c) We calculate first the mass flow rate of cold water in the cold battery. The energy balance is
{\dot{m}}_{a}(h_{2}-h_{3})={\dot{m}}_{w,c b}{\Bigl(}h_{c b,i n}-h_{c b,o u}{\Bigr)}+{\dot{m}}_{c o n d}h_{c o n d}Since the condensate is {\dot{m}}_{c o n d}={\dot{m}}_{a}(\omega_{2}-\omega_{3})=15.6\mathrm{~g/s} we get that
\dot m_{w,c b}=2.94\ \frac{\mathrm{kg}}{\mathrm{s}}and the mass flow rate of hot water in the heat battery
\dot{m}_{w,h b}\bigl(h_{h b,i n}-h_{h b,o u}\bigr)=\dot{m}_{a}\bigl(h_{4}-h_{3}\bigr)\rightarrow\dot{m}_{w,h b}=0.15\ {\frac{\mathrm{kg}}{\mathrm{s}}}The product of the AHU is the air mass flow rate provided in the conditions of state 4. The fuel is the exergy provided to the cold and heat batteries, as well as that provided by the recirculated air. Therefore, the exergy efficiency of the AHU is
\varphi=\frac{\dot{m}_{a}\Big(b_{4}+b_{4}^{c h}\Big)}{\dot{m}_{w,c b}\Big(b_{c b,i n}-b_{c b,o u}\Big)+\dot{m}_{w,h b}\Big(b_{h b,i n}-b_{h b,o u}\Big)+\dot{m}_{a,1}\Big(b_{1}+b_{1}^{c h}\Big)}Applying Eq. (3.37) and Eq. (3.124) for the calculation of the physical and chemical exergy of the air in state 4 gives
b_{4}=0.44\ {\frac{\mathrm{kJ}}{\mathrm{kg\,d\,a}}}\quad b_{4}^{c h}=0.93\ {\frac{\mathrm{kJ}}{\mathrm{kg\,d\,a}}}and for the air in state 1
b_{1}=0.14\ {\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ a}}\ \ \ \ b_{1}^{c h}=0.80\ {\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ a}}On the other hand, applying Equation (3.44), we find that the exergy provided in the cold battery is
\dot{B}=\dot m c\left(T-T_{0}-T_{0}l n\frac{T}{T_{0}}\right)\qquad\qquad(3.44) \\ \dot{m}_{w,c b}\left(b_{c b,i n}-b_{c b,o u}\right)=5.33\:{\mathrm{kW}}and in the heat battery
\dot{m}_{w,h b}\left(b_{h b,i n}-b_{h b,o u}\right)=0.36\mathrm{~kW}Coming back to the expression that reflects the exergy efficiency, we finally get
\varphi=24.1\%