Question 6.3: Let there be an AHU rotary desiccant dryer like the one in F......
Let there be an AHU rotary desiccant dryer like the one in Fig. 6.14 in which the outside air in state 1(T_{1}=T_{0}=28^{\circ}{\mathrm C},\phi_{1}=\phi_{0}=80\%) leaves the dryer in state 2(T_{2}=45^{\circ}{\mathrm{C}},\,\omega_{2}=6\,\mathrm{g}/\mathrm{kg~dry~air}). It then passes through the rotary exchanger where it leaves in state 3(T_{3}=26^{\circ}{\mathrm{C}}). At the outlet of the evaporative cooler it is in state 4\,(\omega_{4}=8\,\mathrm{g}/\mathrm{kg}\;\mathrm{d}\,\mathrm{a}), entering the room to be air conditioned. The return air leaves the room in state 5(T_{5}=25^{\circ}{\mathrm{C}},\omega_{5}=11\ \mathrm{g}/\mathrm{kg}\ \mathrm{d}\ a) increasing its humidity in the evap-orative cooler to state 6(\omega_{6}=13\ \mathrm{g}/\mathrm{kg}\ \mathrm{d}\ a). In the heater, a flow of water of 0.64 kg/s enters at 70°C and leaves at 62°C, giving that heat to the air that passes through the rotary desiccant dryer, where it modifies its state and is expelled to the environment in state 9. The rotary desiccant dryer has losses of 6% and the rotary heat exchanger of 4% with respect to the heat supplied. In both evaporative coolers, the water is injected at a temperature of 12°C. Assuming a constant pressure of 1 bar in the whole installation, and with the rate of mass flow of dry air being 1 kg/s, determine
(a) The temperature and absolute humidity of the air at the inlet and outlet of each component of the AHU.
(b) The irreversibilities in the rotary desiccant dryer and percentage of exergy contributed to the AHU.
(c) The irreversibilities in the rotary exchanger.
(d) The irreversibilities in the AHU and its exergy efficiency.
(a) In order to simplify the calculations and show the operations that are being carried out, we use a specific heat for dry air of c_{p,a}=1.004\ {\mathrm{kJ/kg}}\cdot\mathrm{K} and for water vapour c_{p,v}=1.86\,\mathrm{kJ/kg}\cdot\mathrm{K}. As the vapour pressure at 28°C is p_{s}(28^{\circ}C)=37.83 mbar the absolute humidity of the atmospheric air is
\omega_{1}=0.622{\frac{p_{s}(28^{\circ}{\mathrm{C}})}{{\frac{1,000}{\phi_{1}}}-p_{s}(28^{\circ}{\mathrm{C}})}}=19\;{\frac{{\mathrm{g}}}{{{\mathrm{kg}}\,\mathrm{d}\,\mathrm{a}}}}The states 1(T_{1}\,=\,T_{0}\,=\,28^{\circ}{\mathrm C},\;\omega_{1}\,=\,\omega_{0}\,=\,19\,\mathrm{~g/kg~dry~air})\, and 2(T_{2}=45^{\circ}{\mathrm C},\omega_{2}=6\operatorname{g}/\rm kg \operatorname{d}\mathrm{a}) are thus defined. The humidity in state 3 is the same as that of 2, there-fore 3(T_{3}=26^{\circ}C,\omega_{3}=6\,\mathrm{g}/\mathrm{kg\,d\,a}). To determine the temperature in state 4 we under-take a balance of mass and energy in the evaporative process cooler
{\dot{m}}_{a}\omega_{3}+{\dot{m}}_{w,\,E V}={\dot{m}}_{a}\omega_{4}\to{\dot{m}}_{w,\,E V}=2{\frac{g}{s}} \\ {\dot{m}}_{a}h_{3}+{\dot{m}}_{w,E V}h_{w}={\dot{m}}_{a}h_{4}(T_{4}){\rightarrow}{\dot{m}}_{a}\bigl[c_{p,a}T_{3}+{\omega_{3}}\bigl(l(0^{\circ}C)+c_{p,v}T_{3}\bigr)\bigr] \\ {}+\dot{m}_{w.E V}c_{w}T_{w} \\ =\dot m_{a}\left[c_{p,a}T_{4}+\omega_{4}\bigl(l(0^{\circ}C)+c_{p,\nu}T_{4}\bigr)\right]meaning that the temperature T_{4} is
T_{4}=21.1^{\circ}\mathrm{C}Therefore, the state at the outlet of the evaporative process cooler is 4(T_{4}=21.1^{\circ}{\mathrm{C}},\;\omega_{4}=8\;\mathrm{g}/\mathrm{kg}\;\mathrm{d}\,{\mathrm{a}}). The state of the return air at the outlet of the building is known 5(T_{5}=25^{\circ}{\mathrm{C}}_{},\ \omega_5=11\ \mathrm{g}/\mathrm{kg}\ \mathrm{d}\ \mathrm{a}). To define the state at the outlet of the evaporative regeneration cooler we need, as before, the mass and energy balances
\dot{m}_{a}\omega_{5}+\dot{m}_{w,R E V}=\dot{m}_{a}\omega_{6}\rightarrow\dot{m}_{w,R E V}=2\ \frac{g}{s} \\ \dot{m}_{a}h_{5}+m_{w,R E V}h_{w}=\dot{m}_{a}h_{6}(T_{6})\rightarrow T_{6}=20.1^{\circ}{\rm C}And so 6(T_{6}=20.1^{\circ}{\mathrm{C}},\,\omega_{6}=13\ {\mathrm{g}}/{\mathrm{kg}}\,{\mathrm{d}}\,{\mathrm{a}}). State 7 has the same humidity as state 6 and to know its temperature we carry out an energy balance in the rotary heat exchanger
\dot{m}_{a}(h_{2}-h_{3})+\dot{m}_{a}(h_{6}-h_{7})-\dot{Q}_{L,R H E}=0\rightarrow T_{7}=38.2^{\circ}\mathrm{C}since \dot{Q}_{L ,R H E}=0.04\,\dot{m}_{a}(h_{2}\,-h_{3}). So 7(T_{7}=38.2\ \mathrm{C},\,\omega_{7}=13\ \mathrm{g}/\mathrm{kg}\ \ \mathrm{d~~a}). The humidity of state 8 is the same as that of 7 and to find its temperature we carry out a balance of energy in the heat battery
\dot{m}_{a}(h_{7}-h_{8})+\dot{m}_{f}(h_{f,i n}-h_{f,o u})-\dot{Q}_{L,\,R H B}=0 \\ 1.004(T_{8}-38.3)+0.013\cdot1.86(T_{8}-38.3)=0.64\cdot4.18(70-62)\to T_{8}=59^{\circ}\mathrm{C}Therefore 8(T_{8}=59^{\circ}{\mathrm{C}},\omega_{8}=13\mathrm{~g/kg~d~a}). Finally, to define state 9 we carry out a balance of mass and energy in the rotary desiccant dryer
\omega_{8}+(\omega_{1}-\omega_{2})=\omega_{9}\rightarrow\omega_{9}=26\ {\frac{\mathrm{g}}{{\mathrm{kg\,d\,a}}}} \\ \dot{m}_{a}(h_{1}-h_{2})+\dot{m}_{a}(h_{8}-h_{9})-\dot{Q}_{L.R D}=0\rightarrow T_9=41.0^{\circ}\rm Cso that 9(T_{9}=41.0^{\circ}{\mathrm{C}},\,\omega_{9}=26\,\mathrm{g}/\mathrm{kg}\,\mathrm{d}\,\mathrm{a}).
(b) The heat lost plus the exergy destructions are the total irreversibilities of the rotary desiccant dryer. Since for atmospheric air we have b_{1}=b_{1}^{c h}\,=0, and since the exergy of the air in state 9 forms part of the external irreversibilities, the exergy balance in the rotary desiccant dryer tells us that
\dot{m}_{a}\left[\left(b_{8}+b_{8}^{c h}\right)-\left(b_{2}+b_{2}^{c h}\right)\right]=\dot{I}_{R D}Therefore, using the expressions of physical exergy, Eq. (3.37), and chemical exergy, Eq. (3.123), for humid air 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)and in state 8.
b_{8}=1.54\ {\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ a}}\quad b_{8}^{c h}=0.15\ {\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ a}}The exergy balance gives
\dot{I}_{R D}=0.37\,{\mathrm{kW}}The water used in the evaporative coolers is at 12°C. Without taking into account the electricity consumed by the pumps in the installation, we get that the exergy pro-vided to the AHU is
{\dot{m}}_{f}\left(b_{f,i n}-b_{f,o n}\right)+{\dot{m}}_{w a\ E V}b_{w,\ E V}+{\dot{m}}_{w,R E V}b_{w,R E V}=2.40\,\mathrm{kW}so the irreversibility in the rotary desiccant dryer represents 15.5%.
(c) Undertaking an exergy balance in the rotary exchanger gives
(d) From the exergy balance in the AHU, we have
{\dot{m}}_{f}{\bigl(}b_{f,i n}-b_{f,o u}{\bigr)}+{\dot{m}}_{w,\,E V}b_{w,\,E V}+{\dot{m}}_{w,\,R E V}b_{w,\,R E V}+{\dot{m}}_{a}{\bigl(}b_{5}+b_{5}^{c h}{\bigr)} \\ =\dot m_{a}(b_{4}+b_{4}^{c h})+\dot{I}_{A H U}Calculating the physical and chemical exergy in states 4 and 5, we have
b_{4}=0.08~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}}~~~~b_{4}^{c h}=0.57~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}} \\ b_{5}=0.015~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}}~~~~b_{5}^{c h}=0.27~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}}Therefore, the exergy of the air for the air conditioning of the building is \dot{m}_{a}\Bigl(b_{4}+b_{4}^{c h}\Bigr)=0.65\;\mathrm{kW} and the total irreversibilities in the AHU are
\dot{I}_{A H U}=21.03\;{\mathrm{kW}}The object of the AHU is to obtain the airflow in state 4 for the conditioning of the premises; in short, this is its product. Ignoring the exergy of the water in the evapora-tive coolers, the resource used is the exergy provided by the heat battery, so that exergy efficiency gives
\varphi={\frac{\dot{m}_{a}\Big(b_{4}+b_{4}^{c h}\Big)}{\dot{m}_{f}\big(b_{f,i n}-b_{f,o u}\big)}}=3\%