Question 6.6: In the AHU of an air conditioning installation, a flow of 25......
In the AHU of an air conditioning installation, a flow of 250 m³/h of outdoor air at 38°C, relative humidity 78% and pressure of 1 bar is mixed with a flow of 70 m³/h of recirculated air at 26°C and relative humidity 60%. The air resulting from the mixture is passed through a cold battery, where it is cooled to a temperature of 15°C. In this battery, the fluid that cools the air is a flow of water that enters the battery at 7°C and leaves at 12°C. As part of the vapour condenses, there is then a moisture separator and finally a heat battery, where a flow of hot water decreases its temperature from 55°C to 45°C, with the final air temperature being 21°C and where the heat losses are 30%. Determine
(a) The state of the air resulting from the mixture of the two flows.
(b) The heat exchanged by the air in the cold battery and the quantity of condensed water.
(c) The exergy destruction in the cold battery.
(d) The flow of hot water in the heat battery.
(e) The exergy destruction in the heat battery.
(f) Energy efficiency and exergy efficiency of the AHU and total irreversibilities.
(a) First, we calculate the mass flow of the outside air, state 0, and recirculation air, state 1. The specific volume per unit mass of dry air is obtained from the thermal state equation, since
p V=(m_{a}R_{a}+m_{\nu}R_{\nu})T\rightarrow\nu={\frac{V}{m_{a}}}=(\omega+0.622)R_{\nu}{\frac{T}{p}}Calculating the absolute humidity of the air in states 0 and 1. As p_{s}(38^{\circ}{\mathrm{C}})= 66.33 mbar, and p_{s}(26^{\circ}{\mathrm{C}})= 33.64 mbar we have
\omega_{0}=0.622{\frac{p_{s}(38^{\circ}{\mathrm{C}})}{{\frac{p_{0}}{\phi_{0}}}-p_{s}(38^{\circ}{\mathrm{C}})}}=34\;{\frac{\mathrm{g}}{{\mathrm{kg~d~a}}}} \\ \omega_{1}=0.622{\frac{p_{s}(26^{\circ}{\mathrm{C}})}{{\frac{p_{0}}{\phi_{1}}}-p_{s}(26^{\circ}{\mathrm{C}})}}=13\;{\frac{\mathrm{g}}{{\mathrm{kg~d~a}}}}Using the previous expression for the specific volume we have
\nu_{0}=(\omega_{0}+0.622)R_{\nu}\frac{T_{0}}{p_{0}}=0.94\;\frac{{\mathrm m}^{3}}{{\mathrm k g\,}\mathrm{d}\,a}\quad\nu_{1}=(\omega_{1}+0.622)R_{\nu}\frac{T_{1}}{p_{0}}=0.88\ \frac{{\mathrm m}^{3}}{{\mathrm k g\,}\mathrm{d}\,a}Therefore, the dry air mass flow rates are
\dot{m}_{a,0}=\frac{\dot{V}_{0}}{\nu_{0}}=265.9\;\frac{\mathrm{kg\;d\;a}}{\mathrm{h}}\;\;\;\;\;\dot{m}_{a,1}=\frac{\dot{V}_{1}}{\nu_{1}}=79.5\;\frac{\mathrm{kg\;d\;a}}{\mathrm{h}}Once these values are obtained, we calculate the air resulting from the mixture, which we call air in state 2.
\dot{m}_{a,0}+\dot{m}_{a,1}=\dot{m}_{a,2}{\rightarrow}\dot{m}_{a,2}=345.4~{\frac{\mathrm{kg~d~}\mathrm{a}}{\mathrm{h}}} \\ \dot{m}_{a,0}\omega_{0}+\dot{m}_{a,1}\omega_{1}=\dot{m}_{a,2}\omega_{2}\rightarrow\omega_{2}=29\;\frac{\rm g}{\mathrm{kg\,d\,a}} \\ {\dot{m}}_{a.0}h_{0}+{\dot{m}}_{a.1}h_{1}={\dot{m}}_{a.2}h_{2}\to{T}_{2}=35.7^{\circ}\mathrm{C}b) To calculate the heat exchanged by the air in the cold battery, we first determine the condensed water. Assuming that the bypass factor of the battery is zero, so that all the air is treated ideally and saturated. With 3 being the state of the air at the exit of the battery, since p_{s}(15^{\circ}{\mathrm{C}})=17.06 mbar and \phi_3=1 we have
\omega_{3}=0.622{\frac{p_{s}(15^{\circ}C)}{\displaystyle{\displaystyle{\frac{p_0}{\phi_3}-p_s(15^\circ C)}}}}=10.8~{\frac{g}{k g\,d\,a}}Therefore, the amount of condensed water is
{\dot{m}}_{c o n d}={\dot{m}}_{a,2}(\omega_{2}-\omega_{3})=6.29~{\frac{\mathrm{kg}}{\mathrm{h}}}From the energy balance we get
{\dot{Q}}_{c b}={\dot{m}}_{a,2}(h_{2}-h_{3})-{\dot{m}}_{c o n d}h_{c o n d}and since
\dot{m}_{a,2}(h_{2}-h_{3})-\dot{m}_{c o n d}h_{c o n d}=\dot{m}_{a,2}[c_{p,a}T_{2}+\bigl(2500+c_{p,\nu}T_{2}\bigr)\omega_{2}-c_{p,a}T_{3}+\left(2500+c_{p,\nu}T_{3}\right)\omega_{3}]-\dot{m}_{c o n d}c_{c o n d}T_{3}this means
{\dot{Q}}_{c b}=6.51-0.11=6.40\,{\mathrm{kW}}Although the enthalpy of the condensate has been included in this equation, we see that its value is practically negligible compared to the change of the enthalpy of the air.
(c) Carrying out an exergy balance in the battery and taking into account that the small exergy of the condensate is lost, we have
Calculating each of the terms on the left of the equality. We first determine the mass flow rate of cold water in the battery
\dot{Q}_{c b}=\dot{m}_{w,c}{\Bigl(}h_{w(1{2}^\circ{\mathrm C})}-h_{w(7^{\circ}{\mathrm C})}{\Bigr)}\rightarrow\dot{m}_{w,c}=0.30~\frac{\mathrm{kg}}{\mathrm{s}}We now calculate the terms of the exergy balance. The exergy change of water in the cold battery is calculated using Eq. (3.43) and the physical exergy of the air using Eq. (3.37)
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=m c\bigg(T-T_{0}-T_{0}l n\frac{T}{T_{0}}\bigg)\qquad\qquad(3.43) \\ b_{w,in}-b_{w,\;o u}=2.11\mathrm{~{\frac{kJ}{k g}}}\;\;\;\;b_{2}=0.009\mathrm{~{\frac{kJ}{k g}}}\;\;\;\;b_{3}=0.92\ {\frac{k J}{k g}}The specific chemical exergy of the air in states 2 and 3 is calculated by applying Eq. (3.124), resulting in
b_{h a}^{c h}=0.461(\omega+0.622)T_{0}\left[\frac{0.622}{\omega+0.622}l n\frac{\omega_{0}+0.622}{\omega+0.622}\right]\ +\,{\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_{2}^{c h}=0.05\ \frac{\mathrm{kJ}}{\mathrm{kg}}\ \ \ b_{3}^{c h}=1.49\ \frac{\mathrm{kJ}}{\mathrm{kg}}From the balance equation, we have
{\dot{I}}_{c b}=0.41~{\mathrm{kW}}(c) The humidity of the air at the output of the heat battery, state 4, is the same as in state 3, \omega_{4}=10.8\;\mathrm{g/kg}\;\mathrm{da}. The heat given to the air in the heat battery is
{\dot{Q}}_{h b}=\dot{m}_{a,3}(h_{4}-h_{3})\rightarrow{\dot{Q}}_{h b}=0.59\,k Wwith the mass flow rate of hot water being
\dot{Q}_{h b}={\dot{m}}_{w,\,h}\Bigl(h_{w(55^{\circ}{\bf C})}-h_{w(45^{\circ}{\rm C})}\Bigr)\rightarrow{\dot{m}}_{w,\,h}=50.8\;\frac{\mathrm{kg}}{\mathrm{h}}(d) From the exergy balance we get
{\dot{m}}_{w,\,h}(b_{w,i n}-b_{w,o u})+{\dot{m}}_{a,3}(b_{3}-b_{4})={\dot{I}}_{b c}where
\dot{m}_{w,h}\left({b_{w,i n}-b_{w,\,o u}}\right)=21.8\ \rm W \\ \dot{m}_{a,3}(b_{3}-b_{4})=40.5\;\mathrm{W}and therefore
\dot I_{h b}=62.4\,\mathrm{W}(e) Under the perspective of the First Law, the resources (F) used in the AHU are the energy provided by the heat battery, the cold supplied in the cold battery and the enthalpy of the recirculation air. We will ignore the enthalpy of the air in state 0, since it is ambient air. And so
F=\dot{Q}_{c b}+\dot{Q}_{h b}+{\dot{m}}_{a,1}h_{1}=6.40+0.59+1.31=8.30\,k WThe goal of the AHU, that is, its product P, is to prepare the airflow in state 4. Therefore
P=\dot m_{a,4}h_{4}=4.70\,\,\mathrm{kW}Therefore, the energy efficiency of the AHU is
\eta={\frac{P}{F}}={\frac{4.70}{8.30}}=56.6\%We now determine the exergy efficiency. The exergy contributed to the AHU is
{F}{=}\dot{m}_{w,c}\Bigl(b_{w(7^{\circ}{\mathrm{C}})}-b_{w(12^{\circ}{\mathrm{C}})}\Bigr)+\dot{m}_{w,h}\Bigl(b_{w(55^{\circ}{\mathrm{C}})}-b_{w(45^{\circ}{\mathrm{C}})}\Bigr)+\dot{m}_{a,1}\Bigl(b_{1}+b_{1}^{c h}\Bigr)Calculating each of the terms that form part of the resources used
{\dot{m}}_{w,c}\left(b_{w(7^\circ \rm C)}-b_{w(12^\circ \rm C)}\right)=630~W\quad\dot{m}_{w,h}\left(b_{w(55^\circ \rm C)}-b_{w(45^\circ \rm C)}\right)=22~\mathrm{W}The physical and chemical exergy of the air in state 1 is
b_{1}=0.24~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\,\mathrm{a}}}\,b_{1}^{c h}=1.17\,{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\,\mathrm{a}}}\to\dot{m}_{a,1}\Big(b_{1}+b_{1}^{c h}\Big)=31\,\mathrm{W}Therefore, the fuel contributed to the AHU is
F=683\ \mathrm{W}The product of the installation is the flow of air-conditioned generated, this is
P={\dot{m}}_{a,4}\ {\bigl(}b_{4}+b_{4}^{c h}{\bigr)}where, applying Eq. (3.37) and Eq. (3.124) we get
b_{4}=0.49~{\frac{\mathrm{kJ}}{\mathrm{kg~d~a}}}~~~~b_{4}^{c h}=1.46~{\frac{\mathrm{kJ}}{\mathrm{kg~d~a}}}and so
P=187\,\mathrm{W}In short, the exergy efficiency of the AHU is
\varphi{=}{\frac{P}{F}}=27.4\%so that the total irreversibilities of the AHU represent 72.6% of the exergy contributed, that is, 496 W.