Question 6.11: A commercial premises of 250 m² surface and 5 m in height ha......
A commercial premises of 250 m² surface and 5 m in height has a mechanical ventilation system with heat recovery. The recuperator performs the exchange of heat between the extracted air and the new air, incorporating an impulsion fan and another for the return of the air in the premises, with an electrical efficiency of 92%, and with the section of the duct being 1.2 m² and the pressure losses in the recuperator being 190 Pa. The indoor air temperature is 23°C, and its relative humidity is 55%; the outdoor air temperature is 35°C and its relative humidity is 60%, with the recuperator effectiveness being 70%. The premises have an air conditioning system with a heat pump of COP equal to 3.1. If the environmental pressure is 1 bar, determine
(a) The net energy saving achieved in the recuperator and primary energy saving.
(b) The irreversibilities and exergy efficiency of heat recovery.
(c) The threshold value of the outside temperature, from which the recovery is not thermo-dynamically viable.
(a) For a commercial premises, IDA3 quality indoor air is needed, which means that we should consider 4 m³/person. As there are 1250 m³ this is the equivalent of 313 people. A renewal rate of 8 L/s per person implies a ventilation airflow of 2.5 m³/s. The absolute humidity of the atmospheric air is
\omega_{0}=0.622{\frac{p_{s}(35^{\circ}C)}{\displaystyle{\frac{p_0}{\phi_{0}}}-p_{s}(35^{\circ}C)}}=21.7\;{\frac{g}{k g\,d\,a}}and its density
\varrho_{0}={\frac{1}{R_{a}+\omega_{0}R_{\nu}}}{\frac{p_{0}}{T_{0}}}=1.087\ {\frac{\mathrm{kg~d~a}}{\mathrm{m}^{3}}}Therefore, the mass flow rate of dry air is
\dot{m}_{a}=\dot{V}_{0}\varrho_{0}=2.72{\frac{\mathrm{kg}\ \mathrm{d}\ a}{s}}According to the definition of effectiveness, we have
\varepsilon={\frac{T_{0}-T_{1}}{T_{0}-T_{2}}}={\frac{308-T_{1}}{35-23}}\to T_{1}=299.7\mathrm{K}(26.6^{\circ}\mathrm{C})The energy saving that is achieved in the recuperator, when cooling the outside air to that temperature of 26.6°C is
E S=\varrho_{0}\dot{V}_{0}(c_{p,a}+\omega_{0}c_{p,\nu})(T_{0}-T_{1})=23.83~\mathrm{kW}To find the net energy saving, we calculate the power consumed by the fans. The total pressure they must supply is
\Delta p=\Delta p_{s t a}+p_{d y n}=190+1.087{\frac{1}{2}}\left({\frac{2.5}{1.2}}\right)^{2}=192.3\,\mathrm{Pa}and therefore
\dot{W}_{\nu}=\frac{1}{\eta_{e l,\;m}}2\dot{V}_{0}\Delta p=1.04\ {\rm k W}The net energy saving is
E S_{n}=23.83-1.04=22.79\mathrm{~kW}Using the same efficiency as in the previous example for electricity, taking into ac-count that now the COP of the heat pump is 3.1, we
P E S_{n}={\frac{23.83}{3.1\cdot0.416}}-{\frac{1.04}{0.416}}=15.97\,\mathrm{kW}(b) Carrying out an exergy balance in the recuperator, we have
\dot{m}_{a}b_{2}+\dot{W}_{\nu}=\dot{m}_{a}b_{1}+\dot{I}_{r e c}Calculating the physical exergy of the air in states 1 and 2. We first determine the absolute humidity of the indoor air in the premises
\omega_{2}=0.622{\frac{p_{s}(23^{\circ}{\mathrm{C}})}{\frac{p_{0}}{\phi_{2}}-p_{s}(23^{\circ}{\mathrm{C}})}}=9.7\ {\frac{\mathrm{g}}{{\mathrm{kg~d~a}}}}We now calculate the physical exergy of air in states 1 and 2 by applying Eq. (3.37), in which we do not take into account the effect of the pressure drop, resulting in
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}=0.12~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ a}}\;\;\;\;\;b_{2}=0.24~{\frac{\mathrm{kJ}}{\mathrm{kg}\ \mathrm{d}\ a}}From the exergy balance equation and taking into account that {\dot{m}}_{a}=\varrho_{0}{\dot{V}}_{0} gives
\dot{I}_{r e c}=1.37\,\mathrm{kW}The exergy efficiency of the recuperator is
\varphi=1-{\frac{\dot{I}_{r e c}}{\dot{m}_{a}b_{2}+\dot{W}_{\nu}}}=19.0\%(c) From the perspective of the First Law, the recovery ceases to be viable when there is no net energy saving, that is, when
E S_{n}=\rho_{0}{\dot{V}}(c_{p,a}+\omega_{0}c_{p,\nu})(T_{0}-T_{1})-{\dot{W}}_{\nu}=0In a first approximation, considering that the effectiveness of the recuperator re-mains constant, we get that the limit temperature is
T_{0}=T_{2}+\frac{\dot{W}_{\nu}}{\varepsilon\dot{m}_{a}(c_{p,a}+\omega_{0}c_{p,\nu})}=23,3^{\mathrm{o}}{\mathrm{C}}However, if the analysis is done from the perspective of the First and Second Laws, the recovery ceases to be viable since there is no exergy saving, this is
E x S_{n}=\rho_{0}{\dot{V}}\left(c_{p,a}+\omega_{0}c_{p,\nu}\right)\biggl[T_{1}-T_{0}-T_{0}l n{\frac{T_{1}}{T_{0}}}\biggr]-{\dot{W}}_{\nu}=0Considering as before that the effectiveness of the recuperator remains constant, the temperature from which the recovery is no longer viable is obtained by solving the equation
T_{0}\Bigl[\varepsilon+l n\Bigl[1+\varepsilon\Bigl(\frac{T_{2}}{T_{0}}-1\Bigr)\Bigr]\Bigr]=\varepsilon T_{2}-\frac{\dot{W}_{\nu}}{\dot{m}_{a}(c_{p,a}+\omega_{0}c_{p,\nu})}giving approximately
T_{0}\approx305.5\mathrm{K}(32.4^{\circ}{\mathrm{C}})