Question 6.10: A dwelling has a mechanical ventilation system with heat rec......
A dwelling has a mechanical ventilation system with heat recovery. The recuperator, with its corresponding ports and some filters, allows the exchange of heat between the extracted air and the new air, incorporating an impulsion fan and another for the return of the air of the dwelling, with an electrical efficiency of 90%. The ventilation airflow rate is 1200 m³/h, the section of the duct is 16 dm² and the pressure losses in the recuperator are 150 Pa, with its effectiveness being 70%. The indoor air temperature is 21°C, and its relative humidity is 55%, with that of the outside air being 5°C and its relative humidity 70%. The dwelling has a radiant floor heating system with an air-water heat pump of COP equal to 3.6. 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 exergy saving achieved.
(d) The decrease in the electrical power of the heat pump required when the recuperator is installed
(a) According to the definition of effectiveness, we have
\varepsilon=0.7={\frac{T_{1}-278}{21-5}}\to T_{1}=16^{\circ}C(289\,K)With p_{s}(5^{\circ}{\mathrm{C}})=8.7 mbar, we first calculate the atmospheric air humidity
\omega_{0}=0.622{\frac{p_{s}(5^{\circ}{\mathrm{C}}\,)}{{\frac{p_{0}}{\phi_{0}}}-p_{s}(5^{\circ}{\mathrm{C}}\,)}}=3.8~{\frac{\mathrm{g}}{{\mathrm{kg\,d\,a}}}}As the atmospheric air density is
\varrho_{0}={\frac{1}{R_{a}+\omega_{0}R_{\nu}}}{\frac{p_{0}}{T_{0}}}=1.238\ {\frac{\mathrm{kg~d~a}}{\mathrm{m}^{3}}}the energy saving achieved with the recuperator is
E S=\varrho_{0}\dot{V}_{0}\left(c_{p,a}+\omega_{0}c_{p,\nu}\right)\left(T_{1}-T_{0}\right)=4,589\;\mathrm{W}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_{dyn}=150+\frac{1}{2}\left(\frac{1200}{3600\cdot0.16}\right)^{2}1.238=152.2\,\mathrm{Pa}and therefore the fan power is
\dot{W}_{\nu}=\frac{1}{\eta_{e l.m}}2\dot{V}_{0}\Delta p=112\mathrm{~W~}The net energy saving is
E S_{n}=4,477{\mathrm{~W}}We use the value supplied by IDAE as the coefficient of passage, from primary energy to final; as of 2016, for peninsular electricity, this is 2.403 kWh total PE/final kWh, which is equivalent to saying that the electrical efficiency of the peninsular electrical system is \eta_{e l}=41.6\%. Taking into account that heating generation is carried out with a heat pump of COP = 3.6, we get that the Primary Energy Net Saving is
P E S_{n}={\frac{4.589}{3.6\cdot0.416}}-{\frac{112}{0.416}}=2.795\ \mathrm{W}(b) Since the pressure inside the dwelling is equal to the external pressure, the increase in the exergy of the intake air in the recuperator is
b_{1}=\left(c_{p,a}+\omega_{0}~c_{p,\nu}\right)\left[\left((T_{1}-T_{0})-T_{0}~l n\frac{T_{1}}{T_{0}}\right)\right]=214~\frac{\mathrm{J}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}To calculate the physical exergy of the indoor air, we first determine its absolute humidity
\omega_{2}=0.622{\frac{p_{s}(21^{\circ}{\mathrm{C}})}{\frac{p_{0}}{\phi_{2}}-p_{s}(21^{\circ}{\mathrm{C}})}}=14\;{\frac{\mathrm{g}}{{\mathrm{kg~d}}\,\mathrm{a}}}\,and its density
\varrho_{2}={\frac{1}{R_{a}+\omega_{2}R_{\nu}}}{\frac{p_{2}}{T_{2}}}=1.152\ {\frac{\mathrm{kg\;d\;a}}{{\mathrm{m}}^{3}}}According to Eq. (3.37) we have previously written for state 1, the specific exergy of the indoor air, state 2, is
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}{p0}}\qquad\qquad(3.37) \\ b_{2}=457\;{\frac{\mathrm{J}}{\mathrm{kg\;d\,a}}}From the equation of 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}The mass flow rate of dry air is
\dot{m}_{a}=\varrho_{0}\dot{V}=0.428\ \frac{\mathrm{kg~d~a}}{\mathrm{s}}Therefore, the irreversibilities in the recuperator are
\dot{I}_{r e c}=\dot{m}_{a}(b_{2}-b_{1})+\dot{W}_{\nu}=216\mathrm{{\rm~W}}The product of the recuperator is the flow of preheated air 1, while the fuel is the sum of the exergy of the air in the room in state 2 and the power of the fans. Therefore, the exergy efficiency of the recuperator is
\varphi=1-\frac{\dot{I}_{r e c}}{\dot{m}_{a}b_{2}+\dot{W}_{\nu}}=\ 29.8\%(c) The net exergy saving is
E x S_{n}={\dot{m}}_{a}b_{1}-{\dot{W}}_{\nu}=-20~WWe can see the great difference that exists between the net saving of energy and the net saving of exergy. Although there is energy saving in the recovery, even at relatively high ambient temperatures, at this temperature of 5°C there is a greater consumption of exergy with the recuperator, so the temperature limit from which there is saving of exergy is noticeably lower. However, to understand the situation more clearly we need to continue with the next question.
(d) Since the existence of the recuperator supposes a reduction in the demand of heating of 4589 W, the decrease required in the electrical power of the heat pump is
Therefore, even though in the recuperator the saving of exergy is negative, the pre-heating of the air gives a reduction in exergy consumption of 1255 W.