Question 6.15: In a flat solar collector, the heat extraction factor is FR ......
In a flat solar collector, the heat extraction factor is F_{R}=0.82, the coefficient of heat loss to the environment U_{L}=7.8\mathrm{~W/m^{2}K} and the product of the absorbance of the heating plate of the collector and the transmittance of the transparent cover \big[{\overline{{\tau\alpha}}}\big]\,=0.84, with the collector surface area being 5 m². The water flow rate is 600 L/h, with the collector input temperature being 34°C. In a locality where the average incident radiation is 660 W/m² and the ambient temperature is 10°C, determine
(a) The energy efficiency of the collector and the temperature of the water at the outlet.
(b) The irreversibilities and exergy efficiency of the collector.
(c) Find an expression for the exergy efficiency of the collector as a function of the water inlet
and outlet temperature and the ambient temperature.
(a) According to the expression for the energy efficiency of the collector, Eq. (6.110), we have
\eta_{c}=\frac{Q_{U}}{A_{c}G_{s}}=F_{R}\big[\overline{{{\tau\alpha}}}\big]-F_{R}U_{L}\frac{T_{i n}-T_{0}}{G_{s}}\qquad\qquad(6.110) \\ \eta_{c}=\frac{Q_U}{A_{c}G_{s}}=F_{R}\big[\overline{{{\tau\alpha}}}\big]-F_{R}U_{L}\frac{T_{i n}-T_{0}}{G_{s}}=0.69-6.4\frac{T_{i n}-T_{0}}{G_{s}}If we plot \eta_{c} against (T_{i n}\!-\!T_{0})/G_{s}, a negative slope line F_{R}U_{L}=-6.4 results and with an ordinate in the origin equal to F_{R}\big[\overline{{{\tau\alpha}}}\big]\,=0.69. Under the conditions of the example, collector efficiency is
\eta_{c}=0.69-6.4{\frac{34-10}{660}}=45.7\%The water temperature at the outlet of the collector will be
\dot{m}_{w}c_{w}(T_{o u}-T_{i n})=\eta_{c}A_{c}G_{s}\to T_{o u}=308\mathrm{K}(35.5^{\circ}{\mathrm{C}})(b) The increase in exergy of the hot water generated in the collector is
\Delta\dot{B}=\dot{m}_{w}c_{w}\left(T_{o u}-T_{i n}-T_{0}l n\frac{T_{o u}}{T_{i n}}\right)=84~WConsidering an equivalent temperature of the sun T_{s u n}=5.700\,\,\mathrm{K}, the exergy of the incident radiation is
\left[1-\frac{4}{3}\,\frac{T_{0}}{T_{s u n}}+\frac{1}{3}\left(\frac{T_{0}}{T_{s u n}}\right)^{4}\right]\!A_{c}G_{s}=3,081\,\mathrm{W}Undertaking an exergy balance, we find that the total irreversibilities in the collector are
\dot{B}_{r.s}=\Delta\dot{B}+\dot{I}_{c}\rightarrow\dot{I}_{c}=2,997\,\mathrm{W}so that the exergy efficiency of the collector is
\varphi_{c}=2.7\%(c) The expression for the exergy efficiency as a function of (T_{i n},T_{o u },T_{0}) is obtained from the relationship that links the exergy efficiency with the energy efficiency.
\varphi_{c}=\eta_{c}{\frac{G_{s}}{\dot B_{s}}}{\frac{\Delta b}{\Delta h}}since
\eta_{c}=\eta_{c}(T_{e},T_{0})=0.69-6.4{\frac{T_{i n}-T_{0}}{660}}=0.69-0.01(T_{i n}-T_{0}) \\ \frac{G_{s}}{\dot{B}_{s}}=\frac{G_{s}}{\dot{B}_{s}}(T_{0})=\frac{1}{\left[1-\frac{4}{3}\,\frac{T_{0}}{5.700}+\frac{1}{3}\left(\frac{T_{0}}{5.700}\right)^{4}\right]} \\ \frac{\Delta b}{\Delta h}=\frac{\Delta b}{\Delta h}(T_{i n},T_{o u},T_{0})=1-T_{0}\frac{l n(T_{o u}/T_{i n})}{T_{o u}-T_{i n}}Substituting these relations in the equation that links the energy efficiency and the exergy efficiency, we obtain the expression for the exergy efficiency as a function of (T_{i n},T_{o u },T_{0}).