Question 6.2: Let there be a LiBr/H20 single effect absorption refrigerato......
Let there be a \mathrm{LiBr/H_{2}0} single effect absorption refrigerator, as shown in the diagram in Fig. E.6.1. The generator is driven by a flow of hot gases, while the condenser and absorber give heat to the ambient air which is at 35°C. In the evapo-rator, a flow of water is cooled from 18°C to 12°C. The states and mass flows of the cycle are shown in Table E.6.1. With this information, calculate
(a) The heat exchanged in the generator, absorber, condenser and evaporator and the EER of the unit.
(b) The exergy destruction in the absorber.
(c) The exergy efficiency of the refrigerator
Table E.6.1 Thermodynamic data of the states.
| i | T_{i}(\mathrm{C}) | h_{i}(\mathrm{kJ/kg}) | s_{i}(\mathrm{kJ/kg}\cdot\mathrm{k}) | m_{i}(\mathrm{kg/s}\mathrm{}) | X_{i}(\%\mathrm{LiBr}\mathrm{}) |
| 1 | 105 | 257.63 | 2.246 | 0.171 | 64 |
| 2 | 77.3 | 194.34 | 2.274 | 0.171 | 64 |
| 3 | 58.5 | 194.34 | 2.274 | 0.171 | 64 |
| 4 | 47 | 126.74 | 2.278 | 0.184 | 59.5 |
| 5 | 47.74 | 128.18 | 2.282 | 0.184 | 59.5 |
| 6 | 70.2 | 186.6 | 2.445 | 0.184 | 59.5 |
| 7 | 105 | 2696.9 | 8.448 | 0.013 | _ |
| 8 | 47 | 196.3 | 0.663 | 0.013 | _ |
| 9 | 10 | 196.3 | 0.663 | 0.013 | _ |
| 10 | 10 | 2519.35 | 8.905 | 0.013 | _ |
| 11 | 372 | 3624.34 | 6.307 | 0.034 | _ |
| 12 | 200 | 2308.09 | 4.727 | 0.034 | _ |
(a) The heat given up by the hot gas to the generator is
\dot{Q}_{G}=(\dot{m}_{1}h_{1}+\dot{m}_{7}h_{7})-\dot{m}_{6}h_{6}=44.72\;k Weffectively coinciding with. {\dot{m}}_{11}(h_{11}-h_{12})\,=44.72\,{\mathrm{kW}}
The heat given to the air in the absorber is
\dot{Q}_{A B S}=\left(\dot{m}_{3}h_{3}+\dot{m}_{10}h_{10}\right)-\dot{m}_{4}h_{4}=42.66\,\mathrm{kJ}and in the condenser
\dot{Q}_{C O N D}=\dot{m}_{7}(h_{7}-h_{8})=32.50\ \mathrm{kW}The cold produced in the evaporator is
\dot{Q}_{E V A P}=\dot m_9(h_{10}-h_{9})=30.20\,\mathrm{kW}We can verify that these results are correct through a balance of energy
\dot{Q}_{G}+\dot{Q}_{E V A P}+\dot{W}_{p}=\dot{Q}_{C O ND}+\dot{Q}_{A B S}where the power of the pump is \dot{W}_{p}=\dot m_{4}(h_{5}-h_{4})=0.27\,\mathrm{kW} which can be considered negligible compared to the heats exchanged. As we can see, the energy balance is satisfied.
The EER is
E E R\approx\frac{\dot{Q}_{E V A P}}{\dot{Q}_{G}}=0.67(b) In the absorber, all the heat transferred to the refrigeration circuit is finally dissipated, and the exergy provided to that refrigeration circuit is destroyed. Therefore, we group that exergy with the destructions in the term for irreversibility, resulting in the following equation for the exergy balance
\dot{m}_{3}(b_{3}-b_{4})+\dot{m}_{10}(b_{10}-b_{4})-\Delta\dot{B}_{A B S}^{c h}=\dot{I}_{A B S}Calculating each of the terms on the left of the previous equality
{\dot{m}}_{3}(b_{3}-b_{4})=11.77\,{\mathrm{kW}} \\ \dot{m}_{10}(b_{10}-b_{4})=4.57\,{\mathrm{kW}} \\ -\Delta{\dot{B}}_{A B S}^{c h}={\dot{m}}_{3}R_{M,3}T_{0}\sum_{i}\left(y_{i}l n_{y_{i}}\right)_{3}-{\dot{m}}_{4}R_{M,4}T_{0}\sum_{i}\left(y_{i}l n_{y_{i}}\right)_{4}=0.78\mathrm{~kW}since M_{m3}=0.64.86.84+0.36.18=62.0\ k{\mathrm{g/kmol~and~}}M_{m4}=0.595.86.84+0405.18=58.9\,\,\mathrm{kg/kmol}. And so
\dot{I}_{A B S}=17.12\,\mathrm{kW}(c) To calculate the exergy efficiency, we first determine the exergy of the cold produced
\dot{Q}_{E V A P}=\dot{m}_{w}4.18(18-12)\to\dot{m}_{w}=1.20~\frac{k g}{s}The exergy given to the cooled water is
\Delta\dot B_{w}=2.09{\mathrm{~kW}}As the exergy contributed to the generator is
{\dot{m}}_{11}(b_{11}-b_{12})=28.21\mathrm{~kW}the exergy efficiency of the absorption refrigerator is
\varphi={\frac{\Delta\dot{B}_{w}}{\dot{m}_{11}(b_{11}-b_{12})}}=7.4\%