Consider the absorption refrigeration cycle, shown in the following line diagram, which uses lithium bromide as carrier and water as refrigerant to provide \(1 \mathrm{~kW}\) of cooling. By using steam tables at http://www.ohio.edu/mechanical/thermo/ property_tables/ \(\mathrm{H}_{2} \mathrm{O}\) /index.html and the chart giving the properties of lithium bromide and water, calculate first

a. Heat removed from the absorber

b. Heat removed from the condenser

c. Heat added to the evaporator

d. COP of the cycle 

Then calculate, for a flat-plate collector with an \(F_{R}(\tau \alpha)\) intercept of 0.81 and an \(F_{R} U_{c}\) of \(3 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), the area required for operation in Arizona at noon in August for a 3 -ton unit. Assume that the enthalpy of the water vapor leaving the condenser can be approximated by the equation

\[ h_{v c}=2463+1.9 T_{c} \mathrm{~kJ} / \mathrm{kg} \]

and that the enthalpy of the liquid water is

\[ h_{\ell c}=4.2 T_{\mathrm{c}} \mathrm{kJ} / \mathrm{kg} \]

where \(T_{c}\) is the temperature of the evaporator in \({ }^{\circ} \mathrm{C}\). In the analysis, assume that evaporation occurs at \(1^{\circ} \mathrm{C}\) and condensation at \(32^{\circ} \mathrm{C}\).

Transcribed Image Text:

Superheated water vapor 0.406 g/s 2603 kJ/kg Generator Condenser Solution 74C 32C 9.35 g/s 47.54 47.54 60% LiBr mbar mbar -84 kJ/kg Pump Liquid water 0.406 g/s 134.08 kJ/kg Expansion valve Solution 9.76 g/s 57.5% LiBr -165 kJ/kg 32C 6.57 mbar Absorber Saturated water vapor 0.406 g/s 2464.89 kJ/kg 1C 6.57 mbar Evaporator