Question 15.11: The following data are given for a lithium bromide–water sys......

Parts (a) and (b) will be solved concurrently. Table 15-4 shows a tabulation of data. Properties of pure water were obtained from Table A-1, and solution properties were btained from Chart 5 in Appendix E. Note that Chart 5 expresses concentration in terms of the absorbent lithium bromide.

The high- and low-side pressures are found from the temperatures t_{8}  and  t_{10},  and  i_{8}  and   i_{10} are found in Table A-1. The enthalpy of the pure water vapor at state 7 may be computed by Eq. 3-19, since it behaves as an ideal gas at the low pressure. States 3 and 4 are saturated conditions and may be located on Chart 5 as shown in Fig. 15-33. Enthalpy and concentration values may then be read for states 3 and 4.

i_{\text{v}}  =  i_{g}  +  c_{p\text{v}}  t              (3-19)

For the evaporator

\dot{m}_{9}  =  \frac{\dot{q}_{e}}{i_{10}  –  i_{9}}  =  \frac{10 (200)}{1011}  =  1.98  lb/min

For the absorber

\dot{m}_{6}  =  \dot{m}_{10}  \frac{x_{1}  –  x_{10}}{x_{6}  –  x_{1}}  =  \frac{1.98 (0.60)}{0.05}  =  23.8  lb/min

\dot{m}_{1}  =  \dot{m}_{6}  +  \dot{m}_{10}  =  25.8  lb/min

For the generator

\dot{q}_{g}  =  \dot{m}_{4} i_{4}  +  \dot{m}_{7} i_{7}  –  \dot{m}_{3} i_{3}

=  23.8 (-27)  +  1.98 (1151)  –  25.8 (-35)

\dot{q}_{g}  =  2538  Btu/min

Neglecting the pump work,

COP  =  \frac{\dot{q}_{c}}{\dot{q}_{g}}  =  \frac{200(10)}{2538}  =  0.79

For a generator source temperature of 220 F and a refrigerated medium temperature of 45 F, and assuming an environment temperature of 100 F, the maximum COP becomes

(COP)_{max}  =  \frac{505(680  –  560)}{680 (560  –  505)}  =  1.62

and

\eta_{R}  =  \frac{0.79}{1.62}  =  0.49

Assuming saturated water at 220 F leaves the steam coil,

\dot{m}_{s}  =  \frac{60 \dot{q}_{g}}{i_{fg}}  =  \frac{60(2538)}{965}  =  158 lb/hr