Question 15.10: Consider the cycle of Fig. 15-31 and the following given dat......

(a) Table 15-3 is a tabulation of thermodynamic properties and flow rates for the problem. Figure 15-32 is a schematic i–x diagram showing the state points.

The given data establish all the pressures, and the temperatures t_{3},  t_{4}  and  t_{7} are given. States 3, 4, and 7 are saturated states and can be located on Chart 2 in Appendix E and the concentration and enthalpy values read. States 7 through 12 have the same concentration because no mixing occurs. It is also true that states 1, 2, and 3 have the same concentration and x_{4}  =  x_{5}  =  x_{6} as well. Now states 1, 8, and 12 are saturation states and may be located on Chart 2, making use of the known pressures and concentrations, and i_{1},  i_{8} ,  and  i_{12}   read.  Since  t_{9}  =  t_{8}  –  10   state  9  is  determined  from  t_{9} ,  and  x_{9}   and  i_{9} are read from the chart.

(b) An energy balance between points 8 and 12 in Fig. 15-32 yields

\dot{m}_{8} i_{8}  +  \dot{q}_{e}  =  \dot{m}_{12} i_{12}  =  \dot{m}_{g} i_{12}

\dot{m}_{8}  =  \frac{\dot{q}_{e}}{i_{12}  –  i_{8}}  =  \frac{(100) (200)}{631  –  148}  =  41.4  lbm/min

From Fig. 15-31 it is obvious that \dot{m}_{7}  =  \dot{m}_{8}  =  \dot{m}_{9}  =  \dot{m}_{10}  =  \dot{m}_{11}  =  \dot{m}_{12}.

Mass balance on the absorber gives

\dot{m}_{12}  +  \dot{m}_{6}  =  \dot{m}_{1}

\dot{m}_{12} x_{12}  +  \dot{m}_{6} x_{6}  =  \dot{m}_{1} x_{1}

and

\dot{m}_{6}  =  \dot{m}_{12}  \frac{x_{12}  –  x_{1}}{x_{1}  –  x_{6}}  =  41.4  \frac{0.996  –  0.408}{0.408  –  0.98}

\dot{m}_{6}  =  221.3  lbm/min

\dot{m}_{1}  =  \dot{m}_{12}  +  \dot{m}_{6}  =  221.3  +  41.4  =   262.7  lbm/min

From Fig. 15-32, \dot{m}_{1}  =  \dot{m}_{2}  =  \dot{m}_{3} and  \dot{m}_{4}  =  \dot{m}_{5}  =  \dot{m}_{6}.

(c) The pump work may be expressed as

w_{12}  =  i_{1}  –  i_{2}  =  \frac{(P_{1}  –  P_{2})  \text{v}_{1}}{J}

where J = 778 (ft-lbf)/Btu. The specific volume \text{v}_{1} is an empirically determined quantity and is given in the ASHRAE Handbook, Fundamentals Volume (3). For x_{1}  =  0.408  and  t_{1}  =  79  F,  \text{v}_{1}  =  0.0187  ft^{3}/lbm. Then

w_{12}  =  \frac{144 (30  –  200)  (0.0187)}{778}  =  -0.588  Btu/lbm

\dot{W}_{p}  =  w_{12} \dot{m}_{1}  =  (-0.588)  (262.7)  =  -155  Btu/lbm

i_{2}  =  i_{1}  –  w_{12}  =  -25  –  (-0.588)  =  -24.4  Btu/lbm

The enthalpy, pressure, and concentration at state 2 then establish t_{2}.

The energy balance on the solution heat exchanger yields

\dot{m}_{2} i_{2}  +  \dot{m}_{4} i_{4}  =  \dot{m}_{3}  i_{3}  +  \dot{m}_{5}  i_{5}

\dot{m}_{4}  =  \dot{m}_{5}  and  \dot{m}_{2}  =  \dot{m}_{3}

Then

i_{5}  =  i_{4}  =  \frac{\dot{m}_{2}}{\dot{m}_{4}} (i_{3}  –  i_{2})  =  159  –  \frac{262.7}{221.3} [109  –  (-24.4)]

i_{5}  =  0.64  ≈  0  Btu/lbm

The temperature t_5 is then read from Chart 2 as a function of P_{5},  x_{5},  and  i_{5}.  The  enthalpies  i_{6} =  i_{5} are coincident on Chart 2; however, state 5 is at 200 psia, whereas state 6 is at 30 psia. Since state 6 is subcooled, t_{6}  =  t_{5}. States 9 and 10 are coincident on Chart 2, since the process is one of throttling. Temperature t 10 may be found using the method described in Example 15-8. The energy balance on the evaporator gives

\dot{m}_{10} i_{10}  +  \dot{q}_{e}  =  \dot{m}_{11} i_{11}

\dot{m}_{10}  =  \dot{m}_{11}

i_{11}  =  \frac{\dot{q}_{e}}{\dot{m}_{10}}  +  i_{10}

i_{11}  =  \frac{(200)(100)}{41.4}  +  137  =  620  Btu/lbm

State 11 is a mixture of liquid and vapor, and its temperature is found from Chart 2 to be 40 F.

The pump horsepower requirement may be computed from W_{p} above:

power = \frac{\dot{W}_{p}}{\eta_{m}}  =  \frac{155 (778)}{0.75 (33,000)}  =  4.87  hp

(d) To compute the coefficient of performance it is necessary to establish the generator heat-transfer rate. However, this cannot be done without an analysis of the complete generator–rectifier column dephlegmator unit. To make a reasonable analysis, details of the column design and experimental data are required. For purposes of simplicity it will be assumed that the generator heat-transfer rate is 115 percent of the net heat transfer for the complete rectifying unit. An energy balance gives

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

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

221.3 (159)  +  41.4 (560)  –  262.7  (109)

\dot{q}_{net}  =  33,462  Btu/min

Then

\dot{q}_{g}  =  1.15 \dot{q}_{net}  =  38,480  Btu/min

\dot{q}_{d}  =  38,480  –  33,462  =  5020  Btu/min

Neglecting the pump work,

COP  =  \frac{\dot{q}_{e}}{\dot{q}_{g}}  =  \frac{100(200)}{38,480}  =  0.52

(e) The maximum COP may be calculated from Eq. 15-32. It will be assumed that t_{o}  =  80  F, t_{g}  =  240  F, and  t_{e}  =  0  F were given. Then

(COP)_{max}  =  \frac{T_{e}(T_{g} –  T_{o})}{T_{g} (T_{o}  –  T_{e})}                        (15-32)

(COP)_{max}  =  \frac{460(700  –  540)}{700 (540  –  460)}  =  1.31

and

\eta_{R}  =  \frac{COP}{(COP)_{max}}  =  \frac{0.52}{1.31}  =   0.40