Question 14.3: Lithium Bromide Absorption Cycle A lithium bromide chiller o......

The solution consists of, first, finding thermodynamic properties at all state points 1–5. Second, we use mass and energy balance equations to find the heat and mass flow terms.

The property table is completed as follows:
(a) First, the known temperatures in the second column are entered. The given temperatures within each component are assumed to be the temperatures of fluid streams exiting that component.
(b) Second, pressures are entered in the table. We refer to the saturated steam tables (Table A3) to determine the condenser and evaporator pressures p_{3} and p_{5} since the corresponding temperatures are specified. Consequently, the pressures at all state points are now known.
Note that pressures in the LiBr cycle are sub-atmospheric and small.
(c) Next, the LiBr mass fractions at points 1 and 2 can be deduced from Figure 14.7 as the intersection of the known pressure and temperature lines (the online HCB software can be used for more accurate determination). Of course, for states 3, 4, and 5 consisting of pure water vapor, the mass fraction is zero.
(d) Finally, from Figure 14.7 (or from the online HCB software), the enthalpies at points 1 and 2 are deduced. The enthalpies at points 3, 4, and 5 are easily  determined from the steam superheat and saturated tables (Tables A3 and A4).

At this point, a mass balance is written on the generator, and then each heat ow term can be found. The total mass balance and the LiBr mass balance written on the generator are

\dot{m}_{2} + \dot{m}_{3} = \dot{m}_{1} = 4800  lb_{m}/h      (14.16)

\dot{m}_{1} X_{1} = \dot{m}_{2} X_{2}         (14.17)

From these expressions, the two unknown mass flow rates can be found:

\dot{m}_{2} = \dot{m}_{1} \left\lgroup \frac{X_{1}}{X_{2}} \right\rgroup = 4800 \left\lgroup \frac{0.50}{0.67} \right\rgroup = 3582  lb_{m}/h            (14.18)

\dot{m}_{3} = \dot{m}_{1}  –  \dot{m}_{2} = 4800  –  3582 = 1218  lb/h = \dot{m}_{4} = \dot{m}_{5}     (14.19)

The four heat rate terms can now be found from energy balances on the four components of the AC cycle. The arrows in Figure 14.6 show the direction of the heat flow:

+ 3582 × (-24)  –  4800 × (- 71) = 1656  kBtu/h          (14.20)

\dot{Q}_{cond} = \dot{m}_{3} (h_{3}  –  h_{4}) = 1218  \times (1150.5  –  72) = 1314  kBtu/h               (14.21)

+ 1218 × (1083.3)  –   4800  × (- 71) = 1574  kBtu/h         (14.22)

As a check, by the first law of thermodynamics, since pumping power is negligible,

= 1232  kBtu/h

Finally, the COP, defined as the ratio of cooling produced in the evaporator to the heat added in the generator, is

Comments
The solution to AC cycle problems involves the use of two sets of thermodynamic tables or charts—those for pure water and those for LiBr. Although the reference states for various LiBr charts found in the literature may vary from those in Figure 14.7, the energy flows will be the same. Figure 14.7 has the advantage that all thermodynamic variables are included in a single LiBr chart rather than two charts, as in other references.
Although the COP of an absorption cycle is much lower than that for a VC cycle, remember that the heat input to a power plant that produces the electricity to power VC machines is about three times the power plant’s electric output.
Therefore, the ratio of cooling effect to thermal energy input (to the power source) is not much different for VC and AC cycles