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17-1. What is the COP of an ideal heat-operated refrigeration cycle that receives the energizing heat from a solar collector at a temperature of 70 C, performs refrigeration at 15 C, and rejects heat to atmosphere at a temperature of 35 C?
Solution: Eq. 17-4.
s^ (^ a r)
r s a T T T
T T T
COP
Ts = 70 C + 273 = 343 K Tr = 15 C + 273 = 288 K Ta = 35 C + 273 = 308 K
COP
COP = 1.47 - - -Ans.
17-2. The LiBr-Water absorption cycle shown in Fig. 17-2 operates at the following temperatures: generator, 105 C; condenser, 35 C; evaporator, 5 C; and absorber, 30 C. The flow rate of solution delivered by the pump is 0.4 kg/s. (a) What are the mass flow rates of solution returning from the generator to the absorber and of the refrigerant? (b) What are the rates of heat transfer of each component, and the COPabs?
Solution: Saturation pressure at 35 C water = 5.63 kPa (condenser) Saturation pressure at 5 C water = 0.874 kPa (evaporator)
(a) At the generator, LiBr-Water Solution: Fig. 17-5, 105 C, 5.63 kPa, Refer to Fig. 17-2. x 2 = 70 %
At the absorber, LiBr-Water Fig. 17-5, 30 C, 0.874 kPa x 1 = 54 %
w 1 = LiBr-Water Solution delivered by pump. w 2 = Solution returning from generator to absorber. w 3 = refrigerant water flow rate.
Total mass-flow balance: w 2 + w 3 = w 1 = 0.4 kg/s
LiBr Balance: w 1 x 1 = w 2 x 2 (0.40)(0.54)= (w 2 )(0.70) w 2 = 0.3086 kg/s
Flow rate of solution = w 2 = 0.3086 kg/s - - - Ans. Flow rate of refrigerant = w 3 = w 1 - w 2 w 3 = 0.40 - 0. w 3 = 0.0914 kg/s - - - Ans.
(b) Refer to Fig. 17-6.
Enthalpies: Enthalpies of solution, Fig. 17-8. h 1 = h at 30 C and x of 54 % = -178 kJ/kg h 2 = h at 105 C and x of 70 % = -46 kJ/kg
Enthalpy of water liquid and vapor: Table A- h 3 = h of saturated vapor at 105 C = 2683.75 kJ/kg h 4 = h of saturated liquid at 35 C = 146.56 kJ/kg h 5 = h of saturated liquid at 5 C = 2510.75 kJ/kg
w 3 =w 4 =w 5 =wc
Generator qg = w 3 h 3 + w 2 h 2 - w 1 h 1 qg = (0.0914)(2683.75) + (0.3086)(-46) - (0.40)(-178) qg = 302.3 kW - - Ans.
Condenser qc = wch 3 - w 4 h 4 qc = (0.0914)(2683.75 - 146.56) qc = 231.9 kW - - Ans.
Absorber qa = w 2 h 2 + w 5 h 5 - w 1 h 1 qa = (0.3086)(-46) + (0.0914)(2510.75) - (0.4)(-178) qa = 286.5 kw - - - Ans.
Evaporator qe = w5h5 - w4h qe = (0.0914)(2510.75 - 146.56) qe = 216.1 kW - - - Ans.
COP = qe / qg = (216.1 kW) / (302.3 kW) COP = 0.715 - - - Ans.
17-3. In the absorption cycle shown in Fig. 17-9 the solution temperature leaving the heat exchanger and entering the generator is 48 C. All other temperatures and the flow rate are as shown in Fig. 17-9. What are the rates of heat transfer at the generator and the temperature at point 4?
Solution: Refer to Fig. 17-9. w 1 = w 2 = 0.6 kg/s w 3 = w 4 = 0.452 kg/s
Heat balance through heat exchanger w 3 h 3 - w 4 h 4 = w 2 h 2 - w 1 h 1 w 3 (h 3 - h 4 ) = w 1 (h 2 -h 1 )
Solution:
Use Data of Ex. 17-3 and Ex. 17-2 and Fig. 17-6.
(a) Initial: w 1 = 0.6 kg/s w 2 = 0.452 kg/s w 3 = w 4 = w 5 = 0.148 kg/s x 1 = 50 % x 2 = 66.4 % Enthalpies: Fig. 17-8. h 1 = h at 30 C and x of 50 % = -168 kJ/kg h 2 = h at 100 C and x of 60 % = -52 kJ/kg Enthalpies: Table A- h 3 = h of saturated vapor at 100 C = 2676.0 kJ/kg h 4 = h of saturated liquid at 40 C = 167.5 kJ/kg h 5 = h of saturated vapor at 10 C = 2520.0 kJ/kg
qg = w 3 h 3 + w 2 h 2 - w 1 h 1 = 473.3 kW qc = wch 3 - w 4 h 4 = 371.2 kW qa = w 2 h 2 + w 5 h 5 - w 1 h 1 = 450.3 kW qe = w 5 h 5 - w 4 h 4 = 348.2 kW
q
q COP g
e abs = =
New Solution: When w 1 is reduced to 0.4 kg/s (concentration of solution remains unchanged as first approximation) w 1 = 0.4 kg/s w 2 + w 3 = w 1 = 0.4 kg/s w 1 x 1 = w 2 x 2 (0.4)(0.5) = w 2 (0.664) w 2 = 0.3012 kg/s w 3 = 0.0988 kg/s qg = w 3 h 3 + w 2 h 2 - w 1 h 1 qg = (0.0988)(2676.0) + (0.3012)(-52) - (0.4)(-168) = 315.9 kW qc = wch 3 - w 4 h 4 qc = (0.0988)(2676.0 - 167.5) = 247.8 kW qa = w 2 h 2 + w 5 h 5 - w 1 h 1 qa = (0.3012)(-52) + (0.0988)(2520) - (0.4)(-168) = 300.5 kW qe = w 5 h 5 - w 4 h 4 qe = (0.0988)(2520.0 - 167.5) = 232.4 kW
] Assume: Mean temperature of heating medium in the generator = 120 C. Mean temperature of the cooling water in the absorber and condenser = 25 C. Mean temperature of the water being chilled in the evaporator = 15 C.
New temperature of components: Generator = 120 - (315.9 / 473.3)(120 - 100) = 106.6 C (increase)
Absorber = 25 + (300.5 / 450.3)(30 - 25) = 28.34 C (decrease) Condenser = 25 + (247.8 / 371.2)(40 - 25) = 35.0 C (decrease) Evaporator = 15 - (233.4 / 348.2)(15 - 10) = 11.66 C (increase)
With change in component temperature. Fig. 17-5, 35 C condenser temperature, 106.6 C solution temperature x 2 = 0.70 (increase)
At 11.66 C evaporator temperature, 28.34 C solution temperature x 1 = 0.46 (decrease)
Enthalpies: Fig. 17-8. h 1 = h at 28.34 C and x of 46 % = -158 kJ/kg
h 2 = h at 106.6 C and x of 70 % = -45 kJ/kg
Enthalpies: Table A-1. h 3 = h of saturated vapor at 106.6 C = 2686.2 kJ/kg
h 4 = h of saturated liquid at 35 C = 146.56 kJ/kg
h 5 = h of saturated vapor at 11.66 C = 2523.0 kJ/kg
w 1 = 0.4 kg/s w 2 + w 3 = w 1 = 0.4 kg/s w 1 x 1 = w 2 x 2 (0.4)(0.46) = w 2 (0.70) w 2 = 0.263 kg/s w 3 = 0.137 kg/s
qg = w 3 h 3 + w 2 h 2 - w 1 h 1
qg = (0.137)(2686.2) + (0.263)(-45) - (0.4)(-158) = 419.4 kW
qc = wch 3 - w 4 h 4
qc = (0.137)(2686.2 - 146.56) = 348 kW
qa = w 2 h 2 + w 5 h 5 - w 1 h 1
qa = (0.263)(-45) + (0.137)(2523) - (0.4)(-158) = 397 kW
qe = w 5 h 5 - w 4 h 4
qe = (0.137)(2523.0 - 146.56) = 325.6 kW
0.776 (increase) q
q COP g
e abs = =
New temperature of components: Generator = 120 - (419.4 / 473.3)(120 - 100) = 102.3 C (increase) Absorber = 25 + (397 / 450.3)(30 - 25) = 29.4 C (decrease) Condenser = 25 + (348 / 371.2)(40 - 25) = 39.1 C (decrease) Evaporator = 15 - (325.6 / 348.2)(15 - 10) = 10.3 C (increase)
With change in component temperature. Fig. 17-5, 35 C condenser temperature, 102.3 C solution temperature x 2 = 0.675 (increase)
At 10.3 C evaporator temperature, 29.4 C solution temperature x 1 = 0.4875 (decrease)
Enthalpies: Fig. 17-8.
w 1 x 1 = w 2 x 2 (0.4)(0.475) = w 2 (0.6975) w 2 = 0.2724 kg/s w 3 = 0.1276 kg/s
qg = w 3 h 3 + w 2 h 2 - w 1 h 1
qg = (0.1276)(2684.1) + (0.2724)(-45) - (0.4)(-162) = 395 kW
qc = wch 3 - w 4 h 4
qc = (0.1276)(2684.1 - 152) = 323 kW
qa = w 2 h 2 + w 5 h 5 - w 1 h 1
qa = (0.2724)(-45) + (0.1276)(2522.2) - (0.4)(-162) = 374.4 kW
qe = w 5 h 5 - w 4 h 4
qe = (0.1276)(2522.2 - 152) = 302.4 kW
0.766 (increase) q
q COP g
e abs = =
New temperature of components: Generator = 120 - (395 / 473.3)(120 - 100) = 103.3 C (increase) Absorber = 25 + (374.4 / 450.3)(30 - 25) = 29.2 C (decrease) Condenser = 25 + (323 / 371.2)(40 - 25) = 38.1 C (decrease) Evaporator = 15 - (302.4 / 348.2)(15 - 10) = 10.66 C (increase)
With change in component temperature. Fig. 17-5, 38.1 C condenser temperature, 103.3 C solution temperature x 2 = 0.675 (increase)
At 10.66 C evaporator temperature, 29.2 C solution temperature x 1 = 0.4875 (decrease)
Enthalpies: Fig. 17-8. h 1 = h at 29.2 C and x of 48.75 % = -165 kJ/kg
h 2 = h at 103.3 C and x of 67.5 % = -50 kJ/kg
Enthalpies: Table A-1. h 3 = h of saturated vapor at 103.3 C = 2681 kJ/kg
h 4 = h of saturated liquid at 38.1 C = 159.5 kJ/kg
h 5 = h of saturated vapor at 10.66 C = 2521 kJ/kg
w 1 = 0.4 kg/s w 2 + w 3 = w 1 = 0.4 kg/s w 1 x 1 = w 2 x 2 (0.4)(0.4875) = w 2 (0.675) w 2 = 0.2889 kg/s w 3 = 0.1111 kg/s
qg = w 3 h 3 + w 2 h 2 - w 1 h 1
qg = (0.1111)(2681) + (0.2889)(-50) - (0.4)(-165) = 349.4 kW
qc = wch 3 - w 4 h 4
qc = (0.1111)(2681 - 159.5) = 280 kW
qa = w 2 h 2 + w 5 h 5 - w 1 h 1
qa = (0.2889)(-50) + (0.1111)(2521) - (0.4)(-165) = 331.6 kW
qe = w 5 h 5 - w 4 h 4
qe = (0.1111)(2521 - 159.5) = 262.4 kW
0.751 (increase) q
q COP g
e abs = =
New temperature of components: Generator = 120 - (349.4 / 473.3)(120 - 100) = 105.2 C (increase) Absorber = 25 + (331.6 / 450.3)(30 - 25) = 28.7 C (decrease) Condenser = 25 + (280 / 371.2)(40 - 25) = 36.3 C (decrease) Evaporator = 15 - (262.4 / 348.2)(15 - 10) = 11.23 C (increase)
Take the average: qg = (1/2)(349.4 + 395.0) = 372.2 kW, 104 C
qc = (1/2)(280 + 323) = 301.4 kW, 37 C
qa = (1/2)(331.6 + 374.4) = 353 kW, 29 C
qe = (1/2)(262.4 + 302.4) = 282.4 kW, 11 C
Full load COPabs = 0.
New COPabs:
0.759 (increase)
q
q COP g
e abs = = =
0.189or 18.9%
Reduction inqe = =
Therefore Capacity decrease by less than reduction in solution flow rate (33 1/3 %). Table 17-1. Influence of reduction in solution flow rate of pump Solution concentrate Refrigerating Component Temperature x(gen) x(abs) Capacity COP(abs)
Generator "increase" "increase"
Absorber "decrease" "increase"
Condenser "decrease"
Evaporator "increase" "decrease" "increase"
(b) Initial:
s^ (^ a r)
r s a T T T
T T T
COP
Ts = 100 C + 273 = 373 K
Tr = 10 C + 273 = 283 K
Ta = 1/2(30 C + 40 C) + 273 = 35 C + 273 = 308 K
COPideal = −
COPabs = 0.
New:
w 1 = w 2 + w 3
1 = (w 2 /w 1 ) + (w 3 /w 1 )
1 = (0.67/x 2 ) +(w 3 /w 1 )
-3.3 = (0.67/x 2 )(-55) + (1 - 0.67/x 2 )(2676)
0.67/x 2 = 0.
x 2 = 0.683 = 68.4 %
(a) Mass flashes to vapor = w3/w
w 3 /w 1 = 1 - (.67/x 2 )
w 3 /w 1 = 1 - (0.67 / 0.684)
w 3 /w 1 = 0.0205 kg/kg of solution flowing through generator II - - Ans.
(b) x2 = 0.684 = 68.4 % - - - Ans.
17-7. The combined absorption and vapor-compression system shown in Fig. 17-16 is to be provided with a capacity control scheme that maintains a constant temperature of the leaving chilled water as the temperature of the return water to be chilled varies. This control scheme is essentially one of reducing the refrigerating capacity. The refrigerant compressor is equipped with inlet valves (see Chap. 11), the speed of the turbine-compressor can be varied so long as it remains less than the maximum value of 180 r/s, and the control possibilities of the absorption unit are as described in Sec. 17-11. The characteristics of the steam turbine are that both its speed and power diminishes if the pressure of the supply steam decreases or the exhaust pressure increases. With constant inlet and exhaust pressures the speed of the turbine increases if the load is reduced. Device a control scheme and describe the behavior of the entire system as the required refrigerating load decreases.
Answer:
- If the return water to be chilled reduces, the refrigerating capacity will be reduced.
- For the refrigerating capacity reduced, the steam entering the generator of absorption unit will be throttled to reduce the generator temperature.
- For the vapor-compression unit, the compressor can be controlled by adjusting prerotation vane at the impeller inlet.
- For the entire system with the above control scheme, there is a possibility that the speed of turbine- compressor will increase greater than 180 r/s. So it is better to control only the exhaust pressure by increasing it then throttled before entering the generator of absorption unit. The refrigerating capacity and power diminishes as the exhaust pressure increases with constant supply steam.
17-8. The operating cost of an absorption system is to be compared with an electric-driven vapor-compression unit. The cost of natural gas on a heating value basis is $4.20 per gigajoule, when used as fuel in a boiler it has a combustion efficiency of 75 percent. An absorption unit using steam from this boiler has a COPabs of 0.73. If a vapor-compression unit is selected, the COP would be 3.4, and the electric motor efficiency is 85 percent. At what cost of electricity are the operating costs equal?
Solution:
Let qe = refrigerating capacity = kWh
Operating cost of natural gas = ($4.20 /GJ)(1 GJ / 106 kJ)(3600 kJ / 1 kWh)(qe / 0.73)(1 / 0.75)
= $ 0.0276164qe
Let x = operating cost in cents / kWh
Operating cost of electric motor. = (x / 100)(qe)
Then: (x / 100)(qe) = 0.0276164(qe)
x = 2.76 cents / kWh - - - Ans.
