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CHAPTER 17 - ABSORPTION REFRIGERATION Page 1 of 11 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 atemperature of 35 C?Solution:Eq. 17-4. ( )( ) r asasr TTTTTTCOP −−= Ts = 70 C + 273 = 343 KTr = 15 C + 273 = 288 KTa = 35 C + 273 = 308 K ( )( )( )( ) 288308343308343288COP −−= COP = 1.47 - - -Ans .17-2. The LiBr-Water absorption cycle shown in Fig. 17-2 operates at the following temperatures: generator, 105C; condenser, 35 C; evaporator, 5 C; and absorber, 30 C. The flow rate of solution delivered by the pump is0.4 kg/s.(a) What are the mass flow rates of solution returning from the generator to the absorber and of therefrigerant?(b) What are the rates of heat transfer of each component, and the COP abs ?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 kPax 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/sLiBr Balance:w 1 x 1 = w 2 x 2 (0.40)(0.54)= (w 2 )(0.70)w 2 = 0.3086 kg/sFlow rate of solution = w 2 = 0.3086 kg/s - - - Ans .Flow rate of refrigerant = w 3 = w 1 - w 2 w 3 = 0.40 - 0.3086 w 3 = 0.0914 kg/s - - - Ans . CHAPTER 17 - ABSORPTION REFRIGERATION Page 2 of 11 (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/kgh 2 = h at 105 C and x of 70 % = -46 kJ/kgEnthalpy of water liquid and vapor: Table A-2h 3 = h of saturated vapor at 105 C = 2683.75 kJ/kgh 4 = h of saturated liquid at 35 C = 146.56 kJ/kgh 5 = h of saturated liquid at 5 C = 2510.75 kJ/kgw 3 =w 4 =w 5 =w c Generator q g = w 3 h 3 + w 2 h 2 - w 1 h 1 q g = (0.0914)(2683.75) + (0.3086)(-46) - (0.40)(-178) q g = 302.3 kW - - Ans .Condenser q c = w c h 3 - w 4 h 4 q c = (0.0914)(2683.75 - 146.56) q c = 231.9 kW - - Ans .Absorber q a = w 2 h 2 + w 5 h 5 - w 1 h 1 q a = (0.3086)(-46) + (0.0914)(2510.75) - (0.4)(-178) q a = 286.5 kw - - - Ans .Evaporator q e = w5h5 - w4h4q e = (0.0914)(2510.75 - 146.56) q e = 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 enteringthe generator is 48 C. All other temperatures and the flow rate are as shown in Fig. 17-9. What are the ratesof heat transfer at the generator and the temperature at point 4?Solution: Refer to Fig. 17-9.w 1 = w 2 = 0.6 kg/sw 3 = w 4 = 0.452 kg/sHeat 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 ) CHAPTER 17 - ABSORPTION REFRIGERATION Page 3 of 11 Enthalpies remain unchanged from Ex. 17-4 and Ex. 17-3.h 1 = -168 kJ/kgh 3 = -52 kJ/kgAt point 2, temperature = 48 CFig. 17-8, x 1 = 50 % solution, 48 Ch 2 = -128 kJ/kgw 3 (h 3 - h 4 ) = w 1 (h 2 -h 1 )(0.452)(-52-h 4 ) = (0.6)(-128-(-168))h 4 = -105.1 kJ/kgq g = w 3 h 3 + w5h5 - w2h2w 5 = 0.148 kg/sh 5 = 2676.0 kJ/kgq g = (0.452)(-52) + (0.148)(2676) - (0.6)(-128) q g = 449.4 kW - - - Ans .At point 4, h 4 = -105.1 kJ/kg, x 3 = 66.4 %Fig. 17-8. t 4 = 70 C - - - Ans .17-4. The solution leaving the heat exchanger and returning to the absorber is at a temperature of 60 C. Thegenerator temperature is 95 C. What is the minimum condensing temperature permitted in order to preventcrystallization in the system?Solution: Refer to Fig. 1709.Figure 17-8.At crystaliization, 60 C solution temperaturePercent lithium bromide = 66.4 %Figure 17-5, x = 66.4 %, 95 CVapor pressure = 6.28 kPaSat. Temp. of pure water = 37 C Minimum condensing temperature = 37 C - - - Ans .17-5. One of the methods of capacity control described in Sec. 17-11 is to reduce the flow rate of solution deliveredby the pump: The first-order approximation is that the refrigerating capacity will be reduced by the samepercentage as the solution flow rate. There are secondary effects also, because if the mean temperature of the heating medium in the generator, the cooling water in the absorber and condenser and the water beingchilled in the evaporator all remain constant, the temperatures in these components will change when theheat-transfer rate decreases.(a) Fill out each block in the Table 17-1 with either “increases” or “decreases” to indicate qualitativeinfluence of the secondary effect.(b) Use the expression for an ideal heat-operated cycle to evaluate the effects of temperature on theCOP abs . CHAPTER 17 - ABSORPTION REFRIGERATION Page 4 of 11 Solution:Use Data of Ex. 17-3 and Ex. 17-2 and Fig. 17-6.(a) Initial:w 1 = 0.6 kg/sw 2 = 0.452 kg/sw 3 = w 4 = w 5 = 0.148 kg/sx 1 = 50 %x 2 = 66.4 %Enthalpies: Fig. 17-8.h 1 = h at 30 C and x of 50 % = -168 kJ/kgh 2 = h at 100 C and x of 60 % = -52 kJ/kgEnthalpies: Table A-1h 3 = h of saturated vapor at 100 C = 2676.0 kJ/kgh 4 = h of saturated liquid at 40 C = 167.5 kJ/kgh 5 = h of saturated vapor at 10 C = 2520.0 kJ/kgq g = w 3 h 3 + w 2 h 2 - w 1 h 1 = 473.3 kWq c = w c h 3 - w 4 h 4 = 371.2 kWq a = w 2 h 2 + w 5 h 5 - w 1 h 1 = 450.3 kWq e = w 5 h 5 - w 4 h 4 = 348.2 kW0.736qqCOP geabs == New Solution:When w 1 is reduced to 0.4 kg/s (concentration of solution remains unchanged as firstapproximation)w 1 = 0.4 kg/sw 2 + w 3 = w 1 = 0.4 kg/sw 1 x 1 = w 2 x 2 (0.4)(0.5) = w 2 (0.664)w 2 = 0.3012 kg/sw 3 = 0.0988 kg/sq g = w 3 h 3 + w 2 h 2 - w 1 h 1 q g = (0.0988)(2676.0) + (0.3012)(-52) - (0.4)(-168) = 315.9 kWq c = w c h 3 - w 4 h 4 q c = (0.0988)(2676.0 - 167.5) = 247.8 kWq a = w 2 h 2 + w 5 h 5 - w 1 h 1 q a = (0.3012)(-52) + (0.0988)(2520) - (0.4)(-168) = 300.5 kWq e = w 5 h 5 - w 4 h 4 q e = (0.0988)(2520.0 - 167.5) = 232.4 kW] Assume: Mean temperature of heating medium in the generator = 120 C. Mean temperature of the coolingwater 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)
