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7. Consider a LiBr machine operating according to the following schematics: GENERATOR CONDENSER TEMPERATURE CONSTANT LI CCENTRATION UNES, HEAT OUT NEAT EXCHANGER SOLUTION PUMP RESTRICTOR CONCENSER PRESSURE RESTRICTOR VAPOR EAPORATOR PRESSRE SPLLOVER EVAPORATO CRYSTAL CZanoN LNE LOW TEMPERATURE HEAT OUT EWORATOR AeSoRER GLNERA TEMPERATUE TEMERATUE TEMPERATURE TEMPERATURE he system operates as follows: 1. Refrigeration load = 500 tons II. Evaporator temperature, state 8 = 41.1 °F II. Absorber equilibrium temperature, state 3 = 107.2 °F IV. Actual solution temperature, state 4 = 100.9 °F V. Solution temperature, state 5 = 170.3 °F %3D VI. Solution temperature, state 1 = 209.6 °F %3D VII. Solution temperature, state 2 = 128.1 °F Refrigerant vapor temperature, state 6 = 200 °F Refrigerant temperature, state 7 = 100 °F Refrigerant spill-over rate, state 9 = 2.5% of state 8 Concentrations of solution on a LiBr Duhring's phase diagram per above temperatures XII. Cooling water temperature entering = 85 °F VII. IX. X. XI. Chilled water temperature delta = AT = 54 – 44 = 10 °F XIII. XIV. Cooling water tower water flow rate = 1800 GPM Absolute pressure of generator = 65.9 mm Hg Mass fraction of LiBr in the solution from the absorber, Y, = 0.595 XV. XVI. XVII. Mass fraction of LiBr in the solution from the generator, Y, = 0.646 REFRGERANT VAPON PHESSURE

7. Consider a LiBr machine operating according to the following schematics: GENERATOR CONDENSER TEMPERATURE CONSTANT LI CCENTRATION UNES, HEAT OUT NEAT EXCHANGER SOLUTION PUMP RESTRICTOR CONCENSER PRESSURE RESTRICTOR VAPOR EAPORATOR PRESSRE SPLLOVER EVAPORATO CRYSTAL CZanoN LNE LOW TEMPERATURE HEAT OUT EWORATOR AeSoRER GLNERA TEMPERATUE TEMERATUE TEMPERATURE TEMPERATURE he system operates as follows: 1. Refrigeration load = 500 tons II. Evaporator temperature, state 8 = 41.1 °F II. Absorber equilibrium temperature, state 3 = 107.2 °F IV. Actual solution temperature, state 4 = 100.9 °F V. Solution temperature, state 5 = 170.3 °F %3D VI. Solution temperature, state 1 = 209.6 °F %3D VII. Solution temperature, state 2 = 128.1 °F Refrigerant vapor temperature, state 6 = 200 °F Refrigerant temperature, state 7 = 100 °F Refrigerant spill-over rate, state 9 = 2.5% of state 8 Concentrations of solution on a LiBr Duhring's phase diagram per above temperatures XII. Cooling water temperature entering = 85 °F VII. IX. X. XI. Chilled water temperature delta = AT = 54 – 44 = 10 °F XIII. XIV. Cooling water tower water flow rate = 1800 GPM Absolute pressure of generator = 65.9 mm Hg Mass fraction of LiBr in the solution from the absorber, Y, = 0.595 XV. XVI. XVII. Mass fraction of LiBr in the solution from the generator, Y, = 0.646 REFRGERANT VAPON PHESSURE

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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7. Consider a LiBr machine operating according to the following schematics: GENERATOR CONDENSER -HIGH TEMPERATURE CONSTANT LIBr CONCENTRATION UNES. HEAT OUT HEAT EXCHANGER SOLUTION PUMP RESTRICTOR CONDENSER PRESSURE RESTRICTOR VAPÓN EVAPORATOR PRESSURE SPILLOVER ADsORDER EVAPORATOF -CRYSTALLIZATON LINE HEAT OUT LOW TEMPERATURE IN EVORATOR ABSORBER TEMPERATLRS TEMPERATURE TEMPERATURE GENERATOR TEMPERATURE The system operates as follows: I. Refrigeration load = 500 tons Evaporator temperature, state 8 = 41.1 °F Absorber equilibrium temperature, state 3 = 107.2 °F Actual solution temperature, state 4 = 100.9 °F I. III. IV. V. Solution temperature, state 5 = 170.3 °F VI. Solution temperature, state 1 = 209.6 °F VII. Solution temperature, state 2 = 128.1 °F VIII. Refrigerant vapor temperature, state 6 = 200 °F IX. Refrigerant temperature, state 7 = 100 °F X. Refrigerant spill-over rate, state 9 = 2.5% of state 8 Concentrations of solution on a LiBr Duhring's phase diagram per above temperatures XI. XII. Chilled water temperature delta = AT = 54 – 44 = 10 °F XIII. Cooling water temperature entering = 85 °F XIV. Cooling water tower water flow rate = 1800 GPM Absolute pressure of generator = 65.9 mm Hg XV. XVI. Mass fraction of LiBr in the solution from the absorber, Y. = 0.595 XVII. Mass fraction of LiBr in the solution from the generator, Yg = 0.646 CONCENTRATED
Transcribed Image Text:7. Consider a LiBr machine operating according to the following schematics: GENERATOR CONDENSER -HIGH TEMPERATURE CONSTANT LIBr CONCENTRATION UNES. HEAT OUT HEAT EXCHANGER SOLUTION PUMP RESTRICTOR CONDENSER PRESSURE RESTRICTOR VAPÓN EVAPORATOR PRESSURE SPILLOVER ADsORDER EVAPORATOF -CRYSTALLIZATON LINE HEAT OUT LOW TEMPERATURE IN EVORATOR ABSORBER TEMPERATLRS TEMPERATURE TEMPERATURE GENERATOR TEMPERATURE The system operates as follows: I. Refrigeration load = 500 tons Evaporator temperature, state 8 = 41.1 °F Absorber equilibrium temperature, state 3 = 107.2 °F Actual solution temperature, state 4 = 100.9 °F I. III. IV. V. Solution temperature, state 5 = 170.3 °F VI. Solution temperature, state 1 = 209.6 °F VII. Solution temperature, state 2 = 128.1 °F VIII. Refrigerant vapor temperature, state 6 = 200 °F IX. Refrigerant temperature, state 7 = 100 °F X. Refrigerant spill-over rate, state 9 = 2.5% of state 8 Concentrations of solution on a LiBr Duhring's phase diagram per above temperatures XI. XII. Chilled water temperature delta = AT = 54 – 44 = 10 °F XIII. Cooling water temperature entering = 85 °F XIV. Cooling water tower water flow rate = 1800 GPM Absolute pressure of generator = 65.9 mm Hg XV. XVI. Mass fraction of LiBr in the solution from the absorber, Y. = 0.595 XVII. Mass fraction of LiBr in the solution from the generator, Yg = 0.646 CONCENTRATED
Determine: a. Enthalpy of state 8 vapor, ha (BTU/lbm) b. Enthalpy of state 7 liquid, h, (BTU/lbm) c. Flow rate of refrigerant (Ib/min refrigerant) (i.e. 1.025* 500 TONS * 200 BTU/min / TON)/ (h8-h7) (BTU/lb), where the 1.025 factor is the refrigerant spil- over rate) d. Flow rate of dilute solution (Ib/min dilute solution) (i.e. (1-Y,)X-(1-Y,)(X-1) = 1 affords X = 12.67 Ib/min dilute solution / Ib min refrigerant, thus the Ib/min of dilute solution = 12.67 Ib/min dilute solution / Ib min refrigerant*the answer of part c. e. Flow rate of concentration (Ib/min concentrated solution) (i.e. Ib/min concentration solution = ans. from part d. – ans. from part c.) f. Enthalpy of state 9 liquid, he (BTU/lbm) g. Enthalpy of state 2 liquid, h2 (BTU/lbm) h. Enthalpy of state 4 liquid, ha (BTU/lbm) i. Enthalpy of state 5 liquid, hs (BTU/lbm) j. Enthalpy of state 1 liquid, h, (BTU/lbm) k. Enthalpy of state 6 vapor, h, (BTU/Ibm) I. Absorber heat in (BTU/min) m. Absorber heat out (BTU/min) n. Generator heat out (BTU/min) o. Generator heat in (BTU/min) p. Condenser heat in (BTU/min) q. Condenser heat out (BTU/min) r. Evaporator duty (BTU/min) s. System Heat In = Evaporator duty + Generator load (BTU/min) t. System Heat Out = Absorber load + Condenser Load (BTU/min) u. COP = Evaporator duty/ Condenser Load v. Rate of steam flow to the generator (Ib/hr per ton) (assume 4 °F sub-cooling of the steam condensate in the generator) w. Rate of steam flow to generator (GPM per ton) for hot water at 240 °F with the AT = 54 – 44 = 10 °F water range
Transcribed Image Text:Determine: a. Enthalpy of state 8 vapor, ha (BTU/lbm) b. Enthalpy of state 7 liquid, h, (BTU/lbm) c. Flow rate of refrigerant (Ib/min refrigerant) (i.e. 1.025* 500 TONS * 200 BTU/min / TON)/ (h8-h7) (BTU/lb), where the 1.025 factor is the refrigerant spil- over rate) d. Flow rate of dilute solution (Ib/min dilute solution) (i.e. (1-Y,)X-(1-Y,)(X-1) = 1 affords X = 12.67 Ib/min dilute solution / Ib min refrigerant, thus the Ib/min of dilute solution = 12.67 Ib/min dilute solution / Ib min refrigerant*the answer of part c. e. Flow rate of concentration (Ib/min concentrated solution) (i.e. Ib/min concentration solution = ans. from part d. – ans. from part c.) f. Enthalpy of state 9 liquid, he (BTU/lbm) g. Enthalpy of state 2 liquid, h2 (BTU/lbm) h. Enthalpy of state 4 liquid, ha (BTU/lbm) i. Enthalpy of state 5 liquid, hs (BTU/lbm) j. Enthalpy of state 1 liquid, h, (BTU/lbm) k. Enthalpy of state 6 vapor, h, (BTU/Ibm) I. Absorber heat in (BTU/min) m. Absorber heat out (BTU/min) n. Generator heat out (BTU/min) o. Generator heat in (BTU/min) p. Condenser heat in (BTU/min) q. Condenser heat out (BTU/min) r. Evaporator duty (BTU/min) s. System Heat In = Evaporator duty + Generator load (BTU/min) t. System Heat Out = Absorber load + Condenser Load (BTU/min) u. COP = Evaporator duty/ Condenser Load v. Rate of steam flow to the generator (Ib/hr per ton) (assume 4 °F sub-cooling of the steam condensate in the generator) w. Rate of steam flow to generator (GPM per ton) for hot water at 240 °F with the AT = 54 – 44 = 10 °F water range
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