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7. Consider a LiBr machine operating according to the following schematics: GENERATOR CONDENSER HIGH TEMPERATURE CONSTANT US GNCENTAON LINES, HEAT OUT HEAT EXOHANGER SOLUTION PUMP RESTRICTOR PHESS RESTRICTOR VAPOR O EwonATon PRES SPILLOVER EVAPORATO CRYSTALAON LINE HEAT OUT LOW TEMPERATURE EWPORSTOR AoER TEMPETUHEER TOATURE GENEROR TEMPERURE The system operates as follows: I. Refrigeration load = 500 tons I. Evaporator temperature, state 8 = 41.1 °F III. Absorber equilibrium temperature, state 3 = 107.2 °F Actual solution temperature, state 4 = 100.9 °F Solution temperature, state 5 = 170.3 °F Solution temperature, state 1 = 209.6 °F Solution temperature, state 2 = 128.1 °F IV. V. VI. VII. 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 temperatur Chilled water temperature delta = AT = 54 – 44 = 10 °F Cooling water temperature entering = 85 °F 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 Mass fraction of LiBr in the solution from the generator, Y; = 0.646 VII. IX. X. XI. XII. XIII. XIV. XV. XVI. XVII. twwt PEFRSERANT VAPOR PRESSURE

7. Consider a LiBr machine operating according to the following schematics: GENERATOR CONDENSER HIGH TEMPERATURE CONSTANT US GNCENTAON LINES, HEAT OUT HEAT EXOHANGER SOLUTION PUMP RESTRICTOR PHESS RESTRICTOR VAPOR O EwonATon PRES SPILLOVER EVAPORATO CRYSTALAON LINE HEAT OUT LOW TEMPERATURE EWPORSTOR AoER TEMPETUHEER TOATURE GENEROR TEMPERURE The system operates as follows: I. Refrigeration load = 500 tons I. Evaporator temperature, state 8 = 41.1 °F III. Absorber equilibrium temperature, state 3 = 107.2 °F Actual solution temperature, state 4 = 100.9 °F Solution temperature, state 5 = 170.3 °F Solution temperature, state 1 = 209.6 °F Solution temperature, state 2 = 128.1 °F IV. V. VI. VII. 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 temperatur Chilled water temperature delta = AT = 54 – 44 = 10 °F Cooling water temperature entering = 85 °F 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 Mass fraction of LiBr in the solution from the generator, Y; = 0.646 VII. IX. X. XI. XII. XIII. XIV. XV. XVI. XVII. twwt PEFRSERANT VAPOR PRESSURE

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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7. Consider a LiBr machine operating according to the following schematics: GENERATOR CONDENSER HIGH TEMPERATURE CONSTANT US CNCENTRATION LINES HEAT OUT HEAT EXOHANGER SOLUTION PUMP RESTRICTOR RESTRICTOR VAPOR O PES SPILLOVER EVAPORATO EWPORSTOR eER GONEe TEME E TEATUR HEAT OUT LOW TEMPERATURE TEMPER The 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 Actual solution temperature, state 4 = 100.9 °F Solution temperature, state 5 = 170.3 °F IV. V. Solution temperature, state 1 = 209.6 °F Solution temperature, state 2 = 128.1 °F VI. VII. VIII. 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 IX. X. XI. XII. Chilled water temperature delta = AT = 54 – 44 = 10 °F Cooling water temperature entering = 85 °F XII. XIV. Cooling water tower water flow rate = 1800 GPM XV. Absolute pressure of generator = 65.9 mm Hg Mass fraction of LiBr in the solution from the absorber, Y, = 0.595 Mass fraction of LiBr in the solution from the generator, Y; = 0.646 XVI. XVII. rsad on N
Transcribed Image Text:7. Consider a LiBr machine operating according to the following schematics: GENERATOR CONDENSER HIGH TEMPERATURE CONSTANT US CNCENTRATION LINES HEAT OUT HEAT EXOHANGER SOLUTION PUMP RESTRICTOR RESTRICTOR VAPOR O PES SPILLOVER EVAPORATO EWPORSTOR eER GONEe TEME E TEATUR HEAT OUT LOW TEMPERATURE TEMPER The 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 Actual solution temperature, state 4 = 100.9 °F Solution temperature, state 5 = 170.3 °F IV. V. Solution temperature, state 1 = 209.6 °F Solution temperature, state 2 = 128.1 °F VI. VII. VIII. 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 IX. X. XI. XII. Chilled water temperature delta = AT = 54 – 44 = 10 °F Cooling water temperature entering = 85 °F XII. XIV. Cooling water tower water flow rate = 1800 GPM XV. Absolute pressure of generator = 65.9 mm Hg Mass fraction of LiBr in the solution from the absorber, Y, = 0.595 Mass fraction of LiBr in the solution from the generator, Y; = 0.646 XVI. XVII. rsad on N
Determine: a. Enthalpy of state 8 vapor, h, (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 spill- over rate) d. Flow rate of dilute solution (lb/min dilute solution) (i.e. (1-Y.)X-(1-Y.)(X-1) = 1 affords X = 12.67 lb/min dilute solution / Ib min refrigerant, thus the lb/min of dilute solution = 12.67 lb/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, h: (BTU/lbm) 6. Enthalpy of state 2 liquid, h: (BTU/lbm) h. Enthalpy of state 4 liquid, h. (BTU/lbm) i. Enthalpy of state 5 liquid, h; (BTU/Ibm) J. Enthalpy of state 1 liquid, h. (BTU/lbm) k. Enthalpy of state 6 vapor, h: (BTU/lbm) 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 aut (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, h, (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 spill- over rate) d. Flow rate of dilute solution (lb/min dilute solution) (i.e. (1-Y.)X-(1-Y.)(X-1) = 1 affords X = 12.67 lb/min dilute solution / Ib min refrigerant, thus the lb/min of dilute solution = 12.67 lb/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, h: (BTU/lbm) 6. Enthalpy of state 2 liquid, h: (BTU/lbm) h. Enthalpy of state 4 liquid, h. (BTU/lbm) i. Enthalpy of state 5 liquid, h; (BTU/Ibm) J. Enthalpy of state 1 liquid, h. (BTU/lbm) k. Enthalpy of state 6 vapor, h: (BTU/lbm) 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 aut (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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