SM_15A - Chapter 15 - Section A - Mathcad Solutions 15.1...

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Unformatted text preview: Chapter 15 - Section A - Mathcad Solutions 15.1 Initial state: Liquid water at 70 degF. H 1 38.05 BTU lb m ⋅ = S 1 0.0745 BTU lb m rankine ⋅ ⋅ = (Table F.3) Final state: Ice at 32 degF. H 2 0.02- 143.3- ( ) BTU lb m ⋅ = S 2 0.0 143.3 491.67- BTU lb m rankine ⋅ ⋅ = T σ 70 459.67 + ( ) rankine ⋅ = (a) Point A: sat. vapor at 32 degF. Point C: sat. liquid at 70 degF. P = 85.79(psia). Point D: Mix of sat. liq. & sat. vapor at 32 degF with the enthalpy of Point C. Point B: Superheated vapor at 85.79(psia) and the entropy of Point A. Data for Points A, C, & D from Table 9.1. Data for Point B from Fig. G.2. 512 H B 114 BTU lb m ⋅ = For superheated vapor at 85.79(psia) and S = 0.2223: H D H C = H C 34.58 BTU lb m ⋅ = For sat. liquid at 70 degF: S A 0.2223 BTU lb m rankine ⋅ ⋅ = H A 107.60 BTU lb m ⋅ = For sat. liquid and vapor at 32 degF, by interpolation in the table: Conventional refrigeration cycle under ideal conditions of operation: Isentropic compression, infinite flow rate of cooling water, & minimum temp. difference for heat transfer = 0. (c) The only irreversibility is the transfer of heat from the water as it cools from 70 to 32 degF to the cold reservoir of the Carnot heat pump at 70 degF. Ans. η t 0.889 = η t Wdot ideal Wdot = Ans. Wdot 14.79kW = Wdot mdot Work ⋅ = Work 14.018 BTU lb m = Work Q C T H T C- T C ⋅ = Q C 181.37- BTU lb m = Q C H 2 H 1- = T H T σ = T C 491.67 rankine ⋅ = For the Carnot heat pump, heat equal to the enthalpy change of the water is extracted from a cold reservoir at 32 degF, with heat rejection to the surroundings at 70 degF. (b) Ans. Wdot ideal 13.15kW = Wdot ideal mdot W ideal ⋅ = mdot 1 lb m sec ⋅ = W ideal 12.466 BTU lb m = W ideal H 2 H 1- T σ S 2 S 1- ( 29 ⋅- = 513 S A S vap = S vap 0.2229 BTU lb m rankine ⋅ ⋅ = S liq 0.0433 BTU lb m rankine ⋅ ⋅ = H A H vap = H vap 106.48 BTU lb m ⋅ = H liq 19.58 BTU lb m ⋅ = For sat. liquid and vapor at 24 degF: (Note that minimum temp. diff. is not at end of condenser, but it is not practical to base design on 8-degF temp. diff. at pinch. See sketch.) Point C: Sat. Liquid at 98 degF. Point D: Mix of sat. liq. and sat. vapor at 24 degF with H of point C, Point B: Superheated vapor at 134.75(psia). Point A: Sat. vapor at 24 degF. η 0.75 = Practical cycle. (d) The irreversibilities are in the throttling process and in heat transfer in both the condenser and evaporator, where there are finite temperature differences. Ans. η t 0.784 = η t Wdot ideal Wdot = Ans. Wdot 16.77kW = Wdot mdot H B H A- ( 29 ⋅ = mdot 2.484 lb m sec = mdot H 2 H 1- ( 29- 1 ⋅ lb m sec ⋅ H A H D- = Refrigerent circulation rate: 514 Wdot lost.condenser mdot T σ ⋅ S C S B- ( 29 ⋅ Qdot condenser- = Qdot condenser mdot H C H B- ( 29 ⋅ = Wdot lost.compressor mdot T σ ⋅ S B S A- ( 29 ⋅ = T σ 70 459.67 + ( ) rankine ⋅ = THERMODYNAMIC ANALYSIS Ans....
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This note was uploaded on 01/29/2011 for the course CHEM 101 taught by Professor Brown during the Spring '10 term at The University of Akron.

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SM_15A - Chapter 15 - Section A - Mathcad Solutions 15.1...

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