{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter15_A

# Chapter15_A - Chapter 15 Section A Mathcad Solutions 15.1...

This preview shows pages 1–5. Sign up to view the full content.

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. 595

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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. W ideal H 2 H 1 T σ S 2 S 1 () := W ideal 12.466 BTU lb m = mdot 1 lb m sec := Wdot ideal mdot W ideal := Wdot ideal 13.15kW = Ans. (b) 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. T C 491.67 rankine := T H T σ := Q C H 2 H 1 := Q C 181.37 BTU lb m = Work Q C T H T C T C := Work 14.018 BTU lb m = Wdot mdot Work := Wdot 14.79kW = 596
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 () := mdot 2.484 lb m sec = mdot H 2 H 1 1 lb m sec H A H D := Refrigerent circulation rate: 597

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Wdot lost.condenser mdot T σ S C S B () Qdot condenser := Qdot condenser mdot H C H B := Wdot lost.compressor mdot T σ S B S A := T σ 70 459.67 + ( ) rankine := THERMODYNAMIC ANALYSIS Ans. η t 0.279 = η t Wdot ideal Wdot := Ans. Wdot 47.22kW = Wdot mdot H B H A := mdot 2.914 lb m sec = mdot H 2 H 1 1 lb m sec H A H D := Refrigerent circulation rate: S D 0.094 BTU lb m rankine = S D S liq x D S vap S liq + := x D 0.284 = x D H D H liq H vap H liq := H D H C := S B 0.228 BTU lb m rankine := H B 121.84 BTU lb m = The entropy at this H is read from Fig. G.2 at P=134.75(psia) H B H A H' B H A η + := H' B 118 BTU lb
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 25

Chapter15_A - Chapter 15 Section A Mathcad Solutions 15.1...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online