Chapter9_A

# Chapter9_A - Chapter 9 Section A Mathcad Solutions 9.2 TH:=...

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S2 0.21868 := P2 138.83 := State 3, Wet Vapor at TC: Hliq 15.187 := Hvap 104.471 := P3 26.617 := State 4, Wet Vapor at TC: Sliq 0.03408 := Svap 0.22418 := P4 26.617 := (a) The pressures in (psia) appear above. (b) Steps 3--2 and 1--4 (Fig. 8.2) are isentropic, for which S3=S2 and S1=S4. Thus by Eq. 6.82): x3 S2 Sliq Svap Sliq := x3 0.971 = x4 S1 Sliq Svap Sliq := x4 0.302 = (c) Heat addition, Step 4--3: H3 Hliq x3 Hvap Hliq () + := H4 Hliq x4 Hvap Hliq + := H3 101.888 = H4 42.118 = Q43 H3 H4 := Q43 59.77 = (Btu/lb m ) Chapter 9 - Section A - Mathcad Solutions 9.2 T H 20 273.15 + K := T H 293.15K = T C 20 273.15 + K := T C 253.15K = Qdot C 125000 kJ day := ω Carnot T C T H T C := (9.3) ω 0.6 ω Carnot := ω 3.797 = Wdot Qdot C ω := (9.2) Wdot 0.381kW = Cost 0.08 kW hr Wdot := Cost 267.183 dollars yr = Ans. 9.4 Basis: 1 lbm of tetrafluoroethane The following property values are found from Table 9.1: State 1, Sat. Liquid at TH: H1 44.943 := S1 0.09142 := P1 138.83 := State 2, Sat. Vapor at TH: H2 116.166 := 298

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(Refrigerator) By Eq. (5.8): η Carnot 1 T C T H := η Carnot 0.43 = By Eq. (9.3): ω Carnot T' C T' H T' C := ω Carnot 10.926 = By definition: η W engine Q H = ω Q' C W refrig = But W engine W refrig = Q' C 35 kJ sec := Whence Q H Q' C η Carnot ω Carnot := Q H 7.448 kJ sec = Ans. Given that: η 0.6 η Carnot := ω 0.6 ω Carnot := ω 6.556 = Q H Q' C ηω := Q H 20.689 kJ sec = Ans. (d) Heat rejection, Step 2--1: Q21 H1 H2 () := Q21 71.223 = (Btu/lb m ) (e) W21 0 := W43 0 := W32 H2 H3 := W32 14.278 = W14 H4 H1 := W14 2.825 = (f) ω Q43 W14 W32 + := ω 5.219 = Note that the first law is satisfied: Σ Q Q21 Q43 + := Σ W W32 W14 + := Σ Q Σ W + 0 = 9.7 T C 298.15 K := T H 523.15 K := (Engine) T' C 273.15 K := T' H 298.15 K := 299
(isentropic compression) S' 3 S 2 = H 4 37.978 Btu lb m := T 4 539.67 rankine := From Table 9.1 for sat. liquid S 2 0.22244 0.22325 0.22418 0.22525 0.22647 Btu lb m rankine := H 2 107.320 105.907 104.471 103.015 101.542 Btu lb m := Qdot C 600 500 400 300 200 Btu sec := η 0.79 0.78 0.77 0.76 0.75 := T 2 489.67 479.67 469.67 459.67 449.67 rankine := The following vectors contain data for parts (a) through (e). Subscripts refer to Fig. 9.1. Values of H2 and S2 for saturated vapor come from Table 9.1. 9.9 or -45.4 degC Ans. T C 227.75K = 9.8 (a) Q C 4 kJ sec := W 1.5 kW := ω Q C W := ω 2.667 = Ans. (b) Q H Q C W + := Q H 5.5 kJ sec = Ans. (c) ω T C T H T C = T H 40 273.15 + () K := T H 313.15K = T C T H ω ω 1 + := 300

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Ans. Wdot 94.5 100.5 99.2 90.8 72.4 kW = Wdot mdot H 23 () ⎯⎯⎯⎯⎯ := Ans. Qdot H 689.6 595.2 494 386.1 268.6 Btu sec = Qdot H mdot H 4 H 3 ⎯⎯⎯⎯⎯⎯⎯ := Ans. mdot 8.653 7.361 6.016 4.613 3.146 lb m sec = mdot Qdot C H 2 H 1 ⎯⎯⎯ := H 3 273.711 276.438 279.336 283.026 286.918 kJ kg = H 23 24.084 30.098 36.337 43.414 50.732 kJ kg = H 1 88.337 kJ kg = H 1 H 4 := H 3 H 2 H 23 + := H 23 H' 3 H 2 η ⎯⎯⎯ := H' 3 115.5 116.0 116.5 117.2 117.9
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Chapter9_A - Chapter 9 Section A Mathcad Solutions 9.2 TH:=...

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