SM_9A - Chapter 9 - Section A - Mathcad Solutions 9.2 TH :=...

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H2 116.166 = 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. 6.73): 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 = ϖ Carnot T C T H T C - = (9.3) ϖ 0.6 ϖ Carnot = ϖ 3.797 = Wdot Qdot C ϖ = (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: 258

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(Refrigerator) By Eq. (5.8): η Carnot 1 T C T H - = η Carnot 0.43 = By Eq. (9.3): ϖ Carnot T' C T' H T' C - = ϖ Carnot 10.926 = By definition: η W engine Q H = ϖ Q' C W refrig = But W engine W refrig = Q' C 35 kJ sec = Whence Q H Q' C η Carnot ϖ Carnot = Q H 7.448 kJ sec = Ans. Given that: η 0.6 η Carnot = ϖ 0.6 ϖ Carnot = ϖ 6.556 = Q H Q' C η ϖ = 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) ϖ Q43 W14 W32 + = ϖ 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 = 259
(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

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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_9A - Chapter 9 - Section A - Mathcad Solutions 9.2 TH :=...

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