Chapter5_A - Chapter 5 Section A Mathcad Solutions 5.2 Let...

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η Work Q H = Whence Q H Work η := Q H 1.583 10 5 × kW = Ans. Q C Q H Work := Q C 6.333 10 4 × kW = Ans. (b) η 0.35 := Q H Work η := Q H 2.714 10 5 × kW = Ans. Q C Q H Work := Q C 1.764 10 5 × kW = Ans. 5.4 (a) T C 303.15 K := T H 623.15 K := η Carnot 1 T C T H := η 0.55 η Carnot := η 0.282 = Ans. Chapter 5 - Section A - Mathcad Solutions 5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8) η Work Q H = 1 T C T H = T C 323.15 K := T H 798.15 K := Q H 250 kJ s := Work Q H 1 T C T H := Work 148.78 kJ s = or Work 148.78kW = which is the power. Ans. By Eq. (5.1), Q C Q H Work := Q C 101.22 kJ s = Ans. 5.3 (a) Let symbols Q and Work represent rates in kJ/s T H 750 K := T C 300 K := Work 95000 kW := By Eq. (5.8): η 1 T C T H := η 0.6 = But 123
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Q C 3.202 10 6 × kW = Work Q C T H T C 1 := Work 5.336 10 6 × kW = Ans. Q H Q C Work + := Q H 8.538 10 6 × kW = Ans. 5.8 Take the heat capacity of water to be constant at the value C P 4.184 kJ kg K := (a) T 1 273.15 K := T 2 373.15 K := QC P T 2 T 1 () := Q 418.4 kJ kg = S H2O C P ln T 2 T 1 := S H2O 1.305 kJ kg K = S res Q T 2 := S res 1.121 kJ kg K = Ans. (b) η 0.35 := η Carnot η 0.55 := η Carnot 0.636 = By Eq. (5.8), T H T C 1 η Carnot := T H 833.66K = Ans. 5.7 Let the symbols represent rates where appropriate. Calculate mass rate of LNG evaporation: V 9000 m 3 s := P 1.0133 bar := T 298.15 K := molwt 17 gm mol := m LNG PV RT molwt := m LNG 6254 kg s = Maximum power is generated by a Carnot engine, for which Work Q C Q H Q C Q C = Q H Q C 1 = T H T C 1 = T H 303.15 K := T C 113.7 K := Q C 512 kJ kg m LNG := 124
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Q 15000 J := (a) Const.-V heating; UQ W + = Q = nC V T 2 T 1 () = T 2 T 1 Q V + := T 2 11 0 3 × K = By Eq. (5.18), Sn C P ln T 2 T 1 Rln P 2 P 1 = But P 2 P 1 T 2 T 1 = Whence C V ln T 2 T 1 := S 20.794 J K = Ans. (b) The entropy change of the gas is the same as in (a). The entropy change of the surroundings is zero. Whence S total 10.794 J K = Ans. The stirring process is irreversible. S total S H2O S res + := S total 0.184 kJ kg K = Ans. (b) The entropy change of the water is the same as in (a), and the total heat transfer is the same, but divided into two halves. S res Q 2 1 323.15 K 1 373.15 K + := S res 1.208 kJ kg K = S total S res S H2O + := S total 0.097 kJ kg K = Ans. (c) The reversible heating of the water requires an infinite number of heat reservoirs covering the range of temperatures from 273.15 to 373.15 K, each one exchanging an infinitesimal quantity of heat with the water and raising its temperature by a differential increment. 5.9 P 1 1 bar := T 1 500 K := V 0.06 m 3 := n P 1 V RT 1 := n 1.443mol = C V 5 2 R := 125
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S A 8.726 J mol K = S B 8.512 J mol K = Ans. S total S A S B + := S total 0.214 J mol K = Ans.
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This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter5_A - Chapter 5 Section A Mathcad Solutions 5.2 Let...

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