SM_5A - 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 97
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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 = Q C P T 2 T 1 - ( 29 = 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 P V R T 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 = 98
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Q 15000 J = (a) Const.-V heating; U Q W + = Q = n C V T 2 T 1 - ( 29 = T 2 T 1 Q n C V + = T 2 1 10 3 × K = By Eq. (5.18), S n C P ln T 2 T 1 R ln P 2 P 1 - = But P 2 P 1 T 2 T 1 = Whence S n 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
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SM_5A - Chapter 5 - Section A - Mathcad Solutions 5.2 Let...

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