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Thermodynamics 1 (MEP 261) Quiz 5 ( 20 Marks): Model Answer
Class ZA, Date: MAY 12, 2012
Name: University ID:
7–26 During the isothermal heat addition process of a Carnot cycle, 1400 kJ of heat is added to the working fluid from a source at 437°C. Determine (a) the entropy change of the working fluid, (b) the entropy change of the source, and (c) the total entropy change for the process. 6 Marks
Solution
Solution
1)
1400 kJ Source
437°C
1400 kJ 700 K
1400 kJ 700 K
-2.0 kJ/ K 2.0 kJ/ K
2.0 kJ/ K - 2.0 kJ/ K = 0
2
7–46 A piston–cylinder device contains 1.0 kg of saturated water vapor at 190°C. Heat is now transferred to steam, and steam expands reversibly and isothermally to a final pressure of 600 kPa. Determine the heat transferred and the work done during this process.
From Table A-6, u2 & s2 are evaluated by interpolation
(u2-2566.8 kJ/kg)/(2639.4 kJ/kg - 2566.8 kJ/kg)=(190 kJ/kg -158.83 kJ/kg)/(200 kJ/kg -158.83 kJ/kg) Hence, u2 = 2621.77 kJ/kg
(s2-6.7593 kJ/kg.K)/(6.9683 kJ/kg.K – 6.7593 kJ/kg.K)=(190 kJ/kg.K -158.83 kJ/kg.K)/(200 kJ/kg.K - 158.83 kJ/kg.K)
Hence, s2 = 6.917535 kJ/kg.K
The heat transfer for this reversible isothermal process can be determined from Solution
2)
190°C T1 = 190°C u1 = ug@T = 190°C = 2589.0 kJ/kg
s1 = sg@T = 190°C = 6.5059 kJ/kg.K
p2 =
600 kPa,
T2 = T1= 190°C=190+273= 463K
From Table A-5, For p2 =
600 kPa (0.6 MPa),
Tsat = 158.83°C As T2 (=190°C) > Tsat (=158.83°C), Hence state 2 is superheated steam= (463K)(1 kg)(6.917535-6.5059 kJ/kg.K)= 190.59 kJ
Wb,out = 190.59 kJ- (1 kg)( 2621.77 kJ/kg-2589.0 kJ/kg)= 157.82 kJ
