Chapter_7_2

What is the final temperature of the nitrogen and how

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Unformatted text preview: of 500 K while its volume doubles. What is the final temperature of the nitrogen, and how much work did the nitrogen do against the piston, in kJ/kg? System: The closed piston-cylinder device T Nitrogen system boundary 1 P1 W 2 P2 s P P1 P2 1 s = const. 2 v Property Relation: Ideal gas equations, constant properties Process and Process Diagram: Isentropic expansion Conservation Principles: Second law: Chapter 7 - 45 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-46 Since we know T1 and the volume ratio, the isentropic process, ∆s = 0, allows us to find the final temperature. Assuming constant properties, the temperatures are related by FG v IJ T =T Hv K FG 1 IJ = 500 K H 2K k −1 2 1 1 2 1.4 −1 = 378.9 K = 105.9 o C Why did the temperature decrease? First law, closed system: Note, for the isentropic process (reversible, adiabatic); the heat transfer is zero. The conservation of energy for this closed system becomes Ein − E out = ∆E −W = ∆U W = − ∆U Using the ideal gas relations, the work per unit mass is Chapter 7 - 46 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-47 W = − mCv (T2 − T1 ) W w = = − Cv (T2 − T1 ) m kJ = −0.743 (378.9 − 500) K kg ⋅ K kJ = 90.2 kg Why is the work positive? Extra Assignment For the isentropic process Pvk = constant. Use the definition of boundary work to show that you get the same result as the last example. That is, determine the boundary work and show that you obtain the same expression as that for the polytropic boundary work. Example 7-12 A Carnot engine has 1 kg of air as the working fluid. Heat is supplied to the air at 800 K and rejected by the air at 300 K. At the beginning of the heat addition process, the pressure is 0.8 MPa and during heat addition the volume triples. (a) Calculate the net cycle work assuming air is an ideal gas with constant specific heats. (b) Calculate the amount of work done in the isentropic expansion process. Chapter 7 - 47 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-48 (c) Calculate the entropy change during the heat rejection process. System: The Carnot engine piston-cylinder device. W Air system boundary Q Property Relation: Ideal gas equations, constant properties. Process and Process Diagram: Constant temperature heat addition. T TH 1 QH 2 TL 4 QL 3 s Carnot Cycle Conservation Principles: a. Apply the first law, closed system, to the constant temperature heat addition process, 1-2. Chapter 7 - 48 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-49 Qnet ,12 − Wnet ,12 = ∆U12 Qnet ,12 = mCv (T2 − T1 ) = 0 = Wnet ,12 So for the ideal gas isothermal process, Wnet ,12 = Wother ,12 + Wb ,12 = PdV 1 dV 1 V V2 = mRT ln V1 = 2 z z 2 mRT FG IJ H K kJ )(800 K ) ln(3) kg ⋅ K = 1kg (0.287 = 252.2 kJ But Qnet ,12 = QH QH = 252.2 kJ Chapter 7 - 49 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-50 The cycle thermal efficiency is η th = Wnet , cycle QH For the Carnot cycle, t...
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This note was uploaded on 09/17/2009 for the course MAE 301 taught by Professor Hassan during the Fall '08 term at N.C. State.

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