Chapter 9 - Chapter 9 Second law analysis for a control...

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Unformatted text preview: Chapter 9 Second law analysis for a control volume Read BS, Chapter 9 In this chapter, we will apply notions from control volume analysis to problems which involve the second law of thermodynamics. Recall that the fundamental description of our axioms is written for systems. We simply modify these axioms when applying them to control volumes. We shall omit most of the details of the reduction of the second law to control volume formulation. It is not unlike that done for mass and energy conservation. 9.1 Irreversible entropy production First recall an important form of the second law for a system, Eq. (8.31): S 2 S 1 integraldisplay 2 1 Q T . (9.1) Let us introduce a convenient variable, the Irreversible entropy production : a quantity which characterizes that portion of entropy production which is irreversible. We note that entropy can be produced by reversible heat transfer as well, which we segregate and do not consider here. We adopt the common notation of 1 2 for irreversible entropy production, with units of kJ/K . Our 1 2 is equivalent to 1 S 2 gen of BS, but is more aligned with the notation of non-equilibrium thermodynamics. We give it the subscripts to emphasize that it is path-dependent. Mathematically, we recast the second law for a system by the following two equations: S 2 S 1 = integraldisplay 2 1 Q T + 1 2 , (9.2) 1 2 . (9.3) 265 266 CHAPTER 9. SECOND LAW ANALYSIS FOR A CONTROL VOLUME Clearly, this is just a notational convenience which moves the inequality from one equation to another. On a differential basis, we can say for a system dS = Q T + . (9.4) And for time-dependent processes, we say for a system dS dt = 1 T Q dt + dt . (9.5) Now, let us expand to an unsteady control volume, which should be similar to that for a system, with corrections for inlets and exits. We get dS cv dt = summationdisplay j Q j T j + summationdisplay i m i s i summationdisplay e m e s e + cv . (9.6) Let us study this equation in some common limits. First, if the problem is in steady state, then 0 = summationdisplay j Q j T j + summationdisplay i m i s i summationdisplay e m e s e + cv , (9.7) steady state limit. If it is in steady state and there is one entrance and one exit, then mass conservation gives m i = m e = m , and 0 = summationdisplay j Q j T j + m ( s i s e ) + cv , (9.8) steady state, one entrance, one exit. We can rearrange to say s e s i = 1 m summationdisplay j Q j T j + cv m . (9.9) If there is no heat transfer to the control volume, then s e s i = cv m , (9.10) no heat transfer to control volume, steady state, one entrance/exit....
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Chapter 9 - Chapter 9 Second law analysis for a control...

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