Chapter_7_2

# T m s m s ii k qk ee s gen s2 s1 cv kj

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Unformatted text preview: A general closed system and its surroundings (an isolated system) can be treated as an adiabatic system, and the entropy balance becomes System + surroundings: S gen = ∑ ∆S = ∆S system + ∆S surroundings Control Volumes The entropy balance for control volumes differs from that for closed systems in that the entropy exchange due to mass flow must be included. ∑ T + ∑m s − ∑m s i i k Qk e e + S gen = ( S2 − S1 ) CV ( kJ / K ) In the rate form we have Qk ∑ T + ∑ mi si − ∑ me se + S gen = ∆SCV ( kW / K ) k This entropy balance relation is stated as: the rate of entropy change within the control volume during a process is equal to the sum of the rate of entropy transfer through the control volume boundary by heat transfer, the net rate of entropy transfer into the control volume by mass flow, and the rate of entropy generation within the boundaries of the control volume as a result of irreversibilities. Chapter 7 - 75 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel &amp; M. A. Boles 7-76 For a general steady-flow process, by setting ∆SCV = 0 the entropy balance simplifies to S gen = ∑ me se − ∑ mi si − ∑ Qk Tk For a single-stream (one inlet and one exit), steady-flow device, the entropy balance becomes S gen Qk = m( se − si ) − ∑ Tk For an adiabatic single-stream device, the entropy balance becomes S gen = m( se − si ) This states that the specific entropy of the fluid must increase as it flows through an adiabatic device since S gen ≥ 0 . If the flow through the device is reversible and adiabatic, then the entropy will remain constant regardless of the changes in other properties. Chapter 7 - 76 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel &amp; M. A. Boles 7-77 Therefore, for steady-flow, single-stream, adiabatic and reversible process: se = si Example 7-17 An inventor claims to have developed a water mixing device in which 10 kg/s of water at 25oC and 0.1 MPa and 0.5 kg/s of water at 100oC, 0.1 MPa, are mixed to produce 10.5 kg/s of water as a saturated liquid at 0.1 MPa. If the surroundings to this device are at 20oC, is this process possible? If not, what temperature must the surroundings have for the process to be possible? System: The mixing chamber control volume. 10 kg/s P1 = 0.1 MPa T1 = 25oC 0.5 kg/s P1 = 0.1 MPa T1 = 100oC 10.5 kg/s P3 = 0.1 MPa Saturated liquid Mixing chamber Qsys ? Surroundings To = 20oC Property Relation: The steam tables Process and Process Diagram: Assume steady-flow Chapter 7 - 77 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel &amp; M. A. Boles 7-78 Conservation Principles: First let’s determine if there is a heat transfer from the surroundings to the mixing chamber. Assume there is no work done during the mixing process, and neglect kinetic and potential energy changes. Then for two entrances and one exit, the first law becomes Qnet F V + gz IJ = W + ∑ m Gh + H 2 K 2 i i i i inlets net F V + gz IJ + ∑ m Gh + H 2 K e e 2 e e exits Qnet...
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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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