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Lec 19: Entropy changes, relative pressures and volumes, work

Lec 19: Entropy changes, relative pressures and volumes, work. For next time: Prepare for Midterm 2 on Thursday, November 6th Outline: Isentropic processes for ideal gases Internal reversible work Entropy balance equations Important points:

julian-whitfield
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Lec 19: Entropy changes, relative pressures and volumes, work

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  1. Lec 19: Entropy changes, relative pressures and volumes, work

  2. For next time: • Prepare for Midterm 2 on Thursday, November 6th • Outline: • Isentropic processes for ideal gases • Internal reversible work • Entropy balance equations • Important points: • Try to identify the “governing equations” and not get bogged down in all the special cases • Understand how to use the Tds relationships • Don’t forget to apply the 2nd Law when working problems with entropy or internal reversible processes

  3. Isentropic processes of ideal gases with constant specific heats • Before we had the situation where the specific heat could be considered constant, If this is zero, so that s2 = s1

  4. Isentropic processes of ideal gases with constant specific heats: approximation Then And The value in the exponent is:

  5. Isentropic processes of ideal gases with constant specific heats: approximation So This is only applicable for an isentropic process and cp, cv and k constant.

  6. Isentropic processes of ideal gases with constant specific heats: approximation From the other entropy change equation for an ideal gas with constant specific heats Simplifying, we get:

  7. Isentropic processes of ideal gases with constant specific heats: approximation We can combine the two T2/T1 equations to get: Rearrange: So the polytropic process with n=k is an isentropic process.

  8. TEAMPLAY Assuming constant specific heats, find the final pressure of air which undergoes an isentropic process from P1 = 14.7 psia, 60°F to a final temperature of 620°F.

  9. Isentropic processes For ideal gases with variable specific heats, the calculations are straightforward: Because it is isentropic, s2 – s1 = 0.

  10. Isentropic processes If T2 is needed and T1, p1, and p2 are known, and one may need to interpolate to find T2.

  11. Isentropic processes If p1, T1, and T2 are known and p2 is needed, Notice from this that p2 and p1 can be given as functions of T, for isentropic processes.

  12. Isentropic processes So, the relative “pressure” is defined and tabulated: Note: pr is not a pressure in spite of its name--it has no units. The right hand side is also only dependent on temperature.

  13. Isentropic processes We can use the pressure ratios directly for determining pressure changes in isentropic processes. NOTE: Pr only works for isentropic processes!!! Don’t try to use it for any other process.

  14. TEAMPLAY Use relative pressures in your tables in your books to find the final pressure of air which undergoes an isentropic process from P1 = 14.7 psia, 60°F to 620°F.

  15. Isentropic processes There is a similar relationship for volumes: As with Pr, this only applies to isentropic relationships and the right hand side only depends on temperature.

  16. Work in an internally reversible flow system • Earlier we had • This was true for a quasiequilibrium process.

  17. Work for an internally reversible flow system. • Quasiequilibrium process--one for which departures from equilibrium are infinitesimally small. • All states through which a system passes in a quasiequilibrium process may be considered to be themselves equilibrium states.

  18. Work for an internally reversible flow system • A reversible process must proceed through a series of equilibrium states. • Otherwise, there would be a tendency for the system to change spontaneously, which is irreversible. • Therefore, a quasiequilibrium process is an (internally) reversible process

  19. Work for an internally reversible flow system • Consider an internally reversible steady flow system: • Second law

  20. Work for an internally reversible flow system • The first law (not limited to internally reversible processes at this point) says

  21. Work for an internally reversible flow system • If we do limit it to internally reversible processes, which we will emphasize with subscripts, • Then the heat transfer term can be replaced

  22. Work for an internally reversible flow process • Combining the laws yields • Rearranging

  23. Work for an internally reversible flow process • Now, use a Tds relation: • Tds = dh – vdp

  24. Work for an internally reversible flow process • An expression for internally reversible work in a steady flow process.

  25. Work for an internally reversible flow process • In the absence of KE and PE effects, • On a P-v diagram, P 2 1 v

  26. Work • For open systems, • For closed systems

  27. Compressor work • By using the relationship Pvn = constant (or Pvk = constant) and solving for v = , the expression or as the book uses can be integrated to get the expressions on p. 310

  28. Entropy change for a closed system Entropy trans-fer to the sys-tem via heat transfer; can be + or –, dep-ending on the sign for Q. Entropy change of system as it goes from 1 to 2; can be + or – depending on the other two terms. Entropy prod-uction term: >0 when internal irrerversibilities are present; =0 when no int irr are present; <0 never.

  29. Entropy change of an internally reversible process; heat transfer The area under either the red or blue curves would represent the heat transfer for either process: V=c T 2 P=c 1 s

  30. TEAMPLAY What does a reversible, adiabatic process look like on a T-s diagram??

  31. Entropy production and transfer We can use entropy (the second law) to find the minimum work (associated with a reversible process).

  32. Example Problem Refrigerant 134a is compressed in a piston-cylinder assembly from saturated vapor at 10°F to a final pressure of 120 psia. Determine the minimum theoretical work input required per unit mass of refrigerant, in Btu/lb.

  33. Look at possibilities on a Ts diagram... 2? P = 120 psia T, F 2? 2? P = 26.65 psia 10 1 S Starting at 1, where is point 2? What can we know?

  34. Adiabatic Compression of R-134a The first law gives us We know the process is adiabatic, so q = 0. We ignore KE and PE. Then w = u1 – u2 u1 = 94.68 Btu/lb from Table A-11E.

  35. Adiabatic Compression of R-134a We need to find u2 for minimum work on this compressor. A reversible system will have minimum work done on it. We showed this in ch. 6 for cycles. It is shown again for open systems on pp. 308-309. That analysis can easily be extended to closed systems by replacing h with u. Minimum work corresponds to the smallest allowable value for u2, which we can determine from the second law.

  36. Adiabatic Compression of R-134a • q = 0 because it is adiabatic. • sgen = 0 because the minimum work occurs for a reversible process (frictionless). • So s2 = s1 • For irreversibilities, sgen > 0, which implies that s2 > s1

  37. Adiabatic Compression of R-134a P = 120 psia 2a T, F 2s P = 26.65 psia 10 1 S

  38. TEAMPLAY The reversible situation is s2 = s1 = 0.2214 Btu/(lbm-R) For a simple, compressible system, you now know enough to find u2s and w.

  39. Entropy rate balance for control volumes For a closed system, we had (as one of our forms of the second law) For a closed system, a rate form can be written:

  40. Entropy rate balance for control volumes Now, if entropy is transferred or convected, into and out of the control volume, the equation becomes, for an open system

  41. Entropy rate balance for control volumes Rate of entropy transfer into the CV as a result of heat transfer Rate of entropy transfer into the CV with the mass flow. Rate of entropy transfer out of the CV with the mass flow. Rate of entropy produc-tion due to irre-versibili-ties in the CV Rate of change of entropy within the control volume (CV).

  42. Entropy rate balance for control volumes This can be simplified: Assume steady flow, one heat transfer term , and one exit and one inlet:

  43. Entropy rate balance for control volumes Now, if one divides through by we get

  44. Entropy rate balance for control volumes • This equation is almost identical to the one for closed systems. In that case, the whole system goes between two states, 1 and 2. • For an open system, the change is from inlet to exit, and it is for the change of condition of the mass in the system as it moves from state 1 at the inlet to state 2 at the exit.

  45. Entropy rate balance for control volumes For sgen  0 (irreversibilities present): If the process is also adiabatic, then s2  s1 What can we conclude? An adiabatic, irreversible process will always move to the right on a Ts diagram.

  46. Look at adiabatic compression processes on a Ts diagram P2 T 2a 2s P2 > P1 P1 Forbidden States 1 s

  47. Look at adiabatic expansion processes on a Ts diagram P1 T 1 P1 > P2 P2 Forbidden States 2a 2s s

  48. Isentropic Processes(constant entropy) For real substances, such as water, R-134a, etc., use tabular entropy values or EES. An isentropic compressor has R-134a saturated vapor entering at 10°F and leaving the compressor at 120 psia. What does the process look like on a Ts diagram?

  49. Adiabatic Compression of R-134a P = 120 psia T, F 2s P = 26.65 psia 10 1 S Pt 2 is at s2 = s1 and p = 120 lbf/in2.

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