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

2. For next time: Outline: Important points:
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
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 = 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 = 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 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 = 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 > P1 P1
Forbidden
States 1 s

47. Look at adiabatic expansion processes on a Ts diagram1
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 = psia 10 1 S
Pt 2 is at s2 = s1 and p = 120 lbf/in2.