1. Lec 18: Isentropic processes, TdS relations, entropy changes

2. For next time:
Read: § 7-2 to 7-9
Group project subject selection due on November 3, 2003
Outline:
Entropy generation and irreversible processes
Entropy as a property
Entropy changes for different substances
Important points:
Entropy is a property of a system – it is not conserved and is generated by irreversible processes
Know how to identify an isentropic processes
Know how to use the tables to find values for entropy

3. Recall we had entropy
s2 - s1 =
Units are

4. Let’s look at a simple irreversible cycle on a p-v diagram with two processes
Let A be irreversible and B be reversible

5. Evaluate cyclic integral
Irreversible cycle
By Clausius Inequality
Evaluate cyclic integral
(non-rev)
(rev)

6. Irreversible cycle For the reversible process, B, dS=Q/dT, thus:
Rearranging and integrating dS:

7. Second Law of Thermodynamics
This can be viewed as a mathematical statement of the second law (for a closed system).
Entropy is a non-conserved property!

8. We can write entropy change as an equality by adding a new term:
entropy transfer due to heat transfer
entropy production or generation

9. Entropy generation Sgen  0 is an actual irreversible process.
Sgen = 0 is a reversible process.
Sgen  0 is an impossible process.

10. TEAMPLAY Consider the equation
You have probably heard, “Entropy always increases.”
Could it ever decrease? What are the conditions under which this could happen (if it can)?

11. Entropy transfer and production
What if heat were transferred from the system?
The entropy can actually decrease if
and heat is being transferred away from the system so that Q is negative.

12. Entropy Production
Sgen quantifies irreversibilities. The larger the irreversibilities, the greater the value of the entropy production, Sgen .
A reversible process will have no entropy production.

13. Entropy transfer and production
> 0, Q could be + or –; if –,
because Sgen is always positive.
< 0, if Q is negative and
= 0 if Q = 0 and Sgen = 0.
= 0 if Q is negative and
S2 – S1

14. Isentropic processes
Note that a reversible (Sgen = 0), adiabatic (Q = 0) process is always isentropic (S1 = S2)
But, if the process is merely isentropic with S1 = S2, it may not be a reversible adiabatic process.
For example, if Q  0 and

15. Entropy generation Consider
What if we draw our system boundaries so large that we encompass all heat transfer interactions? We would thereby isolate the system.

16. Entropy changes of isolated systems
And then
But Sgen0. So, the entropy of an isolated system always increases. (This is the source of the statement, ‘The world is running down.’)

17. Entropy It is tabulated just like u, v, and h. Also,
And, for compressed or subcooled liquids,

18. The entropy of a pure substance is determined from the tables, just as for any other property

19. Ts Diagram for Water

20. TEAMPLAY Use the tables in your book
Find the entropy of water at 50 kPa and 500 °C. Specify the units.
Find the entropy of water at 100 °C and a quality of 50%. Specify the units.
Find the entropy of water at 1 MPa and
120 °C. Specify the units.

21. Ts diagrams
Recall that the P-v diagram was very important in first law analysis, and that
Work was the area under the curve.

22. For a Ts diagram Rearrange: Integrate:
If the internally reversible process also is isothermal at some temperature To:

23. On a T-S diagram, the area under the process curve represents the heat transfer for internally reversible processes d

24. Entropy change of a thermal reservoir
For a thermal reservoir, heat transfer occurs at constant temperature…the reservoir doesn’t change temperature as heat is removed or added:
Since T=constant:
Applies ONLY to thermal reservoirs!!!!

25. The Tds Equations

26. Derivation of Tds equations:
For a simple closed system:
dQ – dW = dU
The work is given by:
dW = PdV
Substituting gives:
dQ = dU + PdV

27. More derivation…. For a reversible process: TdS = dQ
Make the substitution for Q in the energy equation:
Or on a per unit mass basis:

28. Tds Equations
Entropy is a property. The Tds expression that we just derived expresses entropy in terms of other properties. The properties are independent of path….We can use the Tds equation we just derived to calculate the entropy change between any two states:
Tds = du +Pdv
Starting with enthalpy, it is possible to develop a second Tds equation:
Tds = dh - vdP

29. Let’s look at the entropy change for an incompressible substance:
We start with the first Tds equation:
Tds = cv(T)dT + Pdv
For incompressible substances, v  const, so dv = 0.
We also know that cv(T) = c(T), so we can write:

30. Entropy change of an incompressible substance
Integrating
If the specific heat does not vary with temperature:

31. TEAMPLAY
Work Problem 7-48

32. Entropy change for an ideal gas
Start with 2nd Tds equation
Tds = dh - vdp
Remember dh and v for an ideal gas?
v=RT/p
And
Substituting:

33. Change in entropy for an ideal gas
Dividing through by T,
Don’t forget, cp=cp(T)…..a function of temperature! Integrating yields

34. Entropy change of an ideal gas for constant specific heats: approximation
Now, if the temperature range is so limited that cp  constant (and cv  constant),

35. Entropy change of an ideal gas for constant specific heats: approximation
Similarly it can be shown from
Tds = du + pdv
that

36. TEAMPLAY
Work Problem 7-62

37. Entropy change of an ideal gas for variable specific heats: exact analysis
Integrating..
To evaluate entropy change, we’ll have to evaluate the integral:

38. Entropy change of an ideal gas for variable specific heats: exact analysis
Evaluation of the integral
And so(T), the reference entropy, is tabulated in the ideal gas tables for a reference temperature of T = 0 and p = 1 atm.

39. Entropy change of an ideal gas for variable specific heats: exact analysis
Only is tabulated. The is not.
So,

40. Entropy change of an ideal gas
Note that the entropy change of an ideal gas, unlike h and u, is a function of two variables.
Only the reference entropy, so, is a function of T alone.

41. Sample Problem
A rigid tank contains 1 lb of carbon monoxide at 1 atm and 90°F. Heat is added until the pressure reaches 1.5 atm. Compute:
(a) The heat transfer in Btu.
(b) The change in entropy in Btu/R.

42. Draw diagram: Rigid Tank => volume is constant Heat Transfer CO:
m= 1 lbm
State 2:
P = 1.5 atm
State 1:
P = 1atm
T = 90oF
Heat Transfer

43. Assumptions Work is zero - rigid tank kinetic energy changes zero
potential energy changes zero
CO is ideal gas
CO in tank is system
Constant specific heats

44. Apply assumptions to conservation of energy equation
For constant specific heats, we get:
Need T2 ––> How do we get it?

45. Apply ideal gas EOS:
Cancel common terms...
Solve for T2:

46. Solve for heat transfer
Now, let’s get entropy change...

47. For constant specific heats:
Since v2 = v1