Thermo lecture notes

1. 1
Property
Relationships

2. 2
dUWQ revrev =−δδ
TdSQrev =δ PdVWrev =δ
dUPdVTdS =−
Apply the differential form of the first law for a closed
stationary system for an reversible process
The T-ds relations:

3. 3
PdvduTds +=
This equation is known as:
First Gibbs equation or First Tds relationship
Divide by T, ..
T
Pdv
T
du
ds +=
Although we get this form for reversible process, we still
can compute ∆s for an irreversible process. This is because S
is a point function.
Divide by the
mass, you get

4. 4
Second T-ds (Gibbs) relationship
Recall that… Pvuh +=
vdPPdvdudh ++=
Take the differential for both sides
Rearrange
to find du
Substitute in
the First Tds
relationship PdvduTds +=
vdPPdvdhdu −−=
vdPdhTds −=
Second Tds relationship, or Gibbs equation

5. 5
T
vdP
T
dh
ds −=Divide by T, ..
Thus We have two equations for ds
T
vdP
T
dh
ds −=
T
Pdv
T
du
ds +=
To find ∆s, we have to integrate these equations. Thus
we need a relation between du and T (or dh and T).
Now we can find entropy change (the LHS of the
entropy balance) for liquids and solids

6. 6
2- Entropy Change of Liquids
and Solids
T
Pdv
T
du
ds +=
Solids and liquids do not change specific volume appreciably
with pressure. That means that dv=0, so the first equation is
the easiest to use.
0
T
du
ds =
Thus For
solids and
liquids
Recall also that For solids and liquids, CdTdu =
T
CdT
T
du
ds ==∴

7. 7






=∆
1
2
ln
T
T
Cs
Integrate
to give…
Only true
for solids
and liquids!!
What if the process
is isentropic?
0ln,0
1
2
=





⇒=∆
T
T
Cs
The only way this expression
can equal 0 is if,
Hence, for solids and liquids, isentropic
processes are also isothermal.
21 TT =

8. 8
3- The Entropy Change of Ideal
Gases, first relation
The entropy change of an ideal gas can be obtained by
substituting du = CvdT and P /T= R/υ into Tds relations:
( )
1
2
2
1
12
υ
υ
lnR
T
dT
TCss v∫ +=−⇒
Tds du pdυ= +
v
dT d
ds C R
T
υ
υ
= +
integrating
First relation
du Pd
ds
T T
υ
= +

9. 9
A second relation for the entropy change of an ideal gas
for a process can be obtained by substituting dh = CpdT
and υ/T= R/P into Tds relations:
( )∫ −=−
2
1 1
2
12
P
P
lnR
T
dT
TCss p
Tds dh vdp= − p
dT dp
ds C R
T p
= −
integrating
Second relation
dh vdp
ds
T T
= −

10. 10
( )
1
2
2
1
12 ln
υ
υ
R
T
dT
TCss v∫ +=−
( )∫ −=−
2
1 1
2
12
P
P
lnR
T
dT
TCss p
 The integration of the first term on the RHS can be done via
two methods:
1. Assume constant Cp and constant Cv (Approximate Analysis)
2. Evaluate these integrals exactly and tabulate the data (Exact
Analysis)

11. 11
Method 1: Constant specific
heats (Approximate Analysis)
First relation






+





=∆
1
2
1
2
lnln
v
v
R
T
T
Cs v
Only true for ideal gases, assuming constant heat capacities
Second relation
Only true for ideal gases, assuming constant heat capacities






−





=∆
1
2
1
2
lnln
P
P
R
T
T
Cs p
( ) ⇒+=− ∫ 1
2
2
1
12 ln
υ
υ
R
T
dT
TCss v
( ) ⇒−=− ∫
2
1 1
2
12 ln
P
P
R
T
dT
TCss p

12. 12
Method 2: Variable specific
heats (Exact Analysis)
 We could substitute in the equations for Cv
and Cp, and perform the integrations
 Cp = a + bT + cT2 + dT3
 But this is time consuming.
1
2
2
1 P
P
lnR
T
dTC
s p
−=∆ ∫We use the
second relation

13. 7.5. Isentropic
Processes of an
ideal gas
13

14. 14
Isentropic Processes of Ideal
Gases
 Many real processes can be modeled as
isentropic
 Isentropic processes are the standard against
which we should measure efficiency
 We need to develop isentropic relationships
for ideal gases, just like we developed for
solids and liquids

15. 15






+





=∆
1
2
1
2
lnln
v
v
R
T
T
Cs v
For the isentropic case, ∆S=0. Thus






−=





1
2
1
2
lnln
v
v
R
T
T
Cv
Constant specific heats (1st relation)
vC
R
v v
v
v
v
C
R
T
T






=





−=





2
1
1
2
1
2
lnlnln
Recall
Recall also from previous chapter, the following relations..…
11 −=−=⇒−= kC/CC/RCCR vpvvp
1
2
1
1
2
−






=





∴
k
v
v
T
T Only applies to
ideal gases,
with constant
specific heats

16. 16
0lnln
1
2
1
2
=





−





=∆
P
P
R
T
T
Cs p
pC
R
p P
P
P
P
C
R
T
T






=





=





1
2
1
2
1
2
lnlnln
k
k
P
P
T
T
1
1
2
1
2
−






=





∴
Only applies to
ideal gases,
with constant
specific heats
Constant specific heats (2nd
relation)
Recall..… or1/ −= kCR v
1
/ p
k
R C
k
−
=

17. 17
Since…
k
k
P
P
T
T
1
1
2
1
2
−






=





1
2
1
1
2
−






=





k
v
v
T
T
and
k
k
k
P
P
v
v
1
1
2
1
2
1
−
−






=





Which can be
simplified to…






=





1
2
2
1
P
P
v
v
k
Third
isentropic
relationship
HENCE