2. : The T-ds relations
Apply the differential form of the first law for a closed
stationary system for an internally reversible
process
δQint rev − δWint rev = dU
δQint rev = TdS δWint rev = PdV
TdS − PdV = dU 2
3. Divide by the
mass, you get Tds = du + Pdv
This equation is known as:
First Gibbs equation or First Tds relationship
Divide by T, .. du Pdv
ds = +
T T
Although we get this form for internally reversible process,
we still can compute ∆s for an irreversible process. This is
because S is a point function.
3
4. Second T-ds (Gibbs) relationship
Recall that… h = u + Pv
Take the differential for both sides
dh = du + Pdv + vdP
Rearrange
to find du du = dh − Pdv − vdP
Substitute in the
First Tds
relationship Tds = du + Pdv
Tds = dh − vdP
Second Tds relationship, or Gibbs equation
4
5. dh vdP
Divide by T, .. ds = −
T T
Thus We have two equations for ds
du Pdv dh vdP
ds = + ds = −
T T T T
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
5
6. Entropy Change of Liquids- 2
and Solids
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
du Pdv Thus For du
ds = + solids and ds =
T T liquids
T
, Recall also that For solids and liquids du = CdT
du CdT
∴ ds = =
T T 6
7. Integrate to T2 Only true
give… ∆s = C ln
T for solids
1 and
liquids!!
What if the process T2
∆s = 0, ⇒ C ln = 0
T
?is isentropic
1
T1 = T2
The only way this expression
can equal 0 is if,
Hence, for solids and liquids, isentropic
processes are also isothermal. 7
8. Example(6-7): Effect of Density of a Liquid on
Entropy
Liquid methane is commonly used in various cryogenic
applications. The critical temperature of methane is 191 K
(or -82oC), and thus methane must be maintained below
191 K to keep it in liquid phase. The properties of liquid
methane at various temperatures and pressures are given
next page.
Determine the entropy change of liquid methane as it
undergoes a process from 110 K and 1 MPa to 120 K and 5
MPa
(a) using actual data for methane and
(b) approximating liquid methane as an incompressible
substance. What is the error involved in the later case?
8
10. Example (6-19): Entropy generated when a hot
block is dropped in a lake
A 50-kg block of iron casting at 500 K is dropped in a large
lake that is at 285 K. The block reaches thermal equilibrium
with lake water.
Assuming an average specific heat of 0.45 kJ/kg.K for the
iron, determine:
(a) The entropy change of the iron block,
(b) The entropy change of the water lake,
(c) the entropy generated during this process.
10
11. (a) The entropy change of the iron block,
T2 285
∆Siron = 50 × 0.45 × ln
= mC ln = −12.65 kj / K
T1 500
we need also to find Q coming out of the system.
Q − W = ∆U Q − 0 = mC (T2 − T1 )
Q = 50 × 0.45(285 − 500) = −4838 kJ
Tsurr= 285 K
(b) The entropy change of the water lake,
T=500K
Q 4838
∆Slake = = = 16.97 kJ / K
Tlake 285
11
12. (c) the entropy generated during this process.
Choose the iron block and the lake as the system
and treat it is an isolated system.
Tsurr= 285 K
Thus Sg = ∆Stot = ∆Ssys + ∆Slake
T=500K
Sg = ∆Stot = -12.6 + 16.97
= 4.32 System
boundary
12
13. The Entropy Change of Ideal- 3
Gases, first relation
First relation
The entropy change of an ideal gas can be obtained by
substituting du = CvdT and P /T= R/υ into Tds relations:
Tds = du + pd υ du Pd υ dT dυ
ds = + ds = C v +R
T T T υ
integrating 2
dT υ2
1
∫
⇒ s2 − s1 = Cv ( T )
T
+R ln
υ1
13
14. Second relation
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:
dh vdp dT dp
Tds = dh −vdp ds = − ds = C p −R
T T T p
integrating
2
dT P2
1
∫
s2 − s1 = C p ( T )
T
− R ln
P1
14
15. 2
dT υ2
s2 − s1 = ∫ Cv ( T ) +R ln
1
T υ1
2
dT P2
1
∫
s2 − s1 = C p ( T )
T
− R ln
P1
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)
15
16. Method 1: Constant specific
)heats (Approximate Analysis
First relation
2
dT υ2 T2 v2
s2 − s1 = ∫ Cv ( T ) +R ln ⇒ ∆s = Cv ln + R ln
T v
1
T υ1 1 1
Only true for ideal gases, assuming constant heat capacities
Second relation
2
dT P2 T2 P2
s2 − s1 = ∫ C p ( T ) − R ln ⇒ ∆s = C p ln − R ln
T P
1
T P1 1 1
Only true for ideal gases, assuming constant heat capacities
16
17. Sometimes it is more convenient to calculate the
change in entropy per mole, instead of per unit mass
T2 v2
∆s = s2 − s1 = Cv ln + Ru ln
T v kJ/kmol. K
1 1
T2 P2
∆s = s2 − s1 = C p ln − Ru ln
T P
kJ/kmol. K
1 1
Ru is the universal gas constant
17
18. Method 2: Variable specific
heats (Exact Analysis)
2 C p dT P2
We use the
second relation
∆s = ∫
1 T
− R ln
P1
Wecould substitute in the equations for Cv
and Cp, and perform the integrations
Cp = a + bT + cT2 + dT3
But this is time consuming.
Someone already did the integrations and
tabulated them for us (table A-17)
They assume absolute 0 as the starting point
18
19. The integral is expressed as:
T2 dT T2 dT T1 dT
∫T1
C p (T )
T
= ∫
0
C p (T )
T
− ∫ C p (T )
0 T
T2
dT
∫ = s2 − s1
0 0
Cp( T )
T1 T
T
dT
s = ∫
0
Where Cp( T )
0 T
is tabulated in Table A-17
Therefore
P2
∆s = s − s − R ln 0
2
0
1
unit : kJ / kg .K P1 19
20. Is s = f (T) only? like u for an
.ideal gas Let us see
Temperature dependence
Pressure
P2 dependence
∆s = s − s − R ln
0
2
0
1
P1
From this equation, It can be seen that the
entropy of an ideal gas is not a function only of
the temperature ( as was the internal energy) but
also of the pressure or the specific volume.
The function s° represents only the temperature-
dependent part of entropy
20
21. How about the other relation
2
dT v2
∆s = ∫
1
Cv
T
+ R ln
v1
We can develop another relation for the
entropy changed based on the above relation
but this will require the definition of another
function and tabulating it which is not
practical. T
dT
?= ∫
0
Cv ( T )
T
21
22. 6-4 Isentropic Processes
The entropy of a fixed mass can be
changed by
1. Heat transfer,
2. Irreversibilities
It follows that the entropy of a
system will not change if we have
1. Adiabatic process,
2. Internally reversible process.
Therefore, we define the following:
22
23. 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
23
24. )Constant specific heats (1st relation
Recall T2 v2
∆s = Cv ln + R ln
T v
1 1
For the isentropic case, ∆S=0. Thus
R
T2 v2 T2 R v2 v1 Cv
Cv ln = − R ln
T v ln = − ln = ln
T
1 1 1 Cv v1
v
2
Recall also from ch 2, the following relations..…
R = C p − Cv ⇒ R / Cv = C p / Cv − 1 = k − 1
k −1 Only applies to
T2 v1
∴ =
T v
ideal gases,
with constant
1 2 specific heats 24
25. Constant specific heats (2nd
)relation T2 P2
∆s = C p ln − R ln = 0
T P
1 1
R
T2 R P2 P2 Cp
ln =
T C ln P = ln P
1 p 1 1
k −1
Recall..… R / Cv = k − 1 or R /C p =
k
k −1
Only applies to
T2 P2 k
ideal gases,
∴ =
T P with constant
1 1 specific heats
25
26. …Since
k −1 k −1
T2 v1 T2 P2 k
=
T v and =
T P
1 2 1 1
k −1
k −1
HENCE v1 P2 k
v =
P
2 1
k
Which can be v1 P2 Third
simplified to… =
v P isentropic
2 1 relationship
26
27. Full form of Isentropic relations of Ideal Gases
k −1 k −1 k
T2 v1 T2 P2 k v1 P2
=
T v = =
v P
T P
1 2 1 1 2 1
Compact form
1− k
Tv k −1
= constant TP k
= constant Pv k = constant
Valid for only for
1- Ideal gas
2- Isentropic process
3- Constant specific heats
27
28. That works if the specific heat constants can
be approximated as constant, but what if
?that’s not a good assumption
We need to use the
exact treatment 0 P2
∆s = s − s − R ln
0
2
0
1 P
1
This equation is a
good way to evaluate P2
property changes, s = s + R ln
0
2
0
1 P
but it can be tedious
if you know the 1
volume ratio instead
of the pressure ratio 28
29. P2
s = s + R ln
0
2
0
1 P
s20 is only a function
1 of temperature!!!
s −s
0 0
P2 Rename the exponential term
2
= ln
1
P as Pr , (relative pressure)
R 1 which is only a function of
temperature, and is tabulated
P2 s2 − s10
0
on the ideal gas tables
= exp
R
P1
s2
0
s2
0
exp
P2 R÷
exp
R ÷ Pr 2
= P2 Pr 2
= =
P1 s10 s10 Pr 1 P1 Pr 1
exp exp
R÷ R÷
29
30. You can use either of the following 2 equations
P2 Pr 2 P2
= s = s + R ln
0
2
0
1 P
P Pr1
1 1
This is good if you know the pressure ratio but how about
if you know only the volume ratio
In this case, we use the ideal gas law
P v1 P2 v2 v2 T2 P T2 Pr1 T2 Pr1 vr 2
1
= ⇒ = 1
= = =
T1 T2 v1 T1 P2 T1 Pr 2 Pr 2 T1 vr1
v2 vr 2
where vr = T / P r ∴ =
v1 vr1
Remember, these relationships only hold for ideal gases and
isentropic processes 30
31. Example (6-10):
Isentropic Compression of Air in a Car Engine
Air is compressed in a car engine from 22oC and 95
kPa in a reversible and adiabatic manner. If the
compression ratio V1/V2 of this piston-cylinder device
is 8, determine the final temperature of the air.
<Answer: 662.7 K>
Sol:
31