Entropy Change
• Property diagrams (T-s and h-s diagrams)
– From the definition of the entropy, it is known that Q=TdS during
a reversible process.
– Hence the total heat transfer during such a process is given by
Qreversible =  TdS
• Therefore, it is useful to consider the T-S diagram for a reversible
process involving heat transfer
T
On a T-S diagram, the area under the
process curve represents the heat transfer
for a reversible process.
S
Example
Show a Carnot cycle on a T-S diagram and identify the heat transfer at both
the high and low temperatures, and the work output from the cycle.
T
QH
1
• 1-2, reversible isothermal heat transfer
QH = TdS = TH(S2-S1) area 1-2-B-A
2
TH
TL
Wout
Win
4
A
S1=S4
3
• 2-3, reversible, adiabatic expansion
isentropic process, S=constant (S2=S3)
• 3-4, reversible isothermal heat transfer
QL = TdS = TL(S4-S3), area 3-4-A-B
QL
B
S2=S3
S
• 4-1, reversible, adiabatic compression
isentropic process, S1=S4
Net work Wnet = QH - QL, is the area enclosed by 1-2-3-4, the
shaded area
Mollier Diagram
• Enthalpy-entropy diagram, h-s diagram: it is valuable in
analyzing steady-flow devices such as turbines,
compressors, etc.
• Dh: change of enthalpy from energy balance (from the
first law of thermodynamics)
• Ds: change of entropy from the second law ( a measure of
the irreversibilities during an adiabatic process)
h
Dh
Ds
s
TdS (Gibbs) Equations
• Since, the area under the T-s line is equal to the heat transfer for a
reversible process, it would be useful to have a relationship between
Temperature and Entropy to obtain the heat transfer,T.
• Such a relationship exists for a closed system containing a pure,
compressible substance undergoing a reversible process
dU = Qrev - Wrev = TdS - PdV
=> TdS = dU + PdV or Tds = du + pdv ( per unit mass)
• This is the well-known Gibbs equation
• Alternatively, eliminate du by using the definition of enthalpy, h = u + pv
=> dh = du + pdv + vdp, => du + pdv = dh – vdp
TdS
Tds = du + pdv
Tds = dh – vdp
Comments Regarding theTdS Equations
Equations relate the entropy change of a system to other
properties, such as enthalpy (h), internal energy (u), pressure
and volume.
• Hence, even though T-ds relations were derived for a
reversible process, they are valid for any process, reversible
or irreversible.
• They are applicable for closed or open systems.
• They are valid for all pure substances, single or multi-phase.
Example – Application of Tds relations
• Consider steam undergoing a phase transition from liquid to vapor at a
constant temperature of 20°C. Determine the entropy change sfg=sg-sf
using the Gibbs equations and compare the value to that read directly
from the thermodynamic table.
du P
ds 
 dv , change from liquid to vapor
T T
1
P
s fg  s g  s f  (u g  u f )  (v g  v f )
T
T
From table A-4:
T=20°C, P = 2.338 kPa, vf = 0.001002(m3/kg), vg=57.79(m3/kg),
uf=83.9(kJ/kg), ug=2402.9(kJ/kg)
Substituting into the Tds relation:
sfg= (1/293)(2402.9-83.9) + (2.338/293)(57.79-0.001002) = 8.375(kJ/kg K)
ug - uf
vg - vf
1/T
P/T
Compares favorably with the tabulated value sfg=8.3715(kJ/kg K)
Entropy Change of an Incompressible Substance (YAC:6-8)
• For most liquids and all solids, the density does not change
appreciably as pressure changes, hence dv  0.
• Gibbs equation states that Tds = du+pdv
 Tds = du where du = CdT, for an incompressible substance and
Cp=Cv=C is a function of temperature only.
• Therefore, ds = du/T = C dT/T
Integrate to determine the entropy change during a process
2
2
dT
T2
s 2  s1   ds   C (T )
 Cavg ln( )
T
T1
1
1
where Cavg is the averaged specific heat of the substance
over the given temperature range
Note: According to the above equation, an isothermal process for a pure
incompressible substance is isentropic.
• Specific heats for some common liquids and solids can be found in
thermodynamic tables such as Table A-14 to A-18
Example – Entropy Change of Incomp. Substances
A 1-kg metal bar, initially at 1000 K, is removed from an oven and quenched by
immersing in a closed tank containing 20 kg of water, initially at 300 K.
Assume both substances are incompressible and Cwater= 4(kJ/kg K), Cmetal =
0.4(kJ/kg K). Neglect heat transfer between the tank and its surroundings.
(a) Determine the final temperature of the metal bar, (b) entropy generation
during the process.
Tm=1000 K, mm=1kg,
cm=0.4 kJ/kg K
Tw=300 K, mw=20 kg,
cw=4 kJ/kg K
Solution
(a) Energy balance from the first law:
DU  Q-W  0, no heat transfer and no work done
DU water  DU metal bar  0, both bar and water reach final temperature Tf
mwcw (T f  Tw )  mm cm (Tm  T f )  0
mw (cw / cm )Tw  mmTm (20)(10)(300)  (1)(1000)
Tf 

 303.5( K )
mw (cw / cm )  mm
(20)(10)  1
(b) No heat trans fer with t he outside Q  0, the entropy
balance of the system Ds  s(generati on)  s g
s g  Ds(water)  Ds(bar)  mw cw ln
Tf
Tw
 mm cm ln
Tf
Tm
303.5
303.5
s g  (20)( 4) ln
 (1)(0.4) ln
 0.928  0.477  0.451(kJ / K )
300
1000
The total entropy of the system increases, thus satisfy the second law
Entropy Change of Ideal Gases (YAC:6-9)
From Gibbs’ equations, the change of entropy of an ideal gas can
be expressed as:
du P
dh v
ds 
 dv 
 dP
T T
T T
For an ideal gas, u=u(T) and h=h(T), du=cv(T)dT and dh=cp(T)dT and
Pv = RT
dT
dv
dT
dP
 R , and ds  cP (T )
R
T
v
T
P
By integration, the change of the entropy is
ds  cv (T )
2
2
dT
v2
dT
P2
s2  s1   cv (T )
 R ln( ) or s2  s1   cP (T )
 R ln( )
T
v1
T
P1
1
1
we need to know the function cp (T) and cv (T) in order to complete
the integration,
Entropy Change of Ideal Gases – Special Cases
The dependence of specific heats on temperature makes the integration more
complex. However, for certain cases, one can make simplifying
assumptions.
Case 1: If specific heats are assumed constant, integration is simplified:
s2  s1  cv ln(
T2
v
)  R ln( 2 ) or
T1
v1
T2
P2
s2  s1  cP ln( )  R ln( )
T1
P1
Note: The above is strictly true for monatomic gases, e.g.He, Ar
It is fairly accurate if the temperature difference is small.
Case 2: Calculate the specific heat at an average temperature, Tavg, and assume
it to be constant. This also allows the specific heat to be taken out of the integral
Note:
This approximation is generally fairly accurate if the temperature
difference is not too large, usually good if DT < few hundred degrees.
Strictly speaking, one should look at the temp. dependence of specific heats for
the particular substance to evaluate the validity of this approximation.
Isentropic Processes for Ideal Gases
If a process is isentropic (that is adiabatic and reversible), ds = 0, s1=s2,
then it can be shown that:
T2
v1 k 1
T2
P2  k 1 / k
 ( ) , and
( )
T1
v2
T1
P1
cp
P2
v1 k
and
 ( ) , where k 
P1
v2
cv
The above are referred to as the: first, second and third, respectively, isentropic
relations for Ideal Gases (assuming constant specific heats).
They can also be written as:
Tvk-1 = constant
TP(1-k)/k = constant
Pvk = constant
First isentropic relation
Second isentropic relation
Third isentropic relation
Example
• Air is compressed from an initial state of 100 kPa and 300
K to 500 kPa and 360 K. Determine the entropy change
using constant cp=1.003 (kJ/kg K)
T2
P2
s2  s1  cP ln( )  R ln( ) if c Pis constant
T1
P1
360
500
s2  s1  1.003ln
 (0.287) ln
 0.279( kJ / kg K )
300
100
• Negative entropy due to heat loss to the surroundings