Transcript Lecture_4_2nd Law an..

The Second Law of Thermodynamics
Consider the following process:
mgh
A rock spontaneously rises by lowering
its temperature such that mCpDT = mgh
so that DU = 0.
Since energy is conserved, this type
of process is not forbidden by the
first law, but we know this never
happens! There is a natural flow of
things or direction for spontaneous
processes to occur.
We also know that various forms of
work can be completely converted
into heat, e.g., rubbing of two rocks
together in a heat reservoir such that
they undergo no temperature change
so that W = Q and DU = 0.
In general work of any kind can be done on a system in contact
with a reservoir giving rise to a flow of heat without altering
the state of a system, W = Q. Work can be converted entirely in
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to heat by a suitable dissipative process.
Heat Engines - The conversion of heat into work
In order to convert heat into work we require a machine that will
consume heat and produce work. The machine itself must not suffer any
permanent change; it must play a passive role in that following the process
it must return to its initial state. The machine must pass through a cycle.
def.
Thermal efficiency,  
work out W

.
heat in
Q1
Applying the first law to the operation of the machine or engine, W = Q1Q2
where Q2 corresponds to any heat rejected from the engine,
Q  Q2
Q
Q1
 1
 1 2
Q1
Q1
Heat engine
system
Q2
W
2
Carnot Cycle
The Carnot cycle is a reversible cycle operating between two temperatures.
A  B: Isothermal expansion adsorbing heat Q1.
B  C: Adiabatic expansion decreasing T from T1 to T2.
C  D: Isothermal compression rejecting heat Q2.
D  A: Adiabatic compression increasing T from T2 to T1.
*Note that if the cycle is operated in reverse  refrigerator.
p
T1
A
Q1
W
Carnot cycle for a gas
W
B
T1
D
Q2
Q1
C
T2
V
T2
Q2
Q2
Since all the steps are reversible DU = 0, W = Q1 - Q2 and   1 
Q1
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The 2nd Law
Kelvin Statement - No process is possible whose sole result is the complete
conversion of heat into work. This addresses the efficiency
of conversion.
Clausius Statement - No process is possible whose sole result is the transfer of heat
from a colder to a hotter body. Spontaneity of processes and the
irreversibility of nature.
Kelvin Statement
Clausius Statement
Carnot’s Theorem:
No engine operating between two given reservoirs can be more efficient than a
Carnot engine operating between the same two reservoirs.
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Proof of Carnot’s Theorem
T1
QC1
WC = QC1 - QC2
QH1
C
H
WH = QH1 - QH2
QH2
QC2
Assume the existence of
a Hypothetical engine
such that,
 H  C , or
T2
WH WC

.
QH 1 QC1
Since the Carnot engine is reversible we can drive it backwards using the
mechanical energy from H. The Carnot cycle can be adjusted (adiabats) so
that in one cycle it uses exactly as much work as H produces.
WC  WH and therefore QC1  QH 1.
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Proof of Carnot’s Theorem
T1
QC1
WC = QC1 - QC2
QH1
C
H
Composite
Engine
WH = QH1 - QH2
QH2
QC2
T2
This is a violation of the Clausius statement
of the second law!
Now consider C and H as
a Composite Engine.
This composite engine
produces no net work but
simply extracts heat from a
cold reservoir and delivers
an amount of heat,
QC1  QH1  0
to a hotter reservoir.
Carnot’s Theorem: any engine  reversible
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Corollary: All reversible engines operating between the same temperature
reservoirs are equally efficient.
Thus the efficiency of any reversible engine operating between the same reservoirs
are equally efficient. For any reversible engine,
Q1
 f T1 ,T2 .
Q2
Qn+2
Cn+2
Tn+2
Wn+2
Qn+1
Tn+1
Wn+1
Cn+1
Qn
Cn
Tn
Consider a series of Carnot Engines
Q
Cn 2 : n 2  f Tn 2 ,Tn1 
Qn1
Qn1
Cn1 :
 f Tn1 ,Tn 
Qn
Qn
Cn :
 f Tn ,Tn1 
Qn1
Wn
Qn-1
Tn-1
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Qn+2
Cn+2
Tn+2
Wn+2
Qn+1
Tn+1
Wn+1
Cn+1
Qn
Tn
Cn
Wn
Qn-1
Therefore:
Tn-1
For the composite engine,
Q
Composite : n 2  f Tn 2 ,Tn1 
Qn1
Then,
f Tn2 ,Tn1   f Tn2 ,Tn1   f Tn1 ,Tn   f Tn ,Tn1 
This can only be true if the f’s factorize such that
f Tn 2 ,Tn1  
f Tn 2  Tn 2

f Tn1  Tn1
Qn 2 Tn 2

The ratio of the temperatures of the reservoirs is equal
Qn1 Tn1
to the ratios of heat exchanged by a reversible
engine operating between the same reservoirs.
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According to Carnot’s theorem and its corollary we can make the following statements:
  reversible
Q2
Qr 2
1
 1
Q1
Qr1
Q2 Qr 2 T2


Q1 Qr1 T1
Q2 T2

Therefore
. Taking the heat entering the system as positive, we can say
Q1 T1
Q1 Q2
Q
Q

 0 and generalizing this result,   0 or   0.
T1 T2
T
T
For any closed cycle,
, where the equality necessarily holds
for a reversible cycle.
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Entropy
def:
We can now define a new variable, the entropy S, by the relation
dS =
Qrev
T
for an infinitesimal reversible change. This definition
holds for reversible changes only. For a finite reversible change of
state, the change in entropy is given by,
2
S 2  S1  
1
 Qrev.
T
.
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Entropy in Irreversible Processes
Since entropy is a state function, the change in entropy accompanying a state
change must always be the same regardless of how the state change occurs. Only
when the state change occurs reversibly is the entropy change related to the
heat transfer by the equation
Q
DS  
.
T
B
X
Consider an irreversible change AB. Construct
any reversible path R thus forming an irreversible
cycle ABRA. For the irreversible cycle the
Clausius theorem says,
R
A
x
Determination of the change in
entropy for an irreversible change
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B

Taking the integral in two parts,
Airrev .
B

i.e.,
Airrev .
Q
B
But

Arev .
T
Q
T
 SB  S A ,


Arev .

Airrev .
dS 
Q
T
dS  0
.
T
T
A


Brev .
Q
T
0
.
by definition of entropy. Thus
B
or
Q
B
Q
Q
T
 SB  S A ,
Thus we have this general result for a differential
irreversible change.
For a thermally isolated system Q = 0 and we have the
general result known as the law of increase of entropy.
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Some Interesting Examples
Isothermal expansion of an Ideal Gas
dU   Q   W  0
dU  TdS  pdV  0
Adiabatic Free expansion of an Ideal Gas
(Joule expansion - no Q or W exchanged
with surroundings)
Combined 1st & 2nd Law
W  RT ln V2 / V1   RT ln  p1 / p2 
DS system  R ln V2 / V1 
DS surr  DS system ; DStotal  0
This is an irreversible process, but we can
always use the combined Law and integrate
from the initial to final state by a convenient
reversible path.
DS system  R ln V2 / V1 
DS surr  0; DStotal  DS system  0
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Isothermal dissipation of Work
P1, V1
Diathermal Walls
Electrical work is dissipated isothermally by heat flow into
a reservoir. There is no entropy change of the system because
it’s thermodynamic coordinates do not change. The
reservoir adsorbs Q = W units of heat at temperature T so its
entropy change is
DS resevoir  DStotal  Q / T  W / T
Reservoir (T)
Adiabatic dissipation of Work
Composite system
P, T1
dS system 
Adiabatic Walls
DQ
Any Reversible Path
T
Electrical work is dissipated in a thermally isolated
system maintained at fixed pressure. The T of the
system increases irreversibly. The coordinates of the
system change from P,T1 to P,T2 . The entropy change
can be calculated by
 Q  C p dT
T2
C p dT
T1
T
dS system  
 T2 
 C p ln    0.
 T1 
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Examples
An inventor claims to have developed
a power cycle capable of delivering a
net work output of 410 kJ for an energy
input by heat transfer of 1000 kJ. The
system undergoing the cycle receives
heat transfer from hot gases at
T = 500 K and discharges energy by
heat transfer to the atmosphere at
T = 300 K. Evaluate this claim.
The thermal efficiency is  
Wcycle
QH

410 kJ
 0.41
1000 kJ
The maximum efficiency of any power cycle is
  1
QC
T
300 K
 1 C  1
 0.40
QH
TH
500 K
No good!
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Different Forms of the Combined 1st and 2nd Law
Using the definition of entropy
dU  TdS  pdV
Since enthalpy is defined as,
H = U + pV
dH  TdS  Vdp
Rearranging these equations and
writing them on a unit mass basis,
Tds  du  pdv
Tds  dh  vdp
We can use these forms to determine
the entropy change of an ideal gas
subjected to changes in p, v, T.
du p
 dv
T T
dh v
ds 
 dp
T T
ds 
We already know that for an ideal gas,
du  cv dT and dh  c p dT and pv  RT
dT
dv
R
T
v
dT
dp
ds  c p
R
T
p
ds  cv
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Entropy Production
B

B
X
Airrev .
B

R
Airrev .
x
Determination of the change in
entropy for an irreversible change
Recall the determination of entropy
change for an irreversible process.

Airrev .
Q
T
T


Arev .
Q
T
B
Q
T
.
 SB  S A ,
It is convenient to define a
quantity  such that
A
B
Q
A


Brev .
Q
T
B

Airrev .
Q
T
A


Brev .
Q
T
 
Def:
 is necessarily a positive
0
quantity called entropy
production.
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Entropy Production
Rewriting the expression
Q
B

Airrev .
T
Q
A


T
Brev .
  ,
The 2 terms on the RHS of the
equation are path dependant.
we obtain,
B

Airrev .
Q
T
 S A  S B  
Q
B

S A  SB  
Airrev .
B
SB  S A 
entropy
change

Airrev .
T
Q
T
entropy
transfer
If the end states are fixed the entropy change
on the left hand side of this equation can be
determined.

The 1st term on the RHS of the
equation is the entropy transfer associated
with heat transfer. The direction or sign of
the entropy transfer is the same as heat
transfer.
The 2nd term is the entropy production term.


entropy
production
 0 irreversibilities present
= 0 no irreversibilities present
 :
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Entropy Production
 can not be less than zero
By contrast the change in the entropy
of the system can be positive, negative
or zero:
 0

S B  S A : = 0
< 0

Here Qj/Tj is the amount of entropy
transferred to the portion of the
boundary at temperature Tj .
On a time rate basis for a closed
system
Qj
dS
  
dt
j Tj
The entropy balance can be expressed
in various forms. If heat transfer takes
place along several locations on the
boundary of the system where the temperatures
do not vary with position or time,
SB  S A  
j
Qj
Tj

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Examples
Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly.
The water undergoes a process to the corresponding saturated vapor during which the
piston moves freely in the cylinder. If the change of state is brought about by heating
the water as it undergoes an internally reversible process at constant pressure and
temperature determine the work and the heat transfer per unit of mass in kJ/kg
At constant pressure the work is simply,
W
 p  vg  v f   170 kJ/kg Table A-2
m
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Examples
Since the process is reversible and occurs at constant temperature
g
Q   TdS  mT  sg  s f 
f
Q
 T  sg  s f   2257 kJ/kg
m
This could also have been calculated our old way
ug  u f 
Q W

m m
Q
  u g  u f   p  vg  v f   hg  h f
m
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Examples
The figure shows a system receiving heat
Q from a reservoir. By definition the reservoir
is free of irreversibilities, but the system is not,
fluid friction, etc. Let’s determine the
entropy change of the system and that of the
reservoir.
For the system,
S2  S1 
Q

Tb
For the reservoir
DS res 
Qres
Q
   0  
Tb
Tb
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Examples
Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly.
The water undergoes a process to the corresponding saturated vapor during which the
piston moves freely in the cylinder. There is no heat transfer with the surroundings.
If the change in state is brought about by the action of a paddle wheel, determine the
net work per unit mass in kJ/kg and the amount of entropy produced per unit mass in
kJ/kg-K.
As the volume of the system increases during this process, there is an energy transfer
by work from the system during the expansion as well as an energy transfer by work
to the system done by the paddle wheel. The net work is evaluate from the change
in internal energy.
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Examples
From the 1st Law, DU = -W.
On a unit mass basis we have,
W
   ug  u f   2087.56 kJ/kg
m
The minus sign indicates that the work input by stirring is greater in magnitude than the
work done by the water as it expands.
The entropy produced is evaluated by applying the entropy balance,
 2 Q

DS   
 0 
1 T

On a unit mass basis,

m
 sg  s f  6.048 kJ/kg-K
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Entropy Diagrams
isentrop
T
isenthalp
isochor
isobar
isotherm
S
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Entropy Diagrams
CCW – refrigeration cycle
T
CW – power cycle
S
Carnot cycle on a T – S
diagram.
Area representation of heat transfer
for an internally reversible process
of a closed system.
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Thermodynamic Potentials
Combined 1st and 2nd Law
Enthalpy is a function of S and p
H  H S , p 
dU = TdS - pdV
U  U S ,V 
 H 
 H 
 dp
dH  
 dS  
 S  p
 p  S
 U 
 U 
dU  
 dS  
 dV
 S V
 V  S
 H 
 H 
 ,
T 
V  


S

p

p

S
 U 
T 

 S V
 U 
p  -
 ,
 V  S
 T 
 V 
   

 p  S  S  p
 T 
 p 
-
  
 V  S  S V
Potential Function in terms of S and p, Enthalpy
Potential Function in terms of T and p,
Gibbs Free Energy
Lengendre Transform  subtract a -d(pV) term
from dU
Lengendre Transform  add a -d(TS) term
to dH
d ( H-TS ) = Vdp - SdT
where G = (H-TS) is the Gibbs Free Energy
and G = G (p, T)
dU + d(pV) = TdS - pdV + d(pV)
d(U + pV) = TdS + Vdp
where H = (U + pV) is the Enthalpy
and H = H (S, p)
 G 
 G 
 dp
dG  
 dT  
 T  p
 p T
 G 
 G 
 ,
S 
 ,V  
 T  p
 p T
 S 
 V 
   

 p T  T  p 27
Potential Function in terms of T and V, Four Fundamental Thermodynamic Potentials
Helmholtz Free Energy
Lengendre Transform 
subtract a -d(TS) term from dU
d(U-TS) = -pdV - SdT = dA
where A = A(V, T) is the
Helmholtz Free Energy
 A 
 A 
dA  
 dV    dT
 V T
 T V
 A 
 A   S 
 p 
S    , p  
 ,
  
 T V
 V T  V T  T V
The Maxwell relations are useful in that
the relate quantities that are difficult or
impossible to measure to quantities that
can be measured.
dU = TdS - pdV
dH = TdS + Vdp
dG = Vdp - SdT
dA = -pdV - SdT
The appropriate thermodynamic potential
to use is determined by the constraints
imposed on the system. For example,
since entropy is hard to control (adiabatic
conditions are difficult to impose) G and A
are more useful. Also in the case of solids
p is a lot easier to control than V so G is
the most useful of all potentials for solids.
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Some important bits of information
For a mechanically isolated system kept at constant temperature and volume
the A = A(V, T) never increases. Equilibrium is determined by the state of
minimum A and defined by the condition, dA = 0.
For a mechanically isolated system kept at constant temperature and pressure
the G = G(p, T) never increases. Equilibrium is determined by the state of
minimum G and defined by the condition, dG = 0.
Consider a system maintained at constant p. Then
DG   SdT
T2
T2
T1
T1
DS   S T dT   C p T d ln T
T2
S T2   S T1    C p T d ln T
T1
T2

 T2


DG  G T2   G T1    T2  T1 S T1    dT  C p T d ln T 
T

T1

 1

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