1 The Second Law of Thermodynamics Consider the following process: A rock spontaneously rises by lowering its temperature such that mC p DT = mgh so that DU = 0. mgh Since energy is conserved, this typ 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 roc together in a heat reservoir such t they undergo no temperature change so that W = Q and DU = 0. neral work of any kind can be done on a system in c a reservoir giving rise to a flow of heat without a tate of a system, W = Q. Work can be converted enti at by a suitable dissipative process.
Transcript
1
The Second Law of Thermodynamics
Consider the following process:
A rock spontaneously rises by loweringits temperature such that mCpDT = mgh
so that DU = 0.
mgh
Since energy is conserved, this typeof process is not forbidden by thefirst law, but we know this neverhappens! There is a natural flow ofthings or direction for spontaneousprocesses to occur.
We also know that various forms ofwork can be completely convertedinto heat, e.g., rubbing of two rockstogether in a heat reservoir such thatthey undergo no temperature change
so that W = Q and DU = 0.
In general work of any kind can be done on a system in contactwith a reservoir giving rise to a flow of heat without alteringthe state of a system, W = Q. Work can be converted entirely into heat by a suitable dissipative process.
2
Heat Engines - The conversion of heat into work
In order to convert heat into work we require a machine that willconsume heat and produce work. The machine itself must not suffer anypermanent change; it must play a passive role in that following the processit must return to its initial state. The machine must pass through a cycle.
def.
Thermal efficiency, .Q
W
in heat
out work
1
Applying the first law to the operation of the machine or engine, W = Q1- Q2
where Q2 corresponds to any heat rejected from the engine,
1
2
1
21 1Q
Q
Q
QQ
system
Q1
Q2
WHeat engine
3
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.
Carnot cycle for a gas
p
V
A
B
C
DT1
T2
Q1
Q2
W
Q1
Q2
T1
T2
W
Since all the steps are reversible DU = 0, W = Q1 - Q2 and1
21Q
Q
4
The 2nd Law
Kelvin Statement - No process is possible whose sole result is the complete conversion of heat into work. This addresses the efficiencyof conversion.
Clausius Statement - No process is possible whose sole result is the transfer of heatfrom a colder to a hotter body. Spontaneity of processes and theirreversibility 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
C HWC = QC1 - QC2 WH = QH1 - QH2
T1
T2
QH1
QH2QC2
QC1
Assume the existence of a Hypothetical engine such that,
.Q
W
Q
W or ,
C
C
H
HCH
11
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.
1 1 and therefore .C H C HW W Q Q
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C HWC = QC1 - QC2 WH = QH1 - QH2
T1
T2
QH1
QH2QC2
QC1
Proof of Carnot’s Theorem
Composite Engine
Now consider C and H asa Composite Engine.
This composite engineproduces no net work butsimply extracts heat from acold reservoir and deliversan amount of heat,
to a hotter reservoir.
011 HC QQ
This is a violation of the Clausius statementof the second law!
Carnot’s Theorem: reversibleengine any
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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 reservoirsare equally efficient. For any reversible engine,
.T,TfQ
Q21
2
1
Qn+2
Tn+2
Wn+2
Tn+1
Tn
Tn-1
Cn+2
Cn+1
Cn
Qn+1
Qn-1
Qn
Wn+1
Wn
Consider a series of Carnot Engines
121
22
nn
n
nn T,Tf
Q
Q:C
nnn
nn T,Tf
Q
Q:C 1
11
11
nnn
nn T,Tf
Q
Q:C
8
For the composite engine, Qn+2
Tn+2
Wn+2
Tn+1
Tn
Tn-1
Cn+2
Cn+1
Cn
Qn+1
Qn-1
Qn
Wn+1
Wn
121
2
nnn
n T,TfQ
Q:Composite
Then,
111212 nnnnnnnn T,TfT,TfT,TfT,Tf
This can only be true if the f ’s factorize such that
1
2
1
212
n
n
n
nnn T
T
Tf
TfT,Tf
Therefore: The ratio of the temperatures of the reservoirs is equal
to the ratios of heat exchanged by a reversible
engine operating between the same reservoirs.
1
2
1
2
n
n
n
n
T
T
Q
Q
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reversible
According to Carnot’s theorem and its corollary we can make the following statements:
1
2
1
2 11r
r
Q
Q
Q
Q
1
2
1
2
1
2
T
T
Q
Q
Q
Q
r
r
Therefore . Taking the heat entering the system as positive, we can say1
2
1
2
T
T
Q
Q
.T
Q or
T
Q result, this nggeneralizi and
T
Q
T
Q000
2
2
1
1
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 = 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,
Qrev
T
2.
2 1
1
.revQS S
T
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Entropy in Irreversible Processes
Since entropy is a state function, the change in entropy accompanying a statechange must always be the same regardless of how the state change occurs. Onlywhen the state change occurs reversibly is the entropy change related to the heat transfer by the equation
.Q
ST
A
B
R
X
x
Consider an irreversible change AB. Constructany reversible path R thus forming an irreversiblecycle ABRA. For the irreversible cycle the Clausius theorem says,
Determination of the change in entropy for an irreversible change
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Taking the integral in two parts,
i.e.,
But by definition of entropy. Thus
or
. .
0irrev rev
B A
A B
Q Q
T T
. .
.irrev rev
B B
A A
Q Q
T T
.
,rev
B
B A
A
QS S
T
.
,irrev
B
B A
A
QS S
T
.Q
dST
Thus we have this general result for a differential
irreversible change.
0dS For a thermally isolated system Q = 0 and we have thegeneral result known as the law of increase of entropy.
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Some Interesting Examples
Isothermal expansion of an Ideal Gas
0
0
dU Q W
dU TdS pdV
Combined 1st & 2nd Law
2112 /ln/ln ppRTVVRTW
0;
/ln 12
totalsystemsurr
system
SSS
VVRS
Adiabatic Free expansion of an Ideal Gas
00
12
systemtotalsurr
system
SS;S
V/VlnRS
(Joule expansion - no Q or W exchangedwith surroundings)
This is an irreversible process, but we canalways use the combined Law and integratefrom the initial to final state by a convenientreversible path.
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P1, V1
Diathermal Walls
Reservoir (T)
Isothermal dissipation of Work
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. Thereservoir adsorbs Q = W units of heat at temperature T so itsentropy change is
T/WT/QSS totalresevoir
Adiabatic dissipation of Work
P, T1
Adiabatic Walls
Electrical work is dissipated in a thermally isolatedsystem 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 changecan be calculated by
T
DQ dS Path ReversibleAny system
2
1
2
1
ln 0.
p
T psystem pT
Q C dT
C dT TdS C
T T
Composite system
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Examples
An inventor claims to have developeda power cycle capable of delivering anet work output of 410 kJ for an energyinput by heat transfer of 1000 kJ. Thesystem undergoing the cycle receivesheat transfer from hot gases at T = 500 K and discharges energy byheat transfer to the atmosphere atT = 300 K. Evaluate this claim.
The thermal efficiency is 410 kJ0.41
1000 kJcycle
H
W
Q
The maximum efficiency of any power cycle is
300 K1 1 1 0.40
500 KC C
H H
Q T
Q T 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,
T
T
d
ds
s dh vdp
du pdv
We can use these forms to determinethe entropy change of an ideal gassubjected to changes in p, v, T.
dh vd
du pd
s dp
s dvT T
T T
We already know that for an ideal gas,
and and v pdu c dT dh c dT pv RT
p
v
dT dpds c
dT dvds c
p
R
R
v
T
T
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Entropy Production
A
B
R
X
xDetermination of the change in entropy for an irreversible change
Recall the determination of entropy change for an irreversible process.
. .
0irrev rev
B A
A B
Q Q
T T
. .irrev rev
B A
A B
Q Q
T T
It is convenient to define a quantity such that
. .
.irrev rev
B B
A A
Q Q
T T
.
,irrev
B
B A
A
QS S
T
is necessarily a positive quantity called entropy production.
Def:
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Entropy Production
. .
,irrev rev
B A
A B
Q Q
T T
Rewriting the expression
we obtain,
.
.
. entropy
entropy productionchange
entropytransfer
irrev
irrev
irrev
B
A B
A
B
A B
A
B
B A
A
QS S
T
QS S
T
QS S
T
If the end states are fixed the entropy changeon the left hand side of this equation can bedetermined.
The 2 terms on the RHS of the equation are path dependant.
The 1st term on the RHS of theequation is the entropy transfer associatedwith heat transfer. The direction or sign ofthe entropy transfer is the same as heattransfer.
The 2nd term is the entropy production term.
0 irreversibilities present:
= 0 no irreversibilities present
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can not be less than zero
Entropy Production
By contrast the change in the entropyof the system can be positive, negativeor zero:
0
: = 0
< 0B AS S
The entropy balance can be expressedin various forms. If heat transfer takesplace along several locations on the boundary of the system where the temperaturesdo not vary with position or time,
jB A
j j
QS S
T
Here Qj/Tj is the amount of entropytransferred to the portion of theboundary at temperature Tj .
On a time rate basis for a closed system
j
j j
QdS
dt T
20
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 thepiston moves freely in the cylinder. If the change of state is brought about by heatingthe 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, 170 kJ/kgg f
Wp v v
m Table A-2
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Examples
Since the process is reversible and occurs at constant temperature
2257 kJ/kg
g
g f
f
g f
Q TdS mT s s
QT s s
m
This could also have been calculated our old way
g f
g f g f g f
Q Wu u
m mQ
u u p v v h hm
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Examples
The figure shows a system receiving heatQ from a reservoir. By definition the reservoiris free of irreversibilities, but the system is not,fluid friction, etc. Let’s determine the entropy change of the system and that of thereservoir.
For the system,
2 1b
QS S
T
For the reservoir
0resres
b b
Q QS
T T
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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 thepiston 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 thenet work per unit mass in kJ/kg and the amount of entropy produced per unit mass inkJ/kg-K.
As the volume of the system increases during this process, there is an energy transferby work from the system during the expansion as well as an energy transfer by workto the system done by the paddle wheel. The net work is evaluate from the changein internal energy.
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Examples
The minus sign indicates that the work input by stirring is greater in magnitude than thework done by the water as it expands.
From the 1st Law, U = -W.
On a unit mass basis we have, 2087.56 kJ/kgg f
Wu u
m
The entropy produced is evaluated by applying the entropy balance,
2
1
0Q
ST
On a unit mass basis,
6.048 kJ/kg-Kg fs sm
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Entropy Diagrams
S
T
isotherm
isentrop
isobarisochor
isenthalp
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S
T
Carnot cycle on a T – Sdiagram.
CW – power cycle
CCW – refrigeration cycle
Entropy Diagrams
Area representation of heat transferfor an internally reversible processof a closed system.
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Thermodynamic PotentialsCombined 1st and 2nd Law
dU = TdS - pdV
VSSV
SV
S
p
V
T- ,
V
U-p
S
UT
dVV
UdS
S
UdU
V,SUU
Potential Function in terms of S and p, Enthalpy
Lengendre Transform subtract a -d(pV) termfrom dU
dU + d(pV) = TdS - pdV + d(pV)
d(U + pV) = TdS + Vdp where H = (U + pV) is the Enthalpyand H = H (S, p)
Enthalpy is a function of S and p
pSSp
Sp
S
V
p
T ,
p
HV
S
HT
dpp
HdS
S
HdH
p,SHH
Potential Function in terms of T and p,Gibbs Free Energy
Lengendre Transform add a -d(TS) termto dH
d ( H-TS ) = Vdp - SdTwhere G = (H-TS) is the Gibbs Free Energyand G = G (p, T)
pTTp
Tp
T
V
p
S ,
p
GV,
T
GS
dpp
GdT
T
GdG
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Potential Function in terms of T and V,Helmholtz Free Energy
Lengendre Transform subtract a -d(TS) term from dU
d(U-TS) = -pdV - SdT = dAwhere A = A(V, T) is the Helmholtz Free Energy
VTTV
VT
T
p
V
S ,
V
Ap,
T
AS
dTT
AdV
V
AdA
Four Fundamental Thermodynamic Potentials
dU = TdS - pdV
dH = TdS + Vdp
dG = Vdp - SdT
dA = -pdV - SdT
The appropriate thermodynamic potentialto use is determined by the constraints imposed on the system. For example,since entropy is hard to control (adiabaticconditions are difficult to impose) G and Aare more useful. Also in the case of solidsp is a lot easier to control than V so G isthe most useful of all potentials for solids.
The Maxwell relations are useful in thatthe relate quantities that are difficult orimpossible to measure to quantities thatcan be measured.
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Some important bits of information
For a mechanically isolated system kept at constant temperature and volumethe A = A(V, T) never increases. Equilibrium is determined by the state ofminimum A and defined by the condition, dA = 0.
For a mechanically isolated system kept at constant temperature and pressurethe G = G(p, T) never increases. Equilibrium is determined by the state ofminimum G and defined by the condition, dG = 0.
