Transcript Introduction to Second Law (contd.) [Lecture 4].

MEL140 (Second Law:
Continued)
Recap on Irreversible process
• A system undergoes an irreversible process, when
– second law demands that reversal of the process leaves a finite trace on the
surroundings. Connection with the Kelvin-Planck statement (board).
OR equivalently.
– Traceless reversal is prohibited by second law (Connection with Clausius
statement)
Ways to look at “irreversibilities”:
Lack of thermodynamic equilibrium between system and surroundings renders (existence
of “driving forces”) a process irreversible:
Lack of thermal equilibrium: finite temperature differences.
Lack of mechanical equilibrium: finite pressure differences (e.g. free expansion).
Chemical equilibrium: e.g. reactions/phase-transformations that complete, diffusion of
dye/ink in water.
General sign of lack of equilibrium: if the system is isolated and observed
instantaneously, processes (internal adjustments) will be found to occur.
Recap on Irreversible process
Ways to look at “irreversibilities”:
General sign of lack of equilibrium: if the system is isolated and observed
instantaneously, processes (internal adjustments) will be found to occur.
During an irreversible process, “internal currents/fluxes” that lead to
dissipation are present due to driving forces either between the
system and surroundings or between parts of a system.
Dissipation can be identified when during a process without thermal
interaction with the surroundings, the sum of the macroscopic
potential and macroscopic kinetic energy of the system decreases.
(energy goes to “microscopic modes”, remember discussion on work
energy theorem)
e.g. mechanical friction, shocks/”explosions”, plastic deformations, “Joule heating
effects” (resistors differs from capacitor/inductor), eddy currents.
Moving from one equilibrium state to another is possible in finite time only by irreversible
(fast) processes.
To show that heat transfer through a finite temperature
difference is an irreversible process
tH
tH
Q1-Q
Q1
W=Q1-Q
H
Q
W=Q1-Q
Q
tC
Violation of Kelvin Planck
statement
Example of irreversibility due to lack of equilibrium:
unrestrained expansion of a gas
A
800 kPa
B
0 kPa
A membrane separates a gas in chamber A from vacuum
in chamber B. The membrane is ruptured and the gas expands
Into chamber B until pressure equilibrium is established. The
process is so fast and the container is insulated enough such
that negligible heat transfer takes place between
the gas and the surroundings during this process.
At the end of the unrestrained expansion process, the gas (system) has
the same internal energy, as it had initially.
Some questions not yet answered
• What kind of engines and refrigerators have the
best possible performance?
• What factor(s) affect the performance of a heat
engine and a refrigerator?
• What is the best possible performance of a
heat engine and a refrigerator?
The Carnot principles
• First Carnot principle: The efficiency of an irreversible
heat engine is always less than the efficiency of a
reversible heat engine operating between the same
two reservoirs.
(Irr<rev)
• Second Carnot principle: All reversible engines
operating between the same two reservoirs have the
same efficiency.
 Rev1 =Rev2
η=
W ork delivered
H eat input from the hot reservoir

W net , out
QH
Proof of First Carnot principle
• Proof by contradiction: Assume Irr>Rev
η=
W net
QH
tH
Q
Q
Wirrev-Wrev
WRev
<Wirrirr
W
<W
Rev
WIrr=Q-QIrr
Rev+Irr
Rev
Irr
QIrr
QRev
>QIrr
Q
Rev>Q
Irr
TtLC
QRev-QIrr
tL
Conclusion: Assumption Irr>Rev is incorrect. Efficiency of a reversible engine is
higher than that of an irreversible engine.
Proof of Second Carnot principle
• Proof by contradiction: Assume Rev1>Rev2.
tH
QQ
Q
WRev1-WRev2
W
WRev2
<WRev1
Rev2<W
Rev1
WRev1=Q-QRev1
Rev1+Rev2
Rev2
Rev2
Rev1
QRev1
QRev2
>QRev1
Q
Rev2>Q
Rev1
TtLC
Conclusion so far: Assumption Rev1>Rev2 is incorrect.
QRev2-QRev1
tL
  Rev 1   Rev 2
Proof of Second Carnot principle
(continued)
• Proof by contradiction (continued): Assume Rev1<Rev2.
tH
QQ
Q
WW
<W<W
Rev1
Rev2
Rev1
Rev2
WRev2-WRev1
WRev2=Q-QRev2
Rev1+Rev2
Rev2
Rev1
Rev1
QRev1>Q
Q
Rev1
Rev1
Tt C
L
QRev2
QRev1-QRev2
tL
Final conclusion: Rev1=Rev2. All heat engines working between the
same reservoirs have the same efficiency.
An important implication of the second
Carnot principle
• The efficiency of a reversible heat engine does not depend
on its working fluid, method of execution of cycle, type of
reversible engine used, amount of heat drawn from or
rejected by the engine etc. It may however depend on a
characteristic of the reservoirs.
• By what characteristic is a reservoir specified?
Ans.: Temperature.
• The only factors that could affect the efficiency of a
reversible engine is, therefore, the temperatures (tH,tL) of
the reservoirs it is connected to.
 rev  f ( t H , t L )
Uses of the second Carnot principle
• To develop a thermodynamic temperature
scale, which is a temperature scale that does
not depend on the properties of a particular
substance.
• To calculate the maximum efficiency of a heat
engine (or maximum COP of a
refrigerator/heat pump).
Empirical and thermodynamic temperature
scales
Empirical temperature scale
Thermodynamic temparature scale
•
•
•
•
A scale that is based on the
measurement of a temperaturesensitive property of a certain
substance (e.g. pressure exerted
by a constant volume of helium
gas, thermal expansion of a
enclosed mass of
mercury/alcohol etc.).
A thermometer reads the
“empirical temperature”.
Notation for empirical
temperature: (t)
A scale that is independent of the
properties of any substance.
• Lord Kelvin was aware of the second
Carnot principle and suggested in1848, that
a thermodynamic temperature scale could
be based on the theoretical consideration
that, during the operation of a reversible
heat engine, the amounts of heat
exchanged between system and the
reservoirs depend only on the
temperature of the reservoirs and not on
the properties of any substance.
Notation for the
thermodynamic
temperature scale
to be developed : (T)
Empirical temperature scales (t)
• Empirical scales are determined through
experimentation with “thermometric substances”.
• Single fixed point scale (>1954)
• Constant volume gas thermometer and the ideal gas
temperature scale.
t ideal
gas
 273.16
p
p tp
The ideal gas temperature scale: a very accurate empirical temperature
scale developed from experiments using the constant volume gas
thermometer
• Step 1: Bring the thermometer in contact with water at triple point
(tp). Measure the pressure ptp.
• Step 2: Bring the thermometer in contact with the body at
temperature T. Measure the pressure p. Calculate
t m easured  273.16
p
p tp
• Now, redo steps 1 and 2, at each instance reducing the number of
moles (mass) of gas used in step 1 (and 2), such that pt=2n
(n=10,9,8,.,3 etc.) mm Hg.
• Perform this experiment with various gases (A,B,C)
•
C Capillary tube
L Lip
M Mercury manometer
Gas A
Gas B
Gas C
tmeasured
At low pressures, ideal gas
behavior is approached by all gases.
ptp
A temperature scale based on the second
Carnot principle
• Since
 rev ( t H , t L )
But
 rev ( t H , t L ) 
W net , out
QH
QL

QH  QL
QH
 f (t H , t L )
QH
 1
QL
QH
QH
 f (t , t
Q
Properties of
according to second law
H
L
)
L
f ( t1 , t 3 )  f ( t1 , t 2 ) f ( t 2 , t 3 )
Q1
Q2
Q2
Q3
f ( t1 , t 2 ) 
 f ( t1 , t 2 )
 f (t 2 , t3 )
Q1
Q3
Q3
f (t 2 , t3 ) 
 f ( t1 , t 3 )
 QH 
T (t H )

 
T (t L )
 Q L  rev
T ( t1 )
T (t 2 )
T (t 2 )
T (t3 )
Constructing a single-fixed-point
thermodynamic temperature scale
• Place one of the reservoirs
 Q1 
T ( t 2 )  T ( t1 ) 

 Q 2  rev
in thermal equilibrium with
water at triple point (tp);
prescribe the value 273.16 to
the constant T ( t1 )
Procedure for calculating
thermodynamic temperature
for t>ttp
for t<ttp
The reversible engine is
operated with a fixed Qtp ; then
Q(t) is measured to obtain T(t)
using:
 Q (t ) 
T ( t )  273.16 

 Q 
 t p  r ev
T is known as the Kelvin/absolute temperature scale
(in either case)
The Kelvin scale is not the only
thermodynamic temperature scale.
• The Kelvin scale calculates the thermodynamic
temperature using :
 Q (t ) 
T ( t )  273.16 

 Q 
 t p  r ev
( Q ( t )  T ( t )   T '( t )  )
Alternatively any monotonic function T’(t) of T(t)
can also be chosen to define a new thermodynamic temperature scale.
 T (t ) 
2
e . g . T ( t ) , T ( t )  273.15, ln 

 27 3.16 