# Optics I

Document Sample

```					 Hong Kong Polytechnic University

Heat & Light (AP306)
Part I: Thermal Physics

Dr.Haitao Huang (黄海涛)
Department of Applied Physics, Hong Kong PolyU
Tel: 27665694; Office: CD613
Lecture Notes can be downloaded from http://ap.polyu.edu.hk/apahthua

Textbook:
Thermal Physics, 2nd Edition, C.B.P.Finn

Assessment Weighting:
Continuous Assessment:      Quiz (and midterm exam):        35%
Final Exam:      65% (Minimum requirement: D)
Minimum requirement for overall grade: D

Heat & Light----by Dr.H.Huang, Department of Applied Physics         1
Hong Kong Polytechnic University                                          Temperature
Some Concepts:                             Wall

System; Surroundings;                                       Piston
Gas system

Boundary / Wall
Surroundings

An equilibrium state is one in which all the bulk physical properties of the
system are uniform throughout the system and do not change with time.
State Variables / Thermodynamic Variables / Thermodynamic Coordinates
Stat Functions / State Properties
Two systems in thermal contact can finally reach thermal equilibrium.
Zeroth Law of Thermodynamics:
If each of two systems is in thermal equilibrium with a third, they are in thermal equilibrium
with one another.
A                B            C

A        B          A       C

Heat & Light----by Dr.H.Huang, Department of Applied Physics                           2
Hong Kong Polytechnic University                                   Temperature
Temperature:
A property of a system that determines whether or not that system is in thermal
equilibrium with other systems.
If two systems have the same temperature so that they are in thermal equilibrium,
this does not necessarily mean that they are in complete or thermodynamic
equilibrium. For this condition to hold, in addition to being in thermal equilibrium,
they would also have to be in mechanical equilibrium and chemical equilibrium.
Indicator Diagram:               P

T1

T2
T3

V

Equation of State:
All the bulk properties can be fixed by specifying two independent state variables.

f  P, V , T   0             PV  nRT

Heat & Light----by Dr.H.Huang, Department of Applied Physics                 3
Hong Kong Polytechnic University                                    Temperature
Temperature Scale:
It is customary to choose a fixed point at which ice, water and water vapor coexist in
equilibrium; this is known as the triple point of water. The temperature at this point is
defined as 273.16K. The gas scale temperature is determined from,
P
Tgas  273 .16
PTP
where Tgas is the temperature of an ideal gas with constant volume at pressure P.
PTP is the pressure of the constant volume gas at the triple point of water. The unit of
temperature on the ideal gas scale is K (Kelvin). The Celsius scale t is measured in
C and is related to the ideal gas scale T by,

 
t  C  T K   273 .15

Homework:        exercises 1.1-1.3

Heat & Light----by Dr.H.Huang, Department of Applied Physics                  4
Hong Kong Polytechnic University                      Reversible Process and Work
Process / End Points / Reversible Process:                         P
P2, V2

Reversible processes are quasistatic processes where
no dissipative forces such as friction are present.
P1, V1
How to recognize reversible process?

V
1    V 
Volume thermal expansivity                    
V    T  P

Linear expansion coefficient            1  L 
                          3
L  T  F

Bulk modulus
 P    1
and compressibility     K  V      
 V T 

L  F 
Young’s modulus       Y          
A  L T

Heat & Light----by Dr.H.Huang, Department of Applied Physics                    5
Hong Kong Polytechnic University               Reversible Process and Work
Example: Calculate the increase in tension F of a wire clamped between
two rigid supports, a distance L apart, when it is cooled from T1 to T2.
Known parameters: cross-section area, Young’s modulus and linear
expansion coefficient
L  F 
 F          F         F             Y          
F  F L, T           dF         dT        dL       dT               A  L T
 T  L       L T       T  L
1  L 
          
 F 
T2                    F   T   L                     L  T  F
F  F2  F1          dT                                1
T1
 T  L              T  L  L  F  F T

 F      F   L 
                   AY
 T  L   L T  T  F

T2  F 
dT   AY  dT   AY T2  T1 
T2
F          
T1
 T  L             T1

Heat & Light----by Dr.H.Huang, Department of Applied Physics                        6
Hong Kong Polytechnic University               Reversible Process and Work
Example: Work done by an ideal gas in a cylinder with a          P
1

frictionless piston. (Work is path dependent.)
Suppose that the pressure is P and the balancing force is F at
the equilibrium state,                                                   2
F  PA                                3

V
Infinitesimal small work done by the gas
dW  Fdx  PAdx  PdV
V2         V2
Total work:   W   dW   PdV
V1         V1

W   PdV  nRT  dV V  nRT ln V2 V1 
V2            V2
Along 12:
V1            V1

Along 13  2:      W=P2(V2-V1)

Heat & Light----by Dr.H.Huang, Department of Applied Physics           7
Hong Kong Polytechnic University            Reversible Process and Work
Sign convention for work:                                                             Movable bar

When the surroundings do work on the system, that
l
work is positive; conversely, when the system does
work on the surroundings, that work is negative.                                 dx
Wire frame
More examples on infinitesimal small work:
The work done to stretch a wire at a tension F though a
infinitesimal distance dx is,                              dW  Fdx

The work done to stretch a surface film (increase the      dW  ldx  dA
surface by dA) is,
In a reversible electrolytic cell, the work performed by   dW  dZ
the external charging circuit is,
For a uniformly magnetized material, the external work
required to increase the magnetic moment by dM in the      dW  B0 dM
applied induction field B0 is,
The work required to increase the overall dipole moment
dW  EdP
of a uniformly polarized dielectric material by dP in an
electric field E is,

Heat & Light----by Dr.H.Huang, Department of Applied Physics                              8
Hong Kong Polytechnic University            Reversible Process and Work
Example: Calculate the work done in changing the state of a compressible fluid from (P1,
T1) to (P2, T2) in a reversible process.
V  V ( P, T )                                                         P 
K  V     
 V T
 V         V          V
dV        dP         dT   dP  VdT
 P T       T  P       K                                               1  V 
         
V  T  P
PV
dW   PdV              dP  PVdT
K

2          P2   PV        T2
W   dW                dP   PVdT
1          P1    K        T1

In the case of an isothermal change of liquid: W 

P1
P2   PV
K
dP 
V
2K

P22  P 2
1    

Homework: exercises: 2.1-2.9

Heat & Light----by Dr.H.Huang, Department of Applied Physics                     9
Hong Kong Polytechnic University          First Law of Thermodynamics
Joule’s Experiment
Determined the mechanical equivalent of heat, that is,
it required 4.2kJ of work to raise the temperature of 1kg
of water through one degree Kelvin.

First Law: If a thermally isolated system is brought from one equilibrium
state to another, the work necessary to achieve this change is independent
of the process used.
There must exist a state function whose difference between the two end points (1
and 2) is equal to the adiabatic work. Such a function is called the internal energy
U
Wadi  U 2  U 1

Heat & Light----by Dr.H.Huang, Department of Applied Physics             10
Hong Kong Polytechnic University            First Law of Thermodynamics
If a system is not thermally isolated, the work done in taking the system between a
pair of equilibrium states depends on the path. The change of the internal energy
can be expressed as,
U  U 2  U1  W  Q
where Q is heat. This is the mathematical statement of the first law. This means
that the internal energy can be increased either by doing work on or by supplying
heat to the system. It is true for all processes whether reversible or irreversible.
In closed systems, heat is the non-mechanical exchange of energy between the
system and the surroundings because of their temperature difference.
The sign convention for Q is that it is positive when heat enters the system. For an
infinitesimal process, the first law takes the form,
dU=đW+đQ
In general, W and Q are path dependent. We call they are inexact differentials.
For a compressible fluid or gas where đW=-PdV for an infinitesimal reversible
process, the first law becomes,
đQ=dU+PdV
In differentiating between heat and work, it is very important to be clear as to what
constitutes the system.

Heat & Light----by Dr.H.Huang, Department of Applied Physics                   11
Hong Kong Polytechnic University           First Law of Thermodynamics
Microscopic View of First Law:
Energy can be added to the system in the form of heat by increasing the
randomness of the motion of the constituent molecules. We can also increase the
energy of the system by performing work. In this way we displace the molecules in
an ordered way.
As a result of quantum mechanics, each of the N particles in the system can exist in
a series of discrete energy levels. If there is a population of ni particles in the ith
energy level i, the total internal energy of the system will be     N
U   ni  i
i 1

Now U can be changed in two ways: either the energies i can be changed, with the
populations remaining the same—this is work; or the populations ni can be changed
with the energies remaining the same—this is heat.

Heat & Light----by Dr.H.Huang, Department of Applied Physics                12
Hong Kong Polytechnic University         First Law of Thermodynamics
Heat Capacity C
the limiting ratio of the heat introduced reversibly into the system divided by the
temperature rise,
 Q  đQ
C  lim     
T 0 T
    dT

C 1 đQ
Specific Heat:   c    
m m dT

Since there are a large number of possible reversible paths between the end points
with a temperature difference T, it follows that there are many possible heat
capacities.
đQ  U 
Heat Capacity at constant volume CV  V                  
dT     T V

dQP dU  PdV  H 
Heat Capacity at constant pressure C P                       
dT    dT     T  P
Enthalpy: H=U+PV

For ideal gas: U=U(T)

Heat & Light----by Dr.H.Huang, Department of Applied Physics             13
Hong Kong Polytechnic University           First Law of Thermodynamics
Free Expansion:                          partition
break
partition
gas

For an ideal gas, there will be no temperature change. Real gases decrease their
temperatures slightly when they undergo a free expansion.
 T 
Joule Coefficient:  J      
 V U

Heat & Light----by Dr.H.Huang, Department of Applied Physics         14
Hong Kong Polytechnic University            First Law of Thermodynamics
Adiabatic Process:                                      P     Adiabatic
 U    dU
CV                         dU  CV dT
 T V  dT

The first law becomes:     đQ  CV dT  PdV                               Isothermal

 V 
C P  đ P  CV  P
Q                                                                   V

dT          T  P
For ideal gas: C P  CV  nR

dT      dV
For a adiabatic process: 0  CV       nR
T       V

After integration:   TV  1  const           PV   const

    n2a 
van der Waals gas equation of state:    P  2 V  nb  nRT

    V  

Heat & Light----by Dr.H.Huang, Department of Applied Physics              15
Hong Kong Polytechnic University                First Law of Thermodynamics
Throttling Process:
We do work on the left-hand side:
0
W   Pi dV  PiVi
Vi

The gas does work on the right-hand side:
Vf
W   Pf dV  Pf V f
0
Consider the gas on both sides as a whole
system
U f  U i  0  PiVi  Pf V f

U i  PiVi  U f  Pf V f

Hi  H f
This is a isenthalpic process.

 T 
Joule-Kelvin coefficient       JK       
 P  H

Heat & Light----by Dr.H.Huang, Department of Applied Physics     16
Hong Kong Polytechnic University               First Law of Thermodynamics
Steady Flow Process:
We treat this unit mass as the system. Suppose the
shaft work done by the turbine is w, we can
summarize the following energy changes:
The internal energy changes by u2-u1.
The kinetic energy changes by ½(v22-v12).
The potential energy changes by g(z2-z1).
The net work done on the fluid is P1v1-P2v2-w.
The heat flow into the system is q.

KE  PE bulk  U  W  Q

        
1 2 v 2  v1  g z 2  z1   u 2  u1  P v1  P2 v2  w  q
2
2
1

        
w  h1  h2  1 2 v1  v 2  g  z1  z 2   q
2
2

This is the general equation for steady flow.

Heat & Light----by Dr.H.Huang, Department of Applied Physics         17
Hong Kong Polytechnic University            First Law of Thermodynamics
Steady Flow through a Nozzle:

for fast flow q=0                      
w  h1  h2  1 2 v1  v 2
2
2   
since w=0     v1  v 2  2h2  h1 
2
2
           1
 T1  P 
Assuming an ideal gas and an adiabatic process          1
T    P 
 2    2

Heat & Light----by Dr.H.Huang, Department of Applied Physics     18
Hong Kong Polytechnic University        Second Law of Thermodynamics
Carnot Cycle:

hot body

Q1

E        W

Q2

cold body

Efficiency:     W Q1
U  Q1  Q2  W  0            W  Q1  Q2

Q2
 1
Q1

Heat & Light----by Dr.H.Huang, Department of Applied Physics   19
Hong Kong Polytechnic University             Second Law of Thermodynamics
The Kelvin-Planck statement:
It is impossible to construct a device that, operating in a cycle, will produce no effect
other than the extraction of heat from a single body at a uniform temperature and the
performance of an equivalent amount of work.
In a concise form: A process whose only effect is the complete conversion of
heat into work is impossible.
Key points:
There is no net change of the working system (substance).
The hot body is only a source of heat. It will not do work on the substance.
There is only a single hot body (or heat source).

The Clausius statement:
It is impossible to construct a device that, operating in a cycle, produces no effect
other than the transfer of heat from a colder to hotter body.

Q     reservo
ir at T

Heat & Light----by Dr.H.Huang, Department of Applied Physics                   20
Hong Kong Polytechnic University                  Second Law of Thermodynamics
Schematic representation of the Kelvin-Planck (left) and Clausius (right) statements.

hot body                      hot body

Q                              Q

W=Q          R

Q

cold body                     cold body

The statements are equivalent.

hot body                                    hot body

Q1                   Q1+Q2                 Q2               Q1

W=Q1                                                        W=Q1-Q2
E                    R                     R                E

Q2                    Q2               Q2

cold body                                   cold body

Heat & Light----by Dr.H.Huang, Department of Applied Physics                         21
Hong Kong Polytechnic University       Second Law of Thermodynamics
Carnot’s Theorem:
No engine operating between two reservoirs can be more efficient than a
Carnot engine operating between those same two reservoirs.

hot body                               hot body

Q1              Q1
Q1              Q1
W                                     W
E                        W           E                R
C

Q2              Q2=Q1-W               Q2=Q1-W        Q2=Q1-W

cold body                              cold body

  C
All Carnot engine operating between the same two reservoirs have the same
efficiency.

Heat & Light----by Dr.H.Huang, Department of Applied Physics            22
Hong Kong Polytechnic University         Second Law of Thermodynamics
Temperature Scale:
The efficiency of a Carnot engine operating between the two reservoirs is
dependent only on the temperatures of the reservoirs. This gives us a meaning of
defining a temperature scale, which is independent of any particular material. The
thermodynamic temperature T can be so defined that T1 and T2 for the two
reservoirs in a Carnot engine are related as,
T1  T2    T
C            1 2                  1
Q2
T1      T1                         Q1

T1 Q1

T2 Q2

Heat & Light----by Dr.H.Huang, Department of Applied Physics           23
Hong Kong Polytechnic University      Second Law of Thermodynamics

Temperature Scale:                                            T1
Q1
For Carnot engine C12, T1 T2  Q1 Q2
C12       W12
For Carnot engine C23, T2 T3  Q2 Q3                           Q2
Composite
T2             engine C13
T1 T3  Q1 Q3                                         Q2

C23       W23
We use the symbol Tg for the gas scale temperature
Q3
and T for the thermodynamic temperature as just
T3
defined. We will show that these two temperatures
are identical.

Heat & Light----by Dr.H.Huang, Department of Applied Physics                 24
Hong Kong Polytechnic University                 Second Law of Thermodynamics

For isothermal bc:       PV  nRTg1
since         đ Q  dU  PdV  PdV
we have
 nRTg1 ln Vc Vb 
Vc               Vc   dV
Q1   PdV  nRTg1 
Vb               Vb    V
Similarly,
Q2  nRTg 2 ln Vd Va 

Q1 T1 Tg1 ln Vc Vb 
  
Q2 T2 Tg 2 ln Vd Va 

For ab and cd adiabatics,         Tg1Vc 1  Tg 2Vd 1         Tg1Vb 1  Tg 2Va 1
Vc Vb  Vd Va

T1 Tg1
                  Tg  const  T                  Tg  T
T2 Tg 2

Heat & Light----by Dr.H.Huang, Department of Applied Physics                 25
Hong Kong Polytechnic University                     Second Law of Thermodynamics
Efficiency:                     Q2    T
C  1         1 2
Q1    T1
It is interesting to note that the efficiency would be 100% were we able to obtain a
lower temperature reservoir at absolute zero. This is forbidden by the third law.
Imagine now that the Carnot engine is run backwards to act as a refrigerator. The
coefficient of performance is defined as the heat extracted divided by the work
expended,
T1
Q      Q2      T2
C  2 
R
                                                Q1
W    Q1  Q2 T1  T2
C     W=Q1-Q2
One interesting application of the Carnot refrigerator is the
so-called heat pump. The efficiency of a Carnot heat pump              Q2
is,            Q       Q          T         1                         T2
C 
HP     1
      1
      1

W         Q1  Q2       T1  T2       1  T2 T1
Since the efficiency of a Carnot heat pump rises as the
temperature difference between the reservoirs decreases,
heat pumps are best used in providing background heating.

Heat & Light----by Dr.H.Huang, Department of Applied Physics                 26
Hong Kong Polytechnic University                 Second Law of Thermodynamics
Otto-cycle:                                                 P
c
ab:     TaV1 1  TbV2 1
Adiabatic
bc:     Q1  CV Tc  Tb                                      b
Q1

cd:    Td V1 1  TcV2 1                                                         d
Q2
a
da:    Q2  CV Td  Ta                                       V2               V1       V

Q2    T  Ta
 1        1 d
Q1    Tc  Tb

Td    Ta V1 1  Tc  Tb V2 1

 1
 V2                     1
 1  
V               1
 1                    rc 1

where rc is the compression ratio (=V1/V2). It is important to have as high a
compression ratio as possible in order to get a high efficiency.

Heat & Light----by Dr.H.Huang, Department of Applied Physics                         27
Hong Kong Polytechnic University                                                                                                                           Entropy
Clausius inequality :                                                                    đQ  0
T
P
f
R2

For a reversible cycle, the equality sign holds.
R1
đ QR                     f   đ QR                    i   đQR  0                                            i
R          T        R1 i
                   T       R2

f        T
V
f   đ QR                          i       đ QR
R1

i           T

R2
f              T
for reversible process R2:

f   đ QR
        
i   đQR
R2 i           T              R2   f       T

f   đQ R                          f   đQR                                                          đ QR
                                
f

R1 i            T         R2 i                    T                           S  S f  S i     
R i         T

This state function is called entropy. For an infinitesimal reversible process,

đQR  TdS

Heat & Light----by Dr.H.Huang, Department of Applied Physics                                                                                     28
Hong Kong Polytechnic University                                        Entropy
Example:
Determine the entropy change of a beaker of water when it is heated at atmospheric
pressure between 20C and 100C.

Example:
Calculate the entropy change of an ideal gas undergoing a free expansion doubling
its volume.

Example:
A beaker of water is heated at atmospheric pressure from 20C and 100C by a
reservoir at 100C. Determine the total entropy change of water and reservoir.

Heat & Light----by Dr.H.Huang, Department of Applied Physics           29
Hong Kong Polytechnic University                                           Entropy

Suppose now the path R1 is irreversible, according to the Clausius inequality,
đQ    f đQ    i đQ
 T  i T  R f T R  0                  P
f
R2
f   đ Q  i đQR  f đQR  S  S
i        T     
R f T
 T
R i
f    i
R1
i
For an infinitesimal part of the process,     đQ  dS
T
V
The equality sign holds if the process is reversible.
For an isolated system, dS  0
or for a finite process, S f  S i  S  0
Principle of increasing entropy:
The entropy of a thermally isolated system increases in any irreversible process and
is unaltered in a reversible process.
For a system thermally isolated from the surroundings: S approaches to a maximum;
For a system totally isolated from the surroundings: S approaches to a maximum
with U remains constant.

Heat & Light----by Dr.H.Huang, Department of Applied Physics                30
Hong Kong Polytechnic University                                                    Entropy
The thermodynamic state of a system can be specified by any pair of independent
state functions. In particular, a state is equally well specified by the pair S and T as
by the pair P and V.                               T
A Carnot cycle shown in a T-S diagram.                             a              b
T1
For any reversible process                                                 Q1
adiabatic
Q
R    TdS                                                              adiabatic

T2               d              c
the net heat absorbed in a Carnot cycle is                                 Q2
given by the area shaded.
The differential form of the first law is,                                                  S
dU  đ Q  đ W
For an infinitesimal reversible process,     đ W   PdV               đQ  TdS
dU  TdS  PdV
TdS  dU  PdV
This is the central equation of thermodynamics, or thermodynamic identity.
Example: Calculate the entropy of an ideal gas.

Heat & Light----by Dr.H.Huang, Department of Applied Physics                            31
Hong Kong Polytechnic University                       Thermodynamics Revision
Ideal Gas:
Equation of state: pV  nRT               Molar specific heat:   c p  cV  R

             
Heat capacity: Q  C T f  Ti  cm T f  Ti    
Internal energy change:    U  CV T

Thermodynamics:
Work by the gas : đW=pdV
 U      S                       H      S 
Entropy: dS  đQ             CV        T                    CP        T    
T                 T V    T V                     T  P   T  P
The first law: dU=đQ-đW           dU  TdS  PdV
The second law: S  0        Principle of increasing entropy
W      Q                                  T2
Efficiency of a thermal engine:         1 2         Carnot cycle:   1 
Q1     Q1                                 T1

Heat & Light----by Dr.H.Huang, Department of Applied Physics                       32
Hong Kong Polytechnic University                  Thermodynamics Revision
Special processes:
free expansion
U=0
throttling process
H=0

Basic concepts: state function; reversible process; irreversible process;
Thermal dynamic potentials (Legendre transformations):
H=U+PV;                F=U-TS;                G=H-TS
Maxwell Relations:
 T      P         T     V         P    S           V      S 
                                                              
 V  S   S V       P  S  S  P      T V  V T         T  P   P T

dU  TdS  PdV         dH  TdS  VdP       dF  PdV  SdT         dG  VdP  SdT

Cyclic Relation

Heat & Light----by Dr.H.Huang, Department of Applied Physics                   33
Hong Kong Polytechnic University

Heat & Light----by Dr.H.Huang, Department of Applied Physics   34
Hong Kong Polytechnic University

Heat & Light----by Dr.H.Huang, Department of Applied Physics   35

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 3 posted: 9/4/2011 language: English pages: 35