Thermodynamic Property Relations
Reading Material: Moran & Shapiro, Chapter 11, Bejan, pp 171-188.
Introduction
•
In Thermodynamics I & II, we discussed the meaning of various fluid properties and
how to use them in solving thermodynamics problems. Some properties (T, P, V, m)
can be measured directly, but others (u, h, s) must be derived. The purpose of this
chapter is to develop the tools necessary to derive the unknown quantities from the
measureable ones—i.e. thermodynamic property relations.
Mathematical Representation
•
The states of simple compressible substances are normally specified by two
independent variables, and other properties are written as a function of those two. In
other words, for three properties x, y, and z, we can write
z = f ( x, y )
•
From calculus, we know that we can write a change in the variable z as
 ∂z 
 ∂z 
dz =   dx +   dy
 ∂x  y
 ∂y  x
2.2.1
where the subscripts denote variables held constant, or
dz = M dx + N dy
2.2.2
 ∂z 
 ∂z 
where M =   , and N =   . In thermodynamics, the derivatives M and N
 ∂x  y
 ∂y  x
represent properties of the substance which can be measured or derived.
•
If we are evaluating a process where x=constant, then
dz = N dy
•
For example, if we choose to express pressure as P = P (T , v) then we have
 ∂P 
 ∂P 
dP = 
 dT +   dv
 ∂T  v
 ∂v T
and if we evaluate the change in pressure during a constant temperature process, then
we have
 ∂P 
dP =   dv
 ∂v T
EXAMPLE:
•
Internal Energy
Let z=u, x=s, and y=v. With these choices, we have
 ∂u 
 ∂u 
du =   ds +   dv
 ∂v  s
 ∂s  v
which gives changes in internal energy with changes
in entropy (heat transfer), and changes in volume
(work).
•
Note that if we compare this to the equation
du = T ds − P dv
we can observe that
 ∂u 
  =T
 ∂s  v
2.2.3
 ∂u 
  = −P
 ∂v  s
2.2.4
and
•
These relations will prove to be useful later when we
need to condense expressions involving derivatives to
known properties.
We can represent the
relationship between
properties z = f ( x, y )
graphically as well as
mathematically. If we choose
P = P(T , v) , for example,
then we have a diagram as
shown. In this figure,
the P = P(T , v) surface has
been "cut" by several constant
P, T, and v planes. The
property derivatives are
represented by the slopes of
the surface in various
directions. For example, the
curve abc represents an
isotherm, and at the point ’b’,
 ∂P 
we define   as the slope
 ∂v T
∂P
along the isotherm.
∂v
•a
b
•
Pressure
•c
Specific
Volume
Temperature
Some Mathematical Theorems
• If we write dz = Mdx + Ndy , then the following important relations can be derived (see
homework problem):
 ∂M

 ∂y

 ∂N 
 = 

 x  ∂x  y
 ∂x   ∂y   ∂z 
      = −1
 ∂y  z  ∂z  x  ∂x  y
• The two relations above were based on z = f ( x, y ) , which represents a dependence of
one variable on two others. We can also derive some useful relations among groups of
four variables, with two still being independent (e.g. P, v, T, and s). Let w, x, y, and z be
our variables, and start with the relations x = f ( w, y ) , and y = f ( w, z ) .
Writing the differentials, we have
 ∂x 
 ∂x 
dx =   dw +   dy
 ∂w  y
 ∂y  w
 ∂y 
 ∂y 
dy =   dw +   dz
 ∂w  z
 ∂z  w
Now, substitute dy from the second equation into the first,
 ∂x 
 ∂x   ∂y  
 ∂x   ∂y  
dx =   +      dw +      dz
 ∂y  w  ∂z  w 
 ∂w  y  ∂y  w  ∂w  z 
Compare this to the the differential for x = f ( w, z ) :
 ∂x 
 ∂x 
dx =   dw +   dz
 ∂w  z
 ∂z  w
Equating the coefficients on dw yields:
 ∂x   ∂y 
 ∂x 
 ∂x 
  =   +    
 ∂w  z  ∂w  y  ∂y  w  ∂w  z
and equating the coefficients on dz gives
 ∂x   ∂y   ∂z 
      = 1
 ∂y  w  ∂z  w  ∂x  w
Now we have four general equations relating derivatives of thermodynamic properties:
 ∂M

 ∂y

 ∂N 
 = 

 x  ∂x  y
 ∂x   ∂y   ∂z 
      = −1
 ∂y  z  ∂z  x  ∂x  y
2.3.2
 ∂x   ∂y 
 ∂x 
 ∂x 
  =   +    
 ∂w  z  ∂w  y  ∂y  w  ∂w  z
2.3.3
 ∂x   ∂y   ∂z 
      = 1
 ∂y  w  ∂z  w  ∂x  w
2.3.4
EXAMPLE: Sound Speed
•
 ∂P 
 . Let’s write it in terms of
ρ
∂
 s
By definition, c = 
2
P, ρ, and u instead. Using equation 2.3.3 above and
letting x=P, w=ρ , z=s, and y=u, we have
 ∂P 
 ∂P 
 ∂P   ∂u 
  =   +    
 ∂ρ  s  ∂ρ  u  ∂u  ρ  ∂ρ  s
Also,
 ∂u 
 ∂u   ∂v 
 ∂u   − 1 
  =     =    2 
 ∂ρ  s  ∂v  s  ∂ρ  s  ∂v  s  ρ 
•
 ∂u 
 = − P , (2.2.4), we have
 ∂v  s
Using the relation 
 ∂u 
P
  = 2
 ∂ρ  s ρ
and finally,
 ∂P 
P
c 2 =   + 2
 ∂ρ  u ρ
 ∂P 
 
 ∂u  ρ
Special Case: Ideal Gas
•
2.3.1
For an ideal gas, P = ρRT .
constant, then we also have
relations, we have
If the specific heats are
du = Cv dT . With these
 ∂P 
P  ∂P 
c 2 =   + 2  
 ∂ρ  u ρ  ∂u  ρ
 ∂P 
P
=   + 2
 ∂ρ T ρ CV
= RT +
•
 ∂P 


 ∂T  ρ

RP
R
= RT 1 +
ρCV
 CV
Using R = C P − CV , and
γ ≡



CP
, we have
CV
c 2 = γRT
Maxwell’s Relations
•
Pressure, volume, temperature, and entropy could be considered the four most basic
thermodynamic state parameters, in the sense that work interactions are related to
pressure and volume (δwrev = Pdv ) , and heat interactions are related to temperature
and entropy (δq rev = Tds ) . Also, all four can be fixed experimentally. Using a set of
isobars, isochors, isotherms and adiabats on a Pv diagram, James Clerk Maxwell
(1831-1879) derived a set of equations relating derivatives of these four properties.
Among other things, these useful relations can be used to relate changes in entropy to
measurable properties.
•
Maxwell’s relations can be derived using calculus and the four energy functions
internal energy (u ) , enthalpy (h ≡ u + Pv) , Helmholtz free energy ( f ≡ u − Ts ) , and
Gibbs free energy ( g ≡ h − Ts ) , as follows:
An energy balance for a simple, compressible substance can be written
du = Tds − Pdv
Combining this equation with the definitions of h, f, and g, yields:
dh = Tds + vdP
df = − Pdv − sdT
dg = vdP − sdT
Now we have four equations in the form of dz = Mdx + Ndy , and we can use 2.3.1 to
relate derivatives of M and N:
 ∂T 
 ∂P 

 = − 
 ∂v  s
 ∂s  v
MR #1
 ∂T 
 ∂v 

 =  
 ∂P  s
 ∂s  P
MR #2
 ∂P 
 ∂s 

 =  
 ∂T  v
 ∂v T
MR #3
 ∂v 
 ∂s 

 = − 
 ∂T  P
 ∂P T
MR #4
Keypoints:
•
P, T, and v can be measured, and s can be derived from the Maxwell Relations.
•
The LHS of #3 and #4 come directly from measurements, and could also be
calculated given an equation of state.
EXAMPLE:
•
Evaluation of Entropy
Let’s take a closer look at how the Maxwell Relations
help in the evaluation of entropy. In other words,
let’s express changes in entropy in terms of the
measurable properties P, v, and T. (We will also need
CP and CV.) We have three possible pairs of
independent variables to choose from: (P,v), (P,T),
and (v,T). Each pair has two partial derivatives of s
associated with it, for a total of six:
 ∂s 
 ∂s 
 ∂s 
 ∂s 
 ∂s 
 ∂s 

  ,   ,   , 
 ,   , 
 ∂P  v  ∂v  P  ∂P  T  ∂T  P  ∂v  T  ∂T  v
•
If we can express each of these derivatives in terms
of P, v, T, and other measurable properties, we will
have complete information on the entropy of a pure
substance.
•
The 3rd and 5th come directly from MR #4 and MR #3:
 ∂s 
 ∂v 
  = −

 ∂P T
 ∂T  P
 ∂s 
 ∂P 
  =

 ∂v  T  ∂T  v
•
The 6th and 4th can be evaluated using 2.2.3 and
2.2.4:
 ∂u 
 ∂u   ∂T 
 ∂T 
  =
 
 = CV 

 ∂s  v  ∂T  v  ∂s  v
 ∂s  v
 ∂u 
  =T
 ∂s  v
T
 ∂T 

 =
 ∂s  v CV
Similarly, using enthalpy,
C
 ∂s 

 = P
T
 ∂T  P
•
The two remaining derivatives can be evaluated using
the previous two relations:
CV  ∂T 
 ∂s 
 ∂s   ∂T 
 =


  =
 
T  ∂P  v
 ∂P  v  ∂T  v  ∂P  v
C  ∂T 
 ∂s 
 ∂s   ∂T 
 = P

  =
 
T  ∂v  P
 ∂v  P  ∂T  P  ∂v  P
Keypoints:
•
All partial derivatives of entropy have been
expressed in terms of P, T, v, and C P , C V .
EXAMPLE:
•
Derivation of Joule’s Law
Previously, we discussed Joule’s free-expansion
experiment, concluding for an ideal gas:
u = u (T )
 ∂u 
  =0
 ∂v T
•
Now let’s use what we have developed about
thermodynamic properties to show this is true from a
more fundamental approach. To do this, let’s express
 ∂u 
  in terms of P, v, and T. Since we know an
 ∂v T
equation of state for ideal gasses (Pv=RT), we can
also use derivatives among P, v, and T.
•
Using 2.3.3 and selecting (v,T) as independent
variables, and (u,s) as dependent, we have
 ∂u 
 ∂u   ∂u   ∂s 
  =  +   
 ∂v T  ∂v  s  ∂s  v  ∂v T
•
We can express the partials on the RHS as
 ∂u 
  =T
 ∂s  v
(2.2.3)
 ∂u 
  = −P
 ∂v  s
(2.2.4)
 ∂P 
 ∂s 

 =  
 ∂T  v
 ∂v T
•
(MR #3)
Substituting these equations into the RHS,
R
 ∂u 
 ∂P 
  = −P + T 
 = −P + T = −P + P = 0
v
 ∂v T
 ∂T  v
Measurable Derivatives
• By conducting experiments in which some property is fixed (e.g. use a constant
volume vessel, or an insulated container), and measuring the change in some other
property (P, T, or v) as heat is added or work is
dT
done, one obtains values for certain derivatives.
These properties, like Maxwell’s relations, can be
used to relate unknown properties to known
properties.
δQrev
• In the following, recall that we are taking all
System
processes to be reversible. Also note that there is a
(m,n)
summary of derivatives on page 24.
Constant Volume Heating (no work)
•
Measure δQ and dT, and define
CV =
δQ
m dT
V
Since there is no work, dU = δQ. The definition of entropy gives δQrev=T dS , and
we have
 ∂u 
 ∂s 
CV = 
 = T

 ∂T  v
 ∂T  v
Special cases:
Often, CV= CV(T) (e.g. ideal gas, incompressible substance), or CV=constant. (perfect gas)
EXAMPLE:
•
CV for an Ideal Gas
In general, for any pure substance, we can write u as
a function of two independent properties, such as
u = u (T , v)
and it follows that
 ∂u 
 ∂u 
du = 
 dT +   dv
 ∂T  v
 ∂v T
•
 ∂u 
 gives
 ∂T v
Introducing our definition CV = 
 ∂u 
du = Cv dT +   dv
 ∂v T
•
In general, we must realize that CV = f (T , v ) . However,
in the case of an ideal gas, we know from Joule’s
 ∂u 
 = 0 , so that
 ∂v T
free expansion experiment that 
du = C v dT
•
Since u is a function only of temperature for an
ideal gas, it follows that CV also must be a function
of temperature for an ideal gas:
CV = CV (T )
for an ideal gas
Constant Pressure Heating
dT
δQrev
constant pressure
reservoir (P)
system
(m,n)
sliding
piston
Similar to CV analysis, measure δQ and dT. Since a change in enthalpy at constant
pressure is equal to the heat addition, we have, in an analogous manner to the CV
analysis,
 ∂h 
 ∂s 
CP = 
 = T

 ∂T  P
 ∂T  P
•
In addition, since the fluid in now also expanding, this configuration yields the
"volumetric expansivity", or "coefficient of thermal expansion", which is important in
bouyancy-induced convection:
1  ∂v 
β≡ 

v  ∂T  P
Special cases:
1
Ideal gas
T
β=0
Incompressible substance
Adiabatic Volume Change (no heat transfer)
β=
dP
system
(m,n)
•
δWrev
The "isentropic expansion coefficient" k relates pressure and volume during an
adiabatic (and reversible) expansion, and is defined as
k≡−
•
•
dV
v  ∂P 
 
P  ∂v  s
This gives the fractional change in pressure for a fractional change in volume during
an isentropic process, and is unitless.
Note that we can use k to express the P-v path along an isentrope as follows:
Let s = s ( P, v) and write
 ∂s 
 ∂s 
ds =   dP +   dv
 ∂P  v
 ∂v  P
•
ds =0, and Maxwell’s relations can be used to convert the derivatives
 ∂v 
 ∂P 
0 = −
 dP + 
 dv
 ∂T  s
 ∂T  s
•
Rearranging,
 ∂P   ∂T 
0 = dP − 
 
 dv
 ∂T  s  ∂v  s
 ∂P 
0 = dP −   dv
 ∂v  s
0=
dP 1 v  ∂P 
−
  dv
P P v  ∂v  s
dP
dv
+k
P
v
In general, k=k(T,P), but if we consider a region where k can be considered locally
constant, then we have
0=
•
Pv k =constant
•
•
In Thermo I, we discussed a relation similar to this for isentropic processes of an ideal
C
gas, where we used k = P . Now we see that the relationship is true in general for
CV
isentropic processes if k is the isentropic expansion coefficient. Furthermore, we can
write (see homework problem):
v  ∂P  C P
k=−  
P  ∂v  T CV
which is simply CP/CV for an ideal gas, showing that the more general case reduces to
what we already know for an ideal gas.
The following sketch shows values of k for steam in the liquid, vapor, and
supercritical regions. Notice that in the ideal gas region (low pressures), k becomes a
function only of temperature.
1200
1.22
1.30
1000
1.24
800
Temperature (°C)
1.26
600
2
1.28
300
10
20
1.30
100
0
saturation line
0.1
1
10
Pressure (MPa)
Special cases:
k=
k=∞
CP
CV
Ideal gas
Incompressible substance
EXAMPLE: Sound Velocity in H2O
•
•
By definition, c
2
=
∂P
∂ρ
We can rewrite this as
50
s
100
c2 =
=
∂P
∂ρ
∂P
∂v
s
=
s

1 ∂ρ
(−1) 2
ρ ∂ρ
s

c 2 = −v 2
or
•
∂P ∂v
∂v s ∂ρ
=
∂P ∂ (1 / ρ)
∂v s ∂ρ
s


s
s
∂P
∂v
s
We can write this strictly in terms of properties as
c 2 = −v 2
∂P
∂v
s
 v ∂P
= vP −
 P ∂v
c 2 = vPk = vBs =



s
v
βs
where
k = isentropic expansion coefficient
Bs = adiabatic bulk modulus
βs = isentropic compressibility
s=0.5053 kJ/kg-K
P (kPa)
5000
•
slope ~ c2
5.628
•
1.006
1.004
•
consider H2O at 35°C:
•
estimate βs from steam table data…
First look up saturated liquid data at 35°C:
P=5.628kPa
sf=0.5053 kJ/kg-K
v (cc/g)
vf=0.001006 m3/kg
Now look up compressed liquid at 5 MPa at the same
entropy:
v=0.001004
Thus
βs = −
•
1 ∂v
v ∂P
≈−
s
1 ∆v
v ave ∆P
= 0.398 x10 −3 ( MPa) −1
s
Finally
c=
v
≈ 1588 m / s
βs
EXAMPLE: Sound Velocity in Ideal Gas
•
Previously, we showed
c 2 = −v 2
∂P
∂v
= vPk
s
where
k=−
•
v ∂P
P ∂v
=−
s
For an ideal gas,
k=
•
v ∂P C p
P ∂v T C v
Cp
Cv
≡γ
Therefore, for an ideal gas,
c 2 = vPγ
or
c 2 = γRT
Isothermal Volume Change
dP
constant
temperature
reservoir
(T)
•
δQrev
system
(m,n)
dV
δWrev
By measuring ∆P associated with ∆v, we can define the isothermal compressibility:
1  ∂v 
Κ≡−  
v  ∂P T
•
•
Like other properties, in general, Κ = Κ ( P, T )
 1  ∂v  
Notice the relationship between K and β  = 
  . Both measure slopes of the
∂
v
T

P 

v=v(P,T) surface, K measuring along an isotherm, and β along an isobar:
Pressure
Volume
Temperature
Special cases:
•
K=1/P
Ideal gas
K=0
Incompressible substance
Now let’s return to Maxwell’s Relations and express the derivatives as properties:
MR # 3:
 ∂s 
 ∂P 
 ∂v   ∂P 
  = 
 = −
   (from the cyclic relationship)
 ∂v  T
 ∂T  v
 ∂T  P  ∂v  T
 ∂s 
  =
 ∂v T
β
Κ
MR #4:
 ∂s 
 ∂v 
−  = 
 = βv
 ∂P  T
 ∂T  P
Keypoint:
• Entropy relations are now written in terms of easy to measure properties of the fluid.
EXAMPLE: Compression of Solid
•
A 1kg block of copper is compressed in a reversible
manner from 0.1 to 100 MPa while the temperature is
held constant at 15°C. Determine the work done,
entropy change, and heat transfer.
Q1-2
W1-2
Copper
at 15°C
•
Solution:
•
Over the range of pressure considered here,
volume expansivity β =
isothermal comp. K =
1 ∂v
v ∂T
= 5 x10 −5 K −1
P
− 1 ∂v
= 8.6 x10 −12 Pa −1
v ∂P T
specific volume v ≅ 1.14x10-4 m3/kg
•
Work during isothermal expansion:
2
W = ∫ Pdv
1
Rewriting isothermal compressibility,
vK dPT = −dvT
(note
Thus
2
W = − ∫ PvK dPT
1
Since K and v are nearly constant,
W =
(
− vK 2
P2 − P12
2
)
∂v
dv
=
)
∂P T dP T
W = −4.9 J / kg
•
Entropy Change
•
Start with Maxwell relation
(i.e. “on system”)
 ∂v 
 ∂s 

  = −
 ∂T  P
 ∂P T
and introduce the definition of volume expansivity to
get
 ∂s 
  = −vβ
 ∂P T
•
Since our process is at constant temperature,
ds
 ∂s 
= −vβ
  =
 ∂P T dP T
or
dsT = −vβ dPT
•
Assuming v and β are nearly constant,
s 2 − s1 = −vβ(P2 − P1 )
s 2 − s1 = −0.5694 J / kg − K
•
Heat Transfer
•
Since the process is reversible and isothermal,
q = T (s 2 − s1 )
q = −164.1 J / kg
•
(i.e. “out of system)
As we did for isentropic processes, we can also express the P-v path of an isothermal
process using thermodynamic properties. To do this, let T=T(P,v) :
 ∂T 
 ∂T 
dT = 
 dv
 dP + 
 ∂v  P
 ∂P  v
Rearranging, we have
0=
dv
dP
+ kT
v
P
where we have defined
kT ≡
as the isothermal expansion coefficient.
v  ∂T   ∂P 

 

P  ∂v  P  ∂T  v
(can you show this?)
Special Cases:
kT = 1 (Ideal gas),
•
kT = ∞
(Incompressible substance)
In general, kT=kT(P,T) , but for a small departure from a given state, kT ≈constant ,
and the equation above can be integrated to give
Pv kT = constant
•
Also, after some manipulation,
•
It can also be shown that
kT =
1
ΚP
C
k
= P
kT CV
Constant Enthalpy Expansion
Very slow
flow (T,P)
•
Porous
Plug
T+dT
P+dP
One additional experimental configuration, important in refrigeration, is flow through
a porous plug. This can be used to measure the Joule-Thompson Coefficient:
 ∂T 
µJ ≡ 

 ∂P  h
•
•
If µJ>0, then the fluid cools as it passes through the plug (i.e. ∆P<0 and ∆T<0), and it
may be used as a refrigerant. If µJ<0, then the fluid heats up as it passes through the
plug and is unsuitable for refrigeration. In general, µJ varies with pressure and
temperature, and the dividing line between the region where µJ>0 and µJ<0 is called
the “inversion line”.
Consider the figure below. The expansion process proceeds from the right to the left.
To the right of the inversion line (e.g. a → b) the temperature increases during the
expansion, and the fluid is not a candidate for a refrigerant in this region. To the left
of the inversion line (e.g. b → c), the temperature decreases during the expansion, and
the fluid can be used for refrigeration. Above the inversion line (e.g. d → e)
expansion produces an increase in temperature regardless of how low the pressure
drops.
e
•
T
d
•
constant
enthalpy line
µJ < 0
µJ > 0
c
•
b
•
a
•
inversion line
P
•
How does this look on an ordinary phase diagram? For a typical simple substance,
(compare to the data for Helium and Nitrogen shown below)
P
inversion curve
• critical point
saturation curve
•
T
This is not universally true, however. For example, Thermodynamics: Foundations
and Applications, Gyftopoulos and Beretta, 1991, gives a figure such as this: (Note
to future editor: I believe this figure is wrong and should be removed.)(Compare to
Miller relation.)
constant h lines
liquid states
P
µJ =-1/ρc
Two-phase
liquid-vapor
states
µJ=Tvfg/hfg
saturation
curve
Inversion
curve
µJ < 0
ideal gas
behavior
µJ=0
µJ > 0
T
•
Notice that the inversion line passes through the points of maximum
temperature on every h-line. Also, at low pressures (e.g. atmospheric)
the inversion line approaches a maximum temperature, above which
the fluid is entirely unusable. Some typical values of Tmax are:
•
Plots of the inversion curve for two common cryogenic refrigerants are
shown below:
Tmax
205
621
659
764
780
Gas
H2
N2
Air
O2
Ar
50
Helium-4
T (K)
(Tc = 5.2 K)
0
(Pc = 2.3 bar)
0
40
P (atm)
700
600
T (K)
µJ < 0
500
Nitrogen
400
(Tc = 126 K)
µJ > 0
300
(Pc = 33.9 bar)
200
100
0
0
100
200
300
400
P (atm)
Keypoint:
• Joule-Thompson coefficient is a property that depends on the state (i.e. P, T)
EXAMPLE: Joule-Thompson coefficient related to PvT data
µ=
∂T
∂P h
•
By definition,
•
To evaluate this, let T=T(h,P) and write
∂T
∂h
P
∂T
=
∂h
P
dT =
•
dh +
∂T
dP
∂P h
dh + µ dP
Now, from the Tds relationship for enthalpy, we have
dh = Tds + vdP
•
Let s=s(P,T), write
∂s
∂T
ds =
dT +
P
∂s
dP ,
∂P T
and substitute into the expression for dh to give
dh = T
•
∂s
∂T
 ∂s

dT + T
+ v  dP
P
 ∂P T

We can substitute this into (a) to yield
dT =
∂T
∂h
 ∂s
T
P
 ∂T
 ∂T
dT = T
 ∂h
∂s
P ∂T
 ∂s
 
dT + T
+ v  dP  + µ dP
P
 ∂P T
 

∂T
 dT +
∂h
P
P
 ∂s

+ v + µ  dP
T
 ∂P T

or, since the second coefficient must equal zero,
∂T
∂h
•


∂s
− v = µ
− T
∂P T
P

From MR #4, we can write
 ∂v 
 ∂s 

  = −
 ∂T  P
 ∂P T
•
Also, from the definition of CP,
 ∂h 
CP = 

 ∂T  P
•
Thus we have
µ=
1
CP
 ∂v
T
 ∂T
P

− v

Keypoints:
•
function of Cp, T, v, P
•
if Pv=RT, µ=0
•
Note that the expression above can be used to measure
CP (indirectly) at high P.
(a)
•
Required:1) measure µ with valve expansion process
2) measure P,v,T data
3) compute
∂v
∂T P
4) compute CP
EXAMPLE: Joule-Thompson coefficient for steam
•
Consider steam expanding from 6 MPa, 400°C, to 2 MPa
•
Steam is SHV, since Tsat(6MPa)=275°C
•
If we assume ideal gas, then µ=0, ∆T=0.
can approximate
µ=
∂T
∂P
≈
h
∆T
∆P
h
•
At P=6 MPa and T=400°C, h=3177.2 kJ/kg
•
At P=7 MPa and h=3177.2, T=407.4
•
At P=5 MPa and h=3177.2, T=392.7
•
Thus
µ=
407.4C − 392.7C
7 MPa − 5MPa
µ = 7.35
•
Otherwise, we
h =3177.2
C
MPa
Thus T drops approximately 29°C in going from 6 MPa to
2 MPa
Summary of Measurable Thermodynamic Derivatives
•
It is possible to express any arbitrary derivative as a function of (P,v,T) and these
measurable parameters:
CV
CP
β
k
Κ
kT
µJ
Specific heat at constant volume
Specific heat at constant pressure
Coefficient of thermal expansion
Isentropic expansion coefficient
Isothermal compressibility
Isothermal expansion coefficient
Joule-Thompson coefficient
Property
P = P ( v, T )
 ∂s 
 ∂u 
CV = 

 = T
 ∂T  v
 ∂T  v
 ∂h 
 ∂s 
CP = 

 = T
 ∂T  P
 ∂T  P
1  ∂v 
β= 

v  ∂T  P
v  ∂P 
k=−  
P  ∂v  s
1  ∂v 
K =−  
v  ∂P T
v  ∂P 
kT = −  
P  ∂v T
 ∂T 
µJ = 

 ∂P  h
Ideal Gas Limit
Pv = RT
Incompressible Limit
v = constant
CV (T )
C (T )
C P (T ) = CV + R
C (T )
1
T
0
CP
CV
1
P
∞
1
∞
0
−v
C
0
Bridgman Jacobian
•
In 1914, P. W. Bridgman expressed all partial derivatives of the most frequently used
properties (P, T, v, s, u, h, f, and g) in terms of the measurable P-v-T relationship and
the three easy to measure derivatives (CP, β , and K). Using the shorthand notation
(∂x ) z
 ∂x 
  =
,
 ∂y  z (∂y ) z
he summarized 336 possibilites in a 28 line table:
[P] (∂T)P = -(∂P)T = 1
(∂v)P = -(∂P)v = βv
(∂s)P = -(∂P)s = CP/T
(∂u)P = -(∂P)u = CP - βPv
(∂h)P = -(∂P)h = CP
(∂f)P = -(∂P)f = -s - βPv
(∂g)P = -(∂P)g = -s
[T]
(∂v)T = -(∂T)v = κv
(∂s)T = -(∂T)s = βv
(∂u)T = -(∂T)u = βTv - κPv
(∂h)T = -(∂T)h = -v + βTv
(∂f)T = -(∂T)f = -κPv
(∂g)T = -(∂T)g = -v
[v]
(∂s)v = -(∂v)s = β2v2 - κvCP/T
(∂u)v = -(∂v)u = Tβ2v2 - κvCP
(∂h)v = -(∂v)h = Tβ2v2 - κvCP - βv2
(∂f)v = -(∂v)f = κvs
(∂g)v = -(∂v)g = κvs - βv2
[s]
(∂u)s = -(∂s)u = β2v2P - κvCPP/T
(∂h)s = -(∂s)h = -vCP/T
(∂f)s = -(∂s)f = βvs + β2v2P - κvCPP/T
(∂g)s = -(∂s)g = βvs - vCP/T
EXAMPLE: Sound Velocity
By definition,
c2 =
∂P
∂ρ s
c2 =
( )
∂P ∂v
∂P ∂ ρ−1
=
∂v s ∂ρ s ∂v s ∂ρ
c2 =
∂P
∂ρ
(−1)ρ− 2
∂v s
∂ρ s
c 2 = −v 2
•
s
∂P
∂v s
Now, from Bridgman table,
(dP )s = − CP
T
(dv )s
•
 ∂v  2 ∂v C 
 2 2 KvC P 
P
= − β v −
 = −  ∂T  + ∂P T 
T






P
T
Thus
c 2 = −v 2
c2 =
(∂P )s
∂P
= −v 2
(∂v )s
∂v s
− v2
2
 ∂v
T  ∂v  
+


 
 ∂P T CP  ∂T  P 
or
c2 =
1
 K β2T 

 −
 v CP 
Ideal Gas Limit:
Pv = RT
β=
K=
1 ∂v
v ∂T
=
P
1
T
1
− 1 ∂v
=
v ∂P T P
C P − Cv = R
•
Thus
c2 =
CP
RT
Cv
EXAMPLES: Bridgman Table
Derivative
∂u
∂s v
Derivative evaluated by the
Bridgman table
Derivative in
terms of
properties
T
(2.2.3)
(∂u )v
(∂s )v
=
Tβ 2 v 2 − KvC P
β 2 v 2 − KvC P / T
=T
∂u
∂v s
(∂u )s
(∂v )s
-P
(2.2.4)
=
β2v 2 P − KvCP P / T
− β2v 2 + KvCP / T
= −P
∂T
=µ
∂P h
1  ∂v
T
C P  ∂T
(∂T )h
(∂P )h

− v
P

=
v − β Tv
− CP
1  ∂v
T
C P  ∂T
∂u
∂T V
(∂u )v
(∂T )v
CV

− v
P

Tβ2v 2 − KvCP
=
− Kv
= CP −
T
∂s
∂T V
(∂s )v
T
(∂T )v
CV
Tβ 2 v
K
β2v 2 − KvCP / T
=T
− Kv
= CP −
Tβ 2 v
K
Clapeyron Equation
•
Relates saturation properties
P
• Critical Point
Liquid
Vapor
T
•
Start with MR #3:
 ∂s 
 ∂P 

 =  
 ∂T v
 ∂v T
•
During phase change, T=constant
•
 ∂P 
In general, during phase change   = 0 . That is, v changes but P does not!
 ∂v T
P = P(T , v)
0
 ∂P 
 ∂P 
dP = 
 dT +   dv
 ∂T v
 ∂v T
dP
 ∂P 
= 

dT
 ∂T v
•
MR #3 thus becomes (after evaluating across phase change)
( s g − s f ) s fg
dP
 ∂s 
=
=   =
dT sat
 ∂v T (v g − v f ) v fg
•
Since Tds = dh − vdP and dP=0,
∆s =
∆h
T
and
dP
dT
•
=
sat
h fg
Tv fg
In general, we can replace ‘f’ and ‘g’ with any phase change (e.g. solid to liquid):
dP
dT
=
sat
(h ′′ − h ′)
T (v ′′ − v ′)
Keypoints: Clapeyron Equation
1
CLAPEYRON EQUATION
Important since it relates 3 measurable properties
i)
slope of Psat-Tsat line
ii)
latent heat of phase transformation (h”-h’) from phase (‘) to phase (“)
iii)
volume change during phase transformation
2 Can be used to calculate one of 3 properties from other 2
3
If phase (“) is vapor, at low pressure the equation can be simplified by assuming
i)
v” >> v’
ii)
v” = RT/P (ideal gas)
yielding
dP
dT
=
sat
hg − h′
T ( RT / P)
This can be integrated to get
 P  h g − h'  T2 − T1 


ln 2  =
P
R
T
T
 1
 2 1 
where we have assumed hg-h’ is constant between states 1 and 2.
Kinetic Theory of Gases--State Equations and Specific Heats
•
The ideal gas equation of state can be derived from the kinetic theory of gases
•
It should be noted that such a derivation has no bearing on classical thermodynamics.
•
Recall, classical thermodynamics
1) does not depend on microscopic structure of matter
2) does not predict anything about microscopic features
•
Why study kinetic theory in a course that focuses more on classical thermodynamics?
1) gives good physical understanding to concepts presented in classical
thermodynamics (i.e. ideal gas assumption, definition of CP).
2) helps one understand why real gases behave differently from ideal gases.
•
Analysis: Let’s consider a monatomic gas and make the following assumptions:
1) The gas consists of an enormous number of molecules. All of the molecules are
identical (assuming a pure gas).
2) The molecules are very small relative to the average distance between them. The
volume occupied by the molecules themselves is a tiny fraction of the volume of
the container.
3) The molecules fly about randomly in all directions—there is no preferred
direction.
4) The molecules do not interact except when they collide.
5) Collisions of the molecules with each other and with the walls of the container are
perfectly elastic.
6) The laws of macroscopic mechanics (Newton’s laws) apply to individual
molecules.
•
According to kinetic theory, the pressure exerted by an ideal gas on the walls of a
container results from the action of a great number of molecules striking and
rebounding.
•
Let’s first look at one molecule in a box, with mass m and velocity components Vx,
Vy, and Vz.
y
V
•
Vx
x
L
z
•
Consider a molecule striking the wall at x=L. When it rebounds, there is no change in
Vy and Vz, but Vx changes to –Vx.
•
The corresponding x-component of momentum changes from mVx to –mVx. Thus,
the magnitude of momentum change is 2mVx.
•
Also, the molecule travels the distance L in time L/Vx. Thus it could cross the
chamber and return in time ∆t=2L/Vx. The number of collisions per unit time
(collision frequency) is
f =
•
Vx
2L
Now, the change in x-momentum per unit time is the product of the change per
collision and the frequency of collision,
2
∆( x − momentum)
 V  mVx
= 2mVx  x  =
time
L
 2L 
•
Thus, the force on the wall is
Fx =
•
mVx2
L
(one molecule)
The force of all molecules is
Fx = ∑
mVx2
L
or, since m and L are constant,
Fx =
•
m
Vx2
∑
L
Define the average V2 of n molecules as
n
Vx2 ≡
then
∑ Vx2
i =1
n
(a)
mnVx2
Fx =
L
•
The pressure is thus
Px =
•
Fx
mnVx2
mnVx2
=
=
area L(area) (volume)
(b)
If it is assumed that the pressure is the same in all directions, and that all directions of
velocity are equally probable,
2
•
mnV y
mnVx2
mnVz2
=
=
P=
(volume) (volume) (volume)
(c)
Vx2 = V y2 = Vz2
(d)
Also, since Vx, Vy, and Vz are components of V,
V 2 = Vx2 + V y2 + Vz2
and it can be shown that
V 2 = Vx2 + V y2 + Vz2
(e)
where V 2 is the average of the squared velocity magnitudes for all molecules.
•
Combining (d) and (e),
V 2 = 3Vx2 = 3V y2 = 3Vz2
and (c) becomes
P=
mnV 2
3(vol)
P (vol ) = 13 nmV 2
•
If we assume the temperature of an ideal gas is proportional to the kinetic energy of
the molecules,
(f)
T=D
mV 2
2
(g)
Proportionality
constant
•
Combining (f) and (g) gives
P(vol ) =
2
 2n 
nT = 
 NT
3D
 3DN 
(h)
where n/N is Avogadro’s number (#molecules/mole).
•
The term (2n/3DN) is a constant, call it R , the universal gas constant.
P(vol ) = NR T
“ideal gas”
•
Following (f), we assumed that T was proportional to kinetic energy of molecules.
•
Conversely, we could have just compared (f) with the experimentally derived ideal
gas law….
P (vol ) = 13 nmV 2
…equation (f)
P (vol ) = NR T
…from experiment
or, equating RHS,
T=
•
1
3
n
mV 2 =
NR
2
3
n
mV 2
V2
m
= constant
NR
2
2
Using the physical constant k ≡ NR (Boltzmann’s constant), we could write
n
3
2
mV 2
kT =
2
Keypoint: Equation (i)
•
Average kinetic energy at a given T is the same for any gas.
•
Thus, heavier gas molecules have lower mean speeds than lighter molecules at the
same T.
•
Now let’s look at Cp and Cv under the kinetic theory.
(i)
•
Internal energy of an ideal gas is sum of molecular kinetic and potential energies
•
Since ideal gas is very dilute, molecules are far apart, and gravitational forces
between them are small.
•
If we change the volume (i.e. density), the distance between the molecules changes,
and the potential energy changes. However, since this change is extremely small for
an ideal gas, this change does not have an effect on internal energy.
•
Thus, for ideal gas, internal energy is only dependent on kinetic energy of molecules.
U = NMu =
where
•
nm V 2
2
u = specific internal energy
(J/kg)
M = molar mass
(kg/kmol)
N = # moles
(kmol)
U = internal energy
(J)
The molar internal energy u = Mu
u = Mu =
n 3
kT = 32 R T
N 2
where we have used (g) from above, and finally,
Cv =
∂u
∂T
v
 ∂Mu 
=
 = 32 R
 ∂T  v
•
Now, what about a diatomic gas?
•
Monatomic gas has 3 degrees of freedom (i.e. one must specify 3 independent
quantities to determine the energy, in this case Vx, Vy, and Vz)
•
Equipartition of Energy Principle:
“Total energy of molecules is divided equally among all degrees of freedom.”
•
Thus, monatomic gas has 3 degrees of freedom, each with 1 R T of energy, for a total
2
energy of
3
2
•
RT
Each additional degree of freedom must contribute 1 R T to total energy. Thus, the
2
molar internal energy is
u = Mu =
f
RT
2
for f degrees of freedom.
•
Also,
f
 ∂ ( Mu ) 
Cv = 
 = R
 ∂T  v 2
k=
 f + 2
C p = Cv + R = 
R
 2 
•
Cp
Cv
=
f +2
f
y
x
Diatomic gases have
z
3 degrees of freedom in translation (rotation about z is not a degree of freedom)
2 degrees of freedom in rotation
2 degrees of freedom in vibration along axis connecting them (z axis)
Note: vibration usually only occurs at high T
7 total
5 total
3 total
(at high T)
(at intermediate T)
(at low T)
At low temperatures:
Cv = 3 R
Cp = 5 R
k=5
At intermediate temperatures:
Cv = 5 R
Cp = 7 R
k=7
At high temperatures:
Cv = 7 R
Cp = 9 R
k=9
2
2
2
2
2
2
3
5
7
4
Cv
R
Vibration
3
2
Rotation
Cv of monatomic gas
1
Translation
0
T
Experimental Values at 1atm, 0° C
Gas
Cv
Cp
Ar
He
Air
CO
N2
O2
12.5
12.7
20.8
20.8
20.8
21.0
20.8
21.0
29.1
29.1
29.1
29.3
From Kinetic Theory (no
vibration)
Cv
Cp
12.5
12.5
20.8
20.8
20.8
20.8
20.8
20.8
29.1
29.1
29.1
29.1
Specific Heat Relations
•
Let’s look more closely at specific heats, considering first some mathematical
relations, and also some data for specific substances.
•
Although for an ideal gas specific heats are a function only of temperature, in general
CP=CP(T,P), and CV=CV(T,P). Let’s consider three important concepts:
1. Variation of CP with P at constant T
2. Variation of CV with v at constant T
3. Relation between CP and CV.
Variation of CP with P at constant T
•
Look at the differential form of the entropy equation from before.
ds =
CP
 ∂v 
dT − 
 dP
T
 ∂T  P
Compare to
dz = M dx + N dy
 ∂M
Using 
 ∂y

 ∂N 
 = 
 , we have
 x  ∂x  y
 ∂  CP
 ∂   ∂v 
= −
 
 

 ∂P  T T
 ∂T  P  ∂T  P
or
 ∂ 2v 
 ∂C P 


T
=
−


 ∂T 2 
 ∂P  T

P
Variation of CV with v at constant T
•
Another form of the entropy equation is
ds =
CV
 ∂P 
dT + 
 dv
T
 ∂T  v
Following the previous approach yields
 ∂2P 
 ∂CV 

 = T  2 
∂
v

T
 ∂T  v
Keypoints:
• Both variations can be calculated given the PvT surface.
• Both = 0 for an ideal gas (Pv=RT).
Relation between CP and CV
•
We have used two forms for the ds equation, one involving CP and the other CV.
Let’s equate them:
ds =
C
CP
 ∂v 
 ∂P 
dT + 
 dP = V dT + 
 dv
T
T
 ∂T  P
 ∂T  v
or
 ∂P 
 ∂v 
T
T


 ∂T  v
 ∂T  P
dT =
dv +
dP
C P − CV
C P − CV
Compare this to the differential we obtain from writing T=T(v,P) directly,
 ∂T 
 ∂T 
dT = 
 dv + 
 dP
 ∂v  P
 ∂P  v
Equating either of the pairs of coefficients yields
2
Tβ 2 v
 ∂v   ∂P 
=
C P − CV = −T 
  
Κ
 ∂T  P  ∂v  T
Keypoints:
2
 ∂v 
 ∂P 
1. CP ≥ CV since 
 ≥ 0 and   < 0 for all known substances.
 ∂T  P
 ∂v T
2. In the ideal gas limit, Cp-Cv =R
 ∂v 
3. For solids and liquids, 
 ≈ 0 ⇒ CP ≅ CV
 ∂T  P
 ∂v 
4. In the incompressible limit, 
 =0 and Cp=Cv exactly. This is true for water
 ∂T  P
at 4°C, where the density is maximum.
Cp Data for Ideal Gases
•
Data for some common ideal gases at low pressures (e.g. atmospheric) are shown in
the sketch below:
60
50
CO2
H2O
Recall, kinetic theory gives
CP
monatomic
diatomic
5
R
2
7
R
2
J
mol • K
= 29.1 J
mol • K
= 20.8
CV
3
R
2
5
R
2
J
mol • K
= 20.8 J
mol • K
= 12.5
Specific Heat Data for Solids
•
Specific heat data for some common solids are sketched below:
30
25
20
CV
kJ/kmol-K
Lead
15
Silver
Copper
10
Aluminum
30
CP
25
CV
20
CP and CV
kJ/kmol-K
15
Copper
10
5
0
•
0
100
Notice some trends:
200
300
400
500
600
Temperature, K
700
800
900
1000
∗ At low temperatures, Cp and Cv are nearly identical.
*
At high temperatures, Cp and Cv diverge, with Cv approaching a nearly constant
value of
CV ≅ 3R ≅ 25kJ / kmol ⋅ K
This is known as the Dulong and Petit value, after the scientists who first noticed this
trend.
*
Curves for different materials show a similar shape, suggesting that all curves
might be made to fit the same functional form (the law of corresponding states).
•
Peter Debye, using quantum statistical mechanics, developed the formula
 T
C V = 3R f 
 ΘD



where ΘD is a characteristic temperature of the substance in question, known as the
“Debye Temperature”, and f is a universal function of the dimensionless temperature
ratio T/ΘD.
1.0
Dulong and Petit
CV/3R
0.5
T3 law
0
0
1.0
2.0
T/θ
• Debye temperatures for some common substances are:
Substance
Aluminum
Cadmium
Calcium
Copper
Diamond
Graphite
Iron
Lead
Mercury
Silver
Sodium
Zinc
•
ΘD(K)
428
209
230
343
2230
420
467
105
72
225
158
327
ΘD(°R)
770
376
414
617
4014
756
841
189
130
405
284
589
The complete expression for the Debye formula is:
  T
C V (T ) = 3R 12
  ΘD




3 ΘD
T
0
∫
x3
ex −1
dx −
3Θ D / T 

e Θ D / T − 1
For high temperatures (T/ΘD>>1), this is approximated by
CV
2

1  ΘD  
≈ 3R 1 − 
 
 20  T  
For low temperatures (T/ΘD<<1), this is approximated by
CV
4π 4
≈ 3R
5
 T

 ΘD



3
This is known as the “Debye T3 Law” and is accurate to <1% for (T/ΘD)<0.1 for
isotropic nonmetals.
Keypoints: Debye formula
• ΘD is the “Debye temperature”.
• f(T/ΘD) is the same function for all substances.
• At high temperatures, f→1, and C V ≅ 3R ≅ 25kJ / kmol ⋅ K
• At low temperatures, Cv∝T3
• The Debye formula is a good approximation, but is not always correct. For
example:
H2O:
CV = 4.5R
Barium: C V = 4.7 R
•
Regarding the divergence of CP and CV, recall that the difference can be calculated as
Tβ 2 v
C P − CV =
Κ
Calculated Property Relations
•
Now that we have developed some mathematical tools and defined some measurable
properties, let’s go about calculating some non-measurable properties (u, h, and s).
Enthalpy Relations
•
Start by writing differential form of h=h(T,P).
 ∂h 
 ∂h 
dh = 
 dT +   dP
 ∂T  P
 ∂P  T
By definition, the first derivative is CP. Use the Bridgman table to express the
second:
(∂h )T − v + βTv
 ∂h 
=
  =
−1
 ∂P  T (∂P )T
Thus
dh = C P dT + [v − β Tv ]dP
or
T2
P2
T1
P1
h2 − h1 = ∫ C P dT + ∫
[v − βTv]dP
Special cases:
Ideal Gas: v-β Tv=0 , CP=CP(T)
T2
h2 − h1 = ∫ C P dT
T1
Incompressible Substance: v=constant, β=0 , CP=C(T)
T2
h2 − h1 = ∫ C P dT + v( P2 − P1 )
T1
•
Now let’s examine h2-h1 in more detail.
Along an isobar, P=constant, dP=0, and
T2
h2 − h1 = ∫ C P dT
T1
Along an isotherm, T=constant, dT=0, and
h2 − h1 = ∫
P2
P1
[v − βTv]dP
Since h is a point function (path independent), the integral ( h2 − h1 ) can be evaluated
along any convenient path. For example, consider T-s space:
P1
T
P2
x
•
1•
2
y
s
The quantity ( h2 − h1 ) can be evaluated along either 1→x→2, or 1→y→2. Along the
former path, we have
0
T2
P1
T1
P1
h2 − h1 = ∫ C P ( P1 , T )dT + ∫ ()dP
T2
P2
T2
P1
+ ∫ ()dT + ∫
[v − βT2 v]dP
0
Keypoints:
• Enthalpy can be evaluated at any state given CP(T) data at any reasonable
reference pressure, and an equation of state.
• Many numerical codes and standard handbooks give CP(T) as a polynomial fit at a
reference pressure of 1 atm.
C PD (T )
=
N
∑ R(anT n −1)
n =1
• Since CP represents a partial derivative of two properties
∂h
, it alone is
∂T P
sufficient to completely define other properties.
Energy Relations
•
We can use the same procedure as for enthalpy to express internal energy in terms of
measurable properties.
u = u (T , v)
 ∂u 
 ∂u 
du = 
 dT +   dv
 ∂T  v
 ∂v  T
The first derivative is CV by definition, the second can be transformed by the
BridgmanTable,
(∂u )T β
 ∂u 
 ∂P 
= T − P = T
  =
 −P
 ∂v T (∂v )T Κ
 ∂T  v
Thus

T2
v2   ∂P 
u 2 − u1 = ∫ CV dT + ∫ T 
 − P dv
T1
v1

  ∂T  v
Special cases:
 ∂P 
Ideal Gas: T 
 = P , CV=CV(T)
 ∂T  v
T2
u 2 − u1 = ∫ CV dT
T1
Incompressible Substance: v=constant, dv=0, CV=C(T)
T2
u 2 − u1 = ∫ C dT
T1
Entropy Relations
•
Following previous derivations for ∆h and ∆u , write
Tds = dh − vdP
This yields
T2
s 2 − s1 = ∫ C P
T1
•
P2  ∂v 
dT
−∫ 
 dP
P1  ∂T 
T
P
 ∂v 
For some equations of state, 
 is often difficult to evaluate. In these cases, it
 ∂T  P
may be easier to express s=s(T,v), i.e.
s = s (T , v)
 ∂s 
 ∂s 
ds = 
 dT +   dv
 ∂T  v
 ∂v  T
ds =
CV
 ∂P 
dT + 
 dv
T
 ∂T  v
or
T2
s 2 − s1 = ∫ CV
T1
v2  ∂P 
dT
+∫ 
 dv
v2  ∂T 
T
v
EXAMPLE: Energy, Enthalpy, and Entropy for the Van der
Waals Equation of State
The P=P(v,T) equation of state for a van der Waals gas is
P=
RT
a
− 2
v−b v
in which R, a, and b are constants.
(a) Prove that if CV is also constant, the “caloric”
equation of state u = u (T , v ) of the same gas is
 1 1
u = u o + CV (T − To ) + a − 
 vo v 
where uo = u (To , vo )
(b) Derive the expression for the entropy function
s = s (T , v) , which is valid under the same conditions.
(c) Combining this last result with dh = Tds + vdP , derive
the corresponding expression for enthalpy, h = h(T , v ) .
Solution:
(a) Energy
Writing the differential of u = u (T , v ) , we have
 ∂u 
 ∂u 
du = 
 dT +   dv
 ∂T  v
 ∂v  T
  ∂s 

du = C v dT + T   − P  dv
  ∂v T

  ∂P 

du = C v dT + T 
 − P  dv
  ∂T  v

For the P(v,T) equation of state given above,
R
 ∂P 

 =
 ∂T  v v − b
and
RT  RT
a 
 ∂P 
−
− 2
T
 −P=
v−b v−b v 
 ∂T  v
=
a
v2
Thus
du = C v dT +
a
dv
v2
Since u − uo is independent of path, we can choose any
convenient path to integrate from (To,vo) to (T,v).
If we choose two segments, one along an isochor from
(To,vo) to (T,vo), and one along an isotherm from
(T,vo) to (T,v), we obtain
T
v
a
To
vo
v2
u − u o = ∫ CV dT + ∫
dv
 1 1
u − u o = CV (T − To ) + a − 
 vo v 
(b) Entropy
The same procedure can be used for deriving the function
s(T,v):
 ∂s 
 ∂s 
ds = 
 dT +   dv
 ∂v  T
 ∂T  v
Cv
 ∂P 
dT + 
 dv
T
 ∂T  v
=
=
Cv
R
dT +
dv
v−b
T
Integrating as before from (To,vo) to (T,v) we obtain
 v−b 
T 

s − s o = CV ln  + R ln
 vo − b 
 To 
(c) Enthalpy
Starting from dh = Tds + vdP and substituting ds from
above and dP from the equation of state, we have:
dh = Tds + vdP
2a 
R
RT
  R
C
= T  V dT +
dv  + v 
dT −
dv + 3 dv 
2
v − b   v − b
(v − b)
v
T

 − RTb 2a 
v 

=  CV + R
+ 2 dv
dT + 
2
v−b

−
(
)
v
b
v 

Integrating as before from (To,vo) to (T,v) we obtain
h − ho = RTo
1 1
b(v o − v )
− 2a −
(v − b)(vo − b)
 v vo
v 

+  CV + R
(T − To )
v−b

Summary of Energy, Enthalpy, and Entropy Calculations



Closed Symple System
Ideal Gas Limit
β

du = cv dT +  T − P dv
K

du = cv dT
du = cdT
dh = c P dT
dh = cdT + vdP
dh = cP dT + (1 − βT )vdP
c
ds = P dT − βvdP
T
c
β
ds = V dT + dv
T
K
Kc
c
ds = P dv + V dP
βT
vβT
Incompressible Limit
cP
R
dT − dP
T
P
c
R
ds = V dT + dv
T
v
cP
cV
dP
ds =
dv +
v
P
ds =
ds =
c
dT
T
Generating Thermodynamic Data Tables
•
Let’s assume the following data are known for a pure substance.
#1) Vapor-Pressure data (i.e. Psat over a wide range of Tsat).
#2) PvT data in the vapor region (e.g. from measurements of P and T in a closed
fixed vessel).
#3) Density of saturated liquid ( vf(Tsat) ).
#4) Critical pressure and temperature. (Pc,Tc).
#5) CPD (T ) at low pressure (1 bar or less).
•
From these data, a complete set of thermodynamic tables for saturated liquid,
saturated vapor, and superheated vapor can be generated.
Step 1:
• Determine P=P(v,T) equation of state for vapor region from #2. (Can piece
together several equations of state if needed.)
Step 2:
• Fit saturation data to Psat=Psat(Tsat). One common fit is
ln( Psat ) = A +
B
+ C ln(Tsat ) + DTsat
Tsat
Step 3:
• Complete Psat using above equation (i.e. fill in between data points).
Step 4:
• Use step 1 to fill in between PTv data points in the vapor region.
• Use steps 1 and 2 to find vsat along vapor side of vapor dome.
T
constant pressure
8
•
•
1
7
•
6
•
•
2
•
3
5
•
constant temperature
•
4
s
Step 5:
• Pick a reference state (“1” in the figure above) and define hf=0, sf=0 at that state.
Calculate hfg from the Clapeyron equation:
h fg
 ∂P 

 =
 ∂T  sat T (v g − v f )
 ∂P 
• 
 comes from differentiating the equation from step 2.
 ∂T  sat
• vg comes from step 4
• vf is known (#3)
• T is independent
• Thus h2 can be calculated:
0
h2 = h1 + h fg = h fg
Step 6:
• Calculate s2 as follows:
• Since
s fg = h fg T ,
0
s 2 = s1 + s fg = h fg T
Step 7:
• Proceed along isotherm to P=P3 (superheated vapor)
• v3 is known from E.O.S.
• Calculate h3 and s3:
P3
h3 − h2 = ∫ [v − β Tv ] dP
(along isotherm)
P3  ∂v 
s3 − s 2 = − ∫ 
 dP
P2  ∂T 
P
(along isotherm)
P2
• Repeat for state 4, where P4 is low enough to assume ideal gas.
Step 8:
• Proceed up P=P4 (ideal gas) to elevated temperature T5.
• Calculate h5 and s5:
T5
h5 − h4 = ∫ C PD dT
T4
D
T5 C P
s5 − s 4 = ∫
T4
Step 9:
• Etc.
T
dT