1 Engineering Thermodynamics and Modeling (7 cr) Fall 2007 Laboratory: Heat Engineering Lecturer: Mikko Helle Preliminary schedule: Tue Wed Thu Sep. 10 - Oct. 19 13-15 13-15 10-12 Oct. 29 - Dec. 07 All lectures are planned to be given in Room 325, Heat Engineering Course literature: R. v. Schalien, Engineering Thermodynamics and Modeling, Heat Engineering Laboratory, Åbo Akademi University, 1985. P. v. Schalien and R. v. Schalien, Teknisk Termodynamik Tillståndsstorheter, 1. ed., Åbo Akademi, 1994. Copies of lecture notes Disposition: About one half of the course consists of lectures and the other half of examples, demos and assignments. Examination: First scheduled exam on ???? 2 Contents: The aim of the course is to make the students familiar with the basics of thermodynamics, and how these can be used for solving some technical problems by mathematical modeling. The use of balances is especially central. The course also presents basic modeling principles and the practical implications of the laws of thermodynamics. Table of contents 1. Macro balances 2. Mass balances 3. Elemental balances 4. First law of thermodynamics 5. Second law of thermodynamics 6. Thermodynamic process modeling 7. Properties of state variables 8. Estimation of specific thermodynamic state variables 9. State diagrams 10. State equation on enthalpy basis 11. Thermodynamic equilibrium and the direction of processes 12. Introduction to thermodynamic modeling 13. Treatment on molar basis 14. Chemical reactions in the balance volume 15. Chemical reactions and production of entropy 16. Exergy 17. Introduction to irreversible thermodynamics 18. Conclusions
63
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1
Engineering Thermodynamics and Modeling (7 cr)
Fall 2007
Laboratory: Heat Engineering Lecturer: Mikko Helle Preliminary schedule: Tue Wed Thu Sep. 10 - Oct. 19 13-15 13-15 10-12 Oct. 29 - Dec. 07 All lectures are planned to be given in Room 325, Heat Engineering Course literature: R. v. Schalien, Engineering Thermodynamics and Modeling, Heat Engineering Laboratory, Åbo Akademi University, 1985. P. v. Schalien and R. v. Schalien, Teknisk Termodynamik Tillståndsstorheter, 1. ed., Åbo Akademi, 1994.
Copies of lecture notes
Disposition: About one half of the course consists of lectures and the
other half of examples, demos and assignments.
Examination: First scheduled exam on ????
2
Contents: The aim of the course is to make the students familiar with
the basics of thermodynamics, and how these can be used for solving
some technical problems by mathematical modeling. The use of
balances is especially central. The course also presents basic
modeling principles and the practical implications of the laws of
thermodynamics.
Table of contents
1. Macro balances 2. Mass balances 3. Elemental balances 4. First law of thermodynamics 5. Second law of thermodynamics 6. Thermodynamic process modeling 7. Properties of state variables 8. Estimation of specific thermodynamic state variables 9. State diagrams 10. State equation on enthalpy basis 11. Thermodynamic equilibrium and the direction of processes 12. Introduction to thermodynamic modeling 13. Treatment on molar basis 14. Chemical reactions in the balance volume 15. Chemical reactions and production of entropy 16. Exergy 17. Introduction to irreversible thermodynamics 18. Conclusions
3
1. MACRO BALANCES
In order to make it easier to understand, analyze and improve
(technical, biological, etc.) processes one may use mathematical
models.
"Model building":
• Choice of suitable model
• Estimation of unknown quantities (parameters) in the model
• Model verification
The procedure is, as a rule, iterative.
Important ingredients in model building are additive quantities, such
as
Mass Amount
Energy Entropy
Momentum Population
These are usually included in the model in the form of balance
equations that express the relation between input and output
quantities as well as accumulation/depletion to/from/within a balance
volume, which is constrained by a balance boundary.
The balances can be studied continuously or for a balance time.
4
Example: Return bottles in the store of a market: In the beginning of
the study (t=0) there are 43221 bottles in the store, and during the day
7562 bottles are returned, while 13452 are sent to a factory. The store
holds 37331 at the end of the day.
Bstart + Bin = Bend + Bout
The system is over-determined, so the inventory (Bend) could have
been determined from a simple “bottle balance”.
Extended treatment: Consider ”birth” and ”death” terms, if a factory
for production of new bottles from glass and worn/broken old bottles
is included within the balance boundary.
Often the produced and destroyed quantities can be merged into a net
production term according to
Bprod = produced - destroyed
The balance equation now takes the form:
Bstart + Bin + Bprod = Bend + Bout
inventory + incoming + net production = inventory + outgoing
at start to volume at end from volume
5
Furthermore, the inventory states at the end and the start can be
combined into an accumulation term ΔB, which expresses the
difference between the end and start state (negative values implying
depletion):
Bin + Bprod = ΔB + Bout
If the study is carried out over a time interval Δt and the terms are
expressed as ΔBi , we get after division by Δt
t
B
t
B
t
B
t
B
ΔΔ
+ΔΔ
=Δ
Δ+
ΔΔ outprodin
which yields (if limiting values 0
lim→Δt
are studied) a flow balance
outprodin Bdt
dBBB &&& +=+
inflow rate + net production rate = accumulation rate + outflow rate
6
2. MASS BALANCES
a. Total balance
If processes that transform mass into energy (nuclear reactions) are
excluded, one may note that a total mass balance lacks the production
term, because mass can neither be produced nor destroyed.
In the form of a flow balance, we thus have
outin mdt
dmm && +=
In steady-state, furthermore, the accumulation term vanishes, i.e.,
0=dt
dm, so the equation reduces to
outin mm && =
7
b. Mass balances in micro scale
Often the mass balance is presented for a (micro) volume element,
dV, under the name of the equation of continuity. This equation is
derived below:
Consider a volume element with cross-section area A and length dz as
in the above figure. Since 0prod =m& we may write
0=∂∂
+∂∂
dzz
m
t
m &
and also
{ 0)(
01
=∂
∂+
∂∂
=∂
∂+
∂∂
==⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
∂
∂+
∂∂
z
w
tzA
m
tAdz
zA
m
t
m
dV dV
ρρρ&&
or in three dimensions
0=∇+∂∂
wρρt
A
dz
m& dzz
mm
∂∂+ &
&
8
with
kjizyx ∂∂
+∂∂
+∂∂
=∇
where i , j and k are unit vectors in the directions of the
coordinates.
Special case: Constant density, ρ
0w =∇=∂∂
+∂
∂+
∂∂
compactlyor 0z
w
y
w
x
w zyx
where wi is the velocity component in the direction of coordinate i.
win
wout
9
c. Partial balances
If only one substance (at a time) is studied, one may apply a partial
balance
out,prod,in, ii
ii mdt
dmmm &&& +=+
Here it is possible that mass of component i is “produced”, e.g., in
chemical reactions.
The equation can also be rearranged as
)( in,out,prod, iii
i mmdt
dmm &&& −+=
Reaction Dynamics Transport kinetics
Note! If partial balances for all k substances are added, the total
balance is obtained, and this contains no production term
∑=
=k
iim
1prod, 0&
10
Example
A river (volume flow rate inV& and salt content x0) flows into a creek
(area A and mean depth z). A dam is constructed to shut the creek out
from the sea, still letting out as much water to the sea as the inflow.
Derive an expression for the time required to lower the salt content of
the creek from x1 to x2. Assume the creek to be completely mixed,
i.e., that it behaves as a CSTR. Use the equation to determine how
long it takes to reduce the salt content from 2 % to 1 % in a 3 km
long, 450 m wide and 15 m deep creek, if the river brings 10 m3/s of
fresh water into the creek. The salt content can be expressed as mass
ratio, i.e., in kg salt/kg solution.
Solution: A partial mass balance for the salt is given by
outout0in
)(xm
dt
mxdxm && +=
If the density, ρ, is independent of the salt content of the water
(which is a justified engineering approximation at a low salt content)
the equation can be divided by a constant density. Note that
VVV &&& == outin and that for complete back-mixing xx =out
)()()(
00 xxVdt
VxdxV
dt
VxdxV −=−⇒+= &&&
The rule of differentiation of a product gives
11
{)(
0
0xxVdt
dxV
dt
dVx −=−
=
− &
Separation now gives
0xx
dxdt
V
V
−−=
&
and after introduction of the integration limits
∫∫ −−=
2
100
x
xxx
dxdt
V
Vτ&
or
02
01
02
01 lnlnxx
xx
V
Az
xx
xx
V
V
−−
=−−
= &&τ
Numerically:
For V = 3000 m × 450 m × 15 m = 20250000 m3 and s/m10 3=V& :
h 6 d 16h 390 s140362301.0
02.0ln s2025000 ≈≈==τ
12
How long would it take if the creek were isolated from the sea?
MB: 2
'
0
in1 mdtmm =+ ∫τ
&
PMB: 22
'
0
0in11 xmdtxmxm =+ ∫τ
&
Combination of MB and PMB gives:
( )
in
in1
02
21in
2in10in11
2
'
0
in1
'
0
0in11
';
in
m
mm
xx
xxm
xmmxmxm
xdtmmdtxmxm
m
&
321
&&
=−−
=⇒
+=+⇒
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
+=+ ∫∫
τ
ττ
After inserting numerical values we get τ’ = 23 d 10.5 h.
Alternative and simpler solution? If the salt concentration is reduced
from 2% to 1%, the total volume should be doubled, i.e.,
. s2025000/' in == VV &τ
13
If there are no chemical reactions (i.e., no “production” of
components) the same equation of continuity as for the total mass can
be applied for each of the components, i.e.,
0=∇+∂∂
iii
twρρ
and if, furthermore, the density is constant
0w =∇=∂∂
+∂
∂+
∂∂
iziyixi
z
w
y
w
x
wor0,,,
14
d. Molar balances
If the partial mass balance is divided by the molar mass, Mi , of the
corresponding substance, we get
out,prod,in, ii
ii ndt
dnnn &&& +=+
since ni = mi / Mi. If the balance volume is V one may, as well, state
the equation in terms of concentrations, by writing
V
nnV
dt
dVcrV iiii
)( in,out, && −+=
where ri denotes the production rate per volume for component i.
sm3
mol)(prod, ==
V
nr ii
&
If V is constant we may divide the equation by the volume,
expressing all terms in the right-hand side in concentrations
)(1
in,inout,out iii
i cVcVVdt
dcr && −+=
As a rule, the conditions (and therefore also the reaction rate) vary
with the spatial coordinate, so micro balances are used in the
derivation:
15
∫+=dA
iii dA
A
n
dVdt
dcr
&1
Special case: Variation in only one spatial coordinate, z
A
dz
in,in& dzz
in
in∂
∂+ in,
in,
&&
For the control volume with dV = A dz we have
Adz
z
An
Adzdz
z
n
dVr
nndV
ii
i
ii ∂∂
=∂∂
=−)/(11
)(1 in,in,
in,out,
&&
44 344 21&&
which, after inserting the definition of molar flux , A
nN i
i
&= yields
dz
N
dt
cr ii
i
∂+
∂=
In general form (three dimensions) the equation can be written
i
i
i t
cr N∇+
∂∂
=
Since the molar flux can be divided into a convective and a diffusive
term
16
iii JNxN +=
where N is the total flux and Ji is the diffusion flux, where
wcN = and c
cx i
i =
Inserting Fick’s law of diffusion, z
cDJ i
i ∂∂
−= , yields
2
2
z
cD
z
wc
t
cr iii
i ∂∂
−∂∂
+∂∂
=
or generally
iii
i cDct
cr 2∇−∇+
∂∂
= w
This differential equation describes the micro balance of mass in a
system with chemical reactions and transport of mass.
17
Some simplified (zero- or one-dimensional) models:
1) Complete backmixing
V
cV
V
cV
dt
dcr iiii
in,inout&&
−+=
If the volume is constant, we get VVV &&& == outin
and, if the residence time is defined as VV &/=τ ,
τin,iii
i
cc
dt
dcr
−+=
2) Plug flow
z
cw
dt
dcr ii
i ∂∂
+=
3) Dispersion model (D = dispersion coefficient)
2
2
z
cD
z
cw
dt
dcr iii
i ∂∂
−∂∂
+=
Note that if all (k) partial molar balances are added, the total balance
is obtained, for which
01
prod, =∑=
k
iii Mn&
holds!
18
e. Extent of reaction and reaction rate
Consider a chemical reaction
2CO + O2 → 2CO2
with the stoichiometric coefficients
2;1;22CO2OCO =−=−= ννν
where the convention is adopted that reactants (term on the left hand
side) have negative and products (terms on the right hand side) have
positive values.
The extent of reaction, ξ , is related to the molar production term
according to
ξν iin =prod, or ξν ddn ii =prod,
By analogy, the production term can be written in continuous form
dt
dn iii
ξνξν == && prod,
where ξ& expresses the reaction rate.
The condition that the sum of the mass production terms be zero now
gives the stoichiometric equation of the reaction:
0
1
=∑=
k
i
ii Mν
19
The volume based reaction rate is, thus, given by
dV
dr
ξ=
where the rate of an individual reaction (for component i) is
rr ii ν=
Several reactions:
If q reactions occur simultaneously within the balance volume, one
has to take the summed effect of them. For the batch-wise and the
continuous cases we get
∑=
=q
jjijin
1prod, ξν and ∑
=
=q
jjijin
1prod, ξν &&
where νij is the stoichiometric coefficient of component i in reaction
j. In multi-component systems, these coefficients may be collected in
a stoichiometric matrix
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=
kqk
q
νν
νννν
1
21
11211
M
O
L
ν
For volume based reaction rates we get the reaction rate of a
component i as
∑=
=q
jjiji rr
1
ν
where rj is the reaction rate of reaction j.
20
f. Molar balance for a reacting system
From the molar balance, here written in vector form
outprodin nn
nn &&& +=+dt
d
we get, after inserting the production term, in steady state
ξνnn &&& += inout
where the vector of reaction rates is given by
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=
qξ
ξξ
&M
&
&
& 2
1
ξ
Example
Consider the system
2C + O2 → 2CO Reaction j = 1
C + O2 → CO2 Reaction j = 2
with the species i = 1 2 3 4
C O2 CO CO2
21
Stoichiometic matrix:
2
2
CO
CO
O
C
10
02
11
12
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛−−−−
=ν
The vector of the molar outflows is therefore
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
++−−−−
=⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛−−−−
+
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
⇒
+=
2in,CO
1inCO,
21in,O
21inC,
2
1
in,CO
inCO,
in,O
inC,
out,CO
outCO,
out,O
outC,
inout
2
2
2
2
2
2
2
2
10
02
11
12
ξξξξξξ
ξξ
&&
&&
&&&
&&&
&
&
&
&
&
&
&
&
&
&
&&&
n
n
n
n
n
n
n
n
n
n
n
n
ξνnn
Reaction 1 2
22
3. ELEMENTAL BALANCES
a. General
Balances can also be written for atoms.
The axiom of the chemical stoichiometry:
I. The atoms are the smallest unit elements
II. Every species holds the elements in given proportions
In chemistry, the proportions of the elements are expressed by
chemical formulae.
Consider a species j with the chemical symbol Fj. If the element i,
with the symbol A i, occurs in species j in the molar proportion aij ,
the chemical formula for the species can be expressed as a product of
symbols
∏=i
aij ijF )(A
Example
Species j = 1 SO3
Species j = 2 H2SO4
with the elements i = 1 2 3
A i = S O H
23
F1 = (S)1(O)3(H)0 so a11 = 1 ; a21 = 3 ; a31 = 0 ;
inflow rate + production rate = accumulation rate + outflow rate of entropy of entropy of entropy of entropy The production term must always be non-negative!
40
b. Elements in entropy balances
Entropy forms with accumulation:
- The entropy of the material: K
J)(== smS
Entropy production:
- Production rate of entropy: K
W)(0prod =≥S&
Entropy forms in transport (across the balance boundary):
Without mass transport:
- With heat flows: K
W)(=
T
Q&
With mass transport:
- Entropy of the substance: K
W)(== smS &&
41
Thermodynamic modeling: Heat power production
Consider a plant for heat power production:
In the furnace, a fuel (e.g., coal) is burnt, which gives an elevated
temperature (T1) that can be utilized through a heat flow ( 1Q& ) to the
power plant, where electrical power (P) is “produced”.
Balance boundary as indicated in the figure gives:
EB: PQ =1&
SB: 0prod1
1 =+ ST
Q &&
or 01
1prod <−=
T
QS
&& !!!!
The outlined power plant is not realizable!
The heat flow cannot completely be converted into mechanical or
electrical power: The system that is studied is named a second order
perpetuum mobile.
Fuel, air
T1
Furnace
P
1Q&
Flue gas/waste
42
How could the system be realized?
Entropy has to be carried out of the system!
Can, e.g., be realized by taking out a heat flow, 2Q& , at T2.
New balances:
EB: 21 QPQ && +=
SB: 2
2prod
1
1
T
QS
T
Q &&
&=+
Elimination of 2Q& from the equations gives the feasibility condition
01
1
2
1prod ≥−
−=
T
Q
T
PQS
&&& or 21 TT >
yielding the power
prod21
211 ST
T
TTQP && −⎟⎟
⎠
⎞⎜⎜⎝
⎛ −=
Conclusions:
- The whole heat flow can never be converted into power
- If the entropy production increases, the power decreases
- The ratio 1/ QP & increases with decreasing T2 and increasing T1.
T1
P
1Q&
Entropy out
43
The maximum value of the power, Pmax , is obtained for 0prod =S&
43421
&
T
T
TTQP
η
⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
1
211max
ηT is often called the thermodynamic efficiency factor.
In practice, the entropy production is considered by introducing
another efficiency factor 0≤ η ≤ 1
maxPP η= so { 1
tot
QP T&
ηηη=
P
1Q&
Steam generator ~
2Q& Flue gas, ash
Turbine
Gene- rator
Fuel
44
Thermodynamic modeling: Refrigeration
A cooling room is to be kept at T1 < T0 (=temperature of the
surroundings). Since heat continuously flows in from the
environment, a heat flow ( 1Q& ) is ”pumped” out of the room by an
apparatus A, which is driven by an electrical power P. Another heat
flow ( 2Q& ) is taken out from the apparatus.
EB: 21 QPQ && =+
SB: 2
2prod
1
1
T
QS
T
Q &&
&=+
Practical condition: T2 > T1.Therefore, P can be seen as a way to
make it possible to satisfy 0prod ≥S& . This condition can be written as
01
1
2
1prod >−
+=
T
Q
T
PQS
&&&
P
T1
Cooling room
1Q&
2Q& 1Q&
T2 A
45
while the power is given by
prod21
121 ST
T
TTQP && +⎟⎟
⎠
⎞⎜⎜⎝
⎛ −=
Conclusions:
- The apparatus lets out more heat than it takes in
- If the entropy production increases, the same will happen to the
power requirement.
- As the entropy production assumes its hypothetical limiting value
of zero, the minimum power requirement is obtained:
⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
1
12
1min T
TTQP &
- The ratio 1/ QP & grows with increasing T2 and decreasing T1.
In practice, the entropy production is considered by an efficiency
factor 0≤ η ≤ 1
ηminP
P =
A system that transports heat from lower to higher temperature is
generally called a heat pump.
46
Example of practical realization:
For a heat pump the important quantity is the released heat flow, 2Q& ,
since this is used for heating. It is given by
12
21prod
12
22 TT
TTS
TT
TPQ
−−
−= &&
The power assumes its minimum value at 0gen =S&
2
122min T
TTQP
−= &
Compressor
1Q&
2Q&
P
B
C
Valve
'2T
'1T
2T
1T
Condenser
Evaporator
47
Heat pumps may use sea or river water, the soil or the rock as heat
sources.
48
Thermodynamic modeling: Heating of a gas in a container with a piston
Consider a container with a (mass- and friction-free) piston,
containing a gas with mass m, pressure p and temperature T. A heat
flow can be introduced at temperature T0. The surrounding pressure is
p0.
Interesting special cases: A. Energy input through a heat flow Q& .
B. Energy input through compression.
Casel A:
If the volume V is constant and the heat is brought into the system
during time t, we have p ≠ constant, T ≠ constant, and W = 0 (i.e., no
mechanical work done at the boundary).
T0
p0
Q&
T , p m
49
EB: dttQQumdttQumtt
)(where)(0
2
0
1 ∫∫ ==+ &&
SB: 2
0
prod0
1
)(smSdt
T
tQsm
t
=++ ∫&
EB says that the internal energy of the substance increases as much as
the heat introduced; by recording the temperature in the container we
may experimentally determine u as a function of T at constant v.
SB imposes constraints on the way in which heat can be brought into
the system.
Casel B:
If the pressure at the balance border is constant, p0, and the piston
moves, both V and T change. A work, W, is therefore done at the
balance boundary (while friction work is neglected here).
EB: ∫∫ +=+=+2
1
0220
1 )(v
v
t
dvpmumWumdttQum &
SB: 2
0
prod0
1
)(smSdt
T
tQsm
t
=++ ∫&
If p0 = p = constant, the EB can be written
)()( 121122 hhmvpuvpumQ −=−−+=
50
Thus, the enthalpy of the substance increases as much as the
introduced heat; by recoding the temperature we may experimentally
determine h as a function of T at constant p.
SB imposes constraints on the way in which heat can be brought into
the system.
51
6. PROPERTIES OF STATE VARIABLES
Before a balanced-based thermodynamic model can be used for
simulation, numerical values of the state variables of the model have
to be inserted. Properties of and interrelation between state variables
are therefore treated in this chapter.
a. General properties
In formulating EB and SB, the quantities h, u and s were considered
as state variables: For a given substance in a given state of
aggregation, these depend solely on the internal state, e.g., on T and
p.
),( Tphh = ; ),( Tpuu = ; ),( Tpss =
Thus, the differences
12 hh − ; 12 uu − ; 12 ss −
The state variables thus have
unique values (at given
internal states) that do not
depend on the way in which
the substance has been brought
into the state in question.
p
T
“1”
“2”
×
×
52
do not depend on the state curve p = p(T) along which the state has
been brought from ”1” to ”2”.
The above reasoning is general; any two state variables (with minor
exceptions) can be chosen to describe the internal state is the
reference level is given for the state variables. One may, e.g., write
),( pshh = ; ),( vpuu = etc.
b. Relation between the state variables
The above mentioned implies that the differential of a state variable,
z, can be written as
ydyxYxdyxXzd ),(),( +=
Changes in z are obtained by integration of dz from an initial state,
(x1 , y1) to an end state, (x2 , y2), i.e.,
∫∫ +=−22
11
22
11
,
,
,
,12 ),(),(
yx
yx
yx
yx
ydyxYxdyxXzz
Generally, the integration path x = x(y) has to be given to solve the
problem, which leads to a line integral.
53
However, since the differences in the state variables are independent
of the integration path, one may first take, say, y to be constant (= y1)
and integrate x from x1 to x2 , followed by integration of y from y1 to
y2 at constant x (= x2).
The fact that the path of integration is irrelevant
),(),( 1122
,
,
22
11
yxzyxzzdyx
yx
−=∫
means that the differential of z is exact, i.e.,
dyy
zdx
x
zzd
xy
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=
Therefore
Yy
zX
x
z
xy
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
;
y
“1”
“2”
×
× 2y
1y
2x 1x
x
54
Since
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
∂=⎟⎟
⎠
⎞⎜⎜⎝
⎛∂∂
∂xy
z
yx
z 22
holds for a continuous and differentiable function, we obtain the
useful relation
yx
x
Y
y
X⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
c. The heat equation
The first and the second laws of thermodynamics can be applied to
derive relations between the state variables. Consider the container
(treated in an earlier example) with the piston and a heat flow as the
system of interest, with the balance equations:
EB: ∫∫ +=+=+2
1
0220
1 )(v
v
t
dvpmumWumdttQum &
SB: 2
0
prod0
1
)(smSdt
T
tQsm
t
=++ ∫&
If the integrals are written
[ ] [ ]∫∫ =−=tv
v T
Qdt
T
tQvvpdvp
0 00
1200
)();(
2
1
&
55
where [p0] and [T0] denote integrated mean values, the equations can
be written
[ ] )()( 12012 vvpmuumQ −+−=
[ ] [ ] prod0120 )( STssTmQ −−=
Elimination of Q gives
[ ] [ ] [ ]prod
012012012 )()( S
m
TvvpssTuu −−−−=−
If the change is assumed to happen along a reversible path, where T0
→ T and p0 → p, and the study is done on differentials, the heat
equation is obtained
dvpdsTdu −= This equation forms the starting point in the derivation of many
relations between state variables in thermodynamics.
56
d. Some relations derived from the heat equation
The definition of the specific enthalpy is
vpuh += so one may write
dpvdvpudhd ++= which, after introduction of the heat equation, yields dpvdsThd +=
Two new state variables are now introduced:
function sGibb'or enthalpy free SpecificsThg −=
function sz'or Helmholenergy internal free SpecificsTuf −=
Combination of these with the heat equation yields
dpvdTsgd +−=
dvpdTsfd −−=
Example: Derive Maxwell’s thermodynamic relations (1.57) from the
above equations!
57
7. ESTIMATION OF THERMODYNAMIC STATE VARIABLES
As was mentioned in the previous chapter, thermodynamic modeling
and simulation require access to numerical values of the state
variables that are included in the model. The literature presents
results from systematical studies of state variables, or required
auxiliary quantities, and gives numerical values in tables, diagrams or
as approximating equations. This chapter describes some ways to use
such information in the estimation of the numerical values of state
variables.
a. Specific heat capacity
As described earlier, the specific internal energy and enthalpy can be
estimated by studying the evolution of the temperature of a substance
in a vessel with a piston that is receives heat, Q.
T0
p0
Q&
T , p m
58
The special case with constant volume (i.e., constant specific volume,
v) gives
)( 12 uumQ −= ; v constant
while the case with constant pressure, p, gives
)( 12 hhmQ −= ; p constant
It is useful to report the results of the experiments in normalized
form, e.g., as the specific quantities (”per mass unit”, usually kg) and,
e.g., as the temperature derivatives of u and h at constant volume or
pressure, respectively:
pressureconstant at capacity heat SpecificKkg
kJ)(
olumeconstant vat capacity heat SpecificKkgkJ)(
==⎟⎠⎞
⎜⎝⎛∂∂
==⎟⎠⎞
⎜⎝⎛∂∂
pp
vv
cT
h
cT
u
The values are often reported in the form of tables expressing the
quantities as functions of temperature at different pressures.
59
b. State equations
In models based on mass, energy and entropy balances it is important
to know the specific volume, v (=1/ρ), since this quantity appears in
many relations between the thermodynamic state variables u, h and s.
Numerous expressions of the type
0),,( =Tvpf
have been developed. This section presents some of the simple or
most important models (mainly for gas-phase species).
A. Constant specific volume (i.e., independent of p and T)
constant=v
This simple approximation can, with some constraints, be used for
solid components and some liquids.
B. Linear temperature dependence
bTav +=
60
Best suited for (solid components and) liquids. The relation tells that
the effect of pressure is negligible. E.g., for water at 100 °C, we have
v (p = 1 bar) = 1,044 dm3/kg and v (p = 200 bar) = 1,034 dm3/kg.
C. Ideal gas law
The ideal gas law expresses the relation between pressure, volume
and temperature for a gas
TRnVp =
where the universal gas constant is given by
Kmol
J3144,8=R
If, instead, the specific volume is used, we have
M
TRTR
m
n
m
Vppv ===
The ideal gas law applies with high accuracy at low pressure or high
temperature. The discrepancy at high pressure, and especially in the
region close to the critical point, can, however, be considerable.
D. Ideal gas law with compressibility
The ideal gas law can be corrected by a compressibility factor, z,
pM
TRzv =
where z can be estimated, e.g., from the relation
61
...)()()(
132
++++=v
TD
v
TC
v
TBz
where the temperature dependent quantities B(T), C(T), D(T),... are
called virial coefficients, which can be found in tables for different
species.
Another possibility is to read z from a diagram. In such diagrams, the
leading thought is that gases at the same reduced states exhibit
similar properties. The reduced states are given by the reduced
quantities, normalized according to
c
rc
r p
pp
T
TT == ; ,
where subscript c denotes the critical point. From compressibility
charts we thus get
),( rr pTfz =
A dilemma is still that one, sometimes, e.g., is the reduced pressure is
unknown, has to carry out laborious iterations in reading the diagram.
Therefore, the reduced specific volume
vTR
Mpv
c
cr =
has been depicted in generalized compressibility charts.
As an alternative, the critical compressibility factor
62
MTR
vpz
c
ccc =
may be applied. Such charts express the relation
),,( crr vpTfz = or ),,( crr zpTfz =
graphically.
For instance, gases with zc = 0,26...0,28 accurately obey the
generalized compressibility chart given i Chemical Engineers’
Handbook by Perry et al. (1963).
E. van der Waal’s equation
The van der Waal’s equation of state is of historical interest, since it
was proposed already in 1873. It was developed to consider both
internal attraction forces between molecules and the fact that gas
molecules (at high pressure) may occupy a considerable part of the
gas volume:
M
TRbv
v
ap =−⎟
⎠⎞
⎜⎝⎛ + )(
2
In the equation, which has its background in the kinetic gas theory,
the former term is the cohesion pressure, a/v2, that reduces the
pressure of the gas (against the walls) due to internal attraction
between the gas molecules. The latter correction, i.e., the covolume,
b, reduces the volume available for thermal motion. The parameters a
63
and b can be determined from the condition that the isotherm that
intersects the critical point in a (v,p) diagram should exhibit an point
of inflection there.
Example: Derive expressions for a and b!
0;02
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=⎟⎠⎞
⎜⎝⎛∂∂
cc TT v
p
v
p
c
c
c
c
pM
TRb
pM
TRa
8;
64
272
22
==⇒
Inflection point
v
p
T = Tc
pc
T > Tc
T < Tc
64
F. The Redlich-Kwong equation
The equation presented in 1949 by Redlich and Kwong strongly
resembles the van der Waal equation, and its parameters can also be
determined by the method illustrated in the above example. The
equation has again attracted recent interest, since extensions of the it
have been found to be able to accurately describe the states of
mixtures. The equation is often written in the form
TbVV
a
bV
TRp
mmm )( ++
−=
where Vm is the molar volume defined by
mol
m)(
3
==n
VVm
G. The Beattie-Bridgeman equation
An accurate relation between pressure, temperature and molar
volume is given by the Beattie-Bridgeman equation (1982)
232)(1
m
m
mm V
ABV
TV
c
V
RTp −+⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
mm V
bBB
V
aAA 1;1 00
The equation holds accurately for densities < 0,8 ρc, where ρc = 1/vc.
65
H. The Benedict-Webb-Rubin equation
An extended version of the Beattie-Bridgeman equation was
proposed in 1940 by Benedict, Webb and Rubin. The expression,
which has eight parameters, is
)/exp(1
1
2
2236
3220
00
m
mmm
mmm
VVTV
c
V
a
V
aTRb
VT
CATRB
V
RTp
γγα−⎟⎟
⎠
⎞⎜⎜⎝
⎛++
+−
+⎟⎠⎞
⎜⎝⎛ −−+=
The equation is generally considered to apply for densities < 2,5 ρc.
Example: Nitrogen gas at temperature 175 K has a specific volume of
3,75·10-3 m3/kg. Estimate the pressure by the ideal gas law, the van
der Waal’s (W) equation, the Beattie-Bridgeman equation (BB) and
the Benedict-Webb-Rubin equation (BWR). Compare the results with
the measured value p = 10 000 kPa. The required parameters are:
W: a = 175 m6 Pa / kg2 ; b = 0,00138 m3 / kg
BB: A = 0,10229 kg m5 /(mol s)2; B = 5,378·10−5 m3 / mol
c = 42 m3 K3 / mol
BWR: a = 2,54·10−6 J m6 / mol3 ; b = 2,328·10−9 m6 / mol2