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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 ;
F2 = (S)1(O)4(H)2 so a12 = 1 ; a22 = 4 ; a32 = 2 ;
If a system with several species is studied, it is often advantageous to
collect the a terms in an atom matrix, A. If the number of species is k
and the number of elements is l, the atom matrix is
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=
lkl
k
aa
a
aaa
1
21
11211
M
O
L
A
Condition on the amount of an element, in(
:
Consider a closed system with hydrogen and oxygen, the elemental
amounts of which are Hn(
and On(
. Assume that these two elements
can appear in the system only in the form of O2, H2 and H2O.
Species j = 1 H2O
Species j = 2 H2
Species j = 3 O2
with the elements i = 1 2
A i = H O
24
F1 = (H)2(O)1 so a11 = 2 ; a21 = 1
F2 = (H)2(O)0 so a12 = 2 ; a22 = 0
F3 = (H)0(O)2 so a13 = 0 ; a23 = 2
yielding the atom matrix
O
H
201
022
O H OH 222
⎟⎟⎠
⎞⎜⎜⎝
⎛=A
The total amounts of the elements H and O are
22 HOHH 22 nnn +=(
22 OOHO 2nnn +=(
and these cannot change even though the composition changes, i.e.,
02222 HOHH =Δ+Δ=Δ nnn
(
0222 OOHO =Δ+Δ=Δ nnn
(
or
022 prod,HprodO,H 22=+ nn
02 prod,OprodO,H 22=+ nn
This equation system consists of two equations and three unknowns,
so one variable can be freely chosen. Therefore, the change in
composition can be described by one independent reaction.
25
In the chemistry, the equation system is given in the form of reaction
formulae, e.g.
OH2OH2 222 ⇔+
but information is strictly spoken superfluous if the equation system
is known!
Since the elements, despite chemical reactions within the system,
cannot be produced, the following constraint is imposed on the
production (rate) of the species:
0=
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
prod,
prod,2
prod,1
1
21
11211
klkl
k
n
n
n
aa
a
aaa
&
M
&
&
M
O
L
or compactly
0nA =prod&
This is an extremely important condition for the chemical micro
structure, which in a nut shell expresses the rule of addition of the
elemental “production” terms.
26
From earlier we know that the vector of molar production rates can
be written as ξνn && =prod , so we now have
0νA0ξνA == or&
since the condition should hold independent of the values of the
reaction rates!
b. Choice of independent variables
The number of “independent variables”, which equals the number of
independent chemical reactions, is given by
)(rank)dim( prod An −= &R
If numerical values for R independent variables are known and the
equation system can be solved, the system is uniquely determined.
Example
Consider a system with k = 5 species: O2 , CO , CO2 , SO2 and SO3.
The atom matrix (with the l=3 atoms O, C and S) is
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛=
11000
00110
32212
A
The number of “free variables” is thus R = 5 − 3 = 2 (see below).
27
The condition that elements cannot be produced can be written
0=
⎟⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
prod,SO
prod,SO
prod,CO
prod,CO
prod,O
3
2
2
2
11000
00110
32212
n
n
n
n
n
&
&
&
&
&
To determine the rank of A and the free variables, the matrix has to
be written in upper diagonal form, e.g., by Gaussian elimination,
which gives
0nA =prod
~~&
(In the above example, the atom matrix was already in ready form for
the sake of simplicity.) After this, the first element that differs from
zero on each row of the matrix is marked (by circles in the example).
These are the base variables of the system, and the number of base
variables is the rank of A.
The remaining components are the free variables.
In the example, the base variables are prod,SOprod,COprod,O 22 and, nnn &&& ,
while prod,SOprod,CO 32 and nn && are free variables.
28
The stoichiometric coefficients of the independent reactions are given
as follows: The modified molar production vector prod
~n& is replaced by
a vector of stoichiometric coefficients, fν~ , where, in turn, the
stoichiometric coefficient, νf , for a free variable, f, is given the value
1 while other free variables are set to zero. After this, the equation
system 0νA =f~~
is solved.
In the example, we first obtain
⎪⎪⎩
⎪⎪⎨
⎧
==−=−=
⇒=
⎟⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
0
1
1
0
1
11000
00110
32212
2
2
2
2
2
SO
CO
CO
21
O
SO
CO
O
νννν
ν
νν
0
i.e.,
2221 COCOO ⇔+
and then
⎪⎪⎩
⎪⎪⎨
⎧
=−=
=−=
⇒=
⎟⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
1
1
0
1
0
11000
00110
32212
3
2
2
2
2
SO
SO
CO
21
O
SO
CO
O
νννν
ν
νν
0
29
i.e.
32221 SOSOO ⇔+
By interchanging the order of the columns in the A matrix it is
possible to find all alternative sets of free variables.
The vectors with the stoichiometric coefficients for the independent
reactions are collected in the matrix
( )Rνννν ~~~~21 L=
which also satisfies the equality
0νA =~~
Example: Demonstrate that the equation holds for the previous
example.
30
b. Application to a system with chemical reactions
Pre-multiplication of the molar balances by the atom matrix A gives
outinoutprodin nAn
AnAnAn
AnAnA
0
&&&321
&& +=⇒+=+=
dt
d
dt
d
The atom matrix thus converts the molar (flow) vector into a (flow)
vector of elements.
This equation can in turn, be combined with information about
reaction rates, which can be expressed with the independent chemical
reactions.
Naturally, for volume-based reaction rates we have a similar
condition,
0rA =prod
where rprod is a column vector with elements ri, i = 1,...,q.
31
4. FIRST LAW OF THERMODYNAMICS
a. Energy balances
(Instead of “heat balances” the term “energy balances” is used to
stress that different forms of energy – not only heat – are considered).
First law: The total energy is conserved. (Processes where mass is
converted into energy, i.e., nuclear reactions, are not considered.)
The first law of thermodynamics is applied in energy balances.
The (batch-wise) energy balance equation is
Estart + Ein = Eend + Eout
energy content + incoming = energy content + outgoing
at start energy at end energy
where the incoming and outgoing terms are given by
∫=end
start
inin
t
t
dtEE & ; ∫=
end
start
outout
t
t
dtEE &
For the continuous case one may write
32
outin Edt
dEE && +=
inflow rate = accumulation rate + outflow rate of energy of energy of energy
The total energy balance thus lacks the production term. However, if
only certain forms of energy (e.g., heat) are considered in a partial
balance, there may be a production term.
b. Thermodynamic state
The internal state of a pure substance is uniquely given by the state
of aggregation, pressure and temperature. Quantities that depend
solely on the internal state are called state variables.
E.g., the specific volume (v = 1/ρ) for an ideal gas can be written as
pM
TRTpv =),(
The thermodynamic temperature, T, is defined as a non-negative
quantity, the zero level of which is a hypothetical asymptotic value
that is called the absolute zero. Therefore, T is also called absolute
temperature.
33
c. Elements in energy balances
Energy forms that can be accumulated:
- Potential energy: zgmE =p
- Kinetic energy: 221
k wmE =
- Internal energy: umU =
Energy forms in transport (across the balance boundary)
Without mass transport:
- Work done at the balance boundary: W
- Heat flow: Q&
- Electrical power: P
With mass transport:
- Flow of potential energy: zgmE && =p
- Flow of kinetic energy: 2
21
kˆ wmE && ξ=
- Flow of internal energy: umU && =
- Flow caused by feed of mass: vpmWf && =
where the last two terms can be combined in
- Flow of enthalpy: )( pvumhmH +== &&&
34
Derivation of some terms:
From the relation
∫=⇒=2
1
l
l
lFlF dWddW
using Newton’s law
;aF m= dt
d wa =
the work required to
• move the mass m from the vertical level 0 to z :
,0
p zgmdlgmEz
∫ ==
• give the mass a velocity w
2
0
21
00k wmdwwmd
dt
dmd
dt
dmE
wwl
∫∫∫ ==== wl
lw
• expand a gas at pressure p in a vessel with a (mass-less and
friction-less) piston against the surrounding p − dp
∫∫∫ ≈−==2
1
2
1
)(0
V
V
V
V
l
VdpVddppldAAF
W
which at constant pressure yields
)( 12 VVpW −=
This work is done by the gas and is thus lost to the environment.
The flow of energy due to feed work of mass correspondingly yields
vpmVpWf &&& ==
35
Thermodynamic modeling: Filling of a gas cylinder
Consider the process of filling a gas cylinder from a larger container.
The initial state of the cylinder is p0 and T1 (=the temperature of the
surroundings) and of the container is T1 and p1 (>>p0).
Assumption: The cylinder is filled rapidly; heat exchange with the
walls is negligible. The container is much larger than the cylinder; p1
is constant.
Balance boundary along the inner surface of the cylinder (A):
33
0
2
2
2200)
2ˆ( umdt
whmum
t
∫ =++ ξ&
where ”0” and ”3” refer to the initial and end states of the cylinder,
and”2” is the state of the gas as it flows across the balance boundary
A.
T1 , p1
m& B
A
36
A mass balance yields
3
0
0 mdtmmt
∫ =+ &
Since p3 = p1 >>p0, we may neglect m0 in comparison with m3. An
energy flow balance for B gives
)2
ˆ(2
2
221
whmhm ξ+≈ &&
where h1 can be considered constant. An energy balance for A now
yields
331 umdtmht
o
≈∫ &
or h1 ≈ u3. The definition of enthalpy gives
3313 vphh +≈
The enthalpy of the gas in the cylinder exceeds that of the container;
the gas (temporarily) assumes a higher temperature.
37
5. THE SECOND LAW OF THERMODYNAMICS
a. Entropy balances
The first law of thermodynamics tells us that energy can neither be
created nor destroyed, but it does not give information on the feasible
ways in which energy can be transported.
According to the second law of thermodynamics only such processes
are (spontaneously) possible where the total disorder increases.
1Q&
1Q& 2Q&
2Q&
1T 2T>
Example: An energy balance
for stationary heat conduction
in a plate allows heat transfer
in either directions,
irrespective of the surface
temperatures: 21 QQ && =
Practical experience: Heat
conduction from higher to
lower temperatures!
38
Classical example: Two gases (at same pressure) in a vessel separated
by a wall. If the wall is removed, the two gases gradually mix: The
disorder increases spontaneously! If the wall would be inserted
again, the likelihood that the two gases would be on separate sides
(like in the initial state) is negligibly small.
The rule that states that changes always happen in a way to increase
the total disorder makes it possible to analyze the direction in which
changes (“processes”) take place and also to calculate equilibrium
states (which are limiting values of the changes).
Disorder is described by a quantity called entropy, S, that expresses
the degree of disorder. A necessary condition for a process to be
possible is that the total disorder of the system increases!
The quantity is used in entropy balances:
39
Sstart + Sin + Sprod = Send + Sout
entropy content + incoming + entropy = entropy content + outgoing
at start entropy production at end entropy
where the in- and outflows are given by
∫=end
start
inin
t
t
dtSS & ∫=
end
start
outout
t
t
dtSS &
For the continuous case we may write
{ out
0
prodin Sdt
dSSS &&& +=+≥
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
c = 0,07379 J m6 K2 / mol3 A0 = 0,10673 J m3 / mol2
B0 = 4,074·10−5 m3 / mol; C0 = 816,4 J m3 K2/ mol2
α = 1,272·10−13 m9 / mol3; γ = 5,3·10−9 m6 / mol2
66
c. Estimation of specific enthalpy
In what follows, both specific heat capacity, cp, and specific volume,
v = v (T, p), are assumed to be known.
The differential dh satisfies
pdp
hTd
c
T
hhd
T
p
p
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=321
To obtain (∂h /∂p)T one may divide pdvsTdhd += by dp at
constant temperature, yielding
vT
vTv
p
sT
p
h
pTT
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−=+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
where the right-hand-side result follows from Maxwell’s relations.
We now obtain
pdT
vTvdTpTchd
p
p
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−+= ),(
A: Constant pressure, p = p0 , gives
∫=−1
0
),( 001
T
Tp TdpTchh
67
Furthermore, if cp is constant:
)( 0101 TTchh p −=− ; cp and p constant
B: Constant temperature, T = T0 , gives
∫⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−=−1
0
001
p
p p
pdT
vTvhh
C: For an ideal gas
pM
TRv = yielding
pM
R
T
v
p
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
.
which inserted into the derivative of h with respect to p gives
0=−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
pM
TR
pM
TR
p
h
T
The specific enthalpy of an ideal gas is independent of pressure.
For an ideal gas with constant cp we thus obtain
)( 0101 TTchh p −=−
68
Enthalpy change at phase change:
As a substance changes states of aggregation its specific enthalpy
also changes. In case of vaporization of a liquid, the enthalpy (heat)
of vaporization is given by
''' hhhk −=Δ
where ’’ and ’ denote the state of the vapor and the liquid,
respectively. For the opposite phenomenon, Δhk is the decrease in
enthalpy associated with condensing the vapour.
By analogy, at fusion of a solid we require the a decrease in enthalpy
(heat) of melting (fusion)
hhhf −=Δ '
where symbols without superscripts refer to the solid phase. For the
opposite phenomenon, Δhf is the decrease in enthalpy associated with
freezing the liquid.
For direct transition from the solid to the gas state the specific
sublimation enthalpy (heat) is
hhhs −=Δ ''
By analogy to the above, there is a corresponding change in enthalpy
as a substance changes its (e.g., crystal) modification.
69
d. Estimation of specific entropy
In what follows, both specific heat capacity, cp, and specific volume,
v = v (T, p), are assumed to be known.
The differential ds satisfies
pdp
sTd
T
ssd
Tp
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=
In order to obtain (∂s /∂T)p one may divide pdvsTdhd += by T
dT at constant pressure, yielding
T
c
T
h
TT
s p
pp
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂ 1
From Maxwell’s relations
pT
T
v
p
s⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
so we have
pdT
vdT
T
pTcsd
p
p
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−=),(
70
A: Constant pressure, p = p0 , gives
∫=−1
0
),( 0
01
T
T
p TdT
pTcss
Furthermore, if cp is constant:
0
101 ln
T
Tcss p=− ; cp and p constant
B: Constant temperature, T = T0 , gives
∫ ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−=−1
0
01
p
p p
pdT
vss
C: For an ideal gas with constant cp:
pM
TRv = yielding
pM
R
T
v
p
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
which inserted at constant temperature gives
0
101 ln
p
p
M
Rss −=−
Changes in the specific entropy of an ideal gas with constant cp are
thus given by
0
1
0
101 lnln
p
p
M
R
T
Tcss p −=−
71
Entropy change at phase change:
As a substance changes states of aggregation its specific entropy also
changes. From pdvsTdhd += we have, since temperature and
pressure remain constant at phase changes (dT = dp = 0):
sTdhd =
or after integration
T
hs
Δ=Δ
where
⎪⎩
⎪⎨
⎧
−−−
=Δnsublimatio ofentropy specific''
(fusion) melting ofentropy specific'
ion vaporizatofentropy specific'''
ss
ss
ss
s
In the same way, as change Δs = Δh / T occurs as a substance changes
its modification.
72
e. Estimation of u, g and f
The specific internal energy is given by
pvhu −=
which gives for the differences
)()( 11122212 vphvphuu −−−=−
If the specific volume is constant we have
∫=−2
1
)(12
T
Tv TdTcuu
By analogy, the changes in g and f are obtained as
)()(
)()(
11122212
11122212
sTusTuff
sThsThgg
−−−=−−−−=−
f. Choice of reference (zero) level
In calculating thermodynamic state variables, the reference (zero)
level can be selected arbitrarily, and a suitable choice (e.g., at 0°C)
may result in considerable simplifications in the calculations: If, e.g.,
the enthalpy of a substance at the state it crosses the balance
boundary is taken to be zero, the corresponding terms in the energy
balance equation can be omitted.
73
g. Summary of equations for an ideal gas with constant cp
M
TRvp =
)( 0101 TTchh p −=− )( 0101 TTcuu v −=−
0
1
0
101 lnln
p
p
M
R
T
Tcss p −=−
Combination yields
)( 0101 TTchh p −=−
− )( 0101 TTcuu v −=− ———————————————
))(()()( 01
0
00
1
11 vp ccTT
M
TR
uh
M
TR
uh −−=−−−4342143421
or
M
Rcc vp =−
Thus, the difference between cp and cv decreases with increasing
molar mass.
74
8. STATE DIAGRAMS
State diagrams are used to illustrate the relation between three or
more state variables. Generally, we have
),( yxzz =
i.e., two state variables, x and y, have to be given in order to calculate
a third (arbitrary) variable, z, if the state of aggregation of the
substance is given. Furthermore, the reference level for the diagram
has to be known.
The state diagram depicts two independent variables (e.g., p and v) on
the abscissa and the ordinate, respectively, while other variables (e.g.,
h and s) are given by sets of curves in the diagram.
a. Setting up p-curves in an h,s-diagram
Consider the state variables of water vapor. After selecting a proper
reference point, where h=s=0, the following procedure may be
followed:
First, assume that p=p0=constant. The specific enthalpies and
entropies at different temperatures can now be determined from
75
∫=−T
T
p TdpTchh0
00 ),(
∫=−T
T
p TdT
pTcss
0
0
0
),(
h and s can be determined for different values of T, and the
corresponding isobar can be drawn:
s s0
T0
° °
°
°
°
°
T1
T2
T3
T4
p=p0
∫T
T
pdTc0
h0
h
∫T
T
p dTT
c
0
76
b. Setting up T-curves in an h,s-diagram
Next, take the temperature to be constant, T=T0. The specific
entropies at different pressures can now be determined from
∫ ⎟⎠⎞
⎜⎝⎛∂∂−=−
p
p p
TdT
vss
0
0
Corresponding values for the specific enthalpies are obtained from dh
= Tds + vdp, which after integration at constant temperature yields
∫+−=−p
p
dppTvssThh0
0000 ),()(
T=T0
T0
° °
°
°
°
°
T1
T2
T3
T4
p=p0
h0
h
∫ ⎟⎠⎞
⎜⎝⎛∂∂
−4
0
p
p p
dpT
v
∫+−4
0
),()( 000
p
p
dppTvssT
° °
°
°
p1 p0
p2 p3
p4
s0 s
77
The procedure is repeated from a new starting point, resulting in a
new isotherm (T=T1), yet another starting point, T2 , p0 , etc.
Finally, points of equal pressure are connected by lines (dashed ones
in the above figure). This gives rise to the isobars of the diagram.
On the next page the h,s diagram for water vapour is presented. Its
point of reference (with h=s=0) is water in liquid form at the triple
point.
T0
° °
°
°
°
°
T1
T2
T3
T4
p=p0
h0
h
° °
°
°
p1 p0 p2
p3
p4
s0
°
° °
° ° °
°
°
T=T0
T=T1
T=T2
s
78
79
9. STATE EQUATION ON ENTHALPY BASIS
In the previous chapter, the thermodynamic state equations have been
written in the form
),( pTvv =
which for an ideal gas gives
M
TRvp =
E.g., for water vapor, we may study how the value of the product pv
changes in a state diagram with the conditions. If ϑ = 300 °C the
diagram gives
kPa 5000at kJ/kg 225
kPa 100at kJ/kg 270
====
pvp
pvp
Thus, the product is 17 % lower at the higher pressure!
Instead of following an isotherm, one may follow an isenthalp
(constant enthalpy line, i.e., dh = 0) which yields
80
kJ/kg 3075 kPa, 5000at kJ/kg 260
kJ/kg 3075 kPa, 100at kJ/kg 270
======
hpvp
hpvp
where the product has decreased by only 4 %.
This motivates the use of the approximation
a
hhvp a−≈
where the parameters ha and a are specific for the species in question.
For water vapor (with h=0 for water in liquid form at 0°C, 101.3 kPa)
we have ha = 1951 kJ/kg and a = 4.27:
⎟⎟⎠
⎞⎜⎜⎝
⎛
−⎟⎟⎠
⎞⎜⎜⎝
⎛
≈⎟⎟⎠
⎞⎜⎜⎝
⎛
kPa100427
1951kJ/kg
/kgm3 p
h
v
81
10. THERMODYNAMIC EQUILIBRIUM AND THE DIRECTION OF PROCESSES
According to the second law of thermodynamics, every balance
volume holds an entropy source, i.e., for all processes where changes
happen the condition 0prod >S& shall hold. As the rate of entropy
production reaches the limiting value of zero, the process “goes out”.
In this point equilibrium conditions have established. Therefore, the
balance equations can tell the direction as well as the end point of the
changes in a system. This will in what follows be illustrated by a
treatment of phase equilibria.
a. Phase equilibria
Consider a substance in liquid form in contact with its vapor phase at
temperature T and pressure p. The phases are separated by a phase
boundary. Thereare three possible cases:
A. Mass flow from the liquid to the vapor
B. Mass flow from the vapor to the liquid.
C. No (net) mass transport.
82
A mass flow of liquid 'm& with specific enthalpy 'h is enters into the
balance volume. A mass flow ''m& of vapor with specific enthalpy ''h
flows out from the volume. To maintain the temperature, a heat flow,
Q& , is introduced.
MB: ''' mm && =
EB: '''''' hmQhm &&& =+
SB: '''''' prod smST
Qsm &&
&& =++
Elimination of the heat flow yields
[ ] )'''('')''''()''(''prod ggmTshTshmST −=−−−= &&&
Liquid
Vapor
'm&
''m&
Q&
83
Since T > 0 we have the condition
.0)'''('' >−ggm& Thus, vaporization occurs if ''' gg > .
For the opposite process, condensation occurs if ''' gg > .
The process goes from higher to lower specific free enthalpy! The hypothetical limiting value with zero entropy production rate
corresponds to ''' gg = , or 0=Δg .
The state curve T = T(p) where equilibrium is attained can be studied
on the basis of the condition
'''''' gdggdg +=+ or ''' gdgd =
From dg = − s dT + v dp it follows that
dpvdTsdpvdTs '''''' +−=+− or '''
'''
vv
ss
dT
dp
−−
=
Inserting Δh = Δs/T finally yields the Clausius Clapeyron equation
)'''( vvT
h
dT
dp k
−Δ
=
84
By analogy, the solid-liquid equilibrium is given by
)'( vvT
h
dT
dp v
−Δ
=
and the solid-vapor equilibrium
)''( vvT
h
dT
dps
−Δ
=
These conditions give the equilibrium curves in a T,p diagram. The
equilibrium (saturation) curves intersect in the triple point (H).
H
Vapor
Solid
Liquid T
p
85
Thermodynamic modeling:
Flow of a gas under expansion in a long pipe
Consider flow of a gas under expansion in a long horizontal insulated
pipe (diameter D, length l) where the pressures after the inlet and
before the outlet are p1 and p2, respectively. Since the pressure drop is
considerable, the expansion of the gas must not be neglected.
EB: 2
ˆ2
ˆ2
222
2
111
wmhm
wmhm ξξ &&&& +=+
SB: 2prod1 smSsm &&& =+ Possible state equations:
M
TRvp = or
a
hhvp a−≈
For volume consistent flow of a fluid in a pipe, the pressure (head)
loss and power loss (caused by internal friction) are given by
loss
2
loss ;2
pVWw
D
lp d Δ==Δ &&ρζ
where ζd is the friction factor of the pipe.
86
Limitations:
The flow must not howl (cause noise) in the pipe, i.e., w < 60 m/s
From this condition, the EB, after division by m& , yields
kg
kJ2
2
)m/s 60(1,1
2ˆ
2ˆ
221
1
22
221 ≈≤−=−ww
hh ξξ
Considering the accuracy of enthalpies read from an h,s-diagram, this
term is negligible. We thus have
.0or021 ≈≈− dhhh
If dh = 0 it follows that T is constant for an ideal gas. This means that
pv is constant for the gas in the pipe, according to both the ideal gas
law and the approximation (1.79).
Conditions
constant
constant
0
prod
≈=
=
=
T
pv
dsmSd
dh
&&
87
In order to utilize the entropy balance, the first equation is written as
dh = T ds + v dp = 0
Furthermore, the entropy production, caused by friction, can be
written as
( ) ( )lossprodloss
prod or pdVSdTT
pdV
T
Wd
T
QdSd Δ=
Δ=== &&&&&
&
Note! In this case the relation between the power and head losses (by
friction) provide a quantitative estimate of the entropy production
rate.
Thus
dlD
wmSdT d 2
2
prod ζ&& =
which, according to the second condition, yields
dlD
w
Tsd d 2
1 2
ζ=
Inserted into the first condition, T ds + v dp = 0, we get
02
2
=+ dpvldD
wdζ
88
Does the friction factor change along the pipe?
Since
⎟⎠
⎞⎜⎝
⎛=⎟
⎠
⎞⎜⎝
⎛=⎟
⎠⎞
⎜⎝⎛==
ηζ
ηζ
νζζζ D
A
mD
v
wwDddddd
&(Re)
the factor depends on the dynamic viscosity, η, only. This viscosity
depends on temperature only, and here the temperature, T, is constant
∴ ζ d is constant in the pipe!
Since pv is constant, we may express the specific volume of the gas
in an arbitrary point in the pipe with pressure p, on the basis of the
state at ”1”, as
p
vpv 11=
The first condition now becomes
dppldvpv
w
Dd −=⎟⎟
⎠
⎞⎜⎜⎝
⎛
44 344 21constant
2 11
2
1
1ζ
After integration, the differential equation yields
22
2
2
2
111
2
1
12
1
ppdpplvp
v
w
D
p
p
d −=−=⎟⎟
⎠
⎞⎜⎜⎝
⎛∫
ζ
89
which can be written as
1loss
211
1
22
21 )(
22p
w
D
l
p
ppd Δ==
− ρζ
where the right-hand-side is the pressure (head) loss calculated on the
basis of the conditions after the inlet, i.e., at “1”.
The equation can also be written as
1
21211loss 2)()(
p
ppppp
+−=Δ
Example: How much air flows through a 300 m long horizontal
insulated 2” steel pipe, if the air enters at 50°C and 200 kPa and the
pipe ends in the surrounding?
90
Thermodynamic modeling:
Flow of a gas under expansion in short nozzles
The task is to develop a model for transport of a gas under radical
expansion in a short nozzle. Since a design with highest possible
velocity of a continuous gas jet from the nozzle is desired, the cross-
section area A2 is chosen as small as possible.
Assume that the state T1, p1 of the high-pressure side and the back
pressure pm on the low-pressure side be given. However, the state at
the outlet, T2, p2, is unknown.
EB: 2
ˆ2
ˆ2
222
2
111
wmhm
wmhm ξξ &&&& +=+
SB: 2prod1 smSsm &&& =+
Basic knowledge of fluid mechanics: p2 = pm.
If ”1” is drawn ”far off” the inlet, we have A1 >> A2 , so w1 << w2.
”1”
”2”
T1
p1
m&
0=Q&
pm
91
The model equations can now be written as
mpp
ssmS
hhw
=
−=
−=
2
12prod
21
22
2
)(
2ˆ
&&
ξ
which gives for the outflow velocity
2212ˆ/)(2 ξhhw −=
yielding the mass flow (”media flow”) rate
2212
2
2
22 ˆ/)(2 ξhhv
A
v
wAm −==&
If the (short) nozzle is designed to avoid strong eddies, we may
assume that the production of entropy in this region is small.
The state change is depicted in an h , s diagram:
92
If prodS& is small, s2 will be close to s1, and, therefore, h2 will not differ
much from 2'h . The difference in specific enthalpy cane now be
written as
)( '
21221 hhhh −=− η
where η2 (=0.92-0.96) is an efficiency factor. Introduction of this
equation “consumes” the entropy balance equation!
The model is now:
),(
ˆ/)(2
m1
'
2
'
2
2
2
2
2
'
2122
pshh
wv
Am
hhw
=
=
−=
&
ξη
p1
p2 = pm
2h2'h
1h
s2 s1
h
s
93
The model can be verified experimentally by comparing measured
pressure drops over the nozzle with simulated values.
Example
η2 = 1, 12̂ =ξ , A2 = 1 cm2, p1 = 2940 kPa, C4001
o=ϑ .
kPa 100mp
kJ/kg
'
21 hh −
m/s
w
/kgm3
2v
kg/s
m&
24.5 44 296 0.114 0.260
19.6 113 475 0.139 0.342
14.7 188 615 0.173 0.356*
9.8 280 748 0.238 0.314
5.0 540 1040 0.393 0.264
*) Maximum value
Comparison between measured and simulated values:
×
×
×
×
× × × ×
× Measured ⎯⎯ Simulated
(p1−pm) / bar
kg/s
m& 0.35
14.7
94
The simulation shows very nice agreement with the experimentally
determined values in the beginning of the experiment. As the velocity
increases with increased pressure drop, a maximum value of the mass
flow is reached. After this point, the experimentally determined mass
flow rate is not any longer affected by the pressure loss.
The model is valid up to a maximum velocity that corresponds to the
speed of sound in the fluid in question (w*=615 m/s in water vapor).
The pressure loss, where the maximum mass flow rate, maxm& , is
obtained, is called the critical pressure loss. The velocity of the jet
cannot exceed the speed of sound in the medium.
The model is valid up to pressure losses of p1 − pcr, and after this the
mass flow rate (and velocity) can be considered constant.
Is the thermodynamic model erroneous? No! Only the assumption p2
= pm is wrong. The pressure p2 will be higher that the back pressure,
and the gas loses a considerable part of its pressure after the nozzle.
Length along the nozzle
w
w*
p
p*
95
Is it possible to utilize the fact that the critical pressure drop cannot
be exceeded?
Yes! For instance, if one wants to maintain a constant supply of a gas
to a process where the back pressure varies.
Is it possible to determine maxm& theoretically?
Consider an isentropic expansion of an ideal gas from p1 to p2. For
this holds (at constant specific heat capacity) dh = cp dT, yielding
2
'
212ˆ/)(2 ξTTcw p −=
Furthermore
)/('
or ,0ln'
ln1
2
1
2
1
2
1
2pMcR
p
p
T
T
p
p
M
R
T
Tcp ⎟⎟
⎠
⎞⎜⎜⎝
⎛==−
so
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛−=−
)/(
1
21
'
21 1pMcR
p
pcThh p
The ideal gas law also gives
1)) /(/(
1
2
1
2
2
1
1
'2
2
1
1
'2
−
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛==
pp McR
p
pMcR
p
p
p
p
T
T
p
p
v
v
Thus, the mass flow can be expressed as
96
)) /(2/(22
constant
ˆ2
1
2
1
2
2/1
221
221
2
22pp McR
p
pMcR
p
p
v
AcT
v
Awm p
−
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛==
4434421
&ξ
The critical back pressure, pcr, is the pressure p2 that maximizes the
mass flow rate, i.e.,
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧ −
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−
⎟⎟⎠
⎞⎜⎜⎝
⎛)) /(2/(22
/max
1
2
1
2
12
pp McR
p
pMcR
p
p
pp
Example: Show that the maximization gives the result
v
p
p
p
c
c
RMc
M
Rc
M
Rc
p
p
p
=−
⎟⎠⎞
⎜⎝⎛
+=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
−= κκ
κ
κ e wher
1
1
2
/
2
22
1
cr
The thermodynamic model that was developed is thus limited to
1
cr
1
2
p
p
p
p≥
while for higher pressure drop we simply have
{ } )(max cr2 ppmmm === &&&
97
The values of the heat capacity ratio of gases vary within κ = 1.000...
1.667, which yields a relatively small variation in the critical pressure
ratio
487.0....607.01
cr =p
p
E.g., for water vapor, the ratio is approximately 0.54.
The speed of sound in a medium is given by
s
pw ⎟
⎠
⎞⎜⎝
⎛∂∂
=ρ
*
From the equation of isentropic expansion of an ideal gas we have
κ
ρρ⎟⎟⎠
⎞⎜⎜⎝
⎛=
1
1pp
which yields
M
RTvp
pp
p
s
p κκρ
κρ
κ
ρρκρκ
ρρκ
κ ===⎟⎟⎠
⎞⎜⎜⎝
⎛==⎟⎟
⎠
⎞⎜⎜⎝
⎛∂∂ − 1
11
1
1 1
or
M
TRw
κ=*
Example: What is the speed of sound in air of room temperature?
How much lower is the speed in −30°C?
98
The ratio between the speed and the speed of sound of the fluid is
called the Mach number
*Ma
w
w=
Example: Show that the speed of sound is reached in the outlet of a
nozzle at the critical pressure drop!
Is it possible to produce supersonic jets?
The nozzle shall be designed to have a convergent part followed by a
divergent one. This construction was invented by Gustaf de Laval in
the 1880’s.
Ma < 1 Ma = 1 Ma > 1
99
At supersonic speed an interesting phenomenon takes place: The gas
accelerates when the cross-sectional area grows after the narrowest
section! By the above equations it is possible to derive the optimum
ratio between A2 and Amin.
Example: The pressure in a reactor has been found to vary between
400 kPa and 500 kPa. Design a nozzle that can provide a constant
mass flow rate of steam, kg/s556.0=m& , at 400 °C into the reactor.
Also choose the steam pressure before the nozzle.
x
w
w*
p
p*
Amin
100
11. TREATMENT ON MOLAR BASIS
Thus far, the thermodynamic quantities have been treated on mass
basis, expressing the state variables as specific quantities. In
chemical thermodynamics, it is useful to describe the system on
molar basis. Thus, the amount of the substance is described by the
molar amount
n = ( ) kmol
Since the ratio between mass and molar quantity is the molar mass
kmol
kg)(== M
n
m
the same relation holds between flow rates
Mn
m=
&
&
Molar quantities will here be denoted by upper-case symbols, with
subscript “m”
Molar internal energy Um = ( ) kJ/kmol
Molar enthalpy Hm = ( ) kJ/kmol
Molar free enthalpy Gm = ( ) kJ/kmol
Molar entropy Sm = ( ) kJ/(kmol K)
Molar volume Vm = ( ) m3/kmol
Molar heat capacity (constant pressure) Cm, p= ( ) kJ/(kmol K)
101
In state equations for gases, the conversion from mass to molar basis
is carried out by replacing R/M by R.
Generally, the conversion from specific to molar quantities is
Ym = M y
e.g., Hm = M h or Sm = M s.
Quantities that depend on the amount of the substance are called
extensive quantities, while independent quantities are named
intensive. Thus, the extensive quantity enthalpy, H, is obtained by
multiplying the specific quantity h by the mass m, or by multiplying
the molar quantity Hm by the molar amount n.
Intensive quantities Extensive quantities
h, Hm , s, Sm , v, u, T, p, etc. H, S, V, U, etc.
From this it follows that the differentials can be written
dU = T dS – p dV
dH = T dS + V dp
dG = –S dT + V dp
dF = –S dT – p dV
102
12. BALANCES IN A CHEMICALLY REACTING SYSTEM a. Introduction
Earlier in the course, systems where chemical reactions occur have
not been treated. Principally, the presented thermodynamic models
are valid, but some aspects need further clarification. E.g., the choice
of zero level of quantities in tables or diagrams should be discussed.
Consider a system in steady-state where hydrogen gas is combusted
with pure oxygen gas.
The changes in composition within the balance volume can be written
O2(g) + 2H2(g) → 2H2O(g)
In accordance with earlier formulas, we have
ξν &&&&& iiiii nnnn +=+= in,prod,in,out,
i,Hin,H 22, Tn&
i,Oin,O 22, Tn&
Q&
outout,H ,2
Tn
outoutO,H ,2
Tn&
outout,O ,2
Tn&
103
which for the present case yields the molar balances
ξ&&& 2in,Hout,H 22−= nn
ξ&&& −= in,Oout,O 22nn
ξ&&& 2in,OHout,OH 22+= nn
The energy balance equation is
QHnHnHn
HnHn
&&&&
&&
+++
=+
out,OH,mout,OHout,O,mout,Oout,H,mout,H
in,O,min,Oin,H,min,H
222222
2222
In the chemical thermodynamics, the molar enthalpies, Hm,i ,are
generally set so that the elements in their stable states at 25°C and 1
atmosphere (101.3 kPa) assume zero values.
In the example the molar enthalpies of the hydrogen and oxygen
flows can therefore be obtained from the general equation
∫ ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛∂∂−+=
pT
K p
ipi dpT
VTVdTCH
,
kPa3,101,298
mm,m,m,
If the component changes state of aggregation or modification,
corresponding transition terms, ΔHm, have to be added.
Often the effect of the pressure is small, so only the temperature
integral has to be taken.
104
As the oxygen and hydrogen are transformed into water vapor there is
an associated energy, which is similar to the heat of vaporization
required to vaporize liquid water. These energy terms are defined for
a given reaction at a constant state x (i.e., at Tx , px) through the
reaction enthalpy
∑=
=Δk
iii xHxH
1,mr )()( ν
If the reaction occurs at the reference level of the table, the reaction
enthalpy is denoted by o
rHΔ .
In the example it was noted that the Hm values of both oxygen and
hydrogen were zero at the zero level of the table. In this case we have
o
OHm,
o
r 22HH =Δ
where o
OH m, 2H is the molar enthalpy of water vapor at the zero state of
the table. Such o
m,iH values are therefore often called standard
enthalpies of formation. Thus, experimentally determined reaction
enthalpies have been tabled as molar enthalpies of formation.
105
A table may have the following appearance:
Species mol
kJ/o
m,iH
H2 0 O2 0
C 0 Fe 0
He 0 Fe2O3 −826.0 H2O(l) −286.0 H2S −20.5 CO2 −393.7
b. Combination of information from different tables
If information about enthalpies of formation is available from
different sources, the quantities can be easily converted to make it
possible to use them in the same thermodynamic model. The
conversion is based on the fact that the reaction enthalpies for all
(possible) reactions have to remain unchanged independent of the
source of the table. If the level of the molar enthalpy o
m, iH is adjusted
by bi we thus have that
∑=
+=Δk
iiii bxHxH
1,mr ))(()( ν
should be constant, or
∑=
=k
iiib
1
0ν
106
Example: Convert the above table to make enthalpies of oxygen in
the form of O2(g), hydrogen in the form of H2O(l) and carbon in the
form of CO2(g) zero in the new table.
For the molar enthalpy of water the shift is kJ/mol286OH2+=b .From
O2(g) + 2H2(g) → 2H2O(l)
we have 02222 HOH =− bb or kJ/mol286
2H +=b , which is also the
new value of the molar enthalpy of the hydrogen gas.
By analogy, we have that kJ/mol7.3932CO +=b and from
C(s) + O2(g) → CO2(g)
it follows that 0CCO2=−bb or kJ/mol7.393C +=b , which is also the
new value of o
Cm,H . The new table takes the form
Species mol
kJ/o
m,iH
H2 286.0 O2 0
C 393.7 H2O(l) 0
CO2 0
1401: Determine o
mH for H2O2(l) in the new table, if for H2O2 we
have kJ/mol 187.96om −=H in the old table (on p. 116).
1402: Determine o
mH for H2S(g) in the new table, if the molar
enthalpy of formation of SO2 in this table has the value zero. In the
old table, we have kJ/mol 296.95o)(SOm, 2
−=gH .
107
c. Mixtures and combustion heat
For mixture of gases and non-polar liquids, the enthalpy of the
mixture can, as a rule, be approximated by a weighted sum of the
pure components in the mixture. For strongly polar liquids, however,
mixing enthalpies have to be considered.
The caloric combustion heat is defined as the energy that has to be
removed when 1 kg fuel at 20°C and 101.3 kPa is completely
combusted with oxygen of the same state, if the combustion products
H2O(l), N2(g), CO2(g) and SO2(g) are also brought back to the initial
temperature and pressure. The value thus equals h° of the fuel at this
state for the case where the standard formation enthalpies of the
combustion products are zero.
The effective combustion heat is defined by analogy, but so that the
water is taken to leave the system as steam, H2O(g).
108
d. The energy balance equation
Consider a control volume in steady state into which a flow of
components, inn& , enters, and from which an outflow of components,
utn& , exits. Several chemical reactions may take place within the
control volume.
The energy balance equation is written, using the molar enthalpy
treated in the previous subsections, as
where mH is a column vector of molar enthalpies as elements. It is
sometimes advantageous to formulate the problem as follows:
By this redistribution of terms and considering inoutprod nnn &&& −= ,
the energy balance can be written as
m,ininHnT& utm,utHnT&
)(min xT Hn& )(mout xT Hn&
[ ])(minm,in xT HHn −& [ ])(moutm,out xT HHn −&
109
[ ] [ ])()()( moutm,outminm,inmprod xxx TTT HHnHHnHn −=−+− &&&
Since the state x (i.e., Tx , px) can, naturally, be selected arbitrarily, we
may use the zero level of the table, T0 , p0 , so
o
m,m, )( ii HxH ≡
The molar production rate vector is
ξνn && =prod
Therefore, the first term in the energy balance equation can be written
as a function of the reaction enthalpies at the zero level of the table
[ ] [ ]omoutm,out
ominm,in
or HHnHHnHξ −=−+Δ− TTT &&&
e. The entropy balance equation
By analogy with the above, the reaction entropy ΔSr(x) for a
chemical reaction is
∑=
=Δk
iii xSxS
1,mr )()( ν
Reaction entropies have been determined experimentally. The molar
entropies o
m, iS for pure substances at standard state (25°C, 101.3 kPa)
have been tabled according to the convention that the entropy of a
(crystalline) pure substance at the absolute zero (0 K) take on zero. At
states that differ from the reference state of the table, the molar
entropies are obtained from
110
∫ ⎟⎠
⎞⎜⎝
⎛∂∂
−+=pT
dpT
VdT
T
CSS
p
ip
ii
,
kPa3,101K15,298
m,m,o
m,m,
If the substance changes state of aggregation or modification, an
associated entropy transition term, ΔSm, should be added.
Note! As the entropy of a gas component in a mixture is calculated,
the pressure integral shall be taken to the partial pressure of the
component!
For a reacting system the terms in the balance equation can be written
By analogy with the treatment of the energy balance equation, the
terms can also here be reformulated into the left hand side
[ ])()( minm,inmprod xx TT SSnSn −+− &&
and the right hand side
[ ])(moutm,ut xT SSn −&
Using the reaction entropy, we can now write
[ ] [ ])()()( moutm,utprodminm,inr xSxx TTT SSnSSnSξ −=+−+Δ− &&&&
inm,inSnT& outm,outSnT& prodS&
111
13. CHEMICAL REACTIONS AND
ENTROPY PRODUCTION
a. Basics of the classical thermodynamics
If the entropy balance equation is written as
tdtd
SdSSS =+− prodoutin&&&
we have
[ ] Sd
Sd
Sd
Sd
tdSS
ie
=+−3214434421
&&prodoutin
where deS is the entropy the system exchanges with the surroundings
(external) and diS is the internal entropy production, for which holds
0≥Sdi
Two important special cases are
Isolated system: 0=Sde which gives 0≥Sd
Closed system: T
dQSde = which gives
T
dQSd ≥
where tdQdQ &=
112
b. The direction of chemical reactions. Affinity. Consider the piston-provided reactor with a gas mixture studied
earlier in Chapter 5. We want to determine the direction of chemical
reactions in it and the associated entropy production. Constant
temperature is maintained in the reactor by a reversible heat flow, Q& ,
from/to the reactor and constant pressure by piston displacement.
NB: ξνnn && == prodtd
d
EB: td
Vdp
td
UdQ +=&
SB: td
SdS
T
Q=+ prod
&&
Elimination of the heat flow yields
td
Vdp
td
Ud
td
SdTST −−=prod
&
Q&
T , p
113
Since the pressure and temperature are constant, the extensive
thermodynamic quantities S, U and V in the reactor change only due
to changes in composition, caused by chemical reactions. Such
changes can be expressed by differentiating the extensive quantities
with respect to the molar amounts, i.e.,
td
nd
nd
Vdp
td
nd
nd
Ud
td
nd
nd
SdTST
ik
i npTi
ik
i npTi
k
i
i
npTi
i
ii
∑
∑∑
=
==
⎟⎟⎠
⎞⎜⎜⎝
⎛−
+⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
1 ,,
1 ,,1 ,,
prod
'
''
&
where '
in denotes all other molar amounts except ni. We now have
∑∑==
−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+−=k
ii
ii
i
ii
k
i
i Gtd
ndST
H
VpUtd
ndST
1m,m,
m,
m,m,1
prod 43421&
with
iii
npTii
npTii
npTii
STHGnd
VdV
nd
SdS
nd
UdU
i
ii
m,m,m,
,,
m,
,,
m,
,,
m,
and
;
'
''
−==⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
μ
where μ is the chemical potential.
If there is only one chemical reaction with the reaction rate ξ& , we
may write
114
ξξμν &&&r
1prod AST
k
iii =−= ∑
=
where Ar is the chemical affinity. Thus, we have
0r≥ξ&A
i.e., the direction of the reaction is such that the reaction rate and
the affinity have the same sign.
At equilibrium
0r =A
The case with several simultaneous reactions gives
0rprod ≥=−= ξAξνμ &&& TTST
where Ar is a row vector with the affinities as elements.
At equilibrium all affinities vanish, i.e.,
0... r,2r,1r, ==== qAAA
For a micro volume the condition can be written
0rprod ≥=TVd
Sd T rA&
115
Note! The constraint that the total entropy production rate be
positive is a necessary but not a sufficient condition for the reacting
system (as will be seen later).
c. Temperature dependence of the equilibrium constant To be able to use the condition
0r =A
that holds at chemical equilibrium, the dependence of the chemical
potential, μ, on the state must be clarified. In accordance with earlier
findings, we have
dpVdTSdG +−=
so
dpnd
VddT
nd
Sd
nd
Gdd
iii npTinpTinpTi ''' ,,,,,,
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
or concisely
dpVdTSd iii ,m,m +−=μ
If the potential is expressed with respect to a reference state (Tx , px
and reference composition of the substance), )(o xiμ , we get at other
pressures
∫+=i
x
iiixi
p
p
dpVxpT ,m
o )(),( μμ
116
This gives us the affinity
∑ ∫∑==
−−=k
i
i
x
ii
k
iiir
p
p
dpVxA1
,m1
o )( νμν
The first term can be written as
)()()( o
r
o
1
o xxxk
i
T
ii μμν Δ==∑=
μν
which, following earlier equations, can be expressed
)()()( rr
o
r xSTxHx xΔ−Δ=Δμ
If the chemical potential for every substance is calculated with the
pure substance as a reference, we have )()( m,
o xGx ii Δ=Δμ . The
free enthalpy is often written in tables with the temperature 25°C and
the pressure 101.3 kPa as reference state.
Consider reactions at T = Tx between components in a mixture of
ideal gases, xi RTpV =,m :
∑
∑ ∫
∑ ∫
=
=
=
−Δ−=
−Δ−=
−Δ−=
k
i x
iix
k
i
i
x
xi
k
i
i
x
iir
p
pRTx
p
dpRTx
dpVxA
p
p
p
p
1
o
r
1
o
r
1,m
o
r
ln)(
)(
)(
νμ
νμ
νμ
117
Ar = 0 at equilibrium, so
)(lnlnln)(
11
o
r xKp
p
p
p
RT
x k
i
i
x
ik
i x
ii
x
−=⎟⎟⎠
⎞⎜⎜⎝
⎛−=−=
Δ ∑∑==
ν
νμ
where the equilibrium constant is defined as
∏=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
k
i
i
x
i
p
pxK
1
*
)(ν
where * denotes the equilibrium state.
In the above formula the )(o
r xμΔ values can be estimated through
reaction enthalpies )(r xHΔ and entropies )(r xSΔ .
If o
iμ should be determined at other temperatures, the condition
xii dTxSxd )()( ,m
o −=μ or xdTxSxd )()( ror Δ−=Δμ
may be used. Combined with the definition
)()()( rr
o
r xSTxHx x Δ−Δ=Δμ
we get
)()(
)( r
o
ro
r xHT
xTx
x
x Δ=∂Δ∂
−Δμμ
The rule of differentiation of a product gives for the left-hand-side of
this equation
)(/)(
r
o
r2 xHT
TxT
x
xx Δ−=
∂Δ∂ μ
118
which, at constant pressure p = px, yields
2
r
o
r )(/)(
x
x
x
x
T
TH
T
Tx Δ−=
∂Δ∂ μ
or
2r )()(ln
x
x
x TR
TH
T
xK Δ=
∂∂
This relation is called the van’t Hoff equation. The equation can be
applied to determine the reaction enthalpy from the temperature
dependence of the equilibrium constant that can be experimentally
determined. The reaction entropy, in turn, is given by
x
xxx T
TTHTS
)()()(
o
rrr
μΔ−Δ=Δ
where
)(ln)(o
r xKRTx x−=Δμ
d. Chemical power production Consider a system where power is produced in a chemical reactor. A
set of reactants, inn& , and products, outn& flow to/from the reactor. A
heat flow and power can also be led out
119
NB: outprodin nn
nn &&& +=+dt
d
AB: 0nA =prod&
EB: outin HQPdt
dUH &&& +++=
SB: outprodin ST
Q
dt
dSSS &
&&& ++=+
In case of steady state we have dt
d = 0. After combining EB and SB
by eliminating the heat flow
( ) ( ) prodoutoutinin STSTHSTHP &&&&& −−−−=
Q&
P
outn&
inn&
120
Different problem formulations:
P as main objective: Power plant, combustion engine, etc. outn& are
waste products (e.g., ash, flue gas).
outn& as main objective: Chemical reactor. The technology for utilizing
P not yet available (Exception: fuel cells).
If P is the primary objective, maximum power is obtained if
0prod =S& , which gives, if the heat flow is taken out at the same
temperature as the products
( ) ( ) ( ) outinoutinoutoutoutinoutin GSTHSTHSTHP &&&&&&& −−=−−−=
If, on the other hand, P is neglected, i.e., the case where outn& is the
main objective, we get
( ) outinoutinprodloss GSTHhSTP &&&& −−==
Thus, the “optimal” process is the same for both formulations!
{ } { }outminat obtained ismax GP &
In a chemical reactor, maximum entropy production, i.e., chemical
equilibrium, can be determined by minimizing the Gibbs energy
subject to the constraint
0nA =prod&
121
This constrained optimization problem can be solved by Lagrange’s
method
{ }prodout
prod
min nAλn
&&&
TG +
If this function is differentiated, at constant pressure and temperature,
with respect to genn& we get after equating it to zero
0λAμ =+ T
The vector of Lagrange multipliers, T
l),...,,( 21 λλλ=λ , where l is
the number of elements, can be eliminated by multiplication of the
expression with the (transposed) stoichiometric matrix, Tν
0μνλAνμν0
==+=
TTTT
321
Finally, this equation system can be solved with respect to unknown
reaction rates or unknown free variables (see Chapter 3).
122
Example
Consider an ammonia reactor, with no power output (P = 0), that
operates isothermally at 500 °C and 30 MPa. The feed flows to the
reactor are nitrogen gas (70 mol/s) and hydrogen gas (200 mol/s) and
a heat flow is taken out reversibly at 20 °C. The below table presents
the entropy production and the lost power, 0prodloss TSP &= .
Conversion
degree of N2, %
Ar,out
kJ/mol
prodS&
W/K
Ploss
kW
3 40.7 146.2 42.8
9 25.2 318.3 93.5
15 17.3 432.3 126.7
21 11.4 509.7 149.3
27 6.5 558.0 163.5
33* 2.1 581.2 170.3
39 −2.1 581.0 170.2
The limit of the process is at Ar = 0 or max{ prodS& }.
123
14. EXERGY Consider a flow medium (of e.g., water vapor) at T, p that is fed into
a power plant in order to produce power, P. The medium is taken out
at the ambient state, T0, p0 (e.g., in the form of water as liquid at 20
°C and 101.3 kPa).
EB: QPhmhm &&& ++= 0
SB: xT
QsmSsm
&&&& +=+ 0prod
Elimination of the heat flow yields
( ) prod00 )( STssThhmP xx&& −−−−=
Q&
pTm ,,&
P
00 ,, pTm&
xT
124
The (theoretical) maximum power, Pmax , that can be produced (as Tx
= T0 and 0prod =S& ) is the exergy
( ))( 000max ssThhmP −−−= &
or, expressed per mass flow, the specific exergy
)( 000max ssThhm
Pe −−−==
&
This quantity is sometimes used as a measure on how valuable the
medium (e.g., steam) is.
The heat flow that is (thought to be) led out, expressed per m& , is
called the specific anergy
)( 00min ssTm
Qa −==
&
&
This heat cannot, however, be utilized, since it is obtained at the
ambient temperature.
It is possible to write exergy balances, which is realized if energy and
entropy balances are combined in a process where “0” is the ambient
state. The exergy balance constitutes a study of Pmax for all energy
flows in the system, while prod0 ST & is the exergy loss. However, the
concept of exergy is basically exactly the same as the combination of
EB and SB applied frequently in the theory and applications of this
course.
125
15. CONCLUSIONS The laws of thermodynamics, and models based on these in the form
of balance equations, have been presented.
The balance equations are fundamental building blocks in
thermodynamic process models.
By combining mass/molar, energy and entropy balances, we may
study, e.g.,
• whether processes are feasible
• the design of ideal and real processes
• systems with chemical reactions and chemical equilibrium
In order to apply thermodynamic process models in problem solving,
the numerical values of associated state variables have to be
estimated. Some common equations of state have therefore also been
treated.
The course has, hopefully, given the students basic knowledge in
thermodynamics and a notion of how the theory can be applied to
solve thermal problems.
Recommended