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.