CHAPTER 3

ELEMENT BALANCES

3.1 THE ELEMENT BALANCE EQUATIONS

As we know, chemical species are not conserved in systems involving chemical

reactions. However, the atoms of each reacting and non-reacting conditions (only

nuclear reactions can change atoms). This is because the reaction process only

involves the regrouping of atoms into different molecular aggregates, but the

identity of each atom remains intact. Hence, the number of moles of each

element must be constant, and consequently the mass of each type of element

remains constant as well (atomic weights are fixed).

Conclusion: For steady-state systems, we can write balances involving the mole

and mass flow rates of every element present and these balances require no

correction for element production or depletion due to chemical reactions.

EXAMPLE 3.1: Propylene C3H6 can be produced by the catalytic

dehydrogenation of propane C3H6. Unfortunately, a number of side reactions

occur which result in the production of lighter hydrocarbon as well as the

deposition of elemental carbon on the catalyst surface (which has to be

periodically removed). In a laboratory-scale reactor, a feed of pure propane is

converted into the following gaseous mixture of product (in mol%):

45% propane

32% propylene

6% ethane

1% ethylene

3% methane

25% hydrogen

Along with some evidence of carbon deposition. Assuming that the reactor can

be scaled-up to process 58.2 mol/day propane. Calculate the rate of carbon

deposition which can be expected.

SOLUTION: Since chemical reactions leading to the above spectrum of

products have not been identified, the element balances are the only ones that can

be formulated for this problem. Luckily, they are sufficient to solve the problem.

From the species listed above, it is obvious that carbon and hydrogen are the only

two types of elements occurring in the system.

C3H8 = 0.45

C3H6 = 0.20

C2H6 = 0.06

C2H4 = 0.01

CH4 = 0.03

H2 = 0.25

1

3

2C3H8

C (Deposition)

REACTOR

H BALANCE:

Using as basis N1 = 58.2 mol/day, the molar input rate of hydrogen is 8 x 58.2

mol H/day (because of 8 hydrogen atoms in the C3H8 molecule). Similarly, the

molar output rate of hydrogen can be calculated. As a result, one can obtain:

8(0.45N2) + 6(0.20N2) + 6(0.06N2) + 4(0.01N2) + 4(0.03N2) + 2(0.25N2) = 8 x

58.2

Or: 5.82N2 = 465.8 then N2 = 80 mol/day

C BALANCE:

(3 x 58.2) = 3(0.45N2) + 3(0.2N2) + 6(0.06N2) + 2(0.01N2) + 1(0.03N3 + N3

N3 is the elemental carbon deposition

Hence, N3 = 5.0 mol/day

Note: (drawbacks)

The element balances seem be simpler at first glance than the species balances.

However they have some serious drawbacks as well:

- The number of element balances is usually smaller than the number of

species balances written for the same problem, consequently, less

unknowns can be determined from them.

- The element balances are often more difficult to solve since they involve a

larger number of unknown variables per equation (they are less likely to

lend themselves to sequential solution).

NOTE: (advantages)

The element balance does not involve reaction rates. As a result, we do not have

to know the rate reaction stoichiometric to solve them. This frequently occurs

with reactions with hydrocarbons, specifically in the combustion of fossil fuels.

However, then some information about the molecular/elemental composition of

feed and/or product streams is required. Elemental balances are also common in

bioprocess engineering where the elemental rather than molecular composition of

biomass is typically available.

THE ATOM MATRIX AND ELEMENT BALANCE EQUATIONS

Here we will study some general properties of the element balance equations.

Lets us define the net output rate of species s for a system involving I-input

streams and J-output streams as:

)13(11

I

i

in

is

J

i

out

iss NNN

Given a system involving S species composed of E elements, let es denote the

number of atoms of element e in one molecule of species s. we shall refer to

individual coefficients es .as atomic coefficients. They can be put into a matrix

known as the atom matrix. In terms of this notation, the net molar outflow rate of

element e with species s is ses N . Therefore, the net molar outflow rate of

element e with all species s (s = 1, 2,S) is:

S

s

ses N1

Since the element are conserved, this sum must be exactly equal to zero:

)23(),....,2,1(01

EeNS

s

ses

If we denote by Ae the atomic mass of element e, then Eq. (3.2) can be

expressed in mass units as:

)33(01

S

s

sese NA

Assuming the above element mass balance equations

)43(01 1

E

e

S

s

sese NA

And noting that by definition the molecular mass of species s:

)53(1

E

e

esss eAM

We easily obtain:

)63(011

S

s

s

S

s

ss FNM

This means that the element mass balances sum to yield the total mass balance.

Thus, of the E element balances and the total mass balance, only at most E

balance equations can be independent.

One can also demonstrate that the element mass balances are homogeneous,

hence the principle of the basis of calculation applies as before.

EXAMPLE 3.2: Formaldehyde, CH2O, is produced industrially by partial

oxidation of methanol, Methanol, CH3OH, with air over a silver catalyst. Under

optimized reactor conditions, about 55% conversion of methanol is attained with

a feed consisting of 40% (mol) methanol in air (O2:N2 = 21:79). Although CH2O

is the main product, some production of CO, CO2 and formic acid, HCOOH, is

unavoidable. The reactor output stream is scrubbed to separate liquid products

from gases.

Assuming that the liquid stream contains equal amounts of CH2O and CH3OH

and 0.5% HCOOH, while the gas stream contains 7.5% H2, calculate the

complete composition of both streams.

CATALYTIC

REACTOR

CO2CO

H2 = 75%

N2

CH2O

CH3OH

HCOOH = 0.5%

H2O

O2CH3OH = 40%

N21

3

2

SOLUTION: The system involves 4 element (O, H, C, N) and 9 chemical

species:

O2, H2, CO, N2, CO2, CH3OH, CH2O, HCOOH, H2O

This results in the following atom matrix:

000002000

011110100

222400020

121120102

22222

N

C

H

O

OHFAFMCONCOHO

M = methanol = CH3OH

F = formaldehyde = CH2O

FA = formic acid = HCOOH

Since no flow was given, a basis of 1000 mol mol/day feed will be chosen, N1 =

1000 mol/day. From the known feed composition, we have immediately:

2

2

3

1

/474

/126

/400

Ndaymol

Odaymol

OHCHdaymol

N

From 55% conversion of methanol:

N3,M = 400(1 0.55) = 180 mol/day

Since N3,F = N3,M, then N3,F = 180 mol/day

In order to formulate the element balances, we have to find the species net flow

rates. These are as follows:

daymolNNN inOout

OO /1261260222

22 075.00075.02 NNNH

COCOCONNN ,22 0,

daymolNN NN /474, 22 2

222

22 0, COCOCO NNN

daymolNN MM /220400180400,3

daymolNN FF /18001800,3

33 005.00005.0 NNNFA

OHOHOH NNN 222 ,3,3 0

The four element balances thus become (see the atom matrix)

Oxygen (first row)

01005.02)180(1

)220(1201.0)126(2

2

222

,33

,2,2

OH

CONCOH

NN

NNNN

Hydrogen (2nd

row)

01005.02)180(2

)220(4000075.020

2

222

,33

2

OH

CONCOO

NN

NNNNN

Carbon (3rd

row)

00005.01)180(1

)220(110100

2

2222

3

,2,2

OH

CONCOHO

NN

NNNNN

Nitrogen (4th

row)

00005.00)180(0

)220(002000

2

2222

3

,2,2

OH

CONCOHO

NN

NNNNN

The above reduces to:

)1(29201.0222 ,33

22 OHCOCO

NNNN

)2(520201.015.02,332

OHNNN

)3(40005.0 3,2,2 2 NNN COCO

In addition for N2 and N3 we have:

N2 = N2,CO2 + N2,CO + 0.075N2 + 474 (4)

N3 = 180 + 180 + 0.005N3 + N3,H2O (5)

This forms a system of 5 linear algebraic equations which can be easily solved

for : N2; N3; N2,CO; N2,CO2; and N3,H2O.

The final answer is:

[N2,CO2; N2,CO; N2,H2; N2,N2] = [33.44; 3.67; 41.44; 474] mol/day

x2 = [0.0605; 0.0066; 0.0750; 0.8578]

[N3,F; N3,M; N3,FA; N3,H2O] = [180; 180; 2.89; 215.67] mol/day

x3 = [0.3111; 0.3111; 0.0050; 0.3728]

Conclusion: Element balances are algebraically more complicated than species

balances they tend to involve more unknowns per balance equations, hence

lead to strongly coupled simultaneous equations.

EXAMPLE 3.3: Sulfuric acid can be prepared by absorbing SO3 in water.

Suppose a weak sulphuric acid are used to absorb a stream of SO3 received from

a catalytic oxidizer. In what proportions should the streams be mixed if a 50%

mol H2SO4 solution is to be produced?

1

2

3

4 H2SO4 = 50%

H2O

SO3

H2SO4 = 20%H2O

H2O

SOLUTION: The system involves 3 elements (S, O, H) and 3 chemical species

(SO3, H2O, H2SO4). Since no flow given, the basis of calculation is assumed N4 =

100 mol/h. it looks like 3 element balances should be enough to calculate 3

unknown flowrates (N1, N2, and N3); note that all stream compositions are

known. The element balances are:

S: 1(0 N3) + 0(50 N1 0.8N2) + 1(50 0.2N2) = 0

O: 3(0 N3) + 1(50 N1 0.8N2) + 4(50 0.2N2) = 0

H: 0(0 N3) + 2(50 N1 0.8N2) + 2(50 0.2N2) = 0

Which simplify to:

0.2N2 + N3 = 50 (1)

N1 + 1.6N2 + 3N3 = 250 (2)

2N1 + 2N2 = 200 (3)

However, these 3 equations are linearly dependent,

Since: Eq. (2) 3 x Eq. (1) = 0.5 Eq. (3)

As such these 3 equations cannot be solved and the problem is actually

underspecified by one. The redundancy arises from the fact that oxygen occurs in

a fixed combination with hydrogen as H2O, and sulphur as SO3. Since H2SO4 can

be viewed as (H2O.SO2), the element balances amount to balancing on just the

two groups of elements, H2O and SO3. This situation can occur frequently

enough that in general the number of independent element balances should be

determined as part of the problem analysis. Since in most instances the

redundancy is not as transparent as is the case in EXAMPLE 3.3, a general

algebraic approach is needed.

1.2 THE ALGEBRA OF ELEMENT BALANCES

From the form of the element balance equations:

),....,2,1(01

EeNS

s

ses

It is apparent that the algebraic properties of element balances can be deduced

from the properties of the atomic coefficients es (i.e. the atom matrix). In

particular, questions concerning the dependence or independence of the set of

element balances can be answered by considering the linear

dependence/independence of the vectors of the atomic coefficients of individual

elements. If we denote ],....,,[ 21 eseee then the set of element

balance equations is independent if and only if the row vectors

),....,2,1( Eee are linearly independent.

EXAMPLE 3.4: The atom matrix for the system considered in EXAMPLE 3.3

is:

220

413

101

4223

H

O

S

SOHOHSO

The three vectors formed by considering individual elements (the rows of this

matrix) are:

1,0,1S

4,1,3O

2,2,0H

Since

O

HS

4,1,3

2,2,01,0,13321

21

It is clear that the three vectors are dependent. Hence, the element balances

formed with these coefficients will be dependent as well. Determining whether

element balances are dependent by inspection can be done only in very simple

cases. Generally, a rigorous procedure is required. A technique based on row

operations (rather than columns as before) can be used. H consist of the

following steps:

STEP A: Divide each entry in the ith row by the ith element in that row. If this

element is zero, go to STEP C.

STEP B: add appropriate multiples of the ith row to each remaining row so that

the ith entry in each of these rows becomes zero. Then, set i = i + 1 and go to

STEP A.

STEP C: if the ith element of ith row is zero, interchange column i with any

column to the right of column i which has a nonzero entry in the ith position. The

RETURN to STEP A. If no column has a nonzero ith entry, interchange row I

with any row below the ith row. The process terminates if all rows have been

processed or of all remaining rows is identically zero. If some of the rows are

zero, then the original set of coefficients is dependent and the remaining nonzero

rows constitute a basis of independent vectors.

EXAMPLE 3.5: The solid fertilizer, urea, (NH2)2CO is produced by reacting

ammonia with carbon dioxide. In a given plant, a mixture of NH3 and CO2

consisting of 33% mol CO2 is fed to the reactor. The product stream is found to

contain urea, water, ammonium carbonate (NH2COONH4) and unreacted CO2

and NH3. Calculate the composition of the product stream if 99% conversion of

CO2 is obtained

(NH2)2CO

H2O

NH2COONH4CO2NH3

REACTOR

CO2 = 33%NH3

Solution: The process involves 4 elements (C, O, H, N) and five chemical

species (CO2, H2O, NH3, (NH2)2CO, NH2COONH4). The atom matrix for this

system is:

22100

64320

21012

11001

322

N

H

O

C

carbamateUreaNHOHCO

First row

STEP A: Divide the first row by 1 (no change)

STEP B: (-2) x (1st row) + (2

nd row)

22100

64320

01010

11001

Since these elements are already zero, STEP B terminates

2nd

row

STEP A: Divide the 2nd

row by 1 (no change)

STEP B: (-2) x (2nd

row) + (3rd

row)

22100

66300

01010

11001

Since these elements are already zero, STEP B terminates

3rd

row

STEP A: Divide the 3rd

row by 3

22100

22100

01010

11001

STEP B: (-1) x (3rd

row) + (4th

row)

00000

22100

01010

11001

Conclusion: The complete set of 4 element balances is dependent. The first 3

balances are linearly independent. Their use will be equivalent to using the

complete set of element balances. The problem, as originally stated, has seven

streams variables (2 inputs, 5 outputs), a specified composition, a specified

conversion and a free choice of basis (no flow was given). Therefore:

DOF = 7 3(balances) 1 1 1 = 1

Thus, the problem is underspecified by one specification. Additional information

is needed to complete the solution.

EXAMPLE 3.6: Suppose a mixture of ethylene and butylene is hydrogenated to

produce a product mixture consisting of ethylene, ethane, butylene and butane. If

the feed rates of C2H4 and H2 are 10 and 7 mol/h respectively, and if the output

rates of C2H6, C4H8 and C4H10 are 4.5 and 3 mol/h respectively. Calculate the

unknown feed rate of C4H8 and the unknown output of C2H4.

C2H4C2H6 = 4 mol/h

C4H8 = 5 mol/h

C4H10 = 3 mol/h

HYDROGENATION

C2H4 = 10 mol/hC4H8H2O = 7 mol/h

SOLUTION: The problem involves two unknown flow rates. On the other hand,

the system involves only two elements (H and C). Thus, the two element

balances onght to be enough to solve the problem (provided the balances are

independent!!).

The atom matrix for the system is:

44220

108642

1048462422

C

H

HCHCHCHCH

Which can be easily reduced to the form:

22110

54321

Obviously, the two vectors are independent.

Conclusion: Both C and H balances can be used.

H balance:

0)03(10

)5(8)04(6104)70(28442

in HCout

HC NN

C balance:

0)03(4

)5(4)04(2102)70(08442

in HCout

HC NN

Surprise!! Still dependent!!

40484282 outHC

in

HC NN

20244282 outHC

in

HC NN

Note that C2H4 and C4H8 are the only species that are not completely specified

(their net flow rates, output input are not known). For the C2H4 and balances to

be independent, the corresponding columns of the atom matrix would have to be

independent as well, which is not the case:

20

42

8442

C

H

HCHC

Because

04

8

2

4)2(

In general, element balances of the species which are not completely specified

are independent if:

- They are independent in the sense of the atom matrix reduction procedure

- The corresponding columns of the atom matrix are also independent

1.3 DEGREE OF FREEDOM ANALYSIS

Single unit systems

- The same for reacting and non-reacting cases

- Stream variables and subsidiary relations are counted in the usual manner

- The number of independent element balances is verified by means of the

matrix reduction procedure

- An arbitrary basis is permitted

Multiple unit systems

- The possibility of using overall element balances (internal species and

chemical reactions need not be accounted for! only the elements and

species in the process input and output streams must be incorporated into

the element balances)

- Independent element balances determined using the matrix reduction

procedure

- In order to avoid redundant balances, in the DOF table itemize which

elements will be balanced in which units

EXAMPLE 3.7: High-sulfur coal is gasified to produce a clean synthetic natural

gas (SNG). Powered coal and steam (0.8 kg H2O per 1 kg coal) are charged to a

reactor which produces a complex mixture of gases including 10% (mol)

methane. The heat for the reaction is provided by burning some of the charged

coal with oxygen introduced directly into reactor. The raw product gas is cooled,

treated for removal of entrained ash and sent to a second reactor, the shift

converter. In this unit, the H2:CO mole ratio of 0.56:1 is increased by converting

62.5% of the CO. the shifted gas, still containing 10% CH4, is next treated for

removal of the acid gases, H2S and CO2. Finally, the residual gas, containing

49.2% H2 on an H2O-free basis, is sent to a catalytic reactor which produces

additional methane. The final H2O-free product gas contains 5% H2, 0.1% CO

and the rest CH4 (%mol). Given a coal which analyses (by weight):

66% C

3% H

3% S

18% Ash

10% H2O

Calculate all flows and compositions in the process based on a feed of 10, 000

ton/day coal and assuming that 25% of the carbon fed ultimately appears in the

product.

DOF ANALYSIS

The process involves 4 elements (H, C, S, O) and ash, an inert aggregate of

inorganic minerals which can be treated as a separate as well as an element. All

units except the methanator involve the common species H2, CH4, H2S, CO, CO2

and H2O whose element matrix is:

121000

000100

011010

200242

22242

O

S

C

H

OHCOCOSHCHH

It can be easily shown that the 4 rows of this matrix are independent. The

corresponding matrix for the methanator is:

1100

0110

2042

242

O

C

H

OHCOCHH

The 3 rows of this matrix are also independent. Consequently, the gasifier, shift

converter and acid gas removal unit each will have associated 4 independent

element balances, while the methanator will have 3. The gasifier will in addition

have associated a balance on the ash aggregate. The process input and output

streams involve the same 4 element, 6 species, as well as the ash. Hence, the

overall balances will also consist of the 4 independent element balances plus the

ash balance.

DOF TABLE

Gasifier Shift

reactor

Acid gas

removal

Methanator Process Overall

balance

No. of ind.

variables

14 12 12 8 30 14

No. of ind.

Balances:

H 1 1 1 1 4 1

C 1 1 1 1 4 1

O 1 1 1 1 4 1

S 1 1 1 - 3 1

Ash 1 - - - 1 1

TOTAL 5 4 4 3 16 5

No. of ind.

specified

composition

5 2 1 2 8 6

Number of

subsidiary

relations:

Stream ratio 1 - - - 1 1

CO conversion - 1 - - 1 -

H2/CO ratio 1 1 - - 1 -

H2/(CO + CH4) - - 1 1 1 -

C efficiency - - - - 1 1

Subtotal DOF

basis (selected)

2 4 6 2 1

1

1

Degree of

freedom

0

The coal feed rate seems to be the most convenient choice for the basis of

calculation.

SOLUTION:

Basis of calculation, F1 = 10 000 kg/h (coal)

Carbon efficiency relation:

)(25.0

66.0

1295.0

1

7

,1

,7 massbyF

N

F

F

C

C

Hence

hkmolN /74.144

12*95.0

1000*66.025.07

Steam to coal ratio:

8.018

1

2

1

2 F

N

F

F

Hence

hkmolN /44.444

18

10000*8.02

Overall ASH balance

0.18F1 = F4 therefore F4 = 0.18 x 10 000 = 1800 kg/h

Overall element balances:

Hydrogen:

0]7[)05.0*2949.0*4(]6[2

]5[2][2][1

10000*03.0

76

,52 2

streamNstreamN

streamNsteamNcoal

SH

Carbon:

0]7[95.0]5[2][12

10000*66.07,5 2

streamNstreamNcoal

CO

Oxygen

-2N3 [stream 3] N2 [steam] + 2N5,CO2 [stream 5] + N6 [stream 6] + 0.001N7 = 0

Sulfur:

-0.03 x 10 000/32 + N5,H2S

With N7 already determined, the carbon and sulphur balances can be immediately

solved:

N5,CO2 = 412.5 kmol/h

N5,H2S = 9.375 kmol/h

N5 = 421.88 kmol/h

Then, the hydrogen balance yields:

N6 = 303.11 kmol/h

And, finally, from the oxygen balance:

N3 = 341.90 kmol/h (molecular oxygen)

With the streams N2, N3, F4, N5, N6 and N7 determined, we can now solve both

the gasifier as well as the methanator. Let us proceed with the gasifier.

Gasifier element balances:

Hydrogen:

-300 [coal] 2N2 + 0.1 x 4N8 + 2N8,H2O + 2N8,H2 + 2N8,H2S = 0

Carbon:

-500 [coal] + 0.1N8 + N8,CO2 + N8,CO = 0

Oxygen:

-2N3 N2 + 2N8,CO2 + N8,CO + N8,H2O = 0

Sulfur:

-9.375 + N8,H2S = 0

N8,H2S = 9.38 kmol/h

There are 6 unknowns and 4 balances. However, there is one more subsidiary

relations:

56.0,8

2,8

CO

H

N

N

And the obvious equality:

0.9N8 = N8,H2 + N8,CO + N8,CO2 + N8,H2O + N8,H2S

By a simple although tedious elimination of unknowns, this can be reduced to:

0.4N8 + 1.56N8,CO = 556.83

1.2N8 = 1144.45

And easily solved to yield:

N8 = 953.70 kmol/h

N8,CO = 112.40 kmol/h

And then N8,CO2 = 342.23 kmol/h, N8,H2O = 331.39 kmol/h, N8,H2 = 62.94 kmol/h

and N8,CH4 = 95.37 kmol/h

Shift converter element balances:

Hydrogen

-4N8,CH4 2N8,H2O 2N8,H2 N8,H2S + 0.1 x 4N9 + 2N9,H2O + 2N9,H2 + 2N9,H2S = 0

Carbon

-N8,CH4 N8,CO N8,CO2 + 0.1 x 4N9 + N9,CO + N9,CO2 = 0

Oxygen

-2N8,CO2 N8,CO N8,H2O + 2N9,CO2 + N9,CO + N9,H2O = 0

Sulphur:

-N8,H2S + N9,H2S = 0

N9,H2S = 9.38 kmol/h

There are 5 unknowns left and only 3 equations still available. However, an

additional conversion relation is imposed:

625.0,8

,9,8

CO

COCO

N

NN

Plus by definition:

0.9N9 = N9,CO2 + N9,CO + N9,H2O + N9,H2 + N9,H2S

From the given conversion: N9,CO = 42.15 kmol/h

Again by elimination of unknown, one can obtain:

N9,H2 = 133.20 kmol/h

N9,CO2 = 412.49 kmol/h

Then, the remaining flows can be calculated

N9 = 953.70 kmol/h

N9,H2O = 261.12 kmol/h

N9,CH4 = 95.37 kmol/h

Acid gas removal element balances:

Hydrogen

-4N9,CH4 2N9,H2O 2N9,H2 N9,H2S + 2N5,H2S + 4N10,CH4 + 2N10,H2O + 2N10,H2 = 0

Carbon

-N9,CH4 N9,CO2 N9,CO + N5,CO2 + N10,CH4 + N10,CO = 0

Oxygen

-2N9,CO2 N9,CO N9,H2O + 2N5,CO2 + N10,CO + N10,H2O = 0

Note that the sulphur balance is not needed since N5,H2S is already known. As a

result, we have now 5 unknowns and 3 equations only. However, one more

specification has not been used yet, namely, 49.2%H2 on an H2O-free basis in

stream 10:

492.0

24 ,10,10,10

2,10

HCHCO

H

NNN

N

In addition, the obvious equality holds:

N10 = N10,CH4 + N1,CO + N10,H2 + N10,H2O

Once more, by elimination of unknowns, one can get:

N10,CH4 = 95.37 kmol/h

N10,CO = 42.15 kmol/h

N10,H2 = 133.20 kmol/h

N10,H2O = 261.12 kmol/h

N10 = 431.84 kmol/h

At this point all the streams have been calculated and there is no need for the

methanator element balances.

1.4 APPLICATIONS OF ELEMENT BALANCES PROCESSING OF

FOSSIL FUELS

Species balances are generally preferable to element balances. However, there is

a substantial body of applications for which element balances are a very logical

choice. These include situations when:

- The exact reaction stoichiometry is known or too complex

- The reactants are a complicated mixture of poorly defined (or undefined)

species.

Such is the case of the chemical processing of fossil fuels: natural gas, petroleum,

oil shale and coal.

FOSSIL FUELS AND THEIR CONSTITUENTS

Fossil fuels are hydrocarbon materials formed in the geological time scale from

the organic remains of plants and animals as a consequence of bacterial action

and/or long term compression and heating, generally under anaerobic conditions

below the earths crust.

FORMATION OF COAL

- Peat, a high-moisture-content material as a result of the decay of vegetable

matter.

- The layer of peat is covered up by water and subsequently layers of debris

and sediment (the rapid decomposition stops and a slower anaerobic change

called coalification starts compression and heating, reduction of moisture

and oxygen content, evolution of CO2 and some CH4).

Depending upon the history p,T conditions, various types of coals are produced.

The stages within the process of coalification of peat are referred to as the rank of

the coal. For example, an anthracite coal will have existed first as peat, then as

lignite, then bituminous and finally in its anthracite form.

Moisture content

in raw state %

wt% on dry basis, ash free

C H O

Wood 20 50 6 42.5

Peat 90 60 5.5 32.3

Brown coal 40 60 60 70 5 >25

Lignite 20 40 65 75 5 16 25

Subbituminous 10 20 75 80 4.5 5.5 12 21

Bituminous 10 75 90 4.5 5.5 5 20

Semibituminous

the sedimentary rock/oil shale and sand or limestone formations (initially as a

water emulsion).

Crude oils from various sites vary in composition. Generally, they are composed

of mixtures of:

- Paraffins, C2H2n + 2

- Naphtenes (cyclic saturated hydrocarbons)

- Aromatics

- Some olefins

- Small amounts of organic compounds involving sulphur, nitrogen and

oxygen.

Typically, crude oils will consist of:

85% C

11% H

4% impurities (S, N, O)

NATURAL GAS

H is usually found above or near oil pools, trapped beneath a hard-cap rock. H is

composed of lighter hydrocarbons. Typically, it consists of:

88% mol CH4

8% mol higher paraffins and H2, N2, and CO2

CHARACTERIZATION OF FOSSIL FUELS

Since most colas consist of a mixture of macromolecules, the only practical way

of describing their composition is to resort to an elemental analysis (ultimate

analysis). This involves a standardized chemical determination of five major

elements present in coal: (carbon, hydrogen, oxygen, sulphur and nitrogen)

together with a lumped determination of inorganic, non-combustible residuals

called ash.

ANALYSIS

- Carbon and hydrogen by combustion of a standard sample and collecting

the products (CO2 and H2O)

- Nitrogen and sulphur by direct chemical analysis

- Ash by weighting the residual after heating the sample to 1340 F in the

presence of oxygen

- Oxygen by difference (least difference)

- Sometimes O and H are reported as combined water (the analytical O

combined with hypothetical H in the stoichiometric ratio of water) and net

hydrogen (any hydrogen over that hypothetical).

- High-sulphur coal analysis must account for the presence of iron pyrites

FeS2, which will appear as Fe2O3 in the ash

COMBUSTION OF FOSSIL FUELS

The combustion of a fossil fuel with air or oxygen is an operation in which the

fuel is oxidized to produce heat and partially or completely oxidized residual

waste products. The primary waste product of combustion is a gas consisting

predominantly of CO2, CO, H2, N2, unreacted O2, and remaining unreacted

gaseous fuel constituents. In addition, if the fuel is solid or liquid, there may

remain non-combustible inorganic material called ash.

The main operational concerns in combustion are:

- What rate of air/oxygen should be used with a given fuel feed rate?

- How efficiently is the fuel being used?

- What is the concentration of any pollutants in the combustion waste gases?

In order to answer these questions, some compositions are measured (e.g. of the

flue gas) and the others are calculated by means of element or species material

balances.

A typical flue gas analysis includes (CO2, CO, O2, possibly H2 and CH4, often

SO2, H2S and N2.

The standard experimental procedure for flue gas analysis (ORSAT analysis)

involves the successive contacting of flue gas with solutions that preferentially

absorb one or more of the constituents. Flue gas analyses are always reported on

a water-free (dry) basis.

THEORETICAL OXYGEN AND THEORETICAL AIR

One major quantity of interest in combustion is the relative amount of oxidizing

agent used to react with the fuel:

- Too little air/oxygen too much CO produced (incomplete burning) + low

conversion

- Too much air/oxygen process too violent difficult to control, a large

fraction of heat generated leaves with the flue gases.

- The amount of air/oxygen should be optimized

As with any reaction, there is a calculatable stoichiometric ratio of oxidizer to

fuel (to completely burn the fuel. In case of O2 being used, that ratio is known as

the theoretical oxygen (moles of O2 per unit of mass of fuel).

Since the chief source of oxygen for combustion is air, the theoretical oxygen is

divided by the oxygen content of air to obtain the theoretical air (21% mol O2

and 79% N2)

For purposes of these definitions it is assumed that

All carbon oxidizes to CO2

All hydrogen oxidizes to H2O

All sulphur oxidizes to SO2

Nitrogen does not oxidize

Oxygen present in the fuel is deducted from the total amount of O2 required to

oxidize the other fuel constituents.

EXAMPLE 3.8: Calculate the theoretical air for 100 kg of a dried coal with

ultimate analysis: C = 71.2%; H = 4.8%; S = 4.3%; O = 9.5%; ash = 10.2%.

SOLUTION: Based on the above composition, the 100 kg coal will contain the

following amounts of combustible materials:

C: 71.2 kg = 71.2/12 = 5.933 kmol

H: 4.8 kg = 4.8 kmol (atomic H!)

S: 4.3 kg = 4.3/32 = 0.134 kmol

Reactions:

kmoltrequiremenOCOOC 933.5222

kmolOHOH 2.14

8.422

122

1

kmolSOOS 134.022

Thus, total O2 required by these reactions:

5.933 + 1.2 + 0.134 = 7.267 kmol O2

From this must be deducted the moles of O2 present in the coal, i.e

2297.032

5.9

16

5.9OkmolOkmol

Thus, the theoretical oxygen is:

7.267 0.297 = 6.970 kmol O2/100 kg coal

The theoretical air is:

coalkg

airkmol

10019.33

21.0

970.6

EXCESS AIR

In most applications, more air than the theoretical amount is sually introduced for

combustion. The extra air supplied above the theoretical amount is reported as

percent excess air:

airltheoretica

airltheoreticapliedactuallyairairExcess

sup%

[Note: Another subsidiary relationship!]

EXAMPLE 3.9: A synthesis gas consisting of (mol%):

CH4 = 0.4%

H2 = 52.8%

CO = 38.3%

CO2 = 5.5%

O2 = 0.1%

N2 = 2.9%

is burned with 10% excess air. Calculate the composition of the flue gas

assuming no CO is present.

CO2H2O

O2N2

COMBUSTORCO2H2CO

CH4O2N2

1

2

3SYN. GAS

FLUE GAS

AIRO2N2

Solution: (Element balance will be used)

DOF table

Number of variables 6 + 2 + 4 = 12

Number of element balances 4 (C, H, O, N)

Number of specifications:

- Gas composition 5

- Air composition 1

% excess air 1

Degree of freedom = 1

A basis of calculation is needed, say N1 = 1000 mol/h

Theoretical air (needed for % excess air). The gas will contain (mol/h):

C: 4(CH4) + 383(CO) + 55(CO2) = 442

H: 4 x 4(CH4) + 2 x 528(H2) = 1072

O: 383(CO) + 2 x 55(CO2) + 2 x 1(O2) = 495

The theoretical oxygen will therefore be:

hOmol /5.462)495()1072(442 221

4

1

The theoretical air will be:

hairmol /4.220221.0

5.462

Actual air (from 10% excess air)

By definition:

1.0

airltheoretica

airltheoreticaairActually

Hence:

Actual air = 1.1 x theoretical air = 1.1 x 2202.4 = 2422.6 mol air/h

This air contains:

N2,N2 = 1914 mol/h (79%)

N2,O2 = 508.8 mol/h (21%)

ELEMENT BALANCES

N: 2N3,N2 2 x 1914 2 x 29 = 0

H: 2N3,H2O 2 x 528 4 x 4 = 0

C: N3,CO2 55 383 = 0

O: 2N3,O2 2 x 1 2 x 508.8 + 2N3,CO2 2 x 55 + N3,H2O 383 = 0

From the first 3 balances, it follows immediately that:

N3,N2 = 1943 mol/h

N3,H2O = 536 mol/h

N3,CO2 = 442 mol/h

Then, from the oxygen balance

N3,O2 = 46.3 mol/h

The composition of the flue gas is therefore:

xCO2 = 0.1490

xH2O = 0.1806

xO2 = 0.0156

xN2 = 0.6548

how do combustion parameters like O2, CO, CO2, flue gas temperature and

smoke relate to combustion efficiency?

If it was possible to have perfect combustion, CO2 would be maximized and O2

would be at zero in the flue gas. Since perfect combustion is not possible due, in

part, to incomplete mixing of the fuel and air, most combustion equipment is set

up to have a small percentage of excess oxygen percent.

On the other hand, the lower the flue gas temperature for a given O2 load the

higher is the combustion efficiency. This is because less heat is carried up the

combustion gases. Smoke is the usual indicator of incomplete combustion in oil

burners (formation of elemental carbon root).

MORE ON STOICHIOMETRY OF COMBUSTION

A chemical equation can easily be derived for complete combustion of any fuel

provided its chemical formula (either structural or imperial) is known:

2222

24

424

wSOOHxCO

Owzyx

SOHC

y

wzyx

For example, 1 mol of methane (CH4; x = 1, y = 4, z = 0, w = 0) requires 2 mol

of oxygen for complete combustion to 1 mol of CO2 and 2 mol of water. If air is

the oxidant, each mol of O2 is accompanied by 3.76 mol of N2. The volume of

theoretical oxygen (at 1 bar and 298 K) needed to burn any fuel can be calculated

from the ultimate analysis of the fuel as follows:

323241245.24/2

3 SOHCfuelkgOm

Where C, H, O and S are decimal weights of these elements (on dry mass basis)

in 1 kg of fuel.

EXAMPLE 3.10: An unknown hydrocarbon gas is burned with dry air. The dry

basis (ORSAT) analysis of the product gas shows 1.5% CO, 6.0 CO2, 8.2 O2,

84.3% N2. There is no atomic oxygen in the fuel.

- Calculate the Carbon

Hydrogenratio in the fuel and speculate what

hydrocarbon (saturated) the fuel might be.

- Calculate the %excess in fed to the furnace.

Note the following:

- Gas analyses are expressed on a volumetric basis which according to: PVi =

niRT translate into molar basis.

- Saturated hydrocarbon: CnH2n+2 (n = 1, 2, 3,).

- For convenience, treat the water of combustion as a separate stream

(because of dry basis used for flue gas)

OHnnCOOnn

HC nn 22222 )1(4

224

FURNACE1

2

3

4

N3x3,CO = 0.015

x3,CO2 = 0.060

x3,O2 = 0.082

x3,N2 = 0.843

N4 (H2O)

N1x1,Cx1,H

N2x2,O2x2,N2

AIR

HYDROCARBON

SOLUTION: Number of elements = 4 (H, N, O, C)

Independence of element balances the atom matrix:

0222000

000200

210020

00011

2222

n

n

H

N

O

C

OOHHCNCOCO

0222000

000200

21010

00011

n

n

n

0222000

000100

21010

00011

n

n

n

01000

000200

21010

00011

11

n

n

n

All 4 balances are independent

DEGREE OF FREEDOM

Number of independent variables 2 + 2 + 4 + 1 = 9

Number of independent element balances 4 (C, H, O, N)

Number of specified flows 0

Number of specified composition 1 + 3 = 4

Number of subsidiary 0

Basis of calculation 1

Degree of freedom 9 9 = 0

The most useful basis of calculation appears to be: N3 = 100 kmol/h

OXYGEN BALANCE:

0.015N3[CO] + 2 x 0.060N3[CO2] + 2(0.082N3-0.21N2)[O2] + N4[H2O] = 0

NITROGEN BALANCE:

2(0.843N3 0.79N2) = 0

CARBON BALANCE:

0.015N3[CO] + 0.060[CO2] x1CN1[HC] = 0

HYDROGEN BALANCE:

2N4[H2O] (1 x1C)N1[HC] = 0

From Eq. (2): N2 = 106.7 kmol/h

From Eq. (1): N4 = 14.9 kmol/h

From Eq. (3): x1CN1 = 7.5 kmol/h

From Eq. (4): N1 = 37.3 kmol/h

And then from Eq. (5): x1C = 0.201 and x1H = 1 0.201 = 0.799

Therefore:

98.3201.0

799.0

1

1

11

11 C

H

C

H

x

x

Nx

Nxfuelin

Carbon

Hydrogen

From the saturated hydrocarbon formula:

nn

Carbon

Hydrogen 22

Hence

101.198.322

nn

n

And most likely formula is:

422 CHHC nn

The theoretical O2 required is:

hOkmolNxNx HC /96.144

799.0201.03.37 2114

111

The theoretical air required is:

hkmol /24.7121.0

96.14

Actual air supply: N2 = 106.7 kmol/h

)arg(%8.4924.71

24.717.106100

100%

elvery

airltheoretica

airltheoreticaairActualairExcess

EXAMPLE 3.11: Combined mass and energy balances for a double-drum coal-

fired boiler

Find:

a) Steam production per 100 kg dry coal

b) Carbon usage efficiency

c) Thermal efficiency of the boiler

d) Overall thermal efficiency of the boiler

e) Theoretical adiabatic flame temperature

Given data:

Coal Cp = 1.09 kJ/kg.K (dry)

Air Cp = 29.2 kJ/kmol.K (25 to 350C)

Ash Cp = 0.96 kJ/kg.K

Table of mean molar Cp values (mass%, dry): C = 73%, H = 0.6%, N = 1.0%, O

= 0.3%, S = 2.1%, inerts = 23%

Total coal moisture as fired = 6.0% (by mass, net) GCV formula (empirical

heat of reaction, assumed temperature independent):

GCV(dry) = 340.9(C*%) + 1323(H%) + 68.4(S%) 15.3(Ash%) 119.9[(O +

N)%], kJ/kg

Where C*% stands for the burnt carbon,

Assumption to be made:

Steady state operations

Complete of S and H in coal

Combustion of C to CO2, but some carbon remains unreacted

Adiabatic operation (no heat losses, perfect insulation from the surrounding)

MASS BALANCING

Net output flowrates

232 21.0010.0: 2 NNNO O

232 79.0,: 22 NNNN NN

22,: 32 COCO NNCO

22,: 32 SOSO NNSO

18

06.0,: 132 22

FNNOH OHOH

)(: 11 massFForNNCoal coalcoal

)(: 4 massFFAsh Ash

Element balances

Choose a basis: hkgFdry /1001

Hence

)(/38.10606.01

1001 wethkgF

012

)06.01)(73.0(

12

30.0:

2

ashashCO FFNC

)1(005718.0025.0 14,3 2 FFN N

01

)06.01(006.02:

2

coalOH FNH

)2(000615.0 1,3 2 FN OH

014

)06.01(01.02:

2

coalN FNN

)3(00006714.058.12 12,3 2 FNN N

016

)06.01(003.0222:

2222

coalOHSOCOO FNNNNO

)4(000351.0

242.002.0

1

,3,3,323 222

F

NNNNN OHSOCO

032

)06.01(021.02:

2

coalSO FNS

)5(00006169.0 1,3 2 FN SO

070.0)06.01(023.0: coalcoal FFInerts

)6(0216.070.0 14 FF

Note: Eq. 6 is mass balance, the molecular mass-whatever it is cancels out)

Solving Eqs. (1) to) (6) at F1 = 106.38 kg/h gives:

F4 = 32.86 kg/h (from Eq. (6))

N3,CO2 = 5.26 kg/h (from Eq. (1))

N3,H2O = 0.655 kmol/h (from Eq. (2))

N3,SO2 = 0.066 kmol/h (from Eq. (5))

Equations (3) and (4) have three unknowns at this stage: N3,N2; N3 and N2.

However, the following holds:

(1 0.01)N3 = N3,N2 + N3,CO2 + N3,SO2 + N3,H2O

0.99N3 = N3,N2 + 5.983

By elimination of variables:

N3 = 2.917 kmol/h

N3,N2 = 21.656 kmol/h

N2 = 27.367 kmol/h

Flow rates in streams 5 and 6 must be obtained from energy balance.

ENERGY BALANCING

Reference state for enthalpy: 1 atm, 25C (became Cp Data given in this stage)

[Energy In in air + coal] + [Heat from combustion] = [Energy transferred to

generate steam]

Energy In in air (sensible heat), at 35C

hkJtCNHN refp /7991)2535(*2.29*367.2735 2222

Energy In in coal (sensible heat) at 35C

Dry coal

hkJtCFHF refdry

p

drydrydry /1090)2535(*09.1*100351111

Coal moisture

hkJHHFHF OHOHOHOHOH /5.266)8.1046.146(*383.6 2535,1,1,1 22222

Total: 1090 + 266.5 = 1356.5 kJ/h

Heat from combustion (assumed temperature independent) %C* in the GCV

formula stands for the burnt carbon.

hkgFcarbonTotal dry /73100*73.073.0 1

This can be treated in the GCV formula as inert ash

Carbon in ash = 0.3 x F4 = 0.3 x 32.86 = 9.86 kg/h

Carbon burnt = 73 9.86 = 63.14 kg/h (63.14%)

Thus, specific heat of combustion (on dry basis):

coaldrykgkJ

GCVHC

/21804)3.01(9.119

86.32*3.151.2*4.686.0*132314.63*9.340

Total heat from combustion

hkJHF Cdry /218040021804*1001

Energy Out in Ash (sensible heat), at 120C

hkJtCFHF refp /6.2996)25120(96.0*857.32120 4444

Energy Out in flue gases, at 150C

hkJ

HHNtCNHN lOHgOHOHi

OHrefpi i

/13815128797109354

150 25 )(150

)(,3,333 22

2

2

Energy transferred for the steam generation

Qs = 7991 + 1356.5 + 2180400 2996.6 + 138151 = 2 048 600 kJ/h

Flow rates F5 and F6 (water demand)

Obviously, F5 = F6 and heat transferred to this stream is

sQHHFHFHF 5665566

From steam Tables:

kgkJliquidCbarH /5.146,35,1 05

kgkJsteamCbarH /2.3229,400,32 06

Thus

hkJF /5.6695.1462.3229

20486006

COAL COMBUSTION EFFICICIENCY (Carbon usage efficiency)

%5.86100100*73.0

857.32*30.0100*73.0100

coalinC

ashinCcoalinCC

Thermal efficiency of the boiler

100

dtransferreHeat

dtransferreHeatT

Heat transferred = 2 048 600 kJ/h

Heat supplied = (sensible heat in coal and air) + (heat from combustion)

= 7991 + 1356.5 + 2180400 = 2189748 kJ/h

Thus:

%6.931002189748

2048600

T

Overall Thermal efficiency

100sup

pliedheatMaximum

dtransferreHeatOT

The heat input if all the carbon had burnt = Maximum CH (combustion)

= (340.9 x 73) + (1323 x 0.6) + (68.4 x 2.1) (15.3 x 23) (119.9 x (1 + 0.3))

= 25315.4 kJ/h

Maximum heat supplied = 7991.2 + 1356.5 + 100 x 25315.4 = 2 540 888 kJ/h

Thus

%6.801002540888

2048600

OT

THEORETICAL FLAME TEMPERATURE, Tf

The theoretical flame temperature results from the energy balance written for the

furnace only (before the flue gas reaches the steam generator).

[Energy In in air + coal] + [Heat from combustion] = [Heat lost in ash] + [Latent

heat to vaporize the coal moisture + combustion water at 25C] + [sensible heat

to raise the flue gas temperature from 25C to Tf]

i

fpi TCN i 25287976.299621804005.13567991 ,3

hkJTCNi

fpi i/215795425,3

This equation is nonlinear because Cpi = f(Tf) and can be solved iteratively. The

suggested method is:

(i) Tf, say Tf = 2000C

(ii) Calculate 8.1054,3 i

pi iCN

(iii)

i

pi

f

iCN

T,3

215795425

(iv) Calculate a new 8.1054,3 i

pi iCN

(v) CTf02064

5.1058

215795425

This is called the theoretical (adiabatic) flame temperature, because no heat

losses to the surroundings were assumed (perfect insulation of the furnace).

Heat Recovery Steam Generator

Double-drum single pressure boiler with natural circulation

Steam

Superheated steam value 5.3 t/h

Superheated steam pressure 1.5 MPa

Superheated steam temperature 375C

Feed water temperature 105C

Condensate

Value 19.8 t/h

Inlet pressure 0.4 MPa

Temperature inlet/outlet 55/75C

The cogeneration unit including a heat recovery steam generator was installed at

the home facility. Besides power and steam generation, it serves as a main

reference unit for prospective customers. It also enables testing of further

technical improvements in real conditions of regular operation. The double-drum

(with tube bundle between drums)is designed as a unit with natural water

circulation. It is seated on the supporting structure above the gas turbine. Starting

from the boiler inlet, flue gas horizontally flows through steam superheated made

by horizontal coils and vertical headers, then through the evaporator bundle

between the drums. In a vertical part of the flue gas duct there are water and

condensate heater bundles with horizontal headers. The flue gas boilers pass is

made of sheet, inner overpressure resisting duct, externally insulated. There is a

silencer and a short stack at boiler outlet. All the heating surfaces are drainable.

Instrumentation and boiler control system is modified to be used for attendance-

free operation and automatic start-ups and shut-downs.

1.5 THE RELATIONSHIP BETTWEN ELEMENT AND SPECIES

BALANCES

Element balance solution is a necessity when

(i) Elemental composition are the only compositions available, or

(ii) No composition data are available and the species present are not

known, or

(iii) Reactions are occurring but the stoichiometry is unknown or very

complex.

Apart from the above, preference is given to the method that offers the best

opportunity of obtaining a unique solution with the last effort and the least

amount of data collection and measurement of stream variables (specified

values). The following rules can be formulated.

Element balances are: ),....,2,1(01

EiNS

i

iei

Species balances are: ),....,2,1(0 SiNi

All the S species balances are independent, however not all E element balances

have to be independent. If is the number of independent element balances

then the two approaches are equivalent if S (since we can write the

same number of independent equations). If S then preference is given to the species balances. Using the matrix reduction procedure to the atom matrix

it can be shown easily that the case S is impossible (we always have

S )

REACTING SYSTEM

Consider a chemically reacting system involving V stream variables and S

species. Suppose that the stoichiometric coefficient of all reactions are known

and out of R reactions are independent. According to the DOF analysis for

reacting system, the system will involve V variables ans S independent

species balances, hence it will have SV . Degrees of freedom. On the

other hand, suppose the system involves E elements and of the element

balance are independent. Then, VDOF . The two approaches are equivalent and can be used interchangeably when (DOF) species = (DOF)

element:

VSV

Or

S

If S , then the species balance balances are preferred, otherwise the element balances should be used.

EXAMPLE 3.12: Nitrogen dioxide, NO2, is produced by the catalytic oxidation

of ammonia at elevated temperatures. The reactor product streams is cooled and

the passed to an absorber as shown in Figure below, in which the nitrogen

dioxide reacts with water to form nitric acid, HNO3. In the absence of oxygen,

nitric oxide (NO) and nitrous acid, HNO2 are also formed. In the gas phase, the

nitrogen dioxide exists in equilibrium with nitrogen tetroxide, N2O4 and the

N2O4/NO2 equilibrium constants are 0.017 and 0.021 m3/kmol for the absorber

gas inlet and gas outlet temperature and pressures. The outlet liquid stream

(stream 4) has a flow rate of 175 kmol/h. three independent chemical reactions

have been identified to occur in the absorber, namely:

12 422 ONNO

22 2322 HNOHNOOHNO

3322 NOHNOHNONO

1 3

42

HNO3 = 125. mole %

HNO2 = 0.30 mole %

H2O

N4 = 175 kmol/h

N2O4NO2NO

N2

Combined (N2O4 + NO2) = 0.005 mole %

021.023

3

2

42

NO

ON

N

NK

N2O4NO2N2

Combined (N2O4 + NO2) = 6 mole %

017.022

2

2

42

NO

ON

N

NK

H2O

In order to design the plant it is necessary to carry out the material balance across

the reactor. Two possible methods are available which could lead to a unique

solution for the flow rates and compositions of the inlet and outlet streams,

namely Species Balances and Element Balances.

Determine which of the two methods is best suited to finding the unique solution.

Write down a complete set of independent balance equations including the

subsidiary relations for the method selected in (a)

SOLUTION:

1 3

42

HNO3 = 125. mole %

HNO2 = 0.30 mole %

H2O

N4 = 175 kmol/h

N2O4NO2NO

N2

Combined (N2O4 + NO2) = 0.005 mole %

021.023

3

2

42

NO

ON

N

NK

N2O4NO2N2

Combined (N2O4 + NO2) = 6 mole %

017.022

2

2

42

NO

ON

N

NK

H2O

(a) (i) Species balances:

Reactions:

12 422 ONNO

22 2322 HNOHNOOHNO

3322 NOHNOHNONO

Number of independent stream variables = 11 + 3 = 14

Number of independent species balances = 7

Number of specified values = (1 flow + 2 comps) = 3

Number of subsidiaries: combined mole%s = 2

Equilibrium K valves = 2

Total = 14

Degree of freedom = 14 (7 + 3 + 2 + 2) = 0

Therefore, a unique solution to the species balances is possible.

(ii) Element balance

210

101

042

002

111

011

001

210

121

002

042

131

011

021

2

2

2

42

3

2

OH

HNO

N

ON

HNO

NO

NO

HONHON

Note: A transpose of the atom matrix is used here which is acceptable with

columns rather the rows.

Number of independent stream variables = 11

Number of independent species balances = 3

Number of specified values = (1 flow + 2 comps) = 3

Number of subsidiary relationships = 4

Degree of freedom = 11 10 = +1

The specified values include a flowrate, so there is no option of a free choice of

basis.

Therefore, the species balances are a better choice than the element balances.

Alternately: S = 7 species

reactionstindependen3

balanceselementtindependen3

437S

Therefore, species balances are preferred

(b) Species balances

Reactions:

12 422 ONNO

22 2322 HNOHNOOHNO

3322 NOHNOHNONO

N2O4 NO2 NO N2 H2O HNO3 HNO2

1 2 3 4 5 6 7

1

22

1

33

142 : rNxNxON

321

22

2

33

22 22: rrrNxNxNO

3

33

3: rNxNO

22

4

33

42 : NxNxN

2

144

52 : rNNxOH

23

44

63 )175*125.0(: rrNxHNO

32

44

72 )175.0*003.0(: rrNxHNO

Subsidiaries:

00005.0323

1 xx

060.0322

1 xx

17.0)(

)(22

2

2

2

42 NO

ON

N

NK

021.0)(

)(23

3

3

2

42 NO

ON

N

NK

A NOTE ON SYSTEMS INVOLVING MULTIPLE REACTIONS AND

MULTIPLE UNITS

In systems involving multiple reactions and multiple units, some species may

form in one unit and disappear in other unit. As a result, those species may be

present neither in the input streams nor in the output streams of the system.

REACTOR I REACTOR IIA, B 1 A, B, C, D 2

3E

4A, B, D, E, F

DCBA FEC

In the above example, if the second reaction goes to completion, the species C is

not going to appear in stream 4. However, the overall balances should be written

(and taken into account in the DOF analysis) for all the species of the system, no

matter whether they are present in the external (input/output) streams or not.

Reason: Even though the input and output flows of these species are zero, the

corresponding overall species balance do involve reactions rates. As such, their

role is to establish some relationships between these rates.

Reactor I Reactor II Process Overall

Variables 6 + 1 10 + 1 12 + 2 6 +2

Balances:

A 1 1 2 1

B 1 1 2 1

C 1 1 2 1!!

D 1 1 2 1

E 0 1 1 1

F 0 1 1 1

SUBTOTAL 4 6 10 6