Handbook_on_Material_and_Energy_Balance_Calculat1

Chapter 6 Reactive Material Balances 325

A very rapid change occurs between 99.8 % and 100.2 % of stoichiometric air. The dramatic
difference in /?CO, /?H2, and p02 from XSF to XSA effects on the tendency for the combustion
products to react with the material being heated. The change in p02 is 11 orders of magnitude,
while that for CO and H2 is about five orders of magnitude. The volume fraction of C02 and H20
remain nearly constant at about 10 % and 20 % respectively.

This shows the importance of close control of the fuel and air flows when the material being
heated is susceptible to oxidation. Where oxidation is a problem, combustion can be confined to
the inside of alloy steel tubes, which radiate heat to the material. The combustion gases are thus
kept completely separate from the material being heated.

Fuel combustion for heating purposes uses excess air (XSA) to assure complete combustion,
which eliminates virtually all of the CO and H2 from the product gas. The fuel is the limiting
reactant for XSA combustion, so if the reactants are well mixed, combustion of the fuel
hydrocarbons is spontaneous, complete and irreversible. If the burner is well designed and the air
and fuel well mixed, 10 % XSA is generally adequate for complete combustion. However, the
recent emphasis on fuel efficiency means that some process operators are striving to use as low as
3 % XSA. The thermal efficiency is improved by minimizing the amount of XSA and enriching
the air with oxygen. Oxygen enrichment also increases the flame temperature. Industrial oxygen
(-96 %02) is used as the oxidant where the hottest flame is desired. Material balance calculations
involving oxygen enrichment can be handled in two ways. First, by keeping the air and oxygen as
separate reactant items. Second, treating the oxidant as a single gas containing a specified volume
or mass fraction of oxygen and nitrogen.

EXAMPLE 6.4 — Combustion of Natural Gas with XSA.

Natural gas (NG) is burned with 15 % XSA using air that has a dew point of 23 °C. The NG
has <jpC02 and <jpC2H6 = 1.0 % (rest CH4). Calculate the volume fraction of moist air in the fuel/air
mixture, and the pounds of moist air/pound of fuel. Use precise values for the <jp02 in air.

Solution. The calculation basis chosen for the first part of the example is stoichiometric
combustion of one mole of fuel with moist air. The air requirement can then be multiplied by 1.15
to get the air for 15 % XSA. The following two reactions express the combustion of hydrocarbons
in the fuel:

CH4(g) + 202(g) -+ C02(g) + 2H20(g) [6.36]

C2H6(g) + З Ж Ш - 2C02(g) + 3H20(g) [6.37]

The <jp02 in moist air at a dpt of 23 °C is calculated from Equation [6.28] as 20.38 %02. The
moist air requirement for Equation [6.36] is then 2(0.98)/0.2038, or 9.617 mole. The moist air
requirement for Equation [6.37] is 3.5(0.01)/0.2038, or 0.172 mole, for an air/fuel mole (or
volume) ratio of 9.79 for stoichiometric combustion. At 15 % XSA, this converts to an air/fuel
volume ratio of 11.26, or <pNG = 8.16 % in the mix.

The basis for the second part of the calculation is one lb of dry air. The mass fraction of С, Н,
and О in the NG was calculated by MMV-C or from Equation [1.13]:

wH-0.2437; wC = 0.7369; wO-0.0194

Application of Equation [6.35] gives 16.75 lb dry air/lb fuel for stoichiometric combustion,
which ratio is 19.26 for 15 % XSA. The correction for moist air is based on Equation [6.29],
which indicates that the mass of moist air is 1.018 times that of dry air. Thus, the combustion
process requires 19.61 lb of moist air/lb of fuel.

Assignment. An industrial heating process uses the above NG with a feed-forward controller to
maintain a mass ratio of air/NG of 20. In the winter, the average dpt is 0 °C, while in the summer
it is 21 °C. Calculate the % XSA for the two different seasons.

326 Chapter 6 Reactive Material Balances

EXAMPLE 6.5 — Effect of Oxygen Enrichment on the Oxidant Requiredfor Complete Combustion.

In some cases, the increase in thermal efficiency obtained by enriching the air with oxygen
outweighs the added cost of the oxygen. Determine how much moist air (dpt of 20 °C) and
industrial oxygen (<jp02 = 0.97) is required for the combustion of one mole of scrubber gas from the
Figure 4.4 flowsheet. Make calculations as a function of oxygen enrichment at 18 % XSO (excess
oxidant). The fuel (scrubber gas) for this example has the following composition:

q>CO = 0.286; <pC02 = 0.460; φΗ2 = 0.192; <рН20 = 0.020; g>N2 = bal.

Solution. The relevant equations for the combustion of CO and H2 of the fuel are:

CO(g) + V202(g)^C02(g) [6.38]

n2(g) + y202(g)^n20(g) [6.39]

On a basis of one mole of fuel, the stoichiometric oxidation of the CO requires 1^(0.286) =

0.143 mol 02, and the stoichiometric oxidation of H2 requires V2(0.192) = 0.096, for a total
stoichiometric 0 2 requirement of 0.239 mol. At 18 %XSO, 0.282 moles of 0 2 are required. This
must be supplied by some combination of moist air and industrial oxygen.

The degree of oxygen enrichment may be expressed as a percentage or fraction of the moist
air (φ02 = 0.2048) replaced by industrial oxygen (<p02 = 0.970). For example at 10 % oxygen
enrichment, the oxidant would consist of a gas having <p(moist air) of 90 % and industrial oxygen
of 10 %. (100 % oxygen enrichment means no air.) The following two equations were used to
calculate the oxidant required for 18 % XSO.

0.282 = - ί ϋ - + .Κ [6.40]
0.2048 0.97

OXNR = -IX IX [6.41]
+ MA

where MA refers to moles of moist air, IX refers to moles of industrial oxygen, and OXNR is the
fractional degree of oxygen enrichment. An Excel worksheet was set up to solve the two equations
for six different degrees of oxygen enrichment. OX refers to the total moles of oxidant (MA + IX).
Figure 6.21 displays the relevant results. Note that with no oxygen enrichment, 1.38 moles of
moist air are required, and with no moist air, 0.29 moles of industrial oxygen are required.

Oxygen Enrichment for Fuel Combustion

1.4 \/>\ IVM1AM

1.2 \ - О -IX

1.0 V A — Δ - OX

О

О 0.8 X L^ 4

0.6 ч ч . ^ - -A- - -

0.4

0.2 . . . r

0.0 a-2 0.2 0.4 0.6 0.8 — — <!>

0.0 degree of oxygen enrichment (OXNR) 1.0

Figure 6.21 Amount of moist air (MA), industrial oxygen (IX), and total oxidant (OX) required
for 18 % XSO combustion of one mole of ironmaking scrubber gas.

Assignment. Calculate the mass ratio of oxidant/scrubber gas for the example.

Chapter 6 Reactive Material Balances 327

6.3.2 Combustion of Liquid Fuels

The refining of petroleum produces a wide variety of hydrocarbon substances for use by
industry and other consumers. Generally, up to 10% of the crude oil may be refined into fuels for
industrial use. Distillate fuel oils consist of the fraction intermediate between kerosene and
lubricating oil. Residual fuel oils consist of viscous products remaining after all of the more
volatile hydrocarbons have been distilled off. ASTM has developed a grading system for fuel oils,
with Grades 1, 2, and 3 used for domestic and small industrial uses. Grade 5 is cracking-still tar,
and Grade 6 is a mixture of residual and tar-like substances. Grade 6 fuel oil is typically for large
industrial and commercial use. Fuel oils have a mass fraction of carbon between 86 % and 89%,
with the rest being mostly hydrogen. The oil also has small amounts of N, S, and ash-forming
ingredients.

The proper combustion of liquid fuels requires that the air and fuel be well mixed. This is
done by atomizing the fuel with steam or air before mixing with the bulk of the combustion air.
Owing to the viscosity of Grade 6 oils, they are heated to as high as 95 °C before being delivered
to the burner. Complete combustion of fuel oil may require a higher degree of excess air than does
natural gas. For well-designed burners, 1 2 - 1 5 %XSA is adequate for complete combustion.

Many of the material balance concepts developed for the combustion of gaseous fuels is
applicable to the combustion of liquid fuels. Section 5.4 discussed the use of stoichiometric
reactions for calculating the oxygen requirement for combustion of fuel oil, and should be
reviewed before proceeding with this Section. Equations [6.32 - 6.35] are also applicable to fuel
oil combustion.

The discussion in Section 5.4 used an approach of writing two separate equations for the
combustion of the С and H fractions of the fuel oil. An alternative approach is to determine an
empirical formula for the fuel oil, and write one equation that embodies the combustion of both H
and С This approach is satisfactory for simple reactant substances like fuel oil if it consists of
only two or three elements. In the original case the oil had wC = 0.87 and wH = 0.13, so the mole
ratio of H/C = 128.97/72.43 = 1.7806. This fuel oil has an empirical molar mass of 13.806, and
one kg contains 72.43 moles of С atoms and 129.0 moles of H atoms. The empirical formula for
fuel oil is CHi.7806,* and gives the following combustion equation:

СН1780б(/) + 1.445202(g) -> C02(g) + 0.8903H2O(g) [6.42]

The stoichiometric combustion of one mole of CHUSOÓ requires 1.445/0.21 = 6.88 moles of

dry air (assuming <jp02 in air to be 21.0 %). Assuming that the Mair is 29.0 g, the mass of
stoichiometric dry air per kg of oil is:

6.88(72.43)(0.0290) = 14.45 kg air/kg fuel oil.

This is in very good agreement with the value of 14.5 obtained from Equation [6.35].

It's sometimes convenient to express the empirical formula for one kg of oil. In the above

case, the oil formula would then be C72.43Hi29. Thus the heat of combustion of one mole of this
fuel oil is the same as the heat of combustion of one kg..

6.3.3 Combustion of Solid Fuels

Solid fuels (particularly coal) play a major role in generating electric energy and as a fuel and
reductant in ferrous extractive metallurgy. Coal is the product of physical and chemical alteration
of plant material laid down millions of years ago. The initial deposit was in the form of a brown
fibrous material called peat, which slowly evolves into lignite, then bituminous coal, and finally
anthracite. The bituminous coals are of greatest interest to the electric utilities because of their
abundance and widespread availability. They are also of great importance to the steel industry
because of their ability to form coke.

The empirical formula could also have been written С0.5б1бН.

328 Chapter 6 Reactive Material Balances

There are two different analytical methods for characterizing the composition of coal, both of
which are reported on a dry basis. A proximate analysis involves heating the coal in the absence of
air to drive off the volatile constituents, and analyzing the residue for carbon and ash. An ultimate
analysis determines the amount of carbon, hydrogen, oxygen , nitrogen, sulfur, chlorine, and ash.
The ash-forming minerals change composition during combustion, but this is usually ignored.
Coal has a large range of compositions. Typical compositions of three different bituminous coals
are shown below.

Proximate Analysis (%) Ultimate Analysis (%)

Coal Type Volatile Fixed Ash Carbon Hydrogen Nitrogen Sulfur Oxygen
Matter Carbon

Low-volatile 16 79 5 85.5 4.8 1.5 0.8 2.6
3 86.5 4.9 1.6 0.6 3.6
Medium-volatile 22 75 7 79.5 5.2 1.5 1.3 6

High-volatile 34 60

Coal is prepared for use by crushing and screening, followed by washing processes to remove
as much of the ash-forming materials as possible. Sulfur, which is present in the coal as pyrite
(FeS2) and an organic form, is only partly removed by washing. For use as a fuel, the washed coal
is then air-dried and pulverized. The combustion of pulverized coal is rapid due to the suspension
of particles in the air, and introduction of the mixture into a high temperature environment. The
combustion air is often preheated to further dry the coal and improve ignition. The release of
volatile matter is practically instantaneous, and the combustion of the remaining carbonized
particles is rapid. Virtually complete combustion occurs with 1 5 - 2 0 % XSA. There are two
main problems associated with the use of coal. One is the generation of ash particles and the other
is the generation of S02. Dust collectors are used to control stack particulate emissions, and
scrubbers may be required to lower the SO2 in the exhaust gas. Other coal-burning emittants of
increasing concern are nitrogen oxides (NOx) and mercury.

The complex composition of coal and the ambiguity of the ash-forming mineral composition
make it impractical to write balanced chemical equations for every coal element undergoing
chemical change during combustion. Instead, the ultimate analysis is used in connection with
Equation [6.35] to calculate the mass of air required. Then an element balance for C, H, O, S, and
N can be made with the understanding that an overall material balance may not close perfectly. If
the amount of halides and alkali metals are important, they can be included in the material balance
if appropriate information is available.

EXAMPLE 6.6 — Stack Gas Composition and Dew Pointfor Coal Combustion with Dry Air.

A high-volatile bituminous coal containing 3.8 % water is burned with 15 %XSA. The
ultimate analysis of the dry coal is 78.3 %C, 5.2 %H, 1.5 %N, 1.4 %S, and 8.2 % 0 . Calculate the
mass and volume (at STP) of dry air required for combustion of 1 lb of moist coal, the stack gas
composition, and its dew point.

Solution. The mass of dry air required for 15 %XSA (%XSA here refers to mass units)
combustion of dry coal is obtained from Equation [6.35]:

(1.15) ШГ = 11.5(0.783) + 34.3(0.052) + 4.3(0.014 - 0.082) = 12.1

lb of coal

The density of dry air at STP is 0.08067 lb/ft3, for an air volume of 150 ft3 dry air/lb dry coal.

However, 1 lb of moist coal contains 0.9634 lb of dry coal, so the combustion requirements are
11.7 lb dry air/lb moist coal, and 145 ft3 dry air/lb moist coal.

The composition of the stack (combustion) gas is obtained by making a ledger for each
reactant element, and distributing amounts of each to the known product species. The amount of
reactant elements is obtained by dividing the mass by the molar mass.

Chapter 6 Reactive Material Balances 329

Quantity of Elemental Reactants to Combustion Process

Element С H S О N Ar total

mass from dry coal, lb 0.7543 0.0501 0.0135 0.0790 0.0145 — 0.911

mass from moisture, lb — 0.0043 — 0.0337 — — 0.038

mass from air, lb — — — 2.7999 9.1367 0.1549 12.092

lb-mol 0.06281 0.05392 0.00042 0.18204 0.65333 0.00387 0.956

The product species from chemical reactions are CO2, H20 and S02, while some additional
H20 comes from coal moisture. The N2 and 0 2 come from coal and air, while the Ar comes from
air. The volume of each species is based on 359 ft3/lb-mol. The volume calculation for all gases

except 0 2 is straightforward. The total О in must be decreased by twice the amount used to form
C02 and S02, and half the amount used by H. The remainder is present as 02. In this example,
precise data was used for the composition of air, but in practice, the Ar is often considered as part

of the "N2", with only a small attendant loss in accuracy.

Quantity of Product Species from Coal Combustion at 115 % Stoich. Air

Species C02 H20 so2 o2 N2 Ar total

lb-mol 0.0628 0.0270 0.00042 0.0143 0.3267 0.0039 0.4350
voi, ft3 at STP
22.55 9.68 0.15 5.14 117.27 1.39 154.8
φ,%
14.6 6.3 0.10 3.3 75.8 0.9 100.0

The/?H20 in the stack gas is 0.063 atm. The dew point can be obtained from Equation [2.15]
as 311 K, or 100 °F. This result identifies an important aspect when sampling stack gases: if the
stack gas cools below the dew point anywhere in the sampling train, some moisture may condense,
thus affecting the accuracy of the analysis.

Assignment. Calculate the stack gas composition and dew point if the coal were burned with air
having a dpt of 72 °F.

6.3.4 Use of Feed-Forward and Stack Gas Analysis for Combustion Control

Fuel and air composition are important criteria for making combustion material balances.
Feed-forward control is commonly used for the combustion process itself. The difficulty with this
method is that other reactions might be taking place in the furnace that affect the atmosphere in the
furnace. Air can leak into the furnace at charge point openings, the composition of the fuel can
fluctuate, moisture or other gases may evolve from the materials being heated, and combustion
gases may react with materials in the furnace. For these and other reasons, control of the furnace
atmosphere may best be done by analysis of the stack gases, with the attendant use of a feedback
controller to regulate fuel and/or oxidant supply.

If the composition of the fuel were accurately known (as would be likely for natural gas or

fuel oil as fuels), it's possible to develop a relationship between the composition of the combustion
gas, the % XSO, and the <jp02 in the oxidant gas. Usually only φ02 in the combustion gas is

calculated because a certain minimum oxygen amount is necessary to assure complete oxidation of

the fuel. Suppose we need to develop a relationship for the XSO combustion of natural gas (NG)

with air having variable oxygen enrichment. Following the procedures described earlier for

making combustion balances, the first step is make calculations on the reactants and products for

stoichiometric NG combustion. For a given NG hydrocarbon, each С atom requires one molecule

of 02, and each H atom requires % molecule of 0lA2 . For the products, each С atom produces one
molecule of C02 and each H atom produces molecule of H20. Table 6.12 shows these

relationships for the predominant NG hydrocarbon species.

330 Chapter 6 Reactive Material Balances

Table 6.12 Oxygen demand and hydrocarbon combustion product amounts for NG combustion.

Species CH4 СгН6 СзНз C4H10

w02/nspecies 2 3.5 5 6.5

n comb, products 3 57 9

In addition to the hydrocarbon combustion products, the combustion gas also contains the
inert substances present in the NG and those present in the oxidant gas. For combustion air with a
dpt of 20 °C, Table 6.11 shows that the oxidant gas (i.e., the humid air) has φ02 = 20.5 %, so the
stoichiometric oxidant gas demand is obtained by dividing the stoichiometric oxygen demand by
0.205. The amount of inert oxidant gas substance is obtained by multiplying the stoichiometric
oxidant gas demand by the factor (1 - 0.205), or 0.795. Generic calculations are made in Excel
with formulas to calculate the stoichiometric amount of oxidant gas required for any NG burned
with any oxidant gas φ02. Table 6.13 shows the results for the combustion of 100 moles (or
volume units) of two different NG compositions using air with a dpt of 20 °C as the oxidant gas.

Table 6.13 Amount of oxidation gas and oxidation gas inerts, and NG combustion product
amount for the stoichiometric combustion of 100 units of NG of Table 6.9 compositions.

Source of NG NG comb. prod. Ox gas amount Ox gas inert amount

NA wellhead NG 306 995 791

LNG 320 1049 834

An NG with more high-molecular-mass hydrocarbons requires more oxidant gas (on a volume
basis) for combustion. This is an important factor as more LNG enters the NG pipeline system.

The next step calculates φ02 in the combustion gas at various values of percent excess air (or
more correctly, percent excess oxidant gas, designated here as % XSO). All of the excess oxidant
gas passes intact to the combustion gas, and since we know the φ02 in the oxidant gas, we can
calculate the amount of excess 02 in the combustion gas. Formulae in Excel calculated the
combustion gas <jp02 at three levels of oxidant gas (p02, at five levels of excess oxidant gas (XSO).
Figure 6.22 shows the results.

Figure 6.22 Oxygen content of the combustion gas as a function of the percent excess oxidant gas
(% XSO), for three different φ02 levels in the oxidant gas.

Chapter 6 Reactive Material Balances 331

The trendlines are nearly linear. If they were truly linear, they should extrapolate to the origin
(zero % 0 2 in the combustion gas when % XSO = 0). However, they do not, thus indicating a small
degree of curvature. Somewhat surprisingly the composition of NG had negligible effect on the
results, so the Figure 6.22 relationship holds for any reasonable NG composition. A regression
analysis of the data set gave the following equation.

%02 in CG = 0.00813(% XSO) + 0.00495(% 02 in OXG) + 0.007222(% XSO)(% 0 2 in XG) + 0.013 [6.43]

where CG refers to the combustion gas, OXG refers to the oxidant gas, and %XSO refers to the
percent excess of OXG above stoichiometric. R2 for the regression fit was 0.9992, which indicates
that the equation reliably represents the data over the range of values used for its generation.

The calculational procedure outlined above used % XSO as the independent variable, and

%02 in CG a the dependent variable. A slight modification of the Excel formulae can calculate the
% XSO needed to give a desired % 0 2 in CG.

0%/ Aγ ςb ηU = ( % O 2 i n C G ) ( N G c p + sogi)(100) L[6.44J1
( % 0 2 in OXG - % 0 2 in CG)(sog)

where NG cp refers to the natural gas combustion products as defined in Table 6.13, sogi refers to
the stoichiometric oxidant gas inerts amount, and sog refers to the stoichiometric oxidant gas
amount, all based on 100 volume (or amount) units of natural gas burned (see Table 6.13).
However, since the relationship between % XSO and % 0 2 in CG has already been established by
Equation [6.43], it's easier to use Goal Seek to determine % XSO from % 0 2 in CG.

One of the complications in controlling the flow of gases to a burner is that the combustion
flow control for many systems consists of a pressure regulator and an orifice meter. This type of
meter introduces a flow change if the fuel gas changes composition. The flowrate of a gas through
an orifice (at constant pressure drop) relative to that of some reference gas depends on the square
root of the density ratio of the gas vs. that of the reference gas. Equation [6.45] shows this
relationship for volumetric flow.

Tr t n t Volume of reference gas

Volume of natural gas = . - [6.45]

^/Density ratio, natural gas/reference gas

For example, suppose the reference gas was pure CH4, which has a density of 0.7518 g/L at
STP. If the fuel gas were to change to a composition of <pCH4 = 93 %, balance C2H6, the density
of the changed natural gas would be 0.7596 at STP. If the volumetric flowrate of CH4 was
originally 100 L/sec, the flowrate of the changed natural gas would be 99.5 L/sec. This different

flow can affect the amount of heat generated by the combustion process. The effect of natural gas

composition changes on flowrate, air requirement, and heat production are covered in detail in

Chapter 10 and in folder NG combust & Wobbe Index on the Handbook CD. The workbook also

has selected Excel programs for making combustion calculations.

You now know how to make a material balance using fuel analysis and instream flow rates
(feed-forward control). Alternatively, feedback control is a useful alternative to feed-forward,
especially if the flow rate of fuel or air is not accurate. Feedback control requires that the
combustion gas composition be used to calculate the oxidant/fuel ratio and the % XSO. Many
times, the combustion gas sample is dried before being analyzed, in which case, care must be taken
to remove the water quickly and completely. If this is not done correctly, water accumulating in
the sampling line can re-enter the sample, dissolve a little C02, and dissolve much of the S02.
Using combustion gas composition to calculate the % XSO is a "backwards"-type calculation, and
therefore requires a different approach than feed-forward. Two procedures for making such back-
calculations are described in Example 6.7.

332 Chapter 6 Reactive Material Balances

EXAMPLE 6.7 — Calculation of % Excess Air from Stack Gas Analysis.

A sample of dried stack gas from a fuel oil-fired furnace was found to contain φ02 = 3.05 %.
The oil analysis was wC = 87.5 % and wH = 12.5 %. What was the % XSA for combustion?

Solution. The arithmetic for using the gas analysis to make a material balance can use an element
balance. A convenient basis is 100 moles of dry stack gas. The air is assumed dry, with 21 %02.
Atomic species are selected for those elements that undergo chemical change, while a molecular
material balance equation is written for the inert species (here, only N2). The material balance
equations are written in a form suitable for solution by Excel's Solver tool, which means setting
the product amount minus the reactant amount = 0. Equation [6.46] relates the amount of H in the
oil to H in the water removed from the stack gas. Equation [6.47] relates the amount of С in the
dry stack gas to the amount of С in the oil. Equation [6.48] relates the amount of О in the water
removed from the stack gas plus the О in the dry stack gas, to the amount of О in the combustion
air. The last equation is self-explanatory. The unit for mass of oil is grams.

H balance: 0 = 2(пЯ20) - 0.125(/woil)/1.008 [6.46]

С balance: 0 = <pCO2(100) - 0.875(/woil)/12.01 [6.47]
О balance: 0 - nU20 + 2(3.05) + 2(<pCO2)(100) - 2(0.21)(wair) [6.48]

N2 balance: 0 - 100(1 - 3.05 - <pC02) - 0.79(rair) [6.49]

These four equations were written in formula format on a worksheet, and Solver found the
following result.

moil, g nan <pC02 wH20
184.4 105.7 0.1343 11.43

184.4 0.875 0.125
■+ -
12.01 4x1.008
The stoichiometric air for 184.4 g of oil is: ??airst = - 0.21 91.2

105 7-91 2
The % excess air is then: % excess air = 100 :
— = 15.9%

91.2

The amount of combustion gas is 111.4 moles, and <jpH20= 10.3 %.

Alternatively, you can use a spreadsheet to set up the material balance in a ledger format for
an arbitrary value of % excess air, and then use Excel's Goal Seek tool to determine the correct
value of % excess air that will give a value of 3.05 % 0 2 in the dry stack gas. This technique uses
Goal Seek to change one of the input variables in order to attain a specified output variable. In
effect, we simulate a feed-back controller by using Goal Seek. The ledger is shown below for a
basis of 1000 g of fuel oil, and the composition of the dry gas calculated. Goal Seek then varied
the cell containing the fractional excess air (upper left-hand corner) until the volume fraction of 0 2
in the dry stack gas reached 0.0305. Goal Seek found a solution at 15.9 % excess air, in agreement
with the solution found by Solver. Figure 6.23 shows a chart of the results of additional
calculations using SuperGS (an Excel program on the Handbook CD).

Reactants for Stoichiometric Combustion Products for XSA Combustion
co2 H20 02
0.159 Fuel oil С H 0 2 N2 N2

Mass, g 1000 875 125 3323 10944 3206 1117 529 12687

Amount, n 201.4 72.8 124.0 103.9 390.7 72.8 62.0 16.5 452.9

φ wet gas — — 0.21 0.79 0.120 0.102 0.0273 0.748
φ dry gas
0.1343 0 0.0305 0.8352

Chapter 6 Reactive Material Balances 333

σев> 3.6 Combustion of Fuel Oil with Excess Air
о % XSA = 5.99(% 02) ■ 2.24
-c2o 3.2 -^^^

fr л^^^

■ö 2.8

с

О 2.4

10 12 14 16 18 20
% excess air

Figure 6.23 Relationship between the % XSA and the 0 2 content of dry stack gas produced by the
burning of fuel oil of wC = 87.5 % and wH = 12.5 %. The relationship follows a near-linear trend
over the range investigated, as shown on the chart text box. The relationship should not be
extrapolated outside the range indicated in the Figure.

Assignment. 1) Calculate the % XSA for the combustion of fuel oil having wC = 0.873 and wH =
0.127 with a dry stack gas composition of <jpC02 = 12.0 %, for two cases: dry air and air having a
dpt of 20 °C. 2) Calculate the relationship between the dew point of the as-produced stack gas and
the % XSA for dry and 20 °C dpt.

The above discussion involved using the composition of the stack gas, or the combustion gas,
as an indicator of the proper air/fuel ratio to assure the desired combustion atmosphere control.
However, it's not always easy to get a good sample of the combustion gas as it leaves the burner.
That's because combustion may not be complete at the sampling point, or the air and fuel might
not have been perfectly mixed before they entered the burner. Also, taking a sample of very hot
gas (-1800 °C) requires a special sampling probe. Sampling the stack gas from a furnace has other
difficulties, such as various gases evolving from the process itself, or air leaks into the furnace.
For processes that require critical control of the furnace atmosphere, the sampling point(s) must be
chosen carefully, and the fuel and air flows measured accurately.

The various relationships between the oxygen content of the air, the composition of the fuel,
and the extent of XSA have been collected into the folder NG combust & Wobbe Index on the
Handbook CD. Formulae were derived that relate the amount of oxidant needed to attain a
specified amount of oxygen in the combustion gas. The worksheet formulae are written in a way
that does not require the use of Goal Seek.

6.3.5 Use of FlowBal for Combustion Calculations

FlowBal is very effective in making all types of combustion calculations. Consider the
combustion of a fuel oil with excess air where steam is used to atomize the oil. One kg of steam is
used per 20 kg of oil, with 350 kg/min of moist air for combustion . The combustion air has a
relative humidity of 60% at 27 °C, and the fuel oil has wC = 87.29%, wH - 12.53%, bai. S. The
splitter fraction to the dryer (S6) is 0.002, and is chilled in the dryer to 0 °C where virtually all the
water vapor condenses. Some of the S02 in the dryer ingas dissolves in the condensed water (S8).
Figure 6.24 shows a sketch of the process.

It's good practice to check that the air is truly in excess even if so stated The equations in Section 6.3.2
verify that 350 kg of air for 20 kg of fuel oil is well in excess of stoichiometric.

334 Chapter 6 Reactive Material Balances

-► Water

Figure 6.24 Flowsheet for combustion process with gas drying before sampling.

Mass units are used for all but S7. One benefit of using mass units is that the pressure and
volume of the various gas streams is irrelevant for a material balance, and for this reason, many
industrial controllers use mass flow measuring devices. Volume units are readily obtainable from
the MMV-C program, or from FlowBal's amount result table. Table 6.10 gave air composition in
mass units. Ignoring the trace of C02, the (humid) combustion air composition is:

wN2 = 0.7464; w02 = 0.2288; wU20 = 0.0122; bai Ar

As a check, the PsyCalc 98 program verified that wH20 = 0.0122 (12.4 gH20/kg dry air).

Equations [6.50 - 6.52] give the chemical reactions taking place in the burner. Owing to the
use of excess air, the three reactions are spontaneous and complete.

С + 02(g) — C02(g) [6.50]
2H + 1/202(g)^H20(g) [6.51]

S + 02(g)-+S02(g) [6.52]

FlowBal wrote 32 equations for 33 unknowns, so one more equation is required. This is the

relationship between wS02 in S8 andpS02 in S7. The equation at 0 °C is wS02, % = 25(pS02,
atm). FlowBal's Insert Equation feature uses a special equation format:

25*{pS7-SQ2> - {S8-SQ2TI

Figure 6.25 shows the FlowBal solution. In some cases (as discussed in the FlowBal User's
Guide), FlowBal may not converge with the default starting estimates, in which case, you may
need to insert better estimates to get convergence.

P (atm) 1.2 1.2 2 1.1 1.1 1.1 1.05 1.05

T(K) 330 310 500 1880 1880 1880 273.15 273.15

Str-unit Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Volume (m3) Mass (kg)

Spec-unit Mass pet Mass pet Mass pet Mass pet Mass pet Mass pet Volume pet Mass pet

Str-name Air Oil Steam BrnrGas ProcGas SmplGas DrySplGas Water

Streams 1 2 3 4 5 6 7 8

Flow 350 20 1 371 370.258 0.742 0.483 0.0556

С 0 87.3 0 0 0 0 0 0

H 0 12.6 0 0 0 0 0 0
s0
0.1 000000
0 2 22.88 0
0 3.651 3.651 3.651 3.741 0

N2 74.64 0 0 70.415 70.415 70.415 82.428 0

Ar 1.26 0 0 1.189 1.189 1.189 0.976 0

H20 1.22 0 100 7.491 7.491 7.491 0 99.9986

C02 0 0 0 17.244 17.244 17.244 12.849 0

S02 0 0 0 0.0108 0.0108 0.0108 0.0055 0.0014

Figure 6.25 Results of FlowBal calculation of excess air oil combustion with steam-vaporized oil.
The condensed water (S8) has about 1 % of the S02 that entered in S6.

Chapter 6 Reactive Material Balances 335

The amount results table was used to calculate that the flowrate of the process gas (S5) was
2880 STP m3/min, with the following volumetric composition.

0 2 N2 Ar H20 co2 so2

3.29% 72.54% 0.86% 12.00% 11.31% 0.005%

Once convergence is assured, you can go back and change any input variable such as mass of
combustion air, and examine the relationship between it and one of the constituents in the sample
gas stream. For example, a material balance can show how much air leaks into the process gas
(leak air that bypasses the burner) by carefully analyzing one of the smaller-amount gases such as
02. FlowBaPs Repetitive Solve tool calculated the φ02 in S7 at different amounts of combustion
air between 350 and 400 kg/min. The mass of air in-leak is based on the burner flow controller
accurately providing it with 350 kg/min of air, so the in-leak air is obtained by subtracting 350
from the SI flow. Figure 6.26 shows a near-linear relationship between the two variables.

Air Leak Based on Combustion Gas Analysis

4.5 5.0

02 in S7, vol. %

Figure 6.26 Relationship between the mass of air leaking around burner and the φ02 measured at
the sampling port on S7. Burner operating at 350 kg/min humid air, 20 kg/min oil and 1 kg/min
steam. Text box equations developed by Excel's Trendline tool.

A combustion control process may require knowledge of content of a certain constituent of a
process gas, but drying the sample before analysis changes its composition. Suppose the operator
needed to control the φ02 in the oil combustion process gas based on the φ02 analysis of the dry
sample gas (S7). FlowBal was set to do a repetitive solve for the composition of S5 as a function
of the total amount of air involved in the combustion of 20 kg of oil and 1 kg of steam. The S5
composition values were converted from mass % units to mole % units by MMV-C, and the S7-
φ02 calculated by a one-equation material balance based on complete removal of water vapor from
S5. The S7-<p02 and the S5-<p02 relationship was linear between 290 and 400 kg/min air flow.
The following equation relates the measured S7-<p02 to the actual process gas φ02:

% 0 2 in S5 = 0.887(% 0 2 in S7)

6.3.6 Trace Combustion Products

Stack gas analysis also plays an important role in determining and controlling the amount of
nitrogen oxides emitted by combustion. These so-called NOx gases are of increasing concern in
setting air quality standards, and their amount is a function of the temperature and composition of
the stack gas. Six oxides of nitrogen exist, four of which are present in negligible amounts in
combustion processes, while two are formed in sufficient quantity to warrant consideration: NO
and N02, designated as "NOx" species. The relevant equations are shown below, and Figure 6.27
has a plot of the equilibrium constant vs. temperature. The figure shows that higher temperatures
increase the amount of NOx. Data were taken from FREED's Reaction option.

336 Chapter 6 Reactive Material Balances

02+N2->2NO; Keq = {рНОУ [6.53]
(P02)(pK2)

02+'/2Ν2^·Ν02; Keq = pH02 [6.54]

(P02)4W:

The small values for Кщ show that there is very little tendency for 0 2 and N2 to react, so NOx
formation is an example of negligible reaction extent. Clearly, temperature is a major factor on the
formation of NO, and is a lesser (but still significant) factor for N02.

1.E-03 /Ceq for NOx Formation
; ! ! ! '=

1.E-04 ^^ , a

I \ !i "П. ш - ■ " * ^ ir ^ *"

a - -o

. cу - * ^^^^

t 1111 ^^
^^^
1.E-05 '

/Ceq — 0 — NO - -a- ■ N02

1.E-06

^f^

'I

1.E-07

f

7

л er n o 900 1000 1100 1200 1300 1400 1500 1600

800

t, °C

Figure 6.27 Values of Кщ for the formation of two moles of NO and one mole of N02 according
to the reactions written above as Equation [6.53] and [6.54]. A logarithmic scale was used for Кщ
because of the large effect of temperature on the Кщ for NO formation.

Another important factor on the equilibrium position is the amount of 0 2 and N2 in the stack
gas; the greater the % XSA, the greater the amount of NOx produced. This is shown by

calculating the amount of NOx formed at 1000 °C by the combustion of natural gas (NG) with

different % XSA. The stack gas composition used for this calculation is the one developed in

Section 6.3.1, and plotted in Figure 6.21. A sample calculation at 10.5 % XSA gives a stack gas
p02 = 0.0177 and/?N2 = 0.707 atm. Since the extent of NOx reaction is very small, the formation
of the NOx gases will not appreciably change the /Ю2 and /?N2, which simplifies the arithmetic.
Equations [6.53] and [6.54] are used:

[ 6 . 5 3 ] : ^ = 8.06xl0-_7/ = (pNOr 0>ror ; pNO =1.0xl0'4 atm
( p 0 2 ) ( p N 2 ) (0.0177X0.707)

[6.54] :Keq = 2.27 x l 0 _ : ) = /?N02 ^N02 ^ N 0 2 =3.4xl0"7 atm
(p02)y[pN2
(0.0177)V0.707 '

Chapter 6 Reactive Material Balances 337

This shows that the NOx at 1000 °C is predominantly NO, and that's still true at 1500 °C.
Temperature has the greatest effect on NOx composition, with % XSA having a significant but
lesser effect (see Figure 6.28). The NOx composition can be lowered by decreasing the stack gas
temperature before discharge, and providing a catalyst for the decomposition of NO to the
elements. Various techniques have been developed to lower the NOx content of a combustion gas,
such as introducing a small amount of urea into diesel engine exhaust (this decreases the/?02).

NO for NG Combustion

800 900 1000 1100 1200 1300 1400 1500 1600

t,°C

Figure 6.28 Composition of NO in stack gas (volume ppm) produced by burning NG with XSA.
Calculations based on equilibrium between NO, N2, and 02. Composition of N02 shows a similar
trend, but is two to three orders of magnitude smaller than NO.

The above NOx calculations were based on equilibrium being reached in the gas phase. For
furnaces where the retention time of the combustion gas is short, equilibrium is not reached and the
NOx composition will be less. Once NOx gases are formed, the equilibrium relationships indicate
that their amount will decrease as the stack gas cools. If cooling takes place too fast for
equilibrium to be attained, the amount of NOx present at elevated temperature will show little or
no decrease as temperature drops.

A considerable amount of effort is being made to design burners that minimize the NOx
generation while still maintaining the thermal efficiency of a process (Kamal et. al. 1999, Besik et.
al. 1998). One option is to operate as close as possible to stoichiometric air even though
unavoidable stream flow fluctuations will cause occasions of less-than-stoichiometric air. The
combustion gas will then contain small amounts of CO and H2 which can be oxidized by the
introduction of air at a point in the furnace where temperatures are lower. Another way to
minimize NOx formation in a furnace is to design the burners and fornace to avoid excessively hot
spots. This can be done by recycling some of the furnace offgas to lower the flame temperature, an
option examined in more detail in Chapter 9. Alternatively, oxy-fuel burners can be used to
minimize the N2 content of the furnace gas. Once NOx forms, catalytic converters can be used to
assist its decomposition as the stack gas cools.

Two other "trace" combustion products are present in XSA combustion products: CO and H2.
Heretofore, combustion reactions using excess air were assumed to be irreversible, so the only
combustion products were H20 and C02. The extremely small amount of CO and H2 was
justifiably neglected in the material balance. However, the assumption of irreversible formation of
H20 and C02 is valid only for flame temperatures below 2000 К where the values of Кщ for the
formation of H20 from the elements, and for C02 from CO, are above 1200. Use of oxygen-

338 Chapter 6 Reactive Material Balances

enriched air for fuel combustion generates a much hotter flame than normal air — often in excess
of 2500 К (Baukal 1998). A similarly-hot flame may be generated by using preheated normal air.

A good way to calculate the amount of CO and H2 in a combustion gas with excess 0 2 is by

calculating the equilibrium "extent of decomposition" of C02 and H20 according to Equations

[6.55] and [6.56], written on the basis of one mole of 02. 0?со)2дю2)

2C02(g) - 2CO(g) + 02(g); Кщ = {pCQ2Y [6.55]

(pH2)2(pQ2) [6.56]
2H20(g) - 2H2(g) + 02(g); Кщ = (рЯ20)2

The value of Кщ and a material balance on H and О can be used to calculate the equilibrium
position, and from that, the extent of the decomposition reaction of C02 and H20.

EXAMPLE 6.8 — Calculation of CO, H2 and NO content in Hot Stack Gas.

Methane is burned with oxygen-enriched air with <jp02 = 50 % to obtain a combustion gas
having <p02 = 3.5 %. The combustion gas temperature varies between 2000 and 2500 K.
Calculate the composition of the combustion gas, and express the results in terms of the extent of
reaction of C02 and H20.

Data. FREED's Reaction tool gave a chart of log(ATeq) vs. \IT for Equations [6.53], [6.55] and
[6.56]. Trendline gave equation parameters as follows:

[6.53] KКeщq= /{Р0v{*^2p)_H{wр,O~Щ);,2т) ч ; log(^eq) = -9440+ 1.326

[6.55] ККщщ- = v(p"C(^pOC7)02(2νp)"20Γ2)ζ / ; log(^eq) = ^ ^ + 8.692

[6.56] KC^ee(*qqn=— V(pf H^("Pp2Uz)tì2/2(0pv)0)22^2)" z / ; log(A:eq)="26,355 +6.088
T

A^eq for Equation [6.55] is larger than that for [6.56], which means that the decomposition
tendency for C02 is greater than that for H20.

Solution. FlowBal is used to make the material balance, and its Repetitive Solve feature is used
for different temperatures. The burner has two instreams and one outstream, eight species, and
four reactions. The process only has one device, so no flowsheet sketch is needed. One atm
pressure is assumed. Figure 6.29 shows the input array.

The FlowBal reaction selection is to carry out R #1 to completion, and use R #2 and R #3 to
define the "back-reaction" of H20 and C02 to their precursors. Finally, R #4 defines the
equilibrium amount of NO. FlowBal wrote nine equations for 12 unknowns, so three more
equations are needed: the Кщ expressions for R #2, 3, and 4. FlowBal creates cell comments with
the inserted equation format as shown below.

{pS3-NO}/v2 - 0-035*{pS3-N2}*10A(-9440/<T-S 3> + 1.326)~|

0.035*{pS3-CO}*2 - {pS3-CO2}A2*10A(-28900/<T-S 3> + 8-692) |
0.035*{pS3-H2}A2 - {pS3-H2O}A2*10A(-26335/<T-S 3> + 6.088) \

Chapter 6 Reactive Material Balances 339

P(atm) 1 1 1 R#1 CH4 + 202 -> C02 + 2H20; Keq = very large
T(K) 300 300 2000 R#2 2H20 -> 2H2 + 02; Keq = 10A-26335/T + 6.088
R#3 2C02 -> 2CO + 02; Keq = 10Λ-28903/Τ + 8.692
Str-unit Amt (kg-mol) Amt (kg-mol) Amt (kg-mol) N2 + 0 2 -> 2NO; Keq = 10A-9440/T + 1.326

Spec-unit Mol pet Mol pet Mol pet
Str-name Methane O-RchAir BrnrGas R#4

Streams I 1 2 3 R#1 R#2 R#3 R#4

Flow 100

CH4 100 -1
0 2 50 3.5 -2
N2 50 ? 1

C02 ?1 -2
H20 ? 2 -2
CO ?
H2 ? 2 2

NO ?

Burner
Instreams Outstreams Reactions

1
2
3
4

Figure 6.29 Input array for methane combustion with oxygen-enriched air. The burner gas was
set to contain 3.5 %02.

Figure 6.30 shows the results at 2000 K. 3.5 % 0 2 in the burner gas is a 9.5 %XSO condition.
The CO, H2 and NO contents are between 0.3 and 0.06 %, so their values could be neglected if the
emphasis is on the S2 and S3 stream flows, or on the composition of C02 and H20. The burner
gas will cool as it transfers heat to the furnace charge, and the CO, H2 and NO contents in the
furnace will revert to "trace" amounts that are negligible for material or heat balance calculations.
The kinetics of CO and H2 formation are rapid, but NO formation and decomposition are often so
slow that equilibrium calculations are not applicable. The objective of the equilibrium calculations
is thus mainly to show the CO and H2 contents, and the conditions that favor or minimize the
tendency of NOx to form. The NO amount may exceed environmental regulation limits.

FlowBal's Repetitive Solve feature calculated the burner gas composition for five
temperatures between 2000 and 2500 K, for 3.5 % 0 2 in the burner gas. Figure 6.31 shows the
results. Clearly, as the combustion gas temperature exceeds 2000 K, CO, H2, and NO are no
longer "trace" species, and their amount is enough to affect all aspects of a system balance.

P (atm) 1 1 1

T(K) 300 300 2000 Stream Amounts (kg-mol)
Str-unit Amt (kmol) Amt (kmol) Amt (kmoll)

Spec-unit Mol pet Mol pet Mol pet Str-name Methane O-RchAir BrnrGas
2
Str-name Methane O-RchAir BrnrGas 3
Streams 1 2 Streams 438.0 538.5
3 1
Flow Flow 0 0
100 438.0 538.5 100
219.0 18.85
CH4 100 0 0 CH4 100 219.0 218.34

02 0 50 3.5 02 0 0 99.30
0 199.69
N2 0 50 40.55 N2 0
0 0.700
C02 0 0 18.44 C02 0 0 0.308
1.289
H20 0 0 37.08 H20 0 0

CO 0 0 0.130 CO 0

H2 0 0 0.057 H2 0

NO 0 0 0.239 NO 0

Figure 6.30 FlowBal results array for the combustion of 100 moles of methane with sufficient

oxygen-enriched air to produce 3.5 % 0 2 in the burner gas. The extent of C02 and H20 reaction is
about 0.006 and 0.0015 respectively.

340 Chapter 6 Reactive Material Balances

Figure 6.31 Equilibrium composition of selected species in burner gas as a function of gas
temperature. Oxidant gas is 50 % 0 2 at 9.5 %XSO. Temperature has a large effect on H20 and
C02 decomposition; both %CO and %H2 increase by a factor of more than 20 over the range.
Temperature has a smaller effect on NO formation; the %NO increased by a factor of 3.
Assignment. Repeat the calculation for 15 %XSO (do not specify the % 0 2 in S3).

6.4 The Production of a Reducing Gas

Specially-prepared gases are used in two main areas of materials processing. First, for the
reduction of a metal oxide, such as the formation of metallic iron from hematite (Ре2Оз). In this
application, a reducing gas is prepared with an oxygen potential lower than that of iron oxide, so
oxygen tends to transfer from the oxide to the gas. Second, as controlled atmospheres for the
bright annealing of metals and the carburization of steel.. For this, the gas must be able to prevent
iron oxidation, and/or establish a certain carbon content of steel. The main components of such
gases are CO and H2, with lesser amounts of CH4.

The largest use of reducing gases is for the reduction of iron oxide. When reduction occurs
below the melting temperature of any solid materials, the process is called direct reduction, and the
iron so produced is called direct reduced iron (DRI). Natural gas (NG) is the main raw material,
and is used in three ways; first as a feedstock for the production of the reducing gas, second as a
fuel for supplying the necessary heat in the reduction furnace and the gas reformer, and third as a
coolant and carburizing agent for freshly prepared DRI. NG cannot be used directly to reduce iron
ore because it decomposes to form soot at a temperature below where iron oxide can be reduced.
The process for preparing a reducing gas from NG is called reforming. Section 6.2.2 introduced
this technology and the relevant reactions.

The main constraint on the specification for NG is its sulfur content (mainly present as H2S);
if above 10 ppm, it can deactivate the reforming catalyst. Techniques are available to remove
sulfur if necessary. .

In order for a reducing gas to be effective, it must contain substantial amounts of CO and H2.
The higher the volume fraction of (CO + H2), the more iron oxide that can be reduced by a given
amount of gas. Therefore, a major part of a gas-based DRI plant is the reformer, which must
produce a large quantity of gas high in CO and H2. This is done by reacting NG (or some other
hydrocarbon feedstock) with oxygen-containing gases so that the desired composition of (CO +
H2) is obtained. This is called catalytic reforming, and is carried out in different ways.

Chapter 6 Reactive Material Balances 341

a) Steam reforming. Paraffinic-bonded hydrocarbon feedstock reacts with steam to form reducing
gas according to:

CnH2n+2 + (n)H20 -> (n)CO + (2n +l)H2 [6.57]

Reforming is carried out over a catalyst above atmospheric pressure, at temperatures between
750° and 1050 °C. Modern plants operating with improved catalysts use only a slight excess of
steam to produce a gas with over 90% (CO + H2). This is referred to as one-step reforming, and up
to 2% unreacted CH4 is likely to be present in the product. Reaction [6.57] is strongly
endothermic, so heat must be supplied for the reformer tubes, and recuperation of heat from the
flue gas is necessary for good thermal efficiency. In one-step reforming of NG, approximately
40% is used for fuel and 60% is reformed. Offgas from the iron ore reduction furnace (sometimes
called spent gas) may also be used as a fuel.

Reforming can also be carried out using partially dewatered spent gas in place of part or
nearly all of the steam. In the case of CH4, the reforming reaction becomes:

CH4+ xH20 + (1 - x)C02 -► (2 - x)CO + (x + 2)H2 [6.58]

In older plants, an excess of steam above stoichiometric is required to enhance the extent of

reaction [6.57] and [6.58]. Unavoidably, the excess H20 remains in the reducing gas, thus diluting
the fraction of (CO + H2). Also, some CO is consumed by reaction with the excess steam:

H20 + CO -> C02 + H2 [6.59]

The reaction represented by Equation [6.59] was introduced in Chapter 5 as the water-gas
shift reaction (WGR), and was used earlier in this Chapter during combustion gas calculations.
The WGR is reversible, and is written in either direction. The WGR plays an important role in
adjusting the composition of the reformer gas during its production, and the reducing gas during
oxide reduction. When excess steam is used, the reformed gas must be cooled to condense the
extra steam in order to obtain a high proportion of (CO + H2), as described in Example 4.15.
However, this step is wasteful of energy, which favors the newer reforming processes that produce
a suitable fraction of (CO + H2) in one step.

b) Partial oxidation. Paraffmic hydrocarbons can be oxidized by supplying just enough oxygen to
form CO and H2: ForCH4:

CH4 + У2О2 -> CO + 2H2 [6.60]

Compared to steam reforming reactions, partial oxidation produces less gas per unit of
hydrocarbon; however, it uses less fuel because Equation [6.60] is slightly exothermic*. Partial
oxidation produces a reducer gas with a higher CO content. Partial oxidation may also be used on
liquid hydrocarbon feedstock, or even coal. As compared to steam reforming, higher pressures and
temperatures are typically required to produce a gas with high (CO + H2). Historically, partial
oxidation has not been widely used in DR processes, but new advances in the technology may
make it more widely used. Another application of Equations [6.59] and [6.60] is in the production
of hydrogen for use in fuel cells for an advanced automobile.

c) In-Situ Reforming. Most gas-based processes take advantage of the natural catalytic activity of
fresh DRI to carry out some reforming in the reduction reactor. Oxygen and makeup natural gas
are injected into the reducing gas just before entry to the furnace. The role of oxygen is to produce
some C02, H20, and heat. Some reforming occurs on freshly reduced hot DRI according to a
combination of Equations [6.57], [6.58], [6.59], and [6.60]. In a recent development, all reforming
is carried out in-situ; no reformer is needed. Offgas from the reduction furnace is tempered by
removing most of the water and C02, and then reheated to about 40°C above the typical reducer
gas entry temperature. A flowsheet for this process was shown in Figure 4.69, and a material
balance for gas tempering was calculated as part of Example 4.15. Oxygen and natural gas are

In fact, some of the "excess" heat allows the addition of a little superheated steam to produce a reformed
gas with more H2.

342 Chapter 6 Reactive Material Balances

added in amounts indicated by Equation [6.60]. In-situ reforming is particularly attractive for
hydrocarbon feedstock with sulfur levels above those permissible for catalytic reforming. The
constant renewal of DRI in the furnace means that fresh uncontaminated catalyst (the DRI) is
always present for the reforming reactions.

When the WGR proceeds close to equilibrium, it's Кщ can be used as a supplementary
restriction when writing material balance equations. Equation [6.59] (the WGR) was introduced
briefly in Section 5.8, and used in a gas reforming example in Section 6.2.2. The assumption of
WGR equilibrium is valid for processes operating above about 800 °C, and is a useful
approximation below that temperature if a suitable catalyst is present. In all cases, it's the best
approximation to gas composition other than analyzing the gas. The Кщ expression is:

Кщ for the WGR (РЯ2)(рС02) [6.61]
(pH20)(pCO)

There is no volume change for the WGR, hence the equilibrium position is not affected by
system pressure. None of the partial pressure terms have exponentials, so the Кщ expression may
use mole or volume units. This greatly minimizes the arithmetic complexity of material balance
calculations involving the WGR. Figure 6.32 shows values of Кщ for the WGR over a range of
temperature. Equation [6.61] is valid regardless of which direction [6.59] is written; if you write
the equation in the opposite direction, change the sign on log(Abq). An expanded figure and Кщ
equation is on the Handbook CD in folder Charts.

Keq for Water-Gas Shift Reaction

CO + H20 -> C 0 2 + H2
Log(Keq) = 1010/f - 1.231 (650 -1150 °C)
Log(Keq) = 1725/T - 1.573 (950 -1450 K)

Log(Keq) = 1 0 2 4 / t - 1.25 (650 -1650 °C)
Log(Keq) = 1650/T - 1.50 (900 - 2000 K)

700 750 800 850 900 950 1000 1050 1100

t,°C

Figure 6.32 Plot of Кщ for the water-gas shift reaction as a function of temperature. Values of
Кщ may be read with satisfactory accuracy from the diagram. Alternatively, the text box
equations provide an analytical representation of the data. The accuracy of the expanded
temperature range equation set is not as good as that for the shorter temperature range.

EXAMPLE 6.9 — Calculation of Reformer Gas Composition.

Natural gas undergoes steam reforming over a catalyst at 950 °C. The NG has the following
composition: <pCH4 = 0.930; <pC2H6 = 0.030; <pC02 = 0.020; φΝ2 = bai. The steam/NG ratio is
1.3. Calculate the volume fraction of (CO + H2) in the reformed gas. What should the steam/NG
ratio be in order to bring this fraction to 0.94?

Chapter 6 Reactive Material Balances 343

Data. The XRCR4 according to Equation [6.58] to produce CO and H2 was measured to be 0.98.
XRC2n6 is one.

Solution. An XRCH4 of 0.98* means that 98 % of the CH4 reacts to form CO and H2. Equation
[6.56] proceeds to completion for C2H6 and almost to completion for CH4. In a second reaction,
some of the remaining H20 reacts according to the WGR to form H2 and C02. We assume the
WGR goes to equilibrium. The basis of the problem is chosen as one mole of NG, so 1.3 moles of
steam are used. Please review Example 5.9 for a different approach.

The strategy is to write element balance equations for C, H, and O, along with A^eq for the
WGR, Equations [6.59] and [6.61]. Excel's Solver tool will be used to solve the equation set. The
following symbols designate the unknowns in the reformed gas: Y, W, M, D, and T refer to the
moles of H2, H20, CO, C02, and CH4.

H balance: 4(0.93) + 6(0.03) + 2(1.3) = 2Y + 2W + 4(T)

С balance: 0.93 + 2(0.03) + 0.02 - M + D + T

О balance: 1.3 + 2(0.02) = M + 2D + W

CH4 extent: 0.93 - 0.98(0.93) = T
Кщ for WGR at 950 °C: YD/MW = 0.68

These five equations were rewritten in a format appropriate for use by Solver, which found
the following solution:

Species CO C02 H2 H20 CH4 N2
Moles 0.929 0.062 2.926 0.287 0.019 0.020

This gives a volume fraction of (CO + H2) = 0.91.

The next part of the question asks for the amount of steam that should be used to give <p(CO +
H2) = 0.94. This requires that the H and О amount balance equations include a variable for the
moles of steam, and an additional equation written to specify the <p(CO + H2):

0.94 = (M + Y)/(M + Y + D + W + T + N2)

Solver found the following solution to this equation set:

Species CO C02 H2 н2о CH4 N2 steam

Moles 0.954 0.038 2.902 0.170 0.019 0.020 1.159

The specified <jp(CO + H2) is attained as a steam/NG ratio of 1.16.

Assignment. Use an atomic species balance to calculate the amount of industrial oxygen (<jp02 =
0.97, rest N2) to add to one mole of the Example NG to give a <jp(CO + H2) = 0.92 when reforming
occurs at 925 °C. Assume X/?CH4 according to Equation [6.60] is 0.985.

Gas reforming is also used in the production of controlled atmospheres for carburizing or
decarburizing steel. There are many ways of preparing gas atmospheres for carbon control, but all
are based on controlling the composition of the gas during some sort of thermal treatment such as
is used in preparing reducing gases. One common technique is to react a mixture of NG and air in
contact with a catalyst at temperatures near 1100 °C, and then quench the gas to prevent
composition changes. This technique is called endothermic gas generation because the overall
process of NG/air reaction requires heat input. Sometimes the quenched endogas is chilled to
remove nearly all the water. The endothermic gas may be stored for later use in the carburizing

The JL/?CH4 would increase with a higher steam/NG ratio, and decrease with a lower temperature.

344 Chapter 6 Reactive Material Balances

process, or it may be sent directly from the generator to the carburizing furnace. Another

technique uses a mixture of methanol (CH3OH) and industrial nitrogen. Upon heating, the
methanol decomposes to (mainly) CO and H2. Since the nitrogen and methanol are not absolutely
pure, small amounts of other gases are also present.

The C-H-0 system has three elements, and six important species, so NIRx = 3. Chapter 5

pointed out that there are various ways to sequence the chemical reactions involved in the

oxidation of NG. Suppose we wish to make a material balance to calculate the products of

reforming NG with 0 2 to give a gas that's largely CO and H2 (but will unavoidably contain some
C02 and H20). An 02/CH4 ratio between 0.55 and 0.6 would accomplish the goal. What three-
reaction set might be best? First, an initial reaction between NG and the 0 2 in air to form CO and
H2. If we assume the NG is all CH4, the first reaction is:

CH4(g) + V202(g) -> CO(g) + H2(g) [6.62]

This reaction is spontaneous and complete, therefore CH4 is limiting. The small amount of 02
remaining will react completely (02 is now limiting) with the products of [6.62]. We can pick the
02 reaction with CO or with H2; here we pick H2. We can then use the WGR to calculate the
reformed gas composition.

y202(g) + H2(g)^H20(g] [6.63]

CO(g) + H20(g) -> C02 + H2(g) [6.64]

This gives a reasonable three-reaction set. At reforming temperatures near 1100 °C, the

extent of CH4 reaction approaches one. At carburizing temperatures near 850 °C, the extent of
CH4 reaction is not one. The amount of CH4 remaining can be estimated from an observed
correlation from gas samples: the product /?CH4xpH20 = 0.00002. This relationship is
reasonably valid between 840 ° - 860 °C. An alternative to using the correlation relationship is to
use the value of Кщ for Equation [6.57] and assume equilibrium is reached.

Owing to the variations in NG composition and the humidity of air, it's not possible to
prepare an endogas with a specified carbon potential by accurately controlling the flow of NG and
air. Instead, the carburizing furnace gas is sampled and analyzed, and small additions of CH4 or
steam can be made to bring the carburizing gas to the proper composition. Over many years of
experimentation and plant tests, a very good correlation has been found between the dew point
temperature of the carburizing gas and the carbon content of different steels. The relationship
between dpt and carbon content is fairly complex, and will not be discussed further here.
Developments in gas analysis have made it possible to measure the mass fraction of С, Н, O, and
N in a carburizing gas, and with certain correlations, the dpt can be calculated from the gas
analysis rather than being measured. Example 6.10 describes this procedure.

EXAMPLE 6.10 — Calculation of dptfrom Gas Analysis.

An endogas atmosphere prepared by reacting NG with air is used to carburize steel at 840 °C.
The mass fractions of C, H, О and N were:

wC = 0.1405; wU = 0.04623; wO = 0.1959; wN = 0.6012

Calculate the partial pressure of each gas species and the dpt of the carburizing gas.

Solution. The sum of the mass fractions does not add up to 1.0, which is probably caused by not
analyzing for Ar. On a basis of one kg of gas, the mass fractions were converted to moles using
the MMV-C program. The moles of N were assumed to be all N2, and the moles of Ar were added
to the moles of N2. The amount balance for each element is shown below, where the abbreviations
are the same as used in Example 6.9.

Chapter 6 Reactive Material Balances 345

C: 11.698 = M + D + T*

H: 45.866 = 2W + 2Y + 4T

O: 12.244 = M + 2D + W

N + Ar: '/2(42.922) + 0.406 = 21.867 = N2
The value of Кщ for the WGR from Figure 6.32 is 0.92, which gives:

0.92 = YD/MW

The relationship for the quantity of CH» present requires that the partial pressures of each species
be used instead of the number of moles. Assuming a total pressure of 1 atm, the following
equation expresses the correlation for the amount of CH4 present:

0.00002
M + D + Y + W + T + N2 M + D + Y + W + T + N2

Solver found a solution as follows:

Species CO co2 H2 H20 CH4 N2
Moles 11.344 22.197 0.4639 0.1361 21.867
Partial pressure 0.2018 0.2181 0.3948 0.0083 0.0024 0.3889

0.0039

Equation [2.15] was used to calculate that the dpt at a/?H20 of 0.0083 atm is 277 K, or 4 °C.
Based on empirical equations for a plain carbon steel, this gas would carburize the steel to wC =
0.73 %.

Assignment. Suppose it was possible to add pure CH4 to the carburizing gas. How many grams of
CH4 should be added to one kg of carburizing gas to lower the dpt to 0 °C?

The choice of methane-reforming reactions is also important when using FlowBal on these
systems. If the XR factor is used, reaction sequencing is important, but not if gas composition
relationships are used. Please see the FlowBal User's Guide and workbook FB Examples.xls for a
natural gas reforming process (worksheet NGRfrm).

6.5 Gas-Solid Oxidation-Reduction Processes

Production technology for the vast majority of metals and ceramics involves at some point a
change in oxidation state, usually reduction of an oxide. A precursor stage may be required to
produce a concentrate material suitable for reduction. An ore may be concentrated by physical
means to produce an oxide or sulfide concentrate, or the impurities in a waste product may leached
away to produce a residue containing metal values. The next step often involves a
pyrometallurgical reaction between the solid and a gas to exchange some element (usually oxygen)
between the phases. There are many applications of pyrometallurgical processing of metal
concentrates at temperatures below where a liquid phase is formed. The first part of this section
describes oxidation during calcination, the second part the gas-phase reduction of iron oxide, and
the last part involves the roasting of sulfide concentrates.

6.5.1 Oxidation-Reduction During Calcination

Calcination is the thermal decomposition of complex compounds to form simple oxides and a
gas (Wikipedia 2010), with possible simultaneous oxidation or reduction. Calcination was
discussed in Section 5.5.4 in connection with the thermal decomposition of Ca(OH)2 to form
CaO(c) and H20(g). The calcination of Ca(OH)2 required only one independent reaction because

* An extra significant figure was included with the balance equations to minimize round-off errors.

346 Chapter 6 Reactive Material Balances

there was no change in oxidation state of calcium. However, the calcination of a substance
containing an element with multiple states of oxidation may require two or more independent
chemical reactions if the oxidation occurs in stages. One independent reaction may suffice if
oxidation is complete in one reactor. Consider the oxidative calcination of siderite, a common
constituent in iron ore. If iron oxidizes completely to Fe203 in one reactor, then only one
independent reaction is needed.

2FeC03(c)+ y202(g) - Fe203(c) + 2C02(g) [6.65]

The reaction is spontaneous, complete, and irreversible in the presence of excess oxygen. It's
also exothermic, so no heat is required once the reaction is initiated. A material balance on the
oxidation of one kg of FeC03 is very simple; the stoichiometric amount of air is 230 L (STP), so
air in excess of 230 L/kg provides the necessary excess 02 to cause X7?FeC03 to be one.

The calcination of ferric sulfate is more complex. Iron and sulfur are already in their highest
state of oxidation, but at elevated temperature, some of the sulfur is reduced from the +6 to the +4
state of oxidation. Three reactions can be written but only two are independent; [6.66] and [6.67]
are chosen arbitrarily. Equation [6.67] was discussed earlier in this Chapter as Equation [6.8] and
[6.17], which you should review before reading on.

Fe2(S04)3(s) -Fe203(c) + 3S03(g) [6.66]

2S03(g)-2S02(g) + 02(g) [6.67]

When Fe2(S04)3(s) is heated in an inert atmosphere, both of the above reactions occur to
generate a mixture of S03, S02, and 02. As temperature increases, the partial pressure of each
species increases. Eventually, at the calcination temperature, the /?S03 + pS02 + p02 will reach
one atm. It's useful to calculate this temperature to help in process design. There are four

unknowns (the partial pressures of each gas and the temperature), so four equations are required.
First, when (and only when) both Fe2(S04)3 and Fe203 are present, pS03 is uniquely determined by
the value of Кщ for Equation [6.66}. Second, the relationship between the partial pressures of the
three gases is uniquely determined by the value of Кщ for Equation [6.67]. Third, the sum of the

three pressures equals one atm. Finally, there is a constraint imposed by the stoichiometry of
Equation [6.67]: pS02 = 2p02* FREED's Reaction tool was used to obtain Кщ expressions for Уз
of Equation [6.66], and [6.67]. Substituting / = /?S03, d = pS02, and x = p02, the four required
equations are:

/ = 10(-9750/749.24) id) Q) = ^(-10,210/74-9.69) d+t + x=\ d=2x
t2

Solver found a solution; the calcination temperature of Fe2(S04)3 is 1005 K.

p&02 pS03 p02 T
0.435 0.347 0.218 1005

The meaning of this temperature is that if the sulfate is calcined by the application of heat
only, its temperature must exceed 1005 К in order to assure complete conversion to Fe203. The
calcination temperature will be lower if the sulfate is heated in a flowing gas of any composition
except the one in the above table because the partial pressure of the calcination gas species are
diluted, and hence complete conversion occurs at a lower temperature.

If ferrous sulfate is calcined in the presence of excess oxygen, iron will be oxidized to the +3
state, and be present either as ferric sulfate or hematite, depending on the temperature. At the same
time, some of the sulfur will be reduced from the .+6 to the +4 state, according to the extent of
Equation [6.67]. If the ferrous sulfate had water of hydration, as would be expected if it came from

* This constraint is valid only because the sulfate is heated in the presence of an inert gas.

Chapter 6 Reactive Material Balances 347

a hydrometallurgical process, the H20(g) is driven off at lower temperatures before the sulfate
begins to react. Three independent reactions define the calcination of solid hydrated ferrous
sulfate. The first reaction is new, while the other two were discussed in the earlier part of this
section. However, if the temperature was high enough, Equation [6.66] would not occur. Equation
[6.68] would be irreversible, and [6.67] reversible.

2FeS04 7H20(c) + Vi02{g) - Fe203(c) + 2S03(g) + 14H20(g) [6.68]

Fe203(c) + 3S03(g) - Fe2(S04)3(s) [6.66]

2S03(g)-2S02(g) + 02(g) [6.67]

EXAMPLE 6.11 — Calcination of Wet Pickling Cake.

Example 4.6 described a process for recovering crystalline FeS04*7H20 from a steel pickling
process. A version of this sulfate is currently used in water purification, but the steel company
would like to recycle it internally. This requires oxidative calcination of the sulfate to produce
hematite. The sulfate filter cake consists of wFeS04-7H20 = 0.86, wH2S04 = 0.07, bai. H20. A
conceptual process uses a propane-fired burner to direct hot gas to a calcining furnace.
Preliminary lab tests indicated that 1000 m3 of hot burner gas would produce about 14 kg of
hematite. The burner operates to produce a gas having 6.0 % 0 2 at 1610 °C. The calcine gas
should also contain about 6 % 02. Design purposes require a mass balance for the production of
500 kg of hematite per minute. Figure 6.33 shows a sketch of the flowsheet. Propane flow is
measured in kg.

Propane r-»__^VVfit filtfir cake J » ■► Calcine gas
Air 7 Hematite
B u r n e r l ?Γϋί ?as _ 4 _ ^ Calciner
(680 °C) V 7 = 500 kg
-HI (1600i«C)J> = 36,000 m3
Air-----
*

Figure 6.33 Flowsheet for the production of 500 kg of hematite per minute by oxidative
calcination of wet ferrous sulfate heptahydrate filter cake. Solid lines indicate solid phase streams,
while dashed lines indicate gas phase streams.

Data. Preliminary lab tests indicated that 1000 m3 of hot burner gas would produce about 14 kg of
hematite. Stream 6 sampling indicates that XRSO3 is about 0.92.

Solution. The text of the problem indicates that Equation [6.68] spontaneous and complete, and
Equation [6.66] does not occur under the specified circumstances. Equation [6.67] may not reach
equilibrium but we know the XRSO^. We needn't specify the physical state of water since we
know the state from the process context, and are not doing a heat balance. Two of the required
independent reactions are then [6.67] and [6.68], and the other two are shown below as Equations
[6.69] and [6.70]. This gives four independent reactions and ten species. Owing to the mixed
units, FlowBal is a good choice for making the material balance.

H2S04(/) - H20(g) + S03(g) [6.69]

C3H8(g) + 502(g) - 3C02(g) + 4H20(g) [6.70]

FlowBal wrote 16 equations for 17 unknowns, so a subsidiary relationship is needed. For this,
we use the X-R tool to indicate the XRSO3 for Equation [6.67]. Figure 6.34 shows the FlowBal
solution for this case.

348 Chapter 6 Reactive Material Balances

P(atm) 1111 1 1 1
T(C) 30 30 30 1600 30 680 680

Str-unit Mass (kg) Volume (m3) Ma SS (kg) Volume (m3) Volume (m3) Volume (m3) Mass (kg)

Spec-unit Mass pet Volume pet Mass pet Volume pet Volume pet Volume pet Mass pet

Str-name Propane Air FltrCake BrnrGas Air CalcGas Hematite
4 7
Streams 1 2 3 5 6

Flow 362 5418 2024 36,000 311 24,209 500

C3H8 100 0 0 0 0 0 0

N2 0 79 0 73.46 79 58.77 0

0 2 0 21 0 2 21 3 0

FeS04(H20)7 0 0 86 0 0 0 0

H2S04 0 0 7 0 0 0 0

C02 0 0 0 10.52 0 7.96 0

S02 0 0 0 0 0 2.29 0

S03 0 0 0 0 0 0.20 0

H20 0 0 7 14.02 0 27.78 0

Fe203 0 0 0 0 0 0 100

Figure 6.34 Results of FlowBal material balance for the wet cake calcination process.

Assignment. Repeat the calculation using fuel oil for the burner that has wC = 87.29%, wR
12.53%, bai. S, using enough burner air so that the flow of S5 is zero.

6.5.2 The Reduction of Iron Ore Concentrate

The production of steel from virgin raw materials always involves gas-phase reduction of the
oxide. In the iron blast furnace the iron oxide concentrate is reduced by reaction with (mainly) CO
in the upper part while the oxide is still solid. The product is melted in the lower part of the blast
furnace, and discharged as a liquid. However there are many cases where the economics favor
carrying out the reduction and melting stages in two devices. When reduction is carried out below
the melting point of any component, it is called direct reduction, and the product is designated
DRI. The DRI is melted and refined with scrap iron in an electric furnace to produce steel.

Most commercial DR processes use a reducing gas consisting of a mixture of CO and H2
prepared by NG reforming, but one process uses a gas consisting predominantly of H2 as the
reducing agent. Reduction takes place either in fluid bed furnaces (three or more in series), or a
vertical shaft furnace. The spent gas is dewatered and recycled along with fresh reducing gas. The
reducing gas usually contains at least a small amount of N2 (and sometimes a considerable
amount), depending on its source. A bleed stream is required to prevent N2 build-up and as a way
to remove lesser amounts of volatile impurities that enter with the concentrate. Thermal energy is
provided by heating the reducing gas, and in some cases, preheating the concentrate or ore. A
typical iron oxide raw material for DR consists mainly of hematite (Ре20з), with small amounts of
various gangue minerals that are not reduced in the process.

The most common DR process uses a moving-bed shaft furnace. Iron ore concentrate pellets
are fed into the top of the furnace through a lock system. DRI is cooled and removed from the
bottom. If desired, some CH4 can be introduced near the point of DRI discharge to convert part of
the iron to iron carbide, Fe3C. A recent development is the discharge of hot DRI, which is
conveyed promptly to an electric melting furnace. Some DRI is discharged hot, then pressed into a
pillow-shaped briquette to seal the pores. This is called hot briquetted iron, or HBI.

The extent of oxide reduction depends mainly on temperature, retention time of solid in the
reactor, and flow ratio of (reducing gas)/Fe203. Test work is required to optimize these variables
with respect to maximizing the production rate of quality DRI. One criteria is that the gas used in
iron production must always have an excess of H2 and CO over that present at equilibrium between
iron and wustite (Feo.9470) . The application of this criteria is discussed in the following sections.

Wustite is a phase of variable O/Fe ratio which is never less than about 1.055. As an approximation, some
texts refer to wustite as having the formula FeOi.06, or even FeO.

Chapter 6 Reactive Material Balances 349

6.5.3. The Chemistry of Fluidized Bed Reduction of Iron Ore by Hydrogen

The H-Iron process uses nearly pure hydrogen to reduce iron ore to iron (DRI) using multiple
fluidized beds in series, with final reduction to iron in the lowest bed. You may wonder why
multiple beds are used instead of a single bed. The answer is that multiple beds increase the
efficiency of hydrogen use. In addition, in a single bed, some of the iron oxide leaves before much
reduction has occurred, while some oxide remains in the bed for a while even after reduction is
complete. Fluidized beds in series allow better control over the retention time so hydrogen
efficiency is improved. We will explore the process by making a material balance on a two-stage
reduction process. The H-Iron process uses at least four beds in series, but using only two
fluidized beds simplifies the arithmetic, while still illustrating the computational techniques.

The top bed is a pre-reduction reactor, and the lower bed is the final reactor where the DRI is
produced. Figure 6.35 shows a flowsheet for the simplified (two-bed) process. The hydrogen-rich
reducing gas is prepared by reforming natural gas with steam in a manner discussed earlier,
followed by separation of the CO, C02 and most of the unreacted steam. The spent gas from the
process (S8 on the flowsheet) is cooled to dewater it, and returned to the reduction furnace (with
more hydrogen) as S4. Some of the FR gas is split to bleed to remove inerts and impurities. The
steam reforming and offgas treatment portions of the flowsheet are not considered here.

Ore „ . ^ Spent PR gas
(Fe203) 1 δ " (H2 + H20)

PR oxide Bleed FR gas
" " ► (H2 + H20)
(Fe304 +
Fe01/M) Split FR gas
(H2 + H20)

! Main FR gas
(H2 + H20)

DRI (Fe + Reducing gas
FeO106) " (H2 + H20)

Figure 6.35 Flowsheet for a two-stage fluidized bed reduction of hematite by a hydrogen-rich gas.
Solid lines represent streams containing solid phases, while dashed lines represent gas phase
streams. All gas streams contain a mixture of H2 and H20.

Several chemical reactions occur in the process. First, consider the final reducer (FR) because

it determines how much H2 is required to produce a certain amount of Fe. The key reaction is the
one that reduces the wustite to Fe, and it may be written two ways:

1.06H2(g) + FeOi.<>6(c) -> 1.06H2O(g) + Fe(c); ^eq = ρΆ2Ολм/рЯ21 M [6.71а]

H2(g) + Fe0.947O(c) -+ H20(g) + 0.947Fe(c); ^eq = pU20/pR2 [6.71b]

Note that the pU20/pii2 in equilibrium with both iron and wustite is the same for either
reaction, but Equation [6.71b] is preferred to avoid exponentials. Figure 6.36 shows a plot of Keq
for [6.71b] over the temperature range used for most reduction processes. (A larger chart and
extended equation range is shown in Handbook CD in folder Charts.) If pH20/pìi2 is that for
equilibrium with iron and wustite, wustite will not be reduced, nor iron be oxidized by the gas.
Therefore, the FR gas must have a />H20//?H2 lower than the values shown on the Figure. For

350 Chapter 6 Reactive Material Balances

example at 700 °C, the equilibrium gas has a/?H20/pH2 = 0.41, so the equilibrium gas has φΗ2 =
71 %. Thus the gas composition in Stream 5 must have φΗ2 >71 % in order for any reduction to
proceed, and for extensive reduction, S6 (and of course S7) must also have <pH2 >71 %. This is
true for any composition of iron oxide solid that enters the FR. For long solid retention times and
rapid kinetics, the FR offgas composition may closely approach 71 %H2.

Fe-FeOx-H2-H20 Equilibria ^
0.65 900

0.6 p H20/p H2 = 8.20x10"4^) - 0.160
p H20/p H2 = 8.20x1 O^T) - 0.384

0.55

Q. 0.5 ^ "^
О
£ 0.45 wustite ^^ HUM
^^
0.4 ^s*

0.35 950 1000
0.3

6()0 650 700 750 800 850
temp, °C

Figure 6.36 Water vapor/hydrogen ratio in equilibrium with iron and wustite. Metallic iron is
stable at gas compositions below the line, and wustite is stable above the line. Text box equations
are linear approximations of/?H20/pH2 vs. temperature (Celsius and kelvin) for the equilibrium.

Next, consider the pre-reduction stage where hematite is being reduced to magnetite and

wustite. Here reduction occurs with a lower <jpH2. A different composition of wustite exists in
contact with magnetite. The value of x in the wustite formula FeOx is greater than 1.06 in contact
with magnetite, and varies with temperature. Figure 6.37 shows relevant data for the

magnetite/wustite equilibria.

A simplified version of the chemical reactions taking place in the pre-reduction stage is:

H2(g) + 3Fe203(c)-+ 2Fe304(c) + H20(g) [6.72]

(4-3x)H2(g) + Fe304(c) -> 3FeOx(c)+ 3xH20(g) [6.73]

Reaction [6.72] is stoichiometrically irreversible, with hematite being the limiting reactant.
Reaction [6.73] proceeds to an extent dictated by various factors mentioned earlier, such as
average retention time in the reactor and /?H20/pH2. Results from a series of bench-scale tests on
the reduction of pure hematite at 900 °C showed that for a feed rate of hematite of 3.00 kg/h and
H2 of 500 L/h (STP), the pre-reduction bed exit gas had <jpH2 of 52 %. The solid product contained
magnetite and wustite, with wustite having an O/Fe ratio of 1.15.

Based on this data, the mass of О in the hematite feed and offgas is:

О in hematite = 3000x3x16 = 901.7g
(2x55.85)+ (3x16)

n . 0.48x500x16

О in gas = = 171.3 g
5 22.415

Chapter 6 Reactive Material Balances 351

FeOx-Fe304-H2-H20 Equilibria

9
8
7

a5

2

1

0
600 650 700 750 800 850 900 950 1000

temp, deg С

Figure 6.37 Water vapor/hydrogen ratio in equilibrium with magnetite and wustite (solid line),
and composition of wustite in equilibrium with magnetite (dashed line). Fe304 is stable above the
solid line, and FeOx is stable below it. There is no simple equation relating temperature and the Y-
axis variables.

The solid product thus contains 2098.3 g Fe and 730.4 g O. A material balance on either Fe
or O, and an overall material balance indicate that the product consists of 1800 g of wustite (of
composition FeOi.15) and 1029 g of magnetite. The extent of magnetite reduction can be calculated
from this data using either mass or amount units. Complete conversion of the hematite to
magnetite (i.e., extent of hematite reaction via Equation [6.71] is one) produces 2899 g (12.52
moles) of magnetite. The next reaction left 1029 g (4.44 moles) of magnetite unreduced.
Therefore, the magnetite reaction extent was 1870/2899 (or 8.08/12.52) = 0.645. The magnetite
reaction extent would go up if the flowrate of H2 increased because the reaction rate is a function
of the φΗ2. Alternatively, the magnetite reaction extent would increase with increased retention
time.

This section started by asking why a single bed wasn't used. The theoretical minimum
amount of pure H2 required to reduce one mole of hematite to two moles of iron in a single bed at
700 °C is 10.32 moles. This is because at equilibrium, the exit gas from the single bed must have a
/?H20//?H2 = 0.41. Three moles of H2 are consumed in reduction, and 7.32 moles remain
unreacted. But suppose a pre-reducer converts the hematite to % mole magnetite; then only 2.67
moles of H2 are consumed to produce the two moles of iron, leaving 6.51 moles H2 unreacted. The
equilibrium amount of H2 required for a two-stage process is then 9.18 moles, vs. 10.32 moles
required for one stage. In practice, the actual amount of unreacted H2 will be more than the
equilibrium amount, but in any case, pre-reduction lowers the overall amount.

6.5.4 Excel Simulation of the Fluidized Bed Reduction of Hematite

This section shows how to make a material balance using the two-stage H-Iron system as the
example. The balance will explore the relationship between three factors: extent of magnetite
reaction in the pre-reducer, the extent of wustite reaction in the final reducer, and the effect of
water-vapor content in the new reducing gas (stream 4). The principle questions are: How much
new gas is required as the extents of magnetite and wustite reduction change? What is the effect of
increasing moisture in the new gas? Remember that wustite has a variable composition, depending
on temperature and which other solid is present. Near 800 °C, Figure 6.37 shows that FeOui

352 Chapter 6 Reactive Material Balances

(oxygen-rich wustite, designated W) is in equilibrium with Fe304. FeOi.06 (designated W)
represents the wustite composition in equilibrium with iron at all temperatures.

Samples taken from a plant operating near 800 °C showed a good correlation between the
extent of wustite reaction {XRW), and the /?H20//?H2 in the reactor. The term gas ratio (GRFR)
indicates the /?H20//?H2 in the FR. Equation [6.74] adequately expressed the plant results for XRW
between 0.86 and 0.98.

GRFR = -TJ2(XRW)2 + 13.17(X7?W)-5.18 [6.74]

The process is simulated by writing a set of feasible chemical reactions and specifying the
extent of reaction of the limiting solid reactant in each case. We also must specify which reactions
occur in each reactor. There are various reaction sequences that could be written for this process,
but the most logical are for sequential reduction of the higher to lower iron oxides, and then to
iron. The following four reactions are assumed to take place in the sequence written. This means
that hematite is reduced to its specified extent via Equation [6.75] before [6.76] starts, etc. The
first two reactions occur in the pre-reducer, and the last three take place in the final reducer. Since
there is no gas volume change for any reaction, streams 4, 5, and 9 of Figure 6.35 all have the same
molar flowrate.

3Fe203 + H2 -+ 2Fe304 + H20 [6.75]

Fe304 + 0.67H2 -► 3FeOL11 + 0.67H2O [6.76]

FeOi.n + 0.05H2 -> FeO106 + 0.05H2O [6.77]

FeOi.06 + 1 -06H2 -► Fe + 1.06H2O [6.78]

Notice the two different forms of wustite in the above reactions: FeOui (W) and FeOi.06
(W). Hematite, magnetite and W are limiting reactants, being completely consumed in the

process. Reaction [6.75] is considered stoichiometrically irreversible, so by definition the
Xi?Fe203 is one in the PR. The XRFQ3O4 via Equation [6.76] in the PR is user-defined, and is
usually between 0.6 and 0.99. The Xi?Fe304 is one in the FR. Equation [6.77] takes place only in
the FR, where XRW* is one. Equation [6.78] takes place only in the FR, and XRW is between 0.86
and 0.98. This means that the DRI always contains some FeOi.06·* Since the X7?Fe203 and XRW
are one, only two reaction extent values are needed to define the system. First the XRFQ304 in the
PR, and second the extent of FeOi.06 reduction in the FR (designated as XRW). These XR values
are determined mainly by the gas composition in each reactor, but temperature, retention time and
type of ore also play a part. Here, temperature and retention time are ignored. XRM is user-
defined and GRPR is calculated by a material balance. XRW is user-defined, but GRFR is calculated

by Equation [6.74].

When plant data isn't available, we must resort to thermodynamic data for the Fe-O-H system
to estimate the relationship between gas composition and extent of species reaction. The
equilibrium constant for Equation [6.75] is a very large number, which explains why Xi?Fe203 is
one in the PR. Кщ for Equation [6.76] is about 3, so Fe304 reduction requires the PR gas to have a
GRPR <3. To provide a reasonable driving force for magnetite reduction, the GRPR (i.e., S8 gas)
needs to be below 2, or even below 1. The equilibrium /?H20//?H2 for wustite reduction is about
0.5 (see Figure 6.36), so the GRFR (i.e., S5 gas) needs to be well below 0.5, even as low as 0.25.

The independent operating variables are set as XRM, XRW, and the volume fraction of water
vapor in the reducing gas (S4). The dependent operating variable is the GRFR as calculated by
Equations [6.74].

The amount variables for the material balance calculation are the moles of the various reactive
species in streams 2, 3, 4, 5, 6, and 8. (Other variables could be chosen). The constraint equations
use the amount variables as described below, in terms of the XRM and XZ?W· The material balance

It is theoretically possible to produce DRI with no residual wustite, but practically, there will always be
some unreduced wustite short-circuiting the fluidized bed, so the DRI is never wustite-free.

Chapter 6 Reactive Material Balances 353

equations for the solid species are simplified by noting two things: the amount of Fe entering and
leaving each reactor is two, and the amount of gas is the same in S4, S5, and (S6 + S7). Symbols
used to indicate the main variables are:.

H = moles of hematite.

M = moles of magnetite.

W = moles of FeOui.

W = moles of FeOi.06.

I = moles of iron.

XRM = extent of magnetite reaction via [6.75].

XRW = extent of wustite reaction via [6.77].

(S2), (S3) etc. = stream numbers.

GRFR = gas ratio value for the FR reactor.

<pH20(S4) = volume fraction of water vapor in the reducing gas (stream 4).

SF = splitter fraction of S5 flow to S6 flow

For the first simulation, the independent operating variables were set as: XRM = 0.65, XRW =
0.964, SF - 0.9, and <pH20(S4) = 0.06. The specified XRM gives GRFR = 0.342 via Equation
[6.74]. The main variables are the composition of six gas streams: S2, S3, S4, S5, S6, and S8 (SI
is defined, the composition of S7 and S6 are the same, and the splitter ratio of S5 to S6 is defined).
Each stream has two species, for 12 unknowns. Material balance, GR and flow equivalence
equations were written around each reactor, with each expression set equal to 0.

1: PR: Fe balance: : 2H - 3M(S2) - W'(S2)
2: PR: M formed - M reduced - M out 2H/3-X7?M(2H/3)-M(S2)

3: FR: Fe balance: 2H - W(S3) - I(S3)

4: FR: W formed - W reduced - W out 2H - (X7?W)(2H) - W(S3)

5: PR: О balance: 3H + H20(S6) - 4M(S2) - 1.11 Wf(S2)

6: FR: О balance: 4M(S2) + 1.11 W(S2) + H20(S4) - H20(S5) - 1.06W(S3)

7: Composition of S4 <pH20(S4) - H20(S4)/[H20(S4) + H2(S4)]

8: Composition of S5 GRFR-H20(S5)/H2(S5)

9: Composition of S6 GRFR-H20(S6)/H2(S6)

10: Splitter fraction of S6 to S7 H20(S6) + H2(S6) - SF[H20(S5) - H2(S5)]

11 : Equivalence of S4, S5 flow H20(S4) + H2(S4) - H20(S5) - H2(S5)

12: Equivalence of S6, S8 flow H20(S6) + H2(S6) - H20(S8) - H2(S8)

The equations were written as formulae in Excel. Solver obtained a solution, with the result
that S4 was 11.81 moles and S6 was 10.63 moles. <pH20(S5) = 25.5 %, while <pH20(S8) = 31.3 %.
SuperSolver is a good way to explore the effect of changing an independent variable on S4 and
other process variables. Typical results are shown in Figure 6.38 and 6.39.

354 Chapter 6 Reactive Material Balances

35.0% Effect of XRM on Process 11.7
11.6
34.5% —D— % H20 in S8 — 0 — S4 flow ^ 11.5
0.95 11.4
CO 34.0% >4^ 11.3
11.2
О 33.5% ^ ^J\ 11.1
** - ^ 11.0 CO
см ^ 10.9
10.8
X 0.7 0.75 0.8 0.85 0.9 10.7
XRM in PR
c^ 33.0%
32.5% - - "

32.0%0.6 0.65

Figure 6.38 Effect of XRM on S4 flow and <pH20 in S8 at XRw = 0.96, <pH20(S4) - 0.06, and SR
= 0.8. Over the same range of XRM, GRPR went from 0.478 to 0.531, and the percent of removed
О occurring in the PR went from 20.8 5 to 26.3 %.

The results show that the majority of chemical work is done in the FR, so conditions in the FR
dominate the process requirements. The least amount of reducing gas is required when XRM
approaches 1. The GRPR is well below the value for magnetite/wustite equilibrium, which may
allow a smaller PR with a shorter residence time. Alternatively, the splitter fraction of S5 to S6
could be increased.

The water vapor content in S4 (φΗ20 in S4) has a large effect on the amount of reducing gas
required (flow of S4). Figure 6.39 shows that a change in S4 composition from 1 % to 9 % H20
requires a 50 % increase in S4 flow.

14 Effect of S4 Composition on Process 36.0%
— o — S4 flow, moles
13 — D - %H20 in S8 35.5%

</> 2% 4% 6% 8% 35.0% CO

Фо 1 2 H20 in S4, vol. % 34.5% CO
E с
5
11 34.0% О

_o 1 0 33.5% X

ч- 33.0%

CO 32.5%

0% 32.0%
10%

Figure 6.39 Effect of S4 composition on S4 flow and <pH20 in S8 at XRW = 0.96, XRM = 0.8 and
SR = 0.8. The simulation shows the benefit of decreasing the <pH20 in S4 by removing as much
water vapor as possible. The <pH20 in S8 decreases because the amount of reducing gas increases
with increased φΗ20 in S4.

All of the above simulations were performed at XRW = 0.96, which is equivalent to wO in the
DRI of 1.20 %. Lowering XRW also lowers the amount of reducing gas, but it also degrades the
quality of the DRI by increasing its oxygen content. You can try your own different simulations
using FlowBal.

Chapter 6 Reactive Material Balances 355

6.5.5 FlowBal Simulation of the Fluidized Bed Reduction of Hematite

The previous H-Iron process simulation required knowing the extents of all feasible chemical
reactions to estimate the composition of the solid-phase streams. This concept is somewhat
idealistic in that some of the reactions assumed to be irreversible might not be. In fact, there is no
good way to know for sure the exact quantities of each solid phase in the PR oxide. An alternate
concept is to ignore the phase assemblage, and simply seek to correlate the outstream stream
compositions. This section shows how to use FlowBal with this concept on an H-Iron process that
operates at a higher temperature. Test work on a purified source of iron ore showed that around
920 °C, the <jpH20 in the PR and FR outstream was well-represented by a polynomial equation for
wO in the PR and FR oxide. These relationships are valid only for the retention times,
temperature, type of ore and pressure that gave rise to the equations.

S8-<pH20, % = -6.085(S2-wO, %)2 + 328.81(S2-wO, %) - 4322 [6.79]

S5-<pH20, % = 0.315(S3-wO, %)3 - 3.565(S5-wO, %f + 14.3(S5-wO, %) + 13.22 [6.80]

The objective is to use FlowBal to calculate the composition of the DRI and other streams for
a specified amount of hematite and reducing gas. The flowrate of hematite was fixed at 170
kg/sec, and reducing gas flow varied between 550 and 700 m3/sec, measured at 1.2 atm and 1200
K. S4 contained <pH20 = 3.0 % and the split fraction of S5 to S6 was 0.85. FlowBal allows a
variety of units for the constituents of a particular stream. Mass pet is a convenient unit for the
solid phase streams, and volume pet for the gas phase streams.

The system chemistry for FlowBal requires two reactions. First, hematite is parsed to the
element atoms in the PR. Second, the oxygen atom reacts with hydrogen in the PR and FR. This
means that the extent-of-reaction concept used earlier is not useful here. Instead, each mole of
Fe203 in SI is converted to two moles of Fe and 3 moles of O. Oxide reduction is simulated by
"removing" О from the solid by reaction with H2, which occurs in both the PR and the FR
according to Equation [6.81]. Underlining О designates oxygen combined with a solid.

H2 + О -> H20 [6.81]

Getting started with FlowBal requires stream numbers and species, and reaction coefficients.
A ? indicates stream flows and composition values that FlowBal is to calculate. Figure 6.40 shows
the initial array along with the defined process units. Note stream units, temperatures and
pressures.

P (atm) 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2

T(K) 310 1190 1185 1200 1185 1185 1185 1190

Str-unit Mass (kg) Mass (kg) Mass (kg) Volume (m3) Volume (m3) Volume (m3) Volume (m3) Volume (m3)

Spec-unit Mass pet Mass pet Mass pet Volume pet Volume pet Volume pet Volume pet Volume pet

Str-name Ore PROxide DRI RednGas MnFRGas SpltFRGas BleedGas SpentGas

Streams 1 2 34 5 6 7 8 R#1 R#2
Flow 170 ? ? 600 ? ? ? ?
-1
Fe203 ? ~ -1
1
Fe ? ? 2
О ??
H2 ? ? ? ? ? 3

H20 3 ? ? ? ?

Pre-Reducer (RX) Final Reducer (RX) FRGa s Splitter (SP)
Instreams Outstreams Reactions" llnstreams Outstreams Split Fract
llnstreams" Outstreams Reactions Heat Heat
2 32 5 6 0.85
~6^ 2 1 45 7?
8 2

Figure 6.40 Input array for FlowBal material balance on two-stage H-Iron process.

356 Chapter 6 Reactive Material Balances

FlowBal wrote 19 equations for 21 unknown variables. The additional two equations are
those relating the PR and FR outstream compositions. Equations [6.79] and [6.80] were written
into the Insert Equation dialog box in the usual FlowBal format. Figure 6.41 shows the solved
material balance. :

I{ S 5 - H 2 0 } - ( 0 3 1 5 4 S 3 4 ) } A 3 - 3 . 5 6 5 4 S 3 - O } A 2 + 1 4 . 3 * { S 3 - O } + 1 3 . 2 )

{ S 8 - H 2 0 } + 6 . 0 8 5 * { 5 2 - 0 } л 2 - :328.81*{S2-0} + 4322 |

P (atm) 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2
1190 1185
T(K) 310 1200 1185 1185 1185 1190
Mass (kg) Mass (kg)
Str-unit Mass (kg) Mass pet Volume (m3) Volume (m3) Volume (m3) Volume (m3) Volume (m3)
PROxide Mass pet
Spec-unit Mass pet DRI Volume pet Volume pet Volume pet Volume pet Volume pet
2 3
Str-name Ore 155.5 122.0 RednGas MnFRGas SpltFRGas BleedGas SpentGas
0
Streams 1 0 97.47 45678
76.46 2.53
Flow 170 23.54 0 600 592.5 503.6 88.9 505.8
0
Fe203 100 0 00000
0
Fe 0 00000

00 00000

H2 0 97 68.35 68.35 68.35 53.77

H20 0 3 31.65 31.65 31.65 46.23

Figure 6.41 FlowBal material balance for one second of operation of a two-stage H-Iron process.
The split fraction of S5 to S6 was 0.85. The PR removed about 28 % of the oxygen from the
hematite, the FR removed about 66 %, and the DRI contained about 6 % of the hematite oxygen.
The O/Fe ratio in the DRI was about 0.090, and wFeOi.06 in the DRI was about 10.9 %, balance Fe.

FlowBaPs Repetitive Solve feature can calculate the composition of the DRI and PR oxide
as a function of the S4 flow. Figure 6.42 shows the results.

PR and FR Oxide Composition

23.65

^ ^ w O in DRI 23.60 ^
—□— wO in PR oxd a?

23.55 ■Окσ
К

23.50 °-
23.45 °

23.40
550 600 650 700

S4 flow, m /sec ambient

Figure 6.42 Composition of oxide streams as a function of reducing gas flow, m /sec at 1200 К
and 1.2 atm. S4-<pH20 = 3.0 %.

The wO in the DRI is affected by two factors. First, the increased flow of reducing gas
removes more oxygen, and second, less oxygen enters with the PR oxide. You can try different
simulations yourself by setting the system up in FlowBal. These results illustrate how FlowBal
can use results derived from plant measurements as part of a material balance simulation of a
process. The results are strictly valid only for process conditions used for the derivation of
Equations [6.79] and [6.80], but in any case, they show how increasing the amount of reduction in
the PR lowers the wO content of the DRI.

Chapter 6 Reactive Material Balances 357

EXAMPLE 6.12 — Simulation of a Pre-Reduction Fluidized Bed Process.

In some cases, feed to a blast furnace or to a direct smelting process may not require highly-
reduced DRI. Such a product is designated pre-reduced iron, or PRI, and is a mixture of wustite

and iron, with wO between about 5 % to 15 %. Figure 6.43 shows a flowsheet for PRI production.

A process design engineer wants know how much of the offgas can be recycled while keeping the
reducer gas from going above 12 %N2. Air leaks in at the rate of 7.0 m3 (actual)/tonne of hematite,
and the new gas contains <jpN2 = 2.5 % and <pH20 = 3.0%. The condenser operates at 32 °C, at
which temperature the /?Η20 = 0.0455 atm. Use FlowBal to make a material balance for the
reduction of 1000 kg/min of hematite to PRI having wO between 12 % and 16%.

Hematite o. * Bleed
Air

Ore O-gas Rcy-gas
Reducer
kL...φΝ2 = 12%
PR Ox <- Rdn-gas Condenser
p H 2 0 = 0.0455
Water
New gas. φ H20 ■ 3%, φ Ν2 = 2.5%

Figure 6.43 Conceptual flowsheet for the production of PRI from hematite at 880 °C. Solid lines
are solid-phase streams, gray lines are gas-phase streams, and a dashed line represents liquid.
Streams 6 and 8 are heated to 887 °C before entering the reducer.

Data. Laboratory tests were made on a single-stage fluidized-bed pre-reduction process at 880 °C
to determine the mass fraction of О in the product as a function of the <jpH20 in the furnace offgas
containing H20, H2, and N2 at 1.3 atm, with <pN2 = 12%. The results were:

<pH20, % 29.4 31.3 33.1 34.2 35.3 36.0 37.0 37.9 38.5 39.8 39.9 40.5
wO, PR, 12.1 12.6 13.6 13.9 14.2 14.6 14.9 15.0 15.6 15.9 16.1 16.3

Solution. The test data were charted to develop a correlation between the two variables. Figure
6.44 shows the results. The data adequately fit a linear model. The rounded-off Trendline
equation is used as one of FlowBal's equations.

PR Oxide Reduction

41

39 %H20 = 2.623(%0) - 2.132
R2 = 0.9919
0)

E> 37
%
с 35
О
.£ 33

31

29
12 12.5 13 13.5 14 14.5 15 15.5 16

%0 in PR oxide

Figure 6.44 Relationship between oxygen content in PRI and the %H20 in the offgas.

Three reactions are required. Reaction #1 parses one mole of hematite to 3 moles of О and 2
moles of Fe. Reaction #2 consumes the oxygen in the air that enters with the ore. Reaction #3

358 Chapter 6 Reactive Material Balances

transfers the oxygen (as O) from the deconstructed hematite to H20, leaving the specified amount
of О in the PRI. Note the oxygen content of the PRI in cell K12. This value will be varied during
repetitive solve to obtain a material balance for a range of wO in the PRI from 12 % to 16 %
(FlowBal automatically calculates the corresponding dependent variable wFe). While we know
that the pH20 in the reduction gas is 0.0455 atm at 32 °C, we do not know the <pH20 because the
stream pressure is not 1 atm, so we enter a ? and write an equation so FlowBal can figure it out.

FlowBal wrote 22 equations for 24 unknowns, so two more equations were needed. First, we
know that the/?H20 in S6 = 0.0455 atm, which we write in FlowBal as:

{pS6-H20} - 0.0455

We do this by checking the box marked "PP" on the Ins Eqn dialog display. This tells
FlowBal that the unit for just that entry is partial pressure, not voi pet. FlowBal will convert this
value to voi pet when writing the equation set. If you wanted to explore the effect of varying the
condenser temperature, you could use Equation [2.15] and pick up the temperature of S6.

Second, we have the relationship from Figure 6.44 between the %H20 in the offgas and the
wO in the PRI. The equation format for FlowBal is:

{S3-H20}-2.62*{S9-0}+2.13

Prior to a repetitive solve, it's good practice to do a single solve to see if FlowBal converges,
and see if the results make sense. The results appeared correct. Repetitive Solve involves
selecting the stream and variable, setting the wO range from 12 to 16 with the number of solves
wanted, and selecting the streams and variables of interest. Figure 6.45 summarizes the results for
five solves.

The results show that large gas flow changes are required for rather modest changes in wO in
PRI. As expected, less new gas is required (S8) as wO in PRI increases, but S8 becomes a lower
fraction of the total gas needed (S8 + S6). The magnitude of these trends depends on two factors.
First, the total amount of oxygen removed from the ore, and second, the change in %H20 in the
offgas. The combination of these two factors means that the trends in stream properties with wO in
the PRI are non-linear.

4000 H2 Reduction of Hematite to PRI 450

3500 ...... 400
350 £
(/> 3000 — S3 -^^_
CЮcOo 2500 г * · -S6 ^"^^^ 300 4o-
<g 2000 *"""— S4
^^^^^^^^ 250 О
о5 1500 mm ^3 / 200
г-
" * **
1000
_ ** ^

500 13% 14% 15% 150
12% 16%
oxygen content of PRI, %0

Figure 6.45 Stream flows for PRI production process at various wO PRI values. Over the same
range of %0 in PRI, the <pN2 in S6 went from 16.4 to 19.3 %. The gas volumes are actual m3/min,
and the flow of water is kg/min.

Assignment. Plant operators decide to produce PRI with wO = 14%. What is the effect on stream
properties for a range of <jpN2 in the offgas between 12% and 24%?

Chapter 6 Reactive Material Balances 359

6.5.6 Shaft Furnace Reduction of Iron Ore Concentrate

Although our discussion started with fluidized bed iron ore reduction, a fluidized bed is not
the most commonly used reactor to produce DRI. Most DRI is produced in shaft furnaces, with
ore pellets fed into the top of the furnace, and DRI withdrawn from the bottom. The shaft furnace
is a counter-current flow reactor, so there is a continuous temperature and composition gradient
from top to bottom of the furnace. The shaft furnace is commonly pressurized to above 3 atm to
improve productivity. Figure 6.46 shows a diagram of a DRI reduction shaft furnace. The gas and
solid temperatures are within 20 degrees of each other in the main part of the shaft, but the gas is
appreciably hotter than the solid near the top. Reduction takes place more rapidly as the ore
temperature goes above about 600 °C and continues until the solid passes into the transition region.
There it is cooled by recycled reducing gas, and discharged. Carbon can be added to the DRI by
adding some natural gas to the coolant gas. Most of the DRI carbon is present as iron carbide:

CH4(g) + 3Fe(c) — Fe3C(c) + 2H2(g) [6.82]

Cool & Store

Figure 6.46 Sketch of shaft furnace for production of DRI.

Part of the reducing gas is prepared by steam reforming of natural gas (please see Example
5.9 and Section 6.2.2). Additional reducing gas is prepared by tempering the exhaust gas from the
top of the furnace (i.e., cooling to remove water and desorbing some of the C02 — please see
Example 4.15), and adding it to the reformed gas. The amount of reducing gas can be increased by
adding oxygen and natural gas to the gas supply line just before it enters the furnace. The ore is
reduced in stages as it descends, from hematite (Fe203) to magnetite (Fe304) to wustite (FeOx) and

360 Chapter 6 Reactive Material Balances

finally to iron. However, these stages overlap, so that a particular particle may have an outer layer
of iron, while wustite and magnetite are present in the center. The final DRI is a mixture of iron,
iron carbide, wustite and gangue minerals.

Since the ore and gas are continuously changing composition in the furnace, no one chemical
reaction can describe the situation at one position in the furnace. An overall balance is made
instead, which only requires information about the composition of the ore and gas in the entry and
exit streams. A common assumption is that the gas phase species are in equilibrium with each
other via the WGR down to about 600 °C. Below that, reduction is slow and the gas composition
changes very little. Thus, the system is best considered as one in which the solid transfers oxygen
to a gas which is close to internal equilibrium, but the gas and bulk solid are not in equilibrium
with each other. In addition, the gas must always be reducing with respect to the iron oxide
present in the ore pellet. We have already seen from Figure 6.36 what the value of/?H20//?H2 must
be in order to reduce wustite. A similar diagram could be constructed for values of pC02/pCO
required to reduce wustite. However, if the WGR is used, only one ratio (the pC02/pCO or the
pU20/pH2) is needed. The critical material balance question is to determine the minimum amount
of reducing gas necessary to reduce a certain amount of ore to DRI. This critical value is a
function of the temperature and H/C ratio in the gas.

Consider a bed of iron oxide exposed to a reducing gas. As О is added to a reducing gas, the
amount of (H20 + C02) increases relative to the entire composition, but there is no change in gas
volume caused by adding O. Thus, the question becomes "how much О can be added to this gas
so that it will always (even if only barely) be reducing to wustite between 600 ° and 900 °C?" This
amounts to calculating a factor that expresses the "degree of oxidation" of the reducing gas as О is
added to it. The appropriate expression is:

* -A .· * v лС02+лН20

Degree of oxidation of gas = X = Г6.831
nCO + nC02 + nH2 + nH20 LJ

Starting with a CO - H2 mixture, X goes from 0 to 1 as oxygen is added. One mole of a CO -

H2 mixture can react with one mole of O, regardless of the initial ratio of CO to H2. The gas ratio

in Equation [6.83] is thus a unique indicator of the "degree of oxidation" of the gas because the

change in X is directly proportional to the number of moles of О transferred to it.

The relationship between gas composition and degree of reduction of oxide in a shaft furnace
can be depicted graphically by use of a Rist diagram (Biswas 1981). The X-axis is the degree of
oxidation of the gas described in Equation [6.83], and the Y axis is the degree of oxidation of the
iron (designated Y), expressed as a ratio of the atoms of О per atom of Fe.

n f -A,- f · v l-5(Fe+++) + F e + + nO r6 841

Degree of oxidation of iron = Y = = L°-Ö4J

Fe rcFe

Figure 6.47 shows a Rist diagram for hematite reduction in a counter-current shaft furnace at

900 °C using reformed natural gas of H/C = 6.08. The dashed line represents conditions for

complete reduction of hematite to iron (YR= 0) with an incoming gas consisting only of CO and H2

(XR = 0). The critical operating parameter is given by the value of X and Y at point W at 900 °C.

The value of X depends on temperature and H/C ratio in the gas as given by the equilibrium
constants of the WGR and pH20/p¥l2 for the iron/wustite reaction. The value of Y is given by the

formula for wustite in equilibrium with iron (Y = 1.056 for Feo.947, but commonly expressed as
1.06). The operating line must pass through or to the left ofpoint W. For a very reducible ore and

long retention times, the operating line can pass very close to point W. The condition of the

operating line passing through W represents the theoretical maximum reduction efficiency of the

gas. Therefore, the value of X at point W is useful as a benchmark for comparing actual to

theoretically possible operating conditions.

The intersection of the operating line to Y = 1.5 gives the value of X in the gas leaving the top
of the shaft. The distance between the operating line and the phase boundary lines on a Rist

Chapter 6 Reactive Material Balances 361

diagram indicates the difference in oxygen potential between the gas and solid. Clearly, the gas
above point W is highly reducing to the solid charge all the way to the top. The critical condition
is at point W; the gas must be able to reduce all (or the vast majority of) the wustite in the lower
part of the furnace, after which it has plenty of "reducing power" for reduction of magnetite and
hematite in the upper part of the shaft. This observation is similar in concept to the two-stage
fluidized bed reduction of iron oxide, where we saw that FR is the critical reactor.

%CQ2 + %H2Q

x = %co2 + %н2о + %co + %н2

Figure 6.47 Rist diagram for a continuously operating shaft furnace with reduction taking place at
900 °C and a reducing gas of H/C - 6.08. The Y-axis is the O/Fe in the solid (so O/Fe = 1.5 for
hematite feed), and the X-axis is the degree of oxidation of the reducing gas. The dashed line is
the operating line for idealized complete metallization of hematite with a reducing gas consisting
only of CO and H2, and reaching equilibrium with wustite and iron at point W, where Y = 1.056.
The solid line represents an operating line for a practical process, producing DRI with YR= 0.06
and using an initial reducing gas of XR = 0.0825. Gas flow is 110% of that required for an
operating line passing through point W. Point W represents conditions of solid wustite of O/Fe =
1.14, and point M represents solid magnetite in equilibrium with wustite. Point M' represents
magnetite in equilibrium with hematite, and point H represents the incoming hematite. The
intersection of the operating line with the top axis represents the degree of oxidation of the gas
exiting the furnace.

The value of X at point W can be calculated by solving four equations. The first equation
specifies the atom ratio of H/C in the ingas to the shaft, which is obtained from the gas analysis .
Two equations are required to indicate the equilibrium gas composition at point W. The fourth
equation specifies a basis; here, one mole of reducing gas is chosen. The letters D, M, W, and Y

H/C is a convenient gas composition parameter because it does not change as the gas takes up oxygen from
the ore.

362 Chapter 6 Reactive Material Balances

refer to the moles of CO2, CO, H20, and H2 in the reducing gas. The small amount of CH4 that
may be present in the reducing gas is ignored.

Hydrogen to carbon atom ratio in the reformed gas: H 2Y + 2W [6.85]
С M+D

Equilibrium constant for the water-gas shift reaction: YxD = Keq [6.86]
MxW

Equilibrium constant for />Η20//>Η2 for iron/wustite equilibrium : —W— = Keq [6.87]

Basis of one mole of gas: M + D + H + W = l [6.88]

Consider the use of a reducing gas of H/C = 6.08. The required values of the two equilibrium
constants were obtained from Figures 6.31 and 6.37 as follows:

t,°C 700 750 800 850

Keq for WGR 1.61 1.31 1.08 0.91

Equil/?H20/pH2 for Fe/wust 0.41 0.45 0.49 0.53

Solver solved Equations [6.85 - 6.88] at each temperature, and X was calculated from
Equation [6.83]. Figure 6.48 shows the results. X increases with temperature, so less gas is
needed to reduce hematite as temperature increases. Higher temperatures also favor reduction
kinetics so long as the pores are not sintered shut and the solids do not agglomerate in the shaft.
The latter two factors limit the reduction temperature to <940 °C.

Gas Ratio X for Iron/Wustite Equilibria Figure 6.48 Value of gas ratio
0.35 X at point W on Rist diagram
as a function of temperature,
X = I^Ox-IO^t)* 0.206 for a reducing gas having H/C
0.34 R2 = 0.9999

0.33 = 6.08. Equation in text box
was obtained from Excel's
0.32 Trendline tool, and represents

0.31 750 t, °C 800 a linear approximation to the
calculated results.
700 85Q

The above relationships can be used to make a material balance on the maximum amount of
hematite that can be reduced by 100 moles of (CO + H2) gas of H/C = 6.08 in a shaft furnace
where the reduction zone temperature decreases from 925 °C at the reduction gas entry point to
900 °C at point W. Under ideal conditions, the gas entering the reduction zone has XR= 0.0 and
the DRI leaving the reduction zone has Y = 0.0. This corresponds to the dashed line in Figure
6.47. Based on the text-box equation in Figure 6.48, X at point W = 0.35. Therefore, 100 moles of
gas can take up 35 moles of О from the entry point to point W. At the same time, one mole of Fe
must deliver 1.056 moles of О to the gas in moving from point W to DRI discharge. Therefore,
33.14 moles of wustite of O/Fe = 1.056 can be reduced from point W to DRI discharge. This is
equivalent to 16.6 moles of hematite/100 moles of reducing gas. As the gas passes up the shaft
from point W to the discharge, it takes up 14.7 more moles of О from the reduction of hematite to
wustite. Thus X at gas discharge = 0.497, in agreement with the intersection of the dashed line
with the top axis of the Rist diagram.

An expanded chart of the equilibrium CO/CO2 and H2/H20 ratios in equilibrium with iron and wustite is
shown in the Handbook CD in folder Charts.

Chapter 6 Reactive Material Balances 363

The equation of the dashed line in Figure 6.47 is: Y = 3.018X. The slope of this line is the

amount of reducing gas required to reduce one mole of FeOi.5.* Therefore, one-half the reciprocal
of the slope is the amount of iron ore of composition Fe203 that can be reduced by one mole of
reducing gas, or 0.166 moles of hematite per mole of gas, which agrees with the previous

calculation.

The shaft furnace operates at greatest efficiency when the operating line passes closest to

point W. However, at the high gas flows used to achieve the desired throughput rate, the gas never

reaches equilibrium with the bulk solid. Hence the operating line of a practical furnace always

passes to the left of point W. In addition, the incoming gas always has some H20 and C02 (XR >
0) and the DRI is never completely metallized (YR > 0). A practical operating line is shown in
Figure 6.47 as a solid line with an incoming gas XR = 0.0825 and DRI having YR = 0.06. The
operating line was drawn at 110 % of the gas that would have been required had the operating line

passed through point W. Another factor not considered in this simplified analysis is the formation
of CH4 in the upper part of the shaft according to Equation [6.89]. Кщ of this reaction increases
with decreasing temperature.

2CO(g) + 2H2(g) - C02(g) + CH4(g) [6.89]

EXAMPLE 6.13 — Material Balance on Shaft Furnace Reduction of Hematite.

A reformed natural gas having a volume pet of 62 %H2, 28 %CO, 4 %H20, 3 %C02, and 3
%N2 is used to reduce hematite in a shaft furnace. The DRI exits the reduction zone at YR = 0.05.
Calculate the maximum theoretical amount of hematite that can be reduced by 100 moles of the
above gas when point W is at 865 °C. Calculate the composition of the top gas if it stops changing
composition at 600 °C.

Data. Figures 6.32 and 6.36 give a value of Кщ of 0.86 for the WGR and Кщ of 0.55 for the
iron/wustite reaction.

Solution. The H/C ratio of the gas is 132/31 = 4.26 and the initial value of X is 7/97 = 0.07216.
The Rist diagram operating line would initiate at YR = 0.05 and XR = 0.07216. A calculation of X
at point W requires solving Equations [6.85 - 6.88] at 865 °C. Solver found the following
solution:

<pH2 = 42.6%; φ Η 2 0 - 2 3 . 4 % ; <pCO2=10.0%; ф С О - 2 1 . 0 % ; <pN2 = 3 %.

X was calculated from the above as 0.344. Therefore, an operating line passing through the
initiating point and point W has the following equation:

Y = 3.70X- 0.217

As mentioned earlier, half the reciprocal of the slope of the operating line is the amount of
hematite reduced by one mole of reducing gas, which in this case is 0.135 mole of hematite
(expressed as FeOi.5), or 13.5 mole per 100 mole of reducing gas. In practice, less than 13.5 mole
would be reduced because the operating line must pass to the left of point W.

The descending solid changes from an O/Fe atom ratio of 1.5 to 1.056 at point W, thus
transferring 0.444 mole О to the gas per mole of Fe entering the furnace. For one mole of reducing
gas in, 0.270 mole of Fe enters as hematite, thus transferring 0.444(0.268) = 0.120 mole of О to the
gas. The value of X then increases to 0.464 in the top gas. Solver was set to solve the top gas
composition at 600 °C using Equations [6.83], [6.85], [6.86], and [6.88], where Кщ for the WGR
= 2.72. The following results were obtained:

<pH2 = 40.5%; φ Η 2 0 - 2 5 . 5 % ; <pC02=19.5%; фСО=11.4%; φΝ2 = 3 % .

If nitrogen or other inerts are present, the reducing gas in this calculation refers to the inert-gas-free
amount.

364 Chapter 6 Reactive Material Balances

Note that over half the gas consists of CO and H2. After removal of most of the H20 and
some of the C02, the gas can be recycled back to the reformer.

Assignment. Draw a Rist diagram for the above process where 11 moles of FeOi.5 are fed to the
furnace per 100 moles of reducing gas. Calculate the composition of the top gas, again assuming
that no gas-phase chemical changes take place below 600 °C, and the DRI has an O/Fe of 0.05.

FlowBal can simulate a shaft furnace reduction process by dividing the shaft into zones. The
simplest case involves two zones in a manner similar to the two-zone H-iron process. Hematite is
reduced to wustite (approximated as FeOi.06 in this example) in a pre-reduction zone (designated
PRZ) in the top part of the shaft furnace, above point W in the Rist diagram Wustite is reduced to
discharge DRI in a final reduction zone (designated FRZ) in the bottom part of the furnace. The
top gas is tempered, dewatered, and de-carbonated, then recycled to the bottom of the furnace.
Additional reducing gas is generated in a gas reformer fed with natural gas and steam.

Figure 6.49 shows the flowsheet for a shaft furnace reduction process. For simplicity, the
flowsheet omits the top gas processing (S9 to S6) and a heat-exchange zone where the top gas
preheats the hematite ore. We also neglect the gas that enters the shaft from the DRI cooling zone.
The goal is to calculate the flow of steam and natural gas needed to carry out the process.

iOre (Fe203) I - -►Top gas Figure 6.49 Simplified two-zone flowsheet for the
1 shaft furnace reduction of hematite by reformed
P re-red'n natural gas. Solid lines represent solid phase
zone streams, while dashed lines represent gas phase
streams. Hematite (SI) is reduced to wustite (S2)
PR oxide r in the pre-reduction (PRZ) zone, and wustite is
(FeOi.oe) reduced to DRI (S3) in the final reduction (FRZ)
Mid-fee zone. The top gas (S9) is cleaned and tempered
i 8 gas (not shown on the flowsheet) and recycled to the
X FRZ (S6). Additional reducing gas (S7) is
prepared by reforming natural gas (S4) and steam
Final red'n (S5). The natural gas consisted of <pCH4 - 98 %,
balance N2. The composition of the recycled top
zone gas is shown in the following table.

DRI (Fe +^ Recycld H20 co2 H2 CO N2
FeOi "" "" Top gas
6.2 5.2 60.1 20.7 bai

I 5 Steam The system is described by four chemical
reactions that take place in the noted reactors.

PRZ: Fe203 + 0.88H2 -»■ 2FeOi.06 + 0.88 H20 [6.90]

FRZ: FeOi.oe + 106H2 -> Fe + 1.06H2O [6.91]

GR: CH4 + H 2 0 - > CO + 3H [6.92]

PRZ, FRZ, and GR: CO + H20 -> H2 + C02 [6.93]

Chapter 6 Reactive Material Balances 365

Equation [6.90] is irreversible in the PRZ. Equation [6.91] has two user-defined parameters.
First, for this example, the extent of FeOi.06 reaction was set as 0.95, which is equivalent to an
O/Fe ratio for the DRI of 0.053 (i.e., Y = 0.053 on the Rist diagram). Second, the value for the
/?H20/pH2 in S8 was set at 80 % of the Кщ value for Equation [6.91] at the S8 temperature to give
adequate distance for the operating line away from point W. Equation [6.92] is irreversible in the
GR. Equation [6.93] (the WGR) reaches equilibrium at 923 К in the PRZ, and at outstream
temperatures in the other two reactors. Кщ for the WGR was taken from Figure 6.32.

FlowBal assigns temperatures to streams, not devices, because a device seldom has one
temperature that characterizes it. In this system, the gas phase is always hotter than the solid phase
in the two zones. The simplifications caused by using only two zones means that we might choose
a reaction temperature different than either the incoming or exit gas stream.

The furnace feed was 350 kg of ore/min, with 430 m3/min (actual) recycling gas flow. The
reducing gas (S8) had a user-defined degree of oxidation of 0.06, but the value of X to the shaft
furnace will be greater than 0.06 because S6 has X = 0.124. The value of X for Rist-diagram
purposes is based on the combined streams S6 and S7, the details of which will come from the
FlowBal material balance.

FlowBal wrote 24 equations for 30 unknowns, so six more are needed. The abovementioned
criteria indicate the need for one XR equation and five related to gas composition. The FlowBal
formatted equations are:

| Final Reduction Zone, Reaction 2, XR-FeOl-06 = 0,95 |
|{S9-H2}*{S9-CQ2> - 2*{S9-CO}*{S9-H20> |

[{S8-H2}*{S8-CQ2} - 0.75*{S8-CO}*{S8-H2O} |
|{S7-H2}*{S7-C02} ■ 0.64*{S7-CO}*{S7-H2O} [

|{S8-H2Q> - 0.48*{S8-H2> |
|({S7-H20}+{S7-C02}) - 0.06*({S7-H2O}+{S7-CO2}+{S7-H2}+{S7-CO})

FlowBal originally did not converge because the WGR direction for one or more streams was
written in a direction opposite to what happens in the reactor. FlowBal cannot find a feasible result
in such a situation when the Assume Non-Negative box is checked on the Solver options dialog
box. When this happens, either uncheck the box or select the built-in FB Solver. This is a
Newton-Raphson program that automatically sets all stream variables to positive values, while
omitting this restriction for the R-R values.

Figure 6.50 shows the results. FlowBal also produces an accessory results table with amount
units (kg-mol) that are better for calculating certain process characteristics. From the amount
table, X = 0.095 for combined streams S6 + S7 entering the bottom of the shaft furnace, and 0.402
for the gas exiting the top of the furnace (S9). 4.384 mole of FeOi.5 are reduced per 21.66 mole of
combined S6 + S7, or 20.66 mole of nitrogen free shaft input gas, so the equation of the operating
line is:

Y = 4.71X-0.39

Compare this to the result of Example 6.13, where one mole of reducing gas reduced 0.270
mole of FeOi.5. For this example, only 0.212 mole is reduced.

There is some flexibility in deciding how much tempered top gas can be recycled to the
furnace; here an arbitrary 430 m3 was selected. Not all of the tempered top gas can be recycled
because there must be a bleed for N2 to prevent it from building up. FlowBaPs Repetitive Solve
feature was used to explore the effect of changing this variable over the range 400 - 490 m3. As
the amount of recycled top gas increases, less reducing gas (S7) is needed, so there is an incentive
to recycle as much top gas as possible, commensurate with bleeding enough top gas to remove the

366 Chapter 6 Reactive Material Balances

inerts to an acceptable level. Figure 6.51 shows the effect of recycled gas flow on X and the slope
of the operating line.

P (atm) 2.7 2.8 2.9 3 3 2.9 2.9 2.8 2.6

T(K) 300 1170 1185 300 800 1230 1250 1185 840

Str-unit Mass (kg) Mass (kg) Mass (kg) Volume (m3) Volume (m3) Volume (m3) Volume (m3) Volume (m3) Volume (m3)

Spec-unit Mass pet Mass pet Mass pet Volume pet Volume pet Volume pet Volume pet Volume pet Volume pet
Str-name Ore PR Oxd DRI NatGas Steam RecyGas RdcngGas MdFceGas TopGas
Streams 1 2 3
4 5 6 7 8 9
Flow 350 319.1 248.5 18.2 59.8 430 329.2 752.3 574.3
Fe203 100
FeO1.06 0 0 0 0 0 0 0 0 0

100 6.42 0 0 0 0 0 0
Fe 0 0 93.58 0 0 0 0 0 0
CH4 0 0 0 98 0 0 0 0 0
H20 0 0 0 0 100 6.2 4.97 22.87 25.26
C02 0 0 0 0 0 5.2 1.00 6.57 13.09
H2 0 0 0 0 0 60.1 71.16 47.65 45.27
CO 0
0 0 0 0 20.7 22.39 18.25 11.73
N2 0 0 0 2 0 7.8 0.48 4.65 4.65

Figure 6.50 FlowBal result for the shaft furnace reduction of hematite using a mixture of
tempered recycled top gas and newly-formed reducing gas from a natural gas steam reforming
device. Note that volume units are actual, so you must convert to STP for comparison purposes, or
use the amount result array.

Simulations of this type of zone-simulated continuous-flow process are improved by
increasing the number of zones. If we were interested in a better simulation that included a heat
balance, we would need a heat-exchange zone at the top, we would use three oxide reduction
reactions, and include the presence of CH4 in the top gas. Equation [6.89] can proceed as
temperature falls below about 900 K, especially above atmospheric pressure.

Shaft Furnace Reduction Parameters 0.100

4.82 0.099

4.80 —o—OL slope 0.098 ^

<D 4.78 —D— X in S6+S7 ^/^ s \ CO
с 0.097 £
= 4.76 ^^' ^}s CO
§■ 4.74 0.096 .E
^/^ s X
О 4.72 0.095
О ^S у
§■ 4.70 0.094

" 4.68 0.093
500
4.66 Y У '
4.64
420 440 460 480
400 recycled gas flow, m /min ambient

Figure 6.51 Effect of variation in recycled gas flow (S6) on the slope of the operating line and the
degree of oxidation in the combined gas instream to the furnace (S6 + S7). An increase in S6
means less reducing gas (S7) is needed, but the total gas flow into the furnace increases. This is
because the increase in X means that more (S6 + S7) is needed.

6.5.7 The Roasting of a Sulfide Concentrate

Many of the transition and heavy elements occur in the earth as sulfides. These sulfides, such
as ZnS, PbS, CuFeS2, and SnS, are concentrated from ore by flotation to produce a concentrate

Chapter 6 Reactive Material Balances 367

containing predominantly one sulfide. If the ore is complex, the concentrate may contain several
sulfides. There are many different methods for the recovery of metals from sulfide concentrates.
In some cases the first step is oxidizing {roasting) the sulfide with air to convert the metal sulfide
to an oxide. This is an exothermic process that requires control of the temperature and roaster gas
composition. If the sulfide is to be roasted to remove all of the sulfur {dead roasted), excess air is
used and the sulfide minerals are the limiting reactants. For partial roasting, air is the limiting
reactant and the sulfide will be only partially oxidized. One way to carry out roasting is by means
of a fluidized bed, as was shown by the flowsheet in Figure 4.3 for a zinc sulfide concentrate.

Partial roasting is sometimes used as a first step in copper smelting. The concentrate is
mainly CuFeS2 and FeS2, along with some gangue minerals. In order to make a matte from the
roasted concentrate, the amount of air is limited so that not all the metal sulfides are oxidized.
When the calcine is smelted, the matte will contain a mixture of Cu2S and FeS containing wCu
between 50 % and 60 %. Figure 6.52 shows a simplified flowsheet for the roast/smelt part of the
process.

-► Roaster gas

Concentrate *\
*\ Roaster / ci,.v s^ -► Smelter gas

► Slag
* Matte

Figure 6.52 Flowsheet for roast/smelt portion of a copper smelter. The reverberatory smelting
furnace requires a burner to smelt the charge.

There is a direct relationship between the amount of roaster air used per amount of
concentrate, and the wCu in the matte. Various reactions can be written to depict the phases
present during chalcopyrite roasting, but for a material balance, the following three roaster
reactions suffice.

2CuFeS2 + 0 2 -► Cu2S + 2FeS + S02 [6.94]

FeS2 + 0 2 -► FeS + S02 [6.95]

3FeS + 502 -+ Fe304 + 3S02 [6.96]

All reactions are spontaneous and irreversible, but only reactions [6.94] and [6.95] proceed to
completion. The extent of FeS oxidation is less than one because air is limiting, and the extent
may be calculated on that basis.

During smelting the Fe304 reacts with FeS to form FeO, which reports to the slag:

3Fe304 + FeS -> S02 + lOFeO [6.97}

Reaction [6.97] is spontaneous but not complete; the extent of Fe304 reaction approaches one
so long as the wCu in the matte stays below about 55 %.

Consider a copper concentrate with wCu = 23.6 %, wFe = 30.1 %, wS = 34.6, balance Si02.
Material balance calculations are simplified by changing the concentrate mass composition to
amount units of the minerals. On a basis of 1 kg of dry concentrate, MMV-C shows that it has
3.7138 moles Cu, 5.3899 moles Fe, and 10.7904 moles S. Mineralogically, the feed to the roaster
contains 3.714 moles CuFeS2, 1.676 moles FeS2, and 1.947 moles Si02. A material balance is
carried out by setting up a ledger in an Excel worksheet to calculate the amount of the various
species formed as a function of the amount of roast air for 1 kg of concentrate. A series of oxygen
consumption calculations was carried out according to the stoichiometry of Equations [6.94 -
6.96], followed by an iron balance calculation for the smelting reaction. Table 6.14 shows the
results ledger for 40 moles of air.

368 Chapter 6 Reactive Material Balances

Table 6.14 Ledger for Roasting and Smelting 1 kg of Copper Concentrate

moles air in 40.0 moles FeS oxidized by remaining 0 2 2.920
moles O2 in 8.400 moles FeS remaining after reax 6.82 2.470
moles CuFeS2 in 3.714 0.973
moles FeS2 in 1.676 moles Fe304 produced by reax 6.82 0.324
moles O2 used by reax 6.80 1.857 moles FeS oxid. by smelting (reax 6.83) 2.145
moles CU2S to matte 1.857 3.714
moles FeS from reax 6.80 3.714 moles FeS to matte 2.145
moles O2 used by reax 6.81 1.676 moles Cu to matte 4.002
moles FeS from reax 6.81 1.676 moles Fe in matte 484.2
moles FeS after 6.80 and 6.81 5.390 48.7%
moles O2 for reax 6.82 4.867 moles S in matte
mass of matte, g

wCu in matte

These calculations show that 6.464 mol of sulfur are oxidized by roasting (60 % of the
incoming) and 0.324 mol by smelting (3.0% of the incoming) to produce a matte having wCu =
48.7 %. 60 % of the incoming iron is removed in the slag .

Once the worksheet ledger is set up, it's easy to make a series of calculations of the wCu in
matte for various amounts of roasting air. Figure 6.53 shows the results, along with the mass of
matte. Considering the uncertainties in concentrate analysis and other simplifications, a linear
relationship adequately describes the effect of amount of air on Cu in matte.

59% Smelting Roasted Copper Concentrate 520
57%Vo \ * %Cu in matte = 0.062(nrT air) - 6.72
55% 500

% Cu in matte co
— matte wt, kg
i480
900 950 1000
volume of air, STP m3/tonne concentrate 1460

440 m
TO

420 E

400
1050

Figure 6.53 Relationship between the amount of roaster air and matte properties for the roasting
of one tonne of dry concentrate, followed by smelting. Results are based on reaction sequence
[6.94-6.97].

In contrast to copper roasting prior to smelting, zinc concentrate requires dead roasting before
further extraction steps. The naturally-occurring mineral form of zinc is sphalerite, ZnS. The ore
contains other metal sulfides such as pyrite (FeS2), chalcopyrite (CuFeS2), and galena (PbS). The
roasting step uses excess air to completely oxidize the concentrate. The ZnO in the calcine is
leached by sulfuric acid and the zinc is recovered by electrowinning. Complete conversion of ZnS
to ZnO is important because ZnS is not soluble in sulfuric acid. Figure 4.3 shows a diagram for
the roasting of zinc concentrate in a fluidized bed.

* A typical copper smelting slag also contains about 1 % Cu, which is ignored here.

Chapter 6 Reactive Material Balances 369

The main goal is complete oxidation of the ZnS to ZnO, but other factors play a part in
producing an easily-leachable calcine. For example, temperature must be controlled within narrow
limits to prevent agglomeration of the calcine which can cause incomplete oxidation. The very
exothermic oxidation of the sulfide minerals requires some way to remove heat from the roaster.
Most of the heat is removed by boiler tubes that line the inside of the roaster. Water is circulated
in the tubes, and flashes to superheated steam. If needed, a small amount of water can be sprayed
into the roaster bed to fine-tune the temperature control. The <jp02 in the roaster gas is also
important to assure complete oxidation, while both temperature and <jp02 are important in
minimizing the tendency to form basic zinc sulfate or ZnS04. The amount of air and its possible
enrichment with oxygen are important for other reasons, such as to minimize the amount of dust
carryover from the roaster to the downstream gas handling equipment. A material balance shows
the relationships between these variables, and establishes conditions for process control.

EXAMPLE 6.14 — Roasting a Zinc Sulfide Concentrate.

How much oxygen-enriched air (containing φ02 = 24 %) should be used per tonne of dry zinc
concentrate to give an offgas containing 3.5 %02? The chemical analysis of the dried concentrate
is given below, which enters moist (with wH20 =11 %). The term "Insol" refers to gangue present
in the concentrate that does not react during roasting nor dissolve in the acid used for chemical
analysis. Temperature-control water spray is added at a rate of 35 kg/tonne of dry concentrate to
maintain a bed temperature near 1150 K..

element Zn Fe Cu s Insol

mass fraction 54.6% 5.6% 3.3% 32.2% 3.8%

Solution. An examination of the concentrate analysis indicates that it does not sum to 100 %,
which is typical of an analysis of such a material. The cause is some combination of analytical
error and the presence of other minor elemental sulfides or oxides. We ignore the error and
proceed.

For a material balance alone, it is not always necessary to determine the nature of the
mineralogical species present in the concentrate; we can treat it as a mixture of metallic elements,
sulfur, and inert gangue. However, an energy balance requires that the mineralogical species be
determined so their heats of formation can be accounted for. This procedure was illustrated in
Example 5.2.

First the water feed is determined. At 11 % water, one tonne of dry concentrate has a mass of
1123.6 kg when entering wet, and another 35 kg of water enters as spray in the plenum, for a total
of 158.6 kg water in. Since water does not react in this system, it passes directly to the offgas.
Next we identify the oxides in their highest oxidation state: ZnO, CuO, Fe203, and S02.* Since we
haven't determined which sulfide species are present, we can only write reactions for the oxidation
of the elements. All of the reactions are spontaneous and complete because oxygen is present in
excess.

Zn + l/202(g) -> ZnO(c) [6.98]

Cu + 1/202(g) -► CuO(c) [6.99]

Fe + 3A02(g) -> !/2Fe203(c) [6.100]

S + 02(g)-*S02(g) [6.101]

As suggested earlier, an equation basis of one mole of the atomic element oxidized makes it
easier to keep track of the requisite amount of oxygen. Two ledgers were set up in an Excel
worksheet to make the calculations. The first two rows of Ledger 1 list values of the mass and

* The highest oxidation state of sulfur is as S03, but the extent of oxidation of S02 to S03 is very small at
typical roaster temperatures.

370 Chapter 6 Reactive Material Balances

amount of each reactant element in 1000 kg of dry concentrate. (The program MMV-C was used
to convert mass quantities to moles). The second two rows list the amount of 0 2 and N2 for the
stoichiometric oxidation of each element, according to the equations written above. The next row
is the amount of stoichiometric oxygen-enriched air. The last two rows display the amount of
excess 0 2 (and associated N2) for each element when using the specified amount of excess air.
These results are linked to Ledger 2, where a bordered cell indicates the fractional amount of
excess air. A change in this cell changes the values in the last two rows of Ledger 1, and most of
the values in Ledger 2. Goal Seek was used to change the value in this box to get φ02 = 3.3% (as
specified in the problem statement), which occurred at 0.168 volume fraction excess air (16.8 %
excess air). The process requires 1660 STP m3 oxygen-enriched air/tonne of dry concentrate.

Assignment. A careful re-analysis of the concentrate indicated that the missing mass is Pb, which
appears in the calcine as PbS04. Repeat the excess air calculation taking Pb into account.

Ledger #1 : Oxidation of Elements in 1000 kg of Dry Concentrate

Zn Fe Cu S Total gas H20 SiO;

Mass, kg 546 56 33 322 — 158.6 38
—
Moles 8.35 1.00 0.52 10.04 — 8.81 —

n 02, stoich. 4.18 0.75 0.26 10.04 15.23 —

«N2, stoich. 13.22 2.38 0.82 31.80 48.23

Ox-Enr air, stoich. 17.40 3.13 1.08 41.85 63.46

«o2,xs 0.70 0.13 0.04 1.69 2.57
8.12
«N2,XS 2.23 0.40 0.14 5.36

Ledger #2: Roaster Gas Produced When Gas has 3.3 %02.

XSA-|-^ 0.168 "| O^ N2 SO^ ϊ ζ θ Total

Moles 2.57 56.36 10.04 8.81 77.77

Volume fract. 3.30% 72.46% 12.91% 11.32% 100.00%

6.6 The Production of Gases with Controlled Oxygen and Carbon Potential

Heretofore, we have used the C02/CO or H20/H2 ratio as criteria for determining the tendency
of a gas mixture to reduce a metal oxide. Figure 6.36, for example, defines the value of the
H20/H2 ratio required to reduce wustite, and a similar calculation can be performed to express the
C02/CO ratio for the same purpose (see expanded charts for this purpose in the Handbook CD in
folder Charts). However, from a more fundamental standpoint, the criteria of whether or not a
particular mixture of gases can reduce an oxide to the metal (or, conversely, oxidize a metal) is the
partial pressure of oxygen, ovp02. This value, the/?02, is an unambiguous indicator of the oxygen
potential of a gas mixture.

A p02 calculation involves а Кщ calculation of a reaction containing 02(g) and at least one
other species containing oxygen. Equation [6.5] is a good example of such an equation. Equation
[6.5a] shows the Кщ expression for Equation [6.5]. The Кщ expression defines the equilibrium
p02 of a gas having a known temperature, pCO and/>C02.

CO(g) + y202(g) - C02(g); ^eq = PC02_ [6.5a]
(PCOyP02

In a similar manner, the Кщ expression for Equation [6.39] defines the/?02 of a gas having a
known temperature, pR2 and/>H20, as shown by Equation [6.39a]:

H2(g) + V*Hg) - H20(g); Кщ = ^ 2 , 0 [6.39a]
(РЯ2)4Р°2

Chapter 6 Reactive Material Balances 371

Owing to the very small value of p02 for a reducing gas, it is sometimes more convenient to
express the oxygen potential Кщ expressions in logarithmic terms, especially since the variation of
log(Xeq) follows a near-linear relationship with reciprocal temperature.

In Section 6.4, we also mentioned that gas mixtures can carburize or decarburize a metal,
which is important in the heat treatment of steel. The carbon potential of gas mixtures containing
C02 and CO can be defined by the Кщ expression for Equation. [6.6].

2 [6.6a]

C(graph) + C02(g) - 2CO(g) Кщ = ^ C ° }

Here, the carburizing/decarburizing potential for the gas is unambiguously defined by the
activity of carbon (aC), relative to graphite as the standard state. When carbon is present, its
activity is one, and the equilibrium position can be defined relative to the pCO and pC02.
Equations [6.5a, 6.6a, and 6.39a], together with an equation for the formation of CH4(g) from the
elements, are sufficient to define the equilibrium position for the C-O-H system. Including
carbon, there are seven species, so NIRx for the C-O-H system is four.*

Calculations involving the carbon and oxygen potential fall into two categories. First, given a
gas composition and temperature, what is the aC andp02? Second, what should the reactants be to
generate a gas with a specified aC and ρ02Ί These two categories are two sides of the same coin;
both require writing material and heat balance equations, and correctly applying Кщ constraints.
This type of calculation involves a heavier dose of chemical thermodynamics that we've heretofore
used, so are not covered in the Handbook. Instead, please see folder Atmosphere on the Handbook
CD, which contains a more complete explanation of the system, and some worked out examples
using Excel, FlowBal, and THERBAL.

6.7 Processes Controlled by Chemical Reaction Kinetics

Many of the reactive processes discussed so far involve chemical reactions that do not
proceed to equilibrium in the time allotted. In such a situation, we typically used data about the XR
of a critical reactant, or some other information from experiments or plant data to bring the DOF to
zero. But in a few cases, the rate-controlling reaction has been studied to the point where we know
the reaction order and the rate constant k. This information can be used with the appropriate rate
equation to calculate a material balance. Please review the principles of chemical reaction kinetics
that were discussed in Section 5.5.2.

Carbon black, sometimes called lampblack, is formed by the non-catalytic thermal cracking of
acetylene (C2H2) according to Equation [6.102]. When a mixture of acetylene and nitrogen is
introduced into a hot reactor, partial decomposition occurs, and the gaseous products carry the fine
particles of carbon out. The outstream is cooled, and the carbon collected in a cyclone. The
reaction kinetics have been studied at various temperatures. At 1200 °C, the reaction is first-order,
with an acetylene half-life of 0.92 min. A continuously-mixed reactor operating at 1200 °C with a
volume capable of containing 12 moles of gas is used to produce carbon black at 1 mol/min. We
seek to calculate the required flow of a feed gas having <jpC2H2 = 20 %, balance N2. One-half mole
of C2H2 decomposing produces one mole of С

C2H2(g)-2C(s) + H2(g) [6.102]

Equation [5.40] is revised as follows:

In ЯС2Н2Ш ln(2) = kty2 [6.103]
nC2ÌÌ2oui

The aC can also be defined by chemical reactions other than Equation [6.6a].

372 Chapter 6 Reactive Material Balances

where tVl is the half-life, here 0.92 min. Therefore, к = 0.75 min \ A steady-state material balance
for acetylene (A) is:

Input flow of A - output flow rate of A - consumption of A = 0 [6.104]

(Fn)(xAin) - (Fut)(xA0Ut) - k(xA0Ut)( V) = 0 [6.105]

In order to produce 1 mol/min carbon, we need to consume Vi mol/min of A. Therefore:

consumption of A = 0.5 = k(xA0Ut)(V)

Solving, (xA0Ut) = 0.0556. We know that the Fn = Fu\ and χΛίη = 0.2. Therefore, the inflow
of gas must be 3.46 mol/min. The instream gas contains 0.692 mole of acetylene, and the
outstream gas contains 0.192 mole of acetylene and entrains one mole of carbon. The outstream
gas composition is 14.4 %H2, 5.6 %C2H2, balance N2.

The above calculation was relatively easy because there was no volume change during
acetylene decomposition, so Fn = F°ut. The arithmetic is basically the same for any type of reactor.
But when a volume change occurs, Fn Φ F°ut, and the calculations require an additional material

balance. For example, in Section 5.5.2, we looked at batch reactor decomposition of ammonia,

where each mole of ammonia decomposing produced two moles of product. Suppose that a stream

of pure NH3 is passed into a continuously mixed reactor at 700 К to produce a protective
atmosphere for annealing an alloy steel. We seek to calculate the required reactor volume to

produce an outgas having <pH2 = 65 %.

Letting A = NH3 and H = Hdi2d, nn'ftfusipe=ci1fy.5(nnAAmm - nA0Ui). Also, the outstream flow Fut = nA0Ut +
2(nAm - nAmX). The problem so we can pick any basis; let nAm = Fn = 1

mol/min. After the calculations, the reactor volume can be sized as desired by scaling. Equation
[6.106] expresses the relationship between nAout and nH°ut.

xHToTuOtUt = \.5-\.5nA0Ut = ол6^5с [6 106]
(2-nA0Ut)

Solving, nA0Ut = 0.235, so /^ut, = 1.765 mol/min. The outstream contains 65.0 %H2, 21.7
%N2, and 13.3 %NH3.

Based on the rate constant value of 0.0025 from Section 5.5.2, Equation [6.107] relates the
reaction rate of ammonia in the reactor to the in-minus-out rate:

k(xA0Ut)(V) = 1 - nA0Ut = 0.0025(0.133) V [6.107]

Solving, V = 2300 moles, or 142 m3 of gas at 700 К and 1 atm. The rate of thermal
decomposition of NH3 is 0.765 mol/min (Xi?NH3 = 0.765), so 1.765 times as much gas exits the
reactor as enters. We can scale the process by knowing that the reactor volume must be capable of
holding 2300 times the instream flow, in mol/min. The material balance arithmetic is simpler and
more generic if FlowBal is used to calculate the stream properties. Then you only need Equation
[6.107] to size the reactor.

A similar calculation for a plug-flow reactor is more complicated than the continuously mixed
or batch reactor because the gas composition changes along the length of the reactor. This means
the gas flow rate and stream composition varies along the length of the reactor.

6.8 The Reconciliation of an Existing Materials Balance

The main objective of making a material balance is to analyze what is happening in a process.
In some cases, a complete analysis-based materials balance is presented to the engineer as the
starting point for making a process analysis. However, such a balance is usually flawed. There is
usually a discrepancy between the overall material balance, or that of one of the important
elements. There is no opportunity to collect data, sample stream flows, or validate the data in any
way. In this section, we will see what can be done to reconcile flawed material balance data, here

Chapter 6 Reactive Material Balances 373

for an iron blast furnace. The results will then be used to calculate the extent of carbon reactions in
the stack and the mass of gas produced.

The iron blast furnace (BF) is the most common method of converting iron ore concentrate
into metallic iron. The BF is a counter-current reactor in which iron oxide is reduced by gas in the
shaft, and melted (smelted) in the hearth to produce a liquid iron-rich alloy (called hot metal as it is
tapped). Approximately 93 % of the world's iron is produced in this way. The main zones of the
BF are the hearth, the combustion zone, and the stack. The main reactions taking place in these
zones are as follows:

Stack Fe203 + 3(CO, H2) -> 2Fe + 3(C02, H20)
С + C02 -► 2CO
С + H20 -> CO + H2
(CaC03, MgC03) -> (CaO, MgO) + C02

Combustion с + v2o2 -+ с о
zone с + н2о -► с о + н2

Oil + 0 2 - + ( C O + H2)

{Hттear^tиh ^j 2C + Si0z2-> Si + 2СО
С + MnO -> Mn + CO

Figure 6.54 shows a sketch of a BF, together with the amount and composition of the main

streams. The gases leaving the combustion zone are entirely CO, H2 and N2. Reduction of iron
oxide occurs throughout the lower and middle portion of the stack to produce C02 and H20. An
important aspect of BF performance is the amount of carbon consumed in the stack by reaction
with C02 and H20 (called solution loss), which generates more CO and H2 for ore reduction. If the
solution loss is too low, then the entire amount of (CO + H2) for ore reduction must be produced in
the hearth. If too high, not enough С reaches the combustion zone. The amount of solution loss is

thus important in optimizing the heat balance for the furnace. A material balance on the BF can

indicate the amount of solution loss and hence indicate if it is at the optimum value for the

particular burden. The mass of the top gas can also be calculated to complete the mass balance.

The mass of the top gas was not listed, probably because it was difficult to measure. Before
using the data to calculate the this and the carbon solution loss, we must check the material balance
to see if it is correct, and if not, how we might find the likely errors and correct them. Since we
have both the mass and composition of most of the streams, there is redundant data which can help
reconcile any discrepancies. The BF has 11 elements and 13 streams, and at first glance, a material
balance calculation would seem to be a very complex task. However, the task is not all that
complex because many elements appear in only two or three of the 13 streams, and several
elements are in species that do not undergo any chemical reactions. An organized approach is to
take the elements (or species if they do not react) stream by stream, and make a ledger to track
their mass in and out.

The first step is to examine the data to see if the compositions add up to 100%. This is best
accomplished in a ledger format (next page), which is shown first for the input materials. The
accuracy of the chemical analysis indicates that the mass of each substance or element may be
rounded off to the nearest 10 kg. The mass of all input species is about 173 080 kg while the
output mass is 59 020 kg plus the mass of the top gas. As will be shown later when the top gas
mass is calculated, the output mass is about 340 kg less than the input mass.

The compositions sum to 100 % for the air, fuel oil and steam, so these values are not shown
in the ledger. This is reasonable for such common and easily handled substances. None of the
other stream compositions add up to 100%, probably owing to the presence of small amounts of
alkalis, phosphorus, zinc, chlorine, and other unreported elements.

374 Chapter 6 Reactive Material Balances

Iron ore (69,800 kg + 2930 kg H20) } Top gas
27.67% C02, 25.00% CO,
91.80% Fe203, 4.84% Si02, 1.25% CaO 4.46% H20, 0.21% H2

0.95% MgO, 0.51% MnO, 0.45% Al203 Dust (1650 kg)
72%Fe203, 13% C,
BOF slag (2240 kg) 3% CaO. 2% MgO
29.15% FeO, 13.30% Si02,
38.80% CaO. 10.50% MgO, Slag (10,380 kg)
3.10% MnO, 4.12% Al203, 0.092% S 39.30% CaO, 37.30% Si02,
10.4% MgO, 9.13% Al203,
Limestone (3760 kg) 1.05% MnO, 1.57% S

54.60% CaO, 0.40% MgO Hot metal (47,260 kg)
94.20% Fe, 4.48% C,
Dolomitic limestone (980 kg) 0.64% Si, 0.56% Mn,
0.062% S
32.6% CaO, 19.1% MgO
Coke (22,970 kg + 1100 kg H20)

91.1% C, 3.88% Si02, 0.24% CaO,
0.15% MgO, 0.15% Mn,
2.42% Al203, 0.85% Fe, 0.82% S

Hot Blast Air (63,960 kg) ^COMBUSTION/

24.0% O, 75.9% N,0.1% H - 1 ZONE 4
Oxygen (3300 kg)
Fuel Oil (1880 kg)

87.3% C, 12.15% H, 0.55% S
Steam (500 kg)

88.8% O, 11. 2% H

Figure 6.54 Mass balance data for the operation of a modern blast furnace for 1 % of the
production of a 24 hour day. Composition units are mass fractions for all substances. The N2
composition of the top gas is obtained by difference. The blast air composition is listed in terms of
elements; the presence of H reflects the moisture in the air. Ar is combined as N2.

Input Streams

Ore Fe203 Si02 CaO MgO MnO A1203 Total % of total
kg 64,080 3380 870 660 360 310 69,660 99.8%

BOF slag FeO Si02 CaO MgO MnO A1203 S Total j % of total
kg 650 300 870 240 70 90 2 2220 99.1%

sCoke Fe Si02 CaO MgO Mn A1203 С 1 Total % of total

kg 200 890 50 30 30 560 190 20,930 22,880 99.6%

Limestone CaO MgO co2 Total % of total

kg 2050 20 1610 3680 97.9%

Dol. lime CaO MgO C02 Total % of total
kg 320 190 460 970 99.0%

Ore water 2930 Coke water 1100

О N H С s Total
Air 15,350 48,550 60 0 0 63,960
Total species mass in:
Oxygen 3300 0 0 0 0 3300
173,080 kg
Steam 440 0 60 0 0 500

Fuel oil 0 0 230 1640 10 1880