Handbook_on_Material_and_Energy_Balance_Calculat1

Chapter 9 System Balances on Reactive Processes 525

the same either way because linear heat content equations are acceptable. But when the heat
balance requires multi-term non-linear heat content equations, selecting ART as an independent
variable can often simplify the number or complexity of equations.

Oxy-Steam Oil Reforming

0.24 0.470
0.465
D Steam о 02 0.460 о
Linear (Steam) Linear (02 )
0.23 E

Ö °·22

E 0.21 steam = -2.91E-04T + 0.582
co
ω 0.20

0.19 <T

0.18 1225 1250 1275 1300 1325 0.445
1200 1350

ART, К

Figure 9.11 Amount of steam and oxygen required for adiabatic reforming of one mole of fuel oil
to produce a reducing gas having a degree of oxidation X = 0.06.

Assignment. Calculate the effect of increasing the steam temperature on the value of X of the
reducing gas when operating at an ART of 1260 K.

EXAMPLE 9.5 — Oxidation ofS02 to SO3 for Sulfuric Acid Production.

Offgas from smelting and roasting processes contains S02 which cannot be discharged to the

environment. The S02 must be oxidized to SO3 for production of H2S04. Please review the

discussion of this process described in Section 9.1.1. After cleaning, the roaster gas is at 400 °C.

It is sent to a catalytic converter which operates nearly adiabatically. Additional air (at 400 °C) is

added to promote the oxidation of S02. (Please review Section 6.2.1 for additional background.)

The catalyst promotes the following reaction:

S02(g) +1/202(g)^S03(g); Keq = /?S03 [9.22]
(pS02)^07

Make a system balance to determine the composition of gas leaving the converter for different
amounts of 400 °C air added to the 400 °C zinc roaster offgas described in Example 6.14 Ptotal =
1 atm. A catalytic converter operating on a slightly different roaster gas has an ART of «600 °C.

Data. The zinc roaster gas composition (slightly rounded off) is:

<pS02=12.9%; φΗ20=11.3%; φ 0 2 - 3 . 3 % ; balance N2

Solution. FREED tables were used to determine heat content data for air and the roaster gases and
the АЯ°ГХ for Equation [9.22], following the procedures outlined earlier for such calculations. The
heat content was a linear function of temperature over the short temperature range of 400 - 600 °C,
and the АЯ°ГХ had a negligible change with temperature within ±25 °C of 600 °C. A summary of
the heat content and heat of reaction data is shown in the next page table.

526 Chapter 9 System Balances on Reactive Processes

#гЯ4оо, = A + Bf, kcal/mol
AВ

Air -3.054 0.00762
Roaster gas -3.354 0.00837

ДЯ°гх at «600 °C = -23.45 kcal/mol S03

Since the formation of S03 is exothermic, the temperature will rise as the reaction proceeds.
According to LeChatelier's principle*, it's extent will be less as temperature rises. Reversibility
can be checked by calculating Кщ for Equation [9.22] at different temperatures near 600 °C using
FREED's Reaction tool. Figure 9.12 shows the results. Кщ varies between about 6 at 630 °C to
16 at 570 °C. Reaction reversibility is therefore verified, withX7?S02 less at higher temperature.

Log Keq for S03 Reaction vs. 1/f

1.3

1.2 so2 + 1ΛΟ2 -» so3
Ф 1.1 Log(Keq) = 2417It - 3.029
§> 1.0
^00*^r
0.9
^&^0"^

0.8 ^ * ^

0.00157 0.00162 0.00167 0.00172 0.00177

1/f, °C

Figure 9.12 Relationship between log(Xeq) near 600 °C for Equation [9.22] and reciprocal
temperature (in degrees Celsius).

From stoichiometry, effective conversion of the S02 requires the 02/S02 ratio in the gas to be
more than 0.5, so clearly the roaster gas requires added air. For a basis of 100 moles of roaster gas,
we need at least 15 moles of air, and probably much more, to increase the extent of S02 oxidation.
To show the effect air has on oxidation of S02, we can calculate the ART and gas composition as a
function of amount of adding between 10 to 150 moles of air.

Before getting into the details of the system balance, it's important to understand what's
happening during the passage of air and roaster gas through the catalytic converter. The mixture of
air and roaster gas must be at least 400 °C on entering the converter to "kindle" the reaction. S02
oxidizes as the gas mixture moves along the length of the converter, with a commensurate rise in
temperature. However, the extent of S02 oxidation (as determined by the value of Кщ) decreases
with increasing temperature. By the time the gas mixture leaves the conversion reactor, it has
reached equilibrium (at the ART). Adding more air increases the reaction tendency but the
products absorb more heat, so the ART is likely to be lower as the amount of air increases. The
system balance results are the same irrespective of what reaction mechanism we choose or what
actually happens in the reactor. Normally we choose mechanism the that simplifies the arithmetic.

The simplest strategy is to heat the roaster gas and air from 400 °C to the ART and then carry
out Equation [9.22] to the extent governed by the value of Кщ at the ART. In essence, the ART is
the basis temperature. Since the ART is not known, we carry it as a placeholder variable in the
system balance equations, and use Solver to find a value of the ART that causes the sum of the
heat effect terms to equal zero. Since we are making the same calculation for multiple air

* Le Chatelier's principle states that the extent of an exothermic reaction always decreases with increasing
temperature.

Chapter 9 System Balances on Reactive Processes 527

additions, we will use SuperSolver for all arithmetic. The steps are outlined below, with H used to
indicate the heat term in kcal units, and Г indicates the ART.

• Step 1. Heat N moles of air from 400 °C to ART.
#i=tf(0.00762T- 3.054)

• Step 2. Heat 100 moles of roaster gas from 400 °C to ART:
tf2=100(0.00837T- 3.354)

• Step 3. Carry out Equation [9.22] at T to form Y moles of S03.
# 3 = Y(-23.45)

• Step 4. Calculate the partial pressure and moles of each reactive gas at the ART. This
requires eight equations, seven for gas species and one for the heat balance (Нг+ H2 + H3 = 0).

To simplify the arithmetic, the amounts of H20 and N2 were combined into one species, "inerts".
The partial pressure of each species is designated by a lower-case letter (d, y, x, and i for S02, S03
02, and inerts), and D, Y and X for moles (we already know I, the moles of inerts, by specifying a
basis of 100 moles of roaster gas and the value for TV). The seven gas species equations are:

d= D x =-D + Y + X + I ; i=l-y-d-x
D+Y+X+I D+Y+X+I

2417 -3.029 X = 16.2 + 0.21W - D -1.5Y; 12.9 = D + Y
10ÄFT

dVx

The first three equations define the partial pressures in terms of moles, the fourth equation
specifies the total pressure, the fifth equation is Кщ, the sixth equation is the 0 2 balance, and the
seventh equation is the S balance. Solver was used first to make sure the equations were written
properly and a solution was feasible. Then SuperSolver was used to calculate the values of the
seven gas variables and temperature for 14 different values of TV, starting at N = 10. Figures 9.13
and 9.14 show the results. The ART is maximized at about TV = 20, but Y increases throughout, у
also maximizes at the same N, while x increases throughout. The converter gas amount increases
from 106 to 244 moles over the range.

Adiabatic Oxidation of Roaster Gas to S 0 3

<

25 50 75 100 125 150
moles air added to 100 moles roaster gas

Figure 9.13 Relationship between ART and amount of S03 formed by oxidation of S02.to the
amount of air added to a zinc roaster gas,

528 Chapter 9 System Balances on Reactive Processes

0.12 Partial Pressure of Gases at the ART

E 0.1 25 50 75 100 125
+ω* moles of air added to 100 moles of roaster gas
о 0.08
i-
<03<>/)/) 0.06
Q. 0.04
<>/
0.02

0 150
0

Figure 9.14 Relationship between the partial pressures of resulting gas species and the amount of
air added to zinc roaster gas.

The results show that the extent of S02 reaction is increased by air addition, but at the price of
increasing gas volume (and a low/?S03). A better way to attain a high conversion of S02 to S03 is
to add only about 25 moles of air, cool the converter gas back to 400 °C, and send it through
another converter. In fact, more than two converters are commonly used, with intermediate
cooling back to 400 °C (or even less if the catalyst is fresh). The ART is less for each succeeding
converter, and with three or four converters in series the final ART may be as low as 450 °C. At
that temperature, Кщ for Equation [9.22] is about 170, and the X/?S02 reaction approaches 1.

Assignment. Calculate the ART for the reaction between an equimolar mixture of 02, S02, and N2
initially at 400 °C. Ptotal - 2 atm.

An interesting application of the adiabatic reaction concept is calculating the AFT in the iron
blast furnace where the hot blast enters the bed of coke carbon (the raceway region of the blast
furnace). The gas so produced is called the bosh gas, and if dry air is used, consists only of CO
and N2. The blast furnace has been mentioned several times already, and was the subject of a
material balance in Chapter 6. Recall that heat for the blast furnace reactions comes from
preheating the blast air and burning coke carbon in the raceway. But coke is expensive, so various
blast modification techniques have been used to minimize the amount of coke required to produce
hot metal. The main blast modifications are:

• Superheating the blast to 1220 °C or more.

• Fuel injection such as natural gas, oil, or coal.

• Oxygen enrichment.

• Steam injection.

Each modification changes the raceway bosh gas temperature. High AFT temperatures are
best for minimizing coke combustion, but if too high can cause early melting of the above-bosh
oxide burden. Each furnace burden has an upper limit on blast AFT, so the objective is to modify
the blast in a way that will decrease coke consumption while maintaining a certain AFT. The AFT
calculation assumes that the coke carbon arrives in the raceway at a temperature somewhat less
than the AFT (here we assume 100 degrees below the AFT), and in the simplest case of air-only
blast, produces only CO. The blast furnace operator can adjust the above four factors to control the
AFT, while at the same time changing the composition of the blast and the coke requirement.

An AFT calculation for a dry air hot blast requires heat content equations for air, C, N2 and
CO. FREED's Graphics tool and Excel's Trendline tool were used over appropriate temperature

Chapter 9 System Balances on Reactive Processes 529

ranges to obtain a linear approximation to heat content data in terms of kcal/kg, for temperature
units of degrees Celsius.

#t-#25, kcal/kg = A + B/

Air N2 H2 CO с
в 0.288 0.314 4.316 0.316 0.522

A -31 -65 -1256 -63 -202

Blast furnace calculations usually use mass units so we adopt a basis of 1 kg of coke С
burned, which in turn requires 1.332 kg of 02. Using wO of dry air = 0.232, the mass of air
required is 5.74 kg. The product mass is 4.41 kg N2 and 2.33 kg CO. The steps in making a heat
balance are: cool the air to 25 °C, cool the С from (AFT - 100) to 25 °C, burn the С at 25 °C, and
then heat the product gas to AFT. The АНсотЪ of С to CO by 0 2 at 25 °C is -2197 kcal/kg С The
sum of all heat effects is set to 0, and the AFT calculated. A sample calculation for air at 1000 °C
is shown below. The heat effect for each step is in kcal, designated H.

1. Cool air: #=-5.74{0.288(1000)-31} =-1475

2. CoolC: #=-{0.522(AFT-100)-202}=-0.522(AFT) + 254

3. Burn С to CO: # = -2197

4. Heat CO to AFT: tf= 2.33{0.316(AFT)-63} = 0.736(AFT)- 147
5. Heat N2 to AFT: H= 4.41 {0.314(AFT) - 65} = 1.385(AFT)-287

6. ΣΗ=0: 3852+1.60(AFT) = 0

Solving, the AFT = 2408 °C. The observed temperature on the inside of the raceway will
naturally be less than this because of other material entering that region, but the important thing is
not so much what the AFT is as how much it is affected by blast modifications, and how much to
change two or more modifications (those that lower AFT balanced against those that raises it) to
bring the AFT back to a desired value. For example, suppose we want an AFT of 2430 °C, which
is to be attained by some combination of adding superheated steam to the blast while increasing the
temperature of the dry air (designated tair). The reaction of steam with С to form CO and H2 is
endothermic, and in addition, consumes some of the С Therefore not as much С is available for
the exothermic formation of CO, and since not as much air is required, less heat is brought in by
the blast air. These deficits must be made up by heating the blast air to a higher temperature than
would otherwise be necessary. In recompense to producing less CO, the addition of steam
produces H2, which is an effective reducing agent for iron oxide.

A worksheet was set up in Excel to make repetitive calculations for different amounts of
added steam. The basis was kept at 1 kg of С burned, so 2.33 kg of CO is produced irrespective of
the amount of steam added. The steam enters at 450 °C and 20 bar. A system balance worksheet
was constructed so that each entry of steam mass resulted in a material balance for the blast and a
value for /air. The steps in the calculation are as follows:

1. Calculate the mass of coke С consumed by steam. (1.50 kg of steam consumes 1 kg of
carbon). Calculate the mass of CO (2.33 kg) and H2 produced by H20/C reaction.

2. Calculate the mass of oxygen and air required for the remaining С Calculate the mass of
CO produced by air oxidation.

3. Cool the air from its hot blast temperature (4ir) to 25 °C.

4. Cool the С from 2330 °C to 25 °C.

5. Cool and depressurize the steam from entry condition to saturated steam at 25 °C using
SteamTab program. (-194 kcal/kg steam)

Although coke is only about 92% C, the heat of combustion of its С is greater than that of graphite C.
Since these two things tend to cancel each other out, we've used graphite data for coke.

530 Chapter 9 System Balances on Reactive Processes

6. Carry out reactions at 25 °C. (+2613 kcal/kg C oxidized by steam).

7. Heat gases to 2430 °C.

8. Sum heat effects and solve for fair. (SuperGoalSeek was used).

Figure 9.15 shows the results of a series of calculations. Going from dry air to 25 g steam/kg
of air requires a 250 °C increase in the blast air temperature. An interesting consequence of adding
steam is that the volume of bosh gas decreases while the volume fraction of reducing gas in the
bosh gas increases. These results are shown in Figure 9.16. A lower bosh gas volume means that
the blast furnace can be driven harder without danger of flooding, and a higher fraction of reducing
gas in the bosh gas means that a slight decrease in coke rate might be possible (in terms of mass of
coke per tonne of hot metal).

Our original calculation basis was a dry air blast, which can only be approximated if the
ambient conditions are a cold winter day, when the amount of water vapor in the air is very low.
On a summer day, air at 30 °C and 50 %RH contains 13.4 g H20/kg dry air. You might be
interested in seeing where this condition lies on the figures below.

Effect of Steam on Blast Air Temperature

1000 1050 1100 1150 1200 1250 1300

blast air temp, °C

Figure 9.15 Required blast air temperature to keep the AFT at 2430 °C when adding steam to the
blast (basis is one kg of coke С burned). The text box equation relates the two variables.

Figure 9.16 Effect of steam addition on bosh gas properties. The volume fraction of the reducing

gases (CO and H2) increases with the amount of steam injected, while the total bosh gas volume
(per kg of coke C) decreases.

Chapter 9 System Balances on Reactive Processes 531

Various commercial software programs are available for making equilibrium material and
heat balances on single devices, and calculating the AFT/ART. We've already mentioned some of
these (THERBAL, HSC, FactSage), and (in addition to FactSage) there is one web-based
calculator for making equilibrium calculations on multi-component systems that proceed to
equilibrium (Navier 2010). The main drawback for all of these programs (except FactSage) is that
you must know which species exist at equilibrium.

9.3.2 ART for Condensed-Phase Reaction Processes

A number of material production processes occur among liquids and solids. For example the
production of intermetallic compounds and cermets can take place by direct reaction synthesis.
This requires that the reaction is sufficiently exothermic to heat the products to a temperature
where rapid reaction occurs. To assure a rapid reaction, powdered reactants are intimately mixed
and pressed into a dense compact. The mixture is heated in an inert atmosphere until reaction
commences. Alternatively, the compact is placed in a ceramic container and a spark or some other
ignition source is directed to a point on the mixture. Once ignition occurs, the reaction spreads
rapidly. The heat loss is small compared to the heat of reaction so it approaches adiabatic status.
For example, consider the formation of titanium aluminide from the elements:

Ti(c) + 3Al(c) -> TiAl3(c); Atf°rx at 298 К - -35 000 cai

The Hi4o0-H29s for TiAl3 is 28 250 cai, so there is over 6 kcal more heat than is necessary to
reach 1400 K, where one would expect complete reaction to have taken place in less than a second
(the melting point of TiAl3 is about 1650 K). This example is oversimplified in that there are other
aluminides which may form in small quantities. If the adiabatic reaction temperature (ART) is too
high, it can be tempered by mixing in a little bit of the reaction product.

Another type of condensed-phase reaction is the Thermit (sometimes called Thermite) process
(Wikipedia 2010), which is used in remote locations to weld steel rails. The Thermit process is
one of several metallothermic reduction processes in current use. It involves the aluminothermic
reduction of magnetite to form iron and aluminum oxide. A mixture of granular aluminum,
magnetite and small pieces of iron is ignited with a magnesium wire, and in a few seconds, a pool
of molten iron is generated and flows from the holding crucible.

Metallothermic reduction is very useful because metals such as titanium, vanadium, niobium,
zirconium and uranium cannot be reduced by such common reducing agents as C, H2 or CO.
Either their oxide (or halide) is too stable, or they form carbides and hydrides instead of metal. A
variety of techniques has been developed to produce these metals, such as electrowinning from a
fused salt, or reduction by an even more reactive metal. Almost all Ti is prepared by Mg reduction
of TiCl4 (the Kroll process—see Example 5.5). The reaction propagates throughout the mix very
quickly. The ART must be above the melting point of at least one of the products so that they can
be easily separated after the mass solidifies.

EXAMPLE 9.6 — Metallothermic Reduction of Uranium Tetrafluoride.

A process is being designed to produce uranium from the fluoride. Mg is the reducing agent
and the products are to be U(/) and MgF2(/)· The reactants are intimately mixed and placed in a
properly sealed container, and the mixture is ignited with an electrical spark (Wikipedia 2010). U
melts at 1408 К and MgF2 at 1536 K, so to obtain both products in a molten state, the calculated
ART should be >1650 К to allow for heat losses to the container. The objective of this example is
to calculate the ART for Mg reduction of UF4 to see if it is above 1650 K, and if not, what
alternatives are possible to "boost" the ART.

Data. All data are from FREED.

* The heat loss is mainly on the outside where the compact is next to the container.

532 Chapter 9 System Balances on Reactive Processes

Solution. The first part of this example is to calculate a heat balance for reactants initially at 298,
and products at 1650 K. If AHprocess is negative, the ART is >1650, and a follow-on calculation will
determine the value of ART. If AHpr0CQSS is positive, the process must be modified to increase the

ART. The Af/°rx for the production of one mole of U at 298 К and the heat contents of the

products at 1650 К are:

2Mg(c) + UF4(c) -> U(c) + 2MgF2(c); Atf°rx = -327.6 kJ [9.23]
#165о-Я298 for two moles of MgF2 = 2(166.05) - 332.1 kJ

H\65o-H298 for one mole of U = 70.1 kJ

A#process = 332.1 + 70.1 - 327.6 = +75.1 kJ

The positive sign on AHpT0CQSS shows that there isn't enough heat from the reaction to bring the
products to 1650 K. There are two ways to obtain additional heat. The first option is to heat the
reactants to some starting temperature before initiating the reaction. The second option is to
substitute some UF6 for the UF4 reactant. The first option will be solved analytically, and the
second option graphically.

An analytic solution uses HT-H29s equations for the reactants Mg(c) and UF4(c). The units are
J/mol of substance.

Mg(c): #т-#298 ( 2 9 8 - 9 2 2 к ) = 21.52Г+ 0.005807(F) + 7950/Г- 6960

UF4(c): #т-#298 (298 - 1309 K) - 123.56Г+ 0.004812(F) + 92 466/Г- 40 370

The units were changed to kJ by dividing the above equations by 1000. Twice the first
equation plus the second equation was set equal to the required AHprocess of 75.1 kJ. Goal Seek was
used to solve the equation to get T = 718 K. Therefore, the first option would be to heat the
reactants to 445 °C before igniting the mixture. This is a reasonable solution to the problem
providing the mixture doesn't ignite spontaneously during heating (before getting to 445 °C).

The second option is a graphical solution. AHpr0CQSS at 1650 К will be calculated for different
amounts of UF6 added. The stoichiometric reaction for production of one mole of U is:

(A)UF4(c) + (1 - A)UF6(c) + (3 - A)Mg(c) -► U(c) + (3 - A)MgF2(c) [9.24]

The first step was to calculate a material balance and АЯ°ГХ for values of A between 1 and
0.75. The next step is to calculate Ηί650-Η29^ for one mole of U and various amounts of MgF2. The
molar heat contents are.

MgF2(/): #165о-Я298 = 166.05 kJ/mol. U(Z): #165ο-#298 = 7 0 Л k J / m o 1

Finally, the АЯргосе88 is calculated. The results for each value of A are shown in the Table 9.7
and plotted in Figure 9.17.

Table 9.7 Products and Reactants to form One Mole of Uranium

A = moles UF4 1 0.95 0.9 0.85 0.8 0.75
moles UF6 0 0.05 0.1 0.15 0.2 0.25
moles U 1 1 1
2 1 2.1 1 2.2 1
moles MgF2 2 2.05 2.1 2.15 2.2 2.25
moles Mg -327.6 2.05 -412.4 2.15 ^97.2 2.25
402.2 -370.0 418.8 ^54.8 435.4 -539.6
АЯ°гх at 298 К, kJ 74.6 410.5 6.4 427.1 -61.8 443.7
40.5 -27.7 -95.9
#1650-#298? kJ

^-^'process? KJ

Chapter 9 System Balances on Reactive Processes 533
Mg Reduction of UF4 + UF6 Mixtures at 1650 К

0.95 0.9 0.85 0.8

mols UF4/1T10I of U produced

Figure 9.17 Plot of AHpr0CQSS for magnesium reduction of mixtures of uranium fluorides to produce
products at 1650 K. The reactants are at 298 K. The process heat is zero at an ART of 1650 K.

As UF6 is substituted for UF4, ЛЯ°ГХ becomes more negative, but there are also more moles of
MgF2 to heat to 1650 K. However, the АЯ°ГХ becomes negative faster than the Hi650-H29s of the
products becomes positive, so increasing the fraction of UF6 has the desired result. The proper
amount of reactants to attain an ART of 1650 К are 0.89 moles of UF4, 0.11 moles of UF6, and
2.14 moles of Mg.

Assignment. Calculate the amount of Ca to use (in place of some of the Mg) to reach a product
temperature of 1750 К using only UF4, with no reactant preheat. Neglect any heat of solution
between CaF2 and MgF2.

9.4 System Balances Using FlowBal

As you've seen from Chapter 8, FlowBal must first calculate the material balance before
being able to calculate a heat balance for the system. In general, we want to use FlowBal to find
process conditions which give a net heat = 0. By doing this, we answer the question: what should
a temporarily-specified stream property be such that the net device heat = 0? However, when the
material and heat balances are coupled, you need to insert one or more placeholder values to obtain
a material balance. FlowBal's heat balance calculator can then find the net device heats for the
solved material balance, but in general they will not be zero. The non-closure of heat balances can
be resolved with FlowBal's Repetitive Solve tool by varying the placeholders to close each
device's heat balance. This procedure is detailed in the FlowBal User's Guide.

Two processes will illustrate these concepts. First, a copper concentrate roasting process with
one reactor and two heat exchangers. Second, a rutile chlorination process which has several
devices. The example discussion assumes that you have read the FlowBal User's Guide, and have
used it on some simple processes.

The first example deals with the management of heat during the sulfation roasting of copper
concentrate. The objective of this process is to convert a chalcopyrite concentrate to CuS04 and
Fe203 for subsequent CuS04 leaching with water. Thermodynamic and bench-scale tests showed
that the roaster outstreams should be at 750 °C, and the roaster offgas should have certain
compositions of 0 2 and S02 to stabilize the CuS04 and to prevent the formation of Fe2(S04)3.
Specifically, the φ02 and <pS02 should be equal, and the Σ(φ02 + <pS02) should be at least 25 % of
the roaster gas. These two restrictions require the use of oxygen-enriched air. The process is very
exothermic, and requires two heat exchangers to manage the roaster/offgas temperature and to
temper the offgas prior to dust removal. Figure 9.18 shows the process flowsheet.

The process involves ten species, of which eight are involved in chemical reactions with five
elements. Therefore, NIRx = 3, which are shown as Equations [9.25], [9.26], and [9.27].

534 Chapter 9 System Balances on Reactive Processes

CuFeS2(c,/) + 3.7502(g) - CuS04(c) + S02(g) [9.25]
FeS2feyr) + 2.7502(g) - УХъОъ{с) + 2S02(g) [9.26]
[9.27]
S02(g) + V202(g)-S03(g)

Sup-htd 9 Warm R-gas WHB water

IHX steam R s t r _ 8 WHB 7 . Sup-htd
(HX) WHB steam
cooling water

2

Concentrate ■> Calcine (to leach)

Figure 9.18 Flowsheet for sulfation roasting of copper concentrate. The roaster has internal boiler
tubes (i.e., an internal heat exchanger labeled IHX) to maintain the roaster outstreams temperature
at 750 °C. A waste heat boiler (WHB) is a separate heat exchanger that cools the hot roaster gas to
warm R-gas (320 °C) for de-dusting in an electrostatic dust precipitator. Solid lines represent solid
phase streams, while dashed lines represent gas phase streams.

Equations [9.25] and [9.26] are spontaneous and complete. Equation [9.27] does not reach
equilibrium; typically, XRS02 = 0.25. Figure 9.19 shows FlowBal's starting array. FlowBal wrote
21 equations for 24 unknowns, so we need three SRs from the constraints described earlier. These
are shown below, in FlowBal format. Figure 9.20 shows the material and heat balance results.

Roaster, Reaction 3, XR-SQ2, (g) = 0.2JT|

{S4-Q2, (g)} - {S4-SQ2, (д)У] |{S4-SQ2, (g)} + {S4-Q2, (g)} - 26 |

P (atm) 1.2 1.2 1.1 1.1 1 50 50 90 90

T(C) 27 30 750 750 320 45 275 45 320

Str-unit Mass (kg) Volume (m3) Mass (kg) Volume (m3) Volume ( 713) Mass (kg) Mass (kg) Mass (kg) Mass (kg)

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

Str-name Cu Cone O-rich Air Calcine Hot R-GasWrm R-gas WHB Wtr WHB Stm Rstr Wtr Rstr Stm

Streams 1 2 34 5 6 7 8 9
1Flow
70 ? ? ? ? 20 | ? 1 60 | ?

CuFeS2, (c,l) 88.3

FeS2, (pyr) 6.1

Si02, (qtz) 3.3 ?

H20, (l,g) ? ? ? ????

N2, (g) ? ??

0 2 , (g) ? ??

S02, (g) ??

S03, (g) ??

CuS04, (c) ?

Fe203, (c) ?

Concentrate Roaster (RX) Heat Roaster IHX (HX) Waste Heat Boiler (HX)
Instreams Outstreams Reactions 2000 Instreams Outstreams Heat Instreams Outstreams Heat

1 31 890 4 5 500
2 42 67

3

Figure 9.19 FlowBal starting array for the sulfation roasting of 70 kg of copper concentrate.
Boxed cells indicate placeholder values that will be changed to close the heat balance. Heat losses
are estimates. Notice the way the HX device stream numbers are sequenced. Corresponding HX
in- and outstreams must be listed on adjacent row cells. For example, WHB instream 4 exits as
outstream 5, and instream 6 exits as outstream 7.

Chapter 9 System Balances on Reactive Processes 535

P(atm) 1.2 1.2 1.1 1.1 1 50 50 90 90

T(C) 27 30 750 750 320 45 275 45 320

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

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

Str-name Cu Cone O-rich Air Calcine Hot R-GasWrm R-gasWHB WtrWHB Stm RstrWtr Rstr Stm

Streams 1 2 3 4 5 67 8 9

Flow 70 67.75 85.80 179.65 114.56 L 20 | 20 | 60 60

CuFeS2, (c,l) 88.3 0 0 0 0 000 0

FeS2, (pyr) 6.1 0 0 0 0 00 0 0

Si02, (qtz) 3.3 0 2.69 0 0 000 0

H20, (l.g) 2.3 0 0 3.80 3.80 100 100 100 100

N2, (g) 0 47.44 0 65.87 65.87 0 0 0 0

02, (g) 0 52.56 0 13 13 0 0 0 0

S02, (g) 0 0 0 13 13 0 0 0 0

S03, (g) 0 0 0 4.33 4.33 0 0 0 0

CuS04, (c) 0 0 62.65 0 0 000 0

Fe203, (c) 0 0 34.65 0 0 000 0

Heat В alance (kcal) Outstreams Reactions ΣΙη EOut Σ Surr Σ Rxn Device Net
Device Instreams 1 -105,196 -135
3 13,005 2 -7096 -1197 28,281 2000 -114,703 -84,557
Concentrate 1 -21 4 15,276 3 -2411
Roaster 2 -114

Roaster IHX 8 -1197 9 43,185 43,185 0 0 41,987
Waste 4 -15,276 5 6426
-15,675 20,392 500 0 5216
Heat Boiler 6 -399 7 13,966

Process Net -37,354

Figure 9.20. System balance for sulfation roasting of copper concentrate. Flow of streams 6 and 8
are placeholder values. The process requires about 68 m3 air (actual) enriched to φ02 = 52.6 %.

The first thing we notice about the heat balance is that it does not close, as expected, because
we just guessed at the S6 and S8 flows. The roaster device heat is a large negative number, hence
our placeholder flow value of 60 kg of water for S8 is much too low. S9 leaves the roaster boiler
tubes with about half the enthalpy needed to bring the roaster net heat to zero, so the flows of S8
and S9 need to be closer to 120 kg. The next point is that the WHB net heat is a positive number,
and is about half of the amount of heat removed by S7, so the S6 flow should be about half of the
placeholder flow of 20 kg. The simplest way to find the correct S6 and S8 flows is to use the
Repetitive Solve feature. First, we need to vary the S8 flow until the Roaster IHX net heat is
equal to but of opposite sign to the roaster device net heat. If you set a target of Roaster IHX =
+84,560 kcal and enter a range of S8 flow (100 and 140) with 3 solves, FlowBal will calculate the
correct flow to give the Roaster IHX net heat of 84,560 kcal. The result is 120.8 kg, but
considering the significant figures in the other data, an integer value (121 kg) is satisfactory for S8.
Now enter 121 for the S8 flow in the input array.

Next, change the S6 flow until the WHB net heat is close to zero. Use Repetitive Solve over
the range 8 - 14 kg with three solves and a target of zero. The WHB heat balance closes at 12.3
kg, so we use 12 kg and insert this back into the input array for the S6 flow. Finally we recalculate
the material and heat balances to be sure the HX stream flows give a correct heat balance. The
correct heat balance is shown below.

Heat В alance (kcal) Outstreams Reactions Σ In I O u t Σ Surr Σ Rxn Device Net
Device Instreams 1 -105196 -135 28281 2000 -114703 -84557
Concentrate 1 -21 3 13005 2 -7096 -2415 87089 0 0 84675
Roaster 2 -114 4 15276 3 -2411

Roaster IHX 8 -2415 9 87089 -15516 14805 500 0 -210
Waste 4 -15276
5 6426
Heat Boiler 6 -239 7 8379

Process Net -92

536 Chapter 9 System Balances on Reactive Processes

There are two things worth mentioning about this example. First, we used FREED's water
data, which assumes ideal behavior for steam. However, at typical boiler pressures, steam is not
ideal. Steam tables can tell us how much error is involved in assuming ideal behavior for steam.
In the WHB case, the steam tables show an enthalpy difference between water at 45 °C and
superheated steam at 50 atm and 275 °C of 630 kcal/kg, while FREED data shows an enthalpy
difference of 680 kcal/kg. These differ by a factor of 1.08, so for "real" water and steam, the S6
flow should be about 13.5 kg. The water/steam data for the IHX differ by a factor of 1.11, so the
S8 flow should be about 135 kg. Second, notice that we didn't write (nor did we need) a reaction
for the vaporization of water from the concentrate or in the HX devices. This is because we used
the (/,g) species descriptor for water, so FREED automatically vaporizes water to steam at 100 °C,
and includes the enthalpy of vaporization as part of the heat balance. If you're simulating a
process where the water vaporizes or condenses at some other temperature, and both water and
steam are present at that temperature, you need to use the proper descriptors and write a reaction.

Our next example is part of the extraction flowsheet for titanium, where rutile (Ti02) is
converted to a volatile chloride. Figure 9.21 shows the flowsheet for a simplified version of the

process. The flowsheet was put together from descriptions in extractive metallurgy books, and

may not be complete. The feed is a mixture of Ti02 and C, chlorinated between 800 and 980 °C.
Briquettes of rutile and carbon are dropped into a vertical furnace, and chlorine gas is introduced at

the bottom. A heat exchanger heats the chlorine to about 580 °C before it enters the bed of

briquettes. The chemical reactions are sufficiently exothermic to heat the charge to the desired exit

reaction temperature (here, 880 °C). The gaseous reaction products pass to a condenser, where

cold TiCl4(/) and a cooling water loop cools the outstreams to about 42 °C. The offgas is treated in
a subsequent step (not shown) to recover excess chlorine and uncondensed TiCl4(g), which are
recycled to the process. Most of the TiCl4(/) from the condenser is split to a heat exchanger, where
it is chilled in a heat exchanger (HX2) for use as a coolant in the condenser. The gas leaving the

chlorinator typically has about 3 % Cl2. The heat loss in the chlorinator is about 1000 kcal. The
heat loss from the other devices is small or even negligible because they operate close to room

temperature. The basis is 50 kg of solid feed.

Chlor У^ . Ice
water
offgas + 10 ColdfiCÌ4
J HX1 water
Ti02 + С Ti02 TÌCI4 ► Offgas HX2
Chlorinator Condenser
9

[2 w V r m ^ H 8 TiCI4to
Chlorine TiC,4
purification

Figure 9.21 Flowsheet for the carbochlorination of rutile. Chlorine (S2) is heated in HX1 by the

chlorinator offgas (S4). Heat is removed from the condenser by recycled chilled TiCl4(/) (S10).
S9 (TiCl4(/) is cooled in HX2 by chilled water. Solid lines depict solid phase streams, dashed lines
are liquid phase, and dotted lines are gas phase.

Equations [9.28] and [9.29] are a set of two independent chemical reactions for the chlorinator
(other two-equation sets are also possible), while Equation [9.30] represents the physical
transformation (condensation) of TiCl4(g). (The normal boiling point of TiCl4(/) is 136 °C.) Both
Equations [9.28] and [9.29] are spontaneous and complete, and in the presence of excess chlorine,
in theory at least, the rutile and carbon are completely consumed.

Ti02(rutile) + C(graph) + 2Cl2(g) -TiCl4(/,g) + C02(g) [9.28]

T\02{rutile) + 2C{graph) + 2Cl2(g) -TiCl4(/,g) + 2CO(g) [9.29]

TiCl4(g) - TiCl4(/,g) [9.30]

The system needs three placeholder values that will be varied to obtain closed heat balances
on the chlorinator, HX1 and the condenser. Once we find correct values for these, we can solve
the heat balance for the correct water flow through HX2 by simple arithmetic because we know the
Cp for water is one kcal k g 1 · deg-1. The first placeholder must be a stream variable that affects

Chapter 9 System Balances on Reactive Processes 537

the heat balance on the chlorinator. One possibility is the %C in the solid feed (SI), which affects
the relative extent of Equations [9.28] and [9.29]. [9.28] is much more exothermic than [9.29], so
the net device chlorinator heat becomes less negative as the %C increases. We could also use the
XRC for Equation [9.28] to adjust the chlorinator net heat; the larger it is, the more heat is
produced in the chlorinator. Finally, we could specify a value for <pCO (or <pC02) in S5. The
larger the <pCO the less heat is produced in the chlorinator. Here we select the %C in S1 as the
chlorinator placeholder variable, set initially (by a guess) at wC = 17 %.

The second placeholder value is the S5 temperature, set initially at 700 °C. The third
placeholder is the S10 flow, set initially at 1000 kg/min. Figure 9.22 shows the starting array for
the stated basis, with placeholder values in boxed cells.

P (atm) 1.3 1.3 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2
25 25 580 880 700
T(C) 42 42 42 42 2
Str-unit Mass (kg) Volume (m3) Volume (m3) Volume (m3) Volume (m3) Mass (kg) Volume (m3) Mass (kg) Mass (kg) Mass (kg)
Spec-unit
Str-name Mass pet Volume pet Volume pet Volume pet Volume pet Mass pet Volume pet Mass pet Mass pet Mass pet
Streams
Flow Solid Feed ColdCI2 HotCI2 HotChlOG WrmChlOG Wrm1TiCI4 CndsrOG Final TiCI4Wrm2TiCI4 Cld TÌCI4
234 7
C, (graph) 1 ? ?? 5 6 ? 8 9 I 10 R#1 R#2 R#3
50 ? ? ? ? 1000
Ti02, (rutile)
CO, (g) 17 I -1 -2
-1 -1
C02, (g) ?
CI2, (g)
?? ? 2
TÌCI4, (g)
TÌCI4, (l,g) ?? ? 1

??3 ? ? -2 -2

?? ? 1 1 -1

? ??? 1

Chlorinator (RX) Heat Chlorine Heater-HX1 (HX) TÌCI4 Condenser (RX) Heat
Instreams Outstreams Reactions 1000 Instreams Outstreams Heat Instreams Outstreams Reactions 300

1 41 2 3 400 563
32 45 10 7

TÌCI4 Liquid Splitter (SP) Liquid TÌCI4 Chiller-HX2 (HX)|
Instreams Outstreams Heat
Instreams Outstreams Split Fract Heat
6 ? 0 9 10 0

?

Figure 9.22 Starting and device arrays for a rutile carbochlorination process. Volume units are
actual, not STP. Boxed cells are placeholder values.

FlowBal wrote 23 equations for 24 unknowns, so we need an additional SR. For this, we need
to tell FlowBal the partial pressure of TiCl4(g) in S4. This SR involves the Кщ value for Equation
[9.30], derived from FREED's Reaction tool. The inserted equation, in FlowBal format, is shown
below. Figure 9.23 shows the material balance result, and Figure 9.24 shows the heat balance.

{pS7-TiCI4, (g)} - 10*(4,99 - 2034/(<T-S 7>+273)) |

P (atm) 1.3 1.3 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2

T(C) 25 25 580 880 700 42 42 42 42 2
Mass (kg) Volume (m3) Mass (kg)
Str-unit M a s s (kg) Volume (m3) Volume (m3) Volume (m3) Volume (m3) Mass pet Volume pet Mass pet Mass (kg) Mass (kg)
/Vrm1TiCI4 Final TiCI4
Spec-unit Mass pet Volume pet Volume pet Volume pet Volume pet CndsrOG Mass pet Mass pet
6 7 8
Str-name Solid Feed Cold CI2 Hot CI2 HotChlOG WrmChlOG Wrm2TiCI4 Cld TiCI4
1094.42 16.54 94.42
Streams 1 2 3 4 5 9 10
0 0 0
Flow 50 20.27 62.84 99.77 84.20 0 0 0 1000 1000 |
0 49.01 0
C, (graph) 17 J о 0 0 0 0 43.20 0 00
Ti02, (rutile) 0 0 0 0 4.95 0
83 0 0 29.73 29.73 0 2.84 0 00
CO, (g) 00 100 0 100
00

C02, (g) 0 0 0 26.20 26.20 00

CI2, (g) 0 100 100 3 3 00

TÌCI4, (g) 0 0 0 41.07 41.07 00

TÌCI4, (l,g) 0 0 0 0 0 100 100

Figure 9.23 Material balance for the carbochlorination of 50 kg/min of feed containing 17 % С

(8.5 kg). For the selected placeholders, 98.6 kg of TiCl4(g) are produced in the chlorinator, of
which 96 % is recovered in S8.

538 Chapter 9 System Balances on Reactive Processes

Heat Balance (kcai) Device
Net
Device Instreams Outstreams Reactions ΣΙη ZOut ISurr Σ Rxn
4 17102 1 -16836 -5174 17102 1000 -18635 -5707
Chlorinator 1 0 2 -1799
3 -5173 3 5173
Chlorine 2 5 13325 -17101 18498 400 0 1797
Heater-HX1 4 0 6 3406
5 -17102 7 110 3 -4879 -9120 3516 300 -4879 -10,182
TÌCI4 10 -13325 8 294
Condenser 6 4205 9 3112 -3406 3406 00 0
TÌCI4 Liq -3406 10 -4205 -3112 -4205
9 0 0 -7317
Splitter -3112 Process Net -21,409
Chiller HX2

Figure 9.24 Heat balance for the carbochlorination of 50 kg/min of feed containing 17 % C, for
selected placeholders.

As expected, none of the net device heats are close to zero. The chlorinator heat is negative,
which indicates that the %C in SI should be higher. The HX-1 heat is positive, so the S5
temperature should be lower. Finally, the TiCl4 condenser heat is negative, so the cold TiCl4(/)
flow (S10) needs to be a lot larger. FlowBal's Repetitive Solve tool found values for all of the
above placeholder values that will bring the first three device net heats close to zero. SI should
contain 19.5 %C, S5 should be 617 °C, and the cold TiCl4(/) flow (S10) should be 2122 kg/min.
Figure 9.25 shows the system balance with updated placeholder entries.

P (atm) [ ]1.3 1.3 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 2
25 25 580 880 617 42 42 42 42
T(C) Mass (kg) Volume (m3) Volume (m3) Volume (m3) Volume (m3) Mass (kg) Volume (m3) Mass (kg) Mass (kg) Mass (kg)
Str-unit

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

Str-name Solid Feed Cold CI2 Hot CI2 HotChlOG WrmChlOGWrmlTiCH CndsrOG Final TiCI4Wrm2TiCI4 Cld ТЮ14

Streams 1 2 3 4 56 7 8 9 10

Flow 50 19.73 61.18 106.96 82.57 2212.86 18.91 90.86 2122 2122
I оC, (graph) 19.5
00000 0 00

Ti02, (rutile) 80.5 0 0 0 0 0 0 0 0 0

CO, (g) 0 0 0 45.38 45.38 0 70.16 0 0 0

C02, (g) 0 0 0 14.46 14.46 0 22.36 0 0 0

CI2, (g) 0 100 100 3 3 0 4.64 0 0 0

TÌCI4, (g) 0 0 0 37.15 37.15 0 2.84 0 0 0

TÌCI4, (l,g) 0 0 0 0 0 100 0 100 100 100

Heat Balance (kcal) Device
XOut ZSurr Σ Rxn Net
Device Instreams Outstreams Reactions ΣΙη
4 16,970 1 -9962 -5036 16,970 1000 -12,906 28
Chlorinator 1 0 2 -2944
3 5036
3 -5036 5 11,531
6 6887
Chlorine 2 0 7 120 -16,970 16,568 400 0 -2
8 283
Heater-HX1 4 -16970 9 6604
10 -8922
TÌCI4 5 -11531 3 -4695 -2610 7007 300 -4695 3

Condenser 10 8922

TÌCI4 Liquid 6 -6887 -6887 6887 00 0
-6604 -8922
Splitter

Chiller HX2 9 -6604 0 0 -15,526
Process Net -15,497

Figure 9.25 System balance for the carbochlorination of 40.25 kg/min of rutile using 76.5 kg/min

of Cl2. Boxed material balance cells originally contained placeholder values. The solid charge
contains 19.5 %C and produces a chlorinator offgas that has <pCO/<pC02 = 3.14. Chiller HX2 heat
balance discussed below.

The process is relatively efficient in the use of Cl2 and the collection of liquid TiCl4. 96 % of
the TiCl4 produced is collected in S8, and only 3.5 % of the new Cl2 (S2) leaves in S7. Most of
this Cl2 can be recovered from the condenser offgas (S7), which has to be treated elsewhere to
recover the uncondensed TiCl4.

Chapter 9 System Balances on Reactive Processes 539

The HX2 TiCl4(/) chiller requires the removal of about 15 500 kcal/min via chilled water.
The split fraction of S6 to S9 is nearly 0.96. This design requires an inordinate amount of cold
TiCl4(/) (2122 kg/min) to satisfy the condenser heat requirement. Also, HX2 requires 15 500/40 =
390 kg/min of chilled water, which has to be produced elsewhere. A better design would be to
insert an air-cooled HX after HX1 to drop the S5 temperature to about 250 °C. This would
decrease the cooling load on the condenser and HX2.

EXAMPLE 9.7 — Lime-Assisted Reduction of Magnetite.

A college professor has proposed a novel process for the carbothermic reduction of magnetite
in a rotary kiln, whereby lime is added to react with the C02 reduction product in an exothermic
reaction to form limestone. In the absence of lime, a considerable amount of carbon must be
burned by oxygen to provide the necessary heat. The proposed advantage is that with lime, less
carbon and heat are required, thus saving energy. The process requires a lower-than-usual reaction
temperature (780 °C) to take advantage of the ability of lime to react with C02, which means the
solids must have a longer-than-average retention time in the kiln. Figure 9.26 shows the flowsheet.
Make a system balance to determine the advantage of lime addition to the process. The products
are mainly iron and limestone, with some unreacted lime and unreduced wustite. In a separate
series of steps (not shown), the solid product is screened while hot to separate the metallic DRI
from the powdered limestone, and melted in an electric furnace. Any un-separated limestone
becomes a furnace flux.

Fe304 + С --► Offgas
CaO +> Solid product
02

Figure 9.26 Flowsheet for lime-assisted reduction of magnetite. Magnetite/carbon pellets are
added to the kiln along with lime. Oxygen is injected below the bed to burn carbon to create heat,
and provide CO for the magnetite reduction.

Data. The critical reaction is the formation of metallic iron from wustite, which requires

<pCO/<pC02 >1.8. Here, we set фСО/фС02 = 2.1 to provide an adequate driving force. (Please see
the Handbook CD, folder Charts, for an expanded chart of the CO/C02 ratio.) The lime enters at
500 °C from a prior calcining step. The reactor heat loss is estimated at 6 kcal/kg of solid feed.
Bench scale tests indicated that the solid product (S5) must contain at least wC =1.5 % to prevent

the reoxidation of Fe by the injected 02. Other tests indicated that the XRFQ0947O in the wustite
reduction reaction was 0.95, and the ХЛСаО in the lime reaction was 0.92.

Solution. The system has nine reactive species and four elements, so NIRx = 5, as shown in the
following reactions Figure 9.27 shows FlowBal's starting array for the process with no lime
added, on a basis of 350 kg of solid feed/min.

Fe304(c,/) + 0.832CO(g) - 3.168Fe0.947O(c,/) + 0.832CO2(g) [9.31]
Fe0.947O(c,/) + C(graph) - 0.947Fe + CO(g) [9.32]

2CO(g) + 0 2 ( g ) - 2 C 0 2 ( g ) [9.33]
C(graph) + y202(g) - CO(g) [9.34]

CaO(c) + C02(g) - CaC03(calcite) r [9.35]

540 Chapter 9 System Balances on Reactive Processes

P(atm) 11111

T(C) 27 500 27 780 780

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

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

Str-name Solid Feed Lime Oxygen Offgas Solid Prdct

Streams 1 234 5 R#1 R#2 R #3 R #4 R #5
Flow ?
350 о I 70 I ?
C, (graph) ? 1.5 -1 -1

CO, (g) ? -0.832 1 -2 1
?
C02, (g) 0.8321 2 -1

0 2 , (g) 95 -1 -0.5

N2, (g) 5?

СаСОЗ, (calcite) ?1

CaO, (с) 100 ? -1

Fe304, (c,l) ? -1

Fe.9470, (c,l) ? 3.1679 -1
? 0.947
Fe, (c,l,g)

Figure 9.27 Starting array for the no-lime carbothermic reduction of magnetite for one minute of
operation. The boxed value is a placeholder for the oxygen flow which will be changed to close
the heat balance. Note entering lime temperature.

The reaction coefficient values need special care when the dealing with substances that have
non-integer formulas. The reaction coefficients for Equation [9.31] (R #1) must be entered in the
starting array to as many significant figures as Excel allows. For the wustite coefficient, enter the
formula "=3/0.947" in the cell for R #1, and use a similar technique for the other R #1 coefficients.

FlowBal wrote 13 equations for 16 unknowns, which requires three SRs. Two SRs deal with
the XRwastitQ and Ji7?lime, and one with the фСО/<рС02 in S4. The three SR equations are shown
below, in FlowBal format. Figure 9.28 shows the material and heat balance for the process.

Magnetite Reducer, Reaction 2, XR-Fe.9470, (c,l) = 0.95 |

Magnetite Reducer, Reaction 5, XR-CaO, (c) = 0.92 |

{S4-CQ2, (g)}*2.1 ■ {S4-CO, (g)} |

P (atm) 1 1 1 11

T(C) 27 500 27 780 780
Mass (kg)
Str-unit Mass (kg) Mass pet Volume (m3) Volume (m3) M a s s ( k g )

Spec-unit Mass pet Lime Volume pet Volume pet Mass pet
2 Oxygen Offgas Solid Prdct
Str-name Solid Feed 0 34 5
0
Streams 1 0
0
Flow 350 0 70 646.0 193.2
0
C, (graph) 25.99 0 0 0 1.5
100
CO, (g) 0 0 0 66.45 0
0
C02, (g) 0 0 0 31.64 0

02, (g) 0 95 0 0

N2, (g) 0 5 1.90 0

СаСОЗ, (cale 0 000

CaO, (с) 0 000

Fe304, (c,l) 74.01 000

Fe.9470, (c,l) 0 0 0 6.32

Fe, (c,l,g) 0 0 0 92.18

Heat Balance (kcal) Reactions Σ In ZOut Σ Surr Σ Rxn Device Net
Device In streams Outstreams | 72,697 2500 -65,912 9132
Magnetite 1 -112 4 49,180 10,537
Reducer 2 0 5 23,517 125,325
-97,014 -153
3 -41 -104,761

Process Net 9132

Figure 9.28 Material and heat balance for the no-lime carbothermic reduction of magnetite.

Chapter 9 System Balances on Reactive Processes 541

The reducing reactor net heat is positive, which indicates that the placeholder value of 70 m3
(actual) 0 2 is too low. The Repetitive Solve tool was used at an S3 flow of 72, 74, and 76, with a
target device net heat of zero. This was reached at a rounded off S3 flow of 73.2 m3 02/min.
Figure 9.29 shows the material balance for this condition, for which the net heat was 84 kcal. This
value is sufficiently close to zero considering the number of significant figures used to express
stream flow, the uncertainties surrounding the XT? values, and the reactor heat loss.

P(atm) 11111

T(C) 27 500 27 780 780

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

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

Str-name Solid Feed Lime Oxygen Offgas Solid Prdct

Streams 1 2 3 4 5

Flow 350 0 73.2 | 660.6 191.7

C, (graph) 26.5 0 0 0 1.5

CO, (g) 0 0 0 66.4 0

C02, (g) 0 0 0 31.6 0

02, (g) 0 0 95 0 0

N2, (g) 0 0 5 1.9 0

СаСОЗ, (calci 0 0 0 0 0

CaO, (с) 0 100 0 0 0

Fe304, (c,l) 73.5 0 0 0 0

Fe.9470, (c,l) 0 0 0 0 6.3

Fe, (c,l,g) 0 0 0 0 92.2

Figure 9.29 Material balance for the lime-free carbothermic reduction of 257 kg of magnetite at
780 °C with an oxygen flow (boxed cell) that closes the heat balance around the reducer.

We now examine the addition of lime on the system balance. The CaO that reacts with C02
has two effects. First, it decreases the amount of carbon necessary to reduce iron oxide, so for a

fixed feed mass of 350 kg (SI), the mass fraction of magnetite in the SI feed will increase.

Second, the formation of СаСОз is exothermic, so less С needs to be oxidized by 0 2 to close the
heat balance. The above steps were repeated with an arbitrary amount of lime (S2) and a

placeholder amount of 0 2 (S3), to make a system balance. The placeholder 0 2 flow was then
adjusted to close the heat balance. Figure 9.30 shows the results for a range of lime additions.

Effect of Lime on Fe304 Reduction

20 40 60 80 100 120 140 160
kg lime per 350 kg solid feed

Figure 9.30 Effect of lime additions on various process parameters for the carbothermic reduction
ofmagnetiteat780°C.

542 Chapter 9 System Balances on Reactive Processes

The results show that lime has a very beneficial effect on the process by decreasing the
consumption of carbon. The offgas volume (S4) extrapolates to zero at 186 kg of lime, but the
lime addition cannot be that high because some amount of offgas is necessary to remove N2 and
other trace impurities. Addition of lime from zero to 160 kg increases the mass of Fe304 reduced
per mass of С by a factor of 2.1.

Assignment. The so-far unmentioned aspect of this process concept is that lime must be produced
somewhere, and it is usually produced by the burning of a fuel. Possibly the offgas from the
reduction furnace can provide enough heat to calcine the limestone used. Revise the flowsheet to
include the calcination of limestone in a reactor that is fired by the offgas from the reduction
reactor. Assume that owing to heat losses in the transfer duct, the recycled offgas enters the lime
calcination furnace at 700 °C

9.5 Quality of Heat and Thermal Efficiency

Various handbooks list the AFT for the combustion of fuels with stoichiometric air because
the AFT is a useful guide for comparing the ability of certain fuels to provide useful heat to a
process. The useful heat of the fuel means the АНсотЪ of the fuel that is available for use in the
furnace. For example, if steel billets are to be heated to 1100 °C, only the heat content of the
combustion gas in excess of the 1100 °C value is transferable for use or loss. Consider the
following three processes.

1. A fuel-fired steel reheating furnace operates at an average temperature of 1250 °C. Billets
are continually pushed into the furnace, and they slowly move from one end to the other
until they reach 1100 °C. Since the stack gas leaves the furnace at 1250 °C, only HAF1~-
Я1250 of the combustion gas is available for heating the billets.

2. Steel billets are stacked in a furnace, and in a batch heating process, they are heated to
1100 °C. As the process continues, the combustion gases leave at increasingly higher
temperatures. The average stack gas temperature is 800 °C. The useful heat is the HA¥T-
Я80о of the combustion gas.

3. Steel billets are heated to 1100 °C in a long tunnel furnace, with the combustion gas
moving counter-currently to the billets. The stack gas is always 100 - 200° hotter than the
steel, so the useful heat is HAFT-HstsLck.

Since the heat content of the combustion gas at the AFT = АЯсотЬ of the fuel, the useful fuel
heat is also equal to the АЯсотЬ - (Hstack-HA¥T). The fuel choice depends on a number of factors,
but thermal efficiency plays a big part in all cases. Let's examine a simple case for steel (assume
pure Fe) heated from 25 °C to 1100 °C using three different fuels: natural gas (assume 100 %
CH4), fuel oil (86 %C, 14 %H, with a LHV of 41 800 kJ/kg), and coke oven gas (<pCH4 = 0.302;
<pCO = 0.055; <pC6H6 = 0.028; <pN2 = 0.045; <pC02 = 0.022; balance H2). The process fuel rates
will be calculated by a heat balance.

The furnace operates continuously at a stack gas temperature of 1250 °C with sufficient air to
give φ02 = 3% in the stack gas. The steel is processed at the rate of 5 kg/sec, with HL = 300
kJ/sec. Figure 9.31 shows a sketch of the process.

Combustion air Comb, gas
(25 °C) (AFT) Stack gas
Fuel Steel billet *(1250°C)
(25 °C) Cold steel
Hot steel heating furnace " (25 °C)
(1100 °C)
CDCDCDCDCD

Figure 9.31 Flowsheet for heating five kg of steel billets per second.

Chapter 9 System Balances on Reactive Processes 543

The first step is a material balance for each fuel for stoichiometric combustion, and then
adjust it to get φ02 = 3.0% in the stack gas. The results are shown in the ledger below, where NG
stands for natural gas and COG stands for coke oven gas.

Amount of Air for Different Fuels to Get φ02 = 3.0 % in Stack Gas

1 mole NG 100 g Oil 1 mole COG

Stoichiometric air moles 9.52 50.63 5.31

Actual air moles 11.28 59.65 6.31

% XSA 18.4% 17.8% 18.9%

The next step is to calculate the A//comb of each fuel at 25 °C. The АЯсотЬ for CH4 has already
been calculated in connection with the text near Equation [9.11] as -802.3 kJ/mol, and the AHcomb
of the oil was stated as -41 200 kJ/kg. The АЯсотЬ of COG was calculated from FREED as -479.1
kJ/mol. The AFT calculation requires data on the HrH25 for each product gas. Three-term
equations for Ht-H25 were derived from FREED data as shown below. The equations are valid
between 1200 ° and 2600 °C.

#t-#25, kJ/mol - At + Bt2 + С

A ВС A В С
0.04131 6.712E-07 -3.94
H20: 0.05612 2.801E-06 -6.97 N2: 0.03363 1.090E-06 -2.72
C02: 1.200E-06 -9.03 02: 0.03397

The AFT is calculated as before, using Equation [9.19]. The table below shows the values for
A, B, and С for the calculated amount of product gas, and the AFT as calculated by Goal Seek.
Note that the COG has a somewhat higher AFT than the other fuels despite the presence of inerts.
This is mainly because of the benzene and the high H2 content. The useful available heat from
each fuel is the difference between the LHV and the НХ250-Н25 for each fuel's product gas. The
thermal efficiency of each fuel (not the process) is the useful heat divided by the LHV, expressed
as a percentage. The thermal efficiency of NG is 34.5 %, that of oil is 35.0 %, and that of COG is
37.2 %, as shown below.

Thermal Efficiency of Three Different Fuels with Stack Gas at φ02 = 3.0 %

Fuel A В С LHV, kJ AFT, °C H\250-H2S //avail ThEff

1 mol NG 0.451 1.319E-05 -59.05 802.3 1814 525.3 277 34.5 %

100 g Oil 2.338 6.174E-05 -303.7 4120 1806 2715 1405 31.1 %

1 mol COG 0.230 5.960E-06 -29.37 479.1 1880 300.9 178.2 37.2 %

The thermal efficiency of the process is based on the useful process heat, which in this case is
the heat used in heating the steel to 1100 °C. Hnoo-H25 for Fe is 734.6 kJ/kg, so at the rate of five
kg/sec, 3673 kJ/sec is transferred to the steel. HL = 300 kJ/sec, so the total heat required is 3973
kJ/sec. The quantity of each fuel is obtained by dividing the per-second requirement by Яауап from
the above table, and converting to practical units. The process thermal efficiency is shown in the
table below. Calculations are based on one second of operation. Gas volumes are STP.

Process Thermal Efficiency (РТЕ), kJ/sec

Fuel //avail #toFe /Zreqd Fuel flow Fuel flow РТЕ
NG 277 kJ/mol 0.321 m3/sec 31.9%
Oil 1405kJ/100g 3673 kJ/sec 3973 kJ/sec 14.3 mol/sec 31.5%
COG 178kJ/mol 0.283 kg/sec 34.4%
3673 kJ/sec 3973 kJ/sec 283 g/sec
0.500 nrVsec
3673 kJ/sec 3973 kJ/sec 22.3 mol/sec

544 Chapter 9 System Balances on Reactive Processes

This shows that the thermal efficiency of the three fuels examined in this case is proportional
to the fuel's AFT, but this might not be true for other fuels. Most of the A//comb is carried away as
sensible heat in the stack gas, which suggests that thermal recuperation is the best way to increase
the thermal efficiency of a process.

The previous example used excess air so as to completely combust the fuel, and in so doing,
obtain the full LHV of the fuel for use in the process. However, the presence of elemental oxygen
in the combustion gas oxidizes the surface of the steel, and by the time a billet leaves the furnace, it
has an iron oxide scale that consumes some of the steel, sometimes pitting the surface. If the
billets are to be extruded, the scale must be removed. If the process is carried out using less than
stoichiometric air, the combustion gas will contain no free oxygen, and the amount of scale will be
much less. With a slight deficiency of air, the combustion gas C02 and H20 will still oxidize iron,
but the kinetics are much less than oxidation with 02. The scaling tendency decreases as the
air/fuel ratio decreases, and eventually, if the combustion gas contains enough H2 and CO, there is
no tendency to scale. Scale will form only after the steel exits the furnace. Section 6.5.3 discusses
the/?H20//?H2 ratio in equilibrium with both iron and wustite, and Figure 6.36 shows the value of
this ratio for temperatures up to 1000 °C. The chart equation is not sufficiently accurate for a
combustion gas at 1250 °C, so we must resort to FREED's Reaction tool (see, however, folder
Charts in the Handbook CD). The Кщ for Equation [6.71b] is 0.82, so from a practical standpoint,
it is probably satisfactory to aim for а /?Н20/рН2 ratio in the combustion gas of one, and tolerate a
slight amount of scale formation. It's useful to calculate the burner requirements in order to get an
AFT of 2000 °C for CH4 combustion such that when the gas is cooled to 1250 °C, it will have a
/?H20/pH2 ratio = 1. FlowBal is a good choice for this type of calculation. A preliminary
calculation showed that for this case, the AFT using normal air at 25 °C is <2000 °C, so either the
air must be preheated, or it must be enriched with 02.

We have already studied the combustion of CH4 with less than stoichiometric air, and know
that in the absence of CH4 as a product, NIRx = 3. The applicable equations are [6.62], [6.63] and
[6.64]. Figure 9.32 shows the FlowBal starting array for the process, with three placeholder values
displayed in bordered cells. First, we look at the heating process using preheated air. For that, we
need to calculate the placeholders for SI flow and S2 temperature. Second, we look at oxygen
enrichment without air preheat. There we calculate the placeholders for SI flow and % 0 2 in S2.

P (atm) 1.1 1.1 1 1 1 1
[ ]25
T(C) 800 2000 1250 25 1100 C H 4 + y202 --> CO + 2 H 2
Str-unit H2 + У2О2 -► H 2 0
Volume (m3) Volume (m3) Volume (m3) Volume (m3) Mass (kg) Mass (kg)
CO 4■ H 2 0 - + C 0 2 + H 2
Spec-unit Volume pet Volume pet Volume pet Volume pet Mass (kg) Mass (kg)

Str-name Nat Gas Air Brnr Gas Offgas Cold Fe Hot Fe

Streams 1 2 3 4 5 6 R#1 R#2 R#3
IFlow
1 ? ? ? 5 ? -1 -1
1 1
CH4, (g) ?

CO, (g) ??

C02, (g) ??

Fe, (c,l,g) ??

H2, (g) ?? 2 -1 1
1 -1
H20, (g) ??

N2, (g) ? ??

0 2 , (g) LZ J -0.5 -0.5

Burner (RX) Heat Billet Heater (RX) Heat
Instreams Outstreams Reactions I Instreams Outstreams Reactions 300

131 343
22 56

3

Figure 9.32 Starting array for one second of operation of a billet heating process using excess fuel
and preheated air. The billet heater heat loss (kJ/sec) came from the original problem definition.
Boxed cells are placeholder values that will be varied to close the device heat balances.

Chapter 9 System Balances on Reactive Processes 545

First, the air preheat temperature case. We seek the air temperature that produces a 2000 °C
combustion gas temperature (i.e., the burner AFT = 2000 °C). This is done by varying the air
temperature until the burner net heat = 0. Next, the volume of NG that brings the billet heater net
heat to zero. This is done by varying the NG flow until the billet heater net heat = 0. The starting
calculation is for the placeholder SI flow = 1.0 m3 NG and S2 temperature = 800 °C.

FlowBal wrote 16 equations for 19 unknowns, so we need three SR equations. First, the
φΗ20/φΗ2 = 1 in the billet heater (S4). Next, equilibrium is reached owing to the high
temperature, so the gas composition conforms to the WGR (R #3 in the starting array). FREED's
Reaction tool was used to obtain а Кщ equation for the WGR valid between 1000 ° and 2000 °C
(the equations in Figure 6.32 have an upper limit of 1650 °C). The three SR equations are shown
below. The net heat for the burner was 1184 kJ, indicating that the 800 °C placeholder
temperature was too low.

|{S4-H20, (g)} - {S4-H2, (g)}J

|{S4-C0, (g)}*<S4-H2Q, (д)}*10л(-1.26 + 1040/<T-S 4 > ) - {S4-C02, (g)>*{S4- H2,Tg»]

Wl|{S3-CO, (gj}*<S3-H20, (д)}*10л(-1.26 + 1040/<T-S 3>) - ■CS3-C02, (g)}*{S3- H2,

FlowBal's Repetitive Solve tool varied the S2 (ordinary air) temperature between 850 °C
and 950 °C for three temperatures, with a target value of zero for the net burner heat. An air
preheat of 944 °C was obtained. A solve with the 944 °C air preheat temperature showed that the
billet heating furnace net heat was a large exothermic value, so the amount of NG should be
decreased. The NG flow was varied between 0.4 and 0.7 to seek a value that brought the net heat
of the billet heater to zero. The NG flow should be 0.417 m3/sec, which is an 02/CH4 molar ratio
of about 1.14. Figure 9.33 shows the system material and heat balances for one second of
operation (5 kg of steel heated) obtained with the correct S2 temperature and NG flow.

P(atm) 1.1 1.1 1 11 1
T (C) 25 944 1100
2000 1250 25

S t r - u n i t Volume (m3) Volume (m3) Volume (m3) Volume (m3) M a s s ( k g ) M a s s ( k g )

Spec-unit Volume pet Volume pet Volume pet Volume pet Mass (kg) Mass (kg)

Str-name Nat Gas Air Brnr Gas Offgas Cold Fe Hot Fe

Streams 1 2 3 45 6

Flow 0.417 | 9.21 25.44 17.04 5 5

CH4, (g) 100 0 0 00 0

CO, (g) 0 0 11.28 10.01 0 0

C02, (g) 0 0 2.47 3.74 0 0

Fe, (c,l,g) 0 0 0 05 5

H2, (g) 0 0 12.48 13.75 0 0

H20, (g) 0 0 15.02 13.75 0 0

N2, (g) 0 79 58.75 58.75 0 0

02, (g) 0 21 0 00 0

Heat Balance (kJ) Device

Device Instreams Outstreams Reactions Σ In EOut Σ Surr Σ Rxn Net

Burner 1 0 3 9520 1 -670

2 -2946 2 -5766 -2946 9520 0 -6574 0

3 -139

Billet 3 -9520 4 5620 3 -71 -9520 9293 300 -71 2
Heater 5 0 6 3673

Process Net 0

Figure 9.33 System balances for heating 5 kg of steel billets per second using ordinary air heated
to 940 °C. The placeholder values for SI flow and S2 temperature were varied to bring the net
device heats close to zero. Volume flows are actual.

546 Chapter 9 System Balances on Reactive Processes

Notice that the heat content of S4 (5620 kJ) is considerably larger than the heat delivered
when S2 enters the burner (-2946 kJ). Therefore, it's feasible to use a heat exchanger to transfer
heat from S4 to S2 to bring the air temperature to 940 °C. Once the warm S4 leaves the HX, it can
be further chilled to condense the water vapor. This permits using the significant chemical fuel
value that S4 has because of the unburned H2 and CO.

The alternative to air preheating to attain a burner AFT of 2000 °C is to oxygen-enrich the
burner air. Again by using Repetitive Solve for four different values between 30 % and 40 % 0 2
the burner heat balance closes when the <p02 ~ 37.5 %,. (Four different 02 compositions were
used because the relationship between % 0 2 and net burner heat is not linear.) A subsequent set of
repetitive calculations found that 0.597 m3 of CH4 produces zero net heat around the billet heater.
As in the air preheat condition, the 02/CH4 mole ratio is about 1.14. Figure 9.34 shows the system
balance results. Using oxygen enrichment requires about 40 % more NG than when using
preheated air.

When comparing the fuel efficiency using ordinary air, preheated air, and oxygen enriched
air, we must not forget to account for the energy required to produce the technical grade oxygen
(«96 %02) used to enrich the combustion air. The energy requirement depends on the oxygen
production rate, but varies between 300 - 400 kWh/tonne of contained oxygen.

P(atm) 1.1 1.1 1 1 11
T(C) 25
25 2000 1250 25 1100

Str-unit Volume (m3) Volume (m3) Volume (m3) Volume (m3) M a s s (kg) M a SS (kg)

Spec-unit Volume pet Volume pet Volume pet Volume pet Mass (kg) Mass (kg)
Str-name Nat Gas Air Brnr Gas Offgas Cold Fe Hot Fe

Streams 1 2 3 4 5 6

Flow 0.597 | 1.81 24.50 16.42 5 5
CH4, (g) 0
CO, (g) 100 0 0 0 0 0
C02, (g) 0 0 16.77 14.88 0 0
Fe, (c,l,g) 0 0 3.67 5.55 0 5
0 0 5
H2, (g) 0 0 0 0 0 0
H20, (g) 0 0 18.55 20.44 0 0
0 62.5 22.32 20.44 0 0
N2, (g)
о С 37.5 38.69 38.69 0 0
02, (g)
Iо 0

Heat Balance (kJ) Device
Reactions Σ In ZOut Σ Surr Σ Rxn Net
Device Instreams Outstreams
3 9412 1 -959
Burner 1 0 2 -8255 0 9412 0 -9412 0
3 -198
20 3 -102 -9412 9215 300 -102 1

Billet 3 -9412 4 5542 Process Net 1

Heater 5 0 6 3673

Figure 9.34 System balances for heating 5 kg steel billets per second using air enriched to 37.6
%02. The placeholder values for SI flow and φ02 in S2 were varied to bring the device net heats
close to zero. As usual with FlowBal, volume flows are actual.

The FlowBal version of the example specified 2000 °C as the burner AFT, but there would be
additional fuel savings if a higher AFT was attained. This is probably not feasible using air
preheat, but it is with oxygen enrichment above 38 %02. For better heat utilization, the steel could
be preheated in a HX using the sensible heat in S4. The danger then is the presence of deadly CO
that requires careful safety provisions. The best idea might be to split part of S4 to a billet
preheater, where it would be burned with excess air to bring the billets to about 700 °C, where
scaling is minimal. The rest of S4 could be used elsewhere in the plant as a heat source.

Chapter 9 System Balances on Reactive Processes 547

Two additional terms are used in connection with process fuel efficiency: critical process
temperature and enthalpy (or heat) quality. A materials process may involve physical or chemical
reactions that take place only above a certain critical temperature. For example the melting of
aluminum requires a furnace temperature well above aluminum's melting point of 660 °C,
typically near 800 °C. Therefore, an aluminum melting furnace requires heat above the critical
temperature of 800 °C. The process is theoretically feasible as long as the temperature of the heat
source is above 660 °C, but as the heat source temperature approaches 660 °C, the residence time
may become excessive, or the volume of the furnace may have to be increased to the point where it
can no longer operate economically. As the temperature of the heat source approaches the critical
process temperature, the quality of the heat approaches zero. The process will not be feasible no
matter how much heat is available at or below 660 °C. Therefore, we must be concerned not only
with the quantity of heat required for a process, but its quality.

We've already seen that the AFT of a fuel is a good (but not precise) indicator of the quality
of heat available for a process. For example, suppose ceramic parts must be sintered at 1200 °C in
atmospheres that must have a controlled oxygen potential. We can prepare such atmospheres by
varying the amount of air used to burn a fuel, such as natural gas. Figure 9.35 shows the AFT for
the combustion of CH4 with air when both reactants enter at 28 °C. The calculations were made
with the equilibrium thermodynamic program THERBAL, which greatly simplifies the arithmetic,
and allows for the presence of slight amounts of CO, H2, and 0 2 at the equilibrium position at high
temperatures. With some effort, the same results could be obtained by using FlowBal. The
highest AFT (1984 °C) was obtained at 97 % stoichiometric air (9.24 mol air/mol CH4), and the
burner gas contained φ02 = 0.30%.

AFT for the Combustion of CH4 with Air

2000 1 a»г 10

iyuu 9
1800
8

Ü 1700 7 o4

tω 1600 I6 CO

Ό 5
< 1500
4с
1400 3О

2

1300 1

1200 I 0
50%
70% 90% 110% 130% 150% 170% 190%

stoichiometric air, %

Figure 9.35 AFT and oxygen content of burner gas as a function of the % stoichiometric
combustion air between 50 % and 200 % for air and CH4 entering at 28 °C.

The theoretical amount of heat available for the sintering process is the enthalpy of the gas
between the AFT and the critical process temperature of 1200 °C. Figure 9.35 shows that in
principle, the burner would be effective for any range of stoichiometric air between 50% and 200
%. However the process efficiency would be very poor at the extremes ofthat range because the
AFT comes so close to 1200 °C.

The available heat for the process is the enthalpy of the burner gases above 1200 °C. We can
calculate that from the net heat for combustion at 1200 °C because the net heat at the AFT is, by
definition, zero. THERBAL was again used for this, with the results shown in Figure 9.36. Any
significant heat loss, plus the thermal requirements of the ceramic parts, probably means that the
maximum theoretical air ranges from about 60 % to 180 % of stoichiometric. The "quality" of the
combustion heat is a maximum at 100 % stoichiometric air, and approaches zero, or even negative,

548 Chapter 9 System Balances on Reactive Processes

quality at the air range extremes. This example illustrates the importance of close control on fuel
combustion air to maximize the quality of the heat. The ЛЯсотЬ of CH4 at 25 °C is -183 200
cal/mol, while at 100 % stoichiometric air the burner has -87 400 cai available for the process,
which is about 48 % of the combustion heat. The rest leaves as sensible heat in the offgas.

Heat Available for Sintering Process

о 70% 90% 110% 130% 150% 170% 190%

"§ -10,000

^ -20,000

"^ -30,000

5 -40,000

О -50,000
| -βο,οοο
S -70,000
a: -βο,οοο

-90,000 J—

50%

stoichiometric air, %

Figure 9.36 Heat available for sintering process operating at 1200 °C as a function of the %
stoichiometric air.

The heat in the process offgas stream is useful if it can be made available to the process.
We've already discussed how a hot offgas can be used to preheat the combustion air, which raises
the AFT. Alternatively, it can be used to preheat the ceramic parts before they enter the sintering
furnace. In the latter case, preheating the parts decreases the thermal demand for the process, thus
requiring less fuel. The offgas produced by using less than stoichiometric air also has chemical
energy in the form of unburned CO and H2, so after using it to preheat the combustion air, it can be
used as a fuel elsewhere.

9.6 System Balances with Heat Exchangers

Earlier sections of this Chapter used relatively simple systems to introduce the principles and
techniques for analyzing heat balances. Heat exchangers were used in some of the examples, and
were evaluated in terms of their ability to increase the thermal efficiency of a process. Heat
exchangers are becoming more common in the process industry as the cost of energy increases.
This section will look at heat exchangers in more complex systems to synthesize the knowledge
gained from this and earlier Chapters. The emphasis in this section is on heat exchangers that are
more integrated in the process flowsheet than in earlier sections. Some of the examples have
external heat exchangers, some have internal heat exchangers, and some have both.

We illustrate the principles on a fairly simple process: the annealing of cast steel to remove
segregation. The process is typical of many that employ heat exchange between a combustion gas
and a solid. The cold steel enters a long reheat furnace (at the cold end, as defined in Section
8.5.1) and is pushed through unto it reaches the hot end. Burners at the hot end produce a
combustion gas that travels counter-currently to the steel. This sort of arrangement has already
been depicted for aluminum in Figure 8.19 and steel in Figure 9.31. Here we look in more detail at
such processes to investigate the effect of oxygen enrichment on the process thermal efficiency and
the use of furnace gas recycle to mitigate the production of NOx in the offgas. Figure 9.37 shows a
sketch of a steel heating process that for calculational purposes has been split at an arbitrary
location for the removal of a portion of the furnace gas for recycle. There is one reactor (the
burner) and two heat exchangers (the annealing and preheat furnaces).

We start our analysis using plain air for the oxidant, and no recycle. The basis is 2 moles of
NG burned, with excess air proportioned to produce φ02 = 3% in the gas streams. We first seek to

Chapter 9 System Balances on Reactive Processes 549

calculate the amount of species in each stream by making a gas-stream material balance according
to the principles outlined in Section 6.3.1. Table 9.8 shows the results.

Recycle Gas

! 6i

PFOG «-1-°--| Preheat L . Л J S W V F Q G 1 Annealing Κ Η Γ { ^ Γ " ! " 0xyAJr
1 fee r P F I G ^ 2 fee Α ξ ^ - J * - - - ^ -
Fe — L - J T c e j i H Tce I 3 ► Hot Fe NatGas

1 ' Warm Fe ' '

Figure 9.37 Flowsheet for the heating of steel to 1050 °C by the combustion of natural gas.

Reheating furnace is a single unit, but has been split for calculational purposes. Both the annealing

and preheat furnace are simple heat exchangers, with all chemical reactions taking place in the

burner. The oxidant may be air or oxygen-enriched air. The temperature of the annealing furnace

in-gas (S7, AFIG) can be decreased to minimize NOx production by recycling a portion of the

reheat furnace gas (S8, AFOG) back to the burner. Dashed lines represent gas phase streams, and

solid lines represent the flow of iron. The reheat furnace heat loss is 20 kJ.

Table 9.8 Stream amounts for iron annealing process with no recycle, based on initial material
balance for a basis of 2 moles of NG burned. Instream temperatures are 25 °C, the iron outstream
is 1050 °C, and the PFOG (S10) outstream is 200 °C. 22.677 moles of air are required to produce
24.737 moles of combustion gas having <p02 = 3%.

t,°C 25 ? 1050 25 25 ? ? ? ? 200

Name CldFe WrmFe HotFe NG OxyAir RcyGs AFIG AFOG PHIGs PFOG

Str# 1 2 34 5 6 7 8 9 10

F, ? ? ? 2 22.677 0 24.737 24.737 24.737 24.737

CH4 0 0 0 1.8 0 0 0 0 0 0

C2H6 0 0 0 0.12 0 00 000

C02 0 0 0 0.08 0 0 2.12 2.12 2.12 2.12
н2о 0
0 00 0 0 3.96 3.96 3.96 3.96

N2 0 0 0 0 17.915 0 17.915 17.915 17.915 17.915
o2 0 0 0 0 4.762 0 0.742 0.742 0.742 0.742
Fe 0 0 00 0 0 0 0 0 0

The next task is to calculate the amount of iron annealed, but here the material and heat
balances are coupled because the iron flow depends on the heat balance. The simplest procedure is
to start by making an overall system balance. Device details are ignored so only SI, S3, S4, S5,
and S10 are considered. The heat units are J/mol for species, and kJ for streams. FREED was
used to develop Ht-His equations for the gas-phase constituents between 25 °C and 2000 °C, Fe
between 25 °C and 600 °C, and other data, as shown in Table 9.9.

Table 9.9 Heat content data for the annealing of iron. Heat content equations are of the form 7/t-
#25 = At2 + Bt + C, J/mol and degrees Celsius.

A В С AtfcombNG

co2 0.00554 44.62 -1790 -807,700 J/mol

H20 0.00555 33.04 -884
N2 0.00225 29.138 -830 Я1050-Я25 for Fe

o2 0 34.91 -2030 39,270 J/mol

Fe 0.0113 24.5 -619

550 Chapter 9 System Balances on Reactive Processes

The overall system balance is straightforward. The S10 offgas (PFOG in Table 9.8) outstream
temperature is 200 °C. Based on the stream amounts in Table 9.8 and the enthalpy equation
parameters in Table 9.9, Я200-Я25 for S10 = 134.0 kJ. Combustion heat is -1615.4 kJ, Я1050-Я25
for Fe (S3) = 39.27 kJ/mol, and HL = 20 kJ. Equation [9.36] calculates the amount of iron
annealed (entering as SI, exiting as S3).

, rr t Δ 1615.4-134-20 „ „

moles of Fe annealed = = 37.2 Г9.361

39.27

The material balance is now complete, and with no recycle, further heat balances are

uncoupled. Our next objective is to calculate the S7 temperature (the combustion gas AFT) by

making a heat balance around the burner. Two moles of NG are burned with 22.677 moles of air

to produce 24.737 moles of combustion gas. Only one heat balance equation is needed:

1. Cool S7 from AFT to 25 °C. H= 0.00007403i2 + 0.7733/ -23.7 - 1615.4 - 0

Solving, S7 is 1807 °C.

We now turn our attention to calculating the temperature profile of the reheat furnace, which
has been split conceptually into two devices, a preheat and an annealing furnace. Suppose we
make a split where the gas temperature (S9) is 452 °C. If the preheat furnace heat loss is 20 % of
the total, it's heat loss is 4 kJ, which leaves 16 kJ for the annealing furnace heat loss. The preheat
furnace heat balance steps are:

1. Cool S9 from 452 °C to 25 °C. # = -341.1 kJ

2. Heat S10 from 25 °C to 200 °C. # = 134.0 kJ
3. Heat SI from 25 °C to t. H= 0.00042055/2 + 0.91182^-23.04.
4. Heat loss: HL = 4kJ.
Solving, SI is225°C.

Figure 9.38 shows the heat balance results around the preheat furnace for several split points
(i.e., at different S9 temperatures).

Preheat/Annealing Furnace Interface

700

Ü -O— Fe (S2) temp
0 600 - D - Gas (S9) temp JOT
ω JOT
3 500
-CvMO. JOT
400
EQCI). 300 XT

-CMD

0E5 200

bCD 100

CO

12 34 56

heat loss in preheat furnace, kJ

Figure 9.38 Gas and steel temperatures at various points along a reheat furnace having stream
flows as described in Table 9.8. Calculations based on a heat balance around the preheat furnace.
The leftmost point refers to the exit of the reheat furnace (i.e., the cold end), where S9 becomes
S10, and SI becomes S2.

The heat balance at the cold end of the reheat furnace (i.e., where the width of the preheat
furnace approaches zero) shows that the outgas (S10) is 175° higher than the incoming steel (SI).

Chapter 9 System Balances on Reactive Processes 551

As the conceptual split point moves to the right, the length of the preheat furnace increases, and the
temperature of both streams increases. In addition, the difference between the gas and steel
increases in going from the cold to the hot end of the reheat furnace. For example, the gas/Fe
temperature difference is 285° at the rightmost point in Figure 9.38, and at the hot end, the gas
temperature (S7) is 1807 °C and the steel temperature (S3) is 1050 °C, for a difference of 757°.
The reason why the gas and iron stream At is greater at the hot end of the reheat furnace than the
cold end is that the thermal mass of the gas phase is less than that of the steel.

There is increasing concern about the emission of certain combustion offgas species,
especially NOx, which is a greenhouse gas, and contributes to smog formation. The effect of
temperature on the increase of NOx formation was covered in Section 6.3.6, which showed NOx is
lower when flame temperature is lower. One way to lower the flame temperature is to recycle a
portion of the reheat furnace gas back to the burner. The recycle gas temperature depends on
where it is taken. The cooler the recycle gas, the less of it that needs to be recycled. From a heat
balance standpoint, as long as recycle does not affect the exit gas temperature (S10), its only effect
is on the burner AFT. The amount of iron that can be heated and the stream temperatures in the
preheat section of the reheat furnace remain the same. The AFT is lowered for the gas entering the
reheat furnace, but the thermal mass of the gas in the annealing section is increased. These two
factors tend to compensate for each other.

Suppose the NOx level in S10 was deemed acceptable when the burner AFT was 1600 °C.
This gives us two new system variables: the temperature and flow of recycle gas (S6). These two
S6 variables are obviously related; as the S6 temperature goes up, so does its flow. Suppose we
make a heat balance around the burner for an S6 temperature of 452 °C, which is the same
temperature we used in our earlier heat balance calculation around the preheat furnace.

1. Cool S6 from 452 °C to 25 °C. H = /^(-13.79) kJ

2. Combustion of 2 moles ofNG at 25 °C. H = -1615.4 kJ

3. Heat S7 from 25 °C to 1600 °C. H= 1403.2 + 7^(56.72) kJ

4. Heat loss. HL = 0 (burner is adiabatic)

Solving, F* = 4.6 moles.

This procedure was repeated for the same set of gas stream (S8) temperatures used in the
construction of Figure 9.38. Figure 9.39 shows the results.

Recycle Gas for Burner AFT = 1600 °C

6.00

J 5.75
о
5.50
5£ 5.25
gas (S6)
moles 5.00

4.75

о> 4.50
4.25
ω 4.00

200 300 400 500 700

recycle gas (S6) temperature, °C

Figure 9.39 Recycle gas (S6) flow required to produce a combustion AFT of 1600 °C. Horizontal
axis represents the temperature of the recycle gas at point of removal from the reheat furnace.
Calculations based on heat balance around adiabatic burner. Note that the S6, S8 and S9
temperatures are all the same.

552 Chapter 9 System Balances on Reactive Processes

The results show how much recycle gas (S6) is required to lower the combustion gas
temperature to 1600 °C as a function of the temperature ofthat gas. As expected, the higher the
recycle gas temperature, the more is required. The leftmost point represents the condition where
the recycle gas is withdrawn from the cold end of the reheat furnace.

The need for NOx mitigation is more important when oxygen-enriched air is used for
combustion. For example, the calculated AFT at φ02 = 30 % in the oxygen-enriched air is about
2370 °C, which is capable of producing a much higher NOx content than using ordinary air with an
AFT of 1807 °C. If recycling the furnace gas isn't practical, the burner can be operated with less
than stoichiometric air, and combustion can be completed by adding air downstream of the point of
entry of the combustion gas. FlowBal was used for making a system balance for this version of the
process, as discussed in the FlowBal User's Guide (worksheet Anneal).

One of the best ways to minimize the use of energy for a process is to capture the energy of
product streams and return it to the process. In the discussion of an aluminum melting process in
Section 8.5, the thermal energy in the melting furnace offgas was used to preheat combustion air or
preheat the ingots. In both cases, the heat exchanger is external to the reaction unit, and requires
ductwork and fans to move a hot gas through the exchanger. The principles of making a heat
balance on a reactive system with an external heat exchanger are the same as those introduced in
Section 8.5, except that a chemical reaction usually has a heat affect of its own, and may change
the composition and amount of a stream.

EXAMPLE 9.8 — Calcination of Magnesium Carbonate.

The calcination* of magnesium carbonate (MgC03) to MgO occurs in a fluidized-bed furnace
heated by combustion of NG (assume all CH4). Figure 9.40 shows a sketch of the process.
Calculate the effect of air preheat temperature on amount of MgC03 calcined, and the temperature
of the furnace gases leaving the heat exchanger. The basis is 1 m3 CH4 burned with 15.5% excess
air, and heat losses of 2620 kJ in the calcination furnace and 1730 kJ in the heat exchanger.

MgC03 Hot Fee Gas Cold Air
(25 °C) (650 °C) (25 °C)

Fluidized Warm Fee Gas
Bed Furnace (T = ?)

MgO
(650 °C)
Burner

NG T Hot Air (T = ?)
(25 °C)

Figure 9.40 Flowsheet diagram for the calcination of MgC03 in a fluidized bed furnace. The hot
furnace (stack) gas is passed through a counter-current heat exchanger to preheat the combustion

air. Solid lines represent solid phase streams, while dashed lines are gas phase streams.

Calcination is a process whereby a complex compound (usually a carbonate, hydroxide or sulfate) is heated
such that the compound decomposes. The calcination temperature is where the sum of the partial pressures
of the calcination gases equals one atmosphere in the absence of reactive gases. The calcination temperature
for MgC03 is 407°C. The procedure for calculating the calcination temperature is given in Section 6.5.1.

Chapter 9 System Balances on Reactive Processes 553

Data. All data is from FREED.

Solution. As usual, the first step is to make (or at least outline) a material balance for the process.
Also, it will simplify the arithmetic to change the basis for the problem from 1 m3 CH4 to 1 mole
since we are mainly interested in temperature and thermal efficiency, which are intensive
quantities. At the end, we can multiply by 44.61 to convert amounts back to the original basis.
The temperature basis will be 25 °C.

We have already examined the standard reaction for the combustion of CH4. From earlier
sections, recall that 9.524 moles is the stoichiometric amount of air for the combustion of one mole

of CH4. At 15.5 %XSA, 11.0 moles of air are required. The standard reaction for the calcination
ofMgC03is:

MgC03(c) -> MgO(c) + C02(g); A#°rxat 25 °C = 118.1 kJ [9.37]

The stack gas from the furnace contains the combustion gas plus the C02 from calcination.
However, since we do not know the amount of carbonate calcined, the two balances are coupled,

and we must make a system balance. We assign the variable X for the amount of MgC03 calcined.

FREED provided molar heat content data between 200 - 700 °C (Д-Я25) for each product
species. The coefficients were calculated using Excel's LINEST function, as shown in the
following table. Considering the variability and imprecision of the heat losses and the small
temperature span, a linear equation was used. For greater accuracy a quadratic equation is better.

Ht-H25 (J/mol) = A* + В

C02 H20(g) N2 o2 MgO Air

A 48.88 37.25 30.75 32.62 48.49 30.57

В -2896 -1604 -1134 -1325 -2560 -980

The first heat balance is written around the calciner. The first step is to burn one mole of

CH4 at 25 °C. The second step is to calcine X moles of MgC03 at 25 °C. The third step is to heat
the products to 650 °C. The fourth step is to cool the preheated air from some temperature /aph to
25 °C. The fifth step is to account for the heat loss (58 800 J for the revised basis).

Step#l: A#comb - -802 300 J

Step #2: A#rx=118 100(X)J

Step #3: Я65о-#25 for N2 = 8.69(18 850) = 163 800 J
Я65о-Я25 for 0 2 = 0.31(19 880) = 6160 J
Я65о-Я25 for H20 = 2(22 610) = 45 220 J
Я65о-Я25 for C02 = (1 + X)(28 880) = 28 880 + 28 880(X) J
Я65о-Я25 for MgO = 28 960(X) J

Step #4: Я25-Д for air - -1 l(30.57/aph - 980) = -336.3(/aph) + 10 780 J

Step #5: Heat loss - 58 800 J

The sum of the above heat effects is expressed as follows:

-336.3(4Ph) + 175 940(X) - 488 660 = 0

The value of X was calculated for various /aph between 100 and 600 °C. The result was
converted to the mass of MgO calcined for 1 m3 CH4, the original basis specification. Figure 9.41
shows the results. At the upper limit on air preheat temperature,* about seven kg of MgO can be
produced per m3 CH4 burned.

The theoretical upper temperature limit on air preheat is 650°C, but no practical heat exchanger would have
such perfect heat transfer capabilities. A reasonable upper limit is closer to 600 °C.

554 Chapter 9 System Balances on Reactive Processes

The second system balance is written around the heat exchanger. As the hot furnace gas
passes through the heat exchanger, it transfers heat to the combustion air and loses heat (38,800 J)
to the surroundings. The value of H6so-H25 for the entering furnace gas increases with air preheat
temperature because it contains more C02 from calcination. The warm furnace gas temperature is
calculated by a heat balance equation developed in steps. The first step is to heat 11 moles of air
from 25 °C to 4Ph. The second step is to cool the stack gas from 650 °C to its exit temperature
from the heat exchanger (tstk out). The third step is to add the heat loss (38 800 J for this basis).

Step #1 : #t-#25 for air = 1 l(30.57iaph - 980) = 336.3(4ph) - 10 780 J
Step #2: Ht-H650 for N2 = -8.69[30.75(650 - /stkout)] = -173 700 + 267.2(4,кои,) J

tft-tf65o for 0 2 = -0.31 [32.62(650 - 4,kout)] = -6570 + 10.1(istkout) J

#-#65o for H20 = -2[37.25(650 - /stkout)] = -48 430 + 74.5(/stkout) J

#-#65o for C02 = -(1 + X)[48.88(650 - /stkout)] = (1 + X)(-31 770 + 48.88(istkout) J

Step#3: Heat loss = 38 800 J

Calcination of МдСОз with Preheated Air

7.5 >— кдМдО/тЗСН4
]— moles МдО/mole CH4
7

6.5

О6
g 5.5
■Осσ 5
Ъгоа 4.5

éfeqО 4

~-L---TK--f^TО) 3.5
mol МдО/mol CH4 = 0.0019(f aph) + 2.78

3

100 200 300 400 500 600
air preheat temperature, °C

Figure 9.41 Relationship between amount or mass of MgO calcined per unit of CH4 as a function
of air preheat temperature. Hot furnace gas used to preheat combustion air. Calculational results
shown in figure box. Text box equation derived using Trendline tool. To change equation to kg
MgO/m3 CH4, multiply by 1.80.

We already have the relationship between X and ίφ from Figure 9.40. The three-step
equation set was entered as formulae in Excel, and the Goal Seek tool was used to calculate ts± out
for the same range of 4ph used previously.

In practical terms, the calciner would aim for a specific MgO production rate, and vary the
fuel and air input in response to the extent of air preheat temperature. This example converted the
original basis to 1 mol of CH4 burned to simplify the arithmetic, but the results can easily be
converted to a more practical basis, such as the amount of CH4 required per tonne of MgO
produced (called the fuel rate). Figure 9.42 shows the results of the heat exchanger heat balance
and the fuel rate.

Chapter 9 System Balances on Reactive Processes 555

Calcination of MgC03 with Air Preheat

600 190
—0—Tstkout
185
—-Q— specific fuel rate
180 О
175ΜΕ

170 -Фк2 Фсос
165

\ 160 Φ
155 «2 X

150 о

-^¥140

200 300 400 500
air preheat temperature, °C

Figure 9.42 Relationship between selected calciner parameters. The solid line shows how the
stack gas temperature (tstk out) varies with combustion air temperature (7aph)· The dashed line shows
the relationship between the /aph and the amount of CH4 required to produce one tome of MgO.

The warm furnace gas temperature shows a linear decrease in temperature with the air preheat
temperature. Even with a highly efficient heat exchanger, the warm furnace gas still exits above
200°C. Preheating the air from 100 °C to 600 °C decreases the fuel consumption about 25%.

Assignment. Calculate the effect of cutting the combustion air to 11 %XSA. What is the effect of
decreasing the heat losses by 25% in the furnace and heat exchanger?

The concepts introduced in Section 8.5 are easily extended to reactors where both heat
exchange and chemical reactions occur internally, such as a shaft furnace, where the gas and solids
flow counter-currently. Limestone burning kilns, the lead and iron blast furnace, and the cupola
are such devices. Generally the overall reactions in these furnaces are exothermic to bring the
reactants to the required temperature and provide for heat loss.

We start with a simplified version of a process for the calcination of limestone in a shaft
furnace. Here we use externally-heated air to provide the necessary heat. It's convenient to divide
the furnace into three regions, as shown in Figure 9.43. The upper and lower heat transfer zones
are pure heat exchange regions where no chemical reaction occurs. The calcination zone is a
chemical reaction zone where heat exchange also occurs. Each of the pure heat exchange zones
operates like a pebble bed heat exchanger, as described in Example 8.8. Also, please look back at
Example 9.8 where MgC03 calcination was discussed. The objective is to show how to calculate
the unknown temperatures and the amount of air required per mole of limestone calcined.

A system balance requires calculation of the unknown temperatures and the amount of air
required per unit of limestone calcined. Calcination takes place between 1150 and 1250 К in the
calcination zone, along with the required transfer of heat from the gas to the solid. Equation [9.38]
shows the chemical reaction for limestone calcination.

СаСОз(с) -+ CaO(c) + C02(g); A#°rx at 1150 К - 40 100 cai [9.38]

556 Chapter 9 System Balances on Reactive Processes

Limestone I ►

-L Top gas

Tin'-stone = 298 К T 9as _ 9
•out -■
Upper heat

transfer zone
| ^__Jjgas _
y - s t o n e = 1 1 5 0 K l 1200 К
^777777777777777777777777^

Tlime = 1250K Calcination JTair=1300K

zone
\V7777777777777777777777/.

Lower heat
transfer zone

Lime Air
T1 -inair = 9
x lime _ 0

»out -■

Figure 9.43 Sketch of stream flows for moving-bed limestone calciner. Hot air flows counter-
currently to the solids flow. The residence time and heat transfer coefficient sets the gas
temperature in the calcination zone to be about 50° hotter than the solids temperature, while the
incoming air temperature will be about 15° hotter than the exit lime temperature. On the basis of
one mole of limestone calcined, the heat loss is 4000 cai in the lower zone, and 3000 cai in the
calcination zone and the upper zone. Solid lines depict solid stream flows, while dashed lines are
gas streams.

In this simplified example, thermal energy for the process is provided by the sensible heat in
the air, which must be heated externally. For limestone calcination in the chemical reaction zone,
all of the thermal energy must be provided by air cooling from 1300 to 1200 K, so the amount of
air in must be much larger than the amount of limestone calcined. This is clearly an unbalanced
heat exchange situation, with air having the greater thermal mass. So FinairCpinair > ^п181опеСр1П181опе,
so the 70Utlime = Г1па1Г - 15°. However, Toutgas » 298 K. FREED data was used to generate Ят-Я298
equation parameters , as shown in the table below.

#т-#298 = АГ+ B7^ + C, cal/mol

AВС СаСОз AВ С
CaO 11.24 0.00071 -3535
Air 6.757 0.00054 -2090 C02 19.99 0.00495 -6540
8.32 0.00237 -2740

Material and heat balance equations were written around each of the three calciner zones for
N moles of air entering at the bottom and one mole of limestone calcined. The top gas consists of
N moles of air + 1 mole of C02. For ease in calculation, calcination is assumed to take place at
1150 K, although any chosen calcination temperature between 1150 and 1250 К will give the same
result. There is one amount unknown (TV) and three temperature unknowns (routhme, Tmm\ and
7"outgas) so four equations are needed. A balance equation is written for each stream, and the two

heat effect terms are set equal to each other.

Upper heat transfer zone: The limestone stream is heated from 298 to 1150 К and the gas
stream cools from 1200 К to routgas:

19.099(1150) + 0.00495(1150)2 - 06.5040023=7N(1[260.70527(1(2r0ou0tga-s) )270Utgas) + 0.0054(12002 -
(^outg ) ) ] +8.32(1200 routgas) +

Calcination zone: The limestone is calcined at 1150 K, and the lime is heated from 1150 to
1250 K. The gas (air) is cooled from 1300 to 1200 K.

40 100 + 11.24(1250 - 1150) + 0.007Ц12502 - 11502) = N[6.757(1300 - 1200) +
0.0054(13002-12002)]

Chapter 9 System Balances on Reactive Processes 557

Lower heat transfer zone: The lime is heated from 1250 К to Tnu,ime and the air is cooled from
Tm™ to 1300 K. In addition, 70Ulitme . 20.

11.24(1250 - rpoutlime) + 0.0071(12502 (TOutlime)2) = ^[6.757(1300 - rinair) +
0.0054(13002 -Ti air)2)]

One way to solve the equation set is to first use Goal Seek on the calcination zone, and solve
for N. Then Goal Seek can be used on the other two zones to obtain values for all variables.
Alternatively, the four equations can be written in appropriate form for Solver. In either case, a
feasible solution was found.

rp lime r-p air X gas N air
tout A in A out 54i V.8i n mol

1290 К 1310K 1143 К

These results show that a very large amount of hot air is required for the process; 421.3 kcal
are required to heat the air to 1310 K. Most of this energy leaves the system as sensible heat in the
offgas (357.2 kcal). The poor thermal efficiency is a result of the oversimplification of the process,
and can be greatly improved by adding an external heat exchanger to recapture heat from the
offgas. The system flowsheet with an external heat exchanger is shown in Figure 9.44. Both gas
streams have about the same thermal mass, so they will approach each other's temperature at the
ends of the heat exchanger. Suppose the air temperature at the hot end of the heat exchanger is 20°
lower than the incoming hot offgas (i.e., 1123 K), and heat loss is 2500 cai. A heat balance around
the heat exchanger shows that the "cool" offgas exits at 342 K, thus most of its thermal energy has
been transferred to the air. However, additional heat must be supplied to heat the air from 1123 to
1310 К before it enters the calciner. The energy requirement after installation of the heat
exchanger is 82.7 kcal as opposed to 421.3 kcal without the external exchanger. This energy can
be supplied for example by injecting a stream of natural gas into a burner that uses air from the HX
to combust the NG.

Limestone (298 K) Hot top gas (1143 K) I

H = 0 kcal H = 357.2 kcal
H ÙHL = 10.0 kcal Air (298 K)
Ж /yVÒkcal

il iILime (1290 K)Calcining HL = 2.5 kcal
/У =«4-12.1 kcalfurnace

Air (1123 К)
H = 338.6 kcal

i_ Cool top gas (342 K)

Air (1310 К) ID 7/"= 16.1 kcal

Hx = 82.7 kcal

H = 421.3 kcal

Figure 9.44 Flowsheet for calcination of one mole of limestone with 54.8 moles of air. Process
has internal heat exchange in the calcining furnace and an external heat exchanger (HX) for
recovering heat from the calciner offgas. The symbol H designates the heat content above 298 K.
Block arrows designate heat loss to the surroundings (HL) or heat input from the surroundings (Щ.
The requisite heat (82.7 kcal) could be supplied by inserting an NG burner into the air stream
between the HX and the bottom of the calciner.

Limestone calcination by the use of heated air alone is not a practical process because of the
extra equipment needed and the difficulty in dealing with the top gas dust. In practice coke is
charged with the limestone. The coke burns in the calcination zone to produce all the necessary
heat. The amount of air required is much less so the lime exit temperature is also much less.

558 Chapter 9 System Balances on Reactive Processes

Another example of a shaft furnace to carry out a chemical process is the reduction of nickel
oxide. Porous NiO pellets are easily reduced by hydrogen to produce metal. Reduction of NiO
starts at about 400 °C and takes place more rapidly above that, but the metallic form should not
exceed 900 °C otherwise the pellets will sinter and clog the discharge opening. Figure 9.45 shows
a sketch of a possible shaft-furnace process.

1NiO pellets Top gas (T = ?)
(T = ?)
±

Upper heat Tsolid * 430°C
transfer zone Tgas « 420°C

V////////////////////////s\

Reduction

zone J^ Tso,id « 900°C
Tgas « 890°C
Ni pellets Lower heat
(T = ?) * H2 (25°C)
transfer zone
L.

Figure 9.45 Sketch of shaft furnace reactor for the H2 reduction of NiO pellets. The objective is
to reduce all of the NiO in the reduction zone. The heat loss in each zone is estimated to be 500

cal/mol of NiO fed to the furnace, or about 1500 cal/mol NiO overall.

The chemical reaction for the reduction of NiO with H2 is: [9.39]
NiO(c) + H 2 ( g ) -> Ni(c) + H 2 0 ( g ) ; ^ e q = pii^°/

Values of Кщ vs. temperature and other thermodynamic data are shown in the following
tables. The reaction is exothermic, and goes almost to completion, with the equilibrium gas having

only about <pH2 = 0.5%. But to provide a suitable driving force for complete NiO reduction, the
top gas from the shaft furnace should have a <pH2 at least 10 times the Кщ value. If necessary, the
unused H2 in the top gas can be burned to preheat the solid feed.

t,°C 500 600 700 800 900
Кщ=рЯ2/рЯ20 273 225 190 163 143

#,-#25 = At + ВГ + C, cal/mol

AВ С А ВС

NiO 13.619 -0.00056 -478 H20 7.756 0.00144 -186
Ni 7.451 -270
Air 6.989 0.00016 -176 н2 6.827 0.00028 -163
—
N2 6.779 0.00071 -169

NiO + H2 -> Ni + H20 АЯгх = At + С -3.104 -870

Before starting any calculations, it's best to review again the features of counter-current heat
exchange, which will take place in the upper and lower heat transfer zones (UHTZ and LHTZ).
We consider first the UHTZ where NiO and H20 are the counter-current phases. A glance at the
FREED tables for these substances shows that the thermal mass for NiO is much greater than that
of H20. The UHTZ is therefore an unbalanced heat exchanger. Recall from Section 8.5.1 that for
an unbalanced exchanger the value of ΔΓ is smaller at the point where the stream with the greatest
thermal mass enters, and approaches zero at that point as either A or U increase without limit.
Therefore, the temperature of the top gas (mostly H20) must be about the same as the entering

Chapter 9 System Balances on Reactive Processes 559

NiO, and must be above 100 °C to avoid condensation of water vapor in the furnace. The reaction
zone (RZ) is nearly balanced.

The consequence of the unbalanced UHTZ is that the solid stream always requires more heat
than the gas stream can deliver. Suppose for example that the solid and gas streams at the top of
the furnace are at 200 °C. A calculation for H45o-H2oo shows that one mole of NiO requires 3310
cai and one mole of H20 can deliver only 2170 cai. Therefore not only are the temperatures of the
top gas and the NiO feed the same, they must be the same as the temperature at the top of the RZ.
This means that the UHTZ is not really a heat exchange zone at all. Both streams have the same
temperature entering and leaving it! Thus the shaft furnace really consists of only two zones, the
RZ and the LHTZ. The UHTZ does not exist; the top of the RZ is at the top of the furnace.

Next we consider the LHTZ, where the Cp for Ni and H2 are about the same, which means that
the LHTZ is close to being balanced. Note that since the Ni product can leave the furnace above
25 °C, there are no apparent problems with nearly-balanced streams in the LHTZ.

The problem of not being able to recover heat from the top gas, and the need to preheat the
NiO means that the shaft furnace may not be a very practical device for carrying out the process.
Nevertheless, it's useful to make a heat balance on the RZ and LHTZ to see if a shaft furnace is
feasible (even if not practical). Since we have quadratic equations for temperature, we need Solver
to do the arithmetic. The basis is one mole of NiO in.

The variables of interest are: nU2 in, H2 temperature in, boundary temperatures of the RZ, and
product temperature. System balance equations were written for the overall process and for the
RZ, and as a check, for the LHTZ. The gas and solid stream temperatures were assumed equal
along the RZ (i.e., excellent heat transfer). Since only two heat balance equation sets can be
written, three of the five variables must be set, and the two others calculated by solving the
equation set when the sum of heat effects = 0. Since there are only two zones, the overall heat loss
was 1000 cai. The abbreviations used to designate the five variables are:

Тц = temperature of hydrogen in;
пц = moles of hydrogen in;
TRZUB = temperature at the upper boundary of the RZ;
TRZLB = temperature at the lower boundary of the RZ;
rNi = temperature of the nickel product.

The equation set for the overall heat balance is shown below. All of the NiO was assumed to
be reduced at TRZUB, which assumption does not affect the heat balance. The Τ^ζυΒ was selected as
the basis temperature. In each equation, Я stands for the heat effect value. The heat balance
closes when ΣΗ = 0. Solver will change the unspecified variables until the heat balance closes.

1. Reduce NiO at 7RZUB: Я = - 3 . 1 0 4 ( 7 ^ ) - 870
2. Cool Ni from 7 W to ΓΝι: H = 0.00016(7Ni2 - Тшш2) + 7.451(7Ni - Т^ив )
3. Heat H2 from Tu to Т^ив: H = /7Η[0.00028(7^υΒ2 - TH2) + 6 . 8 2 7 ( 7 ^ - Гн)]

4. Heat loss: H = 1000

The equation set for the RZ also uses the Τ^ζυΒ as the basis temperature.

1. Reduce NiO at 7RZUB: Я = - 3 . 1 0 4 ( 7 7 R Z U B ) - 870 Тшш)
2. Heat Ni from Т^ш to T*ZLB: Я = 0.000 ЩТ^ъв2 - ^RZUB2) + 7 . 4 5 1 ( 7 ^ -

3. Cool H2 from TRZLB to Г^ив: Я = nu[0.00028(Т^ш2 - TRZLB2) + 6 . 8 2 7 1 ( 7 ^ - TRZLB)]

4. Heat loss: Я = 500

To get a slight surplus of H2 above stoichiometric, пц was set to 1.05. Also, we seek to keep
TRZLB about 880°C and 7RZUB about 430 °C. This leaves 2 variables unspecified: TH and ΓΝί.
However, Solver was unable to find a solution to the heat balance equation sets, indicating that

with the set variable values, no feasible Гн and Г№ exist. In such a case, it is sometimes helpful to
allow Solver to vary more parameters than there are equations just to see if a solution exists at any

560 Chapter 9 System Balances on Reactive Processes

feasible TRZLB and TRZUB. Again no feasible solution was found. An analysis of the Solver results
indicated that the solid and gas streams in the LTHZ were too closely balanced. This meant that
the Ni product was unable to completely remove the heat created by the reduction reaction and the
heat brought in by the NiO feed. If the mass of the solid product were larger, this condition would
be eliminated. This was done by adding more NiO than could be reduced by H2, thus the solid
product had greater thermal mass*. This created a new variable, /7Nio, and an enthalpy term was
added for NiO in the heat balance equation set. nH was eliminated as a variable, and set = 1.

With these modifications and setting TH = 25 °C, rRZLB - 880 °C and TRZl]B = 430 °C, Solver
found a feasible solution at rcNi0, = 1.237 and Г^+^о = 257 °C with a heat loss of 1237 cai.
However, operating the shaft furnace with these conditions has the onerous requirement of
requiring the NiO feed to be preheated to 430 °C. Even worse, the product has unreduced NiO
(xNiO = 0.20). We now know that the constraints imposed by the internal heat exchange
characteristics render a shaft furnace an impractical way to reduce NiO with H2.

A better flowsheet might be obtained by separating the heat exchange zones from the
reduction zone. A sketch of such an arrangement is shown in Figure 9.46. Reduction is carried out
in a fluidized bed (FB) using granular NiO as a feed. Excess air is used to burn the top gas H2 to
preheat the NiO. Since we are using granular NiO, it's not practical to use vertical counter-current
heat exchangers for the entering and exit streams. To assure that no water vapor will condense in
the solids heater, the offgas temperature was set at 175 °C. Basis temperatures were set so as to
simplify the arithmetic: 175 °C for the solids heater and 820 °C for the FB reactor. The basis
temperature for the H2 heater was its temperature, whatever that turned out to be. The two streams
leaving the FB were assumed to be at the FB temperature, and the two streams leaving the H2
heater were assumed to be at the H2 heater temperature.

Offgas ^ Air (25°C)
(175°C)
NiO (25°C)

Top gas (820°C)

Fluid Ni (820°C)
bed
reducer
(820°C)

H2 (25°C) Ni (T = ?)
r?H2 = ?

Figure 9.46 Sketch of flowsheet for fluidized-bed reduction of nickel oxide by hydrogen. NiO is

heated by combustion of H2. The solids heater is a rotary kiln whose heat transfer coefficient is not
as good as a typical heat exchanger. The H2 heater is a small FB where both exit streams are at the
same temperature (i.e., a co-current heat exchanger). The heat loss in the solids heater was 800

cal/mol NiO, in the FB 1000 cal/mol, and in the H2 heater 300 cal/mol NiO.

Alternatively, Ni could have been added to the top feed. However, since the top feed has to be heated, the
feed heater gas atmosphere would have to be neutral to both Ni and NiO, a difficult problem.

Chapter 9 System Balances on Reactive Processes 561

The basis is one mole NiO in and reduced, so one mole of H20 is produced in the FB. The
amount of combustion air = 0.60 mol. A heat balance can be written around each reactor to
calculate the process variables: ин, TH to the FB, and ΓΝί0 to the FB. The only additional
thermodynamic information required is for air, and formation of H20(g). This information is
shown below.

HX-H25 for air = 6.989/ - 176 (cal/mol)

AH°f for H20(g) at 175°C = -58 156 (cal/mol)

Solids heater: 1 mole NiO heated.
1. CoolFBH2Otol75°C: H= 0.00114(1752- 8202) + 7.756(175 - 820)
2. Cool FB H2 to 175 °C: # = (лн - 1)[0.00028(1752 - 8202) + 6.827(175 - 820)]
3. Heat air to 175 °C: H= 0.60[6.989(175) - 176]
4. Heat NiO to ΓΝί0: Я = -0.0056(Г№О2) + 13.619(ΓΝί0) - 478

5. FormH2Oatl75°C: tf = -58 156(ин-1)
6. Heat loss: Я =800

Fluid bed reducer. 1 mol NiO reduced. rNi02)+ 13.619(820 - W )
1. Cool NiO from ΓΝι0 to 820 °C: Я = -0.0056(8202 7H2) +6.827(820-rH)]
2. Heat H2 from Гн to 820 °C: Я = ян[0.00028(8202

3. Reduce NiO at 820 °C: Я = -3.104(820) - 870

4. Heat loss: Я = 1000

Hydrogen heater. ян heated, 1 mol Ni cooled.
1. Cool Ni from 820 °C to TH: H = 0.00016(ГН2- 8202) + 7.451 ( Г н - 820)

2. Heat H2 from 25 °C to Гн: Я = ян(0.000287V + 6.827ГН- 163)
3. Heat loss: Я =300

Solver varied the three unspecified variables until the ΣΗ for all units = 0. Solver found a
feasible solution:

лн =1.103; Гн-401°С; rNl0 = 892°C.

The AFT for the solids heater flame was about 965 °C, and the offgas had φ02 = 4.3%. It
appears that a three-stage NiO reduction flowsheet is feasible and probably practical, other than the
engineering difficulties associated with handling a granular product through the rotary kiln solids
heater and the hydrogen heater.

EXAMPLE 9.9 — Formation of Nickel Ferrite by Spray Roasting.

Spray roasting is a process whereby a solution phase is sprayed into a hot gas in order to
produce a powder of specific composition. Here a mixed nickel-iron acetate solution is spray
roasted to produce nickel ferrite (NiFe204). The ferrite can be sintered in a subsequent operation to
make ceramic parts. wH20 in the mixed acetate solution varies between 0.70 and 0.80. Calculate
the stream flows to maintain the roasting furnace temperature at 475 °C (±25°) with a furnace
atmosphere of <p02 = 5.0 %. The heat loss is 200 kJ per kg of mixed metal acetate. A basic
flowsheet requires a spray roaster, a heat exchanger for the roaster offgas to preheat all or part of
the air, and a burner to provide necessary heat. Figure 9.47 shows the flowsheet.

562 Chapter 9 System Balances on Reactive Processes

Acetate solution

Dilution
Air

Warm Warm

I roaster offgas Process
air
Spray
roaster

Ni ferrite Hot
< roaster offgas

Figure 9.47 Flowsheet for the production of nickel ferrite. Dotted lines represent gas streams,
dashed lines are liquid streams, and solid lines are solid streams. The splitter and burner are used
as necessary to control the spray roaster temperature.

Data. The oxidation of the acetates involves two steps. First, the acetate must be "undissolved"
from solution, and then oxidized to ferrite. The standard reaction for the oxidation of nickel and
iron acetate to form nickel ferrite is:

№(С2Нз02)2(с) + 2Ре(С2Нз02)з(с) + 1602(g) -> NiFe204(c) + 16C02(g) + 12H20(/,g) [9.40]

Based on unpublished data, Δ#°ΓΧ for Equation [9.40] at 25 °C is -6210 kJ, or -9660 kJ/kg of
mixed acetate roasted. Unfortunately, the heat required to separate the dissolved acetate from
solution is unknown, but is certainly very small compared to the chemical reaction of Equation
[9.39], so is ignored. Table 9.10 shows linear heat content equation parameters obtained by fitting
FREED tabular data from 25 °C to 500 °C.

Table 9.10 Heat content equation parameters for and product gases from the nickel ferrite roaster

process. Data for H20 based on liquid water at 25 °C and steam at offgas temperature; the
equation includes the heat of vaporization.

Ht-H25 (kJ/mol) = At + B, kJ/mol

co o2 H20(/,g) N2 2 Air NiFe204

A 0.0465 0.0371 0.0321 0.0327 0.0324 0.196

В -1.88 42.53 -1.25 -1.20 -1.26 -10.77

Solution. One kg of mixed acetate contains 1.556 mol of Ni(C2H302)2 and 3.112 mol of
Ре(С2Нз02)з. The stoichiometry of Equation [9.40] shows that 16(1.556 + 3.112)/3 = 118.5 mol of
air are required to oxidize one kg of mixed acetate. The actual amount of air to give 5% 0 2 in the
offgas depends on the amount of water brought with the 1 kg of acetate. Table 9.11 shows the
values and equations necessary to make a material balance on the roaster. All of the process air
was passed through the heat exchanger.

The spray roaster temperature t is the dependent variable. Since the thermal mass of the
roaster offgas is much higher than that of the process air, t also represents the temperature of the
warm air entering the spray roaster from the heat exchanger. This represents the highest
theoretical recovery of heat, and approaches a practical situation because of the larger thermal
mass of the roaster offgas. A heat loss value of 200 kJ was intended to allow for both heat losses
and any heat exchanger inefficiency.

Chapter 9 System Balances on Reactive Processes 563

Table 9.11 Molar amounts and material balance equations for the oxidation of one kg of mixed
acetate species (4.668 moles). Material amounts for a 25 % acetate solution shown in the third
column. Steps 1, 2, 4 and 5 are independent on initial acetate composition.

process steps equation 25% results
acetate oxidation air 118.5 118.5
acetate oxidation N2 93.6 93.6
166.5
acetate water (100000/% acetate - 1000)/18.02 18.7
oxidation water 18.7 24.9
24.9 303.7
oxidation C02 94.9
total acetate gas oxdn N2 ± acetate wtr + oxdn wtr + oxdn C02 213.4
extra air for 5% 0 2 total acetate gas/3.2
118.5 + extra air
total air

We first examine the range of acetate compositions that will bring the spray roaster product
temperature to 475 °C (±25°) without dilution air or supplementary heat. The system balance steps
are listed below, and summarized in Table 9.12.

Step 1 : Oxidize acetates at 25 °C
Step 2: Heat all water from 25 °C to t
Step 3: Heat C02 from 25 °C to t
Step 4: Heat N2 from 25 °C to t
Step 5: Heat ferrite from 25 °C to t
Step 6: Cool oxidation air from t to 25 °C

Step 7: Cool extra air from / to 25 °C

Step 8: Add heat loss.

Table 9.12 System balance equations for the formation of nickel ferrite by the oxidation of 1 kg of
mixed acetates. Specific values of Я for the 25% acetate solution are listed in the fourth column.

Steps H (Value or Equation) moles H for 25% Ac
1 4.67(-2070) 4.67 -9660
2 185.2
3 [(100/% Ac - 1)/0.01802+ 18.7](0.0371/+42.53) 24.9 6.871/+ 7875
4 24.9(0.0472?-3.02) 93.6 1.175/ - 7 5
5 93.6(0.0321/-1.25) 1.55 3.004/ - 1 1 7
6 1.55(0.196/-10.77) 118.5 0.304/ - 17
7 94.9 -3.839/ + 149
8 -[118.5(0.0324/-1.26)] — -3.075/ + 120
-[(total acetate gas/3.2)(0.0324t- 1.26)]
200
200

The final temperature for the 25 % acetate solution can be obtained from the fourth column
values as 343 °C. Clearly, the 25 % acetate solution is too dilute to allow the spray roaster to reach
475 °C (±25°). The acetate composition that gives a spray roaster temperature near 475 °C is best
determined using t as the independent variable, and % acetate as the dependent variable. The
system balance equations were written to an Excel worksheet, and Goal Seek was used to solve for
% acetate for three temperatures. The results are shown in Figure 9.48

564 Chapter 9 System Balances on Reactive Processes

Acetate Concentration vs. Roaster Temperature
26.6
• 26.5
Ъ 26·4

CO

с 26.3

ü 26·1

к03 2 6 0

25.9
440 450 460 470 480 490 500 510 I
roaster temp, deg. С

Figure 9.48 Relationship between the acetate composition and ferrite roaster temperature. The
relationship is linear. For acetate compositions above about 26.5%, some of the process air must
be diverted around the heat exchanger. For acetate compositions below about 26.0 %, additional
heat must be supplied by the burner.

To control the spray roaster temperature between 450 °C and 500 °C, the solution should

contain between 26.0 and 26.5 % acetate. Below 26 % acetate, additional heat must be supplied by

the burner. Suppose we wanted to maintain the roaster temperature at 475 °C. The combustion of

CH4 with excess air sufficient to provide 5% 0 2 in the flame requires 12.81 mol of air/mol of CH4.
The A//°Comb of CH4 to H20(g) at 25 °C is -802 kJ/mol, and produces 2 mol of H20(g), one mol of
C02(g), 7.52 mol of N2(g), and 3.29 mol of excess air. The heat available for the process is the
АЯ°СОть of CH4 minus the heat content of the product gases at 475 °C. FREED's Reaction tool
showed that under this condition, the heat available for the process is -604.6 kJ/mol CH4. Heat
balance results on the process at different % acetate compositions are shown in the following table.

The trend is linear; for every decrease of one percentage point in acetate concentration below about

26.2 %, 0.80 mol of CH4 must be used.

% acetate 26.24 25.5 25 24.5
H deficit, kJ 0 342 584 837
mol CH4 needed 0 0.57 0.97 1.38

For compositions over 26.5% acetate, the temperature must be controlled by diverting some
of the process air around the heat exchanger (i.e., the splitter removes part of the cold air to enter
the spray mixer). The same amount of air is used, but not all of it passes through the heat
exchanger, thus, in effect, lowering the air temperature entering the spray reactor. Again, suppose
we wanted to maintain the roaster temperature at 475 °C while using acetate compositions above
26.2 %. Each mol of warm air entering at 475 °C brings in -14 kJ. Heat balance results on the
process at different % acetate compositions are shown in the following table. The trend is linear;
for every percentage point increase in acetate composition above about 26.2%, 15 % of the cold air
should bypass the heat exchanger.

% acetate 26.24 26.5 27 27.5
acetate gas, mol 293.2 291.2 287.3 283.5
excess air, mol 91.6 91.0 89.8 88.6
210.2 209.5 208.3 207.1
total air, mol -116 -332 -541
excess heat, kJ 0 23.7 38.6
bypass air, mol 0.0 8.3 0.11 0.19
split fraction cold air bypassing HX 0.0 0.04

Chapter 9 System Balances on Reactive Processes 565

Assignment. Modify the flowsheet to examine if the burner might be eliminated. Consider the
substitution of acetic acid for some of the water. The only purpose of air would then be to supply
heat and maintain the roaster gas at the correct temperature and <p02 of 5%. The molar АЯ°сотЬ of
acetic acid at 25 °C to H20(/) and C02(g) is -874.2 kJ/mol.

9.7 Aqueous Processes

The techniques involved in balancing ionic reactions and making system balances on
hydrometallurgy processes have already been outlined (Sections 5.10, 6.10. 7.11.3, 8.9), and
applied, in most cases, to relatively simple processes. The strategy involved when heat balances
are required on chemically-reacting systems is basically the same as used in the above sections.
First, obtain data on Ai/°form and AG°form for all relevant species. Second, write appropriate Кщ
expressions for the independent reactions. Third, make material balances to find the amount of
each species. Finally, make a heat balance to account for heat that must be removed or added to a
system to maintain the stated temperature. The difficulty for aqueous system heat balance
calculations is when the heat effect changes the temperature. This is because most thermodynamic
data in typical handbook sources is listed only at 25 °C. The effect of temperature on the A^eq
value can be estimated by the van't Hoff equation [6.125], but even when thermodynamic data is
available at higher temperatures, the results can be in error when the solute ion concentrations
reach a point where the ion activities deviate from ideal behavior.

Applications of the above techniques are demonstrated for a relatively simple
hydrometallurgical process involving recycling lead-acid vehicle batteries. When these batteries
have reached the end of their life cycle, they contain a large amount of lead compounds, mainly
lead sulfate, lead dioxide, and elemental lead. The principal impediment for returning the lead-
containing compounds to a conventional lead smelter is the presence of lead sulfate. The first step
in industrial practice is to use an aqueous solution of sodium carbonate to convert the lead sulfate
to lead carbonate: Table 9.13 shows handbook thermodynamic data for the relevant species.

Table 9.13 Thermodynamic data for the formation of selected species from the elements at 25 °C,
kJ/mol.

Species PbS04 PbC03 Na2C03 Na2S04 H20(/) HCO3 C032~ Pb2+ OH' Na+ S042~
AH° -920.0 -699.2 -1130.7 -1387.1 -689.9 -675.2 +0.9 -230.0 -240.3 -909.4
AG° -813.0 -625.5 -1044.4 -1270.2 -285.8 -586.9 -24.4 -157.3 -261.9 -744.5
-237.1 -527.9

Evaluation of an aqueous process starts by writing simple non-aqueous equation(s) to see if
they have a large tendency to proceed in the desired direction. We already know that the process
works, but it's always good to verify starting assumptions. Equation [9.41] at 25 °C shows that the
reaction is exothermic and has a large negative value of AG°. The reaction is essentially
spontaneous and complete, but to assure complete conversion of PbS04 to PbC03, a small amount
of excess Na2C03 is added. In that case, PbS04(c) is the limiting reactant.

Na2C03(c) + PbS04(c) — Na2S04(c) + PbC03(c); AG° = -38 000 J. AH° = -35 600 J [9.41]

The use of excess Na2C03 means the aqueous solution has basic characteristics, so the pH
should be well above 9. We can thus disregard the presence of HS04"(ag) and H+(aq) as
significant participants in the process stoichiometry. In addition, the small Ksp of PbC03(c) means

that its contribution to the carbonate ions is negligible in comparison to ions produced by the

ionization of the excess Na2C03. Therefore, the only ions having significant molality in solution
are HC03~, C032~, OH", Na+, and S042~. Once these ionic molalities are calculated, we can
calculate the very small molalities of Pb2+ and H+.

For a basis of one kg of water (55.5 moles) containing initially one mole of PbS04(c), we add
1.2 moles of Na2C03. This basis helps the arithmetic because the amount of each ion product
equals its molality. One mole of Na2C03(ag) is stoichiometrically consumed to produce one mole

566 Chapter 9 System Balances on Reactive Processes

of РЬСОз(с). The solution then contains one mol of SO^~{aq) which for ion neutrality is balanced
with two moles of Na+(a^). The remaining (excess) Ыа2СОз(с) dissolves to form carbonate ions
according to Equations [9.42] and [9.43].

Na2C03(c) — 2m\aq) + HCOf(aq) + OlT(aq). [9.42]

Na2C03(c) — 2Na+H) + С032"И) [9.43]

From the stoichiometry of Equation [9.42] we note that the molalities of [HC03~] and [OH~]
are equal. For the dissolution of 0.2 moles of Na2C03(c), we note that [HC03~] + [C032~] = 0.2.
Finally, in the presence of water, whose activity is assumed to be one, Equation [9.44] expresses

the equilibrium for all three ions.

H20(/) + C032(aq) - Н С 0 3 " И ) + OST{aq)\ Keq = [HC03~][OH~] = 8.35x10-5 [9.44]
[C032-]

The two stoichiometry relationships and the Keq expression were solved to give [C032 ] =
0.196, [HCOf] = 0.004, and [OFT] = 0.004, for a pH of 11.6. The [Pb2+] can be calculated from
Equation [9.45] because we know [C032~], and the activity of PbC03(c) = 1.

PbC03(c) - ?hI+(aq) + C03z'(aq); Keq = [CCV ][Pb' J = 1.51xl0"13 [9.45]
aPbCOi

Hence, [Pb2+] = 7.7 x 10 13, an extremely low concentration that may allow the Na2S04 to be
recovered from solution and used for industrial purposes.

These results allow an overall heat balance calculation for the process. Table 9.14 shows a
summary table of input and output amounts, together with the AH°form for the designated amount of
each species at 25 °C. No data is entered for ?h2+(aq) because it is present in such a tiny amount
that it has no effect on the heat balance. The overall net heat for the process is -43 kJ. The final
product is a mixture of the appropriate amounts of H20(/), PbC03(c), Na2C03(c), and Na2S04(c).
The overall Cp is 4,420 kJ/deg, which means that under adiabatic conditions, the process would
experience a 10 degree increase in temperature. Notice that the overall АЯ°ГХ for [9.40] is -36 kJ,
which is not much different than for the aqueous process.

In practice, the Na2C03(c) need not be added in excess if the final aqueous phase has more
value when it contains only Na2S04(ag). In that case the limiting reactant for Equation [9.40]
would be Na2C03(c) and a small amount of PbS04(c) would remain in the insoluble material from
the battery leach process. This small amount can be handled without difficulty in a smelter.

Table 9.14 Summary table of standard heats of formation of the amounts of input and output
species, based on data from Table 9.13.

Input mol, species AH, kJ Output mol, species Atf,kJ
1.2 Na2C03(c) -1357 -699
1.0 PbS04(c) -920 1 РЬСОз(с) -577
55.5 H2O(/) -15862 2.4 Na+ -909.4
-132
1.0 S042"
0.196 -3
CO32" -1
-15861
0.004 HCO3

0.004 OH

55.496 H20(/)

Finally, as with all aqueous systems, calculations based on ideal ionic activity can be greatly
in error. Thus, for example, the [Pb2+] may be in error by an order of magnitude. But even if the
lead level is two orders of magnitude higher, it may still be at an acceptable level for industrial
uses of Na2S04. For situations requiring more accurate electrolyte concentration calculations, it's
better to use a commercial thermodynamic program (OH Systems 2010).

Chapter 9 System Balances on Reactive Processes 567

We now extend our treatment of hydrometallurgy to a multi-device system: the extraction of
copper from chalcopyrite, CuFeS2. Chalcopyrite is by far the most common ore mineral for copper
extraction, and is most often extracted by heap leaching, which involves trickling water through a
bed of crushed ore while encouraging a flow of air through the heap. The process is slow, and may
take many months to dissolve the chalcopyrite (and other copper-containing ore minerals).
Metallurgists have sought for years to find a cost-effective way to leach copper concentrate in
tanks at a satisfactory rate, but have been hampered by slow reaction kinetics and several
competing reactions that hinder complete extraction of the copper. However, reducing the particle
size to <2 ц т and using certain halide ions as leaching catalysts may make this concept feasible.
In addition, a temperature of 50 °C or higher may be necessary for reasonable reaction rates.

Assuming that tank leaching might be feasible, the next step is to investigate the chemical
reactions involving chalcopyrite to develop a conceptual process. An ideal process would leach
the iron first, followed by dissolution of the copper species, and then electrowinning copper from
the purified final leach solution. Textbook descriptions (Hayes 1993; Baczek 1981, Wikipedia
2010) indicate that aqueous oxidation in the presence of sulfuric acid can solubilize iron and
copper. The conceptual process requires the iron and copper components to be leached out in
sequence, in at least two stages. The first leaching stage aims to solubilize all the iron from the
chalcopyrite and form chalcocite, Cu2S. The leach solution is separated from the chalcocite, which
is leached in a second step to solubilize the copper. Final copper recovery is by electrowinning.

A real process would be unlikely to operate so perfectly. Some iron compounds would
remain unleached in the first stage, and some copper unleached in the second. Electrowinning isn't
able to extract all of the copper, so a solvent extraction (or other method) would be necessary to
minimize copper losses and prevent the build-up of iron and other impurities. In addition, copper
concentrate contains gangue and other sulfide minerals besides chalcopyrite. However, in an
early-stage feasibility study, we can neglect these factors, and concentrate on the main ones. We
also recognize that the feasibility results are based on assumption of ideal ionic activity.

Process feasibility is evaluated by equation-writing for the main reactions that take place. To
begin with, we use non-ionic species to determine the stoichiometry of the reactions involved and
calculate the amount of species involved in each step. This is the same way we started with our
evaluation of the lead-acid battery leach process. The process concept is to carry out leaching of
the copper in a way that separates it from the iron, and then recover copper by solvent extraction
and electrowinning. A solvent extraction stage, if desired, can increase the copper concentration
for electrowinning, and remove some of the impurities that could otherwise build up and hamper
efficient electrowinning. Figure 9.49 shows the flowsheet for the conceptual process.

H2S04-rich sorn

CuFeS^ i ; slurrw ^^ ± % i
....:
*co2

" -£ Slurry to disposal

Figure 9.49 Flowsheet for a conceptual chalcopyrite leaching process to recover of copper by a
leach/electrowin process. Solid lines represent the flow of solid-phase streams, while dashed and
dotted lines represent liquid and gas-phase streams. Wash water removes all of the leach solution
from the chalcocite during filtration, and a slight excess of the stoichiometrically-required EW
sulfuric acid is recycled to the EW tank. Solubilized iron and excess sulfuric acid are neutralized
by limestone. The leach tank is shown as single device, but in practice would consist of several
leach tanks in series to better control the residence time.

568 Chapter 9 System Balances on Reactive Processes

Equations [9.46] and [9.47] display the overall reactions for the CuFeS2 and Cu2S leaching
stages without accounting for the presence of the various solute or ionic species involved.
Equation [9.48] is the overall reaction for electrowinning, which produces % mole of oxygen at the
anode. Equations [9.49] and [9.50] show the reactions for precipitating the ferrous sulfate and
neutralizing the sulfuric acid. FREED data was used for AH° and AG° at 50 °C in units of kJ. The
negative value for AG° for all but the electrowinning reaction show that they have a tendency to
proceed in the direction written. As expected, the AG° for electrowinning is positive, which is
overcome by imposing an emf to force it in the specified direction. These preliminary results show
that the general process concept is feasible, and justify further calculations.

CuFeS2(c) + 23A02(g) + !/2H20(/) — !/2Cu2S(c) + FeS04(c) + !/2H2S04(/); АЯ° - -1060. AG° = -910.2 [9.46]
V2Cu2S(c) + 1 %02(g) + im2S04(l) - VTH20(1) + CuS04(c); AH° = -469. AG° = -388.9 [9.47]

CuS04(c) + !/2H20(/) - Cu(c) + %02(g) + H2S04(/); AH° = +243. AG° = +206.4 [9.48]
[9.49]
2CaC03(c) + FeS04(c) + lA02{g) + 2H20(/) - 1/2Ca2Fe205(c) + CaS04(H20)2(c) + 2C02(g);
AH° = +36. AG° = -33

СаСОз(с) + H20(/) + H2S04(/) - CaS04(H20)2(c) + C02(g); AH° = -111. AG° = -137.9 [9.50]

FlowBal was used to make a material balance for the process for a basis of leaching one mole
of chalcopyrite. Insofar as possible, the amount of water in each aqueous phase was set as 55.5
moles. This allows the amount of solutes to be both the molality and the amount. Seven process
specifications were imposed:

* The liquid portion of the chalcocite slurry leaving the CuFeS2 leach tank contains one-tenth
the liquid portion leaving as supernatant leach solution.

* Five moles of wash water are used to wash the Cu2S filter cake.

* The Cu2S filter cake entering the Cu2S leach tank contains five moles of water.

* The solution from the filter entering the neutralize tank contains 55.5 moles of water.

* The H2S04-rich solution stream entering the Cu2S leach tank contains 0.6 moles of
H2S04(/), which is 0.1 mole more than stoichiometrically necessary. The stream from the
SX/EW section is free from copper and iron solutes.

* The Cu2S leach solution entering the SX/EW section contains 55.5 moles of water.

* The solid stream leaving the neutralization tank contains one mole of unreacted CaC03(c).

FlowBal wrote 55 equations, and found a solution, as summarized in Table 9.15. Many of the
equations were trivial, such as the amount of Cu2S, copper, and FeS04 produced. The split fraction
of H2S04-rich solution recycled to the Cu2S leach tank is 0.545.

Table 9.15 Summary of material balance flows in and out of each reactive device.

CuFeS2 Leach Cu2S Leach Electrowin Neutralize

Species In Out In Out In Out In Out
0 1
CuFeS2 1 0 Cu2S 0.5 1 CuS04 1 0 CaC03 4 0.5
0 0.1 Cu 0 1 0 2
Cu2S 0 0.5 CuS04 0.6 0 Ca2Fe205 0 0
1.25 55.5 CaS04(H20)2 1 0
H2S04 0.5 1 H2S04 55 H2S04 0.1 1.1 1 0
0.5 FeS04 0.5 3
FeS04 0 1 o2 o2 0 0
0 H2S04 55.5 52.5
o2 2.75 H20 H20 55.5 54.5
o2
H20 56 55.5
C02

H20

Chapter 9 System Balances on Reactive Processes 569

In order to make a more accurate heat balance, we must know the amount of each ionic
species formed in the reaction devices. When chalcopyrite is oxidized in a neutral aqueous
solution, the iron tends to oxidize first, forming Cu2S (chalcocite) and FeS2 (pyrite) as solid
reaction products:

CuFeS2(c) + 2(1 - n)02(g) - nCu2S(c) + ^FeS2(c) + (1 - n)Fe2+(aq) + (1 - n)S042~(aq) [9.51]

In the early stages of oxidation, when the ionic molality is <0.2, Equation [9.51] proceeds
spontaneously and completely to produce near-equimolar amounts of all four reaction products.
Other ionic species such as HS04~(aq) and tt+(aq) also form. Some chalcopyrite may persist until
the last amount of pyrite is leached. Eventually, all of the iron is oxidized to Fe2+(ag) and the
copper remains as chalcocite. When the first stage leach solution contains some sulfuric acid,
chalcopyrite disappears earlier, and only pyrite is present in the last stages of iron oxidation.

While the amounts of species formed during the stepwise oxidation of chalcopyrite is
interesting, we're mainly concerned with the amounts present at 50 °C when all of the chalcopyrite
has been oxidized but none of the chalcocite. Similarly, we are interested in the species present in
the other reactive devices when the reactions have gone to completion as written. We can calculate
the amounts of ionic species the same way we did in earlier sections. We set the amount of water
at one kg (55.5 mol), and disregard the small amounts of water that are consumed or produced.
The species molality is then equivalent to the amount. The ionic amount calculation requires
thermodynamic data for the ionic species, which is shown in Table 9.16.

Table 9.16 Thermodynamic data from HSC for ionic species (kJ/mol at 50 °C) involved in the
copper extraction process.

Species Fe2+(aq) Cu2+(aq) S042 (aq) HS(V(a<?) OW(aq)
-93 65 -918 -889 -234
Δ#° -91.4 65.2 -730.2 -744.2 -151.0
AG°

For the chalcopyrite leach stage, two stoichiometry-based equations plus the Кщ expression

from Equation [9.54] allow a calculation of all species amounts. Equation [9.52] is the H balance,
in which one mole of H2S04(ag) dissociates to U+(aq) and HS04~(aq). Equation [9.53] is the
sulfur balance for the system, in which the products from the CuFeS2 leach tank contain two moles
of sulfur. Equation [9.54] equilibrates the two sulfur-containing ions and the proton; we have seen

this reaction before in Section 6.10.1 during a study of the ionization of sulfuric acid in water.

[H+] + [HSCV] = 2 [9.52]

[S042~] + [HSCV] = 2 [9.53]

SOfyaq) + U\aq) - HS04(aq); Keq = 186= [ H S ° 4 ] [9.54]
[S042-][H+]

Solving, [H+] = [S042~] = 0.101, and [HS04~] = 1.899. The pH of the solution leaving the
chalcopyrite leach stage = 1.0. Equation [9.55] shows the overall reaction stoichiometry at 50 °C

(with rounded off balancing numbers) for the oxidation of one mole of chalcopyrite with an
instream of 56 moles of water and the ionized species from lA mole of H2S04(/) (calculated by the
method outlined in Section 6.10.2).

CuFeS2(c) + 23/402(g) + ΥΆ20{ΐ) + 0.495HSO4"(^) + 0.005SO42"H) + 0.505H+(a<?) -

Fe2+(aq) + ^Cu.S^) + 1.9HS04"(*?) + 0ASO42(aq) + 0.1H+(^); [9.55]

If we neglect heat losses on the chalcopyrite leach tank, and set all instream temperatures at
50 °C, the overall heat effect is based on the AH° for Equation [9.55]. Data for the condensed
phases is taken from FREED, while that for the ionic species is taken from Table 9.16.

ZAtf°form products - -93 + 0.5(-75.6) + 1.9(-889) + 0.Ц-918) = -1910 kJ

570 Chapter 9 System Balances on Reactive Processes

EA//°form reactants = -173.2 + 0.5(-285) + 0.495(-889) + 0.005(-918) = -760 kJ

H for the CuFeS2 leach - -1910 + 760 = -1150 kJ.

The CuFeS2 leach tank heat effect is about 90 kJ more negative than that calculated by
Equation [9.46], which did not take the heat of formation of ionic species into account. For the
chalcopyrite leach, the small difference between the ionic and non-ionic balance results may not be
important. In any case, heat balance closure for the CuFeS2 leach tank requires some combination
of heat removal from the CuFeS2 leach tank and instreams temperature below 50 °C. For example,
31.2 moles of makeup water enter the CuFeS2 leach tank, and this water has an overall Cp of 2.3
kJ/deg. If this water entered at 25 °C, it would require only about 60 kJ, which is not significant in
comparison to the -1150 kJ of "surplus" heat. Even if all of the instreams entered at 25 °C, the ΣΗ
CuFeS2 leach tank would still be >-1000 kJ. Therefore the CuFeS2 leach tank requires a heat
exchanger to maintain the 50 °C temperature.

We now turn our attention to the Cu2S leach tank, where 0.6 mol of H2S04(tfg) was added by
the recycled EW stream, and an additional 0.5 mol was produced. Following the ion-balancing
procedure used for the CuFeS2 leach tank, the entering stream contained [H+] = 0.605, [HS04~] =
0.595, and [S042_] = 0.005. The outstream contained [H+] = 0.0012, [S042] = 0.901, [HS04~] =
0.199, and the pH = 2.9. Equation [9.56] shows the overall Cu2S leach tank reaction.

Cu2S(c) + \lA02(g) + 0.595HSO4~H) + 0.005SO42~O?) + 0.605Η+(α#) — [9.56]
C\x2\aq) + ιΑΆ20(ΐ) + 0.199HS(VH) + 0.901SO42~(^) + 00\2R+(aq)

Again, with all instreams at 50 °C and negligible heat loss, the overall heat effect for the Cu2S
leach tank is the AH° for Equation [9.56]:

ZA//°form products = +65 + 0.5(-285) + 0.199(-889) + 0.90Ц-918) = -1086 kJ

EA#°form reactants = 0.5(-75.6) + 0.595(-889) + 0.005(-918) = -570 kJ

Я for the Cu2S leach = -1086 + 570 = -516 kJ.

The Cu2S leach tank has a net surplus of heat; about half that of the CuFeS2 leach tank. It too
can't be brought into balance by lowering the temperature of the instreams to 25 °C. Again, a heat
exchanger would be required to maintain the specified 50 °C device temperature.

Next we turn to the neutralization tank, where the aqueous products of Equation [9.55] are

neutralized with three moles of CaC03(c) (plus one extra mole to assure complete reaction).
Equation [9.57] shows the overall neutralization reaction:

ЗСаСОз(с) + FQ2\aq) + l.9USOi(aq) + 0.1SO42"H) + 0.1Н+И) + %02(g) + 3H20(/) -

1/2Ca2Fe205(c) + CuS04(H20)2 + 3C02(g) [9.57]

IAtf°form products = 0.5(-2142) + 2(-2023) + 3(-393.5) = -6298 kJ

EA#°form reactants - 3(-1207) - 93 + 1.9(-889) + 0.Ц-918) + 3(-285) - -6350 kJ
H for the neutralization tank = -6298 + 6350 = +52 kJ

The neutralization tank has a slight deficit of heat, which if unconnected, will cause the
outstreams to be below 50 °C. Unlike the leach tanks, the neutralization process temperature may
not be critical, so it may operate satisfactorily below 50 °C.

The last device is the electrowinning device, which receives a solution containing the products

of Equation [9.56] and discharges a copper-free solution containing 1.1 mol of ionized H2S04(ag).
The discharge composition is [HS04~] = 1.094, [S042~] = 0.005, and [H+] = 1.006. We defer
consideration of the electrowinning device until the next section.

Chapter 9 System Balances on Reactive Processes 571

9.8 Electrolytic Processes

Electrical energy has a long history of use in the metals and ceramics industry. The simplest
application technology is called electrothermics, in which heat is provided for melting or chemical
reactions by passing a current through a resistor or striking an arc between electrodes and a charge.
The benefit of an electrothermic processes is that the substance being heated is uncontaminated by
the combustion products normally used to reach high temperatures. We've already looked at the
reactions in an electric furnace steelmaking process. Another interesting early application of
electrothermics is the Acheson process for making silicon carbide, which was in widespread use by
1895. In that process, a mixture of silica sand and carbon are heated to above 1600 °C by a carbon
resistor. The use of arc heating involves the production of a thermal plasma, which is an effective
way of transferring heat to a charge.

This section deals with the electrolytic use of electricity; specifically, the energy required to
produce or refine a metal by passing a current through an aqueous electrolyte bath. The case of
recovering a metal from solution is called electrowinning. Converting an impure to a pure metal is
called electrorefining. In both cases a current is passed through an electrolyte solution containing
immersed electrodes. Pure water has a small dissociation constant, on the order of 10"14 at 25 °C,
so it contains a very low concentration of U+(aq) and 0¥T(aq) ions. Thus pure water is a very poor
conductor of electricity, and it is practically impossible to electrowin hydrogen from it. However,
we can produce an electrolyte that is a good conductor of electricity by maintaining a metal ion
concentration of 1 molai or higher, and adding other electrolytes to an even higher concentration.
This is the type of solution used in most electrolytic processes.

9.8.1 Energy Requirement for Electrorefining

Many applications require metal of a higher purity than can be made by pyrometallurgical
methods. Notable are copper, lead, nickel, tin, and precious metals. These and other metals can be
electrorefined in order to meet purity specifications. The impure metal is shaped into an anode on
the order of 2 - 4 cm thick. The cathode is usually a thin sheet of refined metal, and receives a
deposit from both sides. In copper electrorefining, the cathode may be composed of a thin sheet of
titanium, from which the deposited copper can be easily stripped. The anode reaction consists of
metal oxidation by the removal of electrons. The cation migrates to the cathode under the
influence of an externally-applied voltage, where it is reduced by the addition of electrons. Anode
impurities more noble than the metal being refined do not dissolve, and sink to the bottom of the
cell as a slime. Less noble anode impurities may also become cations, but do not deposit on the
cathode. A small bleed stream of cell electrolyte is treated in a separate circuit to remove these
less noble metals. The electrolyte for copper electrorefining contains about 0.8 molai Cu. To
improve electrical conductivity, sulfuric acid is added to bring it to 1.8 molai H2S04.

The standard reduction potential for copper at 25 °C is +0.34 V, and is not much affected by
temperature. The reversible single electrode potential at the cathode and anode are equal but of
opposite sign, so the theoretical cell voltage E° for electrorefining is zero. However, during
operation, an electrorefining cell will have small concentration gradients at each electrode, and the
electrolyte, electrodes, and connection points all have a small resistance to the flow of current.
Therefore, a typical electrorefining cell might have a voltage drop across the electrodes between
0.20 and 0.25 V. The cation concentration gradient near each electrode is decreased by proper
circulation of the electrolyte, and the electrolyte resistance is decrease by adding an acid.
Electrorefining is thus characterized by the passage of electricity at as high a current density as
possible across a low potential difference.

The amount of metal deposited during electrorefining is calculated by Faraday's law:

current x time/F = number of gram equivalents deposited [9.58]

where current is in amperes, time in seconds and F (the faraday constant) = 96 484. (Please review
Example 1.18.) The passage of one ampere-hr deposits 0.0373 gram-equivalents of metal. For the
deposition of copper from Cu2+(aq), one gram equivalent = 31.77 g, so we expect 1.185 g Cu to

572 Chapter 9 System Balances on Reactive Processes

deposit. In practice, slightly less copper deposits than expected from Faraday's law. This is
caused by a variety of occurrences, such as current leaking to ground, and the dissolution of a bit
of copper in the acid electrolyte. This discrepancy is designated by the term current efficiency,
which for copper electrorefining is about 96 %, leading to a Cu deposit of about 1.14 g per ampere-
hour. On a larger scale 8.44 x 105 ampere-hours are required to deposit one tonne of Cu. The
electrical energy is commonly expressed in kWh. For copper:

Kx8.44xl05 t9·59]
kWh/tonne of Cu = 1/4/W4 ^,.™
LJ
l000xC£/l00

which shows that the energy requirement is directly proportional to the voltage and inversely
proportional to the per cent current efficiency. For a cell voltage drop of 0.23 V, the energy
requirement is 194 kWh/tonne of Cu at 100 % current efficiency.

The passage of current across the resistance voltage drop produces heat, which must be
dissipated to keep the cell at a constant temperature. Consider an electrowinning cell operating at
65 °C with an electrolyte having [Cu2+] = l molai. Equations [9.60] and [9.61] show the enthalpy
change occurring at the anode and cathode:

Anode: Cu -+ C\x2\aq) + 2e~; A//° = 65.0kJ [9.60]

Cathode: Cu2+(^) + 2e~ -► Cu; AH° = -65.0 kJ [9.61]

Thus the anode reaction is endothermic while that of the cathode is exothermic to the same
extent. While there is no net enthalpy change occurring as a result of electrode reactions, the
anode may operate slightly cooler than the cathode. The resistive heating amounts to 700 kJ/kg of
copper. A single electrolytic cell, containing 45 anodes and a similar number of cathodes produces
500 kg/day of copper. This requires the dissipation of 350 MJ/day of heat, which is accomplished
mainly by heat loss from the tank to the surroundings, and the continual flow of fresh electrolyte
that enters at a temperature below the tank operating temperature of 65 °C. The electrolyte flow is
about 30 tonnes/day per tank, or about 60 kg per kg of Cu produced. There's no way to determine
the magnitude of each factor that influences the electrolyte temperature without more information
about the tankhouse operation.

At a typical 95 % CE, Equation [9.59] shows that copper electrorefining electrical energy is
about 205 kWh/tonne of copper. However, energy is also consumed in the electrolytic cleaning of
the bleed electrolyte, the production of some scrap cathode, the resistive losses in the various
terminals, transformers and plant busbars. Energy is also consumed in converting AC to DC, and
pumping the electrolyte from tank to tank. These and other minor losses bring the AC power
consumption to nearly 300 kWh/tonne of copper refined. At a power cost of 50/kWh,
electrorefining costs about l /40/kg.

9.8.2 Energy Requirement for Electrowinning

We can gain a quick appreciation for the possibility of electrowinning a metal from solution
by looking at tables of the standard reduction potentials for cations. If the half-cell potential is a
positive number, the cation will deposit at the cathode in preference to the evolution of hydrogen.
In the case of copper depositing by the reduction of Cu2+(aq), the cathode reduction potential is
+0.34 V, hence copper deposits in preference to any other metal having a smaller half cell
potential, including the "deposition" of hydrogen at its defined zero cell potential. In theory, no
metal having a half cell potential smaller than zero can be electrowon; instead, hydrogen would be
expected at the cathode. However, in looking at extractive metallurgy books, we notice that metals
such as nickel, iron, and even zinc can be electrowon, even though the half cell potentials of
Ni2+(ag), Fe2+(#g), and Zn2+(aq) are smaller than zero. Fortuitously, a considerable hydrogen
overvoltage is required to electrowin hydrogen from an acid solution, hence allowing more
electronegative metals to be electrowon.

The electrolyte entering a copper electrowinning cell from a solvent extraction circuit
typically contains the same concentration of Cu and H2S04 as that of an electrorefining cell.

Chapter 9 System Balances on Reactive Processes 573

About half of the incoming copper is deposited, with the spent electrolyte returned to the SX
circuit to be recharged in Cu. As in the case for electrorefining, a small amount of electrolyte is
bled to a treatment circuit to prevent build-up of impurities. In plants that do not use SX, the
H2S04 molality is much less, and the electrolyte has higher resistivity.

The situation at the anode in electrowinning is much different than in electrorefining because
of the inert anode material, where, oxygen is liberated. Equation [9.62] at 65 °C gives the standard
oxidation potential for the anode half-cell, and the enthalpy change::

H20(/) - m\aq) + V202(g) + 2e~; E° = -1.196 V; Δ#° = 284.6 kJ [9.62]

In terms of chemical species, the overall electrowinning reaction consumes one mole of water,
produces lA mole of oxygen, and regenerates one mole of sulfuric acid:

H20(/) + Cu2+(aq) + S 0 4 2 " H ) — Cu(c) + 1/202(g) + m\aq) + S042"H) [9.63]

The sum of the anode and cathode enthalpy change is +219.6 kJ, so the electrode reactions are
endothermic. The reversible cell emf is the sum of the half-cell potentials of [9.61] and [9.62], or
-0.86 V. As in the case of hydrogen evolution at the cathode, an oxygen overvoltage is required at
the anode to form oxygen gas dimer; this overvoltage is about 0.6 V. The voltage owing to
electrolyte resistivity is about 0.2 V for SX electrolytes, with another 0.1 V for miscellaneous other
tank accessories. The total cell voltage for electrowinning copper is thus in the range 1.8 - 2.4 V.

The net heat Q for the electrowinning cell is the sum of the heat effect for the (reversible)
electrochemical process itself, plus the heat effect for the passage of current across the various
non-reversible cell reactions and resistances. Q for the electrochemically reversible portion of the
production of copper (kJ/mol) is:

ßrev - Δ#° - AG° = 284.6 + nFE = 284.6 + 2(96.48)(-0.86) - 118.7 kJ/mol [9.64]

which is an endothermic process. If the overall cell voltage is 2.0 v, then Q for the irreversible
portion of the production of copper is:

Qm = (0.86 - 2.0) nFE = -220 kJ

The net heat effect of the electrowinning of Cu at 65 °C is thus -101 kJ/mol, an exothermic
effect. The entering (pregnant) electrolyte temperature must then be less than this to maintain an
average electrowinning tank temperature of 65 °C.

9.9 Summary

Large enthalpy changes are often associated with chemical reactions, and in many cases,
chemical reactions provide both the fuel and reactants for a process. This chapter covered the
principles and techniques for making heat balances in closed and open systems, and introduced
techniques for solving system balances.

When the overall heat effect for a conceptual process is exothermic, some way must be found
to remove the excess heat to bring the system into thermal balance. Inert substances may be
introduced into the system, or an internal heat exchanger can be used to remove heat. If the overall
heat effect is endothermic, some way must be found to introduce heat to the system. A fuel may be
burned to preheat the reactants, electrical heat added, or something else added that reacts
exothermically. If all of the АЯ°ГХ for an exothermic process is transferred to the products, the
resulting temperature is called the adiabatic reaction temperature (ART).

Combustion is a special reaction whereby a fuel is oxidized to produce heat. Under adiabatic
(or nearly-adiabatic) conditions, the AH°comb is transferred to sensible heat (Ят-Я298) in the
combustion products. For many fuels, the adiabatic flame temperature (AFT) from combustion
with stoichiometric air is over 2000 °C. The AH°comb for gaseous fuels can be calculated from
A//°form data from thermodynamic tables. For liquid and solid fuels, the АН°сотЪ is obtained by
direct measurement on the fuel, and is listed in typical engineering handbooks. Empirical

574 Chapter 9 System Balances on Reactive Processes

equations have been developed to estimate AH°comh for fuels such as coal and biofuels where data
is lacking. Care must be taken to distinguish the standard state of H20 in using tabulated values of
_#°comb(i.e.,LHVorHHV).

In many cases, the quality of heat required for a process is paramount. The quality of heat
available from a combustion process depends on the sensible heat contained in the combustion
products as compared to the process demands. The higher the AFT, the greater the quality of heat.
The quality of heat is improved by preheating the combustion air, or enriching the air with oxygen.

The utilization of heat can be improved by using insulation, decreasing the amount of excess
air used for combustion, and increasing the throughput of a furnace. The greatest improvement in
thermal efficiency comes from capturing and recycling heat by use of a heat exchanger. Heat
balances on processes having heat exchangers requires that a separate balance be made on each
unit to assure that the anticipated temperatures are reasonable.

For processes involving multiple units, the heat and material balance equations must be solved
simultaneously. A graphical approach is a good way to express the solution to a problem where a
range of conditions is of interest. An analytical approach is better if accuracy is paramount, in
which case, Goal Seek and Solver are often required. Excel's charting and solving tools are very
effective in either case. Solver is very useful in process optimization because a variable can be
constrained to a maximum or minimum.

FlowBal is a useful alternative to writing and solving system balance equation sets. FlowBal
requires sufficient stream data to bring the material balance DOF to 0. The heat balance adds
additional constraints to a coupled system balance, which causes the system DOF to be <0. This is
handled by adding placeholder stream properties, which can be varied by FlowBaPs Repetitive
Solve target feature to bring the net device and net process heat effects to zero, or any other
specified value. Unfortunately, FlowBal does not handle ionic reactions.

References and Further Reading

Backzec, Frank A. et. al., Recovering Copper from Chalcopyrite Concentrate, U. S. Patent
4256553, March 17, 1981.

Baukal, Charles E., Editor, Oxygen-Enhanced Combustion, CRC Press, 1998.

Baukal, Charles E., Editor, The John Zink Combustion Handbook, CRC Press, 2001.

Chemical Equilibrium Calculator, [Online]. http://navier.engr.colostate.edu/tools/equil.html.
2009.

Gaur, S., and Reed, T, Thermal Datafor Natural and Synthetic Fuels, Marcel Dekker, 1998.

Habashi, Fathi, A Textbook of Pyrometallurgy, Metallurgie Extractive Quebec, Quebec, Canada,
2nd edition, 2002.

Habashi, Fathi, Textbook of Pyrometallurgy, Metallurgie Extractive Quebec, Quebec, Canada,
2002.

Hayes, Peter, Process Principles in Minerals and Materials Production, Hayes Publishing Co,
SherwOood, Queensland, Australia, 1993.

OLI Electrolyte Simulation, www.olisystems.com, 2010.

Wikipedia contributors, "Copper extraction techniques", "Adiabatic flame temperature",
Wikipedia, the Free Encyclopedia, December 2010. http://en.wikipedia.org/wiki/Main_Page.