Handbook_on_Material_and_Energy_Balance_Calculat1

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 175

Cell H4: = C4/(0.99*0.024)

Cell 14: =E4 + F4

With these formulae entered, any change in the values in cells C4:E4 results in immediate
recalculation of the dependent variables. (The recalculation assumes that the stream compositions
remain the same).

However, sometimes we wish to make a calculation in the reverse direction. Suppose that F®
+ F4^ is to be 1225 kg/h by changing FA. Since cell 14 contains a formula, 1225 cannot be entered
there. Instead, by trial and error, one can change cell D4 until 1225 appears in cell 14. This occurs
at FA = 186.7. However, a better way is to use Excel's Goal Seek tool. Goal Seek makes the
changes rapidly and displays the result. The Goal Seek dialog box for this situation is shown
below.

Goal Seek |$I$4 " И& Pressing OK causes Goal Seek to iterate until the
11225 value sought is reached. At that point a new screen
Set cell: Z-3i appears, indicating that a solution was found. The value
To value: |$D$4 _ГЗ of 186.7 kg/h appears in cell D4, which is the same as
By changing cell: found by manual iteration. The accuracy of the Goal Seek
tool can be changed by clicking on Tools >Options
Calculation >Maximum change.

OK Cancel

In the current application, Goal Seek had only a small

Goal Seek Statu< Wim advantage over manual iteration because
the sought value was close to the original
one. But in other cases, Goal Seek is much
Goal Seeking with Cell 14 OK more advantageous than manual iteration,
found a solution,

Cancel especially when several variable changes

Target value: 1225 are to be made. Goal Seek will be used
Current value: 1225.0 5tep again in later examples. For further

Pause information about Goal Seek, please
consult Excel help.

Another useful Excel tool is Solver, which will also be used extensively throughout the
Handbook. A User's Guide for Solver is on the Handbook CD in folder SuperSolver. You should
be sure Solver is installed on your PC. We illustrate the use of Solver on a relatively simple
process for mixing five different scrap alloy materials to make up a feed for a stainless steel
melting furnace. The objective is to calculate the required mass of each scrap alloy to produce ten
tons of DL stainless steel. The likely small losses during melting are ignored here. Figure 4.21
shows a flowsheet for the process.

Alloy S3 Alloy S4 Alloy GA Alloy HC Alloy FC
wCr= 0.185 wCr = 0.125 wCr= 0.12 wCr= 0.16 wCr= 0.70
wNi = 0.095 wNi = 0.02 wNi = 0.02 wNi = 0.59 wFe = 0.30
wFe = 0.720 wFe = 0.855 wFe = 0.83 wFe = 0.05
w\N = 0.02 ivW = 0.04 1
iI wMo = 0.16
i.

Melting furnace

Steel DL (FDL = 20,000 lb)
wCr=0.19 wNi = 0.09
wFe = 0.685 wW = 0.02
wMo = 0.015

Figure 4.21 Flowsheet for stainless steel melting process.

176 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

The next step is to see if the process is correctly specified. There are six process streams, and
five components (Cr, Ni, Fe, W, and Mo). However, not all components are present in each
stream, so SV = 22. There are five independent component balances (one for each element), so IB
= 5. The number of independent element compositions is one less than the analyzed components
in each stream, so 1С = 16. Finally, F = 1 because there is one stream flow {F°L = 20 000). This
gives DOF = 22 - 5 - 16 - 1 = 0, so the system is correctly specified.

Next, we set up the five elemental balances, starting with Cr. When using Solver, it's helpful
to write each equation to sum to zero so that all Solver constraints can be set to zero. The constant
in each equation is the mass of the element in steel DL. The five element balance equations are:

Cr: 0.185(SE) + 0.125(S4) + 0.12(GA) + 0.16(HC) + 0.70(FC) - 3800 - 0

Ni: 0.095(SE) + 0.02(S4) + 0.02(GA) + 0.59(HC) - 1800 - 0

Fe: 0.720(SE) + 0.855(S4) + 0.83(GA) + 0.05(HC) + 0.30(FC) - 13 700 = 0

W: 0.02(GA) + 0.04(HC)-400 = 0

Mo: 0.160(HC)- 300 = 0

These five equations were entered into an Excel worksheet, along with a guess of 4000 lb for
each alloy mass. The appearance of the worksheet setup is shown below. The value of the
guessed alloy mass in the present example is not particularly important so long as it is within
reason.

8 Alloy SE S4 GA HC FC

9 Mass, lb 4000 4000 4000 4000 4000

10

11 Crbal 1360

12 Ni bai 1100

13 Febal -2680

14 Wbal -120

15 Mo bai 340

Solver requires that we assign a target value to one cell, and constraint values to other cells.
By setting up each equation (cells C11:C15) to have a zero sum, the target and all constraints are
zero. The setup screen for Solver is shown below.

Solver Parameters

Set Target Cell: ]Solve

Equal To: С Max Г Min ^Value of: F" JClose

[~By Changing Cells: 31 Guess Options
"3 Add
|$C$9:$G$9 Reset All
Change Help
Subject to the Constraints: -
d Delete
$C$12:$C$15 = 0

Solver changes the amount of each scrap alloy mass (C9:G9) until the target and constraints
(i.e., all equations) are equal to zero. Solver found a solution, as shown below (the number format
was reset to an integer value).

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 177

13 Febal 1.461E-09

É Wbal -2.5E-12
Mo bai -1.25E-12

As a check, the sum of alloy scrap amounts = 20 000 lb.

This type of problem does not always have a solution. In fact, there will generally not be a
solution because it's unlikely that the mass of a set of random scrap compositions can be adjusted
to meet a final composition. If there is no exact solution, Solver's result may indicate the closest
possible approach to a solution.

EXAMPLE 4.4 — Leaching of Salt Cakefrom Aluminum Recycling.

Make a mass balance on the salt cake leaching process discussed in Section 4.6.3.

Data. Seven different stream variables were measured, as shown previously in Table 4.2. In
addition, two SR relationships were obtained from historical observations.

Solution. We used this process as an example for constructing a DOF, and in Section 4.7.2,
showed that the process was underspecified (i.e., DOF = +1). The process needs an additional
stream value, or another SR. Suppose we select, for example, the mass fraction of water vapor in
the wet air (w4H20) as an important process variable. Although this variable was not measured in
the original case, we can make a mass balance over a range of arbitrarily selected values of w4H20,
which will show the effect of w4H20 on the other stream variables.

The process description states that the salt cake is leached in hot water. If the wet air
outstream leaves saturated with water vapor, w4H20 might vary between 0.1 and 0.35, depending
on the leach tank temperature. Equation [1.12] can be used to convert wH20 to xH20, which is
equivalent to/?H20 if Ptotal = 1 atm. Table 4.4 shows the />4H20 for a range of w4H20. The tank
temperature will be higher than the wet air dew point temperature because the presence of
dissolved salt lowers the vapor pressure of water.

Table 4.4 pR20 in S4 calculated for a range of w4H20 at P = 1 atm.

w4H20 0.100 0.150 0.200 0.250 0.300 0.350
0.4630
р4Я20, atm 0.1511 0.2203 0.2859 0.3480 0.4070

The system (as initially underspecified) has four unknown flows, and six unknown stream
variables, for a total of ten unknowns. We can correctly specify the system by picking an
arbitrarily selected w4H20, which by difference from one gives w4Air. This leaves us with eight
unknowns, so we must write eight equations. Since we want a solution for the eight-equation set
over a range of w4H20 values, it's better to use Solver. Doing so minimizes the onerous
calculations that are required to solve for six different w4H20. We can write four component
balance equations, two equations to force the sum of mass fractions to equal one (S4 and S5), and
two SR equations. The first SR states that the mass flow of slurry slag (S7) is 1.3 times that of the
slurry liquid (S6). The second SR states that w5NaCl is usually 95 % of w6NaCl.

The best way to use Solver is to write cell formulae so that the left-hand side of each equation
equal to zero. Set the first equation target cell to 0, and set the other seven equation cell
constraints to 0. The equations are writen below in Excel formulae format. The underlined w4H?Q
designates it will be varied in six steps, as shown in Table 4.4.

178 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

NaCl bai: 0 = F ' V N a C l + F2*w2NaCl - (i^*w5NaCl + i^*w6NaCl
H20 bai: 0 = Fl*wxU20 + ^ * ^ 2 Н 2 0 + F3*w3H20 - (F4*w4H10 + ^*νν5Η20 + ^*w6H20)
Air bai: 0 = F ' V A i r - FSv4Air
Slag bai: 0 = F V s l a g - F 7
Iw5: 0 = 1 - (w5NaCl + w5H20)
Σνν6: 0 = 1 - (w6NaCl + 6H20)
SR#l,S6vs. S7: 0 = F7-1.3*F*
SR #2, NaCl in S5 vs. NaCl in S6: 0 = w5NaCl - 0.95*w6NaCl
w4H20 was initally set = 0.1, and Solver varied the initial estimates of the other eight
variables until all equations were satisfied. Solver found a solution for all six ofthew4H20 listed
in Table 4.4. Figure 4.22 shows selected results.

Effect of w H20 on Process Conditions

1300

с 1100 — o — F4 *Δ· 0.4 Qя
—O—F5 .is
ε . .д. . p4H20 0.35 Z
—O— S-5№CI

i? 900 -' dQ 0.3
0.25

. 700

500

300 0.1 0.15 0.2 0.25 0.3 0.35

mass fraction H20 in wet air (w4H20)

Figure 4.22 Mass balance results for salt cake leaching process. Arbitrary values of w4H20 were
chosen to give DOF = 0, and the other eight dependent variables calculated using Solver.
Increasing w4H20 removes more water from the system, which causes F5 to decrease, and w6NaCl
to increase.

Assignment. Calculate the value of F2 that will give w5NaCl = 0.27 when w4H20 = 0.35.

4.8.2 Software for Conversion of Stream Units: MMV-C

Making a material balance on a complex process may require several hours of arithmetic.
One of the more tedious tasks is converting stream units to a consistent basis. For example,
volume units may be used to designate the composition of a gas stream, and mass fraction for an
alloy stream. Other streams may list the mass or volume flow (but not composition). Finally,
sometimes the problem asks for a result in one set of units but calculations may be more easily
carried out in a different set of units. The program MMV-C (this stands for mass mole volume
converter) can make these conversions for you, thus saving a great deal of time and assuring
accuracy. MMV-C also converts a stream composition of species into element compositions. For
example, suppose we wanted to convert an accurate chemical analysis of air from volume fraction

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 179

to mass fraction of species and elements*. The leftmost two columns of Table 4.5 are the original
data, the center column is the mass fraction of species calculated from MMV-C, and the rightmost
two columns are the calculated mass fraction of the elements. The MMV-C program is discussed
in more detail in the Appendix.

Table 4.5 Composition of air in different units.

Species Mol % Mass % Element Mass %
N2 78.09 75.53 N 75.53
20.95 23.15 О 23.18
o2 0.93 1.28 Ar 1.28
0.03 0.05 С 0.01
Ar
C02

4.9 Balances on Systems with Multiple Devices

So far, our examples have been applied to systems with one device, or two devices that were
treated as one. However, most processes involve many devices, possibly hundreds. Figure 4.23
shows a flowsheet for a simple multi-device processes where material generally flows in one
direction. The system boundary could be drawn around all four devices, in which case it would be
treated as a single device. Alternatively, the system boundary could be drawn around each device,
or any combination of devices. If the DOF for the entire system is a positive number, an additional
material balance on one or more devices is required in order to specify the system completely.

In order to be able to design or optimize a multi-device system, the engineer must be able to
specify the flow and composition of many process streams. This requires a more detailed analysis
than heretofore, but the principles are the same. For the process shown in Figure 4.23, four 1-
device balances can be calculated, six 2-device balances, three 3-device balances, and of course
one overall balance. Not all of the 14 balances are independent because the system is completely
specified whenever four distinct types of device balances are made. A consideration of the number
of devices N and number of components С indicates that the number of independent balances is
equal to the number of devices in a process. If each device involves all of the components, then N
x С independent balance equations can be written.

Furnace gas Clean gas Clean gas
Water ►
Scrubber Mixer

Decant Slurry Clean gas Air
solution Incinerator Fuel
Washing/
settling
tank

Sludge

Figure 4.23 Multi-device flowsheet for cleaning a furnace offgas containing an organic dust. The
dust is removed in the scrubber, washed in the tank, and the sludge incinerated.

* The composition of air listed in Chapter 1 was given on a C02-free basis.

180 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

The DOF strategy developed in Section 4.7 for analysis of single-device systems needs to be
updated for multi-device systems. The rules are basically the same, but some adaptations are
necessary. Consider the two device separation process shown in Figure 4.24. A feed stream of
three metals is separated by distillation in two steps,

J^. ivHg = ? r--4--* ^ =?
'λ ivCd = ?
/ A . и^Нд = 9%
F1 = 1500lb/min Still wCd = 71%
ivHg = 22% ivZn = ?
wCd = 32% F3 = ?
Still II
4LJ i^Hg = 4%
wCd = 35% У. F5 = ?
ivZn = ? wCd = ?
wZn = ?

Figure 4.24 Separation of volatile metals by staged distillation. Solid lines represent liquid phase
streams, and dashed lines are gas streams. There is actually a small amount of Zn in S2 and a
small amount of Hg in S5, but these are neglected to simplify the process.

Three material balances can be made on the system. One around Still I, one around Still II,
and an overall balance around both devices. The number of independent material balances for a
system is equal to the number of devices in the system.

It's important to distinguish between the process DOF and the overall DOF. The process
DOF is made by considering all of the process devices and streams. In contrast, the overall DOF
treats all units as one unit, and considers only streams entering and leaving the system. One way to
characterize a system is to start with the individual devices, then combine them to make a process
DOF. Alternatively, if the overall DOF = 0, an overall balance is useful to obtain stream data that
are helpful in making the individual device balances. We will apply these approaches to the metal
separation process shown in Figure 4.24. We start with the individual device DOF calculation.

Still I has three streams, two of which have three components, and one of which has two.
Therefore, SV for Still 1 = 8. There are three components, so IB = 3. Four stream compositions
are specified, so 1С = 4. The flowrate of SI is known, so F = 1. Therefore, DOF for Still 1 = 0.
The balance around Still I can therefore be solved without reference to Still II information.

Still II has three streams, two of which have three components, and one has two, so SV = 8.
There are three components, so IB = 3. Four stream compositions are specified, so 1С = 4. No
flows are specified, so DOF = +1. However, the balance around Still I will give the flow of S3,
which will give DOF for Still II = 0. This indicates that the process has a DOF = 0.

The overall DOF has four streams, one system instream (SI) and three system outstreams (S2,
S4, and S5). Two of these have three components and two have two components, so SV = 1 0 .
There are three components, so IB = 3. 1С = 4 and F = 1, so DOF = +2. The overall system DOF
is underspecified, so we cannot use it alone to gain any information. Clearly, the information
contained in S3 is critical to making a process balance.

The process DOF requires a modified approach compared to a single device flowsheet.
Notice there is a connecting stream (S3) between the two devices, so we cannot simply add up all
of the SV values from the two individual device DOF calculations, otherwise we would double-
count the SV for S3. For the same reason we cannot simply add up the 1С values from the
individual device DOF calculations. However, the number of balances is the sum of the two
device balances, so IB = 6. The process SV =13, which we can determine by going from process
stream to process stream, or subtracting three from the sum of the individual DOF SV values.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 181

Similarly, 1С = 6 and F = 1, so the process DOF = 0. We expected this result from our analysis of
the individual device DOF calculation. Table 4.6 shows the DOF calculation results.

Table 4.6 DOF calculation results for two-stage distillation process.

DOF factors Still I Still II Process Overall
No. of stream variables (SV) 8 8 13 10
No. of individual balances (IB) 3 3 6 3
No. of specified stream compositions (1С) 4 4 6 4
No. of specified stream flows (F) 1 0 1 1
No. of subsidiary relations (SR) 0 0 0 0
0 +1 0 +2
DOF

One of the most important things in calculating the DOF for a multi-unit system is to avoid
double-counting the DOF factors for connecting streams. It's best to calculate the DOF of each
device before calculating the DOF for the process. It's tempting to evaluate a system by making a
process DOF only. However, the individual device DOF calculations clarify the number of DOF
factors involved, and can tell you if any of the devices have a DOF = 0. If so, start the balance
calculations with that device. Finally, each individual device must have a DOF of zero or greater.
If an individual device is overspecified, the process is incorrectly specified.

Now that you know that the process is correctly specified, try making the material balance,
starting with Still I. The S3 flow should be about 1130 lb/min.

EXAMPLE 4.5 — Refining Crude Boric Acid by a Two-Stage Aqueous Process.

Crude boric acid contains limestone as an impurity. The acid is dissolved in hot water and
pure boric acid is precipitated by cooling the decanted leach solution. Figure 4.25 shows the
flowsheet for one minute of operation. A material balance is sought to determine the loss of boric
acid from the process. Calculate the DOF on each device, the process, and the overall system prior
to making the material balance.

Data. Four streams were analyzed, and the water (S2), СаСОз (S5), and boric acid (S8) are
assumed to be pure. Historical data provided information to give four subsidiary relations.

Crude H3BO3 S 3 & S 4 : Aq soKn of H3BO3
F1=24.9kg
1УН3ВО3 = 72.2% ' *ν3Η*ΒΟ* = ? w4H»BO» = ?
i^CaC03 = ? ιν3ΗοΟ = ? w4HoO = ?

Water _ S6&S7:AqsornofH3B03
F2 = 100.0 kg F6 = ? F1 = ?

Stream relations СаС03 6 v Hr >3B03 : ? nи//77HНо3ВBО0о3 := 3.6%
F5 = ? и/7Н20 = ?
1ИН3ВО3 = и/3Н3В03 + 0.3
У * H3B03; F8 = ?
^ 6 Η 3 Β 0 3 = и^Н3ВОз

Γ δ ^ 1.2F4 /r8 = 1.5F7

Figure 4.25 Boric acid purification flowsheet showing measured stream properties and subsidiary
relationships. Dashed lines indicate liquid streams and solid lines are solid streams. The leach
tank slurry consists of S4 + S5, and the precipitation tank slurry consists of S7 + S8. Paired
streams (slurries) are indicated on flowsheet diagram by a cylindrical shape.

Solution. We start the DOF calculation with the individual devices. The leach tank has five
streams (SI, S2, S3, S4, and S5). SV = 2 + 1 + 2 + 2 + 1 = 8. The leach tank has three species, so
IB = 3. SI has one composition specified, so 1С = 1. SI and S2 have their flows specified, so F =

182 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

2. Two of the four SR equations apply to this device, so SR = 2. Therefore, the DOF = 8 - 3 - 1 -
2 - 2 = 0; the leach tank is completely specified. A balance on the leach tank doesn't require any
precipitation tank information.

The precipitation tank has four streams (S3, S6, S7, and S8). SV = 2 + 2 + 2 + 1 = 7. There
are two species, so IB = 2. S7 has one composition specified, so 1С = 1. F is zero, and the other
two SR equations apply, so SR = 2. Therefore, DOF = 7 - 2 - 1 - 2 = 2, so the precipitation tank
is therefore underspecified by two. However, solving the leach tank balance gives the flow and
composition of S3. These two pieces of information add a value to IB and F. Once the leach tank
balance is made, the precipitation tank DOF = 0. The process thus has DOF = 0.

The process DOF requires the modified approach discussed earlier. The connecting stream
(S3) between the two devices means we cannot simply add up all of the SV values from the two
individual devices, otherwise we would double-count the SV for S3. Therefore, we either need to
subtract two from 8 + 7 to give a process SV = 13, or we need to count the process SV values
stream by stream, which gives SV =13. For the process, we must sum the IB terms for each unit,
so IB = 3 + 2 = 5. Similarly, 1С = 2. We count the process flows to get F = 2. We can sum the
individual SR because each is unique to the device, so SR = 4. So for the process, DOF = 13 - 5 -
2 - 2 - 4 = 0, as expected from the individual DOF calculations.

The overall DOF treats the system as one device, with instreams SI and S2, and outstreams
S4, S5, S6, S7, and S8. SV = 11, IB = 3,1С = 2, and F = 2. Only three of the four SR values apply
to the overall system, so SR = 3. The overall DOF is l l - 3 - 2 - 2 - 3 = + l , s o t h e overall system
is underspecified. Table 4.7 lists the DOF calculation results for the individual units, the process,
and the overall system.

Table 4.7 Results of DOF calculation on boric acid refining process.

DOF factors Leach Tank Ppt. Tank Process Overall
No. of stream variables (SV) 8 7 13 11
No. of individual balances (IB) 3 2 5 3
No. of specified stream comp's (1С) 1 1 2 2
No. of specified stream flows (F) 2 0 2 2
No. of subsidiary relations (SR) 2 2 4 3
0 +2 0 +1
DOF

The best way to start the balance calculation is with the leach tank because its DOF = 0. The
CaC03 balance gives F5 = 6.922 kg, so F4 = 5.769 kg. By difference, F3 - 112.21 kg. An H3B03
balance gives w3H3B03 = 15.22 %, so w4H3B03 = 15.52 %. H3B03 and water balances on the
precipitation tank complete the calculation. Table 4.8 shows the results. Values in bold font are
those that either require plant measurement (two stream flows and two stream compositions) or are
known by prior knowledge (the water and CaC03 are pure). The rest are calculated.

Table 4.8 Mass balance for boric acid refining process. Composition units are mass pet.

Stream No 1 2 3 45678

Flow, kg 24.9 100 112.21 5.77 6.92 89.66 9.02 13.53
% H3BO3 72.2 0 15.22 15.52 0 3.6 3.6 100
0 100 0 0 0
% CaC03 27.8 0 0

% H20 0 100 84.78 84.48 0 96.4 96.4 0

The feed contains 18 kg of H3B03 of which 13.5 are recovered as solid pure H3B03 (S8).
However, if the precipitation tank slurry is dried, the product will be 13.9 kg of H3B03, for an
overall recovery of 77 % of the incoming H3B03. The recovery would be much higher if S6 was
used as the leach solution in place of pure water. However, some water would be required to make

up that lost in S3 and S7.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 183

Try setting this system up in Excel so that a change in one of the four stream properties gives
an immediate new system mass balance. If the four SR equations are not affected, one can then
explore the effect of feed composition, water flow, etc. on the expected loss of H3B03 via S4 or S7.

Assignment. The operator needs to produce 17 kg/min of H3B03 via S8, which requires increasing
SI and S2. The liquid stream compositions don't change, but not all of the H3B03 has time to
dissolve in the leach tank. The CaC03 (that exits in S5) exits as a solid containing wH3B03 =
4.5%. Calculate the SI and S2 flow required. First, make a DOF calculation to see if the process
is correctly specified. If not, what reasonable assumption can you make from the original process
that will give you one additional SR?

We saw that when all of the device balances have a DOF > 0, the system cannot be
completely specified by making device balances alone. Consider a three-stage drying process in
which adsorbed water is removed from clay granules, as shown in Figure 4.26. Each dryer
removes half as much water as the one before it.

Vapor Vapor Vapor

A A A

ВI D F

A> Dryer с , Dryer E шw Dryer G,
FA = 8000 kg/h I III
wAH2O = 0.18 P II FG = 6820 kg/h
wGH20 = 0.04

Figure 4.26 Flowsheet for three-stage clay dryer. Stream A consists of wet clay. Each dryer is
hotter than the one preceding it, and each dryer produces half the vapor ofthat preceding it.

A DOF analysis should be made first on each dryer and then the overall system. Each
individual device involves five stream variables and processes two species, hence two independent
balances can be written per device. One specified stream composition is available for device I and
III only. Finally, the requirement that F® = 2F° and F° = 2F* translates to two flow specifications
(i.e., two SR) which can be assigned to the overall system only. A DOF analysis for each device
and the overall system is:

Dryer I: DOF = 5-2—1 — 1=4-1
Dryer II: DOF = 5 - 2 = +3
Dryer III: DOF = 5 - 2 - 1 - 1= +1
Overall: DOF = 7 - 2 - 2 - 2 - 2 = - l

Thus, the overall system (and the process) is overspecified, while each device is
underspecified. The over-specification of the overall system is verified by making a clay balance
and noting that 6560 kg/h enter Dryer I, while 6547 kg/h leave Dryer III. The overall process
would be properly specified by eliminating a composition or flow from stream A or G. For
example by omitting wGH20 from the stream G specification, the process could be balanced by
writing the following equations:

Total mass flow: 8000 = 7^ + 7^ + 7^ + 6820
Flow relationship:
Flow relationship: 7 ^ = VAF*
Dryer I:
Dryer II: 8000-7^ = 7^
Dryer III: pc — р° = pE
7 ^ - 7 ^ = 6820

The algebra for these equations is rather simple, and each stream flow and composition is
easily calculated. A ledger of the complete process balance follows.

184 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

Notice that wGH20 has been calculated as 0.038, which replaces the value of 0.04 given
originally. (Remember that the stated value of wGH20 was changed to a variable to remove the
over-specification.) The stated value of wGH20 = 0.04 gave an extra equation which was
inconsistently redundant, whereas if wGH20 had originally been stated as 0.038, the extra system
equation would have been consistently redundant.

mass flow, kg/h A В С Stream
clay Wet clay Vapor (I) Clavffl DE FG
H20
total 6560 0 6560 Vapor (ID Clay (II) Vapor (III) Dry clay
1440 674 766
8000 674 7326 0 6560 0 6560
337 429 169 260
337 6989 169 6820

mass fraction 0.82 0 0.895 0 0.939 0 0.962
in stream 0.18 1.00 0.105 1.00 0.061 1.00 0.038
clay 1.00 1.00 1.00 1.00 1.00 1.00 1.00

H20
total

One of the difficulties associated with multiple-device flowsheets is that they often generate a
dozen or more simultaneous equations (some non-linear) to be solved. This requires some sort of
equation-solving software. The equation-solving program adopted for use in this Handbook is
Excel's Solver program, which was introduced in Section 4.8.1. Example 4.6 shows the use of
Solver on a two-stage process.

EXAMPLE 4.6 — Recovery ofKMn04 by Evaporation.

A common way to recover a solute from a solution is by evaporating water until the solid
precipitates out. In this example, potassium permanganate is recovered in a two-stage evaporation
process as shown in the flowsheet (Figure 4.27). Most of the KMn04 is recovered as a slurry, but
a small amount leaves the system from settling tank II (stream I), and is recovered by a separate
chemical process. The input KMn04 solution flowrate is 150 kg/h with wKMn04 = 0.134. The
evaporation tanks are at 115 °C, while the settling tank temperature is designated on the Figure.
Each settling tank has wsolid = 0.72. The objective is to calculate the amount of water to be
removed from Evaporation tank I to give a KMn04 loss of 0.3 %.

Data. The solubility of KMn04 is: wKMn04 - 2.39 x W\t2) + 1.13 x 10"3(0 + 0.0278, where t is
degrees Celsius.

Solution. The process consists of five devices with ten well-defined input and exit streams. A

ledger for the system is shown on the next page. Where possible, the mass fraction of KMn04

obtained by the solubility equation was placed on the ledger. It should be apparent that DOF = +1
ofFB9
around Evaporation tank I, but if we pick an arbitrary value For DOF = 0. Then a KMn04 and
total mass balance around that device is simple arithmetic. example, if FB = 90 kg/h, then
wcKMn04
forth, until = 0.335. These results can be used to make balances around Settling tank I, and so
all stream flows have been calculated. F1 is 3.1 kg/h, with a KMn04 loss of 0.43 %. So
one way to solve the problem is to make balances with different assumed values of F®, plot the
results, and find F® graphically for a KMn04 loss of 0.30 %. Some automation of this procedure is

possible using Excel for the arithmetic, but an explicit answer is not attained.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 185

Stream A В D H

mass flow, kg/h 150 ? ? ??
Liquid/gas 0 0 0 0 ?0
Solid 150 ? ? ???
Total

mass fraction
in liquid phase

KMn04 0.134 0 ? 0.083 0.083 0 0.420 0.028 0.028 0.045
H20 0.866 1.00 ? 0.917 0.929 1.00 0.580 0.972 0.972 0.955

Total 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

mass fraction solid 0 0 0 0 0.72 0 0 0.72 0

KMn04 Water Water
solution (115 °C) vapor

FA= 150 kg/h
wAKMn04 = 0.134
wAH20 = ?

Evap'n tank I soln Evap'n tank II soln
wGKMnO4 = 0.42
wcKMn04 = ?
ivcH20 = ?

Heat Decanted sol'n
to final KMn04

recovery

w'KMn04 = 0.028
w'H20 = ?

Settling tank slurry I Settling tank slurry II

FH = ?

w solids = 0.72 wHsolids = 0.72

wE/LKMn04 = 0.083 w solids = ? LKMn04 = 0.028
wE/LH20 = ?
wJ/LKMn04 = 0.045 w H20 = ?

Final slurry w J / Lu Л ,

H 2 0:

Figure 4.27 Two-stage process for the recovery of KMn04 from solution. The KMn04
precipitates from solution as the temperature is lowered in the settling tanks. Double-line arrows

indicate two-phase (KMn04 + solution) streams.

Even though the above procedure is simple and straightforward, there is a better way: write
KMn04 and total mass balance equations for each device, and use Excel's Solver tool to solve for
all flowrate values and wcKMn04 at once. Solver is a very useful and powerful tool for solving
equation sets, as described in section 4.8.1. Instructions for using Solver are on the Handbook CD.

There are 11 unknowns to solve for: nine stream flows, wcKMn04 and wJsolids, which
requires 11 equations. A KMn04 balance equation can be written around each of the five devices,

186 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

overall mass balance equations written around the five devices, and a fractional KMn04 loss
equation written for stream I, for a total of 11 equations. The equation format sets all terms equal
to zero, which simplifies the Solver setup. The term "ET" means evaporation tank, and "ST"
means settling tank. The ledger is shown on the next page.

The equations were written on a worksheet in column D, formatted to equal zero (hence the
Set Target Cell and Constraints were all set = 0). For example, the KMn04 ET I: mass balance
equation was written in the formula bar for cell D24 as follows:

=D23*0.134-F23*N23

Cell D23 contained the numerical value for FA (150 kg/h), cell F23 contained the value for F°
and cell N23 contained the value for wcKMn04. Initial guesses of 50 kg/h for the nine flow
variables were entered into row 23, along with an initial guess of 0.50 for wcKMn04 and wJsolids.
On the Options dialog box, the sought-after values were set as non-negative. Solver found a
solution (by changing the row 23 values) as shown below.

KMn04 SM: KMn04 ET I: 150(0.134) - 7^(wcKMn04) = 0

KMn04 ST I: 150(0.134) -/^(0.083) -7^(0.72 + (1 - 0.72)(0.083)) = 0

KMn04 ET II: i^(0.083) - 7^(0.42) = 0

KMn04 ST II: 7^(0.42) - 7^(0.028) - i^(0.72 + (1 - 0.72)(0.028)) - 0
7^(0.72 + (1 - 0.72)(0.083)) + 7^(0.72 + (1 - 0.72)(0.028)) - ^(wJsolids + 1 - wJsolids)(0.045) = 0
ΣΕΤΙ: \50-(FB + Fc) = 0
I S T I : Fc-(FD + FE) = 0
ΣΕΤΙΙ: FD-(FG + F¥) = 0
EST II: FG-(F*l + F*) = 0
ZSM: 1^ + 1^-^ = 0

KMn04 loss: 0.003-^(0.028)/(150(0.134) - 0

сD E F G H I JКL M N О

22 FA F* / * F° F* F* F° F" F1 F1 wcKMn04 w1 solids

23 150 100.9 49.1 24.8 24.3 19.9 4.9 2.7 2.2 27.0 0.410 0.730

Note that a total of 120.8 kg/h of water (FB + F*) is removed during the process, 84 % of
which is removed from Evaporation tank I.

Assignment, a) Use the stream flow values to calculate the mass flow of KMn04 in the liquid
portion of each stream, and the solid portion of each slurry stream. Fill in the ledger with the
appropriate values. How much water must be removed from the final KMn04 slurry (stream J) in
order to produce a dry solid product? b) Repeat the example for a stage I temperature of 30 °C and
a stage II temperature of 5 °C.

It is clearly a matter of choice whether to use a hand-calculation technique or an Excel tool to
solve the material balance equations for a process. In the beginning, it's probably best to avoid
over-using Excel because hand calculations may give a better understanding of the process details.
However, as the number of devices and species increase, hand calculations become very tedious
and prone to error. Excel is then recommended for at least the arithmetic. If repetitive calculations
are needed, try setting the material balance equations in Excel as a chain-of-equations format so
that changing a parameter will result in a new result for all stream variables. Excel's Goal Seek
and Solver tools are sometimes the only feasible way to handle a complex flowsheet.

The next example illustrates the procedure for making a material balance on a system more
complex than heretofore. The technique for dealing with complex systems is to simplify them, as

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 187

outlined in Section 4.6. Here, the system was structured so that portions of the flowsheet could be
examined independently, and the balance equations solved without using Solver.

EXAMPLE 4.7 — Removing Dust and S02from a Roaster Gas.

A pollution control residue contains heavy metal sulfides, which are being disposed of as
hazardous waste. A process is being designed to treat these residues by roasting them in air to
convert the sulfides to oxides, from which the metals can be recovered. The roasting process
produces a gas having a dust content of 50 g/m3 at STP. The volume fraction of constituents in the
gas phase is:

<pS02 = 0.120, φΗ20 = 0.080, φ02= 0.080, balanceN2.

The roaster gas contains too much dust and S02 to be discharged to the environment. This is
to be corrected by scrubbing the dusty gas to remove the dust and most of the S02, and recover
these later. The dusty gas is scrubbed with pure water at 10 °C, and the slurry filtered to remove
all of the dust. The filter cake is dried and sent to a metal recovery step. The dryer gas contains a
weight fraction of H20 of 0.50. The filtrate contains dissolved S02 that must be removed and
recovered. The S02 will be removed from the filtrate by vacuum extraction at 10 °C (possibly
with some air to assist removal of S02). The objective is to lower the S02 in the clean gas, while
producing a high-S02 gas that can be more easily treated to remove the S02. Make a material
balance for a system designed to remove 70 % of the S02 from the roaster gas in the scrubber, and
explore the effect of device pressure and amount of air from stream H on overall recovery of SO2
in stream M. Each device pressure is 1.00 atm except for the S02 desorption device, which is a
variable. Assume dry air contains only N2 and O2.

Data. The solubility of S02 in water at 10 °C is expressed in terms of mass of S02/mass of water
as a function ofpS02 in atm, according to the following equation:

£>S02 = 5.66(mass S02/mass H20)

The small solubility of S02 is deemed not to appreciably affect the vapor pressure of water.

Solution. The process contains mixed units (volume and mass), which complicates the mass
balance equations. This requires conversion to a consistent set of units before drawing and
labeling the flowsheet. In addition, the problem statement does not indicate a flowrate of roaster
gas, so we are required to select a basis, from which the process can be scaled as needed. A
convenient process basis is 100 m3 STP roaster gas*, which amounts to 44 620 moles. The MMV-
C program can convert the original roaster gas composition to other units, as shown in Table 4.9.

Table 4.9 Gas composition in roaster gas (stream B). 100 m3 (STP) basis.

volume, m3 moles mass, kg mass fraction

N2 72 3213 89.99 0.6331

o2 8 357 11.42 0.0804

н2о 8 357 6.43 0.0453

so2 12 535 34.30 0.2413

total 100 4462 142.14 1

dust 5
total mass of stream В 147.14

During scrubbing, some of the water vapor and S02 will move from the gas to the liquid
phase. The vapor pressure of water can be obtained from Equation [2.15] as 0.0122 atm, and the

An alternate basis could be 100 moles of roaster gas, and then make the material balance in amount
units rather than mass.

188 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

distribution of S02 between water and gas is obtained from the relationship given earlier. The
design engineer sketched a flowsheet for a conceptual process as shown in Figure 4.28. To avoid
overcrowding in the labeling of the streams, the flowsheet is split in half, and the streams labeled
in Figures 4.29 and 4.30.

Water Cake Gas to
S02 recovery
Cake
Air dryer M

S02 High S 0 2 Vacuum
desorb gas pump
reactor
solution
ΗΤΑΪΓ

Filtrate

Figure 4.28 Flowsheet for roaster gas scrubbing and S02 treatment. Dusty furnace gas is
scrubbed with water, which absorbs some of the S02. The slurry is filtered and the filtrate is
vacuum extracted to desorb most of the SO2. The filter cake is dried and the offgas is mixed with
the high S02 gas and sent to a S02 recovery process. The clean scrubber gas is vented, and the low
SO2 solution could be recycled to the scrubber.

Water _ ! 2 - /nc = ? Clean gas
m -г
wAH20 = 1 Scrubber wcN2 = ? wc02 = ?

Roaster В wcS02 = ? wcH20 = ?
gas
Filter

m B = 147.14 kg mF = ? cake
m dust = 5 kg Slurry
mDвgas = 142.14 kg mFliq = 5 kg
wB/GN2 = 0.6331
mFdust = 5 kg
W B / G 0 2 = 0.0804
v//LS02 = ?
wB/GS02 = 0.2413 wF/LH20 = ?

wB/GH20 = 0.0453 E = 9

m

► vy S 0 2 = ?

wEH20 = ?

Figure 4.29 Roaster gas scrubber/filter portion of flowsheet for offgas cleaning process. Assumed
basis is 100 m3 (STP) roaster gas. m refers to stream mass, while w refers to species mass fraction
in gas or liquid phase. A double arrow designates two-phase flow.

The next step is a DOF analysis around each device. We ignore the dust flow because its
amount is specified on entry and exit streams. For the scrubber, SV = 1 1 (one for stream A, four
for stream B, four for stream C, and two for stream D). The number of independent component
balances is four (N2, 02, H20, and S02). There are three independent stream compositions (all for
stream B), and one amount specification (stream В where we selected a basis). Finally, there are

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 189

three subsidiary relations (vapor pressure of water, gas/liquid distribution ratio of S02, and
specified fraction of S02 absorbed by gas). This gives a DOF = 0. he DOF around the scrubber
has SV = 8, ICB = 3, ISC = 1, and SR = 2 (all dust to filter cake and mass fraction of solid in filter
cake = 0.5). DOF = +2, so the filter is underspecified. However, solving the scrubber balance
defines stream D, which then brings the filter DOF to 0.

Now we can write the relevant equations for the scrubber device, where m,Al:i^q refers to the

mass of the liquid phase (containing dissolved S02) in stream D.

S02 balance: 34.30 = wDS02(mDliq) + wcS02(mc)

S02 split: 7/3 = [wDS02(mDliq)]/[wcS02(mc)]

S02 solubility: [wcS02(mc)]/[wDH20(mDliq)]=^S02/5.66

pS02 relationship: pS02 = /7S02/E«gases in stream C.
H20 balance: mA + 6.43 = wDH20(mDliq) + wcH20(mc)

/?H20 relationship: /?H20 = 7?H20/Xftgases in stream С = 0.0122 atm.

Solving the first two equations gives wcS02(mc) = 10.3, and wDS02(mDliq) = 24.0. In other
words, there are 10.3 kg of S02 in the clean gas and 24.0 kg in the slurry. Next, we write thepH20
relationship for stream С as:

0.0122 = (wCH20 fwCH2OxwC)/18.02
x mC)/18.02 + 89.99/28.01 +11.42/32.0 +10.3/64.06

This equation may be solved to obtain wcH20(mc) = 0.830 kg. The value of mc is thus 112.54
kg. Table 4.10 shows the ledger for stream C.

Next we use the/?S02 relationship:

10.3/64.06
P 2 ~ 10.3/64.06+ 89.99/28.01+ 11.42/32.0+ 0.83/18.02
which givespS02 = 0.04258 atm. The S02 solubility relationship is:

10.3/[wDH2O(mliq)] = 0.04258/5.66

Table 4.10 Gas composition in stream C.

N2 volume, m3 moles mass, kg mass fraction
72 3213 89.99 0.7995
нsoo22о2 8 357 11.42 0.1015
1.05 47 0.844 0.0075
Total 3.6 161 10.30 0.0915
3777 112.55 1.0000
84.65

Thus wDH20(mliq) - 1370 kg. A water balance gives mA = 1364 kg, and wD = 1399 kg. wDliq
is 1394 kg, with wDS02 = 0.0172. The complete ledger for the scrubber balance is shown on the
next page.

The balance arithmetic for the filter is trivial, with m = 1389 kg and m 10 kg. We can
now turn our attention to the rest of the flowsheet, as shown in Figure 4.30.

190 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

A Scrubber streams D
Scrub water Slurry
ВС
1364 Roast gas Clean gas 1394
0
phase mass, kg 0 142.1 112.5 5
liquid 5 0 1399
gas 1364
dust 147.1 112.5 0.9828
total 0.0172

mass fraction
in liquid phase

H20
S02

mass fraction
in gas phase

o2 0.6331 0.7996
0.0804 0.1015
Hso202 0.0453 0.0074
0.2413 0.0915

кmF = 10kg Filteгrгсcake
II F
mrliq = 5kg
m Fdust = 5 kg +> Dry cake
wF/GS02 = 0.0172| mLdust = 5 kg

v//GH20 = 0.9828 Dryer gas К

'· 0.767 mK = ? Gas to S 0 2
wG02 = 0.233 wKN2 = ? ivK02 = ? recovery (M)
ivKS02 :
wKH20 = 0.50 wMN2 = ?
wM02 = ?
Filtrate ► S02 High-S02 < Ш wMS02 = ?
mE = 1389kg desorb wMH20 = ?
reactor gas
wES02 = 0.0172
I -► L o w - S 0 2
wEH20 = 0.9828 j mJ = ? solution
Air- H wJN2 = ?
wJ02 = ? m
m =9 w S02 = ?
wHN2 = 0.767 wJS02 = ?
wJH20 = ? w'H20 =?

w"02 = 0.233

Figure 4.30 Flowsheet section for processing the filtrate and filter cake produced by the section

shown in Figure 4.29. Pressure of all devices is 1 atm except for the S02 desorption device, which
is below atmospheric pressure.

Starting with the cake dryer, and again ignoring the defined solid (dust) flow, the DOF is

calculated as follows: SV = 8 (2 for stream F, 2 for stream G, and 4 for stream K); IB = 4 (N2, 02,
H20 and S02); 1С = 3 (1 each for streams F, G, and K), F = 1 (stream F); and no SR, so DOF - 0.

The cake dryer equations are:

S02 balance: 0.0172(5.00) = wKS02(mK)

H20 balance: 0.09828(5.00) = 0.50(mK)

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 191

0 2 balance: 0.233(mG) - wK02(mK)

N2 balance: 0.767(wG) = wKN2(wK)

Composition relationships: wK02 = 1 - 0.50 - wKN2 - wKS02

These are solved by algebraic manipulations to give the cake dryer ledger. Stream E is
included for completeness:

Cake dryer streams

E FGL К
Filtrate Dryer gas
Filter cake Air Dry cake
1389 ...
phase mass, kg — 5 —— 9.828
liquid —
1389 — 4.828 — —
gas 9.828
dust 5—5
total
10 4.828 5

mass fraction 0.9828 0.9828 — ——
in liquid phase 0.0172 0.0172 — — ...

H20 — — 0.767 — 0.3768
— — 0.233 — 0.1144
so2 — — — 0.500
— — 0 — 0.00875
mass fraction 0
in gas phase
NT

o2

H20

so2

The next device to be abdadliatniocnedofisaitrh(ei.eS.,0m2 Hd=eso0r)p. tiHonereexwtreachtaovre. In the first instance, we
examine the device without SV =7 (two for stream E,

zero for stream H, two for stream J, two for stream L, and the pressure variable for stream J), IB =
2 (H20 and SO2), 1С = 1 (one stream E), F = 1 (stream E), and SR = 2 (the ρΐί20 relationship and
the gas/liquid S02 split relationship). Therefore DOF = +1, so the system is under-specified. The
following equations must be solved to balance the S02 desorption extractor.

S02 balance: 0.0172(1389) = 23.9 - wLS02(mL) + wJS02(mJ)

H2O balance: 0.9828(1389) - 1365.1 - wLH20(mL) + wJH20(mJ)

/?H20 relationship: pH20 = (wH20/Zwgases in stream J)P = 0.0124 atrn, or:

0.0122 = fwJH2OxmJ)/18.02
(wJH20 x wJ)/18.02 + (w3S02 x wJ)/64.06

pS02 relationship: pS02 = (nS02fLngases in stream J)P, or:

^S02 = (wJSO2xmJ)/64.06
(wJH20 x mJ )/18.02 + (wJS02 x mJ )/64.06

S02 solubility: wLS02(mL)/[wLH20(wL)] =pS02/5.66

Before attempting to solve these equations, it is important to consider the range over which
the variables can exist, and to consider ways to simplify the equations. These considerations are:

• The S02 solubility equation can be used to calculate an upper limit on pS02. The incoming
filtrate (stream E) contained 23.9 kg of S02 and 1365.1 kg of water. Had a gas been present in
stream E, the pS02 would have been 0.0991 atm. Therefore, the practical range for pS02 in

192 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

stream J is between about 0.09 and 0.01. In addition, the/?H20 is 0.0122 atm, so the practical
range for P in stream J is from about 0.02 to 0.10 atm.

• The value of wLH20 will be close to 1, and the value of mL will be between about 1360 and
1385.

• The two equations relating pS02 and pH20 can be simplified by noting that the ratio of the
partial pressures of two gases is equal to the ratio of the moles of the gases.

• Since we have no specified composition data for the two output streams, it is easier to use
component mass as a variable in the equations rather than the product of composition and
stream mass.

These considerations allow a simplified four-equation set (with five unknowns), and a
solution method that doesn't require any special software.

S02 balance: 23.9 - mLS02 + mJS02
H20 balance: 1365.1 - mLH20 + mJH20
pR20/pS02 equation: 0.0122/pSO2 = (wJH2O/18.02)/(wJSO2/64.06)
S02 solubility: mLS02/mLH20 =/?S02/5.66

Collecting terms yields one equation with two unknowns, which verifies that the desorber is
under-specified. The equation set (and device) is completely specified if any unknown value is
selected. Table 4.11 shows Excel calculations for a range of pressures, with some parameters
plotted in Figure 4.31.

A number of interesting observations can be made from the balance. First, note the upper

limit on/>S02 of 0.0991 atm, which is the/>S02 in the incoming filtrate. Thus, the upper operating
pressure of the device is 0.1113 atm, which equals pS02 + pH20 (as discussed in the equation

resolution guidelines). At that pressure, no S02 is extracted, and stream I has the same
composition as stream E. As the lower limit of 0.0122 is approached, virtually all of the S02 is
extracted, but at the cost of extensive vaporization of water to stream J. Operation between these

limits shows that lower pressure extracts more S02, but decreases the composition of S02 in the
extracted gas by water vapor dilution.

Table 4.11 Results of desorber calculations at various values of total pressure.

P, atm 0.02 0.04 0.06 0.08 0.1
0.0076 0.0276 0.0476 0.0676 0.0876
pS02, atm 0.13% 0.49% 0.83% 1.52%
wLS02 68.5% 88.8% 93.2% 1.18% 96.2%
w3S02 1356.8 1369.6 1375.7 95.1% 1386.1
mh 32.21 19.43 13.34 1381.0
mj 38.0% 69.0% 79.3% 7.99 2.88
92.4% 72.2% 52.0% 84.5% 87.6%
<pJS02 31.8% 11.6%

% S02 extracted

A similar set of calculations can be made with different amounts of air added via stream H.
Since the oxygen and nitrogen pass through unreacted, it is simpler to consider air as an inert gas,
and write a mass balance equation for mJair. A solution of the resulting equation set requires
specification of two variables; here, P was set as 0.04 atm, and values of A were chosen between 0
and 10 kg. Table 4.12 shows the results, and selected results are plotted in Figure 4.32.

We now turn our attention to the mixer where streams J and К are mixed to yield stream M.
The mass balance for this device is trivial because both input streams are specified. The value of
mH = 9.83 kg, half of which is water. Table 4.13 shows the results for stream M when the desorber
device is at 0.04 atm, with different mounts of air added.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 193

Performance of S02 Desorber vs. Total Pressure

100% Lг1, - ,,,,,rh 1.80%
T
SQJnU%/o XS. f. : -O- i ■ ■ * * 1.60%
и

• * * ^p .^r( 1.40%

oU /о * ** \ „ . - - о

оос 7f ПU0/ Лo Ls х-- τ ^ 1.20% 5
OU /o 4^
j11.00% =с3
ГС 0.80% £
•gE OU/o
40% pH20 Δ N. 0.60% (/>
Ф
^^^ _^r 4 0.40%
Oi nU°//o. *
D wJS02 ^4
Z9ПU0/Ло - Δ - <pJS02
—-У — % ЯП? PYtrartpH 0.20%
1l UП/%o
—0—wLS02
0% — 0.00%

0.02 0.04 0.06 0.08 0.1
Ρ total, atm

Figure 4.31 Extraction of S02 from filtrate to desorber gas as a function of total device (and
stream J) pressure. The double vertical line at P = 0.0122 atm represents the lower limit of

desorber pressure. The 0 symbol represents the mass fraction of S02 in stream L. The Δ symbol
refers to the volume fraction of S02 in stream J (the gas). The x symbol refers to the fraction of
S02 extracted from stream E (the incoming filtrate) to the gas. The D symbol refers to the mass
fraction of S02 in the gas.

Table 4.12 Effect of air amount (via stream H) on desorber performance at P = 0.04 atm.

Air, kg 0 1 3 6 10
wLS02 0.49% 0.43% 0.36% 0.29% 0.24%
wJS02 88.8% 83.5% 75.3% 66.1% 57.3%
1369.6 1368.5 1366.8 1364.9 1362.9
mL 19.43 21.5 25.2 30.1 36.1
! 0.0276 0.0246 0.0205 0.0166 0.0133
0.0030 0.0071 0.0110 0.0143
pS02m, atm 0 61.5% 51.1% 41.4% 33.4%
pAir, atm 69.0% 7.5% 17.9% 27.6% 35.6%
0.0% 75.2% 79.4% 83.3% 86.6%
<pJS02 72.2%
<pJAir

% S02 extracted

Table 4.13 Effect of air amount on mixer stream M at P = 0.04 atm.

Air, kg 0 1 3 6 10
mMN2 3.7 4.467 6.001 8.302 11.37
mu02 1.12 1.36 1.82 2.52 3.45
mMS02 17.34 18.0 19.0 19.9 20.7
mMH20 7.09 7.47 8.15 9.11 10.32
19.43 21.5 25.2 30.1 36.1
J 29.26 31.36 35.04 39.94 45.93
59.3% 57.4% 54.2% 49.9% 45.1%
mmM 50.6% 52.5% 55.4% 58.1% 60.4%
18.6 20.1 22.8 26.7 31.6
wMS02

% input S02 to stream M
Vм, m3 at STP

194 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

Performance of Desorber with Air at P = 0.04 atm

100% 1

90% —0—voi % S02
- D - voi % Air
80% JL -ώ- — ^—_LΔΛ — - — £ — % S02 extracted

8 70%
n 60%
Ё 50%

.g 40%

« 30%
и——-I—20%

10%
r"-°"

0%

46 10
air added, kg

Figure 4.32 Extraction of S02 from filtrate to desorber gas with the aid of air from stream H. The
addition of air assists in the extent of S02 extraction, but at the cost of dilution of the S02 in the
desorber gas. The desorber gas contains a volume fraction of H20 of 31 % throughout.

The material balance results on the overall process can be summarized:

• Between 50 and 60 % of the roaster gas S02 is recovered as an S02-rich gas (stream M). Air
assists in the recovery of S02 but with a penalty of dilution of the S02 in stream M.

• The small amount of S02 in stream L may permit it to be recycled to the scrubber as stream A,
thus minimizing the treatment cost of stream L.

• More S02 could be recovered from the roaster gas if the scrubber temperature were lowered, or
the device pressure increased. Another possibility might be to use counter-current gas/liquid
flow in the scrubber to increase the solubility of S02.

• The small amount of S02 in the dryer gas (stream K) adds very little to the overall S02
recovery, and may best be mixed with stream C. This would increase the composition of S02
in stream M.

Assignment. Suppose that the air for stream H entered at 30 °C at a relative humidity of 70 %.
How would that affect the mass balance for the S02 desorber device?

4.10 Extension of Excel's Calculational Tools for Repetitive Solving

Section 4.8.1 described the use of Excel's two main calculational tools: Goal Seek and
Solver. Goal Seek is a single-use program, so that it has to be invoked repeatedly in order to
obtain multiple answers. As with Goal Seek, Solver is a single-use program, so when answers are
sought for a range of input conditions, the answers for each Solver use must be copied to a
different location on the worksheet before the next Solver use.

Both Excel tools were modified to do repetitive solving while systematically changing one
primary and one secondary variable. The number of solves is the product of the number of
primary times the number of secondary variables. Both programs (SuperGS and SuperSolver) are
on the Handbook CD, and are described in the Appendix.

4.10.1 SuperGS

Goal Seek is a convenient way to solve equations that have complex functions of one variable.
For example, the log(pH20) for saturated steam may contain three or more terms involving Γ, plus

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 195

a constant. Given/?H20, what is 77 Another example involves calculating the P-V-Tproperties of
a gas that can accurately be described by the van der Waals (vdW) equation of state. SuperGS is
useful when making a series of such calculations. Consider the situation when a hot roaster gas is
cooled in a waste-heat boiler. Water is circulated through tubes that line the boiler and is
vaporized to steam by the heat transferred from the roaster gas to the water. For design purposes,
we need to know the velocity of steam traveling through the boiler tubes. Equation-fitting over a
range of steam pressures and temperatures (ChemicaLogic 2009) typical of the boiler device gave
values of the vdW equation of state.

1 0.07057p [4.1]

RTp 1-0.018198/7 RT

The density p is kg/m3, and R is 0.0046152 m3 · bar/(kg · K). According to the Gibbs phase
rule for steam alone, D = 2, so we may arbitrarily fix two of the three intensive properties of
slightly-superheated steam, and use the equation of state to calculate the other. Equation [4.1]
allows the steam density to be calculated if (say) the P and Гаге set.

Suppose we wanted a density calculation for ten different steam P, T combinations, as shown
in the table below.

P,bar 2 3 4 5 6 7 8 9 10 11
Γ,Κ 400 408 416 424 432 440 448 456 464 472

SuperGS is set up as rows of temperature, pressure, and density, with all terms put on the
right-hand side of the equal sign. Goal Seek was then employed to change the user-defined
estimated density value of 1 in cell AI35 until the value in the formula cell (AI37) equaled zero.
At 2 bar and 400 K, Goal Seek found a density of 1.1075 kg/m3, in very good agreement with the
steam table value. Excel's fill command was used to fill the equation across the other nine
columns, with all initial estimates at 1. SuperGS was then opened, and information entered in the
dialog box as shown below.

AH Al AJ AK AL AM AN

33 P, bar 2 3 4 5 6 7

34 T, К 400 408 416 424 432 440

35 density 1.10748 1.64378 2.16706 2.67643 3.1711 3.65044

36

37 vdW eqn 6.6E-06 1.3E-08 4.8E-08 1.2E-08 8.4E-07 9.6E-09

38 SuperGS I (EquatiorΊ I
row
|39~ Set rancie* I CATCT7«CARC47I 1- ^

40
41

42 (* Tovaluiv. | 0

Γ43" ■Г1 ^31 T Density
row
[44~ С от rang«
45 By varying rang«l\ 1 5AIS35:SARS35-|\. ^.

\W MWhile varying als< и
[AT
this range, one a
48 a time
49
50 Г" Use prior result as starting estimate

Γ5Ϊ" Heb Cancel Ok

|52~ шллшш . t

SuperGS calculated all ten densities, seven of which are shown above. Since there was only
one set of variables, we didn't need to use the "While varying also .. " feature.

196 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

4.10.2 SuperSolver

Suppose we wanted to confine our attention to saturated steam alone, which is the VLE line
on the P-V-T surface. (Recall that this subject was discussed in connection with Figure 2.3). D is
now equal to one, so we can calculate the density from P or T alone. First, however, we need an
additional relationship specific to the VLE line. Saturated steam density data were obtained from
the SteamTab program and fitted to an equation in terms of temperature, using Excel's Trendline
tool.

T= 39.745(lnp) + 387.06 [4.2]

The density of saturated steam as a function of pressure may be calculated by simultaneously

solving Equations [4.1] and [4.2]. Equation [4.2] (the VLE equation) was written in row 38 with

all terms on the right-hand side of the equal sign, so we can use the same objective and constraint

for both equations. As a check on the validity of the data entry, Solver by itself was used to
change T and p at P = 2 bar until the vdW and VLE equation cells were zero, and found correct
values of T and p. (SuperSolver requires the use of Solver on one data set to establish a model.)

SuperSolver was now opened, and told to change the temperature and density while
constraining both the vdW and VLE equations to equal zero. SuperSolver found a solution for all
ten values of P, as shown below. Rows 33, 34, and 35 now contain the VLE data for saturated
steam. Alternatively, you can set the VLE density, and ask Solver to change P and T. An
interesting sideline is that over much of the VLE, pressure is a linear function of density.

AH Al AJ AK AL AM AN АО AP AQ AR
33 P, bar 2 3 4 5 7 8 9 10 11
34 T, К 392 407 418 426 6 439 444 448 452 456
35 density 1.133I 1.649 2.158 2.662 433 3.664 4.162 4.660 5.156 5.651
36 3.164
3 7 vdW eqn -2E-09
38 VLE eqn -2E-1 0 -8E-10 -2E-09 -1E-09 -1E-09 -3E-09 -4E-09 -9E-10 -1E-09 -3E-08
39 -8E-C)9 -1E-08 -5E-08 -2E-08 -2E-08 -5E-08 -7E-08 -2E-08 -2E-08

Γ40" SuperSolver J£l
41
— Initial Solver Mod.*e...■l Solve
\42
43 Set target:cell: [ Close

| Д4А4 Equ<alto: Г

[4ж6 By changing cells:

47 [SÄIS34: SAI S3 5

П48" ■Su■bje■ct■to■th■e c■on■str■ain■ts:■ ■ ■ ■ ■ ■ I
49
50 IsAlSBS = 0 лЛ

\5Ϊ тттЛ
52
53 ' В Help
54
55 Target range Secondary var at)les

56 | SAI S3 7: SAR $37 -J —|
|57"
|58~ I Use previous results as starting estimates

59
|6(Γ

61
62

Both programs will be used in working out examples from this point forward. To prepare
yourself for these program's use, please read the User's Guide for SuperGS and SuperSolver.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 197

4.11 Special Multiple-Device Configurations I — Recycle and Bypass

There are three special process configurations that occur often in material processing
flowsheets: recycle, bypass, and counter-current phase flow. A recycle stream may be used to
separate out an unreacted fraction or a catalyst, and return it to the main reactor, or to increase the
concentration of a component in a stream passing on to another device. The concept of recycle
was discussed in connection with Example 4.7, in which stream I could be recycled as scrubber
water to the roaster gas Scrubber. Also, stream I in Example 4.6 could be recycled to Evaporation
Tank I.

Figure 4.33 shows a system flowsheet having both bypass (D) and recycle (G) streams.
Recycle stream G is shown returning directly to Reactor I. However, be aware that sometimes a
recycle stream is mixed with an entry stream before entering a device. If stream G were mixed
with stream A for example, a mixer for the two streams could be added before Reactor I, and the
mixed stream properties calculated.

f F

A Splitter

Reactor 1 С E Reactor II 1
m
В » Mixer
»
D

Figure 4.33 Process flowsheet showing bypass and recycle streams. A portion of stream С
bypasses Reactor II, and stream G is recycled from Reactor II to Reactor I.

A bypass stream around a reactor may be used to modify a downstream device that requires
an unprocessed component. Moreover, even if there is no chemical advantage, both types of flow
may be employed to improve the thermal efficiency. A counter-current process is a way to
increase process efficiency by directing the reactants in opposite directions. Each of these
configurations poses no special analytical difficulties, but may increase the difficulty of the
algebraic solution. This section will discuss balances on systems that contain recycle and bypass
streams. Section 4.12 will discuss counter-current flow systems, which require some new
techniques.

EXAMPLE 4.8 — Absorption of HCl

An intermediate process gas from the production of refined silicon contains variable amounts
of H2 and HCl, but normally (jpHCl « 0.37. Much of the HCl gas is absorbed in water at 25 °C in
an absorption device. The flowrate of the input gas is normally 100 m3/s STP, with the absorption
device water flow set to remove HCl so that the discharge gas is at the maximum allowable content
of φΗΟ = 0.0014. The absorption device water flow is kept constant at this flow level.
Occasionally, the flowrate of the incoming gas is above 100 m3/s, in which case the discharge gas
would have <pHCl > 0.0014. To prevent this, some of the discharge gas is diverted to a bypass
scrubber filled with an absorbent that removes 95 % of the HCl and 90 % of the water. First,
calculate the amount of absorption device water (F°) required for normal gas flow. Next, calculate
the flowrate of other gas streams as a function of FA between 100 and 120 m3/s ifF° is fixed at the
value calculated for FA = 100 m3/s. Figure 4.34 shows a sketch of the system. Assume P = 1 atm.

Data. The following table gives the vapor pressure of HCl and H20 in equilibrium with an
aqueous hydrochloric acid solution at 25 °C.

198 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

Mass fract. HCl 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26
^HCl, mm Hg 0.0145 0.0316 0.0685 0.148 0.32 0.68 1.49 3.2
/>H20, mm Hg 19.3 18 16.7 15.3 14 12.6 11.3 10

Solution. As in Example 4.7, we must deal with both pressure and quantity units, which makes the
material balance equations more complex. The first step is to plot the vapor pressures of HCl and
H20 in terms of more useful units, as shown in Figure 4.35.

Water Absorbent

IB 1

Input A „ Absorption Splitter Bypass Mixer Discharge
gas unit scrubber
FA = ? jjjf^gas
φΑΗ2 = 0.63 (25 °C) Spent absorbent
ФАНС1 = 0.37 FH = ?
φΗΗ2 = ?
Aqueous φΗΗΟΙ = 0.0014
HCl фНН20 = ?

Figure 4.34 Flowsheet for the scrubbing of HCl from process gas. The HCl is absorbed in water
in the absorption device, and if necessary, removed to a even level in the bypass scrubber in order
to attain <jpHHCl in the discharge gas of 0.14 %. The normal flow of input gas is 100 m3/s, STP.

0.005 Vapor Pressure of Gases at 25 °C

0.004 — 0 — p H C I , atm
*+гE■=о» 0.003 — D — p H 2 0 , atm

δ 0.002 pH20 = 0.036 - 0.0879w HCl
LJ^I ÌA ол * л " 7 \ 38.5wHCI
0.001
pHCI = (1.89x10 )e

0 <>— 0.16 0.2 0.24
0.12 mass fraction of HCl

Figure 4.35 Vapor pressure of species in equilibrium with aqueous hydrochloric acid. Excel's
Trendline tool was used to develop mathematical relationships between the vapor pressure in
atmospheres and wHCl. A linear equation adequately represents the /?H20, while an exponential
equation was required forpHCl. The resultant equations are shown in text boxes on the diagram.

The second step is to calculate the wcHCl in the aqueous phase (stream C, hydrochloric acid)
in equilibrium with a gas ofpHCl = 0.0014 atm. This is done by rewriting the/?HCl equation from
Figure 4.35 in logarithmic form. The gas is assumed to be ideal, and since P = 1 atm, <pHCl =
pRCl

/n(0.0014) = -15.482 + 38.5(wcHCl); wcHCl = 0.2314
The above value of wcHCl is used to obtainpOU20 = 0.01564 atm.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 199

The third step is to make a mass and amount balance* around the absorption device for an
input gas flow of 100 m3/s. The FA is 4.461 mol/s and therefore the nAU2 = 2.811 and wAHCl =
1.651 mol/s. The mA is 65.85 kg, from which we obtain wAHCl = 0.9140 and wAH2 = 0.0860. The
72DHC1 and «DH20 in the gas leaving the absorption device is related to the partial pressure and
mole fraction of the gases as given by the following two equations:

*°HC1 = рЯС\ = 0 . 0 0 1 4 = —и ПDHCl D
,D

П Н2+П НС1+П НЮ

* D H 2 0 =/?H2O = 0.01564 = — П H2°

П Н2+П HCl +П HlO

Since nAH2 = «DH2 = 2.811 we have two equations with two unknowns to solve.
Alternatively, we may use either one of the above two equations, and the ratio of WDHC1/T?DH20 =
0.08951. In either case, we obtain

rcDH20 = 0.04473

л°НС1 = 0.004004

Thus, F° in different units are: 2.860 mol/s, 6.618 kg/s and 64.10 m3/s. The gas has wDHCl =
0.02206 and wDH20 = 0.1218.

The normal-condition HCl mass balance around the absorber device is made to calculate the
mass flow of stream C:

mAHCl = wDHCl(mD) + wcHCl(wc)

mA(wAHCl) = 65.87(0.9140) = 0.02206(6.618) + 0.2314(wc)
mc = 259A kg/s of aqueous HCl

The H20 mass balance around the absorber device is used to calculate the stream В flow.

mBH20 = wDH20(wD) + wcH20(mc)

mBH20 = 0.1217(6.618) + 0.7686(259.4)

mBH20 = 200.2 kg/s

Therefore, under normal conditions, 200.2 kg/s of water enter the absorber device (via stream
B). Leaving the device are 259.4 kg/s of aqueous HCl (stream C) and 64.12 m3/s gas (stream D).
You should create a ledger around the absorber device in amount and mass units to summarize
these results, and check for errors.

The calculational method used above was a step-by-step solution of appropriate equations for
calculating a mass balance around the absorber device for one set of input conditions. The mBH20
we sought is the amount required to absorb HCl such that the bypass scrubber was not needed to
meet the <pHHCl specification of 0.14 %. However, the problem requires a mass balance for a
range of FA values, which makes the step-by-step technique rather tedious. When multiple
solutions are required, a different solution method is preferred. Here we will show how
SuperSolver can be used on the system. SuperSolver was developed specifically for applying
Solver to multiple sets of non-linear equations. It was introduced in Section 4.10 and is described
in more detail in the SuperSolver User's Guide.

in For a constant mBH20 = 200.2 kg/s, any increase in FA above 100 m3/s will cause an increase
F°9 F°9 wDHCl and wcHCl. A mass balance study of the effect of increasing FA requires writing

six equations for the various interrelated parameters. The first two equations are the mass balances

* It's convenient to use a mass balance for the aqueous phase and an amount (mole) balance for the gas
phase.

200 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

for HC1 and H20. The next two equations equate the ^DHC1 andpDH20 to the xDHCl and xDH20.
The last two equations relate the pOUCl and р°Н20 to wcHCl. The equation set was written as

formulae in Excel, and the SuperSolver tool used to solve for the six unknowns for five different
of FA tvhaeluaequfoeroums BpHh2a0seo, fm20c0r.2efekrgs/st.o
values refers using the previously determined As a reminder,
wcHCl to the mass fraction of HC1 in the mass of the
aqueous phase, and nOU20 and /7DHC1 refer to the moles of H20 and HCl in the exit gas.

HC1 balance: m AHC1 =F x 0.37x36.46 = / H C 1 X / + W D H C 1 X Ъ6М
H20 balance: 22.414

mBH20 - 200.2 = (1 - wcHCl)(mc) + «DH2O(18.016)

Л°НС1 = - nDHCi+nDH20
3.0924 +

p н2о= —
3.0924+ nD HCl+nDH20

^DHCl = 1.89xl0-7(e38'5xwCHC1)

/?DH20 = 0.036 - 0.088(wcHCl)
The results of a series of calculations for FA between 100 and 120 m3/s are shown in the
ledger below, and in Figures 4.36 and 4.37. The ledger values for stream A are based on the value
for V^ (top row), <pAHCl - 0.37 and φΑΗ2 -0.63. The values in the bordered section of the ledger
are the results from using SuperSolver on the above six equations.

Stream Stream Values

Function 100 105 110 115 120
VA 4.461 5.354
65.854 4.685 4.908 5.131 79.025
A 2.811 3.373
1.651 69.147 72.440 75.732 1.981
n 5.666 6.799
A 60.188 2.951 3.092 3.232 72.226
0.0860 1.733 1.816 1.898 0.0860
m 0.9140 0.9140
«AH2 200.2 5.949 6.233 6.516 200.2
«AHC1 259.4 271.0
0.2314 63.198 66.207 69.216 0.2642
ATT 0.00400 0.01698
0.04470 0.0860 0.0860 0.0860 0.04378
wmAHHC2l 0.001400 0.004945
0.01563 0.9140 0.9140 0.9140 0.01275
ATT 2.859 3.434
200.2 200.2 200.2
wwAHHC2l
mв 262.4 265.3 268.2

1 mс 0.2400 0.2484 0.2565

wcHCl 0.00586 0.00846 0.01206
«DHC1
0.04465 0.04447 0.04417
wDH20
xDAEHCl 0.001951 0.002690 0.003668

xDAEH20 0.01488 0.01414 0.01343

n 3.002 3.145 3.289

In order to attain the specified value ofpHHCl of 0.0014 atm, some fraction of the absorber
output gas (stream D) must be sent to a bypass scrubber (stream E) which removes 95 % of the

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 201

HCl and 90 % of the water. A sample calculation shows that if one mole of gas enters the
scrubber, 1 - 0.95(0.002703) - 0.9(0.01414) - 0.98471 moles exit. Therefore, F* = 0.98471(7^).
For example, at FA of 110 m3/s, wFHCl will be 0.05(0.002703) - 0.000135 moles and / H 2 0 will be
0.10(0.01414) = 0.00141 moles.

Aqueous Phase (Stream C) Properties 0.27
272
0.23
О 268 ^—mass flow of С 120
] — mass fract. HCl in С
'S 264
£ о)
°(/) о) 260

ti)
CO

E

256 105 110 115
100 flow of stream A, m /s

Figure 4.36 Properties of the aqueous phase (stream C) produced by the absorber device as a
function of the amount of input gas at STP.

Gas Phase (Stream D) Properties

80 ri 1.6%
1.2%

Ф 1 f"""- l 0.8%

60 105 110 115 0.4%
100 stream A flow, m3/s
0.0%
120

Figure 4.37 Properties of output stream D from the absorber device as a function of the flow rate
of input gas at STP.

A calculation of molar flowrates of streams E, F, G and H requires five equations because an
additional variable must be introduced for the amount of HCl in stream F. The first two equations
below are the amount balances for HCl around the splitter and scrubber. The third equation
equates the /?HHC1 to the xHHCl. The fourth equation relates the amount of stream F relative to
stream E after 95 % of the HCl and 90 % of the H20 are removed. The fifth equation is the
amount balance around the mixer. Values of я°НС1 and xDHCl were taken from the ledger over
the range of F \ The setup is shown below.

Splitter: «DHC1 = {F* + i^)(xDHCl)
Scrubber: i^(jcDHCl)) - 0.95(/^(xDHCl)) = nFUCl

/?HHC1 = 0.0014 = xDHCl(FG) + KFHCl
Fn

202 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

Scrubber action: F* = 0.98471(7^)
Mixer balance: p* = !& + !?
SuperSolver solved for the five values of V^ to obtain the four molar stream flows and ирНС1.
The stream flow values were converted to volume units by MMV-C and plotted in Figure 4.38.

Stream Flowrates

80 -i

70 \ -^=— - A - """

100 105 110 115 120
stream A flow, m3/s

Figure 4.38 Flowrates of selected streams as a function of the input flow. The flowrates of
streams H and E are not plotted because they lie so close to the lines for streams D and F
respectively.

Assignment. An alternative to using an absorbent in the scrubber is to chill the bypass gas stream
to 0 °C, which will cause some aqueous HC1 to condense. Recalculate the flowsheet stream flows
to determine if using a chiller instead of the absorbent scrubber could be effective. The following
equations give thej^HCl and/?H20 (units of mm Hg) in contact with aqueous acid at 0 °C.

Log(pHCl) = 21.0OHC1) - 5.53

/?H20 = -8.4(wHCl) + 4.60

We now turn our attention to processes that have recycle streams. As mentioned earlier, there
are several reasons for using recycle streams in a process. One reason is that a recycle stream can
increase efficiency by returning unreacted or partially reacted material back to an upstream device
for re-processing. The following example shows how recycling is used in aqueous processing to
improve the quality of a product that is precipitated from a solution.

EXAMPLE 4.9 — Preparation of a Pigment Precursor.

Ferrous sulfate is a raw material for the manufacture of paint, refractory, and cosmetic
pigments. It is produced by the action of sulfuric acid on iron. Coarse ferrous sulfate heptahydrate
is recovered from the acid solution by an evaporation-crystallization process, as shown in Figure
4.39.

Water is removed by evaporation and the solution is chilled to precipitate the heptahydrate.
(A mole of dry heptahydrate weighs 1.83 times as much as a mole of FeS04, and contains 0.83 kg
of H20 per kg of FeS04). The filter removes the coarse crystals, but the fine ones pass through and
are recycled along with the filtrate to the evaporator. The coarse crystals are purified elsewhere,
and it is desired that the solution remaining in the filter cake (i.e., the filtrate liquid) have wH2S04
<20 %. The table below lists the process conditions. Calculate the composition and flow rate of
all streams.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 203

New feed solution flowrate Feed composition Water extraction rate

3000 kg/h wAFeS04= 0.300 1000 kg/h
wAH2S04= 0.040

Coarse crystal composition Filtrate (recycle) composition

wEFeSCv7H20 = 0.750 wFFeSCv7H2O = 0.10

н2о| Concentrated Crystallizer Coarse
Evaporator solution and crystals
Feed solution Filter

Recycle solution
and fine crystals

Figure 4.39 Flowsheet for preparation of coarse ferrous sulfate heptahydrate (FeS04'7H20) by
recycling fine crystals to increase their particle size.

Solution. Figure 4.40 shows a redrawn flowsheet to indicate the process conditions, with a mixer
added to the feed solution stream. The FeS04*7H20 is deemed to dissolve in the mixer to produce
a homogeneous liquid stream B. Calculations are carried out to a larger number of significant
figures than justified to be sure the balance closes.

FA = 3000 kg/h Secondary Water Concentrated
wAFeS04 = 0.30 feed solution jf vapor solution

wAH2S04 = 0.04 F c H 2 0 = 1000 kg/h

wAH20 = ? Crystallizer
and
A 4 Mixer Evaporator Filter
New Π Γ 3
feed wDFeS04 = ? Coarse crystals
iv FeS04'7H20 = 0 wDH2S04 = ?
solution wDH20 = ? FE = ?
wBFeS04 = ? wBH2S04 = ? wEFeS047H20 = 0.75
wE/LFeS04 = ?
wBH20 = ? F

Recycle solution & fine crystals

FF = ? wFFeS047H20 = 0.10 wE/LH2S04 = ?

wF/LFeS04 = ? wF/LH2S04 = ? wF/LH20 = ? wE/LH20 = ?

Figure 4.40 Flowsheet for the production of ferrous sulfate heptahydrate. Stream flow units for F

are kg/h. The term wFeS04*7H20 refers to the mass fraction of heptahydrate in the stream
designated by a superscript letter. The terms wFeS04 and wH2S04 refer to the mass fraction of
species in the liquid (solution) phase, designated by superscript letter L and the stream letter.

Double-line arrows indicate two-phase streams.

A DOF analysis is omitted here because the system actually has a chemical reaction — the
precipitation and formation of FeS04#7H20. We finessed this reaction by simply stating the
relative mass of H20 and FeS04 in the heptahydrate. In place of a DOF analysis, let's try to write
enough equations to solve all of the mass balance unknowns.

An overall mass balance is a good start because there is only one system feed stream and two
system outstreams, and two of these three streams have defined flow. The feed stream flow (FA)
minus the water vapor flow (F0) indicates that FE = 2000 kg/h, of which 1500 kg/h is the solid

204 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

heptahydrate and 500 kg/h is liquid. We can write two more independent balances, of which H20
and H2S04 are the simplest:

H20 balance: 0.66(3000) - 1000 - (0.83/1.83)(1500) = wEH2O(500)

H2S04 balance: 0.04(3000) = wEH2SO4(500)

This gives wEH20 = 0.5993, and wEH2S04 = 0.24*. By difference, wEFeS04 = 0.1607.

Next we turn out attention to the crystallizer/filter, where we have a considerable amount of
information about both device outstreams. Three equations were written for species balances and
one for a mass balance.

H2O balance: wDH20(i^) = 0.5993(500 + 0.97^) + 0.83(1500 + 0.1/^)/1.83

H2SO4 balance: wDH2S04(i^) = 0.24(500 + 0.97^)

FeS04 balance: wDFeS04(^) = 0.1607(500 + 0.9i^) + (1500 + 0.1/^)/1.83

Mass balance around crystallizer/filter: F* + 2000 = F°

Unfortunately, these four equations are insufficient to solve for the five variables. The system
is underspecified, and no further progress can be made without more information. The most likely
item that can be controlled by the system operator is the flow of a stream, such as the recycle flow
(F*). We can learn something about the process by selecting a range of values of F* and noting the
effect on other parameters. The results of the above balances can be used to calculate the flow and
composition of the other streams, as shown in the following ledger for F* = 800 kg/h. Remember
that the removed water (F0, not shown) is 1000 kg/h, and the filter cake is 75% FeS04*7H20.

Ferrous Sulfate Heptahydrate Production Streams

Stream AВ D EF

New feed 2nd feed Cone soln Filter cake Recycle

phase flow, kg/h

liquid 3000 3800 2800 500 720

solid — — — 1500 80
800
total 3000 3800 2800 2000

mass fraction 66.00% 64.42% 51.71% 59.93% 59.93%
in liquid phase 4.0% 7.7% 10.5% 24.0% 24.0%
30.0% 27.9% 37.8% 16.1% 16.1%
H20

H2S04
FeS04

Solver was used to calculate all stream values for F* of 600, 800 and 1000 kg/h. The effect of
changing F* is on streams В
composition as a function of F¥. and D. Figure 4.41 shows the trends in FeS04 and H2S04
A change in F* of ±10 % has only a small change in composition

of solutes in streams В and D (and of course no change in solute composition in the other streams).

Assignment, a) Calculate the maximum allowable wAH2S04 so that the fluid portion of stream E
would have wEH2S04 = 0.15. b) For the case where wEH2S04 = 0.24, and F* = 800 kg/h, a filter
cake washing device is to be added. The cake is washed with a small amount of stream A, which

is set to bypass the rest of the devices. The discharge filter cake from the washing device has

wFeS04-7H20 - 0.75, and the liquid portion of the cake has wFeS04 - 0.18 and wH2S04 = 0.12.
How much of stream A is required for washing the cake? Rebalance the process for this condition,

with the filtrate from the washing device assumed to be free of solid, and returned to the mixer.

* Unfortunately, the wEH2S04 >0.20 for the specified condition, hence the filter cake will require more
washing than originally intended in order to produce an acceptable product.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 205

12% 40% Ù ^ _ _ _ _
11%
i 10% ~~ —■ Ύ § 35% —О—FeS04inB
— 0 — H2S04inB — D — F e S 0 4 in D
Or* — D — H 2 S 0 4 in D aО
E
8 зо%

25%

600 700 800 900 1000 600 700 800 900 1000
flow of stream F, kg/h flow of stream F, kg/h

Figure 4.41 Change in solute composition in streams В and D caused by a change in recycle flow
(stream F).

The use of recycle streams can cause problems in some processes by allowing objectionable
amounts of species to build up to a point where they cause problems if not removed. This was
seen in the above example where the wH2S04 increased. One way to prevent an impurity from
accumulating past a set limit is to add a bleed stream* to a device where the composition of the
impurity is a maximum. An example of a bleed stream is in the electrorefming of copper, where
the small amount of nickel in the anode passes into the electrolyte, but does not deposit on the
cathode until a certain level is reached. To prevent nickel contamination of the copper, a small
amount of electrolyte is bled from the system and stripped of copper and nickel. The purified
solution is then returned to the system. The presence of a bleed stream does not introduce any new
difficulty in the balancing strategy other than increasing the amount of arithmetic.

4.12 Special Multiple-Device Configurations II — Counter-Current Flow

The efficiency of a process can be enhanced by moving the two reacting phases in opposite
directions. Typical examples are washing processes (where an impurity in a slurry is removed by
washing with a solvent), leaching processes (where a portion of a solid dissolves), and moving-bed
reduction furnaces (where solid material is reduced by a gas). Another important class of counter-
flow processes is solvent extraction (SX), which employs two circuits. A metal is extracted in one
circuit from an aqueous phase to an organic phase, and is stripped out of the organic phase into an
aqueous phase in a second circuit. Both circuits usually employ counter-current flow. The result
is a more concentrated, less contaminated solution of a metal in the aqueous phase, which makes
the metal easier to recover.

Washing often consists of two paired devices, called mixer-settlers, or mixer-separators.
Washing occurs in one device, and phase separation in the other. The mixer-settler shown in
Figure 4.42 is often depicted as a single device rather than showing them separately. This
arrangement was mentioned in Section 4.7.4 and shown in Figure 4.19 where leaching was carried
out in the mixer device (Figure 4.20).

Sometimes the liquid fraction of a slurry contains an amount of solute that either prevents
simple disposal, or has too much value to throw away. In section 4.7.2, we showed a process for
"washing away" solutes from a slurry. There are three general types of processes for washing a
slurry containing a saline solution. First, a single mixer-settler like the system shown in Figure
4.42: Second, multiple mixer-settlers where only the slurry stream moves from one set of devices
to the other. Third, multiple mixer-settlers where all streams move counter-currently from one set
of devices to the other.

* Another term used to describe the same thing is г, purge stream.

206 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

Slurry Water Mixer overflow

Clear Figure 4.42 Mixing device (mixer) for
leaching/washing connected to a settling
device (separator) for separating the liquid
and solid phases. The settling device is often
called a thickener.

We can compare the three washing types my making a mass balance for each. Suppose for
1000 lb of slurry is washed with 3000 lb of water in all three cases. The feed slurry has a solids
mass fraction of 0.5, and the liquid phase in the feed slurry has a salt mass fraction of 0.070.

The simplest technique is the single-stage process, sketched adjacent. The double-line arrows

designate two-phase streams (i.e., slurry).

Feed slurry Wash water The feed slurry enters with 500 lb of solids,
(1000 1b) (3000 lb) 465 lb of H20, and 35 lb of NaCl. Washing
is done with 3000 lb of water. A salt balance
Wash solution Washed for the device shows that the mass fraction of
(3000 lb) ^ slurry s a ^ *п the solution phase is 0.0101, so the
(1000 lb) washed slurry contains 5.05 lb of salt, which
is 14 % of the salt present in the original

slurry.

However, within the accuracy of the original problem statement and typical mass balance
analyses, it is common for washing problems to simplify the arithmetic a bit by assuming that the
mass of the solution and water are the same. This assumption is reasonable only when the wash
solutions are dilute, as in this case. With this assumption, the salt loss in the washed slurry is 5.0
lb, and the mass fraction salt in the wash solution is 0.10.

Next, consider a three stage washing process shown in Figure 4.43. This flow arrangement is
sometimes called cross-current flow. Each device consists of a mixer and a separator. The feed
slurry has the same dissolved salt composition as the above single-step example. Each device is
fed with 1000 lb of water, and the wash solution is collected into a single stream. A DOF analysis
for the overall system shows SV = 9, IB = 2, 1С = 1, and F = 5 (only five of the six flows are
independent, since the sixth is obtainable from the other five). The DOF = +1, therefore the
overall system is underspecified, so we have to write balances on the individual devices. A salt
balance is sufficient to specify the system.

Feed slurry Wash water Wash water ИWash water
(1000 lb) || t (1000 1b) (1000 1b) (1000 lb)

В

M/S I M/S II M/S III Washed
slurry
Wash^. W
solution (1000 1b)

Solution mixer

Figure 4.43 Three-stage cross-current washing process for salty slurry. Wash solution flow is
indicated by solid-line arrows, with slurry flow as double-line arrows. Each slurry stream contains
500 lb of solution and 500 lb of solid.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 207

Salt balance (lb) on M/S I: 0.07(500) - >ЛаК(500) + >Лак(1000)

Salt balance (lb) on M/S И: >Лак(500) = wnsalt(500) + wnsalt(1000)

Salt balance (lb) on M/S III: wnsalt(500) = winsalt(500) + wmsalt(1000)

Remember that the term wsalt refers to the mass fraction of salt dissolved in the liquid phase
for streams out of a particular device. These equations were solved in sequence, with the results
displayed in the following ledger:

Cross-current Salt Slurry Washing Streams
c +x D+W
Stream A B+Y Z

phase mass, kg New feed M/S I out M/S II out M/S III out Wash sol'n
liquid
solid 500 1500 1500 1500 3000
total 500 500 500 500 0

1000 2000 2000 2000 3000

mass fraction 0.07 0.02333 0.00778 0.00259 0.01123
in liquid phase

NaCl

The salt content in stream D is 1.3 lb, or 3.7 % of the incoming salt*. This is substantially
better than the single-stage washing using 3000 lb of water that retained 14 % of the original salt.
(An interesting optimization problem would be to determine the amount of wash water for each
device (totaling 3000 lb) that would minimize the salt retention in stream D).

Now consider a three-stage counter-current washing process as shown in Figure 4.44. This
type of flow arrangement is sometimes called continuous counter-current decantation, or CCCD.
The slurry feed rate and composition are the same as above. A DOF analysis for the overall
system shows that it, like the cross-current arrangement, is underspecified. Therefore, a salt
balance must be written for each device.

Feed slurry m Wash water
(1000 lb) IN
в M/SII T W (3000 lb)
Wash Washed
solution
M/S I M/S III =► slurry

(1000 1b)

Figure 4.44 Three-stage counter-current washing process for salty slurry. Wash solution is
indicated by solid-line arrows, with slurry as double-line arrows. Each slurry stream contains 500
lb of solution and 500 lb of solid.

Salt balance (lb) on M/S I: 0.07(500) + wnsalt(3000) = ^salt(500) + >Лак(3000)

Salt balance (lb) on M/S II: >ЛаЩ500) + wlnsalt(3000) = wnsalt(500) + wnsalt(3000)
Salt balance (lb) on M/S III: wnsalt(500) - wnisalt(500) + winsalt(3000)

Overall salt balance: 0.07(500) - wWtßOOO) + winsalt(500)

Any three of the above equations may be selected for the equation set, but sequential solving
of these equations is not possible. The solution method requires either manipulating the equations
algebraically or using Excel's Solver tool. Here, Solver found a solution to the equations, as
shown in the next-page ledger results:

Here again, the assumption is made that the weight of solution is the same as the weight of water.

208 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

Wash solution Z contains 3000(0.01162) = 34.9 lb of salt. The discharge slurry contains 0.14
lb salt, for a retention to the slurry of 0.04 % of the 35 lb of salt that entered in stream A. This is a
substantial improvement over the cross-current washing process, and is an indication of the high
efficiency of counter-current processing.

Counter-current Salt Slurry Washing Streams

Stream A B+Z C+Y D+X

New feed M/S I out M/S II out M/S III out

phase mass, kg

liquid 500 3500 3500 3500

solid 500 500 500 500

total 1000 4000 4000 4000

mass fraction 0.07 0.01162 0.00189 0.00027
in liquid phase

NaCl

The overall efficiency improves in a diminishing way with the number of stages. Figure 4.45
shows the results of a salt balance for slurry washing as a function of the number of stages. A
linear relationship is obtained when the logarithm of the amount of salt retained in the discharge
slurry is plotted against the number of stages. This makes it easy to estimate the number of stages
needed for any specified washing performance. Only two calculations are needed for any specified
condition, from which an equation can be derived to apply to any number of stages.

As a reminder, during the preceding calculations a simplifying assumption was made that the
weight of solution was equal to the weight of water. The validity of this assumption depends on a
low concentration of solute and/or the accuracy required.

Performance of Counter-Current Washing Figure 4.45 Performance
10 of counter-current washing
circuit as indicated by the
0.01 number salt weight present in the
of stages liquid portion of the
discharge slurry. Entering
salt mass = 35 lb. The
results follow a logarithmic
relationship.

EXAMPLE 4.10 — Removal ofCuS04from a Pollution Control Residue.

A pollution-control residue containing copper oxide was leached with sulfuric acid to produce
a slurry having a pulp density of 2.20. The wCuS04 in the solution phase of the slurry is 0.160.
The slurry is washed counter-currently in three stages with water. The slurry moving between
stages and discharge has the same volume fraction liquid as the entering slurry. The volumetric
flow of liquid moving counter-currently to the slurry is the same as the volume of wash water.
How much wash water should be used such that the discharge slurry contains 1.0 % of the
incoming amount of copper? The slurry solids have a density of 5.0.

Data. The approximate specific gravity of CuS04 solutions is:

/?CuS04 solution = 1 + wCuS04

Solution. Heretofore, we used a mass basis for material entering and traversing the system, but the
sometimes the flow between mixer-settler devices is specified on a volume basis. This occurs
when a pump is set for a certain volumetric flow. The assumption of conservation of volume is

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 209

good when the solution is dilute, but in this example, the incoming solution is not dilute, having a
density of 1.16. Even though a constant-volume flow is specified between devices, volume is not
necessarily conserved, so at some point, at least one stream volume must be left unspecified. In
other words, it is not possible to specify arbitrarily the volume of every stream of a process.

For streams where the volume is specified, the mass balance equations require a volume-to-
mass conversion. Please review Example 1.12 for conversion arithmetic between density, volume,
and mass of a slurry. Since no basis was specified, a volume basis of 1000 L input slurry is
chosen, which has a mass of 2200 kg.

The first step is to calculate the mass of each phase in the feed and discharge slurry. For the
feed slurry:

Mass balance: mass of liquid + mass of solid = 2200 kg

Volume balance: volume of liquid + volume of solid = 1000 L.

Recognizing that the volume is equal to mass/density, the above two equations can be solved
for the mass and volume of each phase of the feed slurry.

Feed slurry mass, kg volume, L mass fraction
solid 1354 271 1.0 solid
liquid 846 729
total 2200 1000 0.16CuSO4
0.3845 liquid

The feed slurry liquid contains 135.4 kg of CuS04 and 711 kg of H20. The discharge slurry is
specified to have the same liquid volume as the feed slurry (729 L), and its density is taken as
1.002. The discharge slurry therefore consists of 1354 kg solids, and 730 kg of solution (729 L)
containing 1.4 kg of CuS04 (wCuS04 = 0.001918) and 729 kg H20. The liquid portion of streams
В and С will also flow at 729 L. Similar considerations show that streams X and Y have a
volumetric flow equal to W. However, the volumetric flow of stream Z will be left as a variable
because of the prohibition against specifying the volumes of all streams. Figure 4.46 shows the
flowsheet.

Feed slurry m mw Wash water
FA = 2200 kg (1000 L) Fw = ?
wAsolid = 0.6154 II M/S II

и ^ С и 8 0 4 0.160 l '

Wash M/S I M/S III
solution

Washed slurry
FD = 2083 kg (1000 L)
wDsolid = 0.650
wD/LCuS04 = 0.00192

Figure 4.46 Flowsheet for washing leached CuS04 from insoluble residue. The values for wsolid
refer to mass fraction in the stream, while w#/LCuS04 refers to mass fraction in the liquid portion of
the # stream. Double arrows refer to slurry flow, and single arrows to solution flow. The solution

flow in the slurry streams is 729 L.

As in the salty slurry case (Figure 4.44, the overall system is underspecified, so we write
individual device mass balances. Note that F refers to the mass flow of the liquid phase.

CuS04 balance on M/S I: 135.4 + wII/LCuS04(FY) = wI/LCuS04(/^ + / ^ )
CuS04 balance on M/S II: 0.001918(Fx) + wI/LCuS04(7^) = wII/LCuS04(^ + FY)

210 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

CuS04 balance on M/S III: wII/LCuS04(i^) = 0.001918(Fx) + 1.4

Overall CuS04 balance: 135.4 = 1.4 + wI/LCuS04(^)
Overall mass balance: W + 846 = 730 + F1

We now need to replace the mass flow term with an equivalent volume flow term. For each
stream, F = Vp = V{\
(729 L). The flow of +strweaCmusSX04)a.ndInYad=dWitio, na,nwd eF1knisowtemFpfoorratrhiley liquid flow in each slurry stream
left as an unknown because the

liquid flow from at least one stream must remain unspecified.

M/S I: 135.4 + wII/LCuS04(W)(l + wII/LCuS04) = wI/LCuS04(7^)

M/S II: 0.001918(W)(1.001918) + wI/LCuS04(729)(l + wI/LCuS04) =
(wII/LCuS04)(W + 729)(l + wII/LCuS04)
M/S III: wII/LCuS04(729)(l + wII/LCuS04) = 0.001918(W)(1.001918) + 1.4
Overall CuS04: 135.4 = 1.4 + wI/LCuS04(7^)
Overall mass: W +116 = F1

These equations can be simplified somewhat to yield:

135.4 + W(C + C2) = B(W +116) + 729(B + B2)
0.001922W + 729(B + B2) = (W + 729)(C + C2)
729(C + C2) - 0.001922W + 1.4

134 = B(W+116)

where W = mass (or volume) of wash water, В = wI/LCuS04 and С = wII/LCuS04.

Any three of the above four equations can be selected for solving, and the fourth used as a
check. The arithmetic requires use of Excel's Solver tool. Each equation was rewritten such that
the left-hand side was zero, and the appropriate constraints were added to the Solver interface
screen. Table 4.14 shows the results. The volume of streams W, X and Y is 3063 L. The volume
of stream Z is 3050 L, which (as mentioned earlier) had to be calculated from the mass balance.

Table 4.14 Mass balance for CuS04 washing circuit (kg)

mass fraction of CuS04 M/S I M/S II M/S III
mass of wash liquid out 0.04215 0.00990 0.00192
mass of CuS04 in wash liquid
mass of slurry liquid out 3179 3093 3069
mass of CuS04 in slurry liquid 314 30.6 5.9
760 736 730
32.0 7.3 1.4

The results show that in order to remove 99% of the incoming CuS04 from the leach slurry,
F w = 3063 L. This dilutes wzCuS04 to 4.2 %. Because recovery of copper from such a dilute
stream requires additional treatment, it would be interesting to see the effect of additional washing
stages on copper level. Additional mass balance calculations were performed for four and five
stages, with the results shown in Figure 4.47. The copper content of the wash product solution
increases nearly linearly with number of stages. The result illustrates a classic economic trade-off:
the increased cost of adding and operating more stages against the lowered cost of treating a more
concentrated CuS04 solution.

Assignment. The slurry (streams B, C, and D) contains an equal volume of liquid and solid.
However, with increased settling time, or the use of a cyclone, it is possible to decrease the liquid
volume to half the volume of the solid. Calculate the amount of wash water required, F1, and
wzCuS04.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 211

CuS04 Leach Circuit Performance
3500

wash water
— I D - % CuS04

number
of stages

Figure 4.47 Effect of number of stages on 99 % recovery of incoming CuS04 from leach residue
slurry. The D points refer to the mass % of CuS04 in the wash solution stream Z, and the 0 points
refer to the volume of wash water in stream W.

Another class of counter-current processing is solvent extraction (also known as liquid-liquid
extraction). A solvent extraction (SX) circuit consists of a series of mixer-settlers through which
streams of aqueous and organic liquids flow counter-currently. In the loading (or extraction)
devices, a metal ion preferentially moves from the aqueous phase into the organic phase, such that
the concentration of the metal in the final organic phase is higher than in the entering aqueous
phase. In separate stripping devices, the metal is removed from the organic phase by reversing the
extraction reaction. The organic phase is then returned to the extraction circuit to recover more
metal. The organic extractant for copper is typically an oxime dissolved in kerosene. A two-stage
solvent extraction (SX) system is shown in Figure 4.48. Solvent extraction is used extensively in
the production of copper (-2.5 megatonnes of copper per year), and in other applications such as
waste minimization, recycling, and treatment of hazardous or radioactive effluents. Another
different application is the use of a slag or flux to extract an impurity from a molten metal.

Rich aqueous M/S #1 w M/S #2 Lean aqueous
w
<
Rich organic 4
Lean organic

Figure 4.48 Arrangement of mixer-settler devices for solvent extraction circuit. Liquids flow
counter-currently. Aqueous flow indicated by solid arrows, and organic by dashed arrows

The objectives of SX are two-fold. First, the concentration of the desired metal(s) is increased
several-fold, thus making it easier (and more economical) to recover it in metallic form. Second,
the organic extractant can be selected to favor extraction of the desired metal(s) while rejecting the
undesired ones. In Example 4.10, solvent extraction could be used to increase the mass fraction of
CuS04 to assist in the electrowinning of copper. Iron compounds are common impurities in
copper systems, and are leached if present in the solid. A properly selected solvent would be
highly preferential to copper (typically 500:1 Cu:Fe) while rejecting iron, thereby enhancing the
purity of the solution used for copper electrowinning and enhancing the energy efficiency of the
process.

The basis for copper solvent extraction for materials processing is the exchange of hydrogen
and metal ions between phases:

CuS04(aq)+ 2RH(org) — U2S04(aq) + R2Cu(org)

where R represents an organic molecule.

212 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

During loading, the organic phase transfers hydrogen ions to the aqueous phase in exchange
for metal ions. During stripping, the opposite occurs. Thus, the pH of the aqueous phase
decreases as metal is removed. The important factors involved in making a mass balance on a
solvent extraction circuit are the distribution ratio for metal concentration in organic/aqueous, and
the flow ratio of organic/aqueous phases. Typically, the flow is in volumetric units and
concentration in density units. The volume of the organic and aqueous streams is assumed
constant in SX calculations, even though this is only an approximation.

The most extensive use of SX is in copper production because of the trend away from
pyrometallurgical copper processing. The solutions produced by most leaching operations are too
dilute in copper (1 - 6 kg Cu/m3) and too impure (1 - 10 kg Fe/m3) for direct electrowinning.
Industrial copper electrowinning requires an electrolyte with >35 kg Cu/m3 and an acid content of
about 150-200 kg H2S04/m3. SX is used to concentrate the copper while rejecting the iron. In
summary, the organic extractant phase is:

(a) loaded with Cu from a weak H2S04 pregnant leach solution;
(b) separated from the pregnant leach solution;

(c) contacted with strong H2S04 electrolyte and stripped of most of its Cu.

The different H2S04 strengths of pregnant leach solution and electrolyte make the process
work. The most modern copper SX plants use only one SX stage, but Figure 4.49 shows a sketch
of a two-stage SX circuit for copper to illustrate multi-stage SX systems. Representative
conditions for operations at U.S. plants are shown below, in units of g/L.

Cu Pregnant leach sol'n. Raffinate Pregnant electrolyte Stripped electrolyte
1.5-3.5 0.1-0.4 30-45 25-40
H2SO4 1-4 155-165 170-190
3-6

Copper Waste
ore
Leaching Aqueous raffinate
to leach circuit
Pregnant Extractor Aqueous Extractor
leach solution II (impure, nearly free
of valuable metal)
(impure, dilute)
Spent electrolyte
IИ from electrowinning

Pregnant electrolyte Organic T (strongly acidified)
Organic J
to electrowinning (loaded)
(high in Cu) (stripped) i

Stripper Aqueous Stripper
I II

Electro- "► Copper
winning

Figure 4.49 Leach solution containing copper and iron enters Extractor I, and becomes
successively depleted in copper in passing through the extractors. Normally 90 % of the copper
and less than 0.1% of the iron is extracted. The copper leaving Extractor II is not lost because the
raffinate is recycled to the leach operation. The copper stripped from the loaded organic
replenishes the copper deposited during electrowinning. Small bleed streams are used to prevent
buildup of impurities. Dashed line is organic and solid line is aqueous phase.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 213

The distribution of copper between the organic and aqueous phases in both sides of the circuit
is affected by several factors, such as pH, copper concentration, and concentration of oxime in the
solvent. A distribution curve is determined experimentally by mixing different amounts of
aqueous, organic, and CuS04, adjusting the pH, and sampling the layers after equilibrium is
reached. A plot of the results is called a McCabe-Thiele diagram, which will have different lines
for different copper additions and pH. Each line is called an isotherm, and represents the expected
performance of the extraction circuit when operating at equilibrium in each stage. A McCabe-
Thiele diagram for a later example is shown in Figure 4.50, along with some construction lines
showing a graphical representation of the mass balance. An equation representing the isotherm is
shown as a text box on the diagram. A similar isotherm can be drawn for the stripping circuit, but
is not shown in this example.

The SX plant output is based on the demands of the electrowinning plant. The pH of the
pregnant leach solution (PLS) and the flows of each phase are adjusted to leach and extract an
amount of copper just matching the capacity of the electrowinning plant. The SX plant
specifications can be calculated by a mass balance on each of the devices if the concentration of
copper in the PLS, spent electrolyte (SEL) and pregnant electrolyte (PEL) are known.

Consider an SX plant designed to feed 100 tonnes/d of copper to the electrowinning plant.
The flows of organic and aqueous phases in the extraction side are the same. The distribution of
copper between the organic and aqueous phases is expressed by:

Extraction distribution equation: kg/m3 Cu organic = 3.79(kg/m3 Cu aqueous) *

Stripping distribution equation: kg/m3 Cu organic = (kg/m3 Cu aqueous - 20.6)/10.6

These distribution equations were developed from laboratory data on one organic reagent and
over a rather narrow range of pH, so are not generally valid outside the range of conditions used
for their determination.

For simplicity, the mass balance equations for each device are expressed in units of kg Cu/m3,
at a flowrate of 1 m3/h for the organic and aqueous phases in the extraction side. The volume
flowrate of aqueous in the stripping side (designated E) is dependent on the mass balance
calculation. Once made, the mass balance results can be scaled for the specified copper production
rate. A DOF analysis for each device and the overall process is as follows:

• There are 10 streams, and each stream has 2 components: Cu and "the rest of the stuff
(either organic or aqueous). Therefore, SV for each device is 8, and overall, the SV = 20.

• For each device, IB = 3 (Cu, organic, or aqueous). However, for the overall system, IB is not
12 as one might think, but 11, because the organic flow is continuous around the system, so
only 3 of the 4 organic component balances are independent. This is because the 4th balance
is obtained by difference from the other 3. The number of flows specified is 2 (organic and
aqueous in the extraction side), so F = 2 for each device and overall.

• The number of restrictions for each device is 1, since the distribution equations allow us to
represent the copper concentration in the organic in terms of copper in the aqueous. So SR =
1 for each device, and SR = 4 for the overall system.

The DOF table for the devices and system is shown below. The system would be specified if
three more data items were provided.

Stream Variables Extractor I Extractor II Stripper I Stripper II Overall
Components 8 8 8 8 20
Flows 3 3 3 3 11
Restrictions 2 2 2 2 2
1 1 1 1 4
Degrees of Freedom 2 2 2 2 3

214 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
Referring to Figure 4.49, a mass balance for copper is written around each device for 1 h of

operation. A and R refer to the copper concentration (kg/m3 Cu) in the aqueous and organic phases
respectively, E refers to the flow of the (aqueous) electrolyte, PLS refers to the kg/m3 Cu in the
pregnant leach solution, and SEL and PEL refer to the kg/m3 Cu in the stripped and pregnant
electrolyte, respectively. Note that PEL is A111.

Extractor I Cu in Cu Out
Extractor II A' + R1
Stripper I PLS + R"
Stripper II A" + R"
A' + RIV
PEL(E) + R111
AIV(E) + R1 AIV(E) + RIV
SEL(E) + Rni

The following subsidiary relationships relate the aqueous to organic stream compositions,
based on the distribution equations:

R^J^A1)0·217

Rn = 3.79(An)0·217

Rin = (PEL-20.6)/10.6

RIV = (AIV-20.6)/10.6

Thus, there are 10 stream concentrations and one phase volume E to be determined, and eight
equations. If the PLS, PEL and SEL are specified, the equation set can be solved. These
relationships were appropriately written for Excel's Solver tool, with cells for PLS, PEL and SEL
to be filled in as desired. One set of mass balance equations was solved for SEL at 34 kg/m3 Cu,
PEL (A111) at 46 kg/m3 Cu, and PLS between 2.6 and 3.0 kg/m3 Cu. The results for PLS = 2.8 are
shown in the following table. SO stands for stripped organic, LO stands for loaded organic, E is
the electrolyte volume, and O/A (strip) refers to the flow ratio of organic to aqueous phases in the
stripping side.

A1 A" (raff) AIV R1 (LO) R" RIII RIV (SO) E O/A (strip) Cu recv'd. Cu recv'd.
1.486
0.2544 37.73 4.130 2.816 2.392 1.584 0.2095 4.77 90.9 % 2.55 kg

For an electrowinning plant requiring 100 tonnes of copper daily, the flow rate of PLS should
be 1630 m3/h, and the flow rate of electrolyte 340 m3/h.

The extraction results can be expressed graphically on a McCabe-Thiele diagram, as shown in
Figure 4.50. This diagram shows the isotherm obtained by the extraction distribution equation as a
heavy solid curve. The dash-double-dot line is the operating line, and has a slope equal to the O/A
ratio for extraction (1.0). The horizontal dashed line segments represent the change in copper
concentration in the aqueous phase (PLS at 2.8 to 1.49, then to raffinate at 0.26). The vertical
dashed line segments represent the change in copper concentrations of the organic phase (LO at
4.13 to 2.82).

A variation in PLS copper will change all of the mass balance results. Simulations were
performed with the SuperSolver tool for four different values of PLS, with the results shown in
Figure 4.51. As the PLS went from 2.6 to 3.0, the fraction of copper recovered from the PLS went
from 93.7 % to 89.2 % (but remember, this copper is not "lost" since the raffinate is returned to the
leach plant). If the PLS copper increases while the electrowinning plant is operating at maximum
capacity, the flow of PLS must decrease.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 215

5.0 McCabe-Thiele Diagram for SX of Copper

4.5 LJ-Ì-*-—1
**0*0000*0000^'
\ .XI

4.0 Γ^

3.5

:>■■"' 1
1»
3.0 X"—
о •к"«•**

.£ 2.5 »·

5 2.0 / *>* · \s- Cu in organic = 3.79(Cu in aqueous) '

I/ ! ! 1 ! : 1 : :: ■
I// J*\ \^
1.5 1^**

1.0

0.5 1 Ii i
0.0 1
1 ;

1 ii1 1

■ 0.5 1.5 2.5
Cu in aq

Figure 4.50 McCabe-Thiele diagram for a two-stage process for solvent extraction of copper from
a pregnant leach solution. The heavy curved line is the isotherm for certain unique conditions of
pH, organic reagent, and copper levels. The dash double-dot line is the operating line with slope of
1.0, and the dashed line represent changes in concentration of copper in the two phases as
extraction occurs.

The number of stages for copper extraction is seldom more than two, and the most modern
plants use just one because the raffinate is recycled back to leach more copper, so the unrecovered
copper is not lost. In other applications, there may be several stages to increase metal recovery.
The most efficient system for SX is a tall column in which droplets of one phase sink while the
other phase rises. For a sufficiently tall column and high surface area between the phases, this
arrangement approaches the extraction efficiency of an infinite number of stages. Unfortunately,
column contactors do not handle solids well and are thus not used in conjunction with large ore
leaching operations.

216 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

5.4 Mass Balance on Cu SX Circuit 2.6

5.2 - О - О/А Strip σ>
- a — kg Cu recovered
5o. ж

• 5.0 2.5 I
2.4 EОоФо>з>
<
2.3
О

2.4 2.5 2.6 °" '" 2.7 2.8 2.9 2.2
3

Figure 4.51 Effect of changing the PLS from 2.6 to 3.0 kg/m3 Cu. An equation fitting the line
expressing kg Cu recovered is: kg Cu = 0.718(Cu in PLS) + 0.528.

4.13 Using FlowBal for Material Balance Calculations

Heretofore, calculations have been made using Excel, its calculational tools (Goal Seek and
Solver), and enhancements of these two programs (SuperGS and SuperSolver). However, even
with these tools, the arithmetic of setting up and solving a material balance on a multi-device
process is tedious. We now introduce another computational aid, FlowBal, which is a non-
sequential material and heat balancing program. The program (FlowBal.xls) is on the Handbook
CD, together with a User's Guide and many examples of its application. A separate Excel 2007
version is also included FlowBal.xlsm). A brief description of the program is given in the
Appendix. FlowBal's advantage is most obvious when the stream data is given in mass or real
volume units. FlowBal converts the units to amount, makes the calculations, and returns the result
in both the original units and as amounts, or as mass, if so selected.

FlowBal requires a different way of looking at a process, and time and effort are required to
learn the program. In some cases, the process information needs editing before it can used. Using
FlowBal requires the same understanding of degree of freedom, process chemistry and material
balance concepts required by the other Excel tools. In addition, if FlowBal's starting estimates for
stream flow and composition are too far from the actual values, Solver may not find a solution. If
this happens, enter better starting estimates and try again.

We introduce FlowBal by applying it to three systems similar to those examined in earlier
sections of this Chapter. The first system is a two-stage process for splitting a sulfide roasting gas
to a stack and an acid plant stream. The second system is a multi-stage process for sludge washing
with recycle streams. The third system shows how FlowBal treats paired streams to simulate
multi-phase streams. A good way to get practice with FlowBal is to use it on some of the
Handbook examples worked out earlier in the Chapter.

It's important to read the first part of the FlowBal User's Guide before studying the following
examples because the example description assumes that you have read the Guide, and have tried
using FlowBal on a couple of the easy examples in the Guide. What's more, the User's Guide
assumes you have read enough of the Handbook to understand the basics of material balances.
Before reading these examples, open FlowBal and review the first LiB04 example worksheet.
Then, with FlowBal open, try working the following three FlowBal examples as you read the text.
FlowBal uses the database program FREED, which is on the Handbook CD. FREED is described
briefly in the Appendix, and in more detail in its user guides.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 217

4.13.1 FlowBal Example #1: Mixer/Splitter

The first FlowBal example is a material balance for a process to split the offgas from a copper
concentrate roaster into two streams. Figure 4.52 shows the flowsheet. The plant has a permit that
allows emission of 36 tonnes/day of S02 to a tall stack. Roaster gas in excess ofthat limit must be
sent to a sulfuric acid plant for complete conversion and recovery. Complete oxidation of S02 at
the acid plant requires that the amount of 0 2 and S02 in S5 be the same. The material balance
objective is to calculate the fraction of SI that is split to the stack and the flowrate of air to the gas
mixer. The flowsheet is very simple: only two devices, four species, and five streams. The
pressure is 1 atm throughout, and the stream temperatures are shown below.

Stream number 12 34 5

Stream temperature, К 550 530 530 300 490

Roaster gas Stack gas
F1 = 1000 m3
Acid plant gas
p S 0 2 = 0.113 F5 = ?
φΗ20 = 0.084
φΝ2 = ?

Figure 4.52 Flowsheet for splitting offgas from a copper concentrate roaster to a stack gas and
sulfuric acid gas streams. Flow rate is per minute.

It's a good idea to calculate the DOF before starting FlowBal. First, consider the splitter
device — the composition of S2 and S3 are the same as SI. If the mass of S02 in S2 is stated, the
splitter is completely specified (DOF = 0). Therefore, it is incorrect to state two species amounts
for S2 because that would cause DOF for the splitter to equal - 1 , which would over-specify the
splitter. The gas mixer has DOF = 1, which is converted to zero by stating that <p02 = <pS02 in S5.
This brings the system DOF = 0.

The first step is to insert a blank worksheet into the FlowBal workbook and name it something
like RstrGas. Sketch the flowsheet entirely in widened row 2. Next, check the New/Edit box on
the floating FlowBal toolbar, and add information to the Process Parameters input screen,
including a brief description of the process, and the number of species (here, four). Enter the
stream data for the five streams (stream number, name, and quantity or composition units for the
species in that stream). Volume percent species composition is used for streams 1, 3, 4, and 5, and
mass as the species composition unit for S2. Click on Apply Adds/Edits to enter the information
on the worksheet and click OK/Exit to return the worksheet with a starting array displayed.
Finally, enter the chemical formula for each species via the keyboard in column B.

Click the Add/Edit Devices button and add a splitter device. Leave the split fractions at ?
because they're not stated variables. Add a mixer, select streams, then click Apply Adds/Edits.
The worksheet display generated after these steps is called the empty (initial) setup array.

Add known flowrate and stream composition data to the empty setup array, with ? entered for
unknown values. On the selected basis, the mass of S02 in S2 is 25 kg/min. The filled setup array
is shown below. The Str-unit and Spec-unit cells should not be changed on the worksheet, but all
other cells are editable. A ? means that the value is not explicitly stated and remains to be
calculated by FlowBal. A blank means no amount ofthat species is in the stream.

There are 15 unknowns listed (four stream flows and 11 compositions). Although the <pN2
values are readily obtained by subtracting from 100, it is better to let FlowBal calculate them.
When FlowBal calculates the unknown stream composition by subtracting from the total (one or
100), it displays the result in an italic font in the filled input array. Since we can calculate two ?

218 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

stream compositions that way, there are only 13 real unknowns (the four stream flows and nine
compositions) to be calculated by FlowBal for a material balance.

P (atm) 11111

T(K) 550 530 530 300 490

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

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

Str-name RsterGas StackGas SplitGas Air AcdPltGas

Streams 1 2 3 4 5
? ? ? ?
Flow 1000
? ?
S02 11.3 25
? ? ?
H20 8.4
N2 ? ? ? ? ?
02 21 ?

Gas Splitter (SP) Heat Gas Mixer (MX)
Instreams Outstreams Split Fract I Instreams Outstreams Heat

12? 35
3? 4

Before going on to solve the system, look carefully at the way FlowBal treats certain system
characteristics. First, note that the species unit for S2 is a quantity, not a composition. It's
absolute mass, not mass percent or mass fraction. Second, note that the stream unit for streams 1,
3, 4, and 5 are volume. FlowBal uses the entered stream temperatures and pressures, not STP, so
the volume of streams 1, 3, 4, and 5 is at actual conditions.

Click Read Data, and FlowBal will write equations in terms of the unknown stream
properties. Some of the equations are species material balances around a device, while others sum
the compositions. FlowBal writes 14 equations for the 15 unknowns, creates starting estimates for
the unknowns, and fills in the φΝ2 values for SI and S4. For the 15th equation, enter the stated
relationship φ02 = <pS02 in S5 with the Insert Equation tool, to give DOF = 0. This equation
appears as a comment in cell E36:

[{S5-SQ2} - {S5-Q2} |

Now click Solve to invokes Solver, which finds a solution for the 15 equations, as shown in
Figure 4.53. FlowBal also displays two other result arrays: an amount array and an element
balance. The last two entries in the Results column (cells C35:C36) are the split fractions for the
splitter exit streams. S2 contains 15.6 % of the amount (or mass) of SI.

One of the major advantages of using FlowBal for material balance calculations is that once
the system is set up and a balance is reached, it can be changed to examine how one variable
affects the others. Suppose we wanted to know the mass of S02 emitted by the stack over a roaster
gas flowrate from 900 to 1100 m3/min, at the split fraction to S2 of 0.156. Click on Erase
Previous Results, enter this split fraction on the worksheet (cell D17), enter a ? in place of the 25
mass amount for S02 in S4, and click Read Data. Now use FlowBaPs Repetitive Solve feature
for five solves. FlowBal creates a table, a chart, and fits a linear (three or less solves) or quadratic
equation (four or more solves) to the solved points. The following table is derived from that
operation.

Sl-Flow,m3/minat550K 900 950 1000 1050 1100

S2-S02, kg/min 22.5 23.8 25.0 26.3 27.5

According to the FlowBal generated equation, the mass of S02 in S2 equals 0.025(vol. SI
flow). This simple relationship is the result of fixing the split fraction.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 219

P(atm) 11111

T(K) 550 530 530 300 490

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

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

Str-name RstrGas StkGas SpltGas Air AcdPlntGas

Streams 1 2 3 4 5

Flow 1000 107.9 813.4 247.8 1156.7

S02 11.3 25 11.3 0 7.35

H20 8.4 5.23 8.4 0 5.46

N2 80.3 77.68 80.3 79 79.85

0 2 0 0 0 21 7.35

Stream Amounts (kg-mol)

Str-name RstrGas StkGas SpltGas Air AcdPlntGas
Streams 1 2 3 5
4
Flow 22.157 3.453 18.704 10.065 28.769
S02 2.504 0.390 2.114 2.114
H20 0.290 1.571 0 1.571
N2 1.861 2.773 15.019 0
02 17.792 7.951 22.970
0 0 2.114 2.114
0

Figure 4.53 FlowBal output arrays for roaster gas split process. Top half of output in original
entry units. The air requirement is 248 m3/min, measured at 300 К and 1 atm pressure. <jp02 =
<pS02 = 7.35 % in S5 as indicated by the bordered cells. FlowBal calculates the split fraction in the
last two cells of the Results column. The split fraction of SI to S2 = 0.156.

4.13.2 FlowBal Example #2: Evaporation/Condensation Process

A material balance often involves the condensation or evaporation of a species. Chapter 2
covered the concepts involved in the vaporization and condensation of substances, especially
water. Equation [4.3] shows the type of equation used to express the phase change for the
evaporation of water. FlowBal uses descriptors to distinguish between the physical forms of a
substance, and phase transformations are classified as "reactions" for FlowBal purposes*.
Equations [2.15] or [2.16] can be used to calculate the/?H20(g) in equilibrium with H20(/).

H2O(0 - H20(g) [4.3]

FlowBal creates one additional system variable for each device reaction, designated on the
worksheet as R-RN, where the N descriptor refers to the reaction number in the input array. The R-
R term is a measure of the rate of progress of the reaction, or the molar rate of reaction. Thus R-
Rl refers to the molar rate of reaction #1 as entered on the FlowBal worksheet. This term will be
discussed in more detail in Section 5.5.1, or see the FlowBal User's Guide.

Consider a process for the drying of a sulfide filter cake in a tunnel kiln. Example 2.7
discussed drying of clay in a single device, and Figure 4.26 discussed drying in a series of staged
separators. Figure 4.54 shows a tunnel kiln simulated by using three-stages. The filter cake moves
through the system on pallets that can be heated from below to provide the necessary thermal
energy for drying. The temperature of the dry pyrite should remain below 350 К to avoid incipient
oxidation, and dried to a wH20 of 1.5 % (a slight amount of residual moisture minimizes the lifting
of dust into the air stream). Furthermore, the air flow must be sufficient to maintain 70% RH in S7
to provide an adequate driving force for water evaporation in Dryer I. The air temperature is
assumed to drop uniformly from its entering temperature of 350 К to 320 К in S7, while the pyrite
temperature increases uniformly from 300 to 330 K. This gives a constant 10 degree temperature

* The systems in this Chapter have physical (i.e., phase change) reactions, but not chemical reactions.

220 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

difference between the pyrite and air streams leaving each dryer. The process basis is 100 kg/min

wet pyrite having wH20 = 12.0 %. Finally, each dryer is assumed to remove half the water
contained in the entering pyrite, which brings the dry pyrite to wH20 = 1.5 %.

Dry air. Moist air.
7 F7 = ?
F1 = ? m 1
^Н20 = ?
pH20 = 0.0154 Dryer III Dryer II Dryer I
2 Wet pyrite.
Dry p y r i t e . 8 F2 = 100kg
F8 = ? kg Λ wH2O = 0.12

ivH20 = 0.015

Figure 4.54 Flowsheet for three-stage simulation of continuous pyrite drying process. Dashed
lines (odd-numbered streams) indicate gas flow, while solid lines (even-numbered streams)
indicate pyrite flow.

The first step is to calculate the/?H20 for S7 (at 320 K) from Equation [2.15]; at 100 %RH,
<pH20 = 10.1%. At 70 %RH, <pH20 in S7 should be 7.07 %. The next step is to fill in the input
array with the flow and composition of all streams, and the reaction coefficients. For this example,
it's best to take species formulas from the database to be sure the descriptors match between the
database list and the input array. Since only moisture is transferred to the gas phase, we can use
FREED's empirical formula for air. FlowBal treats phase transformations as reactions, so enter
specifications for all three devices (reactors) and indicate the existence of a reaction in each device.
The input array is shown in Figure 4.55. Note that the R #1 column indicates the evaporation of
one mole of moisture from liquid water to vapor as indicated by Equation [4.3].

There are 11 unknown stream variables (not counting the ones that can be obtained
immediately by difference), and three R-R values (R #1 in each reactor). FlowBal wrote 14
equations for these 14 unknowns, and Solver found a solution, as shown in Figure 4.56. The
process requires 286 m3/min of air at 350 К (223 m3/min at STP). The air requirement is very
sensitive to the temperature profile along the drying system. If all stream temperatures were
increased by 10°, at 100 %RH, φΗ20 = 16.5% at 330 K. At 70 %RH, φΗ20 in S7 should be 11.55
%. This requires 154 m3/min of air at 360 К (117 m3/min at STP). The lower air flow rate
minimizes the carryover of dust from the drying pyrite.

P (atm) 11111111

T(K) 350 300 340 310 330 320 320 330

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

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

Str-name DryAir WetPyr Air-1 Pyr-1 Air-2 Pyr-2 MoistAir DryPyr

Streams 1 2 3 4 5 6 7 8 R#1
? 100 ? ? ? ? ? ?
Flow
? ? 7.07 1
H20, (g) 1.54

H20, (liq) 12 6 ? 3 1.5 -1
? ?
N1.580.42, (g) ?

FeS2, (pyr) ? ? ? ?

Instreams Dryer III (RX) Heat Dryer II (RX) Heat
Outstreams Reactions I Instreams Outstreams Reactions
1
6 31 3 51
8 46

I Dryer I (RX) Heat
I Instreams Outstreams Reactions

241
57

Figure 4.55 Input array and device display for the three-stage drying of pyrite.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 221

P (atm) 11111111

T(K) 350 300 340 310 330 320 320 330

Str-unit Volume (m3) Mass (kg) volume (m3) Mass (kg) volume (m3) Mass (kg) Volume (m3) Mass (kg)
Spec-unit Volume pet Mass pet Volume pet Mass pet Volume pet Mass pet Volume pet Mass pet
Str-name
DryAir WetPyr Air-1 Pyr-1 Air-2 Pyr-2 MoistAir DryPyr
Streams 1 4 5 7 8
2 3 6

Flow 285.6 100 279.6 93.6 275.7 90.7 276.6 89.3

H20, (g) 1.54 0 2.29 0 3.84 0 7.07 0

H20, (liq) 0 12 0 6 0 3 0 1.5

N1.580.42, (g) 98.46 0 97.71 0 96.16 0 92.93 0
FeS2, (pyr) 0 88 0 94 0 97 0 98.5

Dryer III Dryer II Dryer I
R-R1 R-R1 R-R1
0.0767 0.1607 0.3543

Figure 4.56 FlowBal results array for the drying of pyrite. The air flow values are actual, not
STP. The R-R values are the molar rates of water evaporating in each device.

4.13.3 FlowBal Example #3: Systems with Multi-Phase Streams

Processes sometimes contain streams with two or more phases, such as liquid plus solid, or
gas plus solid. When the stream entity is a slurry (an aqueous solution and a solid), the usual
practice is to specify the composition of the aqueous phase separately from the composition of the
solid phase*. FlowBal requires that the aqueous phase be designated as a stream separate from the
solid phase. The solid and aqueous phases move as a single entity, so the solid and aqueous
streams are said to be paired streams. Please review the User's Guide for a more complete
discussion of the treatment of multi-phase streams.

The example for clarifying the paired-stream concept is a version of a process we've looked at
several times already in this Chapter: the leaching of a soluble salt from an insoluble solid
(Sections 4.6.3, 4.7.2, 4.7.4, 4.12, and Example 4.4). Here, a salty dross from an aluminum
recycling plant is leached to separate the soluble chlorides from the insoluble substances. The
leachant is mainly water, but contains some salt. There are no chemical reactions taking place.
The objective is to determine the overall percent recovery of NaCl and KC1 from the dross +
leachant to the supernatant liquid. Figure 4.57 shows a sketch of the flowsheet.

Leachant В 1Thin slurry
Dross
Leach Thickener
tank -► Supernatant liquid
Thick slurry

Figure 4.57 Flowsheet for dross leaching process. The leached dross (as a thin slurry) is passed to
a thickener (sometimes called a settling tank) where the insoluble portion of the dross is removed
as a thick slurry (please review Figure 4.19 with slightly different terminology). The supernatant
liquid (sometimes called the decant liquid) goes to an evaporation pond where the salt crystallizes
out and is recycled. The dashed lines indicate liquid phase, the solid line indicates solid phase, and
the double line indicates a slurry stream. Since slurry streams have two phases, they should be
recast as two (paired) streams for calculational purposes. Streams С and E are really a "pair" of
streams (one solid, one liquid). Stream D and the liquid portion of stream E have the same
composition as the liquid portion of stream C.

Here we use a loose definition of a phase. The term "solid phase" in reference to FlowBal refers to any
mixture of solid substances. This situation is preferably called a "solid phase assemblage".

222 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

The composition of the leachant and dross is listed below. The mass ratio of liquid to solid
phase in the thick slurry is 0.925.

Leachant: w H 2 0 - 9 4 . 5 5 % ; wNaCl = 3.69%; balance KCl

Dross: wNaCl = 38.65 %; wKCl = 24.66 %; wAl = 13.63 %; balance A1203

The flowsheet was redrawn to show the slurry stream as what it really is — two paired
streams, and to correctly designate the thickener for what it really is — a splitter for the liquid
phase. Figure 4.58 shows the revised flowsheet with labeled input stream compositions. Please
see the text that describes Figure 4.15 and 4.16 for further clarification of the paired-stream
concept.

The DOF should be calculated first to be sure the system is correctly specified. Starting with
the leach device, SV = 12, IB = 5, 1С = 5, and F = 2, so DOF = 0 for the leach device. For the
splitter device, SV = 9, Ш = 3,1С = 0, and F = 0. There is a special splitter SR of (N - 1)(S - 1) =
2. This gives DOF = +4, so the splitter underspecified. For the overall process, SV = 18, IB = 8,
1С = 5, and F = 2. SR for the splitter = 2, and there is one additional SR because we specified the
flow ratio of the two thick slurry (paired) streams. This gives DOF = 0, so the system is properly
specified.

Leachant F'sliMOkg/h

wH20 = 94.55% 1 Thin ^Supernatant
slurry liquid
wNaCI = 3.69%; wKCI = ? Liquid
4 splitter ► Thick
— 2 * l Leach slurry
Dross F2 = 550kg/h
tank
wNaCI = 38.65%

wKCI = 24.66%

wAI = 13.63%; wAI203 = ?

Figure 4.58 Redrawn flowsheet for FlowBal purposes. Compare to Figure 4.57. The thickener is
simulated as a liquid-phase splitter. The "Thin slurry" (streams 3 and 4) is former stream C, and
the "Thick slurry" (streams 3 and 6) is former stream E. Each slurry stream is simulated as two
paired streams, one liquid and one solid. This is the way FlowBal tracks the composition of the
liquid and solid phase. Stream pairing is depicted diagrammatically by enclosing the streams
within a cylindrical shape.

The input array display is shown below. The splitter fractions were left as unknowns.

P(atm) 111111

T(K) 300 300 300 300 300 300
Str-unit Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg)
Spec-unit Mass pet Mass pet Mass pet Mass pet Mass pet Mass pet
Str-name Leachant Dross ThnSlrSId ThnSlrLiq SpntntLiq ThkSlrLiq

Streams 1 2 3 4 5 6
? ? ? ?
Flow 1240 550

H20 94.55 ???

NaCI 3.69 38.65 ???

KCl ? 24.66 ???

Al 13.63 ?

AI203 ??

Leach Tank (RX) Heat Liquid Splitter (SP) Heat
Instreams Outstreams Reactions Instreams Outstreams Split Fract

13 45?
24 6?

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 223

There are 15 "true" unknowns in the array table, and two unknown splitter Split Fractions.
FlowBal wrote 16 equations, leaving the user to supply the 17th. For this, write an SR equation for
the splitter device L/S ratio between streams 3 and 6, and insert the equation via the Insert
Equation feature. Even though a flow ratio was listed for the SR, don't use a ratio in any FlowBal
equation. This is because if Solver tries a zero for a denominator value during the solve, it will
cause a #DIV/0! error and immediate exit from Solver. Convert ratios to products for FlowBal
equations (or for that matter, for any Solver equation). The SR equation is best written as:

0.925*{S3-Flow} - {S6-Flow}

The mass unit result table is displayed below. The dross has mNaCl = 212.6 kg and mKCl =
135.6 kg. The leachant has mNaCl = 45.8 kg and wKCl = 21.8 kg. The supernatant liquid
contains wNaCl = 227.9 kg and mKCl = 138.9 kg. This is about 88 % of the total input NaCl and
KC1. Stream 6 has 14 % of the salt originally present in the dross. More of the salt could be
recovered in the supernatant liquid if the L/S ratio of the thick slurry could be decreased, or if it
was filtered, or washed counter-currently in a separate operation.

P (atm) 1 1 1 1 1 1

T(K) 300 300 300 300 300 300

Str-unit Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg)

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

Str-name Leachant Dross ThnSlrSId ThnSlrLiq SpntntLiq ThkSlrLiq

Streams 1 2 3 4 5 6

Flow 1240 550 201.8 1588.2 1401.5 186.7

H20 94.55 0 0 73.82 73.82 73.82

NaCl 3.69 38.65 0 16.27 16.27 16.27

KCI 1.76 24.66 0 9.91 9.91 9.91

Al 0 13.63 37.15 0 0 0
о
AI203 0 23.06 62.85 0 0

4.14 Continuous-Mixing Devices

The passage of material through any type of device may involve one or more of the following
four mechanisms: displacement, mixing, short-circuiting and stagnation. An example of perfect
displacement is plug flow, such as solid objects moving through an annealing furnace. In that case,
each object remains in the reactor for the same amount of time. Plug flow is approached in many
materials processing systems such as shaft furnaces and trough reactors.

In contrast, many other processes are carried out in batch or semi-continuous devices. Even
steady-state processes can be temporarily transient, such as during startup, change from one steady
state to another, or fluctuations in raw material properties. Even then, an adequate understanding
of the process can often be attained by taking averages over long time periods. Where this is not
adequate or feasible, we must consider accumulation, flow variations and composition change with
time as factors in the balance. Here we will look at some simple examples of steady-state and
transient behavior in well-mixed reactors where time is an important independent variable.

4.14.1 Steady-State Processes

Consider a mixing tank fed with a stream of water and a small amount of fine solid material.
The powder and water are assumed to be perfectly mixed immediately after they enter the tank.
Under these conditions, each increment of feed is uniformly distributed throughout the tank the
instant it enters; the mass fraction of powder throughout the tank and in the effluent is exactly the
same. Such a device is called a continuously-stirred tank reactor, or CSTR. A similar device for
gas-solid reaction processes is a fluid-bed reactor, where granular material is agitated by a rising
gas. The condition of continuous flow and uniform mixing results in different retention times for
each particle of powder or molecular entity of the liquid. The mean retention for residence) time
(Θ) is given by:

224 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

θ = quantity in system
rate of treatment

This gives an indication of the typical retention time of a particle in the reactor, although the

actual retention time varies from zero to infinity. Since the extent of a reaction is usually

dependent on the length of time that substances are in

Powder [ contact, some reactants exit unreacted, while others have

Water с undergone complete conversion.

Any consistent set of units may be used to calculate Θ.
For example, if the mass of material is expressed in
kilograms, the rate of treatment might be kg/h, and Θ would
be in hours. The sketch to the left shows the condition of
perfect mixing, where the actual retention time for a given
particle can be anything. The mean retention time for this
case is V/v, where V is the volume of the tank, and v is the
volumetric discharge flow rate. To further clarify the
concept of retention time, let us assume we are able to
identify certain otherwise-identical particles by color or magnetic property. Suppose that at time t
= 0 we introduce n0 identifiable particles as a replacement for the same number of ordinary
particles flowing into the tank. Assume the time required to make this addition is extremely short.
Let n equal the number of identifiable powder particles in the tank at time / (n = n0 at t = 0). Then
the concentration in the tank and the effluent at any time t = nIV. The number leaving the tank in a
time increment dt is -dn, which must equal the product of the effluent volume vdt.

-dn = —v-dt [4.4]
V

Integrating, and remembering that n = n0 at / = 0,

n^t = -v-t [4.5]

eУ

n„

Since Vlv is the mean retention time Θ defined earlier, and nt/n0 may be defined as fT9 the
fraction of the newly added particles remaining in the system at time t, then:

fr=e -t/ [4.6]

Thus/ varies from 1 at t = 0 to 0 at t = oo. Accordingly, if a definite retention time is required
to complete a chemical reaction, some amount will pass through only partially reacted while
another amount is retained longer than necessary. If we are interested in the fraction of particles
that remain in the tank for a certain span of time, we calculate the difference between f for the
upper and lower time limit, and express the difference as a percentage of particles that have a
certain retention-time span.

Better control over the retention time is attained by placing multiple CSTRs in series. The
outlet from the first reactor becomes the inlet to the second, and so on. As the number of CSTRs
increases, the system approaches plug flow. The equation for two CSTRs in series is:

fr ■\Atie)e /θ [4.7]

where both reactors have the same Θ.