Chapter 3 Statistical Concepts Applied to Measurement and Sampling 125
2n, and the total number of effects that can be estimated are T - 1 (these include main effects, two-
factor interactive effects, three-factor interactive effects etc. up to the «-factor effect).
Before moving on, let's look at this example as a regression problem. Actually, we can
perform the regression two different ways. If we use the original data to estimate the model yield
= b0 + b{T + b2? + b3C, then we get the formula yield = -21 + 0.93T + 3.5P + IOC. In this case,
the interpretation is that each increase in temperature of 1 °C between 50 °C and 80 °C results in
an increase in yield of 0.93%; an increase in pressure of 1 atmosphere results in an increase in
yield of 3.5%; and an increase in catalyst loading of 1% results in an increase in yield of 10%. In
this interpretation, the intercept of-21 is meaningless.
On the other hand, we can do regression where each independent variable is recorded as either
+1 or - 1 . The resulting formula is yield = 67 + \At + 1.75/? + 5c. We used lower-case letters here
because the meaning of the variables has changed. For instance, t no longer measures temperature
directly; / = -1 corresponds to T = 50 °C, while t = +1 corresponds to T = 80°C. Thus, an increase
of 1 unit in t causes an increase of 14 % in yield, so an increase in t from -1 to +1 increases yield
by 28 %. This is exactly the effect of temperature that we found before. Note also that an increase
of 1 unit in / corresponds to an increase of 15 °C in T, and 14/15 = 0.93, so the physical
interpretation of the original model can be recovered from this second regression. The
interpretation ofp and с are similar. Worksheet TPC shows the regression tool calculations.
In either case, the model as a whole is statistically significant (p-value < 0.01), as are the
coefficients for T (or t) and С (or c) (again, /rvalues < 0.01). However, the coefficient for P (or p)
has ар-value of 0.088, between 0.05 and 0.1. This suggests strong evidence that temperature and
catalyst loading affect yield, but only marginal evidence that pressure does. Also, each model has
R2 = 0.99. Recall the interpretation of R2, which says here that 99% of the variation in yield can be
explained by the variables T, P, and С We will return to this point shortly.
First, let's look in some detail at a two-level, four-factor experiment for polishing a ceramic.
An engineer identifies four potentially important factors that affect the quality of a polishing
process: speed of the polisher ( ф т ) , the feed rate of the raw material (cm/sec), the depth of
polishing (cm/sec), and the type of abrasive used. Her quality response variable is a "coefficient of
surface roughness" (R). She sets up a two-level experiment, using the levels of Table 3.24. Since
there are four factors, each with two levels, she makes 24 = 16 experimental runs. For
convenience, we use the following variables: W = speed, X- feed rate, Y = depth, Z = abrasive
and R = roughness parameter.
Table 3.24 Factors in the polishing experiment.
Factor Low level (-1) High level (+1)
Speed 16 24
Feed rate
Depth 0.001 0.003
Abrasive 0.01 0.02
Carborundum Diamond
There are 15 possible measurable effects with these runs. Four main effects (speed, feed rate,
depth, and abrasive); six two-factor interactions (speed*feed rate, speed*depth, speed*abrasive,
feed rate*depth, feed rate*abrasive, and depth*abrasive); four three-factor interactions (speed*feed
rate*depth, speed*feed rate*abrasive, speed*depth*abrasive, and feed rate*depth*abrasive); and
one four-factor interaction (speed*feed rate*depth*abrasive). Her results are in Table 3.25.
Readers who want more Excel guidance than appears here are referred to worksheet "Polishing"
for additional explanation.
The first step is to use contrasts to estimate the size of each of these 15 effects. Although
tedious, it is not difficult to create columns of ±l's for each of the interactions. This is simply a
matter of writing Excel formulas to multiply appropriate columns for the original factors; as usual,
you can write them for the first row of the matrix, then copy them down to complete the column.
126 Chapter 3 Statistical Concepts Applied to Measurement and Sampling
Once this is done, we use =SUM(e/fecf column * roughness column)/8 to calculate the effect
sizes. The denominator is 8 because there are 8 pairs in each contrast. Table 3.26 shows the
results.
Table 3.25 Results of the polishing experiment.
w -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 1 1 1
X -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1 1 1 1 1
Y -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1
z -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1
R 44.7 35.7 43.8 30.6 44.8 36.2 44.1 31.9 49.0 40.9 55.1 42.8 50.2 41.9 55.7 41.8
Table 3.26 Table of contrasts for the polishing experiment.
zEffect: W X Y WX WY WZ XY XI YZ WXY WXZ WYZ XYZ WXYZ
-0.25
Effect size: 8.2 0.5 0.3 -10.7 -0.05 3.05 0.05 -0.2 -0.05 -2.2 -0.45 -0.4 -0.25 -0.1
The second step is to use Excel's Regression tool to perform the regression. The Input Y
range is the column of results; the Input X range is the matrix of ±l's we constructed for the
contrasts calculation. Selected elements of the result appear as Table 3.27.
Table 3.27 Selected regression results for the polishing experiment using all effects.
Regression Statistics 1
1
Multiple R 15
R Square 0
Adjusted R Square 16
Standard Error
Observations
Intercept Coefficients P-value
W 43.075 #NUM!
X 4.1 #NUM!
Y 0.25 #NUM!
Z 0.15 #NUM!
WX -5.35 #NUM!
WY -0.025 #NUM!
WZ 1.525 #NUM!
XY 0.025 #NUM!
-0.1 #NUM!
xYZz -0.025 #NUM!
-1.1 #NUM!
WXY -0.225 #NUM!
WXZ -0.2 #NUM!
WYZ -0.125 #NUM!
XYZ -0.05 #NUM!
WXYZ -0.125 #NUM!
Three things are noteworthy about these results. First, each of the coefficients is exactly half
the size of the effects calculated using contrasts. Using W for example, the reason for this is that
the effect from contrasts is the change in roughness as we go from W = -1 to W = +1, an increase
of 2, while the coefficient from regression gives the change in yield for an increase of only 1.
Second, R2 is listed as exactly 1 (which is correct) but i?2acy is listed as 15, which is impossible.
What's happening here is that we have 16 observations (= experimental results), which we are
Chapter 3 Statistical Concepts Applied to Measurement and Sampling 127
trying to explain with 15 variables (= effects), plus an intercept. In the same way that two points in
geometry exactly define a straight line and three non-collinear points exactly define a plane, there
is an exact linear formula explaining the 16 results using 15 effects. Thus, R2 = 1, which says our
model has explained all variation in result. However, adjusted R2 should be between 0 and 1, just
like R2, so there's no chance that it could equal 15. Either Excel has a programming error or more
likely, the Regression tool is inapplicable in this instance.
Third, the ^-values for the coefficients are all listed as #NUML This occurs for the same
reason, and is the result of Excel trying to divide by 0.
It may seem like a good thing to have a model like this with no error, but in fact, it is not. In
particular, a model such as this typically has very poor predictive capabilities, because it has built
in all the experimental errors for the 16 runs actually made. Any future experiments will be
predicted in a way that assumes that exactly the same experimental error will occur in the future
experiments as occurred in the past.
To avoid the problem noted above, this time we use only the main effects and two-factor
interactions for our regression. Selected elements of the result appear as Table 3.28.
Table 3.28 Selected regression results for the polishing experiment, using main effects and two-
factor interactions.
Regression Statistics
Multiple R 0.998734953
R Square 0.997471507
Adjusted R Square 0.99241452
Standard Error 0.630872412
Observations 16
Intercept Coefficients P-value
W 43.075 1.24889E-11
4.1 1.57373E-06
X 0.25
0.15 0.1737975
Y -5.35 0.385246741
-0.025 4.18704E-07
z 1.525 0.880257705
wx 0.025 0.000200834
-0.1 0.880257705
WY -0.025 0.553922295
-1.1 0.880257705
wz 0.000932289
XY
xz
YZ
The eleven-factor (ten-effect) model is excellent. R2 and 7?2adj both exceed 99%. The/?-values
for the coefficients W, Z, WY, YZ, and the intercept are below 0.05. However, the intercept is an
estimate of the roughness when each factor is at the average of its two levels (e.g., when speed is
/4(16 + 24) = 20 rpm, etc.). We are justified in ignoring the intercept here for two reasons: first,
it's not an effect, which is what we are trying to estimate; and second, it's not clear for this
particular example what it means when the intercept is measured at the "average" of carborundum
and diamond for the abrasion factor.
The ten-effect model showed that only two main factors were deemed significant: speed W
and abrasive Z. Increasing W by 1 is the same as increasing the speed by 4 rpm, so the physical
meaning of the coefficient for W94A, is that the roughness parameter increases approximately 1 %
for each increase of 1 rpm in speed. The physical meaning of the coefficient for Z, -5.35, is that
128 Chapter 3 Statistical Concepts Applied to Measurement and Sampling
the parameter decreases by 10.7 % when we switch from carborundum to diamond. The reason we
have to double 5.35 is that the change from carborundum to diamond corresponds to a change of 2
(-1 to +1) in Z. Factors X (feed rate) and 7 (depth) were not judged significant by themselves, but
Y was judged significant with W and Z, while X was not a factor in combination with any of the
other three.
Factorial designs are often used as screening experiments, intended to more to focus on which
factors and interactions are important in determining the size of the response, rather than on how
much the response changes. The idea is that further, more focused experiments can be used for the
latter purpose.
In the examples above, we used the statistical significance, or lack thereof, of the regression
coefficients to tell us that. However, there is a simple graphical technique that can be used to
quickly find which factors and interactions seem to have "large" effects, and that can be used with
any number of effects at once. The idea is that, if none of the effects affected the response, then
the estimated effect sizes would have a normal distribution. Therefore, we can make a normal
distribution plot of the estimated effect sizes and identify those points that do not seem to follow a
normal distribution. The effects associated with these points are the ones that have unusually large
effect sizes. Figure 3.38 is the desired normal distribution plot. Note that we used all 15 main
effects and multiple-way interactions in a way that we could not when using regression to find
which factors are significant. Again, consult the "Polishing" worksheet for Excel work.
Normal Distribution Plot of Estimated Effects for
Ceramic Polishing
10
о
Ф5 о
N
'Φ
ёй о*■» о
Ф "5
5
« -ю о
Ф -6-4-2 0246 8
-15
8
theoretical effect sizes if no factor is significant
Figure 3.38 Normal distribution plot of estimated effect sizes for the polishing example.
The graph shows that the central points form a straight line. These points correspond to
effects that are following a normal distribution, and hence effects that are probably not important.
If you look at the leftmost two points and rightmost two points, however, you see that the actual
effect sizes fall far from the straight line of the others. This tells us that these four points
correspond to effects whose size is so large than they are unlikely to be accidentally that big. That
is, these four effects are the ones most likely to have significant influence on the response.
Looking at the data, we can see that the two points on the left correspond to effects Z and YZ, while
the ones on the right correspond to W and WY. These are the same four effects identified as
significant by regression in Table 3.28 above, so the two methods give consistent results.
We close this subsection by returning to the problem we started with, namely trying to
determine the optimum yield of a reaction. Recall that we described the application of OFAT to a
process that had a response surface "peak", and elliptical yield contour lines (Figure 3.37). How
would we design a full-factorial experiment for this situation?
Chapter 3 Statistical Concepts Applied to Measurement and Sampling 129
A simple two-level (2 x 2) full-factorial design matrix for the process is clearly unsuitable
because of the strongly curved response surface. The simplest reasonable full-factorial matrix is at
three levels for each factor (temperature and time). This is a 3 x 2 matrix, for a total of nine
experiments. Clearly, some knowledge of the reaction properties is required to avoid picking
levels that are so far away from the optimum yield that they adversely affect the results. The
experimental runs were conducted in random order, with the results shown in Table 3.29, and
described in more detail in worksheet "OFAT".
Table 3.29 Yield results for two-factor three-level matrix full-factorial experiments.
time, min (t) 8 16 24 8 16 24 8 16 24
temp, °C (7) 220 220 220 245 245 245 270 270 270
yield, % 39 54 58 64 85 78 67 74 33
Multiple-level models are rather more complex than the two-level models we've discussed so
far. Here we simply use the Regression tool to determine equation coefficients involving the
factors, their quadratics, and their products. There are at most eight effects involving time (t) and
temperature (7): /, T9 tT, t2, T2, Λ , 772, 7V. As before, the regression tool will not allow the
determination of eight effects (and a constant) from nine data points. Therefore, we follow the
prior procedure of ignoring the highest-level effect (7V) for the first regression fit. Table 3.30
shows the results for a seven-effect equation.
Table 3.30 Regression results from full factorial yield data from Table 3.29. Eight equation
coefficients were sought.
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.999948002
R Square 0.999896007
Adjusted R Squa: 0.999168053
Standard Error 0.5
Observations 9
ANOVA df SS MS F
Regression 7 2403.75 343.3928571 1373.571429
Residual
Total 1 0.25 0.25
8 2404
Intercept Coefficients andarci Ern tStat P-value
18.23333319 90.49167 0.201491846 0.873421136
T -0.019999999 0.735951 -0.027175718 0.982703658
-150.0225 5.592514 -26.82559102 0.023720824
t 1.18984375 0.066535 17.88302833 0.035562067
tA2 -2.39871E-12 0.001497 -1.60271E-09
TA2 1.17225 0.043328 27.05527518 1
-0.00578125 0.000271 -21.36195996 0.023519632
Tt -0.00215 8.66E-05 -24.82606157 0.029779822
TtA2 0.025629349
TA2t
We look first at the P-value column to see if any terms lack statistical significance, and note
that the T and Γ2 terms have a high P-value (designated by bordered values). According to the
discussion on hypothesis testing (Section 3.4.3), a variable is not significant at the 95 % confidence
level unless the P-value is <0.05. Therefore, these two terms are the least significant, and we
should perform regression again without them. Table 3.31 shows the results.
130 Chapter 3 Statistical Concepts Applied to Measurement and Sampling
Table 3.31 Regression tool results from full-factorial 3 x 2 yield data from Table 3.29. Six
equation coefficients (for five effects) were sought.
SUMMARY REGRESSION OUTPUT
Regression Statistics
Multiple R 0.99994253
R Square 0.99988506
Adjusted R Square 0.99969349
Standard Error 0.30348849
Observations 9
ANOVA df SS MS F
Regression 5 2403.72368 480.7447 5219.514286
Residual
Total 3 0.27631579 0.092105
8 2404
Coefficients ,tandard Erro tStat P-value
Intercept 13.33333333 0.76376262 17.45743 0.00040966
t -149.345526 1.2261428 -121.801 1.22015E-06
tA2
1.169695724 0.01449674 80.68683 4.19588E-06
Tt
TtA2 1.169486842 0.00981126 119.1984 1.30182E-06
TA2t
-0.0056990 5.7566E-05 -99 2.27199E-06
-0.0021500 1.9868E-05 -108.214 1.73974E-06
Elimination of the T and Ί2 terms has slightly improved 7?2adj, and all P-value terms are now
well below 0.05. Equation [3.24] adequately expresses the percent theoretical yield.
% Yield = -149.35(0 + 1.170(/2)н- 1.1695(7)(0-0.00570(7)(/2)-0.002150(7^X0 + 13.33 [3.24]
Solver was then used to find the values of T and / that gave a maximum yield. The result was
86.7 % yield at 17 minutes reaction time and 249 °C. These results are very close to the
parameters for maximum yield obtained by visual inspection of Figure 3.37. The OFAT technique
missed the maximum by 3 minutes reaction time and 11 °C, while involving more runs.
The drawbacks of OFAT should now be apparent. Of course, OFAT experimentation does
not always lead to incorrect or suboptimal results. This example was deliberately skewed to
illustrate the inadequacies of OFAT testing, and point out the advantages of factorially designed
experiments.
3.5.2 Fractional Factorial Design
In practice, we are often interested in the effect of many factors. For example, in the nickel
precipitation example discussed earlier, the hydrometallurgist might want to know the effect of
more than three factors. However, as more and more factors come into consideration, the total
number of runs grows exponentially. A complete 5-factor two-level design requires 32 runs and a
6-factor two-level design requires 64 runs. In typical industrial settings, it is often very expensive
and time-consuming to conduct all the runs, in which case, a fractional factorial design (FFD)
might be a suitable choice.
As the name suggests, in an «-factor fractional factorial design (FFD), we only conduct a
fraction of all the 2n runs. If we conduct half of all the runs, the design is called a half-fraction
factorial design; if we only conduct lA of all the runs, the design is called a quarter-fraction
factorial design; and so forth. Since, Vi = 2~\ a half-fraction factorial design is often described
Chapter 3 Statistical Concepts Applied to Measurement and Sampling 131
using the shorthand a "2n x design". Similarly, a quarter-fraction factorial design would be a "2n 2
design", and so forth. Smaller fractional designs are possible.
FFD can also be used in screening experiments, to identify the few dominant factors from a
larger number of possible factors. The FFD results are then used to design a second factorial set to
look at these factors in greater depth. This technique is called sequential experimentation, and is
especially important where one embarks on a new course of experimental work where little is
known about the effects of any of the factors.
At this point, you might wonder how fractional factorial design is feasible. As we already
learned, in order to estimate all the main and interactive effects for an л-factor design, we need all
the 2n runs. If we conduct fewer runs, we are going to be unable to estimate some quantities. In
fractional factorial design, we cannot estimate all the effects, which is the price we are going to
pay for the effort that we saved. Therefore, we must exercise care to determine which half, or
which quarter, of our runs can give us the maximum amount of information.
There are two main principles underlying FFD. The first one can be called the diminishing
principle, and is purely empirical. The second one can be called half-matching principle, and is
rigorous mathematically. Let's look at the diminishing principle first. Table 3.32 shows a
summary of all the estimated effects from Example 3.22.
Table 3.32 A table of all main, two-way, and three-way effects from the nickel precipitation
example.
T P С TP PC ТС TPC
28% 3.5% 10% 0.5% 0.5% 3% 0.5%
Each main effect is larger than any of the two-factor effects, and each two-factor effect is at
least as large as the three-factor effect. In addition, when we looked at the regression model for
this system, a model based on T, P, and С alone, without any of the interaction effects, explained
99% of the change in yield. Therefore, the main effects seem to be much more influential on yield
than any interaction effects.
This empirical observation is generally true in practice. Usually, main effects are more
important than two-factor interactions, two-factor interactions are more important than three-factor
interactions, and so on. Because of this principle, it is more important to estimate lower order
effects than higher order ones.
This leads us to the most common use of fractional factorial designs. FFD's are used most
commonly to screen for which main effects, and possibly two-way interactions, have the largest
effect on the response variable. Thus, we assume that the multi-way interactions are small, and
focus our experimental design on trying to be sure that the main effects and possibly the two-way
interactions are unambiguously estimated. Having these results, the engineer can design more
focused experiments to understand the important factors so identified. For instance, in the example
above, further experimental attention might be focused on better understanding just temperature
and catalyst loading, since those effects are over twice as large as the next largest effect.
The half-matching principle is the second principle underlying FFD, and is related to a
property of any complete design matrix. In Table 3.33, pick any pair of columns, and check to see
in how many rows the two columns agree in sign, and in how many they disagree.
Regardless of which pair of columns you picked, you should have found that two of the rows
had both - ' s , two of the rows had both +'s, and four of the rows had one - and one +: that is, four
of the rows match in sign, and four don't match. This should always be true of any expanded
factorial design matrix with any number of factors, as long as all factors have just two levels. For
any pair of columns, half the signs will agree, and half the signs will disagree.
132 Chapter 3 Statistical Concepts Applied to Measurement and Sampling
Table 3.33 Expanded design matrix for our example.
run T P с TP PC тс TPC
1 ---+ + + -
2 - - ++ - - +
3 - +- - - + +
4 - ++- + - -
5 +- - - + - +
6 +- +- - + -
7 ++- + - - -
8 ++++ + + +
Now we put these two principles together to determine which of our T runs to make, and
which to omit. Pick a main effect, and pick a higher-order interaction not containing that effect.
For example, if our factors were temperature (T), pressure (P), catalyst loading (C), stirring rate
(S), and reaction time (H), we might pick as our main effect T and as our higher order interaction
PCSH. The equation T = PCSH will be called the generating relationship for this fractional
factorial design. We will talk more about picking a generating relation later, but for now want to
focus on the results of our choices.
We will construct our fractional factorial design by omitting from the design matrix any row
where the sign for T is different from the sign for PCSH. Table 3.34 has sixteen rows,
corresponding to sixteen experimental runs, one for each remaining combination of factors. Please
see worksheet "TPCSH" for details.
Table 3.34 Partial design matrix for experiment with factors T,P,C,S,H with T = PCSH.
Run T P с s H PCSH TCSH
2 -1 -1 -1 -1 1 -1 -1
3 -1 -1 -1 1 -1 -1 -1
-1 -1
5 -1 -1 1 -1 -1 -1 -1
-1
8 -1 -1 1 1 1 -1
-1
9 -1 -1 -1 -1
12 -1 -1 1 1
14 -1 1 -1 1
15 -1 1 1 -1 -1
17 -1 -1 -1 -1 -1
20 -1 -1 1 1 -1
22 -1 1 -1 1 -1
23 -1 1 1 -1 -1
26 -1 -1 1
27 -1 1 -1
29 1 -1 -1
32 1 1 1
Since we removed all rows where T Ф PCSH, it's no surprise that the T and PCSH columns in
the remaining design matrix are exactly the same. The practical implication of this is that we can't
tell the difference between the effect of T and the effect of PCSH, since we have the same sign for
each row for both of them. Since we noted above that multi-way interactions are likely to be very
small, what we've done is give up any chance of measuring PCSH as separate from T.
Chapter 3 Statistical Concepts Applied to Measurement and Sampling 133
The reason we're doing this is because of practical constraints. You use fractional factorial
designs when you don't have the time and money to do a complete factorial experiment. It's a
question of controlling what you give up. In this case, we give up the ability to measure PCSH.
In fact, since we've cut 16 runs out of our original 32, we've given up the ability to measure
16 effects. In Table 3.34, for example, the columns for P and TCSH are also identical. Thus, we
won't be able to tell the difference between the effect of P and the effect of TCSH. The technical
term for this is confounding P and TCSH. Again, though, since TCSH is a four-way interaction, it
is assumed negligible, and whatever effect is observed will be attributed to P.
Here's a simple technique to determine which factors will be confounded with which in this
example. We're going to be multiplying the algebraic variables T, P, C, S, and H, with the special
rule that anything squared simplifies to 1. For example, (TPC)(TCSH) = T2PC2SH = PSH.
Begin by multiplying the original two effects you chose to confound, in this case (T)(PCSH)
= TPCSH. Now, for any other factor, that factor will be confounded with its product with TPCSH.
For example, the factor TPC will be confounded with (TPC)(TPCSH) - SH. Table 3.35 shows the
15 effects that can be measured with these 16 runs, as well as the effect confounded with each.
Table 3.35 Table of effects and confoundings for the experiment with factors T, P, C, S, and H,
with defining relation T = PCSH.
Effect T P С s H TP TC TS
Confounded Effect PCSH TCSH TPSH TPCH TPCS CSH PSH PCH
Effect TH PC PS PH CS CH SH
Confounded Effect PCS TSH TCH TCS TPH TPS TPC
Confounded effects are often called aliases of simpler effects, and each column in the above
table would be called an alias set. Thus, CSH is an alias for TP, and {TP, CSH} is the alias set of
TP. In theory, TP could be considered an alias for CSH, but the whole idea of fractional factorial
design is that the simpler effect is more likely to be "real" than the more complex effect.
You may have noticed a potential short cut in the above table: there are just five factors, and
if a factor appears in a given effect, it doesn't appear in the confounded effect, and vice versa.
Unfortunately, that's not generally true, as we'll see in the next example.
This design is considered a "good" design for a 25_1 FFD. The reason for that is all the main
effects, and all the two-way interactions, are confounded only with higher order interactions. Since
we are assuming that all the higher-order interactions are negligible, this means that we can
measure all the main effects and two-way interactions unambiguously. There are tables published
of standardized "good" choices for 2n'k designs in a variety of sources, one of which can be found
in (NIST-FFD 2006). We'll work an example to show how to use this webpage by designing a 25"2
fractional factorial design with the same factors, T, P, C, S, and H. If you click on the "2Ш5~~2" link
on Table 3.17 on the reference webpage, look halfway down for the line:
DEFINING RELATION = 1= 124= 135 = 2345
Here the letter "I" is used because of a convention in the field of mathematics that underlies
how the defining relations are chosen. What's important to us is that, in terms of our
multiplication of variables, I behaves like a 1. Since we have consistently listed our factors in the
order T-P-C-S-H, we'll take T = 1, P = 2, etc. Then the defining relation becomes I = TPS = TCH
- PCSH.
The notation 25-2 tells us that we want a quarter FFD (since T1 = VA). This means that we will
need to find not one pair of columns, as above, but two pairs, and keep only those rows of the
design matrix that have both pairs of columns equal to one another. Given the defining relation
above, there is any number of ways to do this. We will multiply the first equation I = TPS by T,
134 Chapter 3 Statistical Concepts Applied to Measurement and Sampling
giving T = PS, and the last equation I - PCSH by P, giving P - CSH. This means that, of the 32
rows in the complete design matrix, we will use only the eight that have T = PS and P = CSH.
Table 3.36 shows these eight rows.
Table 3.36 A design matrix for a 25~2 fractional factorial design.
Run T P с s H PS CSH
4 -1 -1 -1 1 1 -1 -1
7 -1 -1 1 1 -1 -1 -1
10 - 1 1 -1 -1 1 -1 1
13 -1 1 1 -1 -1 -1 1
17 -1 -1 -1 -1 1 -1
22 -1 1 -1 1 1 -1
27 1 -1 1 -1 1 1
32 1 1 1 1 1 1
To find which effects are confounded with others, we note that with only eight runs, there can
be at most seven different effects measured. Thus, each effect is confounded with three others.
Since our defining relation is I = TPS = TCH = PCSH, we find confoundings by multiplying an
effect by TPS, by PCSH, and by TCH to find the three other effects confounded with it. Table
3.37 shows the confoundings.
Table 3.37 A table of confoundings for a 25-2 FFD.
Effect T P С s H PC PH
Confounded PS, TS, TPCS, TP, TPSH, TCS, TcSsH,,
Effects TPCSH, CSH, PSH, РСН, PCS, SH,
TPCH TH TCSH TC TPH TPC
CH
As expected, a 25"2 design is not as successful as a 25"1 design in preserving information.
Each of the two-way interactions in the 25-2 design is confounded either with a main effect or with
another two-way interaction. Thus, only the main effects can be measured unambiguously. In the
25"1 design, each two-way interaction could be measured unambiguously, assuming higher order
interactions are insignificant. However, by using only eight of 32 runs in a 25-2 design, it's not
surprising that we have reduced discrimination from the sixteen rows of a 25"1 design.
In the last two columns of Table 3.37, we've picked PC and PH arbitrarily as the "official"
effects of their alias sets. We could just have easily chosen SH and CS respectively. However,
because we always favor simpler effects in FFD's, it would have been inappropriate to use TCS or
TPH as the effect of their alias set, or TSH or TPC as the effect of theirs. For the same reason, for
example, we prefer T to PS or CH in the first column.
We made two other arbitrary choices. One choice was how we picked the pairs of columns
we would require to be equal from the defining relation I = TPS = TCH = PCSH. We could have
multiplied any two of these by any effect they contain to create the two equations that tell you
which rows in the design matrix to keep or discard. For example, we could instead have multiplied
the second equation I = TCH by C, and the third I = PCSH by SH, giving С = TH and SH = PC as
the defining relations. The relations С = TH and SH = PC will give you a design matrix with the
same eight rows as the design matrix above. However, in some situations (e.g., Example 3.23
below) how you make this choice will result in different rows being chosen. In general, the
recommendation is to choose relations at random, and multiply them by random factors.
The other choice we made was when we identified T = 1, P = 2, etc. For this example, it was
this choice that specified which eight rows of the entire design matrix would be retained in the
fractional design. For example, if we had picked T = 5, P = 2, С = 4 and so on, a different set of
Chapter 3 Statistical Concepts Applied to Measurement and Sampling 135
eight rows would have resulted. The recommendation is to randomly choose the identification of
specific factors with numbers, to ensure you don't always use the same rows in a fractional
factorial design. If you always use the same sets of rows with the same factors, you'll never test
most combinations of factors, and could potentially omit important ones without knowing it.
On the same note, if at all possible, the experiments described by each row should be run in
random order. Don't always do the experiment for the first row in the design matrix, followed by
the experiment for the second row, and so forth. Doing the rows in random order reduces the
possibility that some unforeseen influence caused by the order of the experiments could affect your
results. For example, suppose each run of the experiment requires human recalibration of
instruments. It's not unreasonable that, as the day wears on, human fatigue may cause slight errors
in instrumentation. If you run the experiments in the "standard" order, the rows appear in the
design matrix, all the experiments with the -1 level for the first factor are run first, followed by all
the experiments with the +1 level. That would mean that the measurements for the +1 level of the
first factor are generally less reliable than are those for the -1 level. If there's a consistent bias, the
effect of the first factor could be significantly under- or over-estimated.
EXAMPLE 3.23 — Effect of a Fractional Factorial Design: How Much Information is Lost?
Make a 24"1 FFD for the polishing example. Write a table of confounded effects, estimate the
effect associated with each alias set, and compare these estimates with the estimates obtained from
the full design. Are there any differences in which effects are statistically significant?
Data. See Table 3.25.
Solution. From (NIST-FFD 2006), we click on the "2iV4-1" link in Table 3.17 to find a designing
relation I = 1234. Assigning factors randomly, we let W = 1, Y = 2, Z = 3, and X = 4. We choose
W randomly to multiply the relation by W, and get the resulting pairs of columns W = YZX.
Unlike the previous example, here it is the choice of multiplier, not the choice of factor
assignment, which is critical in choosing rows of the design matrix for the FFD. The design matrix
and alias sets for this design are shown in Table 3.38. For Excel work, again consult the
"Polishing" worksheet.
Table 3.38 Design matrix for a 24"1 design for the polishing example.
Run: w X Y z R
1 -1 -1 -1 -1 44.7
4 -1 -1 1 1 30.6
6 -1 1 -1 1 36.2
7 -1 1 1 -1 44.1
10 1 -1 -1 1 40.9
11 1 -1 1 -1 55.1
13 1 1 -1 -1 50.2
16 1 1 1 1 41.8
Table 3.39 Alias sets for a 2 design for the polishing example.
Effect W X Y z WX WY wz
Alias XYZ WYZ wxz WXY YZ xz XY
The last three columns of Table 3.39 tell us immediately that we will not be able to estimate
the two-factor interactions meaningfully. We can deal with the first four alias sets by assuming
uniformly that three-factor interactions are insignificant, so we can assume that the main effects
can be estimated unambiguously. However, focusing in the last alias set of our example, suppose
we estimate WZ. Since it's confounded with XY, we won't know whether the effect we see is
136 Chapter 3 Statistical Concepts Applied to Measurement and Sampling
"really" the effect of WZ or XY or some combination. Further, since they are both two-factor
interactions, we have no criterion to judge one to be more likely to be insignificant than the other.
Thus, in conclusion, it will only be possible to estimate the main effects unambiguously. To
estimate an effect for each alias set, we use contrasts, resulting in Table 3.40.
Table 3.40 Estimation of effects for a 24_1 design for the polishing example.
Effect Fractional factorial design Full factorial design
W Estimated effect size Estimated effect size
X 8.1
Y 0.25 8.2
-0.1 0.5
z -11.15 0.3
wx -2.25 -10.7
3 -0.05
WY -0.15 3.05
WZ 0.05
Although the estimates vary between the fractional and full factorial estimates, it is generally
true that large estimates in one design correspond to large estimates in the other, and the same is
true for small estimates. The exception is the estimate of WX, which is large in the fractional
design, but negligible in the full design. This brings us back to our point above: in fact, WX is
confounded in this example with YZ, and in the full design, YZ was estimated to be -2.2. Thus,
the fractional design cannot distinguish between the effect WX and the effect YZ.
We use regression to determine which factors are statistically significant. If we include
elements of all seven alias sets in the regression, we will have the same problem that we had in
Table 3.27 when we tried to use regression with too many independent variables. Therefore, we
only regress on the four main effects. Table 3.41 shows the results.
Table 3.41 Results of regression model for main effects for a 24"1 design for the polishing
example.
Intercept Coefficients P-value
42.95 3.53143E-05
W 4.05 0.033384932
0.125 0.915435092
X -0.05 0.966090123
-5.575 0.01422315
Y
z
We can see that the same two factors as before (W and Z) are significant, but since their p-
values are closer to 0.05 than they were for the full factorial design, there is less evidence for their
significance. Of course, this makes sense since we have less information in a fractional design.
Assignment, (a) Create a 25"1 fractional factorial design for the effect of impurities on the strength
of a cement (see worksheet "Cement", and Exercise 3.24 for more information). Write down the
design matrix and the alias sets for the resulting design. Use the fractional design to estimate
effect sizes. Can you still estimate all two-factor effects unambiguously? (b) Create a 25'2
fractional factorial design for the cement example. Write down the design matrix and the alias sets
for the resulting design. Use the fractional design to estimate effect sizes. Can you still estimate
all two-factor effects unambiguously?
In closing, we note that there is a special set of unusually efficient fractional factorial designs
used especially for screening purposes. These are called Plackett-Burman designs, named after
their inventors (NIST-PB 2006). A Plackett-Burman design can be used (a) if you know in
advance that you're only interested in measuring main effects, not any interactions at all; and (b) if
Chapter 3 Statistical Concepts Applied to Measurement and Sampling 137
the number of experiments can be a multiple of four. If these are true, then a Plackett-Burman
design will use even fewer runs than a standard fractional factorial design. They are particularly
useful to in screening experiments to identify the most important variables, after which closer
attention can be paid to their effects. Plackett-Burman designs do not have defining relations, and
are described differently than the FFD's described earlier.
O'Keefe and co-workers have used Plackett-Burman techniques in several of their
metallurgical research projects. The range of variables was selected by first carrying out
preliminary laboratory tests. An eight-run screening design studied the effect of five variables on
the sulfation of zinc from electric arc furnace dust with three additional tests at the center points of
the variables (Chaubal et. al. 1982). An eight-run design (eight main and three center-point runs)
was used to study how six variables affected the removal of cobalt from an electrowinning zinc
electrolyte (Blaser and O'Keefe 1983). Four variables were studied with an eight-run design (plus
three center point runs) to develop mathematical equations for predicting polarizing behavior of
addition-free zinc sulfate electrowinning electrolytes (Sing and O'Keefe 1985).
3.6 Summary
The measurements needed for making a material or energy balance come from plant samples
or laboratory experiments, and inevitably contain error. A statistical analysis of such data helps us
understand its reliability and quantify the amount and source of the error. Statistics can tell us how
confident we should be that the data we took represents reality. Our first step in data analysis used
descriptive statistics, with charts to visualize important relationships and trends, and calculate
certain values to summarize the data. We used Excel's statistical tools to present and organize the
data. Our next step used inferential statistics to draw conclusions about the process or population
being studied. We estimated parameters based on assumptions about the type of distribution, and
tested various hypotheses to see which type of model was best. Models can be used to optimize a
process or express a set of tabular data as quantitative equations.
A great deal of attention was directed to using Excel 's Trendline tool, the Regression tool, and
Solver to find equation parameters for a model. The best type of model is based on a
phenomenological basis. An empirical model is chosen when there is no clear understanding of
the physical or chemical basis for a process or a set of observations. When an equation for a data
set has parameters inseparable from the variables, Solver plus Excel's SSD tool can find the
parameters.
The Chapter concluded with a brief review of experimental design techniques. Statistically
designed experiments (i.e., factorially designed experiments) give the most reliable data with the
least amount of test work. A full-factor design allows the estimation of every possible factor
effect. When faced with several possible factors that might influence the outcome, a partial
factorial design can screen for the most important factors. The NIST e-Handbook was used as a
primary reference and source of examples throughout the Chapter.
References and Further Reading
NIST-Gen 2006. NIST/SEMATECH e-Handbook of Statistical Methods. The e-Handbook is
available on-line at http://www.itl.nist.gov/div898/handbook/. It's also available (free) on a CD as
NIST Handbook 151, which also contains the program Dataplot (NIST Handbook 148). The
chapter titles of the e-Handbook are:
1. Exploratory Data Analysis.
2. Measurement Process Characterization.
3. Production Process Characterization.
4. Process Modeling.
138 Chapter 3 Statistical Concepts Applied to Measurement and Sampling
5. Process Improvement.
6. Process or Product Monitoring and Control.
7. Product and Process Comparisons.
8. Assessing Product Reliability.
NIST-Cer 2006. http://www.itl.nist.gov/div898/handbook/eda/section4/eda42al .htm.
NIST-RFM 2006. http://www.itl.nist.gov/div898/handbook/pmd/section6/pmd642.htm
NIST-FFD 2006. http://www.itl.nist.gov/div898/handbook/pri/section3/pri336.htm. Also, see
section 5.3.3.4.7 for Table 3.17.
NIST-FFD 2006. http://www.itl.nist.gov/div898/handbook/pri/section3/pri3347.htm.
NIST-PB2006. http://www.itl.nist.gov/div898/handbook/pri/section3/pri335.htm.
Bevington, Philip R., and Robinson, D. Keith, Data Reduction and Error Analysis for the Physical
Sciences, 3rd Edition, McGraw-Hill, 2003.
Brereton, Richard, Chemometrics: Data Analysis for the Laboratory and Chemical Plant, John
Wiley & Sons, 2003.
Decoursey, William, Statistics and Probability for Engineering Applications: with Microsoft
Excel, Butterworth-Heinemann (Paperback), 2003.
Dretzke, Beverly J., and Heilman, Kenneth A, Statistics with Microsoft Excel, 3rd Edition, Prentice-
Hall (Paperback), 2004.
Frye, Curtis, Statistical Analysis with Excelfor Dummies, John Wiley & Sons (Paperback), 2005.
Levine, David M., Ramsey, Patricia P., and Smidt, Robert K. Applied Statistics for Engineers and
Scientists: Using Microsoft Excel and Minitab, Prentice-Hall, 2000.
Montgomery, Douglas C , Runger, George C , and Hubele, Norma F., Engineering Statistics, 4th
Edition, John Wiley & Sons, 2007.
Montgomery, Douglas C , Design and Analysis of Experiments, 6th Edition, John Wiley & Sons,
2005.
Orris, J. Burdeane, Basic Statistics: Using Excel and Megastat, McGraw-Hill College (Paperback),
2006.
Rosenberg, Kenneth M., The Excel Statistics Companion CD-rom and Manual, Version 2.0
Thomson Learning, (Paperback), 2006.
Taylor, John R., Introduction to Error Analysis: The Study of Uncertainties in Physical
Measurements, 2nd Edition, University Science Books, 1997.
Triola, Mario F., Elementary Statistics Using Excel, 2nd Edition, Pearson Education, 2004.
Wikipedia contributors, "Descriptive statistics", Inferential statistics", "Normal distribution",
"Histogram", "Scatterplot", "Confidence interval", "Regression analysis", "Design of
Experiments", "Factorial experiment", "Response surface methodology", "Plackett-Burman
designs". Wikipedia, thefree encyclopedia. December 2006.
http://en.wikipedia.org/wiki/Main_Page.
Exercises
3.1 Examine the worksheet "WO" in workbook StatTools.xls. Use Excel to find the mean,
standard deviation, and variance of the %0-in-tungsten data. Write a sentence explaining the
meaning of the mean and standard deviation in that context.
Chapter 3 Statistical Concepts Applied to Measurement and Sampling 139
3.2 The temperature attained by mixing two reactants (initially at 20 °C) is measured 30 times,
and the following data are obtained.
28.3 29.0 27.9 28.9 28.5 28.4 28.5 28.7 28.5 28.1
28.0 28.1 28.5 28.7 29.5 28.4 28.0 28.0 28.4 27.6
29.0 28.3 28.3 28.5 27.9 28.4 28.1 29.2 27.9 28.6
(a) Find the mean, median, standard deviation, and variance of these 30 temperatures. Write
a sentence interpreting the meaning of the standard deviation.
(b) Make a histogram of the data. Describe the resulting shape.
(c) What percent of the temperatures fall above 28.0? Answer the question both by counting
by hand and by using Excel. Are the answers different? Do they both make sense?
3.3 Consult the "FeMn" worksheet in StatTools.xls, which lists wMn in 199 samples of ferroalloy
produced at a smelter.
(a) Find the mean and standard deviation of these measurements.
(b) What percentage of the data fall within one standard deviation of the mean? That is,
what percentage of the data is in the interval (x-s,x-\-s^7 Is this consistent with the wMn
data being normal? Why/why not? Hint: you can answer this question using the
PERCENTRANK function.
3.4 Consult the "Bleedoff worksheet. The data refers to the number of hours between critical
pressure bleedoffs in a chemical reactor; the data shown are the 255 measurements corresponding
to one year's operation of the reactor.
(a) Make a histogram of this data. Would you describe it as symmetric? Tailing off to the
left? Tailing off to the right?
(b) Find the mean and median for the data. Looking at your histogram, why do you think the
mean and median are so different? Hint, remember the mean acts like a center of mass.
3.5 An annealing process discharges hot objects that must be cooled before packaging. The
cooling system is a fan blowing air over the conveyor belt that leads to the packager. An engineer
takes a random sample of 500 objects out of a day's production run of 18,000 and measures their
temperature with a digital thermometer as the objects enter the packaging machine. She records
the instrument reading, which is displayed to the nearest whole °C, for each sample object. The
measured temperatures in her sample ranged from a low of 80 °C to a high of 91 °C (see worksheet
"Cooler"). The resulting frequency distribution is shown in the table at the top of the next page.
Temperature Measurements of 500 Objects Using Cooler
Temperature, °C 80 81 82 83 84 85 86 87 88 89 90 91
# of objects 3 7 21 42 79 113 92 60 48 20 10 5
(a) Make a histogram from this frequency distribution.
(b) Make another histogram using bin widths of 1.75, starting the first bin at 80. How much
does this change the shape of the histogram? Why?
(c) Are the cooler data well fit by a normal model? Justify your answer.
(d) The package will be damaged if items enter it at a temperature above 92 °C. Regardless
of your answer to (c), assume a normal model for the data. According to that model, what
percentage of the items will enter the package at 92 °C or above?
3.6 Refer again to the "WO" worksheet in workbook StatTools.xls.
a) Is the data normally distributed? Answer the question using a normal distribution
plot.
140 Chapter 3 Statistical Concepts Applied to Measurement and Sampling
b) Is there significant evidence at the 0.05 level that the mean of the process producing
this sample is different from 55.5? Justify your answer.
c) What assumptions did you need to make to do part (b)? State them explicitly and
check those that you can check.
3.7 Refer again to the "FeMn" worksheet in workbook StatTools.xls.
(a) Are the data normally distributed? Justify your answer.
(b) Find a 95% confidence interval for the data. State any assumptions you make explicitly
and check those that you can. Write a sentence explaining the meaning of the confidence
interval you constructed.
(c) Find the number of measurements needed to find a 99% confidence interval with width
±0.05 about the mean, assuming the standard deviation remains the same.
3.8 Refer again to the "Bleedoff" worksheet in workbook StatTools.xls.
(a) Are the data uniformly distributed? Check by using a uniform distribution plot.
(b) Find a 90% confidence interval for the data. State any assumptions you make explicitly
and check those that you can. Write a sentence explaining the meaning of the confidence
interval you constructed.
3.9 Consult worksheet "NG" in workbook StatTools.xls. Assume that a different natural gas
company also uses a normal distribution to model their heating value measurements, but that their
normal curve had a mean of 1075 and a standard deviation of 15. They then take an additional
measurement. Find the following probabilities: (/) the probability that this measurement is below
1075 (why does this answer make sense?); (if) the probability that this measurement is below
1047; (Hi) the probability that this measurement is above 1080; and (iv) the probability that this
measurement is between 1047 and 1080. Do the final calculation two ways: once using the
method of the example above, the second time using your answers to (//) and (Hi).
3.10 Consult worksheet "NG" in workbook StatTools.xls. The Handbook text stated that if the
natural gas company measured the heating value of their gas at a rate of 4 times per day, they
would only get a measurement below 1000 Btu/ft3 about once every 115 years. Perform the
calculation to verify this statement.
3.11 Consult worksheet "NG" in workbook StatTools.xls on the heating value of natural gas.
(a) Is there significant evidence at the 0.05 level that the true heating value of the gas is
different from 1050? At the 0.10 level? Does it make sense that as the level of significance
increases, the chances of finding a difference increases too? (Hint: yes.) Why?
(b) Suppose we still had a sample mean x = 1048.1 and a sample standard deviation of s =
11.0, but that our sample size was much larger, say, n = 1200. Would there now be
significant evidence at the 0.01 level that the true heating value of the gas is different from
1050? Does it make sense that as the sample size increases, the chances of finding a
difference increases too? (Hint: yes.) Why?
(c) Suppose now the sample size is 12000 and the sample mean is 1049.7. Is there
statistically significant evidence at the 0.01 level that the true heating value of the gas is
different from 1050? Looking at the difference between 1049.7 and 1050, is there any
practical significance to this difference? Should you give difference answers to these two
questions? (Hint: yes.)
3.12 In the introduction to the natural gas example in section 3.2.2, it was stated that the producer
had a target heating value of 1050 Btu/STP ft3. However, the mean of their 120 sample
measurements was 1048.1 Btu/STP ft3. Is there significant evidence at the 0.01 level that they
have the wrong heating value?
3.13 Based on the sample of 480 measurements on the CerStr worksheet, is there significant
evidence at the 0.05 level that the mean strength of all instances of the ceramic in question is
different from 640?
Chapter 3 Statistical Concepts Applied to Measurement and Sampling 141
3.14 Return to the "Polishing" worksheet. We will be working with the full factorial design data,
rather than the fractional design. Our goal is to compare the analysis of the data using only the
main effects, versus the analysis using both main effects and two-factor interactions.
(a) Use the Regression tool to estimate the effect sizes of the main effects only. Be sure to
check the "Residuals" box in the Regression dialog box. State and interpret R2.
(b) Repeat for a model using both the main effects and two-factor interactions.
(c) Based on i?2adj, which model is better?
(d) Compute and interpret 95% confidence intervals for the residuals of each model.
(e) Is there significant evidence at the 99% level that the larger model is more precise than
the smaller model? Use an FTEST on the residuals of the respective models to answer this
question.
3.15 In the "Cooler" example, the engineer hoped that enough data were collected so that a 95%
confidence interval had a width of ±0.2 °C about the mean. Did she succeed? Assuming the
standard deviation remained the same with a larger or smaller sample size, how many
measurements should he have made to achieve his desired width?
3.16 In an effort to check the adequacy of their measurement system, the natural gas producer
from Section 3.2.2 sent 18 samples to an external lab to have them measure the heating values to
compare with the company's measurements. The results, in Btu/STP ft3, are shown below.
Original 1058 1056 1058 1038 1054 1028 1054 1042 1048
Remeasured 1053.0 1052.0 1053.8 1042.6 1051.2 1037.7 1050.8 1043.5 1048.9
Original 1050 1044 1050 1048 1062 1036 1050 1038 1050
Remeasured 1049.4 1048.0 1048.0 1048.5 1056.5 1041.6 1049.0 1044 1049.4
(a) Assume that the company's measurements serve as a reference. Is there evidence of
systematic error on the part of the external lab? Does the external lab have the same precision
as the original? Use a 95% level of significance. Use Example 3.10 as a guide.
(b) Assume instead that the original measurements are NOT a reference or standard. Can
you tell which lab is more accurate? Why or why not?
3.17 A student was set a task to evaluate the differences, if any, in two methods of measuring the
specific heat capacity of metals. He measured the specific heat capacities of the following
elements or alloys: Al, Be, Brass, Cu, Au, Fe, Li, and Hg. His results (next page) were:
Metal Al Be Brass Cu Au Fe Li Hg
Method 1 0.893 1.828 0.363 0.396 0.140 0.439 3.624 0.144
Method 2 1.095 2.013 0.559 0.633 0.267 0.662 3.778 0.220
(a) Can you detect a difference in accuracy or precision between these two methods? Use a
90% level of significance.
(b) The above measurements were compared to the reference value for each substance:
Metal Al Be Brass Cu Au Fe Li Hg
Reference 0.900 1.824 0.377 0.385 0.129 0.444 3.582 0.139
Is there evidence of systematic error for either method? Again, use a 90% level of
significance. Is there an apparent contradiction between the results in part (a) and the results in
part (6)? How do you explain the different results?
3.18 A foundry produces a copper-nickel alloy by separately melting pure copper and pure nickel
in large crucibles, then pouring the molten metals into a holding furnace. The mass of each metal
is determined by weighing the full crucible, pouring the metal out, and weighing the empty
crucible. The full and empty crucibles are weighed on different scales. The crucible + molten
142 Chapter 3 Statistical Concepts Applied to Measurement and Sampling
copper weigh 2550 ± 50 lb, and the empty crucible weighs 350 ± 5 lb. The crucible + nickel weigh
5150 ± 80 lb, and the empty crucible weighs 480 ± 5 lb. Calculate the mass of Cu-Ni alloy in the
holding furnace, and its uncertainty: mass of alloy = ? ±? What is the uncertainty in the %Ni in
the alloy?
3.19 Consult the "WatrVap" worksheet.
(a) Use the Trendline tool to find a quartic (degree 4 polynomial) model for the heat of
evaporation of water as a function of temperature. Set the intercept to the correct value at 0
°C. Use Equation [3.21] to calculate 7?2асу·
(b) Use the Solver tool to find a model of the form A//vap = a(374.15 - t)b, where / is the
Celsius temperature. Calculate i?2adj.
(c) Make a table with the following columns: temperature T, T2, \IT and Tm where T is
degrees kelvin. Use the Regression tool to find a model with input X range equal to these
four columns, and Input Y range equal to the column of Ai/vap.
(d) Use the Regression tool to find a quadratic/quadratic rational function model for АЯуар.
(e) Which of these models do you prefer, and why?
3.20 A sample of Mo02 was reduced by a mixture of CO and H2 until the fractional reduction X
reached 0.8880. The X vs. time data are shown below.
time, min 0 5 10 20 40 80 150 200 300
X 0 0.0643 0.0880 0.1253 0.1586 0.2353 0.3047 0.3634 0.4218
time, min 400 600 800 1000 1200 1400 1600 1800 2000
X 0.4927 0.5808 0.6166 0.6771 0.7367 0.7772 0.8310 0.8723 0.8880
(a) Find the best polynomial model for time as a function of reduction. Use the equation to
estimate the time required for 95 % reduction. Is it better to express time vs. a polynomial of
X or vice versa? Justify your answer.
(b) See if the pore reduction model (equation below) can describe the reduction of M0O2. If
so, calculate the time required for 95 % reduction.
£=l-3(l-X)2/3 + 2(l-X)
(c) Use the Solver tool to find the best parameters of the form t = a + 4b + cX . Compute R2.
Which do you prefer, your answer here or your answer to (a)? Why?
3.21 The following table shows the results for Test #2 of the NiO reduction example (worksheet
"NiORedn"), except that the value of X = 0.5886 has been changed from 950 to 800 seconds (note
border).
time, min 0 96 190 280 375 482 550
X 0 0.0751 0.1493 0.2205 0.2797 0.3519 0.3910
time, min 660 770 800 1158 1295 1476 1720
X 0.4404 0.5011 0.5886 0.6659 0.7090 0.7658 0.8292
Considering the new point as part of the combined three-test datatile, is this point now an
outlier? Justify your response.
3.22 A Montana mining operation recovers a platinum-metals-group concentrate and carries out
various smelting and leaching operations to recover the platinum and palladium content. One
byproduct is an iron-containing solution that cannot be discharged unless the iron content is below
10 ppm. The dissolved ferrous sulfate is oxidized to ferric sulfate, and then precipitated as jarosite
by adding ammonium hydroxide according to the following reaction:
3Fe2(S04)3 + 2NH3 + 12H20 -> 2NH4Fe3(OH)6(S04)2 + 5H2S04
Chapter 3 Statistical Concepts Applied to Measurement and Sampling 143
The plant wished to find the multivariate operating parameters to lower the iron in solution to
less that 10 ppm. A full-factor design was carried out in the laboratory with the following results.
ppmFe pH Temp, °F Time, hr
62.5
9.43 3.5 210 4
250
250 2.9 210 8
3.65
3.73 3.5 170 8
43.48
2.55 3.5 170 4
2.9 170 4
2.9 210 4
3.5 210 8
2.9 170 8
(a) Use the Regression tool to find the effect of each variable, and the three two-factor
effects. Compare the predicted and actual ppm Fe values. Determine which factors are not
statistically significant.
(b) Repeat the regression analysis without the statistically insignificant factors, and see if the
Ä2adj is improved. Compare the significance F and the factors-values for the two cases.
(c) Recommend the most appropriate time, temperature, and pH for minimizing the iron
content of the jarosite discharge stream.
(d) Use the equation developed in part (b) to make a plot of temperature vs. pH for a
discharge iron content of 10 ppm.
3.23 One treatment of hazardous waste such as municipal solid waste incinerator fly ash is to mix
it into cement to physically prevent its release into the environment. However, it is not known
how different common constituents of fly ash affect the strength and durability of the resulting
cement. The interaction of different constituents is also of concern. A factorial experiment was
designed using two levels of five different heavy metal contaminants (listed on the "Cement"
worksheet as factors A through E). The response is the strength of the resulting cement.
(a) Estimate the size of the main effects and the two-factor interactions using contrasts.
(b) Estimate the size of the main effects and the two-factor interactions using regression.
(c) Identify the main effects and the two-factor interactions for which there is statistically
significant evidence of a change in cement strength.
CHAPTER 4
Fundamentals of Material Balances with Applications
to Non-Reacting Systems
For all except nuclear processes, atoms can neither be created nor destroyed. This principle is a
simple statement of the law of conservation of mass, and is the fundamental principle for making
material balances. In the design stage of a process, an exact material balance can be made because
certain assumptions are made about the extent of reactions, the loss of minor elements (which are
often ignored in a preliminary design), and the accuracy of measurements on feed rates. When
making a material balance on a new (but not yet operating) process, the process engineer will
certainly need some expert advice on these topics. In the operating stage, sampling and
measurement errors are present, and redundant information may be present. In that case, the
balance may not close, so some way must be found to reconcile the balance, such as by ignoring
the least-accurate redundant measurements. This chapter introduces the principles and techniques
for making material balances on processes where chemical reactions do not take place. However,
this Chapter does include phase transformations, such as the evaporation of water. We also
introduce the Excel add-in programs MMV-C, FlowBal and FREED as a way to manipulate data
and make material balances within an Excel spreadsheet. The techniques developed in this chapter
are basic to all subsequent material balance calculations.
4.1 System Characteristics
For material balance purposes, a system is most conveniently defined as a volume in space
that is of interest for a particular study or analysis. Some references designate the system as a
control volume. The boundary of the system is completely arbitrary, and defined to best suit the
purpose of the analysis. Everything outside the boundary is called the surroundings. The system
contains a quantity of material and an assembly of equipment. Material, heat and work may be
allowed to cross the boundary, and the mass of the system may change with time. A system is
completely specified when it is at rest, and any two intensive properties and one extensive property
are specified.
There is no precise definition of the word "process", but here we take it to mean a set of
procedures used to produce a product. A system may consist of a single process that takes place
in a single piece of equipment, or multiple processes taking place in many pieces of equipment.
Figure 4.1 shows a schematic diagram of a system for refining steel scrap. Here the system
boundary is drawn to include several processes, but it could have been drawn around the scrap
mixer or the pump and degassing furnace instead. In some cases, it is convenient to draw sub-
system boundaries around one or more individual processes.
The process in Figure 4.1 is an open system, where material and energy can flow across the
boundary. A closed system allows energy to pass but not material. An isolated system is one that
is impervious to either type of flow. The type of system may change with time, necessarily being
open in the beginning for entry of material, and closed during operation.
The material flowing into, through, and out of a system is transported by streams, which are
arbitrarily designated for the purposes of the analysis. A stream is material in transit, during
which it undergoes no change in composition, mass, or temperature. Based on the way the system
boundary is drawn, the input streams (often called instreams in this Handbook) in Figure 4.1 are
labeled A, B, C, E, and F. The output streams (or outstreams) are G, J, and L. The transfer
streams are D, H, and I, and stream К is an accumulation stream. Other stream types (not shown)
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 145
are recycle and purge streams. A recycle stream returns material from one part of the system to
another, while a purge stream is an outstream that commonly removes trace or minor constituents
from the system. Stream material may be homogeneous or heterogeneous, may be described
according to the type of conveyance, or may simply be a mixture of everything entering or leaving
a system. For calculational convenience, it is desirable to keep different materials as separate
streams wherever possible.
Flux #1 Flux #2 System
/ boundary
Scrap #1 Mixed scrap Melting Slag,
Scrap #2 furnace G ; gas,
Scrap #3
; dust
У^ Electricity
Steel
Pump
Degassing -► Steel
furnace
Figure 4.1 Schematic diagram of a system showing material and energy transfers, and material
accumulation. Each lettered arrow indicates a stream.
Examples of some of the more complex stream materials found in materials processes are:
• A bulk raw material such as an ore. This is usually a very heterogeneous material, and
likely consists of several minerals; it may be wet, and has air amongst the particles. It
often enters the system through an open conveyor.
• A sludge or dust from some other process or from another part of the existing process.
This is likely to be a stream of variable composition, and be enhanced in elements that
enter the some other part of the process in very dilute form.
• A dusty offgas containing hazardous or restricted materials. The composition and flowrate
of such a stream may require particular attention if the amounts of certain substances are
controlled by regulation.
• A gas containing combustible substances, which if not carefully confined, could attain
explosive compositions.
• A waste material of no economic value that may be disposed of by stockpiling, so long as
the level of toxic contaminants is below certain limits.
4.2 Process Classifications
Processes can be classified as batchy semi-batch, continuous, steady-state or transient. Before
working on the material balance, you should classify the process according to these definitions.
• Batch processes. Raw materials are added to a vessel at the beginning of the process, and
kept there until the desired final state is reached. An example is the metallothermic
146 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
reduction of uranium fluoride by calcium. Another example is the production of
intermetallic compounds by direct-reaction synthesis.
Semi-batch processes. Some reactants are put into a vessel, and then other substances are
added steadily until the desired final state is reached. This is a very common type of
process in the extraction industry. One example is the passage of a drying gas over a batch
of wet sludge. Semi-batch processes are used in the production of many metals, and the
sintering of ceramic parts.
Continuous process. The in- and outstream material flows continuously throughout the
duration of the process. Continuous processes are common in the chemical industry, such
as in the refining of petroleum and the production of industrial chemicals. Examples from
the materials industries are the production of Portland cement in a rotary kiln, the
electroplating of zinc on steel, and the production of glass.
If a continuous process is operating such that the main variables (temperature, pressure,
composition, flow rates) are not changing with time, the process is said to be operating at steady
state. This classification holds even if there are minor fluctuations in conditions about a steady
average state. If process conditions are changing with time, then transient or unsteady-state
conditions are said to prevail. Continuous processes are usually run as close to steady state as
possible, but transient conditions are often unavoidable during start-up or shutdown. An example
is the continuous casting of a metal, in which the reservoir of molten metal is periodically filled
from the production line, but the molten metal is steadily poured into a mold. The casting process
may continue steadily for several days or weeks until a new alloy is called for, or the equipment is
shut down for repair.
4.3 Flowsheets
A written description of a process can contain all of the essential information needed for
making a material balance. However, a textual description is not a good way to present
information that contains a lot of numerical data and technical information. A better way is to
present the same information diagrammatically, showing points of entry and removal, the relative
position of equipment, and the path of material from one piece of equipment to another. Such a
diagram is called a flowsheet, and consists of boxes or other symbols to represent process devices,
and lines with arrows to represent streams. For example, please see the flowsheet diagram in
Figure 4.1 for the refining of steel scrap.
The detail or style of a flowsheet depends on its use. Each style of flowsheet has its place,
and is useful in process analysis insofar as it provides understanding. If the main purpose is to
give a general idea of a process, the minor streams and process devices are omitted, and technical
information is generally absent. Figure 4.2 shows a typical indicative flowsheet for the production
of gold from ore. If the emphasis is on the type of equipment being used and the way material is
transported around, each device will be drawn in detail to show the features of the equipment, but
very little information may be provided on process details such as stream material composition or
flow rate. Figure 4.3 shows a flowsheet for the roasting of zinc sulfide concentrate, which is
carried out in a fluidized-bed roaster. Finally, most useful for our purposes is a flowsheet that has
streams labeled with the type of information that helps solve complex material balances. Figure
4.4 shows a process-oriented flowsheet for the direct smelting of iron ore to a molten product that
contains about 95% Fe and 5% С Notice the detail attached to this flowsheet as compared to the
other examples.
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 147
NaCN Return sodium Zinc
cyanide solution powder
Ground V7
gold ore
1=
Solid to
Air \ 7 waste
Sale
Figure 4.2 Typical indicative flowsheet for the production of gold from ore.
Hi-Zn Con Water S'htd. Steam A Water s.htd steam .Water
i Feed Bin / Spray 385°C, 45 bar 100X 385oC 4 5 b a r M00°C
4
Gas
350°C
S'htd. Steam
385°C, 45 bar
Water
100°C
Water, Water, Calcine to
100°C 50°C Leach
200°C
Figure 4.3 Flowsheet for fluid bed roasting of zinc concentrate using shapes to describe the
various devices. Dashed lines indicate the flow of gas, and solid lines the flow of solids. Calcine
is the oxide product of roasting. A waste-heat boiler (designated W-H) recovers heat from the gas
to generate steam. The ESP is an electrostatic precipitator for final dust removal.
148 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
Reduction Gas 1590 kg pellets
180 kg flux
1400 Nm3. 900 °C
4.0 %N2, 42.8 %CO
26.3 %C02, 12.1 %H20, 14.7 % H2
Fan Gas
Cooler —, 1940 Nm3, 100 °C
4.0 %N2, 36.1 %CO
33.0 %C02, 5.3 %H20 Ш Red'n %$
shaft
21.5 % H ;
Cyclone Scrubber
\j Sludge
60 kg dust Pellets 1360 NrrT
(28 % reduction) 20 °C
600 °C
600 °C 4.2 %N2
28.6 %CO
46.0 %C02
19.2 %H2
2.0 % H20
Smelter Gas Process Basis:
Medium volatile coal;
1400 Nm3 40% post-combustion;
90% heat transfer efficiency;
1760 °C water-gas shift equilibrium
4.2 %N2 at 550 °C in shaft & cooler.
48.6 %CO
One tonne hot metal
20.6 %C02 94.8 %Fe, 5.0 % C , 0.1 %S
9.0 %H2 1600°C
17.6 % H 2 0
75 kg dust
50 Nm
Figure 4.4 Simplified flowsheet for the AISI-DOE process for direct smelting of iron ore. Post-
combustion refers to oxidation of CO and H2 in the upper half of the smelting vessel. Heat transfer
efficiency refers to the fraction of post-combustion heat that is transferred to the molten bath. 80
% of the water vapor from the top of the shaft furnace condenses in the scrubber. The process
underwent extensive pilot plant testing, but was never commercialized.
The term device is used to denote a piece of equipment used in a process. Process devices are
often categorized by function. The main devices used in this Handbook are:
• Mixers, whose function is to blend material from two or more streams to form one new
stream. No chemical reactions take place in mixers. The new stream will contain the sum
of the masses of the streams being mixed. For incomplete mixing or complex blending
processes, it is better to represent the process by two or more mixers in series.
• Splitters, whose function is to split stream material into two or more streams. No phase
changes or chemical reactions take place in splitters. The composition of each outstream is
the same as the composition of the instream.
• Separators, whose function is to remove all or part of a substance from an inlet stream by
physical methods. Separation is commonly by phases, such as separating dust from air or
liquid from solids, and is seldom complete. No chemical reactions take place in
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 149
separators. Separators may also carry out splitting, but it is preferable to separate these
operations into different devices. Notice that a separator is a "reverse mixer".
• Reactors, whose function is to carry out chemical reactions. Chemical reactions involve
the separation and combination of molecules and elements to form new chemical
substances.
• Streams, which are flow devices that carry material from one device to the other. Typical
flow devices are pipes, ducts, conveyor belts, and troughs. By definition, material in a
flow device is inert — the device allows no phase transformation, chemical reaction or
energy transfer. If, in a real process, reactions or energy transfer takes place in a flow
device, some other suitable device should be inserted in the stream, with (inert) streams
connecting it to other devices. Streams are indicated on flowsheets as arrows, whose
format may differ to indicate the phase nature of material being transported. Streams are
differentiated by assigning a letter or a number to each stream. For material balance
purposes, transit material having multiple phases should be separated into single-phase
streams and paired with the other single-phase stream(s). This way of designating a flow
device as carrying two or more streams is discussed in detail in Section 4.7.
• Heat exchangers, whose function is to transfer heat energy to or from stream material, or
to exchange heat with the surroundings. No material is transferred from one stream to the
other.
• Controllers, whose function is to sense some stream or substance property, and to make
changes in flow, energy input or removal, or something else that will bring the property to
a set point. Flowrate controllers have the sensor and control device on the same stream.
Feed-forward controllers have the sensor on an adjacent stream and the control device on
the controlled stream so as to maintain a certain relationship between the flows of the two
streams. Feed-back controllers have a sensor at a downstream location, and operate the
control device at an upstream location.
In the most fundamental sense, it's possible to use just one general-purpose device for every
piece of equipment in a process, and then write balance equations that characterize the device's
purpose. However, making a distinction between devices helps in setting up the balance equations.
On that basis, the five fundamental devices involved in heat and material balances are streams,
mixers, splitters, heat exchangers, and reactors. We omit separators from this list because a
separator is nothing more than a mixer operating in reverse; the material balance equations are
comparable. The device definitions are not rigid, and other texts may classify devices differently
The concept of a "reaction" is also treated slightly differently in this Handbook. For example,
even though phase transformations are not reactions, they should be carried out in a reactor device.
This Chapter includes material balance calculations in systems with phase transformations but
excludes chemically-reacting systems. Figures 4.5 - 4.8 show some examples of flowsheets using
the above device descriptions.
Natural Gas Mixed gas
mixer
9 a s 92 % CH4 61.2 % N 2 to burner
8 % C2H6 31.1 % 0 2
7.1 % CH4
0xygen 17W 0.6 % C2H6
3 % N2
Air
79 % N2
21 % 0 2
Figure 4.5 Flowsheet showing the use of a mixer device to prepare combustion gas for burner.
150 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
Slurry Separator Filtrate
85 % H20 98 % H20
15%Si02 2 % Si02
11 % H 2 0
89 % Si02
Figure 4.6 Flowsheet showing the use of a separator device (here, a drum filter) to separate solid
and liquid phases. Notice that a separator is a "reverse mixer". Separation is not complete.
Moist Dry
air Dew point
Dew point a i r
35 °C
Dehumidifier 3°C
Water
Figure 4.7 Flowsheet showing use of a separator device to remove moisture from air. A phase
change must be written in a reaction format if a heat balance is to be performed.
Iron Reduced
ore 95 % Fe30 Reduction 90.9 % Fe iron
reactor
5 % Si02 3.7 % FeO
5.4 %Si02
Reducing
Spent
gas 75 % H2 60 % H2 gas
7 % H20 22 % H20
14% CO 10% CO
4 % C02 8 % C02
Figure 4.8 Flowsheet showing the use of a chemical reactor device to reduce iron oxide.
In a practical situation, it may be difficult to categorize processes exactly as described above.
The device classifications are guidelines for drawing flowsheets in a way that will assist the
material and energy calculations. Sometimes one device may be conveniently substituted for two
or more. For example, when three streams are mixed before entering a reactor, you can omit the
mixer and put all three streams directly into the reactor. You lose information about the
composition of the mixed streams, but the material balance requires less arithmetic to solve.
4.4 The General Balance Equation
Consider the dehumidification of a process gas using a separator, as depicted earlier in Figure
4.7, and in detail in Figure 4.9. The steady state inlet and outlet gas flows (volume) were
measured, the mass flow rate of water was measured, and stream samples taken. The results are
indicated on the diagram. An examination of the overall mass flow indicates clearly that less mass
is leaving the process than going in. What could cause this discrepancy?
• The flow meter on one of the streams is faulty.
• There is a leak somewhere, and water or gas is leaving undetected.
• The walls of the separator are absorbing something from the feed stream material.
• The gas analysis is wrong
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 151
The conversion of gas volume to mass assumed ideal gas behavior, but the gas doesn't
behave ideally.
The process may not have been at steady state. Some flow rates were varying with time,
and streams are not customarily sampled at the same moment.
Ice ЮОкд/hr
273 К 100%Н2О
Process gasiookg/hrj 80.5 kg/hr Dry gas
365 К 61.3 %N2 76.1 % N2 274 K
13.7 % 0 2
11.1 % 0 2 Dehumidifier 10.1 %C02
8.5 % C02 0.1 % H20
19.1 % H20
118.9 kg/hr Water
99 % H20 274 К
1%C02
Figure 4.9 Flowsheet for the dehumidification of a process gas by cooling with ice.
In analyzing a process, we must account for every factor that could cause a discrepancy in the
material balance. If the measurements and calculations are correct and leaks have been eliminated,
then the only explanation left is that some part of the input materials are remaining in the device.
Since total mass is conserved, a general mass balance on the overall process can be written:
Input - output = accumulation
The mass balance for an individual substance in a system where chemical reactions occur is:
Input + generation - output - consumption = accumulation
This is the general balance equation for mass and energy, and it applies to a single device or
an entire system. Two types of balances are recognized. The first is generally applied to a
continuous process (i.e., no accumulation) using flow terms such as kg/min, or L/s. This is the
case for the process shown in Figure 4.9. The second is generally applied to a batch process
between two intervals of time, using quantity of mass rather than flow rate as the unit designation.
4.5 Material Balances on Simple Non-Reactive Systems
Making a balance on a very simple process is straightforward. If the process is simple
enough, you can almost do it without a calculator. We begin with three simple situations with the
goal of developing a procedure that can be used in future, more complex processes. The first
example is for a continuous distillation process, using a separator (i.e., a "reverse mixer") for the
process device. The second example is a batch process for mixing alloys for a brass melting
furnace. The third example is a semi-batch process for the vacuum refining of lead.
EXAMPLE 4.1 — Distillation of a Cd-Zn Alloy.
Figure 2.5 showed that the difference in the vapor pressures of cadmium and zinc might allow
the two metals to be separated by distillation. We now apply the law of conservation of mass to
explore such a process. An alloy with a mass fraction of 50 % cadmium (wCd = 0.50) enters a
distillation device (a separator) at 1000 kg/h and exits as two streams. At steady state, the gas
stream contains 454 kg/h of Cd, and the liquid stream contains 472 kg/h of Zn. Write elemental
balances on each constituent and prepare a complete mass balance.
152 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
Solution. Always start by drawing a flowsheet for the process, labeling the device and streams
with information provided, and designating each stream with a number or letter. Figure 4.10
depicts the flowsheet with labels for the device and streams.
Original alloy (stream A) Distillation - > Vapor (stream B)
m A = 1 0 0 0 k g ; wACd = 0.50 device mBCd = 454kg; mBZn
-► Zn-rich alloy (stream C)
mcZn = 472kg; mcCd = ?
Figure 4.10 Flowsheet showing distillation device for refining a Cd-Zn alloy. The basis is one
hour of time. The symbol m refers to the mass, and the superscript letter to the stream. For
example, mBZn refers to the mass of zinc in stream B. Recall that w is the symbol for mass
fraction, so wACd refers to the mass fraction of cadmium in stream A.
By definition, a steady-state process has no accumulation, so the mass balance equation takes
the simple form: input = output. On a basis of 1 hour of operation, the mass balance equation for
each element is thus:
Cadmium balance: 500 kg = 454 kg + mcCd
mcCd = 500-454 = 46kg
Zinc balance: 500 kg = 472 kg + mBZn
mBZn = 500-472 = 28kg
Total mass balance: 1000 = 454 + 46 + 472 + 28 = 1000. Check •
So the vapor stream В contains 90.8 % of the entering Cd, and wBCd = 0.942. The liquid
stream С contains 94.4 % of the entering Zn and wcZn = 0.911. A good way to display the mass
balance is in a ledger format, which contains the mass and molar flows, and the mass and molar
compositions. A ledger for presenting results gives a very clear picture of the mass, amount, and
composition of each stream. Where appropriate, the volume and volume fraction of each stream
constituent can be added. The best place to create a ledger is in a spreadsheet if that's where your
arithmetic calculations are being made.
Stream
AВ С
mass, kg
Zn 500 28 472
Cd 500 454 46
total 1000 482 518
amount, kmol 7.65 0.43 7.22
Zn 4.45 4.04 0.41
Cd 12.10 4.47 7.63
total
mass fraction 50% 5.8% 91.1%
Zn 50% 94.2% 8.9%
Cd 100% 100.0% 100.0%
total
amount fraction 63.2% 9.6% 94.6%
Zn 36.8% 90.4% 5.4%
Cd 100.0% 100.0% 100.0%
total
Assignment. Add volume values to the ledger for stream В if it is at 120 kPa and 700 °C.
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 153
EXAMPLE 4.2 — Charge Calculation for Feed to a Brass Melting Furnace.
The preparation of an alloy for casting is done by feeding controlled amounts of raw materials
to a charge hopper until it has the correct quantity for the melting furnace. The charge is put into a
furnace, and energy is added until the alloy is melted and the final temperature is reached. Here,
3600 kg of pure zinc, 14 000 kg of pure copper, and 8500 kg of recycled brass are placed into the
charge hopper for the next melting cycle. The mass fraction of copper in the scrap brass is 0.79.
Figure 4.11 shows a sketch of the hopper charging process. Calculate the mass of mixed scrap in
the hopper and its overall composition, and express the answer to 3 significant figures.
Zinc A
3600 kg; wAZn = 1.00
Copper Mixed scrap
to melting furnace
14,500 kg; wBCu = 1.00
Q
Scrap brass
8500 kg; wcCu = 0.79
Figure 4.11 Flowsheet for batch preparation of feed to a brass melting furnace.
Solution. The process device is a batch mixer fed until the desired amounts have been added.
Each stream is labeled with known information. It's often helpful to add the label "wcZn = ?" to
stream С even though we know that wcZn = 1.00 - wcCu. Using similar logic, you may want to
label stream D with "wDZn = ?". While filling the scrap hopper, the mass balance equation is:
input = accumulation
The hopper contents are subsequently emptied into a furnace.
accumulation = output
Start with a total mass balance, then do the element balances.
Total mass balance: 3600 + 14 500 + 8500 = mO= kg mixture produced
mD - 26 600 kg
Copper balance: 14 500 + 0.79(8500) = (26 600)(wDCu)
wDCu = 0.798 = 79.8%
This tells us what we need to know about the furnace feed, including the wDZn. However, a
zinc balance will serve as a check:
Zinc balance: 3600 + 0.21(8500) = (26 600)(1 - wDCu) = (26 600)(wDZn)
wDZn = 0.202 = 20.2 %. 1.00 - wDZn - wDCu = 0.798. Check S
Try making your own ledger and labeling the flowsheet diagram with the results.
Assignment. Owing to an error in the raw material scales, all input weights were in error by 1 % in
the same direction. What effect will this have on the composition of the furnace feed?
EXAMPLE 4.3 — Vacuum De-Zincing of Lead.
During the refining of lead, zinc is deliberately added to react with and remove silver. After
that, the lead typically has a mass fraction of zinc of 0.55 %. A process is tested for removing the
zinc in a vacuum furnace to a level of 0.03 %. The impure lead (called bullion) is poured
continuously in a thin sheet across from a water-cooled panel. Zinc, which has a much higher
vapor pressure than lead, evaporates from the bullion and condenses on the cool panel. Refining
continues until a layer of condensed metal of mass 2240 kg has deposited on the panel. The
154 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
process is stopped, the panel is removed, a fresh panel is inserted, and the process begins again.
Make a mass balance on the process to calculate the mass of bullion that can be refined in each
step. A chemical analysis of the deposit on the panel gives wPb = 5.20 %, the rest being zinc.
Solution. This is a "reverse mixer" device using distillation to separate two metals, and is sketched
in Figure 4.12. The condensate may be considered an accumulation during the continuous flow of
the bullion. The mass balance equation is:
input = output + accumulation
Bullion Vacuum -► De-Zn lead
m A = ? wAZn = 0.55% furnace m B = ? wBZn = 0.030%
-► Condensate
m~ = 2240kg иЛРЬ = 5.20 %
Figure 4.12 Sketch of distillation device for refining lead bullion.
This example differs from the previous two in that the input stream is not completely defined.
Start by making a total mass and a zinc mass balance.
Total mass balance: mA = mB + 2240
Zinc mass balance: 0.0055(mA) = 0.0003(mB) + 0.948(2240)
This gives two equations with two unknowns, which is more complex than the previous two
examples. Writing a lead mass balance equation won't help because it provides no useful
information; the total mass balance equation is just the sum of the lead and the zinc mass balance
equations. However, the two equations are not complex, so simple algebra can solve them.
Bullion: mA = 408 240kg
De-zinced lead: mB = 406 000 kg
Thus, the lead refining process can continue until about 410 tonnes of bullion have passed
through the vacuum furnace, after which the process must be shut down to replace the
condensation panel. The ledger balance is shown below in both mass and amount units.
Stream Stream
A (Bullion) В (Lead) С (C'dnsate)
A (Bullion) В (Lead) С (C'dnsate)
amount, kmol
mass, kg Zn 34.34 1.87 32.49
Pb 1959.44 1958.87 0.56
Zn 2245 122 2124
total 1993.77 1960.74 33.05
Pb 405995 405878 116
total 408240 406000 2240
mass fract. 0.55% 0.03% 94.8% amount frac. 1.7% 0.10% 98.3%
Zn 99.45% 99.97% 5.2% Zn 98.3% 99.90%
Pb 100.0% 100.0% 100.0% Pb 100.0% 1.7%
100.0% 100.0%
total total
Assignment. Set up a template in Excel to make mass balance calculations on the lead refining
example, such that if the user enters the zinc content of the bullion, condensate, and refined lead in
selected cells, the mass of the bullion (mA) and the mass of the de-zinced lead (mB) will be
calculated.
4.6 Strategy for Making Material Balance Calculations
The general procedure for making material balances is always the same: write enough
balance equations for the number of unknowns, and then solve the equations. However, as the
Of course, it is highly recommended that a lead balance be made to check the answer.
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 155
problems become more realistic they become more complex, and it is not always easy to set up the
equations in the proper form (called the setup), or to solve them even if they are written properly
(called the resolution).
When making a material balance around a single device, we write a species or element
balance around the device, hoping to have as many equations as unknowns. If not, we need
additional information about the device. If we get it (by making assumptions or taking more data),
we solve the equation set to get the material balance. This procedure is more complex for systems
with multiple devices. One method is to solve each device's material balance, one device at a
time, until the entire system is balanced. This is called the sequential modular technique, and is
used by some of the commercial flowsheet modeling programs (Metsim 2009). The easiest type of
system to balance with this technique is an unconstrained system. Such systems have all device
instream flows specified, splitter fractions defined, and some information given about the species
reaction extent for each reaction. Constrained systems, however, have certain instream
compositions unspecified, may have stream flows expressed as ratios, and may use equilibrium
constant expressions to define reaction extent. In this case, the sequential modular technique
requires repeating the device-by-device balance calculation many times until convergence is
reached. Recycle streams can slow the approach to convergence.
An alternate technique, called the simultaneous solution method, solves all of the system's
material balance equations at once. The latter technique depends on readily-available equation-
solving software, and is the approach used in this Handbook. Commercial solving programs are
available when the simultaneous solution method is to be applied to large (>200 equations)
systems (FrontLine Systems 2009).
Each approach has its advantages and disadvantages. In larger systems, such as whole plants,
neither technique may be suitable, in which case the system must be divided into manageable
segments. In either case reasonably good starting estimated are desirable.
Most of the systems described in the Handbook consist of three or fewer devices, and fewer
than six species. By following certain guidelines, and with a little experience, you can successfully
tackle such systems in their as-presented form. However, a few more-complex systems are
presented that aren't easy to figure out if tackled as a whole. In that case, it's best to use an initial
oversimplification method. Instead of making a balance around the as-presented system, start with
at most two devices, and ignore substances present in small amounts. The knowledge gained on an
oversimplified system will help as you add devices and substances. A key advantage with this
method is that it generates good starting estimates for the unknown stream properties in a stepwise
manner. Whatever the system complexity, it helps to follow certain guidelines in setting up and
solving material balance equations.
4.6.1 Guidelines for Setting up a Materials Balance
1. Draw a flowsheet. Convert the text-form of a problem into a diagram; use consistent
symbols for different types of process devices (see Section 4.3 for a description of the
basic types of devices). Write the values and units of all known stream variables on the
diagram, and assign algebraic symbols for unknown variables. Write a one- or two-word
description of the stream, and assign each stream a letter or number. This Handbook often
designates streams by the term S#, where # refers to the stream number. Write out any
unusual specifications on the diagram close to the stream where the specification applies.
2. Decide what kind of process is taking place. Is it a steady-state (continuous) process?
Batch? Does something vary with time (unsteady-state)?
3. Set up a ledger. Put the same information on the ledger that you put on the flowchart. Set
up spaces for information that you will calculate and fill in later. Don't do any simple
arithmetic or calculate any unknowns at this point.
156 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
4. Look up or obtain any necessary information that's not included in the problem
statement. This may include the relative atomic and molecular mass numbers, conversion
factors, densities, vapor pressures, equilibrium constants, and so forth.
5. Adopt a set of consistent units for the calculations. If the problem statement contains a
mixture of volume, mass, and amount-of-substance units, it's desirable to convert all units
to either mass or amount. In the absence of chemical reactions, mass units are often
preferred. When chemical reactions are involved, converting all units to amount (moles)
can make the arithmetic simpler. This is because the moles of substances combine in
simple proportions, so it can be easier to estimate the values of unknown stream properties.
Once the material balance is solved, the units can be converted with the help of software
programs. .
6. Choose a basis. Often the problem statement suggests a basis, but sometimes the
resolution is much easier if a different basis is selected. If a stream amount or flowrate is
given, that's usually the best basis to pick. If no amounts are stated, assume one,
preferably for an input stream with no or the fewest unknowns. Selecting a value like 100
kg or 100 moles for a stream amount may help with the arithmetic.
7. Decide how many equations should be written to balance the process. The maximum
number of independent balance equations that can be written for a non-reactive system
equals the number of chemical entities in the input and output streams. However, there are
usually other unknowns and other information. It is usually not obvious just what is
known and what is provided, so a formal procedure for making this decision is desirable.
This is a degree of freedom (DOF) analysis, which will be discussed in Section 4.7.
8. Check which variable values you can determine "by inspection". Examine your ledger
for any simple relationships that you can solve in your head, and enter the result on the
ledger sheet. For example, if the flow of a two-phase stream is 200 kg/h, and information
is given about the mass fraction of one of the phases, the other mass fraction (and the flow
of both phases) can be obtained by difference. Include on the ledger any other simple
pieces of information that are in the problem statement (such as that the flow of one stream
is twice that of another), and express one of the composition variables in each stream as
one minus the sum of the others.
4.6.2 Guidelines for Resolving a Set of Equations
1. Write the equations in an efficient order. Write first those equations that have only
one unknown, and these in an order that will let you solve the second one from the
answer to the first, and so on. Next, write pairs of equations that contain two variables,
and so on. If you end up with simultaneous non-linear equations, look back at your basis
to see if picking a different one will help. If you have too many equations, one or more
might be redundant. If one equation can be obtained from two or more of the other
equations, it's not an independent equation, but it can be used as a check. If you are
provided with more data than necessary to solve the material balance, the data are
redundant. If the data are inconsistently redundant, you must disregard one of the values
in order to close the balance.
2. Solve the equations. If you are able to obtain solutions to one or two equations, see if
the answers can be used in subsequent equations to make them one-variable equations.
If the equation is complex (i.e., contains various powers of a variable that can't easily be
separated), try using Excel's Goal Seek tool. If several equations must be solved (linear
or non-linear), try using Excel's Solver tool. Please review your Excel manual or
Excel's on-screen help file for assistance in using these tools. Some additional help and
special software for using the Goal Seek and Solver tools are on the Handbook CD, and
repetitive versions are described briefly in the Appendix. After the equation(s) are
solved, check to be sure all requested problem variables are calculated.
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 157
3. Check your answer. If possible, use one of the dependent equations to check if the
answer is correct. Fill in the ledger sheet as a further check.
4. Scale the answer. If you used a different basis than set out in the problem statement,
scale your result up or down to the original basis.
5. Present your results. Prepare a table or chart showing the results. A chart is
particularly useful if the results are required over a range of input values.
6. Interpret your results. Summarize the findings in a useful way, and point out any
unexpected or noteworthy result.
4.6.3 Objectives of a Material Balance
Engineers have different objectives in making a material balance for a process. In the
simplest case, the purpose may be to track the loss of just one substance, or to examine how
different process conditions affect the loss of that substance. When there are large penalties for
emitting hazardous substances, or failing to adequately purify a product, all stream flows may be
measured, and streams completely analyzed. This is called a sample-based (or analysis-based)
material balance. It has its purpose, but the effort of analyzing the flow and composition of every
process stream is expensive, and may be counter-productive.
It's reasonable to ask: why bother making material balance calculations when we can analyze
every stream to get a balance? Is there an advantage in sampling some streams and calculating the
properties of the rest? Certainly it's possible to sample every stream and obtain a sample-based
material balance. However, some streams can be very difficult to sample. A satisfactory material
balance closure isn't possible when a stream's mass flow has an uncertainty of 10 %, or when the
analysis of some substances is omitted.
The problem with a sample-based material balance is illustrated for a process to recover salt
from an aluminum dross treatment plant. A salt flux is added to collect the non-metallic
impurities. The resulting salt cake must be treated to recover the salt so it can be used again. This
is done by leaching the salt cake in hot water as shown in the Figure 4.13 flowsheet.
Salt cake 1/ ч4
Leach sol'n '—►[ f ►Wet air
L - - - - - - Salty sol'n
Leach tank
м R
л u· * · 3 : 7 Slag slurry
Ambient air I ►* J
Figure 4.13 Flowsheet for the processing of salt cake from an aluminum refining process. The
NaCl in the salt cake (SI) is dissolved in the leach tank, producing a salty solution (S5) that can be
evaporated to recover the NaCl for re-use. The salt cake slag discharges as a slurry, shown as S6
and S7. Solid lines indicate solid phase streams, dashed lines indicate liquid phase streams, and
dotted lines indicate gas streams.
Table 4.1 shows sampling results taken over a short time of plant operation. We don't know
how the streams are sampled, how long it takes to obtain all the samples, or how the samples are
analyzed. But however obtained, the stream data constitute an analysis-based material balance.
This type of material balance is quite different from the ones made in previous sections — physical
sampling completely replaces material balance calculations.
A cursory check of the data shows that the results do not conform to the principle of
conservation of mass. For example, less mass comes out than enters, and more slag exits than
enters. An analysis-based material balance almost never shows mass conservation because of
inaccurate stream flow measurement, analytical errors, and fluctuations in process chemistry and
flow during the period when samples are taken. Nevertheless, it's easy to see that about 26 kg/min
of NaCl leaves in the salty slurry (via S6). Thus only about 90 % of the salt is recovered (in S5)
158 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
for re-use. Even though the mass balance doesn't close, it does serve to keep track of the flow (and
loss) of NaCl as process conditions change. Such results can sometimes be used to indicate how
the system responds to various changes.
Table 4.1 Stream data taken during operation of salt cake processing operation. Flow units are
kg/min, and chemical composition is mass fraction. Figures in italic font are obtained by
difference.
Stream name Salt cake Leach soln Amb air Wet air Salty soln Slry liq Slry slag
Stream no. 1 2 3 4 5 6 7
335 348 98 141
Flow 415 1260 0 0 1435 0
NaCl 0.644 0.104 0 0 0.257 0.267 1
Slag 0.325 0
H20 0.031 0 0.0105 0.0804 0 0
Air 0.896 0.9895 0.9196 0.743 0.733 0
0
0 0 0
One of the drawbacks of an analysis-based material balance is that it can't be used to predict
the effect of a change in input stream flow or composition on the other stream properties. We
don't know, nor can we elucidate from the sample data, how the process responds when an
instream is changed. Furthermore, the non-closure of the mass flows confirms the difficulty of
getting accurate stream flow measurements. This is not surprising; slurry flows are especially
difficult to measure because the flow sensor is easily clogged or eroded by solid particles. It may
be better to omit flow measurements from difficult-to-sample streams, trust the stream analysis,
and use a calculated material balance to determine the un-measured stream flows. In addition, a
calculated material balance will conform to the principle of conservation of mass. However, there
is some minimum number of flow measurements or stream analyses that are required to calculate
all the rest. This minimum number may be further reduced if the process has reliable subsidiary
relationships that hold over a range of operating conditions, or a controller is present. The best
strategy to obtain mass balance closure is to make the fewest possible stream measurements, make
then on streams that can be reliably sampled, and calculate the rest.
Pursuant to that strategy for the dross leaching process, a query to the operator elicits that he
had the greatest difficulty in measuring the outstream flowrates. However, it's relatively easy to
obtain good data on the instreams, and to measure the proportion of solids in the discharge slurry.
Historically, the mass flow of slurry slag (S7) is about 1.3 times that of the slurry liquid (S6). It's
also relatively easy to get a sample of the slurry liquid (S6) for chemical analysis. Historical data
shows that the NaCl composition of S5 is usually 95 % of that of S6, probably owing to a small
amount of leach solution that overflows the tank before complete mixing.
The easiest (and most reliable) stream data to obtain is thus the flowrate of the three
instreams, the NaCl and slag composition of the salt cake, the NaCl content of the leach solution,
and the moisture content of the ambient air. So the question is: are these seven measurements,
plus the two historical relationships, enough to allow the other stream properties to be calculated?
Answering this question requires a special evaluation to determine the number of independent and
dependent variables for the process. This requires calculating the degrees of freedom for each
device of a process, as described in the next Section.
4.7 Degree-of-Freedom Analysis
It is sometimes difficult to determine whether the stated problem values are sufficient to allow
a complete balance. Fortunately, there is a way to determine this before spending hours on a
problem that has no solution. The method is to determine the degree of freedom (DOF) of the
problem, and if it is zero, the problem is conceptually solvable. In the simplest terms, we know
that it is necessary to have as many independent equations as there are unknowns. An independent
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 159
equation is one that cannot be derived from two or more other independent equations. If the
number of independent equations = number of unknowns, the DOF is zero.
In practice, we often end up with more or less equations than unknowns. If (Nequations >
Nunknowns), we have one or more redundant equations. If one equation is perfectly determinable
from two or more of the others, it is called a consistent redundant equation, and one could choose
any adequate subset of Nequations for solution. If a solution cannot be found using all of the
equations, then we have one or more inconsistently redundant equations. This can happen for
example when the mass flow of each stream into a steady-state system is measured, as is the total
mass flow out, and they are not equal. We may have to make a judgment call about which flow
value is the least accurate, or if the error is caused by excessive round-off of significant figures.
If Nequations < Nunknowns, either something has been overlooked, or a reasonable assumption
must be made. Before starting calculations, it's essential to determine beforehand if Nequations =
Nunknowns. A DOF analysis is a systematic procedure for determining this. The alternative to the
DOF method is to write material balance equations for each substance or element, and write
equations for all other process relationships, and then see if Nequations = Nunknowns. However, the
drawback of this method is that it is easy to inadvertently write one or more non-independent
equations, which creates the impression that a sufficient number of independent equations have
been written.
4.7.1 DOF Concepts
The degree of freedom concept was first discussed in terms of the Gibbs phase rule, Section
2.2, where the symbol D was used to indicate the degree of freedom. A system is completely
defined when a sufficient number of state variables have been specified, in which case D = 0. The
system is invariant, and all unspecified state variables can, in principle, be calculated by means of
one or more equations of state.
The degree-of-freedom concept is also applicable to material balance calculations, albeit from
a different perspective. For material balances, as with phase equilibria, we seek a relationship
between the number of process components and composition variables that will tell us if the
system is correctly specified (i.e., DOF = 0). The principle of conservation of mass is not satisfied
unless the system is correctly specified. Calculating the DOF requires us to define some additional
terms. First, a stream variable is a uniquely identifiable physical entity within a stream where the
sum of all such entities constitutes the stream. Very often the entity is a chemical species, which is
a substance with a defined chemical formula. Examples of species are chemical elements such as
Zn(c,/,g), chemical compounds of invariable composition such as HCl(g), compounds of variable
composition such as solid wustite (FeOx) where the value of x must be specified, or mixtures of
species in defined elemental amounts. An example of the latter type is fuel oil, whose composition
may be represented by an empirical formula of the type C10H17S.05. Sometimes the entity is an
agglomeration of chemical species whose overall composition may or may not be defined.
More fundamentally, the stream variable may be a component, which we defined earlier in
connection with the elaboration of the Gibbs phase rule (Section 2.2). Recall that the concept of a
component is mathematical since the number of components represents the minimum number of
independent entities which are needed to generate all entities in all streams of a process, taking into
account any intrinsic relationships which exist amongst the stream species. In the absence of
chemical reactions, components are usually chemical species, but may also be groups of species of
known composition, or even an assemblage of substances of unknown composition. When phases
of constant composition (even if of unknown composition) are present, the phases, rather than their
constituent species, are the components. One independent material balance can be written for each
process component. Thus identification of the number of components is fundamentally important
knowledge which underpins the creation of a material balance for any process.
As an example of the distinction between a species and a component in a non-reactive system
is a simplified version of air. Suppose air is used to carry off water vapor from water. We may
160 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
choose to consider air as two species, N2 and 02, or as a component having the formula N1.5sO0.42·
In the latter case, the gas outstream consists of two components, i.e., two stream variables, water
vapor and air.
We also need to further emphasize the distinction between a flow device and a stream, a
difference we first introduced in Sections 4.1 and 4.3. A flow device transfers material into and
out of a system, or between devices in the system. The material in transit may consist of multiple
phases such as a dusty gas, which consists of two components. We could designate this (dust +
gas) material as a stream, but if we did, we would have to report the chemical analysis of the dust
portion separately from the gas portion. (The two phases must be separated before analysis
anyway.) However, calculations are easier if we regard the dusty gas as consisting of two streams,
one for each phase. The flow device thus carries two streams, a gas phase stream and a solid phase
stream*. The composition and flow rate of each stream (or phase) are separately reported. This is
the way we handled the slurry from the salt leach process (Figure 4.13); the slurry flow device was
deemed to carry two streams, one solid and one liquid. Despite the advantages of treating each
phase as a separate stream, some textbooks may treat two-phase materials-in-transit as one stream.
You should generally avoid this practice.
4.7.2 DOF Calculation Strategy for a Single Non-Reactive Device
The salt cake example discussed in Section 4.6.3 posed a fundamental question for material
balance calculations: what is the minimum amount of information required to calculate a material
balance? The salt cake example listed the seven most-easily measured stream properties, and
asked: are those measurements, plus two relationships based on historical operation, sufficient to
define the unmeasured stream flowrates and compositions? Table 4.2 shows a set of seven stream
measurements taken from the process.
Table 4.2 Sampling results from salt cake leaching process. The four mass fraction measurements
in italic font were obtained by difference. A ? indicates that a substance is present in the stream,
but no values are available.
Stream name Salt cake Leach soln Amb air Wet air Salty soln Shy liq Shy solid
Stream no. 1 2 3 4 5 6 7
Flow, kg/min 355 ? ? ? ?
405 1245 ? ?
wNaCl 0.635 0.0995 0.0113 ? 1.00
wSlag 0.334 0.9887 ? ? ?
wH20 0.031 0.9005
wAir
The question of sufficiency of information to make a material balance is answered by
calculating the DOF for the process. If the DOF is zero, we have just the right amount of
information. If the DOF is not zero the process is incorrectly specified for making a material
balance. The degree of freedom calculation of a one-device non-reactive process requires four
steps.
• Calculate the total number of independent stream variables (SV). This is the sum of the
number of components over all streams. If a device had three streams, each of which had
three components, SV would equal 9. The salt cake process has four components, two of
We must be careful when describing constituents as phases. The gas constituent is a phase, but the solid
constituent is certainly more that one distinct phase. A better term for the solid material is "solid phase
assemblage", or just "the solid". When the solid has constant composition as it passes through a device
(usually the case in a non-reactive system), it's acceptable to designate it (collectively) as "the solid phase".
A mixture of miscellaneous solids may be considered a component if the mixture does not change
composition.
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 161
which are species (NaCl and H20). Table 4.2 shows seven streams and fourteen stream
variables, so SV = 14.
Subtract the total number of independent balance equations (IB). This is the number of
components present in the system, which in the salt cake process is four. Note that an
independent balance can be written for each component, or for the total mass plus the
number of components - 1.
• Subtract the number of independent stream variables that are specified. This is the number
of independent compositions (1С) given in the problem, plus the number of stream flows or
amounts (F). To be independent, a stream composition must not be obtainable by
difference from other compositions. For example, stream 1 has the composition of NaCl
and slag listed, so the composition of water can be obtained by difference, and hence
wH20 is not an independent composition. Thus 1С = 2 for this stream. A stream of a
single component (such as stream 7) does not have an independent stream composition.
The salt cake process has 1С = 4 and F = 3.
• Subtract the total number of subsidiary relations (SR). Subsidiary relations include the
value of some variable that allows you to write an additional independent equation,
information on how certain process variables are related (ratio of flows, for example),
some physical property relationships (an equation relating the mass and volume), a
stoichiometric relationship (important in reacting systems, which will be discussed in
Chapter 5), or an energy relationship (to be covered in Chapter 7). An example of a
typical subsidiary relationship is the vapor pressure of water (which can be calculated if
the temperature is specified). The salt cake system has two SRs. One relates the flow
relationship between S6 and S7, and the other relates wNaCl in S5 to wNaCl in S6. The
SR item may be difficult to get right when making a DOF analysis.
• The number of subsidiary relations for a splitter requires some additional comment. First,
each specified split fraction constitutes a SR, so specifying the mass, amount or volume
flow of any N - 1 output streams is an SR. Second, owing to the relationships between the
compositions of the split streams, the number of special splitter restrictions is (N - 1)(S -
1), where N is the number of branches from the splitter, and S is the number of species.
The salt cake process doesn't have a splitter.
Once you have correctly determined the factors that define the device's degree of freedom,
calculate the DOF as follows:
DOF = S V - I B - I C - F - S R
The DOF for the salt cake process = 1 4 - 4 - 4 - 3 - 2 = + 1 .
If the device's DOF is positive, the problem is underspecified, and not all of the variables can
be calculated (but some useful information may still be obtained). A complete balance requires
either more information, or a reasonable assumption about a variable. Since the salt cake process
is underspecified, some other stream flow or stream variable must be measured, such as the >vH20
in S4. If the objective is to determine an optimum value for a variable, a device may be
deliberately underspecified. In that case, an objective variable is varied over a range to see if the
dependent variable can be maximized or minimized. If the DOF is negative, the device is
overspecified, and one of the redundant items of information must be disregarded. If the redundant
information is consistent with the rest of the information, the balance will come out the same
regardless of which item is disregarded. If not, there may be a way to assess the reliability of one
or more items, and the least reliable disregarded. Finally, if the DOF is zero, the device is
correctly specified, and a unique solution is possible. Each of these cases will be illustrated in the
following three sections. We will defer the mass balance calculation for the salt cake leaching
process to Section 4.8.1 as Example 4.4. Section 4.9 shows how to apply the DOF concept to
multi-device systems.
162 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
4.7.3 A Washing Process Having Zero Degrees of Freedom
There are a number of processes that involve a separation step. Dust may be separated from a
gas, or sediment may be separated from a slurry. This section deals with the slurry washing part of
the Bayer process. The Bayer process is used to recover aluminum oxide (A1203) from bauxite ore.
The first step is to dissolve the alumina with NaOH to produce soluble NaA102. The residue
(called red mud) is a mixture of insoluble impurity oxides and leach solution, and leaves the
leaching step as a slurry. The liquid portion of the slurry contains valuable NaA102 as a solute, so
it is important to separate the solution from the mud as effectively as possible before the mud is
discarded. This is done by pumping the slurry to a washing tank and washing with water or a
recycled solution (which may contain some suspended particulates and ions). The mud slurry is
discarded after several washings and settlings. Figure 4.14 shows a sketch of the flowsheet with
one washing. We want to make a material balance on the red mud washing device.
Notice that we don't know the composition of the insoluble oxide leach product (the red
mud). For a non-reactive material balance, we don't need to know it. Red mud is a mixture of
mineral species, but since it does not change composition in the washing device, it's treated as a
component. It's important to recognize that a group of species can be treated as a single
component when the group passes through a device without changing composition.
Bauxite NaOH solution Red mud slurry Wash wate*
Leach settling Wash settling
tank tank
Decanted sol'n Decanted^ Washed mud
to Al recovery sol'n to A?
recovery
Figure 4.14 Sketch of the flowsheet for leaching bauxite and washing red mud. The decanted
solution from both settling tanks is passed to another part of the plant where the NaA102 solute is
precipitated as Al(OH)3. The dashed line shows the sub-system boundary where the material
balance will be made.
The sub-system of interest here consists of the separator (washing) device inside the system
boundary. The red mud slurry enters at the rate of 1000 lb/h, with a solids mass fraction of 10 %.
The red mud composition is undefined, and does not change. The liquid part of the inlet slurry
consists of water containing two soluble species: wNaOH = 11 %, and wNaA102 = 16 %. The
wash water stream contains wNaOH = 2.0 %. The decanted solution stream (free of solid) contains
wH20 = 95 % and the washed mud slurry contains 20 % solids. We want to calculate the recovery
of NaA102 in the decanted stream. The main point of this section is to show how to apply the eight
guideline steps (Section 4.6.1) for setting up the balance equations, and the five guideline steps for
resolving the set of equations (Section 4.6.2). It's important to determine if there are any
substances that do not change composition during washing, such as (here) the red mud solid.
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 163
1. Draw and label a flowsheet. This is done in Figure 4.15, based on the system boundary
drawn in Figure 4.14, and is labeled with the information given in the problem statement. Each
stream has been given a letter (А, В and С are instreams, while D, E and F are outstreams). The
stream flowrate is designated with the italic letter F. The mass fraction of each stream constituent
is labeled with the italic letter w (mass fractions are preferable to percentages in the arithmetic).
Superscripts to F and w identify the stream (F® means the flowrate of stream B, for example).
Specified data is labeled on the diagram, and the unknown stream variables are noted as ?.
Although the red mud slurry enters via a single flow device (such as a pipe), we treat it as if it
consisted of two distinct streams traveling together (A and B). We call these paired streams.
Similarly, the washed mud consists of paired streams E and F. Paired streams are indicated by a
cylindrical shape on the flowsheet diagram.
Wash sol'n ! wcNaOH = 0.020
F c = ? с; (иЛн2о = ?)
3feRed mud slurry JL D Decanted solution
FB = 100 Ib/h (solid)" Washing/ FD = ?
FA = 900 Ib/h (liquid)" wDH20 = 0.95
settling
wANaOH = 0.11 wDNaOH = ?
wANaAIO2 = 0.16 tank
(wAH20 = ?) wDNaAI02 = ?
1 Washed mud w H20 = ?
^ P f ^ N! E FE = ? EwENNaaOAHI02= ? ?
=
^7?
4FF
Figure 4.15 Flowsheet diagram for the washing of red mud to recover NaA102. Solid line arrows
represent solid streams, while dashed line arrows represent liquid streams. A cylindrical shape
indicates paired streams, w for species (with a superscript stream letter) refers to their mass
composition in the liquid phase. The parenthesis around a mass fraction term indicates that it can
be calculated immediately by difference (i.e., it is not an independent parameter).
A consistent stream labeling procedure is used throughout this Handbook. First, an italic font
is used to designate the composition unit, such as w for mass fraction, φ for volume fraction, or x
for mole fraction. (Please see Figure 1.2 or the inside of the Handbook back cover for composition
and quantity terminology.) Next, a superscript letter is used to designate the stream. Thus, the
mass fraction of NaOH in stream С would be designated wcNaOH. (The superscript stream letter
may be omitted when there is no ambiguity about which stream is meant.) Finally, the letter F is
used to indicate stream flowrate, and thus requires units to be specified. Therefore the flow of
stream A is designated as FA = 900 lb/h. The time unit may be omitted if it is consistent
throughout. This labeling system will become more familiar as you work through the Chapter
examples.
2. Decide what kind of process is taking place. This is a continuous steady-state process.
The process device is a (phase) separator, but the liquid and solid phases are incompletely
separated because the washed mud still contains liquid and solid .
3. Set up a ledger. The initial ledger should include only the specific information provided
in the problem statement, and nothing that is the result of any calculation. The unknown
information should be indicated by a question mark. The ledger is shown in the following table. It
is immediately clear that some spaces marked with a ? can be filled in "by inspection". For
example, since the mass fractions must add up to 1, wH20 in streams A and С can be filled in
Very often, the outstream liquid phase in the decant liquid and the slurry have the same composition. If so,
the washer/settler is really a mixer with a splitter on the liquid outstream. The splitter is inherent (but
obscure) in the mixer, and splits the outflow liquid into two same-composition streams (D and E).
164 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
immediately. Similarly, the flow of stream F must be 100 lb/h. Other values are similarly
"evident". We could fill these in straightaway. However, it's better to leave these "by inspection"
values out of the first ledger, and fill them in after calculating the DOF.
Aв Stream
Slurry Liq Slurry Sid сD E F
Wash Wtr Decntd Sol'n WshMd Liq WshMd Sid
mass flow, lb/h FA = 900 — Fc = ? FD = ? FE = ?
liquid —
solid ^в=100 F¥ = ?
total FMud = FE + F¥
=T-r Slurry _ 1000
mass fraction 0.11 0.02 ? ?
in liquid phase 0.16 —? ?
NaOH ? ? 0.95 ?
NaA102 1.00 1.00
1.00 1.00
H20
total
4. Look up needed information. There isn't anything else needed.
5. Adopt a set of consistent units. Since the flows of streams AA and В are the only ones
specified, and the units are in lb/h, we will use lb-mass as the unit set for the problem.
6. Choose a basis. One hour of operation.
7. Decide how many equations need to be written. A degree-of-freedom analysis will tell
us how many independent equations are required. This cannot be seen just by inspection of the
ledger. The first part of the DOF analysis is to determine the number of stream variables. Streams
В and F have one component, streams B, D and E have three, and stream С has two, for a total of
13 stream variables. The number of distinct components present is four: H20, NaOH, NaA102,
and solid (the red mud). Hence, four independent balance equations are available. The number of
specified compositions is four (wANaOH, wcNaOH, wANaA102, and wDH20). In addition, we have
two specified flows (FA = 900 lb/h and F® = 100 lb/h), for a total of six specified stream variables.
Based on this information alone, we make a preliminary DOF calculation:
DOF = 1 3 - 4 - 6 - S R - 3 - S R
The system was described in the title of this section as having zero degrees of freedom, so
obviously there must be three subsidiary relationships. One subsidiary relationship is that
w°NaOH = wENaOH. A similar subsidiary relationship is that w°NaA102 = >vENaA102*. (A
similar relationship for wH20 is redundant, since the sum of composition variables must equal 1).
Our third relationship is that the washed mud is 20 % solid (7^ = 4F*). These three relationships
give DOF = 0 and with these the system is correctly specified (unless there is a fourth subsidiary
relationship we haven't noticed). Therefore, it's possible to write three independent component
balances, a total mass balance, and three subsidiary relationships.
8. Calculate variables "by inspection" and update ledger. In step 3, we mentioned that
values such as wAH20 and wcH20 are obvious because all other compositions are known. Enter
these values, and mark clearly the remaining unknowns. The updated ledger shows nine unknown
variables (four flow variables and five composition variables), and are shown in larger bold type.
In step 7, we identified the nature of seven equations, so we need two more. We note that the sum
of compositions for streams D and E must sum to one, which gives us the two equations we need.
This assumes that the washer/settler device is composed of a mixer and a liquid splitter. If this is not true,
one other SR is required.
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 165
АВ Stream
Slurry Liq Slurry Sid
СD E F
Wash Wtr Decntd Sol'n WshMd Liq WshMd Sid
mass flow, lb/h 900 100 Fc FD FE
liquid 1000 — ... FF
solid
total 5FF
mass fraction 0.11 0.02 M>DNaOH wENaOH
in liquid phase 0.16
— H>DNaA102 wENaA102
NaOH 0.73
1.00 0.98 0.95 wEH20
NaA102
1.00 1.00 1.00
н2о
total
This completes the eight guideline steps for calculating the DOF and setting up the balance
equations. Next, we turn to the four steps for resolving the equations, displaying the results in a
ledger (step 5), and interpreting the results (Step 6).
1. Write the equations in an efficient order. We can write five component balance
equations (total mass, water, solids, NaOH, and NaA102), but only four are independent because
there are only four distinct species. Nevertheless, writing all five may be useful if for no other
reason than one can be used as a check. Here, we temporarily skip the water balance equation.
The equations are not complex, so the order is not particularly important. The five component
balance equations are:
Total Mass: 1000 + 7^ = .F0 + 7^ + 7^
Solids: 100 = f*
Water: 0.73(900) + 0.98(7^) = 0.95(7^) + wEH20(7^)
NaOH 0.11 (900) + 0.02(7^) - w°NaOH(7^) + wENaOH(7^)
NaA102: 0.16(900) = wDNaA102(7^>) + wENaA102(7^)
Three subsidiary relations express the fact that the compositions of NaOH and NaA102 in
streams D and E are the same, and that the stream E flow is four times the stream F flow:
NaOH: wENaOH - w ^ a O H
NaA102: wENaA102 = wDNaA102
7^ = 47^
Finally, we recognize that the sum of compositions in streams D and E must sum to one:
0.95 + w°NaOH + Л а А Ю 2 = 1
wENaOH + wENaA102 + wEH20 = 1
2. Solve the equations. The equation set is fairly straightforward, and can be solved using
step-by-step algebra. First, from the solids balance, 7^ = 400 lb/h. This should be substituted into
the total mass, NaOH and NaA102 balance equations (remember, we skipped the water balance
equation). This gives six equations and six unknowns. Next, replace 7^ by (F0 + 500), which
gives five equations and five unknowns:
99 + 0.02(7^) - м ^ а О Щ Т ^ + 500) + wENaOH(400)
144 = wDNaA102(7^ + 500) + wENaA102(400)
166 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
wENaOH = мЛчГаОН
wENaA102 = Л а А Ю 2
0.05 = w ^ a O H + w°NaA102
Collecting terms gives two equations and two unknowns:
99 + 0.02(7^) = w ^ a O H ^ + 900)
144 = 0.05(7^ + 900) - Л а О Щ / ^ + 900)
These two equations can be solved to calculate
F° - 6600 lb/h; F° = 1100 lb/h
The rest of the unknowns are easily obtained by substitution in previous equations:
w°NaOH = wENaOH = 0.0308
м/°ЫаА102 = wENaA102 = 0.0192
3. Check your answer. This is done with the water balance equation. wEH20 = 0.95 as
expected.
4. Scale the answer. The basis wasn't changed, so this isn't necessary.
5. Present your results. An updated ledger is shown below, along with a component
material balance.
mass flow, lb/h A Stream D
liquid Slurry ВС Washed mud
solid Wash water Decanted sol'n
total 900 400
100 6600 7100 100
1000 —— 500
6600 7100
mass fraction 0.11 0.02 0.0308 0.0308
in liquid phase 0.16 — 0.0192 0.0192
NaOH 0.73 0.98 0.95 0.95
NaA102 1.00 1.00 1.00 1.00
H20
total
component flow 99 132 219 12
in liq. phase, lb/h 144 — 136 8
NaOH 657 6468 6745 380
NaA102 900 6600 7100 400
H20
total
6. Interpret the results. 144 lb of NaA102 enters the process in stream A, 136 lb are
recovered in stream D, and 8 lb are lost in stream E. This is a 94.4 % recovery of NaA102. If the
washed mud is dried, its mass will be 120 lb, and will contain 6.7 % NaA102.
The discharge slurry (washed mud) in the above example contains a considerable amount of
liquid. If loss of material in the washed slurry needs to be minimized, it's better to filter the final
slurry to produce a filter cake low in liquid content, and di filtrate liquid that is extremely low in
solid content. The filtrate always has a little solid because of leakage around the edge of the filter
cloth, cracks or pinholes in the cloth, and extremely fine particles that pass through the cloth pores.
If the amount of solid in the filtrate is very small, it may be ignored in the material balanced.
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 167
Figure 4.16 shows a sketch of a filtration system for a slurry derived from leaching a smelter
dust. We again treat the entering slurry as two distinct streams, liquid and solid phase streams.
Two streams that travel together can be called paired streams, which helps to clarify their
relationship. Flow amounts are kg/min.
F 1 = 500 kg siurry " F T ) ; "i filtrate L/S ratio = 140
filter cake L/S ratio = 0.18
wH2S04 = 1.4% F2 = 200 kg
wCdS04 = 3.6% wPbSO4 = 38.0%
wZnS04 = 8.4 % wCdS04 = 0.7 %
иЖ20 = ? wZnS04 = 0.8 %
wFe304 = ?
Figure 4.16 Filtering a slurry from a dust leaching process. Solid lines depict solid streams, and
dashed lines depict liquid streams. Paired streams indicated by cylindrical shape.
The filter system has five species; there are no reactions, and no composition changes.
Therefore, the system has only two components', the solid substance (a mixture of 4 solid species,
called a phase assemblage) and the liquid phase (a solution containing four species).
What looks like a single device here (a filter) really isn't. The process splits each component
into two parts; stream 1 is split into streams 3 and 5, and stream 2 is split into streams 4 and 6. In
reality, the filter is really two devices (each a splitter), one for the solid instream and one for the
liquid instream. It's important to develop the analytic ability to discern when something drawn as
a single device is really made up of two or more devices.
The DOF is calculated by first noting that SV = 6 because there are six streams, each with one
component. IB = 2 because there are only two components. There are no independent
compositions because each stream is only one component, so 1С = 0. However, two stream flows
are defined, so F = 2. Finally, SV = 2 because the stream flow ratio for the filtrate and filter cake
streams is defined. Therefore:
DOF=6-2-2-2=0
This system is so simple that we can skip the starting ledger. There are four unknowns: the
flow rate of streams 3, 4, 5, and 6. The simplest set of four equations is the two component mass
balance equations and the two flow ratio equations. The equations were written in a format that set
the left-hand side equal to zero.
Liquid component balance: 0=Fl - F3 -F5
Solid component balance 0 = F2 - F4 - F6
Filter cake flow ratio: 0 = F5 - 0.18(F*)
Filtrate flow ratio: 0 = F5 - 140(7^)
Substituting 500 for Fx and 200 for F2 gives an easy set to solve. Table 4.3 displays the
results. The values should be rounded off to the correct number of significant figures before a final
report.
Table 4.3 Mass balance results for slurry filtering operation. Flows are kg/min.
Stream name Slry Liq Slry Sid F-Cake Liq F-Cake Sid Filtrate Liq Filtrate Sid
Stream # 2 4 6
1 3 5
Liq flow, kg — — —
Sid flow, kg 500 35.40 464.60
200 198.68 3.32
— — —
168 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
It's possible to make a mass balance by selecting certain individual species instead of
selecting components, but the cake and filtrate flow ratio equations must always be part of the
equation set. For example, ZnS04 and CdS04 can be chosen as component species, as can ZnS04
and H2S04. However, mass balance equations using H2S04 and H20 are not independent, so DOF
= +1, and the system is underspecified. To avoid this complication, the best, and simplest,
selection is the liquid and solid component balances.
4.7.4 A Washing Process Having a DOF = +1
Rutile (titanium dioxide, Ti02) is produced for pigment use by the chloride process. It
contains salt, which must be removed to a low level before the rutile can be marketed. The feed to
the washing/settling tank consists of slurry containing Ti02 and salty water. wTi02 in the slurry =
40 %, while wNaCl in the liquid phase of the slurry = 20 %. The washed rutile exits as a slurry
containing wTi02 = 60 %, and is dried in a follow-on step. The decanted wash solution is not
solids-free; it contains wTi02 = 1.0 %. The plant is designed to produce 4000 kg of dry solid,
which means that the total mass of the dry solid will consist of 4000 kg of a Ti02/salt mixture (the
salt comes from the liquid phase, and remains in the solid when it is dried). A material balance can
show us how much wash water is required to attain the desired specifications. The process is
steady-state.
The solution will follow the same steps as in Section 4.6.1. First, draw the flowsheet and
label the streams (Figure 4.17). The specification for the product is labeled next to stream F.
WBH20 : В Wash water
Rutile С Decanted
slurry solution
(wNLH20 = ?) Water vapor FE = ?
v^NaCI = 0.20 E wEH20 = 1
wATi02
Dryer Dry rutile FF= 4000 kg/h
►
v/NaCI = ?
wFTi02 = ?
Figure 4.17 Flowsheet for washing rutile pigment. Single arrow line is for one-phase streams,
and double arrow lines are for two-phase (liquid + Ti02) streams. Values for wTi02 are for the
stream, while values for wNaCl and wH20 are for the liquid phase as designated by a superscript
stream L term. Parenthesis around the water mass fraction term for stream A indicates that its
composition term is determinable by difference.
In Section 4.7.1 we avoided using two-phase flow entities as streams. However, since some
texts do not follow this practice, we abandon it (temporarily) here to illustrate how to designate
two-phase flows as streams. A second superscript letter is used for two-phase streams to designate
the phase. For example, stream D is a slurry with a solid mass fraction of 0.60, so it is written
wDTi02 = 0.60. To designate the composition of NaCl in the liquidphase of stream D, it is written
wD/LNaCl. The ledger is shown below with the dryer omitted. Streams E and F can be calculated
easily once we know the properties of stream D. Notice that now we don't have a specific letter to
use to designate the phase flow in a two-phase stream.
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 169
Aв Stream с
Rutile slurry Wash water Decanted sol'n D
Washed rutile
mass flow, kg/h
?
liquid ? FB ? 0.6(FD)
solid 0.4(FA) 0.01(FC) FD
total FA FB Fc wD/LNaCl
w H20
mass fraction
1.00
in liquid phase
NaCl 0.20 0.00 wc/LNaCl
H20 ? 1.00 wc/LH20
total 1.00 1.00 1.00
The basis for the problem is 4000 kg of slightly-salty dry rutile, as specified. The number of
stream variables is the total number of species present in each stream:
Rutile slurry (stream A) 3
Wash water (stream B) 1
Decanted solution (stream C) 3
Washed rutile (stream D) _3
Stream variables 10
The number of distinct species present in the system is three: T1O2, NaCl, and H20, hence
three independent balance equations can be written . The number of specified compositions is
four: two for the rutile slurry, one for the washed rutile and one for the decanted solution. There is
one specified flow (slightly-salty dry rutile) of 4000 kg/h, and one subsidiary relationship (the
composition of NaCl in the liquid phase of stream С and D must be equal)**. The DOF analysis is:
DOF=10-3-4-l-l=+l
This shows that the system is underspecified. This means we cannot make a balance on the
system without specifying some other process variable. However, we can learn a lot about the
system by making a material balance as a function of some process variable, such as the flow of
wash water (FB). Owing to the large number of calculations that are required for this situation, a
spreadsheet solution is preferred. In addition, it turns out that the arithmetic is slightly easier if a
new basis is selected: FA = 10 000 kg/h. An updated ledger (containing values calculated "by
inspection" is shown on the next page. This shows that four equations are required for every value
of F® selected. The equations are:
Total mass balance: 10 000 + F® - F° + F®
ТЮ2 balance 4000 = 0.01(7*0 + 0.6(7*0
Salt balance 0.2(0.6)(10 000) = 1200 = wc/LNaCl(0.99/^) + wD/LNaCl(O.47*0
Subsidiary relationship wc/LNaCl = wD/LNaCl
A sample calculation for FB = 35 000 kg/h will be shown, then the results of the spreadsheet
calculation for F® in the range 30 000 to 70 000 kg/hr. The first two equations in efficient order
are as follows:
(0.6)(10 000) + 7** = T^ + 7*0
4000 = 0.01(7*0 + 0.6(7*0
* One may write a balance for each of these three species, or for any two plus a total mass balance.
** This assumption is usually, but not always, valid. If not valid, the problem must explicitly state it.
170 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
This allows the elimination of F® so that F 0 can be expressed in terms of FB:
{2000 + 0.6(FB)}/0.59 = 7^
The sample calculation at FB of 35 000 kg/h gives Fc = 38 980 kg/h. F° is calculated as 6020
kg/h. The Ti02 balance equation above gives FcT\02 = 390 kg/h, and Р°Т\02 = 3610 kg/h.
Stream
ÄВ С D
Washed rutile
Rutile slurry Wash water Decanted sol'n
0.4(FD)
massflow,kg/h 0.6(FD)
liquid 0.6(FA) FB 0.99(FC) FO
solid 0.4(FA) — 0.01(FC)
total FA= 10 000 FB Fc
mass fraction
in liquid phase
NaCl 0.20 0.00 wc/LNaCl wD/LNaCl
wD/LH20
H20 0.80 1.00 wC/LH20
1.00
total 1.00 1.00 1.00
The next equation is obtained by combining the salt balance and the subsidiary relationship:
1200 = wc/LNaCl(0.99/^ + 0AF°)
This allows us to calculate wc/LNaCl = wD/LNaCl = 1200/(40 998) = 0.0293 kg/kg solution.
The rest of the mass flow values are calculated in sequence by multiplying the composition terms
and the solution stream flow values. A ledger for FB of 35 000 kg/h is shown below, where nearly
10 % of the incoming Ti02 is lost to the decanted solution. Clearly, the number of significant
figures reported in the ledger is more than can be justified by the accuracy of the reported
composition terms. However, carrying extra figures avoids excessive round off errors, and helps
assure that the arithmetic is correct. The correct number of significant figures should be adjusted
in making a final balance report.
Stream
AВ С D
Washed rutile
Rutile slurry Wash water Decanted sol'n
2406
mass flow, kg/h 3610
6017
liquid 6000 35 000 38 594
Ti02 4000 — 390
total 10 000 35 000 38 983
mass flow of 1200 0 1130 70.4
species, kg/h 4800 35 000 37 464 2336
NaCl
water
mass fraction 0.20 0.00 0.0293 0.0293
in liquid phase 0.80 1.00 0.9707 0.9707
1.00 1.00
NaCl 1.00 1.00
H20
total
Values from the above ledger can be used to calculate F* = 3610 + 70 = 3680 kg/h, and FE =
2336 kg/h. The wFNaCl = 1.9 %.
Figure 4.18 shows the results of a spreadsheet calculation for a range of values of F®. The
loss of pigment in the decant solution is probably unacceptable, and a discharge stream with 1 %
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 171
solids and 1.5 to 3 % salt might likely violate some environmental regulation. The pigment loss
could be decreased by enlarging the volume of the wash settling tank, but the salt content of the
discharge stream could only be decreased by having two or more washing/settling tanks in series,
with the wash water and slurry flowing counter-currently. This would also lower the salt content
of the recovered pigment. Another possibility is to use a filter on stream D.
Pigment Washing Process 2.5%
2.0% о
20%
18% со
to 16% %Ti02lost E
CM 14% - -D - % salt in Ti02
0.0%
о 12% 40000 50000 60000 70000
10%
8%
6%
30000
wash water, kg/h
Figure 4.18 Results of spreadsheet calculation on effect of wash water flow (stream B) on
pigment washing process. Slurry flow (stream A) into system is 10 000 kg/h.
It is easy to scale the basis to the original requirement of 4000 kg/h solid product (stream F)
by applying a proportion factor to all stream flows. For example, the stream F flow is 3680 kg/h.
Thus, multiplying the stream flows by 1.087 will give the originally-specified dry rutile flow of
4000 kg/h.
To test your understanding of the above calculations, calculate the amount of wash water for
stream В such that 4000 kg of dry pigment (stream F) contains 1.3 % NaCl.
4.7.5 A Leaching Process Having a DOF = -1
Dust from a magnetic particle processing plant contains Fe304 and FeS04. The dust
production rate is expected to fluctuate between 150 and 200 kg/h. The FeS04 is water-leached to
separate the soluble and insoluble fractions. The leach tank product is passed to a settling tank,
where most of the insoluble fraction settles and is withdrawn. The dust entering the leach tank has
wFe304 = 0.85. The leach solution stream has wFe304 = 0.02, and wFeS04 = 0.01 in the liquid
phase. The input streams to the settling tank consist of the output stream from the leach tank plus a
stream of water. The slurry stream from the settling tank has wFe304 = 0.40. The decant solution
has wFe304 = 0.01 and wFeS04 in the liquid phase = 0.024. Tank size indicates that the flow of
water plus leach solution can be up to 1250 kg/h, and effluent guidelines indicate that the total
discharge of FeS04 should not exceed 25 kg/h. A mass balance is required to determine the
maximum dust processing rate.
One way to approach the problem is to determine if the upper limit of 200 kg/h of dust can be
processed without exceeding the abovementioned limits. Figure 4.19 shows the flowsheet, which
you will note is similar to that shown in Figure 4.14. The system boundary is drawn around both
process devices, so we are not concerned with the flow of intermediate stream X. The flowsheet is
redrawn to show stream properties and stream data in Figure 4.20. A ledger for the system is
shown below Figure 4.20.
172 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
System
boundary
Figure 4.19 Process device diagram for leaching and washing dust.
Leaching solution Wash water
FB = ?
В вс
wBBFE e304 = 0.02 The sum ofF and F must
wB/LFeS04 = 0.01 wcH20 = 1 not exceed 1250 kg/h.
(wB/LH20= ?)
Decanted
Dust
FA = 200 kg/h Leaching FE = ? solution
wAFe304 : 0.85 and
(wAFeS04 = ?) wEFe304 = 0.01 ThemE/LFeS04
settling
wE/LFeS04 = 0.024 must not exceed
25 kg/h.
(wE/LH20 = ?)
Washed Fe304 FD = ? wDFe304 = 0.40
wD/LFeS04 = ?
wD/LH20 = ?
Figure 4.20 Flow diagram for leaching and washing Fe304. Liquid plus solid streams are
indicated by double-line arrows. The values of wFe304 refer to stream composition, and the values
of w#/LFeS04 and w#/LH20 refer to liquid phase composition in stream #. Parentheses around mass
fraction terms indicate a composition term that can be determined by inspection. The two system
restrictions are given in text boxes.
Aв Stream D E
Dust Leach solution Washed Fe^04 Decant solution
с
mass flow, kg/h ? ?
Wash water 0.4CFD) 0.01(FE)
FeS04 ? ?
0 FD FE
Fe304 0.85(200) 0.02(FB) 0
total 200 FB Fc
mass fraction — 0.01 0.00 0.024 ?
m liquid phase — ? 1.00 ?
?
FeS04 1.00 1.00 1.00 1.00
н2о
total
Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 173
As before, certain values can be calculated "by inspection". For example, w FeS04 must
equal wE/LFeS04. Unknown quantities are represented by symbols. The revised ledger shows
these updates.
The degree of freedom for the system is calculated next. The ledger and flow diagram show
that there are two stream variables in stream A, three in stream B, one in stream C, three in stream
D, and three in stream E. This gives SV = 12. Three independent balances can be written, one for
Fe304, one for FeS04, and one for water (or total mass flow). Six independent stream
compositions and one stream flow have been specified, and there are three subsidiary relationships:
FB + FC< 1250, wD/LFeS04 = wE/LFeS04 and mE/LFeS04 - 25 kg/h. Thus:
DOF = 1 2 - 3 - 6 - 1 - 3 = - -1
Stream
AВ С D E
Dust Leach solution Wash water Washed Fe^Oi Decant solution
mass flow, kg/h
FeS04 30 0.01(0.98FB) 0 0.024(0.6FD) 0.024(0.99FE)
Fe304 170 0.02(FB) 0 0.4(FD) 0.01(FE)
total 200 FB Fc FD FE
mass fraction
in liquid phase
FeS04 — 0.01 0.00 0.024 0.024
0.976 0.976
H20 — 0.99 1.00 1.00 1.00
total ___ 1.00 1.00
Therefore, the system is overspecified. Nevertheless, we can determine the maximum rate of
upper limit of FB + F^ aonfdFwA haesnanwEu/nLkFneoSw04n.=
dust processing, which will be at the 1250 and set the value 25 kg/h. So we
will make a balance with FB + Fc = The new ledger
is shown below.
A Stream E
Dust ВСD Decant solution
0.15(FA) Leach solution Wash water WashedFe-tC^ 25
0.85(FA) 0.01(FE)
mass flow, kg/h 0.01(0.98FB) 0 0.024(0.6FD)
FeS04 FA 0.02(FB) 0.4(FD) FE
Fe304 FB 0 FD
total 1250 -FB 0.024
0.976
mass fraction — 0.01 0.00 0.024 1.00
in liquid phase 0.99 1.00 0.976
1.00 1.00 1.00
FeS04
H20
total
The five equations to solve are: FA+1250 = FO + FE
Total mass: 0.15(FA) + 0.0098(7^) = 0.0144(7^) + 25
FeS04 balance: 0.85(FA) + 0.02(7^) = 0A(F°) + 0.01(1?)
Fe304 balance: F* + Fc= 1250
Stream В and С flow:
Mass flow of FeS04 in stream E: 25 - 0.024(0.99^) = 0.02376(7^)
174 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
The method of solution is to first calculate 7^ from the last equation as 1052.2 kg/h.
Substitute this into the other four equations, and calculate each stream flow by algebraic
substitutions. The ledger below shows the mass balance for the system, which has small errors
introduced by rounding off during calculations. The maximum amount of dust that can be leached
is 189 kg/h (stream A).
Stream
A ВСD E
Dust
Leach solution Wash water Washed FenQ4 Decant solution
28.4
mass flow, kg/h 160.6 2.2 0 5.6 25
4.4 0 154.8 10.5
FeS04 0 215 1028 227 1017
Fe304 189 222 1028 387 1052
н2о
total
To test your understanding of this example, calculate the mass balance for leaching 189 kg/h
of dust if ϊ° = 1000 kg/h. Note that mE/LFeS04 Φ 25 kg/h, and may exceed the limit. Also, F* +
7 ^ ^ 1250 kg/h.
4.8 Using Excel-based Calculational Tools to Solve Equations
One of the main objectives of a material balance is to see how a difference in one variable
affects all the others. Sometimes the balance equations can be set up in a spreadsheet so that
changing one variable recalculates the material balance. In some cases, you can intentionally use a
circular reference, which causes Excel to repeatedly iterate until the formulae converge. Please
consult your Excel help file for advice on using this technique. Where this setup is not possible,
you can use one of Excel's built-in calculation tools: Goal Seek or Solver. These two tools were
customized for making repetitive material balance calculations (Super Goal Seek and Super
Solver). Another program (MMV-C) can convert material balance results from one set of units to
another. Finally, a process simulation tool was developed as an add-in to Excel (FlowBal). These
and other calculational tools are included on the Handbook CD. This Section introduces the first
three tools; FlowBal will be introduced in Section 4.13. The Appendix gives a brief description of
these programs and other equation-solving tools.
4.8.1 Goal Seek and Solver as Calculational Aids
In some cases, the material balance equation set can be structured so that finding the value of
one of the stream variables results in a solved equation set. In that case, Excel's Goal Seek tool
can find the value that solves the set. The sought-after value might be an input or an output stream
variable. To illustrate, we'll redo the Fe304 wash balance example with Excel, which will decrease
the round-off errors. The results are shown below.
с D Hi Jr VJT H 1
FA
m FeS04 189.4 FB Fc FD FE FB + F€
25.0
221.3 1028.7 387.2 1052.2 1250.0
Values in the first three columns are designated as independent variables in that no formula
was written in the cell. In other words, cells C4:E4 contain the values 25.0, 189.4, and 221.3. The
other four cells (F4:I4) contain formulae based on the mass balance equations written in the above
example. Only four balance equations are needed (instead of five) because there are now only four
dependent variables. The four balance equations are:
CellF4: =G4 + H4 D4 E4
CellG4: = (0.15*D4 + 0.0098*E4 C4)/0.0144
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