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Let us see further the problems on material balance in reduction smelting. In an earlier
lecture, I have given the introduction to reduction smelting and the relevant information
to solve material balance problem in
reduction smelting. In that lecture also, I have solved one problem on lead. Here, some
more problems on reduction smelting. For example, production of lead in a lead blast furnace.
Problem two: the charge for
blast furnace or lead blast furnace is given. It consists of roasted ore, pyrites, cinder,
coke and flux. All of you know flux is used to slag to remove the impurities.
As the result of smelting or reduction in smelting in blast furnace, matte of this composition
is produced. It is said no lead is lost anywhere rather in slag or gases. Normally, there may
be some loss of lead in slag
or gases. Depending on the problem, you have to consider, if it is given; if it is not,
you can ignore it. Now, the gases contained here in the ratio
CO is to CO 2 is 1 is to 1 by volume. You have to calculate A, B and C. Now, in the
B part, this is said - the proximate composition and as I have said at many
occasions that proximate analysis consist of minerals. That means you have to report
the presence of mineral not as individual elements. That is what the proximate analysis
means. Volume of blast as such you
have to deal.
Now, the problem 3 - again there is a different type of problem here; some more complication
has been introduced say the burden which consists of sinter, skimming, limestone, cinder and
coke. Skimming in
fact is the product of lead blast furnace which is again recycled in order to recover
lead. As you see in this particular problem, skimming which is obtained contains significant
amount of lead and lead oxide. So, there is a huge loss of lead. So, many
a times in the plant, to improve the economy of the blast furnace smelting of lead ore
- the skimmings are recycled to recover lead. I mean you may recycle in the blast
furnace as a feed or elsewhere. The whole idea is that you do not want to lose lead.
So, here sinter is a roasted 150 tons, then skimmings, limestone, cinder and coke - they
are all the charge and all these 1 2 3 4 5 constitute and the term is called burden.
Burden consists of all these things. Now, it
is further said, the conditions of slag and the components which are there - it is given
to some extent - it is said slag contains 10 parts Fe O to 7 parts Si O 2. It is also
said, you neglect lead in slag and gases.
Further conditions are imposed of the copper charged, two-third enters into matte and one-third
into lead bullion. Remember, you should also understand that
the product of smelting of lead ore is not pure lead; it is an impure lead which is called
lead bullion. When you say lead bullion, it contains several impurities: say
copper is one impurity, silver is another impurity, antimony and so on, depending upon
the elements which are present. These are all the impurities. It is called as lead bullion.
It is said one-third enters into lead
bullion; all sulphur goes into matte, all silver enters into lead bullion.
In order to get lead, you have to further refine and there is a very detailed refining
sequence of production of refined lead from lead bullion. Several steps are involved for
refining of lead bullion to get pure lead.
Now, again here, the gases carry 1.5 parts of CO and 1 part of carbon dioxide. So, you
have to make charge balance. With the term charge balance, you must be very clear what
is being required here. You have to calculate the input and output
both; that is what the balance means. Then, percentage composition of all products and
percentage of carbon burnt at the tuyere when 90 percent of carbon of coke is
burnt to CO and 10 percent to CO 2. So, this point has to be kept in mind; under that condition
you have to solve the problem.
Now, problem 4 - a roasted lead ore is smelted in a blast furnace with enough calcium carbonate
to make a slag of 18.5 percent calcium oxide, coke is 16 percent of the roasted ore and
analyzes 90 percent
carbon and 10 percent Si O 2. The composition of roasted ore is given; of the lead charge
- 5 percent is lost in dust and flue. In earlier problem, we said no loss; now some losses
are also there. So, this problem
illustrates that and 8 percent enters into the matte. Of the copper charge, 50 percent
enters into the matte and rest enters with lead bullion. 10 percent of sulphur enters
into gases. Per thousand kg of roasted ore,
you have to calculate amounts of lead bullion, matte and calcium carbonate.
These are the problems that I have read quite slowly so that you can read it and you can
understand the problem. The answers are given as usual; you should not see the answers before
you make a solution.
Here, I will proceed for the solution of these problems. Let us take problem number 2. I
will first of all calculate amount of lead charge. I hope you must have read the problem.
I am straight away proceeding
with the solution. Amount of lead charged - that is you have to see from all sources.
The source of lead, it is only the roasted ore from where your lead entering.
So, amount of lead charge that is equal to say -310 upon 223 into 207 plus -you have
lead sulphide also, which is 19 percent; so, that is 190 upon 239 into 207. So, that makes
around 452 kg of the lead which is
being charged. Now, lead in matte as per the problem - lead in matte which is 0.02 into
452 is 9.046 kg as per the statement of the problem. So, the amount of lead produced -- as
there is no loss anywhere. So,
the amount of lead produced will be 452 minus this, so, the answer would be 443.26 kg. That
is the answer when it says that you have to calculate the amount of lead produced.
Now, calculate the matte. Matte contains a lead sulphide; it contains Cu 2 S and it contains
Fe S. What you have to do here? You have to make the total balance. First of all, you
make the sulphur balance.
Sulphur charged - you have to see that from all sources- you have to consider sulphur
charge and in this problem that is equal to 31.44 kg; total sulphur is being charged.
Now, you know amount of lead in the matte is 9.046 kg. From there you calculate Pb S;
then see how much amount of sulphur is combined with lead, and then you know the copper. How
much rest of it? The
amount of copper, you calculate in terms of Cu 2 S. Then the amount of Pb S will be equal
to 10.44 kg, Cu 2 S will be 30 kg and Fe S will be 66.1 kg.
I hope you can calculate. So, we can calculate amount of Pb S. From amount of Pb S, you will
be calculating how much sulphur with Pb S. Cu 2 S, you can calculate and balance sulphur
with that will be iron
and that is where you can calculate the amount of Fe S.
Next you have to calculate say for example, amount of slag. The slag will contain Si O
2, that will also contain Fe O and it will also contain calcium oxide. If you consider
all sources - 400 plus 42 plus 19.8, that
is equal to 461.8 kg -that is the Si O 2. Then F e O will be equal to iron charge minus
iron in matte that will be the iron in slag and Fe O would be 72 upon 56. That amount
is equal to 521.28 kg. Now, about
calcium oxide, it is straight way is equal to 123.2 kg. Amount of slag would be 1106.28 kg.
Now, all that is required is you have to proceed as per what is given in the problem exactly
which amount is going where. You have to be careful while converting, say, if you are
considering the amount of Fe O
and if you are making iron balance, then do not forget to convert to Fe O.
So, these are certain things that one should keep in mind while solving the problems. Many
a times over the years I have seen that the student might commit a mistake in converting,
say iron to Fe O or Si to Si O
2, whatever. It depends upon the elemental balance you do or you can directly do the
Fe O in the slag and for that you have to develop an appropriate skill for that.
Here what we are doing simply - we are balancing the input and output and in between input
and output where the material is going where - you should read the problem very carefully.
For example, in the Fe O
calculation, whatever iron you have charged, part of the iron is going into the matte,
only rest iron is going to slag. So, that point is to be clear because it is a material
balance, input, output and in between input
and output, where it is going? What? You have to consider that and then there should be
no problem. Next you have to calculate volume of gases.
Again here, there is a skill involved and method involved. I mean method will vary from
whichever way I do and how you do. I will follow this way. Moles of carbon
is equal to 13.35, then depending on the ratio, moles of carbon for CO will be equal to 13.35
upon 2 and that of CO 2 will also be 13.35 upon 2, because it is 1 is to 1 ratio.
Then, I can find out the moles of oxygen for CO 2 and moles of oxygen for CO. Moles of
O 2 for CO 2 and CO will be 10.013 kg moles. Now I have to do oxygen from charge because
that much amount of
oxygen will not be derived from the blast. So, oxygen from charge -- that means I have
to see CO 2 from PbO, I have to consider O 2 from Fe O, I have to consider plus O 2 from
Fe 2 O 3 and do not forget to
consider O 2 from CaCO 3. So, if I make all this calculation then I
will be getting oxygen in the charge that is 4.919 kg moles. So, oxygen derived from
the blast will be 10.013 minus 4.19. So, the amount of air amount will be equal to 10.013
minus 4.919 divided by 0.21 into 22.4, so, the amount of blast that will be required
will be 543.66 meter cube at 1 atmosphere and 273 Kelvin.
This is about the say some steps for solving problem number 2. Now, let us go to the problem
number 3. In problem number three again, you have to do the charge balance of the furnace.
I am proceeding to
calculate the charge balance. Solution for problem number 3 - first I will calculate
amount of lead bullion. Lead bullion consists of lead plus silver
and plus copper. So, this point has to be noted that all silver is entering into the
lead. Now, we have to calculate the total amount of lead charged from several sources.
That will come around 77.98 tons. Now, copper charged - again you have to look at the sources
of copper from where it is entering into furnace. The only source is cinder which contains 2
percent copper
sulphide and cinder is 50 tons and another source is sinter also. You have to see from
all sources; you have to take into account the copper which is entering into the system.
That way, the copper charge will be equal to say, 1 upon 100 into 150 into 128 upon
160 plus 2 upon 100 into 50 into 128 upon 160. I mean you have to convert also. You
have to excess given accordingly
copper So, that will be coming out to be equal to
2 tons - that means total copper you are charging is 2 tons. Now, you have to see sulphur charged
and again you have to make some calculations. So, sulphur charge will
be 3.62 plus 0.95 plus 0.5. I have omitted some steps I think they are very easy and
you can come to this figure. You have to consider from all source from where sulphur is entering
- that is equal to 5.07 tons -
that much amount of sulphur charge. Now, you have to calculate iron charged from
all the sources from where iron is charged. I see in the problem, the source of iron is
sinter that is Fe 2 O 3 which is 19 percent and cinder which also contains 90
percent Fe 2 O 3. So, you have to consider both the sources. I have just calculated and
I will leave the calculation for you. That will come out to be 51.45 tons.
Now, you have to calculate the charge balance; so matte and everything we have to calculate.
Copper in matte will be equal to 4 by 3 tons as per the condition of the problem and copper
in lead bullion is equal to
2 by 3 tons, that 2 tons will be divided accordingly. So, sulphur used for copper to make Cu 2 S
- because in the matte, the sulphur will not be free; it is combined with the Cu 2 S. So,
sulphur used for copper is
equal to 1 by 3 tons. Now, say Si O 2 in slag is equal to 40.75 tons; Fe O in slag will be 58.21 tons. So, Fe in slag is equal
to 45.27 tons. Now, why I am doing this thing because first of all I must know how much
amount of iron is
entering into the matte then I can calculate. So, iron in slag I know; now iron in matte,
I can calculate. Iron in matte will be equal to 6.18 tons.
Now, I can calculate Fe S in matte. So, Fe S in matte will be 9.71 tons. I can calculate
sulphur with lead in matte; that will be equal to 5.07 minus 1 by 3 as I have earlier calculated
minus 3.53. So, that will be
equal to 1.21 tons. Now, I can calculate weight of Pb S; that will be equal to 9.04 tons because
you have to multiply by 239 divided by 32 because the sulphur with lead in matte that
is 1.21. Accordingly, you will
get 9.04 tons. Now, I can calculate lead in matte. Lead in
matte will be 7.83 tons; it is simply 207 by 239. So, lead in matte I know. Now, I can
calculate lead in lead bullion; that will be equal to 77.98 minus 7.83- that will be
equal to 70.15 tons. In order to calculate lead in lead bullion, you have to make all these exercise because some
amount of lead in lost in the slag or iron is entering into matte and slag and so on
and to come up
with the lead sulphide and so on. So, straightaway you would not to able to calculate what is
the lead in lead bullion. That is why this exercise has to be done.
Now, copper in lead bullion is 0.67 tons and silver in lead bullion will be 0.34 tons.
Now, I can calculate weight of bullion or amount of bullion. Weight of lead bullion
-I have to sum total, and that will be equal to
71.16 tons. Then we can just calculate the percent. Therefore, lead will be 98.58 percent,
copper 0.94 percent, and silver 0.48 percent. This is just percentage wise lead bullion.
Now, we can calculate say weight of matte. Weight of matte is equal to 20.42 tons. You
have to sum total and if you say percentage wise, let me give some other color,
so Pb S is equal to 44 point 27 percent, Cu 2 S -8.18 percent and Fe S -47.55 percent.
That is what the weight of matte. Next is weight of slag, slag weight is 115.94 tons
and percentage wise, it has Fe O, it has Si O 2, it has calcium oxide. So, Fe O is 50.21
percent, Si O 2 is 35.15 percent and
calcium oxide is 14.64 percent. Now, in this problem you should also understand what are
the inputs and what are the outputs. The output is one lead bullion, matte and slag.
Since the charge does not contain arsenic, there will not be any spice. You remember
in earlier lecture, I have said that the output of the less blast furnace consist of slag,
matte, spice and lead. Since the charge
does not contain any arsenic, there will not any spice in the output of blast furnace.
The rest you had seen that whatever I said in the introduction of reduction smelting
for lead, you are seeing that the outputs are
slag, matte and lead bullion and also outputs you will also we have to find out. The problem
says that percentage of carbon burnt at the tuyere when 90 percent carbon of the coke
is burnt to CO and 10 percent
to CO 2. Now, I am giving you the solution; but then
you have to think how to approach this particular problem. My suggestion would be, in order
to solve this particular part, that is part c of problem 3, I will like you that
not to see the solution, think over the problem and then arrive at your own solution and then
compare with what I have done. Because I have also done with a long time
and long exercise and long thinking, I would like you to see how such problems should be
tackled. What I am doing now, since nothing is known;
I will be doing only oxygen balance and carbon balance- that is the key. So, let us take
it now - say x ton mole of oxygen is supplied through tuyeres. Now,
remember why I have taken this approach, because from the blast furnace, lead is smelting.
It is clear that whatever air you supply, it is used for oxidation of carbon only in
this particular system. That is the key to approach this particular
problem, because whatever air you supply, it is used for oxidation of carbon. I am considering
now, say -y ton mole of CO and CO 2 at the exit. I do not know any
other information except this, because it says 90 percent carbon to CO and 10 percent
to CO 2; rest information I do not have. So, I cannot make any nitrogen balance in order
to calculate the volume of the
blast. Nothing I can do. This is what the information I have.
Now, I have to do oxygen balance. I will write oxygen balance; see the oxygen balance now
- O 2 from charge, because you know a part of the O 2 is also coming from the charge
from Fe 2 O 3 from oxidation,
Fe O and so on; whatever the Pb O is there in the charge. So, oxygen is also coming.
So, accordingly the oxygen which is blowing through the tuyere will be required less than
what oxygen is contained in the
charge. The system has its own oxygen. That point is to be clear.
Oxygen from the charge minus oxygen in slag, because it is following by oxidation, plus
oxygen supplied through air
will be equal to oxygen in exit gas. This balance we should be able to make in order
to solve
this particular problem. You have to consider the oxygen which is present in the system
which is introduced in the form of the charge, also the oxygen in the slag, Fe O and all
these things although has to be
considered. So, if I do that, then I form the balance;
which is 1 minus 0.4 plus x -please do the calculation I have given enough hint to you
- that will be equal to 1.5 y upon 2.5 into 1 by 2 plus y upon 2.. That is my equation
1. Next, let me do the carbon balance. Do not forget to take oxygen from CaCO 3.
Carbon from coke plus carbon from CaCO 3 will be equal to carbon in the exit gases; because
carbon is not being accumulated. At the steady state material balance, input should be equal
to output. So, that will
be y upon 2.5 plus 1.5 y upon 2.5 and this is my equation number 2. If I make this equation
in kg mole, that will be 27 by 12 plus 0.1425 equal to y.
Now, please think it over before you see the solution. I mean do not see the solution.
First try to understand the problem and try to solve and see how to approach such problem.
I mean this is a very tricky
problem.
Now, I can calculate from here y by equation 1 and 2, y will be equal to 2.3925 ton moles.
I hope y, we have considered as CO plus CO 2 and x is equal to 1.075 ton moles. If you
recall x was oxygen supplied
through tuyere or oxygen from air, whichever way you want to understand. So, now I have
to find out carbon burnt for CO. Carbon burnt for CO will be equal to 0.9 into 1.075 into
2. So, carbon burnt for CO
will be 1.935 ton moles and carbon burnt for CO 2 will be 0.1075 ton moles.
So, total carbon in tons will be equal to 24.5 tons. So, percent carbon burnt at the
tuyere
will be equal to 24.5 upon 27 into 100; so, that will be equal to 90.7 percent and that
is the answer for this. Now, that is
what the answer for the part c of the problem number 3. Here, it says that the charge balanced;
so, ultimately you will be putting in the answer, the charge balance. All the weights
inputs are given. So, in the
output, you will be writing the weights of matte, weights of slag, weights of lead bullion,
then carbon burnt at the tuyere and all these things will comprise of the charge balance
of the process. That is what the
problem number 3.
Let us see the 4th problem. 4th problem is where you have to find out amount of lead
bullion, matte and calcium carbonate. Let us make a box of material balance. This is
what a fuel box balance. Let us take the
basis of calculation, say, 1000 kg. 1000 kg is the lead ore that we are using. 1000 kg
lead ore- the composition I am writing, Pb O is given 25 percent, Pb S - 18 percent,
Fe 2 O 3 - 22 percent, Cu 2 S - 2 percent,
Si O 2 - 29 percent and calcium oxide is 4 percent; that is what the ore.
This is the lead ore. Coke is 160 kg and carbon is equal to 90 percent and SiO 2 is equal to 10 percent. The problem further says
that pure calcium carbonate is charge; pure calcium carbonate is the flux, I mean
the agent to flux. It further says you produce lead bullion and lead bullion should contain lead and copper.
That is what the output I am giving. Then next output you have is the matte and matte
will contain Pb S,
Cu 2 S and Fe S. Then, slag and it is said slag is having 18.5
percent calcium oxide. That is what the problem is said. Now, you have to find out the amount
of lead bullion, matte and calcium carbonate in this particular problem.
Now, they are all smelt in the blast furnace and these are the outputs that are given.
Now, to calculate lead bullion, we first of all, calculate lead charged. Amount of lead
charged will be equal to 250 upon 223 into
207 plus 108 upon 239 into 207. Mind you, here I have taken the atomic weights
of lead 207, then sulphur I will be taking 32, iron I will be taking 56, copper, if at
all needed, I will take 64, silicon if needed, I will take 28, calcium if needed, I will
take 40 and that is all and oxygen of course, its atomic weight is 16. These are the atomic
weights that I will be using to solve this particular problem.
So, amount of lead charge will be equal to 232 plus 156 and that makes 388 kg; that is
the amount of lead you will charged. So, loss of lead in fumes and gases or whatever will
be equal to 388 into 0.05; that is
equal to 19.4 kg. Lead in matte, as it is said, it is equal to 388 into 0.08, that is
equal to 31 kg and therefore, Pb S in matte can easily recalculate 35.84 kg.
That is one compound or one part of matte, which is Pb S. Now, let us calculate Cu 2
S. So, for Cu 2 S, the copper charged is equal to 20 into 128 upon 160 that come around 16.
So, copper in matte, as
problem says is equal to 8 kg. Therefore, amount of Cu 2 S is equal to 10 kg. This is
another component of the matte and rest copper is in bullion.
So, amount of lead bullion now we can calculate. Amount of lead bullion will be consisting of lead plus copper; so,
this will be equal to 345.6 kg. That is the amount of lead bullion. You can just sum total,
lead
and the copper, which is entering into the bullion.
Let us calculate now, matte and calcium carbonate. We have calculated only lead bullion. Now,
in order to calculate matte, we have to do some exercise. First of all, we have to calculate
how much amount of
sulphur is charged. Sulphur charged will be equal to 180 into 32 upon 239 plus 20 into
32 upon 160 and that makes 28 kg. That much of amount of sulphur is being charged.
Sulphur in matte is equal to 25.2 kg. Now, we have to find out sulphur with Pb S of matte and sulphur
with Cu 2 S of matte. That means, first you have to find out sulphur with Pb S of matte
and then you have
to find out sulphur with Cu 2 S of matte. How much lead is entering into the matte;
how much copper - that calculation we have done earlier, and total sulphur which is with
Pb S and Cu 2 S is equal to 6.8; here
it is 4.8 and that is equal to 2, of course in kg. They are all in kg.
Sulphur with iron in matte, of course, you have to subtract- that will be 18.4 kg. That
much amount of sulphur with iron in matte. From here, amount of Fe S will be equal to 18.4 into 88 divided by 32,
so,
amount of Fe S will be 50.6 kg.
We are in a position to write down the matte. So, matte amount would be Pb S, then it has
Cu 2 S, then it has Fe S. Pb S, as we have calculated, 35.84, Cu 2 S, 10.00, Fe S is
50.60. Mind you, they are all in kg.
In percent wise, the percent is 37.16, 10.37 and 55.47 that is what the matte comprises
of. Now, we have to calculate the amount of slag
in order to calculate amount of calcium carbonate; because we cannot calculate calcium carbonate
unless you know the calcium oxide. For that purpose, first you
have to calculate the amount of slag and then calcium oxide, then calcium carbonate.
So, the slag will have Si O 2 from all sources - that will be 306 kg. This is the amount
of Si O 2. Then you have to calculate Fe O. Fe O would be iron charged minus iron in matte.
Do not forget to convert it into
Fe O - that will be 72 by 56. So, iron charge is 154 kg minus 32.2 into 72 by 56. So, the
amount of Fe O will be equal to 156.6 kg. Once you know the amount of Fe O, now I can
calculate the weight of slag. Weight of slag will be 306 plus 156.6, because it says, 18.5
percent is the CO in slag. So, 100 minus 18.5 percent will be sum total of
Si O 2 and Fe O. So, if I divide by 81.5 into 100, then the weight slag will be equal to
567.6 kg. Once again, in order to calculate the amount of calcium carbonate which is being
asked in the problem, you must
be wondering why I have calculated Si O 2 and Fe O 3, because the problem says that
the slag contains 18.5 percent calcium oxide. So, unless you calculate the amount of slag,
you cannot calculate the amount of calcium carbonate. If you have another way of calculation
I do not know. Think. Weight of slag I got now. Now, I can calculate
calcium oxide in slag. Calcium oxide in slag will be 105 kg. Again, you have to do here
calcium oxide balance; because, if you read the problem, the problem says the roasted
ore has some four percent calcium
oxide. So, calcium carbonate, Ca O from CaCO 3 will be 105 minus 40 - that will
be 65 kg. As such, amount of CaCO 3 will be equal to 116 kg. This is how the problems are to be tackled. Now, here
in this particular
session, what I have done, I have illustrated the different ways of problems and different
methods to approach the problem. However, it is intended that you develop your own method
of solution of the problem
without seeing the solution and of course, without seeing the answers.