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Having done this mineral processing, that is the first step towards metal extraction
either from pyro metallurgy of hydro metallurgical route. Now, we are in a position to go ahead
with the various unit processes
of metal extraction. Here, first I will deal with the pyro metallurgical and then I will
touch upon hydro metallurgical extraction unit processes also.
Now in this, calcination is the first step and that is use for several purposes. So here
I am giving you first of all the material and heat balance problem in calcination then,
I will proceed with little bit amount of
concept and then straight away I will enter into the solution of these problems.
So, the first problem: calculate the heat energy required to calcine 1000 kilogram of
limestone of composition, 84 percent of calcium carbonate, 8 percent Mg C O 3 and 8 percent
H 2 O which is charged at 298
kelvin. Lime is discharged at 1173 kelvin and gases leave at 473 kelvin.
I mind you, this limestone is not yet used as such in various steel making processes
or in non ferrous metal extraction process as you will see in the solution of the problem;
that the decomposition of limestone is
associated with a large amount of absorption of energy is highly endothermic reaction.
Therefore, use of lime requires that limestone should be calcine and therefore, this problem
is included in order to appreciate the fact that charging of lime after calcination of
lime stone is very important step that
is the objective of this particular problem. Second, it is desire to produce 10 kg mole
lime from calcination of calcium carbonate, calcium carbonate is pure in a rotary kiln,
while in the first problem it was only having 84 percent calcium carbonate; in the
second it is 100 percent calcium carbonate. Rotary kiln is very often used to produce
lime from Calcination of limestone. Rotary kilns are very long kiln and rotate 2 to 3
degree from the axis of the horizontal axis of a one side, the feed enters and from
other side, the calcine material discharges; around 24 meter long this kiln and they are
frequently heated by an externals source of energy and they are use for example, oil or
producer gas. So, this problem is
concerned about that. So, you see the producer gas of composition
this CO2, O2, CO and nitrogen is combusted with 20 percent excess air to obtain the desire
temperature in the kiln, because normally the temperature inside the kiln
is very high and it may go of the order of 1400 or 1500 degree Celsius.
The limestone and air are supplied at 298 kelvin, where as the producer gas is heated
to 900 kelvin, lime is discharged at 1200 kelvin and gases at 500 kelvin. Now, the fact
that they are discharging at low
temperature then, the kiln they are is occurring some sort of heat transfer before the product
and the byproduct leave the rotary kiln, this is the problem number 2.
Problem number 3: In the electrolysis, anhydrous alumina is required. Now, the concept of anhydrous
alumina is bauxite after leaching, it gives you it gives you hydrous alumina that is Al
O H 3. Now, as an electrolysis of Al O H 3 is not
possible, because then only hydrogen will evolve. So for that, it is required that the
Al 2 O 3 which is used for electrolysis to produce aluminum it should be anhydrous,
since there should be no water into the Al 2 O 3 which you are going to charge in the
all around cell. So, in the electrolysis anhydrous alumina
is required. Now, for your information in this anhydrous alumina, the water is most
if chemically combined. So, it is not possible to remove the water just by heating by
100 or 200 degree centigrade; you have to go to 1200 or 1300 or even 1400 degree Celsius
to remove the chemically combined water from the product which you obtain after the Bayer's
process. It is for this purpose, a large amount of
energy is required to produce an anhydrous alumina, in order that alumina can be electrolyze
to produce aluminum. For this purpose, Al O H 3 is calcine at 1700 kelvin in a
rotary kiln. A kiln receives a damp filter cake of Al O H 3 analyzing this 55 percent
Al 2 O 3 and 45 percent total H 2 O, free and combined of which the proportion of combined
water is very high and produce
pure Al 2 O 3 as a solid product. The fuel consumption is estimated to be 0.2
kg of fuel oil of composition, 84 percent carbon and 16 percent hydrogen per kg of aluminum
you see. There is also required a source of energy and therefore, the
energy calculation is a very important step in all most all pyro metallurgical extraction.
Say, you are using around 0.2 kg, per kg alumina; that is around 200 kg for 1 ton of alumina,
in order to have that anhydrous alumina. Assume complete combustion and heat loses, 10 percent
of heat input, 20
percent excess air is use for combustion or you have to find out, volume of gasses wet
and dry composition and perform the heat balance. Now, this is a very important step.
Fourth, calculate the minimum amount of fuel to produce thousand kg Al 2 O 3 use a data
given in problem 3. Now, here the whole idea is to know, what is the minimum fuel? The
minimum fuel is that fuel at
which, the heat input becomes equal to heat output.
So, these are the problems that I have selected. Now, here some of the data you will be requiring
to calculate the material heat; especially to calculate the heat balance. These are the
data and these are the further
data . Now here, what I have done? I have given you the heat content at the temperature
at which the problem is there. That means for example, calcium oxide is leaving
at 1200 kelvin, so data are given according to that. However, I will request you that,
please try to calculate by using the c p value of respective of calcium oxide
of C O 2, N 2, O 2 or whatever the product which are being discharge, that will give
you an additional practice. Whether you can calculate the heat content from c p value
or not, this all requires a patience to see
that you can do integration properly. However, this makes your life easier but, see that
also with the by using the c p calculation. Latent heat is also given and here I have
given some of the references. As usual here I have given the solutions for these problems.
Now, let us go for the solution of these problems. First of all, let me tell you, what is calcination?
In fact calcination, this also referred as a calcining, it is a thermal treatment process
apply to ores and other solid
materials; why, in order to bring thermal composition or thermal decomposition, because
you want to decompose. For example, calcium carbonate to calcium oxide or C O 2 or Mg
O C O 3 and so on. Second
possibility could be phase transition, third removal of volatile fractions; these are
some of the objectives of calcination however, you include more.
Now, the temperature involved say, below the melting point of the components of the raw
material. That means you treat them in the solid state; so product is also solid. However,
if the volatile components like
C O 2 or H 2 O they will be leaving the system. Product of calcination is sometimes called
calcine, now the furnaces which are used they are called rotary kiln. Very commonly rotary
kiln is employed for production of cement.
Cement industry rotary kiln is a very big application, also calcination of limestone,
also the removal of water from alumina and so on. Then fluidized bed reactors, multiple
hearth furnaces
and the fuel or fuel, let
another you can put source of energy. Source of energy, in some case it is the fossil
fuel that is could be oil natural gas or there could be secondary fuel which are derived
from fossil fuel and normally sometimes producer gas is used, as you have
seen in the problem. Oil, natural gas all are the sources of energy;
so it is also energy consuming unit process. Some of the applications of calcination: one
particular application is to produce cement - now here the calcination is one
of the important steps, where you get calcium oxide and that is further mix to produce cement.
So, in cement industry the decomposition of limestone is a very important in your process.
Second application is decomposition of hydrated minerals. Now, this decomposition is done
for one purpose could be we want to make say Al 2 O 3 refractory.
Here, bauxite is the material to produce Al 2 O 3 refractory. Now, bauxite also contains
the water which is chemically combined, so straight away you have to calcinate in order
to produce an alumina, which is
free from water and mind you, alumina is very important effective material in all pyro metallurgical
extraction industries. One of the purposes could be produce alumina
refractory and another purpose could be to produce alumina for electrolysis. As all of you know, because that Al 2 O 3 should
be anhydrous in the sense of
anhydrous otherwise, there is evolution of hydrogen and you will be consuming a large
amount of electrical energy to produce aluminum if you do not remove water from alumina.
Third application could be decomposition of volatile matter, which is contained in petroleum
ore. Another application could be heat treatment to effect phase transformation; however in
a strict sense, you do not
call it to be calcination but, since you are applying the heat below the melting point
of the material. You may class each under as a calcination
heat treatment and you may not class, it does not matter but, all that you require a solid
state transformation from room temperature to another temperature. These are
the certain applications of the calcination. Now, having given those things, let us straight
away go to the problem to the solution of problem number 1.
Now problem number 1, we have to calculate the heat energy to calculate 1000 kg limestone
and composition everything is given to us.
So, the reaction that takes place is C a C O 3 - solid that is equal to C a O - solid
plus C O 2 - gas, this is the reaction or this is the decomposition reaction. Now, we
have 840 kg C a C O 3, we are taking say
1000 kg limestone as the basis. The basis is as already being given in the problem 1000
kg limestone, 840 kg C a C O 3, 80 kg Mg C O 3 and 80 kg water, so this are given to
us. We can straight away calculate the calcined
product. Now before that, what I have used? I have use atomic weight of calcium as 40,
carbon as 12, oxygen as 16 and magnesium as 24. So, from here I will be
getting say, kg moles of calcium carbonate that will be equal to straight away 8.4, kg
moles of Mg C O 3 that will be equal to 0.952, and kg moles of water that will be equal to
4.444. Once you know this, now the calcine product
immediately we can write down, say calcium oxide and magnesium oxide. So calcium oxide
straightaway 1 kg mole 1 kg mole and 1 kg mole C a O will be 8.4 kg
moles, Mg O will be 0.952 kg moles. Similarly the reaction for Mg C O 3 will be Mg O plus
C O 2, solve. We have the calcine product, now the gases that will form because the gases
will also form, so we have
to know the gases. Gases that will be forming the C O 2, C O
2 is coming because of the decomposition and the sources of C O 2 is C a C O 3 and Mg C
O 3, so you have to add both of them and H 2 O these are the two forces.
H 2 O is coming from the limestone composition of gases, C O 2 that will be equal to 9. 352
kg moles and H 2 O that will be equal to 4.444 they are in kg moles. Now, we have to calculate
heat energy, so we
have to calculate now heat of decomposition. The values I have given, I will straight away
write down the heat of decomposition that will be equal to 382186 kilo Calorie; that
is the heat of the composition.
Then I have to find out now, sensible heat in products and gases that is all, that is
what I have to find out and for that and if I sum total of all these, that is what the
energy I require, if I want to decompose
calcium carbonate. So sensible heat for example, in C O that
will be equal to 8.4 kg moles, integrate it from 298 to 1173 kelvin, 49.662 plus 4.519
into 10 to the power minus 3, T minus 6.945 into 10 to the power 5, for T square
this I have to integrate to dt. Now, please do this exercise, however if you
try you can get those values also in the reference which I have given but, then they are intermediate
temperature where you will not find the value, so it is better to
develop the practice of integration. So if you do that, then the calcium oxide
will be equal to, the content in calcium oxide 89547 that is kilo calorie; this is the heat
content or I might have written H 1173 minus H 298 in calcium oxide that is what
well. Similarly, I can find out for Mg O that will
be equal to now 0.952, again I have to integrate from 298 to 1173, 48.995 plus 3.138 into 10
to the power minus 3, T minus 11.715 into 10 to the power 5, T square
integrated d t. So I get now, H 1173 minus H 298 for Mg O that will be equal to 9542
kilo calorie. Then I have to calculate, now the gases are
discharging at 473 kelvin; so I write H 473 minus H 298, for C O 2 - the moles C O 2 are
9.352, I have to write it c p value 298 to 473, 44.141 plus 9.037 into 10 to the
power minus 3 T, minus 8.535 into 10 to the power 5 upon T square, integrate to dt. So
that will be dt here, this value is coming 16252 kilo calorie.
Be careful while calculating the heat content in water vapor, so we are starting 298 is
our basis. First, we will calculate H 2 O - liquid 298 that is equal to H 2 O - liquid
373 kelvin and this value is 6001 kilo calorie.
Do not forget to it; let us in theta by operation. So, another state is H 2 O - liquid 373 kelvin
that will be equal to H 2 O vapor at 373 kelvin, all of you know at the latent heat temperature
change does not occur;
so this value is 11158 kilo calorie. Now, I have to heat the H 2 O - vapor 4.444
that will go from 373 because 100 kelvin already has been taken into account to 473, 30 plus
10.711 into 10 to the power minus 3, T minus 0.335 into 10 to the power
5 upon T squared dt and this value is 3643 kilo calorie.
The heat energy required as asked in the problem, I think by now all of you can guess you have
to sum total all the energies, because the heat of decomposition that is also endothermic
reaction. So, there will be
large amount heat absorption plus all the heat carried out. All that heat if you do
not supply, the decomposition of calcium carbonate will not occur.
So all that you have to add and the addition brings 518329 kilo calorie, now I have used
1 kilo calorie, that is equal to 4.186 kilo joule and that gives me the 2.17 into 10 to
the power 6 kilo joule and that is the
answer for problem number one. That is the way, you will be proceeding to
solve. My appeal to you again that, please solve these problems on your own without seeing
the solution, because if you look to the solution you will be carried away by
my way in which I have done. I would like you to develop your own ways and see that
you come off with the innovative ways. I am sure that there will be definitely alternative
way to solve the problem and who
knows that your way or your way of solution may be much better than what I have done,
so please do yourself that is an important thing.
Now, second problem: in a second problem, you have to produce 10 kg mole of lime and
so on. So, what we have to do? We have to calculate the amount of producer gas. First
thing, how will you calculate the
amount of producer gas? Because producer gas amount is being asked, so the only way to
calculate the amount of producer gas is do the heat balance.
If you do the heat balance because all the energies coming from the combustion of producer
gas; so if you do heat input and make it equal to heat output, now only variable in your
equation would be the amount
of producer gas and you can solve that way.
So, heat balance is the key to solve problem number 2 and now I will just make a box for
you so that I mean, I will find it very convenient when I make a box and try to put all the values
over here. It is not
necessary that you also do in the same way. So what I will do, what we are doing? Say
10 kg mole of calcium carbonate they are entering at 298 kelvin right.
It is fired with producer gas and producer gas is entering at 900 kelvin. Mind you, the producer gas bring sensible
heat into the furnace remember, so the composition of producer gas is given; C O 2, it has
oxygen, it has carbon monoxide, it has nitrogen. Now C O 2 is 7.2 percent, oxygen 1.6 percent,
carbon monoxide 16.6 percent, and nitrogen is 74.6 percent.
Now problem say further, that air is 20 percent excess; that means you have use 120 percent
theoretical air and the supply temperature of air is 298 kelvin, that is what is given.
On the output side, here lime which
is 10 kg mole and say lime is discharge at a temperature of 1200 kelvin and the gases, they discharge
at 500 kelvin. The gases would contain, as you can see in the problem it will contain,
C O 2; now since excess
air is used, it will also contain excess oxygen and since air is used, it will also contain
nitrogen. In this situation, you have to find out the
amount of producer gas; all that you have to do the heat balance. Here I like to tell
you that, this is again, first of all you have to make the material balance of this
combustion. For your information, I have also made one
video course on fuel furnace and refractory and there, very detail calculations from combustion
in different lectures are given. The problems are also given to illustrate,
how to calculate. So, if you find any difficulty in calculating the amount of air which is
required for the combustion, I will request you to see those video lectures on fuel furnace
and technology they are also
available. So, please for the details you can see those lectures on combustion and you
may get feel how to calculate. I have also solved several problems over there. So coming
back to this problem, let us see
that Y kg mole is required for the required producer gas.
Now, first of all we have to calculate the calorific value of producer gas. The illustration
of calorific value also, I have given in the earlier ports on fuel furnace and refractory.
So in fact, in case of combustion the
gaseous fuel, combustion the C O, hydrogen and hydro carbon they are the combustible
component. C O 2, N 2 they are the diluents and they do not take part in the combustion,
so mind you that is an
important thing. So, the calorific value
of producer gas that will consists of only when C O is completely combusted to C O 2.
So, the definition of calorific value is the amount of heat energy released when 1 kg or
1 meter cube
of gas or solid fuel is burnt completely, the products of combustion is complete that
is an important thing for the detail you can see I already referred the video lecture on
fuel furnace and refractory. So, the calorific value of producer gas would
be equal to 11271 kilo calorie into Y because, Y is the amount of kg mole of the producer
gas; that you have taken for combustion purposes. Now straightaway, I can take the material
balance. Now, if I do the material balance then amount of C O 2 that will be equal to
10 plus 0.238 Y. 10 is coming from 10 kg mole calcium carbonate will also give you
10 kg mole of carbon dioxide. Nitrogen: all the nitrogen of producer gas will be available
in the gases or the output that plus the amount of air or nitrogen, of the air. So that will
be equal to 0.746 Y plus 0.302 Y. Then the excess oxygen- Now remember here,
producer gas also contains oxygen. So while calculating amount of oxygen that is coming
from air you have to separate the amount of oxygen that is already
available in the system. So, that will be equal to 0.0134 Y.
So, once you have done material balance. Now you have to do the heat input and heat output;
heat input one is the sensible heat. Heat of decomposition though it is in endothermic
but, it is also as a heat input.
So, I will straightaway write down the values, so sensible heat in producer gas that is equal to 4538 Y kilo calorie, sensible
heat in air is 0 because, air is supplied at 298 kelvin, so I am not writing.
Now, heat of decomposition of calcium carbonate, you have to calculate from the value C a C
O 3, C a O plus C O 2 product minus reaction and so on that already I have illustrated
in my thermo chemistry
lecture. So, that will be equal to 427500 kilo calorie, mind you this is an endothermic
reaction, this not exothermic. Heat carried or taken by calcium oxide, because
it is discharge at 1200 kelvin that is equal to 108000 kilo calorie. Now, heat carried
by flue gases that will be equal to 19870 plus 1978.5 Y.
Now, all that you have to do the heat balance. Now, the heat balance would be say 4538 Y
plus 11271 Y minus 427500 that is equal to 108000 plus 19870 plus 1978.5 Y.
So I can calculate now Y that will be equal to 40.15 kg moles and that will be equal to
899 meter cube express at 1 atmosphere and 273 kelvin. So that is equal to 2964 meter
cube at 1 atmosphere and 900 kelvin,
so this is all about problem number 2.
Now, let us quickly go through problem number 3. In problem number 3, we have a rotary kiln;
this is a rotary kiln. Now in the rotary kiln, the flow is counter current. So here, wet
cake which contains 55 percent
Al 2 O 3, 45 percent H 2 O. Now, this H 2 O is mind you is chemically combined and it
enters a 298 kelvin, I always prefer to write all the details of the problem on this material
balance figure. Now the flue gases, they leave at 800 kelvin.
Now here, the fuel oil is supplied for combustion. So, fuel oil which contains 84 percent carbon
and 16 percent hydrogen is supplied a 298 kelvin.
Here, then supply 20 excess air or you can also call 120 percent theoretical air this
is also supplied at 298 kelvin, when? You have calcine alumina which is at 1000 kelvin,
so that is what is given to us. Now as usual, we have to perform the heat
balance. Well first off all, you have to do the material balance without that you can
do nothing. So, I will just proceed to solution. Now here well. Let me tell another
thing here; the wet cake, it goes in this direction which is the direction of flow of
wet cake and this is the direction of flow of flue gases that is how the flue gases are
leaving. So, they are in the counter current mode.
So, the basis of calculation
is 1000 kg calcine Al 2 O 3. Now, amount of fuel oil is 200 kg is already given. The various
calcination reactions is 2 Al O H 3 that is equal to
Al 2 O 3 plus 3 H 2 O, this is the calcination reaction. Then, we have combustion reaction
is C plus half O 2 that is equal to C O 2. Now remember, if nothing is given then always
assume there is a complete combustion and the product of complete combustion are very
clear carbon is C O 2, H is H 2 O. 2 H plus half O 2 that is equal to H 2
O. So, we have assumed complete combustion.
So, let us now calculate volume of H 2 O in flue gas that will comprise of volume of hydrogen
that is combustible to H 2 O plus the water which is removed from the cake, so that will
be equal to 1376.4 meter
cube. Similarly, we have C O 2 in the flue gas that
will be 314 meter cube. Then, since we are using 20 percent excess oxygen, so there will
also be excess oxygen; so excess oxygen in the flue gas that will be equal to
98.6 meter cube, Nitrogen in the flue gas that will be equal to 2225 meter cube.
Now, I tell you once again over here that it is at most important, that you should be
very thorough in doing material balance because, without material balance we cannot heat balance
at all. If you have any
problem you have done while doing the material balance that this problem will be carried
over till the end of the heat balance because, you have to multiply all the values of the
amount with the heat that has been
produce. So, if you do a mistake in material balance and if you get the solution say after
having ten fourteen steps then all steps will be wrong. So it is a transferred error, so
you have to very careful while doing
material balance problem that is very important. So here, we have to calculate volume of flue
gas. So, immediately we can calculate volume of flue gas, you have to calculate wet and
dry; so wet volume is 4014 meter cube. Now, you have to calculate
composition - the composition wet basis and you have to also calculate composition on
dry basis, so we have C O 2; we have O 2; we have nitrogen and we have H 2 O.
So, if you do this calculation is I have already given in the answer 7.8 percent, 2.5 percent 55.4 percent
and 34.3 percent. Now these volumes are all on volume percent. Dry that will be 11.9 percent,
3.7 percent
and 84.4 percent. Mind you, all these volumes they are express at 1 atmosphere and 273 kelvin
because I have taken 1 kg mole is 22.4 meter cube, so that is what this calculation of
the composition.
So heat balance you have to do it; heat input one will be by combustion. So you have to calculate C
plus O 2 C O 2 and H 2 plus half O 2 H 2 O. So, this heat input by combustion that will
be equal to 2414120
kilo calorie. Then, heat of decomposition of calcium carbonate because this is an endothermic.
If I do combustion as positive then, this will be equal to negative 238137, this is
the heat input. Now, heat output will also be in kilo calorie.
Flue gases that will be equal to 1362135 then Al 2 O 3 it is being discharged at 1000 kelvin
it will be 183431 then heat losses are given as 10 percent of the input. So
here, the input will be 2175983, so heat losses will be 10 percent of that 217598. Then, we
will be having surplus that will be 412819, so if the sum total all, it will again be
equal to 2175983. So, that is how you will be doing the heat
balance for the problem number 3. Now, quickly illustrate the problem number 4. Now in problem
number 4, all that you have to find out the minimum amount of fuel for
calcination in problem number 3. As I noted in problem number 3, that there
is an excess amount of heat and all the amount of heat is coming from combustion of fuel.
So that means, the minimum amount of fuel will that fuel at which the heat
input becomes equal to heat output and that value will be your amount of fuel that is
required. So, again you have to do the heat balance.
If I say, if you have x kg fuel, from here you have start that is unknown, so x kg fuel
your carbon will be 0.84 x hydrogen will 0.16 x. So, the products of combustion that will be equal to 0.07 x
that is C O 2 and
0.08 x H 2 O mind you, they are in kg moles. This is in kg.
So, once you know the products of combustion, then you have to calculate the amount of flue
gas; so flue gas composition C O 2 that is equal to 0.07 x H 2 O 0.08 x plus 45.45. Oxygen
0.022 x and nitrogen
that is equal to 0.496 x. Now, one have to do the again heat balance; heat input heat
output, so this heat input and heat output I have already done in problem number 3. So,
if you balance both then, the only
unknown will be x and this x will come out to be equal 144.4 kg. Now prior to that, you
have to perform heat balance and in heat balance you will be including the term like heat of decomposition, heat
taken by
flue gas; now heat taken by flue gas will be in terms of x.
Then heat taken by Al 2 O 3 if you do all that and you do the balancing then the answer
will be 144.4 kg and that is what about the calcinaion.