# Mod-01 Lec-24 Lead Smelting Material Balance

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