Average Molecular Weight of Mixed Stream

This problem shows a simple material balances. We have a mixture of methane and air and it is capable being ignited if the moles percent is between 5 and 15 percent. So we have a mixture that contains 9 mole percent of methane in air so that is a flammable mixture, and it is flowing at a rate of 700 kg/hr, and we want to dilute it with pure air, reduce the methane to the lower flammability limit. In other words to 5 percent, and again these are mole percents. Find the flow rates of the air coming in and the exit stream, and find the mole fractions of N2 and O2 exiting the systems. So the first thing we are going to do is draw our pictures. So here is our system, and we have coming in a flow rate of 700 kg/hr and this is 0.09 moles of methane per mole of mixture. So it is 0.91 air. We have a second stream coming in that is completely air, and finally we have a stream that is 5 mole percent methane which means 95 percent air. So the first thing we have to look at is what are out unknowns, and we certainly don't know the amount of air coming in or the exit stream. In addition though the stream here is an unknown because all though we are given a mass flow coming in we have mole percents of methane and air. We don't know this stream here and we don't know this stream here. So actually we have 3 unknowns and my the way what we could do is we can take the mole percents in the first stream and change them to mass percents. The reason not to do that is coming out we are given mole percents in addition we have to find mole fractions of N2 and O2. So we have our 3 unknowns. Those three unknowns n1, n2, and n3. We can do 2 independent species balances or one species balances and one overall balance. In addition calculation we can use, and this calculation is from 700 kg/hr to a molar flow rate using an average molecular weight. So we have zero degrees of freedom, which means we can then solve this problem. So the first thing to do is find the average molecular weight. So we take out 0.09 mole percent of methane and multiply it by molar mass of methane which is 16 g/mol and we add 0.91 percent of air and molar mass of air is 29 g/mol, ans we we come up with an average molecular weight of 27.8 g/mol, and that makes sense. You know that is is going to be closer molecular weight air of because 91 percent air. So now we take our 700 kg/hr we divided it by 27.8 times 10 to the minus 3 kg/mol, and lets work with kmol rather then moles other wise we are going to have really large numbers. So one kmol has 100 moles and so we come out with 25.2 kmol/hr, which is our n1. Now that we have solved for n1, We can go ahead and do material balances in order to find n2 and n3. So we can do a methane and air balance. We can do an overall and methane balance. We can do an overall and air balance. Lets start by doing the overall balance. 25.2 plus n2 equal n3, and given the choice between the air and methane balance. If you look at the problem, so lets go back up and look at this picture what you will notice is if you do a methane balance. methane only appears in two streams n1 and n3, and we already know what n1 is. So it is easier to do a methane balance then an air balance. Even though we could do either one and get the same answer. So the methane balance there is 9 percent methane in the 25.2 kmoles coming in and there is 5 percent of methane in n3. So that allows us to easily solve for n3. So our n3 equals 45.4 kmol/hr, and now n2 equals 20.2 kmol/hr. So the next thing, that our questions ask is how do we find the mole fractions coming out in the stream, of both oxygen and nitrogen. So the first thing we have to do because the oxygen and nitrogen are in the air is we have to find the moles of air exiting. So that is just 0.95 times n3, which is 45.4 kmol/hr, and that is 43.1 kmol of air. SO once we find the moles of air exiting we can find the moles of O2 and N2 that are exiting. How do we do that, well 79 percent of air is N2 and 21 percent of air is O2, and that is in mole fractions. So 0.79 of 43.1, which is are the kmol of air exiting equals 34 moles of N2 and 43.1 minus 34 leaves us 9.1 moles of O2, and finally we find the final mole fraction and remember our yO2 which is the mole fraction of oxygen is the number of moles of O2 over the total moles not moles of air but total moles, which is 9.1 divided by 45.4, which is 0.20. yN2 we find the exact same way 34 divided by 45.4, and that equals .75. This corresponds .20 and .75 to the 0.95 mole percent of air in the final stream. So that is a check to make sure that we have done it correctly