What we basically have is a semi-batch operation, in which we have some calcium chloride that
is selectively absorbing water out of the air. So we are removing water from a particular
stream, and that stream has an unknown inlet molar flow rate, but the composition is known. It
is 4.0 mole percent of water. Out of the system we have an unknown flow rate, and we have
an unknown composition, or mole fraction of water. So I'll call that xH2O, and what we want to use material
balances in order to answer this problem . The first part of the problem is what is the value
of n dot. Okay so we can do this most easily by just doing a water balance, and so the
amount of water that is accumulated inside the system. In the semi-batch operation, that
is absorbed in this calcium chloride we will call nH2O and that is equal to the amount
that came in during the five hours. So 0.040 times n dot times 5 hours minus the total
that came out, and what you are given in this problem statement that 25 percent of what
came in did not come out. So therefore 75 percent of the water that went in did come
out. So if we can just take this 75 percent number and multiply it by the number that
is the the left of this. So 75 percent of what comes in, comes out. Close the bracket
there. So write now we want to know how many moles of water we have that are being absorbed
by the calcium chloride. So we are given that there is a gain in mass. From 60 grams to
78 grams. So a gain of 18 grams. So 18 grams of water.We just need to convert to moles.
It so happens that 18 grams equals 1 mole. So we have accumulated 1 mole here, and that
is equal to the term on the right. So we can go through and solve for n dot, and that is
fairly simple algebra, and what we should get is that the total molar flow rate of the
system is 20 moles per hour, which is choice D. So lets go to the next part of this problem,
and I will put this up here. So basically the same problem with the same flow diagram,
but now we are asking what is the mole fraction of water in the outlet stream, and to do this
one you will need to do 2 mole balances. We will need both an overall balance and a water balance.
So let's start with the overall, which basically includes the air plus the water. So the overall
balance again our accumulation term is equal the the number of moles of water, because
water is the only thing accumulating in the system, and this is equal to the total
molar flow rate in over the time in which is occurs. So that is 20 moles per hour that
we just solved for that is coming in to the system, and the accumulation is over 5.0 hours.
Alright so we have a total of 100 moles coming in those 5 hours. We can then subtract from
that our unknown number of moles or rate of moles per hour coming out times again 5.0 hours. So
now we know again that the nH2O is 1 mole. That is how much accumulated. So our
only unknown is n dot out, and n dot out you solve for that and it ends up being equal
to 99 over 5, which is 19.8 moles per hour. Alright so we lose 0.2 moles per hour. So over
5 hours we lose of a total of a mole. Which makes sense, right, because we have one mole
that is accumulating over this period. Okay, and now the last thing we need to do is our
water balance, and we are going to write that a different way this time than last time,
but our water balance we can write that 1 mole of water that is accumulated is equal to 0.04, which is
our molar percentage of water in the inlet stream times 20 moles per hour, over the 5
hours of accumulation time, and then for our outlet term, we have our unknown mole fraction
of water x, and I think I called that xH2O on the previous slide, and we know that the
total molar flow rate is 9.8 moles per hour, and again accumulates over 5 hours, and now
again we have a single unknown, xH2O. So we can solve this problem easily enough
and we determine that it is 3.0 percent and that is answer B, and really we could have
come to this conclusion without going through the detailed balance because we know that we
are only losing a small amount of our total stream. Since it is a dilute stream with water,
and we we are only going from 20 moles per hour to 19.8 moles per hour as we go from
inlet to outlet. Alright, and we also know that we are losing 25 percent of our water. Alright
so if the total flow rate is barely changing and we can essentially neglect how much
it's changing and we are losing 25 percent of our water, which means we should go from 4
parts in a 100 to 3 parts in a 100. With that 100 again being roughly constant. We see it
goes from 100 to 99 here. So this answer that we get mathematically is also an answer that
just make sense intuitively when we look at these numbers. So you can actually come up
with the solution in class without going through the details of the calculation.