Name: 單一組織、部門的自由度分析 (Degree of Freedom Analysis on a Single Unit)
Uploaded: 2021-01-14T06:29:56.000Z
Duration: 7 min 30 s
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This problem is a degree of freedom analysis on a column which we will draw like this,

and we have one flow rate that we are going to call “n_dot 1”. That is coming in as

kg/hr. The mass fraction of B is 0.03 and the rest going into this stream (stream C).

Then we have a second stream coming in, which is 5300 kg/hr and it consists of A and B, and

note my nomenclature it’s totally up to you. How you want to do it, but when I put

in the mass fractions I start it with the stream number like x1 and then followed by

what the species is. So exiting the column we have a mass flow rate. I am going to call

“m3”, and it consists of only A. Then from the middle of the column I have a mass flow

rate of 1200 kg/hr. It has a mass fraction of 0.70 of A. Then some mass fraction of B,

and some mass fraction of C. From the bottom of the column we have some flow rate that

we are going to call “m5”, and it has a mass fraction of 0.6 of B, and the rest of

C, and we are going to do a degree of freedom analysis on this system. To see how many Degrees

of Freedom (DOF) we have, or if any information needs to be supplied in order to solve for

our unknowns. So the first thing we are going to do is count up the unknowns. So take a

look and if we look at the flow rates we have m1, m3, and m5 that we don't know. We look

at the mass fractions we have x1_c, x2_ a, x2_b, x4_b, x4_c, and x5_c. So when we count

them up we find that we have 9 unknowns. So let’s see how many independent species balances

we can make. We have 3 species A, B, and C. So we can make 3 independent species balances,

which would be A, B, and C. Instead we can do an overall balance and two of the species

balances, but we can't do all 4: overall, A, B, and C, because they are not independent.

In addition we have some extra information and that is we can sum up all the mole fractions

in stream 1 and that has to equal 1. All the mole fractions in stream 2 and that has to

equal 1. All the mole fractions in stream 4 and that has to equal 1, and finally all

the mole fractions in stream 5 have to equal 1. Why did I not do that for stream 3? Well

if you go back and look at the picture what you see is that stream 3 is completely A.

So there is no other mole fractions. So now let’s look at this. We have 9 unknowns we

subtract from that our 3 independent species balances. We subtract from that our 4 pieces

of extra information, and we end up with 2 degrees of freedom (DOF). That means that

in order to solve this problem we need two more pieces of information. Now one thing

that we can do to make it easier is use those constraints that some of the mass fractions

have to equal 1 to already solve for some of our unknowns. So we will start with x1_c

and again let’s go to this picture, and if you notice that we have 3% B. Since C is

the only other one, this has to equal 0.97. In stream 2 we know neither of the mole fractions

but we can write x2_a in terms of x2_b. Now when we look at stream 4, you will notice

that we have A, B, and C, but we can write for either x4_b or x4_c in terms of the other

unknown. So let’s write that as x4_b equals 1 minus 0.7, which is the mass fraction of A

minus the mass fraction of C. So this equals 0.3 minus x4_c, and finally in stream 5 we

can figure out x5_c by just subtracting that 0.6 from 1, and this has to equal 0.4. So

we have solved for 4 of our unknowns. So the number of unknowns has become 5. Does this

mean that the number of degrees of freedom has changed, and the answer is no because we

have used our 4 pieces of extra information so we have 0 of those. So now we have 5 minus