A small still is separating propane and butane at 135 C, and initially contains 10 kg moles of a mixture whose composition is x=0.30 (x=mole fraction butane)additional mixture(xf=0.30) is fed at the rate of 5 kg mol/hr. if the total volume of the liquid in the still is constant, and the concentration of the vapor from the still(xd)is related to xs as follows:
xd= xs/(1 xs)
How long will it take for the value of xs to change from 0.3 to 0.4? What is the steady state value of xs in the still(i.e, when xs becomes constant?

Anar22, is there supposed to be some sort of mathematical sign between the 1 and the xs in the vapor concentration formula?

For example, xd = xs/(1+xs)?

No

I'm sorry. But I don't understand what u want.
The question come like this without any things more, or any details.

the concentration of the vapor from the still(xd)is related to xs as follows:
xd= xs/(1 xs)

No? Really? So you're saying that xd=1!! Because that's what xd= xs/(1 xs) works out to be. If that's true, xd is not related to xs in any way. It's simply constant. I have a hard time believing that's what the problem is supposed to read.

i'm sorry. but i don't understand what u want.
the question come like this without any things more, or any details.

Maybe you missed the fact that I added a plus (+) sign in the formula?

xd = xs/(1+xs)

Your original question does not have a plus sign or any other sign there. It just says

xd= xs/(1 xs)

Now do you see what I'm getting at?

oops!
now I got your point
it is
xd = xs/(1+xs)

Unfortunately, I'm probably not the right person to be answering this question since I'm not a chemical engineer, and I can't say that I've ever solved a mass balance equation in my life. ;)

I can take a stab at it, but I make no guarantees that I'm right.

First, I have some assumptions: From the wording of the problem, it sounds like there's only one output from the system. This is contrary to what little I thought I knew about continuous distillation; I thought there would be two outputs (the top, which is more highly concentrated with the more volatile butane; and the bottom, which is has a higher concentration of the less volatile propane). Anyway, if that was the case, there's clearly not enough information given to answer the question. So it must be like I said; there's only one output.

I'm also assuming that when they say the "volume" in the still is constant, they really mean the molar volume of 10 kmol remains the same. Otherwise, if molar volume of the mixture had to change as the concentration changed to maintain constant physical volume, that could make this problem very complicated indeed!

Okay, now on to my version of the solution:

I think it's much easier to answer the second part first. Since there's only one input and one output, and the volume in the still remains constant, the flow rate in must be equal to the flow rate out. Therefore, once we've reached equilibrium that means the concentration in must be the same as the concentration out. Thus we can say:

x_f=\bar{x_d}

x_f=\frac{\bar{x_s}}{1+\bar{x_s}}

0.3=\frac{\bar{x_s}}{1+\bar{x_s}}

\bar{x_s}=\frac{0.3}{0.7}\approx 0.42

where I put a bar over xs and xd to mean that I'm talking about the equilibrium values.

As for the actual dynamic equation, I guess we should write out the material balance:

\text{input = output + accumulation}

So over an infinitesimally small time, dt, we have:

\text{input}=R_f x_f dt = 5(0.3)dt=1.5dt

\text{output}=R_d x_d dt =5\(\frac{x_s}{1+x_s}\)dt

\text{accumulation = input - output} = 5\(0.3-\frac{x_s}{1+x_s}\)dt

where R represents the material flow rate. We can now calculate the change in concentration dxs in the still:

dx_s=\frac{\text{accumulation}}{\text{total molar volume}}=\frac{\(5\(0.3-\frac{x_s}{1+x_s}\)dt\)}{10}=0.5\(0.3-\frac{x_s}{1+x_s}\)dt

Finally, dividing both sides by dt, we should be able to get the differential equation for xs:

\frac{dx_s(t)}{dt}=0.15-0.5\(\frac{x_s(t)}{1+x_s(t)}\)

Unfortunately, I don't have a clue how to solve that differential equation. :(

If I abandon the differential equation approach and just use intuition to solve the problem, I'd say that the concentration is probably going to be some sort of decaying exponential function

\tilde{x_s}=ae^{-bt}+c

where the tilde means that it's my guess at xs, not any rigorous mathematical solution. We can use boundary conditions to solve for all the unknown variables, a, b, and c.

We know from above that x_s(\infty)=\bar{x_s}=\frac37. Plugging that into our guessed equation, we find that our unknown c is 3/7.

We also know that x_s(0)=0.3. Plugging that into our guessed equation (along with our value of c) gives a=-\frac{9}{70}.

Finally, we can find the decay constant b by computing the derivative of the concentration at t=0.

\frac{dx_s(0)}{dt}=0.15-0.5\(\frac{x_s(0)}{1+x_s(0)}\)=0.15-0.5\(\frac{0.3}{0.3}\)=\frac{9}{260}

Now we can plug that into our guessed equation:

\tilde{x_s}=ae^{-bt}+c

\frac{d\tilde{x_s}}{dt}=-abe^{-bt}

\frac{d\tilde{x_s}(0)}{dt}=-ab

\frac{9}{260}=\frac{9}{70}b

b=\frac{7}{26}

Thus, I would guess the final equation would be

\tilde{x_s}=-\frac{9}{70}e^{-\frac{7t}{26}}+\frac37

Solving this for xs=0.4 gives

t_{0.4}=\frac{26}{7}\ln\(\frac92\) \approx 5.6 \text{hours}

I don't think it's exactly correct, but it's the best I can figure. The correct answer is probably some sort of sum or series of exponentials with different decay constants.

If you find out the correct answer (and how to get it), can you post it here?

Please can you send me your email

I prefer not to share my email, but you can send me a private message here on AMHD if you need to.

OK. I will