If all K values are larger than 1 for a multicomponent mixture, then the only possible phase is superheated vapor. Not clear on your statement about quality 75% meaning K-values are equal to 3. K-values are dependent on temperature and pressure for the species and in their simplest form give a relationship between vapor and liquid equilibrium (K=y/x). If all k-values are above 1, then the mole fractions cannot sum to 1 and thus there cannot be two phases.
That is an interesting approach that I need to investigate some more. One question that comes to my mind is the the RR equation does not appear to enforce the required condition that the sums of the mole fractions in each phase must equal one. It only enforces that these sums are equal (not necessarily one). I suppose (or at least guessing) that the K functions are robust enough to not let this happen ?
Its just a mass balance, the only real valies are those who satisfy the sum=1 condition, if you apply newthon-rhapson method you would need to keep iterating till you get the condition satisfied.
but what happens to the volume of the steam formed? For example: I inject 1liter of water and end up with 30% vapour. If I use Ideal gas law: I get more than 1 litre of gas. Though this is physical arent the mass of fluids conserved anymore?
Its a higher volune,but a waaay lower density ,remember that a gas and a liquid have different densities so thats why ,unless you are talking about a nuclear reactor the mass balance is always satisfied.
Good question. What is happening here is that we are trying to get the equation in terms of V/F instead of F/V because V/F is used more often. We also switch the first and second terms on the right hand side. So, for the z_i term, F/V = 1/(V/F). The x_i term is a little trickier. One way to explain it is that we multiplied both terms of the coefficient by (V/F)/(V/F). [And that's okay since it's just 1.] So then you end up with the written answer after a bit of rearranging. Does that help?
We used them off a DePriester chart. However you can also use vapor equilibria charts like those seen in refining handbooks
If all K values are larger than 1 for a multicomponent mixture, then the only possible phase is superheated vapor. Not clear on your statement about quality 75% meaning K-values are equal to 3. K-values are dependent on temperature and pressure for the species and in their simplest form give a relationship between vapor and liquid equilibrium (K=y/x). If all k-values are above 1, then the mole fractions cannot sum to 1 and thus there cannot be two phases.
That is an interesting approach that I need to investigate some more. One question that comes to my mind is the the RR equation does not appear to enforce the required condition that the sums of the mole fractions in each phase must equal one. It only enforces that these sums are equal (not necessarily one). I suppose (or at least guessing) that the K functions are robust enough to not let this happen ?
Its just a mass balance, the only real valies are those who satisfy the sum=1 condition, if you apply newthon-rhapson method you would need to keep iterating till you get the condition satisfied.
so if were given our k components do we just add them all up to make it ki?
Very useful, Thanks!!!
Can you make a video on equilateral traingular diagram of singke stage equilibrium calculation
In this case, would the component that has a higher bubble point pressure be the more volatile component (rich in vapor phase)?
Congratulations, great video!
Does an isothermal flash drum exist? what is the general mechanical design for it?
Dennis, you can assume K_i = y_i/x_i=Psat_i/Ptot==Roult's Law Valid. Understand?
Yep,one its the Henry's law and the another its from Raoult's Law.
Do I have to use the Newton-Raphson method to get a result from the final equation?
Yes, its pretty easy to do so.
Wonderful explanation
are there any books that you would recommend for design and sizing of flash tanks (e.g. residence time, diameter and height, material selection, etc.)
its sounds complicated,but its not
but what happens to the volume of the steam formed? For example: I inject 1liter of water and end up with 30% vapour. If I use Ideal gas law: I get more than 1 litre of gas. Though this is physical arent the mass of fluids conserved anymore?
The derivation is based on a mole or mass balance, not volume.
Its a higher volune,but a waaay lower density ,remember that a gas and a liquid have different densities so thats why ,unless you are talking about a nuclear reactor the mass balance is always satisfied.
Were can i find the k-values?
Can i assume k_i=p_i_sat/P as a start guess and then iterate k_i=y_i/x_i?
There is something wrong in the Precautions section, isn't that?
Thank you
What exactly happened algebraically if one pauses the video at 5:00 ?
Good question. What is happening here is that we are trying to get the equation in terms of V/F instead of F/V because V/F is used more often. We also switch the first and second terms on the right hand side. So, for the z_i term, F/V = 1/(V/F). The x_i term is a little trickier. One way to explain it is that we multiplied both terms of the coefficient by (V/F)/(V/F). [And that's okay since it's just 1.] So then you end up with the written answer after a bit of rearranging. Does that help?
how to find k with tank temperature and pressure?
You can find the K - values on a DePriester chart, or many other vapor-liquid equilibrium charts.
great video
good one
Where can i look up K values?
Calculate i think
You can find the K - values on a DePriester chart, or many other vapor-liquid equilibrium charts.
needs to be simplivied,for everyone,free from oil babys