secondary condenser

[finechem]

e.g., high vapor pressure solvent distillation, primary and secondary condenser - how can the vapor flow from each condenser be calculated @ respective discharge temperatures(no non-condensable flow). For instance, 10,000#/hr THF boil-up rate @ atmospheric pressure (T=150F), what happens when I cool the stream to 100F (partial pressure 5.2 psia) and then to 32F (partial pressure 0.9 psia). How can I find the vapor flowrate from the primary and then the secondary condenser?

Thanks for your help.

 

[Montemayor]

Chem:

Please tell us your scope of work and basic data. Some things don't make sense. We don’t know what operation you are running and why you require 2 condensers in series instead of one. This is not normal but you might be proposing a retrofit (or already have one) with a second vent condenser. If you are going to use a vent condenser, you normally design it for the ultimate temperature you are going to use – the 32 oF. That’s done to capitalize on the economy of scale when you do all the condensing in one condenser. It costs more to do it with two condensers in series – both in capital and operating costs. A lot of students write posts very similar to this type of situation because it is a situation favored by profs for “typical” homework problems that are not practical in industry. I’ll dedicate some time and words to this post, but I expect a response because some of the things you write as a Chemical Engineer don’t make any sense and need some clearing up:

1) The system you describe seems to be a distillation, stripping, or boil-up of pure Tetrahydrofuran (THF). You state that you’re generating a vapor rate of 10,000 lb/hr at atm. pressure and 150 oF and that checks OK. But then you say that there are no non-condensables, so therefore, we are forced to assume the THF vapors are 100% pure.
2) If the system is one of pure THF, then by definition there are no partial pressures involved. What you must mean is VAPOR PRESSURE – not partial pressure; and this checks with THF vapor pressure being 5.2 psia at 100 oF and pure. By definition you have pure THF, there can be no partial pressures involved in this closed system.
3) However, if you have your primary condenser operating at 5.2 psia and your secondary operating at an even lower system pressure (0.9 psia), why would you be concerned with vapor flow? What isn’t handled by one unit is handled by the other.

Your overheads system is blocked to the atmosphere for all practical purposes. It has to be, since it is a pure THF system and at partial vacuum. And by this fact, there is NO FLOW from the 0.9 psia secondary condenser out to atmosphere (14.7 psia). Here, you are facing a formidable problem of trying to get THF vapors from 0.9 psia up to 14.7 psia and out to the atmosphere. Why you would be trying to do this is beyond me. Instead, what nature will be trying to do is push 14.7 psia atmospheric air into your secondary condenser system (& the rest of your system).

The vapor flow that goes from your primary condenser to your secondary condenser is what is left over after the primary unit does its condensing job. With a pure THF system, the vapor flow(s) stops at the secondary condenser. Only you know the duty that the primary unit does since you have kept that condensation load to yourself and haven’t shared it with us. We can’t guess that figure and I’m assuming that the condensing duty is only latent heat and not subcooling. But only you know that for sure. That’s one of the reasons that what is normally done in this case is that one condenser is designed to handle all the overheads – and that is the latent heat required to condense the total 10,000 lb/hr.

As you can see your presentation is full of questions and holes. I’ll await your response.

 

[rmw]

Art, this sounds like homework for Chemical Engineering 101 to me.

rmw

 

[StoneCold]

chem101
He may be splitting the vapor flow rates up due to utility requirements. If you do part of the condensing with cooling water then the design water temperature could be 100F. Then the secondary condenser could run on a chilled water system to get to 32F. That seems logical to me. However if it is a real distillation Art is right that he has simplified things a lot or hidden some important facts from us. But just off the cuff the amount condensed in the first condenser must be 10,000#*(14.7-5.2)/14.7 and the amount condensed in the seond condenser must be 10,000#*(5.2-.9)/14.7 and the amount uncondensed is 10,000# 0.9/14.7. Ignoring the real fact that there are incondensables.

Awaiting comments
StoneCold

 

[DickRussell]

StoneCold, I don't follow your duty calculation. Please explain.

Chem101, we're still waiting for more detail. We also take issue with your remark about there being no non-condensables. As Art pointed out, the system must be designed for air leakage. The amount won't be known with great certainty, but there are general guidelines for approximating leakage for "commercial tight systems" (i.e. well-built), usually a function of operating pressure and system volume.

We also need some input on how the column and its pressure profile is defined. You gave atmospheric pressure at the bottom of the still, 5.2 psia at (out of?) the first condenser, and 0.9 psia out of the second condenser. Having 9.5 psi drop over the column and condenser seems rather high for a vacuum system. Then you have 4.3 psi drop over the second (vent) condenser; this is both excessive for a vacuum system and excessive for a vent condenser, which is just supposed to be doing a cleanup job, squeezing a bit more condensable material out of the vent gas; pressure drop for a vent condenser normally is fairly low. A vacuum pump or ejector must be presumed present.

Given the information you presented, with pressures as "partial" pressures, we can make some sense of this if we take them as such (the DIPPR database gives 5.33 and 0.93 psia for vapor pressure at 100 and 32 F, respectively; your numbers presumably are supposed to be partial pressures), the remainder being the air leakage. Further, assume some total pressure profile (or even isobaric operation - simplistic), and an amount of air leakage. Then the amount of THF in the vapor from either condenser is found as air amount times ratio of partial pressures, assuming no absorption of air into the liquid.

Tell us more, then we can help more.

 

[finechem]

Thank you all for your comments - this is exactly the direction I wanted to take, however I must apologize as I have been in a PHA almost non-stop since posting and only now can reply - I hope that I still have your interest. To start, this is my first post on an engineering forum so I was not sure what to expect (not to mention I picked a poor handle...) - seems I have come to the right place. Now for the missing details (no, not a homework assignment). Actual solvent is methyl tert butyl ether, not THF (used THF as it is more common) - and for this discussion we can use 10,000#/hr boilup rate, but really I am looking more for a discussion of what is happening within my condenser - the calcs can follow later. I have a primary condenser - stonecold is correct, existing equipment and on cooling water to reduce load on already taxed chiller, and a secondary on 50/50 EG, operating at approx. 0F. I have a nitrogen purge in vent downstream of secondary condenser but no purge on pot or column (no in-leakage since process is at atmospheric, i.e., no/very little differential pressure to drive a leak). I have struggled with issues similar to those raised by montemayor in that as my temp is reduced so then is my partial pressure, and as in the case of a closed system, so would my total pressure be decreased. Then one of two things are happening at the outlet of my primary, either 1) there is nitrogen present from start-up (equipment was not evacuated prior to start up so the system was N2 full, and this accounts for my partial pressure, or 2) the pressure decrease across the exchanger will create vacuum such that N2 will be sucked in from my vent (in this case the vent is to the thermal oxidizer - recall I have N2 purge on outlet of secondary). However this is not exactly what I am after. I understand Raoult's Law and I see that environmental regs calculate emission based on non-condenable flow, however this just is not the case. If I were to cool this saturated MTBE vapor (@ atmospheric boiling point) to 80F (assume infinite ht exchange area, at 100F) I would cool the vapor past its dew point such that I condensed a certain proportion of the stream. Furthermore I believe that if instead of 80F I cooled to 30F I would condense a higher fraction of that stream (again using infinite ht trans area - condenser duty would be fixed as that ht trans required to condense that fraction of vapor from the stream). Now is this true? If yes then I am at the root of condenser sizing - the first condenser will condense that fraction from one temp to the lower discharge temp based on total boilup and then the secondary condenser sizing would be based on that vapor flow leaving the primary condenser (still a little unsure with presence of non-condensables) at the discharge temp of the primary to the discharge temp of the secondary. To complete these calcs I used something very similar to comments by stonecold where the fraction condensed was related to the drop in partial pressures across a temperature drop (multiplied to first boilup rate for primary condensing and then multiplied by remaining vapor from primary for the secondary rate. E.g., 10000#/hr MTBE from atmospheric BP 138F to 80F (partial press 4.4psia), then fraction (same since 100% MTBE) condensed would be (14.7-4.4)/14.7*10000=7006#/hr. This means that 10000-7006=2994#/hr vapor @ 80F passes through my primary. Then fraction condensed in the secondary would be (4.4-1.2)/4.4*2994=2177#/hr (note I divided by 4.4 instead of 14.7 since x fraction is 100% MTBE). Then my vapor outlet from the secondary would be 817#/hr. Does anybody agree? Am I rambling? Again, thanks for all of your comments - this time I will answer any questions promptly...

 

[25362]

To chem101, as I see it, upon partial condensation the mol fraction of MTBE in the remaining (equilibrium) vapors would be 4.4[÷]14.7 ~ 0.3.

This converts easily to ~ 57.3 % mass, when the other component in the mix is totally non condensable nor dissolvable nitrogen.

But then you need to know how much nitrogen is there to estimate the mass of MTBE in the vapor. If the nitrogen flow is, say, 42.7 lb/h, the MTBE vapor content would be 57.3 lb/h.

That is if I'm not wrong...

 

[DickRussell]

Is this a continuous process, with "steady-state" feed and distillate rates? What is the function of the column in your process? What is the feed composition (continuous process or the charge to a batch distillation)? Have you obtained pressure readings at the top and bottom of the column and at the exchanger outlets?

If continuous, and your column and exchanger pressure drops are reasonably small so that the column is close to atmospheric throughout (thus with negligible air leakage), then any nitrogen present prior to startup would be swept out rather quickly with the vent stream as you boil up MTBE. DIPPR database lists MTBE NBP as 131 F. If you cool a pure MTBE vapor stream down to 80 F, you will get all liquid and no vapor for the vent condenser to handle. The nitrogen purge will serve to maintain column pressure, presumably. If you do not have a vacuum pump or ejector, there is no way you can go vacuum due to flow and suck in nitrogen from the vent line. If you have high pressure drops, pressure will simply back up from the downstream equipment and become substantially higher upstream of the condensers, not atmospheric.

If indeed you have a significant duty in the vent condenser, meaning you have a substantial amount of MTBE in the vapor downstream of the main condenser, then you must have something else entering the column (with the feed or fed separately). It would be something noncondensable or at least very low boiling. You can approximate how much there is, relative to the MTBE. As 25362 points out, you need to know how much of that something else is present to quantify the MTBE flow from either condenser. At 80 F, the MTBE vp is 5.18 psia (what's your source of vapor pressure for MTBE?), so it is indeed only about 1/3 of that vapor. The "something else" is the other 2/3.

Once you identify what that something else is and how much of it is present, then you can do your condenser duty distribution calculations. If that other stuff has very low flow, then practically all your condensing is done at 131 F, with a sharp dropoff at the end of the curve until you pinch off at 80 F. The vent condenser similarly will squeeze out a bit more MTBE as it, too, pinches off approaching the brine temperature. If there is a huge flow of a low-boiler/noncondensable in the feed, then the situation is reversed, with very little condensing in the first condenser and the main load dumped on the vent condenser. As your noncondensable flow goes way up, you reach a point where you can cool but not condense anything in either condenser. I gather your situation is not that extreme, and that the noncondensable flow is small enough so that you do have some condensing going on in both exchangers. But you must have something else present in the feed(s) to the column, and how much determines everything else.

Perhaps you could get flow and temperatures for the coolant on both exchangers, to back out duties. Then indirectly you could calculate how much you have in lights.