In the realm of solvent removal, I was gonna run a still under vacuum. Keeping the cooling as a constant, will my solvent take off rate run in a linear fashion to kw of heat input + kw of vacuum power?
Example (not real numbers)
10kw heater is 5 litres per hour solvent takeoff.
10kw heater + 2kw vacuum pump is 6litres per hour solvent take off?
Is that right? Or does vacuum increase solvent take-off in a non-linear fashion?
Edit I’m aware at a certain point that the cooling will need to drop in temperature to condense liquid, and the kw will get made up there. Keeping the cooling constant (I have a resevoir) I’m curious how the thermodynamics works here.
you are not changing the energy required to vaporize.
just the temperature at which it occurs.
so the energy saved is only the specific heat of ethanol x the temp differential you gain by pulling vac.
there should be a table that gives the boiling points at various vac levels fairly accessible.
if you’re still supplying the same 10kw of heat, 10kw divided by that “saved” energy x your non-vac production rate should give you your “extra” production. give or take.
You’re correct to a first approximation. The energy required to induce phase change actually does change slightly at increasing vacuum levels, but the increase is less than the energy saved by not having to raise the temp as much.
Example with completely made up and wrong numbers to illustrate (for an equivalent mass/volume of fluid in each scenario) :
At atmospheric pressure, you spend say 5kw raising the temp of your fluid, and 10kw phase changing it.
At vacuum level of N mbar, you spend 2.5kw raising the temp, and 11kw phase changing it. My rudimentary understanding of thermodynamics suggests that means you’re probably using at least 1.5kw of vacuum pump to make it all happen, so the net energy budget is the same, ish, but we’re shifting the energy requirement to different places.
By adding 2kw of vacuum pump, my completely wild ass guess would be that you might see 0.5-1kw of heat equivalent increased throughput.
Clear as mud?
(I’m pretty sure that he is referencing total power draw of the vac pump to get kw)
All of the experimental data I’ve seen does show an energy requirement change as well. It’s small, but it’s there. I’d be happy to be proven wrong if you have contrary published data available.
My opinion is that it’s easier to change the question you’re asking. Phase changes are expensive, no matter how you slice it. That’s just thermodynamics.
In fact, the enthalpy of vaporization actually decreases as pressure rises; it has the opposite relationship that the boiling point does. This is why at high temperatures and pressures supercritical fluids occur: the enthalpy has become zero.
Where does vaporization happen? At the surface…but every time a bubble rises to the top from the bottom or source of heat it breaks the surface tension for a bit speeding up vaporization for a quick sec…make that faster…weaken the surface tension…help the bubbles move and clump together…use other means to create bubbles
That is just not the case. Vaporization occurs throughout the bulk of the medium, though of course you observe most bubbling near the heat source. Once a bubble has been formed, vaporization has already occurred, whether or not the bubble has actually exited the liquid surface yet. And the energy for a bubble to break surface tension is completely minute compared to the energy input to actually convert the liquid to gas to make the bubble.
