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Vacuum Reducer
fast, energy efficient method of reducing sauces and soups | |
When making sauces and soups, it is often necessary to get rid of excess water; to condense, or, as the saying goes to "reduce" the concoction. This requires you to boil off water until the mixture is concentrated enough. Imagine the energy required (and it takes hours to do it, constantly stirring,
so that the sauce doesn't burn and stick to the bottom of the pot)
Solution: a heavy-duty potlid with a rubber sheet laminated on the bottom, to seal any sized pot/pan (also heavy-duty), incorporating a small vacuum pump.
Plug it in, toss the exit tube into an empty cup or something to catch the water, and turn the heat to "simmer"(low).
As the air is evacuated, the water boils at a much lower temperature, the excess water vapour is drawn out of the pot and condenses in the tube or cup when it hits normal atmospheric pressure, and you're piece de resistance is finished in 5 minutes instead of 5 hours and *none* of the contents of the pot have burnt.
Mind, this is *not* a method of cooking, and it certainly isn't a way of making a good cup of tea; simply a better way to reduce stocks and sauces.
Rotary vacuum evaporator
http://www.labrecyc....html?CategoryID=38
Available second-hand [MaxwellBuchanan, Dec 10 2007]
VacuuKettle
VacuuKettle#1149099570
[xaviergisz, Dec 12 2007]
Things I Wish I Had in my Kitchen
http://www.nielsenh...rchives/002670.html
Also mentions this. [jutta, Dec 14 2007]
More on the Rotovap
http://ocw.mit.edu/...Materials/index.htm
Video #6 [quantum_flux, Dec 30 2007]
Another form of vacuum reducer
Kirby_20Lipo_20Attachment
[normzone, Jan 09 2008]
[link]
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Good idea. You could just buy a rotary
vacuum evaporator (link). I'm sure Heston
Blumenthal has several... |
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ooh, I want one of those...high-class 'still replete with blinkenlights. |
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Wikipedia states that M.Blumenthal uses a vacuum jar to create larger-than-normal bubbles in dishes such as souffles... wouldn't you have to eat it in a vacuum jar also ? else the souffle would just collapse as air pressure pushed the bubbles back to their original size |
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Cool, it's like an inverse pressure cooker. |
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This could make some sauces, desserts etc possible that are currently quite difficult. Often the time taken to reduce a sauce will essentially "overcook" it. |
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One thing, but - //Imagine the energy required // - I'm not absolutely certain that you get to save any energy using this, however. Someone flame me if I'm wrong, but won't you be thoroughly cooling the liquid as you draw off the vapour? And won't this corresponding heat extraction be roughly equal to what you save by having it run at lower temperatures? |
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Or am I having a very long day? My point being - I thought the heat of vapourisation (by far the higher percentage of the energy used up in "boiling") was the same irrelevant of operating temperature. |
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//but won't you be thoroughly cooling the liquid as you draw off the vapour?// |
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well, I'm sortof counting on a "roiling boil": the water forming steam bubbles mostly at the bottom of the pot (on top of the hot element of the stove), not on surface evaporation. |
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As far as how much energy I have to dump into the system to get this to happen, I'm not a thermodynamics engineer, and the last 20 minutes spent googling '"latent heat" pressure temperature water' have proven unfruitful: all the answers state unequivocally "100C and 1 atm. pressure" and give no other references. |
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... a little help from more knowledgeable members ? |
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I used to work with high vacuum systems (mass spectrometers); yes, you'll cool the liquid but it will still boil. Ice will sublime .... a basic vacuum pump capable of pulling 10^-2 pascals is cheap enough, and will vac out most things pretty fast. |
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You need a non-return valve in the line to the pump, the low volatility vacuum oil's not very pleasant stuff. |
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//I thought the heat of vapourisation...
was the same irrelevant of operating
temperature.// |
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That is an intriguing point. In theory,
you're trying to rip molecules out of the
liquid, and it should take the same energy
however you do it. However, a vacuum
takes the departing molecules out of the
way, so they don't just go back into the
liquid. Arrgh - I don't know! Excellent! |
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// Arrgh - I don't know // |
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Who are you, and what have you done with the real [MaxwellBuchanan] ? |
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I like the speed, and the not overcooking aspects of this. Strong work. Have a bun. You will note that I have gently drizzled it with a new sauce I have developed using your invention. It has celery! |
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celery EEEEEEWWWWWW HORRIDDD celery |
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great idea, pastry from me |
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IIRC there are several terms you use to determine the energy required to boil a liquid. One of these is the heat of vapourisation, others are for the kinetic energy (heat) of the gas, there are some entropy terms, etc. I had thought the heat of vapourisation was a constant for a given substance. |
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But it's a long time since I've dealt with these calcs. |
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//will vac out most things pretty fast// Yeah, but will it "boil" off a quantity of liquid water quickly? |
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//yes, you'll cool the liquid but it will still boil// |
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Most evaporation takes place at the bottom of the pot, heated by the stove element. |
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//I had thought the heat of vapourisation was a constant for a given substance// |
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My (current) problem is that the references I've seen unequivocally state "100degC" for latent heat of evaporation (2k'ish Joules/cm^3) which leads me to believe that a different temperature (ie: specifically a lower temperature) would have a different figure (hopefully a smaller one). |
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//evaporation takes place at the bottom of the pot// |
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Evaporation is a process that only occurs at the surface of the water. It is wholly dependent upon a difference between the current atmospheric conditions and the maximum vapour pressure for water, which is about 4.5 or 4.6mm Hg at STP, from memory. |
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As for this exercise... The continued removal of vapour by use of a vacuum pump is going to make the equation very complex. In theory, the triple point for water is 0.1degK, so continued reduction in temperature and pressure and boiling point are theoretically possible almost all of the way down to absolute zero.
The reality is that foodstuffs have to reach a median temperature of about 62degC to be regarded as cooked. That would define the lower physical limit of any pressure reduction we could apply here.
To put that in perspective, the boiling point of water atop Mt Everest is approx 71degC, whilst pressure there is about 26kPa, vs 101.325kPa at sea level and 25degC. |
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If the heat of vaporisation and the vapor pressure of a liquid at a certain temperature is known, the normal boiling point can be calculated by using the Clausius-Clapeyron equation thus: |
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TB = (R[ln(P0)-ln(101.325)]/ÄHvap = 1/T0)^-1 |
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where:
TB = the normal boiling point, K
R = the ideal gas constant, 8.314 J · K-1 · mol-1
P0 = is the vapor pressure at a given temperature, kPa
101.325 = atmospheric pressure, kPa
ÄHvap = the heat of vaporization of the liquid, J/mol
T0 = the given temperature, K
ln = the natural logarithm to the base e
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I attempted this equation, only to realise we were moving the goalposts in relation to vapour pressure, constantly. In short, my immediate reaction: I don't think this will boil off water any faster, but for the fact that vapour pressure is reduced by a small amount. |
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Actually, because of the thermal inertia of the system, it equilibrates fairly fast, given that the pump has a reasonably constant characteristic and the volume of material to be reduced is significant compared to the capacity of the pump. |
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You're quite right; it's a surface effect, the loss of volatiles occurs at the liquid/gas interface, causing the development of a "skin" of dewatered material on the surface. If the liquid isn't mechanically stirred, then the rate of extraction rapidly slows, since there's no "boiling" to churn the liquid as there is in a conventional pan. |
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Putting an ice cube in a high-vacuum chamber, pumping it down and watching the cube just "vanish" is quite amusing (if something of a misuse of many thousands of pounds of equipment as a mere source of rather childish entertainment) |
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On balance, a [+] for the intriguing idea. |
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If the container is perfectly insulated, then yes, no boiling and just a surface effect. |
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[MaxB] - try this idea and see if it helps the confusion. If you put an ice cube on the counter, it melts, right? And you didn't add any energy to it, thus violating thermodynamics all to heck and back in the process. No, you didn't, you just neglected to consider that your counter, and the remainder of your house, are being heated to around 295 K just because you don't like trying to stay comfy at anything less than the mid-200's and you forget that you're paying for it. |
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So, your pot of stuff that was wont to bubble into vapor formerly did so at 373 K, and at a rate entirely dependent upon how fast you pumped excess heat into it, will now do so at a lower temperature, and still entirely dependent upon how fast you pump heat into it - but if you lower the boiling point to "room temperature" that "excess heat" may begin to come from ambient heat you paid for from your central heating, rather than that you paid for from your hotplate. (As it loses energy, if you feel the container, it will feel "cold". And if you leave your hand there, the boiling increases, and your hand starts to get cold because now you, fueled by the twinkies you ate, are a heat source.) |
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If the "excess heat" is allowed to make its way to the bottom of the container, it'll still boil at the bottom and bubble up nice as you please. Since [FlyingToaster] is still planning to add heat at the bottom, there is no problem there. |
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So anyway, in net, there isn't a reduction of energy requirement - [Custardguts], your thermodynamics reflex is on the money - but the temperature point is lowered, and does answer to [FlyingToaster]'s spec of not burning the ingredients. The speed aspect may not be quite so valid - as [MaxwellBuchanan] hints, you're just making sure the vapor pressure of the air above doesn't rise up to equilibrium, which speed-wise would place it on par with an open pot boiling with a small fan blowing air over it. |
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Last week the wife and I took twenty hours to boil down a batch of apple butter. With the lid off, the stuff spattered all over the room; with it on, all the water went right back into the sauce. I'm going to try building one of these. |
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I expect to see one of these on one of those '[INSERT COUNTRY NAME HERE] Inventor' shows. |
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Have we figured out yet whether it will actually work? |
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-_- define "work"... I've stuffed enough thermodynamics in my face to figure out that it will. |
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Note that the steam being pumped out at 60C (which I've been using strictly as an example) will immediately condense into water... at 100C !!! |
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It will NOT condense at 100c. Boiling point at STP is actually about 99.6C, despite the myth. |
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Well, it won't save any energy. You might look at it as the difference between pushing the water out with heat, or pulling it out with a pump--either way you have to do the work. |
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My rule of thumb is that it normally takes five times as much energy to boil water away as it does to raise it to the boiling point from room temperature. This vacuum pump might save the 1/6th of the total energy that is used for initial heating, but it probably has internal losses to more than make up for that. |
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If you really worked at it, you could design a system that recycled the energy of the recondensing vapor, for either method; heating or vacuum pumping. But that would be complicated, and isn't in the idea as described. |
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This idea would tend to prevent burning of the sauce, as everything would be happening at a lower temperature. But it would also prevent the development of tasty combinations and caramelisation. |
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It would be very interesting to compare the flavour of two sauces, one produced by conventional heating and the other by vacuum reduction. |
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I'm sorta agreeing with [Maxwell] on the energy breakdown - but I wonder if using the energy to do mechanical work would result in less environmental heating. |
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I don't have the math at my fingertips to even begin this kind of calculation. I can say it would be strange to have a vacuum pump on your list of cooking apparatus. |
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You'd still be able to achieve carmelization, but you'd have to be hovering right over the sauces as they cooked - there wouldn't be the extra water on hand to slow down the sauce going too far. |
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//So anyway, in net, there isn't a
reduction of energy requirement// Aha
- thank you [lurch], it all makes sense
now. |
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//Who are you, and what have you done
with the real [MaxwellBuchanan] ?// I'm
sorry. I was trying not to appear
omniscient, in case people started to
think I'm a smug bastard. Somebody
told me that faults are endearing. It
won't happen again. |
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How would you stir these sauces, while they are locked inside the vacuum apparatus? |
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No need. A rotary evaporator rotates (the
clue is in the name), so the sauce is
constantly churned. |
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If you were using an alternative format,
just use a magnetic stirrer. |
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Incidentally, rotary evaporators are not
tremendously fast. Their main advantage
is their operating at low temperatures. |
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[wagster] - sp. "schloop" (If
you don't believe me, search
the halfbakery. Actually, do
that anyway - it's good
stuff.) |
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Someday, we are going to need
a whole category for
astronautical cooking utensils
and methods. With an infinite
supply of vacuum, this is
going to work very well
indeed; just need to make sure
in zero-g the sauce stays in
the pan and doesn't
get ejected into independent
orbit. |
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In space, no-one cooks over an open flame. |
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OASN: Can you use a rotovap for purposes of mercury evaporation used in gold panning? |
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There are a lot of flavors which are blunted and changed by cooking. Some for the better, some for the worse. There was an article on Cooks challanging the idea that the wine you cook with should be the one you drink with dinner - it turns out that many of the subtler qualities of fine wine are destoyed by cooking, and many of the unappealing attributes of cheap wine are destroyed by cooking, and so one might as well cook with the cheap stuff and save the good stuff. Likewise subtle vegetable and fish flavors can be destroyed by cooking. I understand one cannot cook with stevia leaf because heating destroys the sweetness. |
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I think a lot of these subtle (and off) flavors would be retained with this method. It would be a whole new deal. |
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// With an infinite supply of vacuum, this is going to work very well indeed // |
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Yes, but maybe a bit too well. You've paid a huge price with your crude chemical-fuelled reaction engines to haul water into space, and then you just (literally) blow it out the door ...... what's wrong with this picture ? Better to use the abundant solar energy to drive a vacuum pump, and recover the water for re-use. |
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[baconbrain] has the best answer so far.
//I thought the heat of vapourisation... was the same irrelevant of operating temperature.//
ans: it goes up slightly with lower temperature boiling. Check the CRC rubber handbook tables. |
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A pressure cooker could be rigged to a vacuum pump to realize your invention. Hopefully it would not implode , so avoid the cheap Chineese knock-offs. |
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If your system is thermally insulated, the heat of vaporization will be taken from the sauce which has a chilling effect. If enough liquid is forced to evaporate, boiling ice cubes (the triple point {UnaBabba]) will be reached. Continued insulated sauce reduction results in the entire substance freezing into a solid, and direct sublimation begins. Umm... frozen custard. |
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//ans: it goes up slightly with lower temperature boiling. Check the CRC rubber handbook tables.// |
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Wow. That has got to take the world award for using the most obscure reference for a simple property. |
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The Wiki entry for Enthalpy of vaporization states that it is essentially independant of temperature, but does vary with pressure. I think this is a simplification to avoid working out the enthalpy of the gas changing with pressure. I remember doing calcs at uni where all these values were separated out, and the heat of vaporisation was a constant, not a function. I digress. Regardless, the enthalpy is essentially constant with pressure and temperature. |
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So, once again, this does not really save you any energy. |
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If you wanted to have a low pressure environment for cooking, but didn't need to draw off excess liquid, you could use a pressure cooker with a one-way valve. |
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As the contents of the cooker heat up the gas inside expands and exits from through the one-way valve. When the cooker is cooled back to room temperature (or below) a low pressure environment is formed. |
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Theoretically, the energy saving here would amount to 1 calorie (4.1855J) per cc for each degK below the STP boiling point (99.6C or 372.75K ** usually accepted to be 100C or 373.15K) achieved through pressure reduction. |
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Therefore, assuming we have reduced the pressure to that on top of Mt Everest, or about 25% of STP then we would save 28.6 calories per cc of water we boil away. In the grand scheme of things this represents about 4.67% of the total energy needed to take a cc of water from 25C / 298K to evaporation. |
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It will save almost 5% of cooking energy, notwithstanding the confounding variable of water vapour increasing the pressure of our vacuum as it evaporates and before it is whisked away... or any heat conduction or convection from our apparently ideal cooking pot. |
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//the world award for using the most obscure reference//
If I've said it once, I've said it a million times, don't exaggerate. |
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CRC Handbook of Chemistry and Phisics, 59th Edition 1978-1979, Steam Tables pages E18-E23. (Properties of Saturated Steam and Saturated Water) |
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Basically, when water changes to vapor at a low pressure, it's going to occupy a larger volume, thus more randomness.
Let's not forget that the heat of vaporization for all liquids tends to zero at their critical point temperature. |
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P.S. I like warm custard better. |
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Aren't we just going to have a nice tall glass of water to go with this anyways? |
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