Challenge Questions

I gave you a GA. You said just about the same thing I was going to say, only in fewer words. Good Blogg'in.

Toomuchfun

P.S. Of course, the "near-perfect vacuum" requested by the challenge is not necessary - a pressure of about 60 torr would be low enough to minimise the chances of your scalding yourself, and would be high enough to prevent the other thermal effects I mentioned.
However, not being a physiologist, I'm not too certain what would be the other effects of placing your whole hand in such a vacuum when the rest of you was exposed to even half a standard atmosphere - but I'm certain that it would at best be mighty uncomfortable, and I wouldn't be very surprised if it caused quite severe bruise-blistering.

In a pressure cooker, the temperature is a bit higher than 100º C- that's the whole point of pressure cooking.

The question states that the orifice is open. The temperature of the water inside the pressure cooker only goes above 100 degrees when the stopper is placed on top of the orifice and the pressure subsequently increases inside.

By the way, here is a nice video demonstration of water boiling at 40 degrees C :-

http://www.instructables.com/id/Vacuum-Chamber-Project-2-Boiling--Room-Temperatu/

Because the orifice is small and the heated area large, you could get significant over-temperature inside the pressure cooker. In fact, I've been distracted on occasion and when I returned seen quite a long plume in which it seemed that the steam was still condensing. However, the conditions you assume are probably those intended for the challenge.

Happy New Year all.

Interesting question here, remember that the steam inside the cooker is at satuation where most of the steam is in liquid form, but about 100 C. Now when the steam escapes it is still in liquid form immediately at the orifice but farther away it will change to gas form and in doing so some of the energy will be converted to achieve that. Secondly a liquid offers greater rate of heat conduction than a gas, a lot like standing naked in 0 C weather or swimming in 0 C water. The liquid form will conduct heat much faster than the gas.

So I propose that the answer here has more than one solution including also that the farther away from the orifice you get the more the steam has to disperse.

Read all about it:

Theory:Latent heat Practice: steam scald

I am assuming this is about Bernoulli's Principle and any pressurized gas escaping through a small nozzle will have a corresponding reduction of temperature due to the drop of pressure.

If memory serves, this simply boils down (no pun intended) to the old PV=nRT equation (the ideal gas law).

You can and may have already noticed this effect when you use an air tank to blow up a balloon. The nozzle will get ice cold.

Repeating the experiment in a vacuum will only increase the temperature drop differential because the delta pressure between inside the vessel and outside the vessel will be one atmosphere greater.

Blinding us with science? (look who's talking). But don't forget the latent heat portion - if the pressure inside the cooker in the vacuum were to be > 1-atmosphere, the internals of the stream would not cool down anything like as much as would air.

The boiling water inside the pressure cooker is in the liquid-vapor region, where the temperature of the vapor inside the cooker depends on the mixture state pressure.

In the first case when you close the orfice by your hand, you feel a high temperature of the boiling water which is corresponding to the high pressure inside the pressure cooker( the temperature in this case is high because the pressure is high ).

In the second case ( when you put your hand 5 inches away from the orfice), the water vapor pressure in this state is equal to the atmospheric pressure and the corresponding boiling temperature in this case is low ( maximum 100 degree cetigrade directly after the orfice and decreases when you go away from the orfice) so, at adistance 5 inches away from the orfice you will not get any burn , because the teperature will be less than 100 degree centigrade.

In the second case ( when the cooker in a vacum chamber) , the water inside the pressure cooker will boil its self at the atmospheric temperature without adding heat to it, and you will not get any burn in both cases

sure... there will be a burn in shorter distance, but not in 5 inch distance.

under normal condition, fresh water will boil at 100 deg C. when u boil Fw, ur giving it heat.what ur doing is breaking up it's molecule bond / F.wtr = H2O. hence transformation frm liquid to gas state. ur are actually producing errr.... H2 + O.. or H + O2, ..cant remember. the steam that we see (white cloud) is actually condensate - gas back to liquid form. this happens when the H and O molecule re-bond back i.e becoming f.w droplets. hence,sure when u put ur hand directly at orifice u'll burn, due to the heat energy the gas contain. once it escapes to dense air (the gap 5 inch), the gas molecule rapidly cool (here all hell breaks loose, those H n O molecule will hit each other + the cool H and O2 in the air) and try rebond again, most bond is not strong enough thus breaks up and dissipate. those which made it will become water droplets (which is no longer hv enough heat to burn ur hand) and you'l have that feeling of wetness on ur warmph on ur palm.

In Vacumm state, u'll burn at 5 inch gap. in vacum Fw will boil at lower temperature, dont remember exactly.. 30 to 40 deg C? (same amount of heat - given to same amount of time both condition), by end of time, the pressure in the cooker is much higher than above even given same amount of heat. the pressure jet of gas is much higher than 5 inch compounded more when theres no cool air break the jet stream (vacumm).

the question is now will u see same 'steam cloud' or not......or just heat... errrrrrr.. should be but will dissapears quickly... hmmmm.

thks/rgds

oh shoot, change my answer 4 vacumm part.. . No youl'll nt get burned.. cause the boiling temperature has been lowered.

I guess I didn't flunk Jr. Hi chemestry for no reasin. I thought steam was still H2O but in a gaseous state the same a ice is H2O in a solid state. -- JHF

Ouch. Each molecule of water vapour consists of two hydrogen atoms and one oxygen atom - i.e. H₂O. No breaking of the basic H₂O bond here. In liquids you get complexes consisting of multiple molecules that are constantly reforming. Water is a bit special in this respect, in that a small proportion of breaking complexes will temporarily produce an ionic component; this is often represented as 2H₂O -> HOOH- and 2.H+. So, contrary to what you wrote, water vapour is essentially composed of H₂O molecules, whereas liquid water is a mix of H₂O complexes (= N.H₂O) and HOOH- and H+ ions.

It's all about something called the heat of vaporization. When a gas condenses to a liquid there is a tremendous transfer of heat. When the high pressure, hot water vapor leaves the confinement of the pressure cooker, it encounters the much lower ambient pressure of the kitchen. At this point, the gas molecules coallesce into liquid droplets, absorbing huge amounts of heat from the air. At a location five inches from the cooker, you are feeling mist, not steam.

By the way, this phenomenon is the basis for refrigeration and air conditioning.

I'll assume were using an electric element to heat the water. Otherwise gas would not ignite since there is no oxygen. So that would eliminate one scenario of being burned. Unfortunately, even if it were heated, the boiling point would not matter so much since the water is already changing over to vapor (due to vacuum) and would exit at a very cold rate. You might say you would get burned at the 5 inch point since the cooling effect of the vacuum would be greater (than 1 inch) and cause a freezing burn. My problem would be getting someone to actually put their hand in a vacuum chamber.

Heat source and heat sink,

Heat source is pressure cooker orifice that is point of escape of heat and earth is heat sink. Sink is too big in compare to source and so the heat travel distance is not much before discipation.

There will be temperature gradient and that gredient will not be more than 4 to 6 inch from the source and that is what you are seeing.

Masyood

My assumption is that there is choked flow at the escape orifice. Depending on how much heat is being added to the pressure cooker will determine the pressure inside the vessel and this will determine the flow rate through the orifice. The reason a steam kettle's whistle increases in pitch as it is left on the heating element must be because the pressure inside increases and the choked flow rate increases. A 3 mm orifice seems fairly small for a 2 liter pressure cooker.

Since the steam inside the pressure cooker is at a higher pressure than the outside, adiabatic cooling occurs as the steam expands. This, combined with mixing with air, lowers the temperature.

If we put the pressure cooker inside a vacuum, the same phenomena will occur. There will still be choked flow at the orifice and the steam will cool as it expands. Assuming the pressure cooker is getting the same heat input, the difference will be that the water will start to boil at a lower temperature. Because of the orifice, the temperature and pressure inside the pressure cooker can still increase. Since the heat loss from the pressure cooker due to convection in a vacuum will be zero, for a given wattage of heating element more energy will be going into the water. The steam temperature might still get to be quite hot - hot enough to burn your hand. Since the steam will be expanding into a vacuum, there will be a much greater adiabatic cooling effect.

Just wish I had time to refresh my mind on heat transfer, choked flow, and adiabatic expansion. It should be possible to do fairly accurate predictions of the maximum steam temperature for a given heat input and a prediction of the temperature of the expanded steam jet for both the atmospheric pressure and vacuum cases.

Thanks for a good challenge.

Jim

Well, as some have alluded to - whether it burns or not depends on the pressure inside the pressure cooker.

Assuming the pressure cooker pressurises & only then vents the steam off, then the water will boil at or above 100degC the "steam (water vapour) will be very hot and - YES you will get burned.

If you assume that the 3mm orifice is premanently open (and the pressure cooker does not pressirise) then the water aill also be subject to the vacuum and will boil at a very low temperature. The "steam" (water vapour) will not be hot enough to burn, so _ NO you would not get burned.

Simple really!

Tony

1)a the steam escaping expands and cools (Joule Thompson effect) and b) air is entrained in the steam jet and mixes, cooling the combined steam-airstream some more.

2) Without air to entrain, you would think --ouch and a nasty burn. However, the venting stram will instantly expand from say 30 psia to 0 psia and drop in temperature to freezing.

Nasty freezer burn?

This is a transient. It takes time for the steam to disperse and reach equilibrium conditions. In the mean time, the pressure in the jet of steam is NOT zero. The H2O molecules will eventually push apart due to the internal pressure of the steam jet and the pressure will decline. The work done to disperse the gas will cause the temperature to drop.

Obviously, if you measure a short distance, say 1 cm from the orifice the steam pressure is still high and the temperature is high. If you go a large distance, say 10 m then the jet is likely to have dispersed with low pressure and low temperature because of the work done on the gas to disperse it.

So, the question is, what are the transient conditions 5 inches from the orifice, presumably with at least 1 atmosphere and 100 degrees C inside the pot?

Got an interesting story. Years ago I knew a fellow who worked aboard a large Naval vessel in the propulsion area. This is during the era where they still used steam based drive systems, not diesel propulsion. These ships used superheated steam and it was generally known that you never walked along the main steam lines without waving a stick up and down in front of you. If the end of the stick fell off, you know there was a leak and being as you can not see superheated steam leaks, this was the most effective way to find them and also to protect yourself.

Well long story short, one of the man's fellow workers forgot his stick one day and ended up getting cut in half by a leak. The steam cut into the men and cauterized the wound so as there was no blood. Of course the man did not survive. High pressure steam is a different animal than saturated steam.

Yes, burn in "normal" conditions at steam outlet.

Yes, "bearable" conditions if hold hand 5 inches from steam outlet in normal conditions. (Previously described dispersion effects.)

Yes, burn if touch water vapour exiting boiling pressure cooker inside vacuum chamber.

Most perople are looking at the source and thinking "heat", but what about considering the effects on the "hand" (Or sensor as someone else has already suggested) in the vacuum chamber.

If the water vapour is CONDENSING onto the hand/sensor, then the condensate will be transfering thermal energy to the hand and the hand will get hotter purely due to the phase change from vapour to liquid. (An amount in the order of the heat of vapourisation of the water amount that condenses.)

The heating effect will be offset by the evaporation caused by the vacuum and so the answer to the question "Does it burn" depends totally on the steady state condition of condensation and evaporation that is achieved. For me, the question as posed cannot be answered as it is not clear whether the pressure cooker maintains power input or whether the vacuum alone is the source of forming the water vapour.

Stange things happen inside vacuum chambers. We vacuum deposit aluminium and often see scorched surfaces where the metal has been deposited onto parts with ambient temperatures of 20Deg C and fixtures of the same temperature.

I suggest that the steam in high vacuum condition will not condense (at all) and therefore will carry only the much smaller sensible componant of energy to the hand. Meanwhile, the hand is blistering and freezing at the same time from the vacuum and the sensible heat of the hand boiling off it's own internal water and then subliming from the frozen hand to chill it still further. 5 inches of space should have a huge effect of dispersion of the pressure relieved upon exiting the vessel.

I suspect this level of vacuum would actually drain all body fluids into the vacuum or if the entire body is in there it would froth and boil in vacuum like any food product in a freeze dryer when it hasn't been prefrozen.

The difference of metal with high condensation temperature and water vapor stream of sub zero C condensation temperature predicts a very different consiquence in this high vacuum space.

When you are in free air then the steam escaping from the orifice will get spread in the air and so there is less chance of getting hurt.However in case of vacuum chamber there is no medium to carry steam particles ,that's why it will not spread and directly point on the hand and get hurt.

If we assume choked flow in the first case (expanding to atmosphere) then the pressure of the boiling water must be 1/,58 atm (assuming n of 1.135 for saturated steam) which is 1.747bar; a temperature of 116ºC; an enthalpy of 2700 kJ/kg; and an entropy of 7.172 kJ.kg ºC.

The expansion through the escape orifice exit can be considered to be isentropic and so the exit conditions P= 1.013bar; entropy=7.172 (unchanged); quality 97% dry; Temperature 100ºC

Under the assumption of choked flow,these are the exit conditions regardless of whether the discharge is to 1atm air - or vacuum. Also the mass flow is independent of the exit "back pressure"

So when exiting to atmosphere the steam cools at constant pressure by losing heat to the air and mixing. Therefore you get scalded at the exit point, but after 5" the stream has cooled.

When exiting to a vacuum there is no heat loss, and all of the kinetic energy of the steam is converted to heat. The temperature rises during this stage, and when the stream is brought to rest, the whole process has been one of constant enthalpy (no heat has been added or subtracted) and therefore it will finish at 0 bar with an enthalpy of 2700kJ/kg which is a temperature of 106ºC. So when exiting to vacuum, you get scalded at both the exit point, and after 5" of travel (in addition to the other problems you have from sticking your hand into a vacuum).

QA. I believe that the pressure determined by the choked flow is the minimum pressure for choked flow. If the internal temperature (and pressure) is increased, the choked flow rate will also increase. If a fixed heat input is involved, an equilibrium condition will be reached where the energy losses due to heat transfer plus the energy losses associated with the escaping steam will equal the energy input. Since the convection will be zero for the vacuum case, I contend the temperature and pressure inside the pressure vessel will be higher that for the atmospheric case if the heat input is the same. Also, won't the expansion of the steam into the vacuum result in a substantial cooling of the steam?

Thanks,

Jim

I suspect that Slide Ruler and others who have given similar responses are right. Compare the pressure cooker in the vacuum chamber to a rocket engine firing in space. In both cases, you have expanding gases escaping through a narrow opening.

For the rocket, these escaping gases provide thrust. If the gases just simply expanded in all directions in the vacuum of space, how could they provide thrust?

With this in mind, let's reconsider the pressure cooker in the vacuum chamber. If the opening is located above or near the center of the lid, the escaping gas will tend to keep the cooker in place. I believe most pressure cookers do not have a vent in this location. If the opening is located near the rim, it will push the cooker off the heat source, and once the heat is removed, the boiling slows down to a point where the gases no longer provide thrust. For your hand to be scalded, you would have to have it in place when the gases start to come out, and they will push against your hand as well as burn it.

However, the pressure cooker is given as having a volume of 2 liters. That is rather small, just over 1/2 gallon. Most pressure cookers I have seen are at least twice that size. I haven't checked out what the dimensions would be for a 2 liter pressure cooker, but I imagine it would not be more than 3 - 4 inches high. To hold your hand 5 inches from the vent without stooping over or kneeling, the heat source and pressure cooker would have to be on a stand or table. If the cooker is on a stand, it might fall to the floor before it has a chance to burn your hand.

Hi Jim#

The pressure ratio between the pressure of the (choking - moving at the speed of sound) fluid in the orifice, to the (stationary) fluid in the supply tank; is given by (2/(n+1))^(n/(n-1)) where n is the adiabatic index (taken as 1.135 for wet steam). This comes out to 58% for the case chosen and is generally about 50%.

Another way of saying this is that the choking pressure will be about 50% of the tank pressure. So I'm afraid I don't understand what you mean by it being a minimum.

A good analogy for choking is to think of the behaviour of a stream flowing over a sluice situated at the head of a waterfall A certain mass (kg/min) will flow over the sluice depending upon the upstream water depth. It doesn't matter how high the waterfall is, the flow over the sluice is unaffected. The flow will however change if the upstream water depth changes.

The choking in a nozzle is similar, in that if the back pressure on the nozzle is less than "critical choking pressure" then the mass flow is independent of the back pressure.

You are quite correct in stating that the heat rate balances the losses to dictate the upstream conditions, but lowering the receiver conditions from 1atm to 0 atm does not change the conditions in the orifice as long as they are choked there.

The expansion of the steam to vacuum takes place via a "shock wave" (similar ti a supersonic aircraft), which is adiabatic (no heat leaves the steam) but massively frictional (kinetic energy is turned to heat-content of the steam). Thus there is an increase in entropy and enthalpy. When there is no kinetic energy in the stream of steam, the enthalpy will be the same as when the process began, that is the same as it was at approx 2atm and 106ºC

Hope this helps you to understand my reasoning.

SLR

I must be being dim. I thought that choked flow gave a pressure ratio, not an absolute value. If I'm right, the internal pressure (for choked flow) when the pressure cooker is in a perfect vacuum would be 0/0.58.

What am I missing (or misunderstanding)?

Choked flow does of course imply a specific pressure ratio between the stagnant supply pressure fluid and the sonic speed choked flowing fluid.

What I did was find the minimum vessel conditions for choking when exhausting to 1atm which is 1/.58 =1.724atm. Using this the Tsat =116ºC.

This implies a heat rate input.

For the vacuum condition, I have assumed the same heat rate input so that the only change is the atmospheric back pressure is removed. Since the flow is choked the vessel / nozzle components are unaffected (see my reply to Jim#) but a shock wave is established in the expanding steam leaving the nozzle.

I believe the intent of this challenge is to explore "choked flow", and how it is dependent on the upstream conditions but not the downstream conditions once it is established. I chose the minimum conditions to establish choking with atmospheric back pressure.

Thanks. You've made me think some more. This is an awful long way from any of my areas of expertise, so even this is open to correction

So I tried some sums. If we assume the pan is heated on a standard hob, I don't think the power is quite sufficient for the output stream to be sonically choked when venting to atmosphere (but maybe my assumptions are incorrect - most likely the 2-mm diameter of the vent?).
The latent heat should be about 2.2kJ/gm, the sonic velocity (hot steam rather than room-temperature air) about 500-m/s, and a density of 1gm/litre. So, assuming that all the energy went into latent heat, the jet jet would require
2.2 x 5x10⁴ x 0.1²x∏ / 1000 ≈ 3.4 kW.
I believe this to be about thrice what you might expect to go into the pan when heated by a hotplate or induction heater.

Of course, on its own this does not invalidate your conclusions abut burning yourself - the mass of steam leaving the same orifice into vacuum will be about 1/3 of that given above, and this would be choked at any external pressure below 1/3 of an atmosphere. That gives an exiting steam temperature of about 70-degrees - still plenty hot enough to cause steam burns.

Now, I'm even less certain what happens at 5". Clearly, the velocity of molecules perpendicular to the direction of the stream will cause it to fan out, which means that (at this distance) vibrations perpendicular to the local flow will be substantially reduced - so we have a fast-moving stream of cool steam molecules. Stopping the stream will convert the linear kinetic energy to heat - if it was a gas, that would raise the temperature from perhaps 100K to about 200K, but being water vapour I imagine that it would rise above freezing.

So, although I don't (yet?) agree with your conclusions, I've given you a G.A. for a truly thought-provoking contribution.

Hi FYZ

Thanks for the GA

Your sums on the amount of power, (to maintain the choked condition I assumed) are quite valid (although you took 2mm for the orifice and the value given was 3mm). Since I won't be contributing much to the next challenge, I re-worked my calcs limiting the power to 1kW. The result is given below

Vessel: Psat =0.66bar;Tsat =88.4°C;Enthalpy liquid Hliq = 370kJ/kg; Hgas = 2657kJ/kg; Entropy S = 7.499 kJ/kg °C.

Throat: Pt =0.383bar; Tt = 74.8°C; Ht =2570kJ/kg; quality q =97.22%; density ρ =0.247kg/m^3 Velocity V = √(2*1000*(Hgas - Ht)) = 417m/s

Area 3mm dia = 7.07E-6m^3; Allowing for vena contracta (0.6) Aeff =4.24E-6

Mass flow = Aeff*ρ*Velocity Mflow =0.000437m/s

Evaporation rate = Mflow*(Hgas-Hliq) = 1kW (the desired condition).

With nothing to slow it down in the vacuum, the stream maintains its kinetic energy and its temperature of 74.8°C. If it's brought to rest, its temperature rises to Tsat = 84°C. High enough to give a nasty scald (nevertheles the least of your problems at this point). Another consideration is that approx 3% by mass of the stream is in liquid form, so that will probably survive the 5" journey to your hand

SLR

OK, I missed the 2-mm - so maybe I was understating my case. (But perhaps we should take into account that the narrowest part of the jet is usually smaller than the physical orifice (because the vapour has to converge on the orifice in the first instance).

The reason I still don't forecast burning at 5" distance is that the density of the steam near your (distant) hand is much reduced. So, although every molecule has the original kinetic energy, the temperature at which it will condense is quite low. Because of the spreading, molecules in the same area will also move more-or-less in the same direction, so the kinetic energy is evidenced as velocity of motion rather than as being hot gas. The low condensation temperature means that you can ignore that it is water vapour (at least until it is cool enough that condensation no longer causes problems), and treat it as a high-velocity stream of cool gas. But I could still be wrong - and I don't plan to try it, even using sensors - as all that water would play havoc with my vacuum pump.

I can't believe that water droplets which make up 3% the mass flow and are travelling at 400m/s will not survive a 5" journey lasting .3 milli-secs. Nor can I believe that the temperature of these droplets will be significantly different from the 75ºC at which they leave the orifice. I am still convinced that these conditions would result in burning.

Surface tension will certainly prevent the droplets boiling from inside.

But my argument was not that the droplets will be cool, merely that your skin will already be frozen and your blood boiling, and a droplet will flatten as it hits your skin so that it only heats quite a shallow depth. So only the average power really matters, and the average power available to overheat your skin (due to cooling the liquid droplets) will be approximately 3%*1kW*(85-45)*4.2/2200 or 2.3-Watts. Admittedly, this will be over an area of only about 1-sq.cm, but I imagine that evaporation into the low pressure will easily exceed this.

However, I did make a more significant mistake - unlike subsonic winds where the predominant effect is to increase the rate of heat transfer, supersonic winds result in heating. So the energy available from the steam will be about 1kW*(85-45)*1.5/2200 or 27-Watts, with (at a rough guess) half of that being transferred to your hand.
I don't have the figures for evaporation rates from a wet surface at 45^OC into a pressure of 0.04 Atm, but I imagine that the total of 16 Watts/sq-cm will be enough to overcome this

Of course, we already agreed that other effects will already have caused ice burns and various other issues)

I think that we need to understand the difference between steady state vacuum conditions and the transient that is represented by this problem. In the limit as time progresses, the stream of steam and water vapor will expand and eventually settle in at steady state outer space conditions (assuming a perfect vacuum). However, how is that going to happen just 5 inches from the orifice with this stream of steam shooting out at you? This plume is still a bounded mass with internal pressure. Actually, since there will be a constant mass in a given volume of the stream immediately after the orifice (even though there is mass flow through this volume), inside this volume there will be a steady pressure distribution as long as the flow continues. The previous calculations show the conditions inside this plume.

As an example, solar flares coming off of the sun have temperatures of millions of degrees. Just because it exits into a vacuum does not mean that it instantaneously settles into deep space conditions. Eventually the gas does cool and expand into infinity as it flows along due to the work done by the internal pressure of the gas plume pushing it out in all directions, and energy lost due to radiation.

It's all a matter of scale. Ignoring radiation etc, when the plume has spread to 10x the original area, the total energy will be unchanged, but the temperature will be much lower. Depending on how much interaction there is between the molecules, this could be adiabatic expansion or simply that the spread represents originally locally random directions of motion now being correlated with position in the plume. The reason that this is significant over such a short distance in the stream from the pressure cooker is that the temperature may have been 85-degrees inside the cooker, but longitudinal expansion within the nozzle results in velocity increase and initial cooling, and the outflow is conical, so lateral expansion then converts some of the random thermal motion to systematic lateral stream velocity. (I suspect that the spread to 1-sq.cm is an underestimate, but don't have the tools to check).

Strictly speaking, we should use the standard for pressure cookers.

The U.S. Department of Agriculture standard for pressure cookers is one atmosphere (15 psi) over atmospheric pressure (i.e., 2 atmospheres inside the pressure cooker). This corresponds to 125 degrees C at sea level. Therefore, for a typical pressure cooker, the inside conditions are 125 C in earth's atmosphere at sea level, and one atmosphere and 100 degrees C if the ambient is a perfect vacuum.

Consequently, the delta P should always be considered as 1 atmosphere.

Because there is no atmosphere that the plum must move through when it is in a vacuum, it will spead out some, but in general go straight to the hand at high velocity over a distance of 5 inches, deaccelerate and convert the kinetic energy to heat.

Those are designs standards for when the cooker is sealed. The challenge gives an unsealed pressure cooker - so the limit is power in, not the pressure limit set by the cap.

But the challenge does not specify a limit on power. For a constant differential pressure of one atmosphere in both cases, the power required will be approximately the same in both cases, which would appear to be the more logical approach to analyze this.

"But the challenge does not specify a limit on power"
The original situation is a "two litre pressure cooker boiling in your home kitchen". For such a small pot in a domestic environment, 1-kW nett seems a not unreasonable limit.

"For a constant differential pressure of one atmosphere in both cases, the power required will be approximately the same "
For equal power, the pressure difference will always be larger when the external pressure is lower. I'll take two cases:
If the flow is "choked" when the pan is vented to atmosphere, the absolute pressure inside it will not change as you reduce the external pressure.
For any other case, the density of the steam reduces as the pressures reduce. That means that, to get a a fixed mass of steam through the nozzle, it will need to have a higher velocity. The kinetic energy carried by the steam is then provided by the pressure differential; as the distance is constant, the pressure difference is proportional to the final velocity - and therefore (approximately) inversely proportional to the exit pressure from the nozzle. In practice, there are all sorts of modifying factors, including that latent heat varies with temperature, and that a larger proportion of the steam will condense when the output pressure is lower - but the trend is clear.

I agree.

A pressure cooker has a relief valve keeping a constant differential pressure. In this case, the 3 mm orifice was specified, so the constant differential is not maintained.

Large gas burners on a stove can put out considerably more than 1 kW.

Indeed they can - but how much of this energy can be delivered to a 2-litre pressure cooker? (I think the diameter must be only about 15-cm).

N.B. gas would be a bit of a problem in the vacuum, so I assumed electrical heating

If you feed an oxidizer with the gas, it burns just as well! Then we can have tea and crumpets on the moon.

If you put the pressure cooker in a vacuum the water will "BOIL" turn from a liquid to a vapor at a much lower temprature.

The temperature inside the pressure cooker (saturation temperature) is mostly determined by the heat rate of heat going into the cooker. Any temperature can be maintained, any saturation temperature pressure relationship simply by adjusting the rate of heat transfer into the pressure cooker.

Come on there is obviously no pressure in your job. Giv e a job conulting I'll be in your top 5% Eric Burbidge 562-440-3580

I think the answer is actually "You will not get a burn in either case."

If the pressure is 0.1 psia (near perfect vacuum) the water boils at 35 degrees farenheit. If you assume that the pressure builds to 1psia in the pressure cooker, the temperature is just above 101 degrees which is hot, but probably not hot enough to do any damage.

Regarding the "official" answer - part of the last paragraph looks to be nonsense. "no change in pressure because there is no expansion due to changes in pressure". The only way that the water can boil at a temperature above (say) 40^OC is if the pressure inside the pan corresponds to that boiling temperature. So there is a pressure change as the water exits the pan. What is true is that there is nothing present in the vacuum to absorb the kinetic energy of the steam molecules, so their rms speed will not change. At a distance you will have molecules of steam all moving in roughly the same direction, and with a relatively uniform velocity. In other words, this will be a fast-moving but cool stream. When it hits a cooler surface, some may stick and some may bounce off in random directions. In the first case, all the kinetic energy will be converted back to heat - however, being water in a vaccuum it will soon evaporate - but overall it will transfer kinetic energy to the surface. In the second case, the effect will depend on the angular distribution of the bounce; assuming that the molecular weight of the materials of the surface are not too dissimilar to that of the steam, I think in practice that about half the kinetic energy will be lost to the surface. However, we can see that in neither case will the steam transfer any latent heat, so it will be nothing like having your hand in a jet of hot steam. Whether the surface will get warmer or cooler overall depends on the amount of other water in the skin that is evaporating into the vacuum. (It fair makes your blood boil).

Conclusion: we don't really have enough imformation to say whether water-vapour would cause skin to become hotter than normal blood temperature at a distance in the vaccuum - it depends at least in part on design of the nozzle and the rate of boiling. But the combined effects would be at equivalent to more than deep burns in any case, so I wouldn't recommend anyone to undertake the experiment.