Stratification

[Phoenix]

Thanks again Civvy for your succinctness and accuracy.

Sam,

I really haven't a clue what you're on about.  It may be me, it may be you, but our paths are at right angles and I cannot see us finding common ground.  Here are some comments on your brief treatise.

Even within an open space stratification occurs, (it can,agreed, but not always) with cold air remaining above, (only if we're talking about climatic systems - which we're not) (provided the space is large enough relative to the energy).

Take for example a fire in the open air, on a still day; perhaps a bonfire in a field or park. The hot smoky gasses rise, (true the more dense molecules fall) (more dense molecules?! Why are we getting involved at the molecular level?  Are you talking about the mixture of gases and air?), and a convection current is set up (Are you sure? The phrase 'convection current' is normally taken to mean a cyclical current driven by convective forces - for illustration, put the phrase into Google images.  In the open, even on a still day, I think we'd be likely to witness vertical convection, certainly, followed by probable stratification and dispersion of the gases.  I don't think we'd see the typical cyclical current taking place as we might if the fire was in a room.  Maybe you mean 'convection.'). As the hot smoky gasses rise the heavier particulates (is that a posh word for particles?) reach a point where their buoyancy is outweighed by their density ('buoyancy is outweighed by their density,' you say.  Is buoyancy not a product of relative densities?  If you'll agree that buoyancy is directly related to density how can it become 'outweighed' by the very characteristic that is creating it?). Therefore the more dense particulates (there's that word again) and then more dense molecules (we're looking at the density of molecules again!) start to form layers. (The smoke can be observed to form layers or stratify) the atmosphere above the smoke is at the same temperature as it always was (who  said it was otherwise?). If this were not the case then all smoke from all fires would always reach space (?!??).
 
The other factor in the problem inside a building, is the movement energy (Kinetic) and the mixing into the plume of hot smoky gas stream of the surrounding air; especially relevant to this discussion  if the plume is deflected from the vertical to the horizontal, and then possibly passing a ceiling well.  Hot gasses from a jet engine do not rise until they have lost their kinetic energy of lateral movement (I think Civvy has amply demonstrated that this is nonsense). Some surrounding atmosphere is drawn into the plume by Bernoulli’s principle (Are you sure you have the correct effect?  Has the Venturi effect anything to do with it?)  , true, causing eddies and partially cooling the jet stream. However, the surrounding atmoshere is the same temperature it always was; (albeit rising slowly by the radiation from the hot gas stream). There is no convection as the lateral movement is greater than the movement caused by relative density. (Hence the aeroplane analogy).  (So is that why bricks float?)

Stu

[Phoenix]

Your monkey and hunter experiment assumes both the target and gun are falling simultaneously.

Not at all.  The gun is stationary (apart from a bit of recoil, maybe).  The bullet is falling.  Did you mean bullet?

Stu

[SamFIRT]

Phoenix

I believe your comments on my post only go to show further that my original post was correct. .........Bernoulli’s principle......... Has the Venturi effect anything to do with it? ....Hummm well I would think so. Perhaps Bernoulli would think so too. 

Convection "heat transfer within the atmosphere involving the upward movement" or "circulatory movement in a liquid or gas, resulting from regions of different temperatures and different densities rising and falling in response to gravity"

Hot gasses from a jet engine do not rise until they have lost their kinetic energy of lateral movement (I think Civvy has amply demonstrated that this is nonsense).

I think Newton would disagree. As would Bernoulli.

I take your point about the bullet falling and not the gun. 

[Wiz]

What all the above indicates to me is that smoke stratification is difficult to predict, explain and prove. This is obviously why BS5839-1 skims over it and provides no specific recommendations, as such.
It appears to be one of those 'what if' scenarios that includes so many variables which can therefore produce various differing results depending on which of the numerous variables slightly alters.
It would help if there were at least some basic 'rules of thumb' that could at least provide some small guidance, such as something like 'the likelihood of smoke stratification at heights less than 6m above the seat of a fire are minimal and should be ignored'. But no-one has mentioned anything like this.

[CivvyFSO]

Where is our resident genius BLEVE when you want him?

[Phoenix]

There is a 'rule of thumb' method, Wiz, but you probably won't like it because it's not something you can do in your head.  The maximum expected height to which smoke will rise in an axi-symmetric plume above a fire is given (in m) by:

z_max = 5.54Q_p^1/4(ΔT/Δz)^-3/8

Δt/Δz is the temperature gradient over the height of the atrium in Km^-1.

Q_p is the convective heat output of the fire in kW.

The model has supposedly been reasonably well validated.  I find that it always gives pretty high values for z_max, sometimes they feel, intuitively, a little too high.  Also, establishing the temperature gradient can be problematic.

But there you go, that's the published guidance.

Stu

[CivvyFSO]

WOT NO BERNOULLIS?

[SamFIRT]

zmax = 5.54Qp1/4(ΔT/Δz)-3/8

Very impressive

Do you mean Qp =  KW or KW per time frame because if not this is only a snapshot and Heisenberg’s uncertainty principle must apply

Also,

establishing the temperature gradient can be problematic

. I should say that is an understatement. Nice looking formula though.

Who did the maths? Why 5.54, ¼ and -3/8 ?

But there you go, that's the published guidance

. By whom?

Also. How does the formula take into account the fuel type and therefore the HSG type being produced? How does it take into account the efficiency of combustion and whether the fire is burning in fuel or ventilation controlled conditions? Final question (for today) how does it relate to smouldering. Either in totality or pre or post flaming?

[CivvyFSO]

if not this is only a snapshot and Heisenberg’s uncertainty principle must apply

Quantum physics is nothing to do with it at this scale. Stop trying to be clever.

[SamFIRT]

Stop trying to be clever

Oh dear am I being told off?  That used to happen at school as well. In exactly the same way.    Usually when the teacher could not answer a question. How that brought back memories. 

Heisenberg’s uncertainty principle applies to Newtonian physics as well as quantum. Ie …the position and the velocity of an object cannot both be measured exactly, at the same time.

Arguably it applies to philosophy as well.

But never mind I can see I am being irritating. Just like I was at school when I kept asking questions.  And I thought this was a forum for understanding and knowledge exchange  .

End of posts 

[Phoenix]

Weren't we talking about smoke?

[CivvyFSO]

Ok, It's late and I have nothing better to do.

Who did the maths?

Some bloke named Gunnar Heskestad I think.

Why 5.54, ¼ and -3/8 ?

Because that is what that clever bloke named Heskestad found.

By whom?

James A. Milke in the SFPE Handbook, but I believe that the main credit should go to Heskestad.

(Especially since the SFPE handbook seems to have an error as it is ^-3.8 in there, not ^-3/8.)

Also. How does the formula take into account the fuel type and therefore the HSG type being produced? How does it take into account the efficiency of combustion and whether the fire is burning in fuel or ventilation controlled conditions?

First of all, forgive me if I am being stupid but I have no idea what you mean by HSG.

It clearly does not concern itself with the fuel type, fuel amount or efficiency, All of those things would help to determine Q.

A useful analogy here would be as follows:

[Conversation in a pub]
Sam's mate: I worked out that my car does about 50mpg
Sam: But you would need to know about the aerodynamics of the car, the octance of the fuel, the rolling resistance, the weight carried in the car and loads of other things!
Sam's mate: No, I just filled it up, then the next time I filled it up, looked how many miles I had done and looked at how many gallons it took to fill up, divided the miles by the gallons, and it came to 50.
Sam: But what about when it is cold petrol doesn't burn as efficiently, surely you had to calculate that?
Sam's mate: No mate.
Sam: What about other loads, like having the air-con on? That alters the fuel consumption too!
Sam's mate: It does 50mpg on average, I worked it out.
Sam: But what about air pressure? What about the weight difference as you use more fuel?
Sam's mate: I am going home.
Sam: In your car?
Sam's mate: Yes.
Sam: Don't believe your sat-nav.
Sam's mate: Why?
Sam: Because Heisenberg's uncertainty principle proves that there is no way that a sat-nav can actually know where you are AND the speed you are going!
Sam's mate: I think I need some new friends.

Anyway..... The equation is usually used for smoke calculations in shopping malls and atria, the types of places where stratification could be a problem, therefore the required inlet air is calculated so ventilation controlled fires are not a problem. If you have a ventilation controlled fire then stratification is probably less of a problem than unburnt gases. And rather than force everyone to calculate stuff from first principles, some kind people such as Mr Milke and Mr Heskestad sometimes try to simplify things for us common people.

Final question (for today) how does it relate to smouldering. Either in totality or pre or post flaming?

It doesn't.

To be honest, if you are asleep in a room with such a slow burning fire, which never reaches a detector, and it kills you, I am afraid that you have just been very very unlucky.

[BLEVE]

WOT NO BERNOULLIS?

Well some one is talking a load of bernoullis, that's for sure and it aint Civvy

To be fair, I reckon John Klote should also get a mention as he had some contribution to Mr Milke

[BLEVE]

.

[Wiz]

.

Well that's pretty clear, Bleve!