temperature on the other side of fire wall 1

[bmagdalena]

Hi all,

I was tasked with calculating temperature on the other side of fire wall if the fire breaks down. Normally Im doing structual calculations thats why I need some help. I have sample calculations and "heat transfer" by Mills.
I have 2 mobile facilities that are next to each other, about 9" gap. I need to calculate what temperature will be on the other side of the wall, if there is fire on one side. If we have to we can make gap between them to be about 2feet.
based on the sample calculations we assume that fire will have temp of 1200K, first wall has (2) 12gage steel walls with 2" insulation inside. assumption is that fire will emitate radiation heat of 117.6kW/m2 (that is calculated from Stefan-Boltzman law), and then that heat was used to calculate temp on the other side based on conductuion law. what I dont understand insulation resistance was ignored and resistance of steel, which is almost nothing, was used to get temp on the other side.
anyone have any idea?

 

[corus]

If you have a layer of insulation then this will have a high thermal inertia, taking time for the heat to conduct through. As such you should look at the transient change in temperature. In the worst case you'd have steady state conditions. Normally steel has a very low thermal resistance compared with insulation and it's often excluded from the calculation, though I can't see why the overall resistance isn't used, ie. the sum of x/k.A terms, as would apply in your case.

corus

 

[IRstuff]

Once you've included the insulation, you may find that your calculation underestimates the heat flow, since it appears that you are only using radiated emission.

Assuming 2inches at 0.1W/m-K insulation, I get the outside wall temperature around 180°C in steady state, but the accuracy is pretty low without a more detailed analysis.

TTFN

FAQ731-376

 

[bmagdalena]

I might be doing something wrong then.
when I calculated temp on the other side of wall I got T2=T1-(q*l/k)=1200-(117600*0.0508/0.17)=-33941K wich is of course unresonalbe number.
also if fire emitts 117.6kW/m2 can I treat this as heat exchange between fire and the wall. based on book heat exchange by radiation is q=J-G. so it would appere that you cant do that.

 

[IRstuff]

But, the point is that the 117.6 kW/m^2 assumes a 1200K outside wall, so you've essentially answered the question by stipulating the answer and ignoring the insulation.

Your equation is nonensical, since you're equating the radiated energy from the surface to the conducted energy through insulation, which is, of course, absurd, since the surface temperature, when accounting for insulation, is nowhere near 1200 K.

TTFN

FAQ731-376

 

[bmagdalena]

I guess I have problem with applying the process of radiation, conduction and convection to my problem. because the only known I have is temperature of the fire 1200K and temp on the other side of second wall 300K.
So what should I do to get initial heat transfer to get temp T2?

 

[IRstuff]

see:

http://files.engineering.com/getfil...8be2-45f4-9318-294905e42a70&file=hot_shed.pdf

or

http://files.engineering.com/getfil...e538-454e-bfa1-9e532aa0574b&file=hot_shed.mcd

TTFN

FAQ731-376

 

[bmagdalena]

Thanks IR. I don’t have MathCAD available so I went through the cacls and now I have few questions.
Tw and Ta are shown in K, and should be in C, right, since you add Tb to both of them.
Tw is temp on the wall on fire side, and Ta is on the other side of the exposed wall?
Heat transfer hamb=5W/m^2K - where this is coming from?
Then based on the equations I see we are using convection process between fire and wall, then we ignore completely conductivity in the wall, and then we use radiation and conductivity in the 9in gap and again ignore conductivity through the other wall.

Is that correct?

 

[bmagdalena]

one more why Tw is 160K , and the answer is Tw 154K?

 

[IRstuff]

> Delta temperatures don't care about absolute, so only the radiation equation needs 273.15K added

> 106K is the guess temperature that the solver starts with

> No, you read the equation incorrectly. Conduction THROUGH the wall = convection + radiation.

TTFN

FAQ731-376

 

[zekeman]

From what I can see, worst case, ignoring convection I got thermal resistances
R12=1/3, R34=1/3 and R40=1/3 ( in BTU,ft hr F units) and since the effective radiation resistance is about 1/15 I get that the outer walls are about 900 K and the inner wall of the second building is about 600 K.
Looks like you may have structural problems at 900 K.

A more accurate analysis which includes convection would mitigate these results. By how much? . You may have to consult the literature for this since convection is complex and air gap plays a strong role.

 

[zekeman]

Error
My bad. I had an error in the insulation resistance.
Should be
R12=R34=3, R40=1/15
Makes a difference. Then the outer walls of the 2 buildings absent convection come close to 750 K.

The assumption of convection /radiation for an open space(R40=0.33) between them yields an outer firewall temperature of 390 K
which is on the low side of the actual.

 

[bmagdalena]

IR,
I understand your eq now.
but still, I did heat transfer 7 years ago, and just putting eq and pluging the numbers dosent explain to me how the process is going. so the only think I can get from eq is that we can compare heat from conduciton from wall adjacent to fire to the heat fransfer via radiation and conduction in the 9in gap. But how this gives me temperature T4?

zekeman,
can you explain what you are using and where? is that 750K my T1?

 

[IRstuff]

It's a heat transfer process, and you have conservation of energy, therefore, the heat flow across each boundary must be equal to the heat flow between boundaries.

The issue is whether a 9" gap supports conduction, since it's larger than many furnace vents, but an extremely large area and height wall might wind up doing a large amount of conduction.

The worst case would be to assume that there is no convection, which means that all of the heat going through the first wall, by definition, must go through the second wall. In that case, Zekeman's answer of 750K is similar to what I get: T2=757K T3=742K.

That is not surprising, since the problem is symmetrical with no external losses, so the temperature in the air gap would be expected to be roughly (1200K+300K)/2 = 750K

TTFN

FAQ731-376

 

[corus]

Conduction through air is generally negligible. With such a large gap you might consider natural convection, but in the worst case neglect this. At such high temperatures the dominant heat flow in the gap will be via radiation. You really need to look at this as a 1D problem with heat flow to the flame and across the gap by radiation. You'll also need to estimate the emissivity of the surfaces too. For this non-linear problem you need something more than hand calcs.

corus

 

[LOUDOG]

If you are doing spacing calculations to stop the spread of fire you are probably doing it wrong. Why are you doing this calculation.

 

[bmagdalena]

in oilfield industry constructing typical fire wall with gypsum board is not practical, they just dont do it. modules that my company did will be placed ajdacent to the big pipebarn. since pipebarn is different category we need 2hr rated wall. we cant test the wall, so we are trying to calculated the temp on the other side that would symulate tests done according to ASTM 119. So purpose of wall in not to stop the fire but stand for 2 hr.

 

[sailoday28]

Is the space between the walls closed at the top and or bottom?
If there is no chimney effect, then convection should be up from the hot (flame) side and down on the cooler (warm) surface.

 

[bmagdalena]

gap between facilities will be encolsed, so yes top and bottom will be closed

 

[IRstuff]

OK, so no convection. That would mean that your non-fire module will get exceedingly hot.

TTFN

FAQ731-376