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temperature on the other side of fire wall 1

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bmagdalena

Structural
Jun 22, 2009
57
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?
 
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yes that is true. that why we have that gap between walls that we can make bigger if nessesary. so theoretically, we have to proof that with air gap between walls temeperature T4 will not get bigger than 300K +250K. where 300K is temp on the inside of the module To.
im still missing what is driving force for heat exchange. and everytime I think I understand new aspect comes up and blows my theory up.
 
Even in the worst case, where T3 is nearly 750K, since the thermal resistance R34 is 10 times the resistance R40, T4 will be (T3-T0)*1/10+300=345K
 
zekeman, where you get from that resistance R34 is 10 times resistance R40?
 
Here is a Googled reference

Natural convection in an air layer enclosed between two vertical plates with different temperatures



E.R.G. Eckertf†, a and Walter O. Carlson‡, a

aHeat Transfer Laboratory, University of Minnesota, Minneapolis, Minn., USA

Received 22 January 1960. Available online 27 February 2003.
 
IR thanks a lot

now I know how to take is appart and it make sense to me.
 
bmagdalena (Structural)
19 Feb 10 16:10
zekeman, where you get from that resistance R34 is 10 times resistance R40?


From my first post, I indicated that R12=R34=3, R40=1/15
and since T3 is about 750 (see IRstufis post for verification),
the differential temperature between is T4-T0=750-300=450 so invoking the seies drop like an electrical circuit
T4=T0+(T4-T0)*1/15/(1/15+3)= 300+450*1/45=310
which corrects an error I made earlier, making it even smaller.
The 1/15 and 3 derive from the inside film and the insulation resistance as IRstuff showed.
 
IRstuff
got one more question?
Convection in air gap.I dont understand your equation for hair(Tm). I looked in "Heat transfer" by Mills, and I didnt find this eq.
I found relationship to calculate hc based on Nu, Ra, and Pr number. but when I run numbers for Ra number assuming deltaT=1200-300, then Ra came to be more than 10^7 which is limit to apply the Nu equations so Im stuck again.
 
Based on your response that the gap was covered, I assumed that there was no convection, and that the only heat transfer is conduction through the air.

If no fresh air is added to the gap, then that's a reasonable approximation.

TTFN

FAQ731-376
 
indead, the gap will be enclosed, although it might not be seal tight. it just going to be light gage flashing attached to both buildings.
that is why I was thinkging about natural convection in gap itself.
 
Since the walls are ostensibly isothermal, there needs to be an opening at the top to get decent convection. Otherwise, it's just random swirling air, and you might as well model it as conduction. The answer will only differ by a few degrees.

TTFN

FAQ731-376
 
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