Radiative heat transfer

[DieterVk]

Hello,

I'm having difficulties calculating the time to reach a certain temperature caused by radiation. I cannot find any method to get time into my calculations.

Any help?

Specific problem:
Double-walled tank with the outer wall at 700 °C. How long does it take for the inner wall to reach 500 °C?
I've calculated that the conduction trough the walls is so rapid that this does not has to be calculated.
Also the convection of the air between the walls is negligible. So I guess the inner wall is only heated due radiation. Correct?

Thanks!

 

[LittleInch]

Wow, that's some temperature. I'm not sure "tank" adequately covers that sort of temperature

Time is a factor of heat input and mass of metal - heat capacity of the metal. as you say you can assume that the entire wall heats up at the same rate. Caron steel is 0.49 Kj/kg/K.

So at a fairly basic level if you have 10kg of steel and Input of say 1kW (1000J/sec) it will take 49 seconds to go up 10 degrees C assuming no losses from the lump of metal

The other factor is whether any heat then "escapes" from the inner wall or do you ignore those losses?

I'm surprised you think convection isn't a factor care to share how you came to that conclusion?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[DieterVk]

Well,

For calculating the overall heat transfer coefficient:
1/U = 1/h + D/k
there isn't much air between the walls, and only natural convection occurs. So let's assume
h = 10 W/m²K (which is an overestimation)
D = 0.02 m
k= 20 W/mK (for steel)

1/U = 0.1 + 0.001 = ... so the walls are thin enough to rule out the conductivity.

Now q = U (T1-T2) + o-(T1^4-T2^4)/(2/e-1) (which is the conduction/convection plus the radiation part)
o- = Boltman constant = 5.67*10^-8
e=0.85 for steel

q = 10*(1000-800) + 5.67E-8*5,904E+11
= 2000 + 33475
So because the radiation part is much larger, we can rule out the convective part.

I know it is an estimation, but that's okay.

 

[DieterVk]

Furthermore,
I assume a fire under a tank. I assume the tank fails when the inner wall reaches 500°C. That's why I need to calculate the time for this to happen.

I assume the surface as infinite, so no masses or surfaces.

The tank is filled with liquid at atmospheric temperature. How can I incorporate this? How can I calculate the time?

 

[davefitz]

The liquid in the tank will also absorb some of the heat, so it will help to cool the walls of the inner tank. If the net heat flux from 700C to 500 C is in excess of the "liedenfrost flux" ( aka DNB heat flux) then a vapor layer will form between the inner tank inside wall surface and the liquid's ability to absorb the heat will be greatly reduced.

"In this bright future, you can't forget your past..." Bob Marley

 

[LittleInch]

I can see what you're trying to do, but you can't assume the area to be infinite. In any case radiation is done per m2, just look at 1 m2, you know the thickness of the plate - gives you a mass and hence time.

however if the inner tank is filled with liquid then I think you're doing this all wrong.

Fire cases are about the change in pressure and potential boiling / over pressure of the tank.

Normally if you have an intense fire under a piece of equipment, you assume it will fail at some point. What you don't want it to do is explode and create further hazards.

You then bund the tank so if the contents come out and catch fire you limit the area of the fire.






Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[DieterVk]

Yes, the inner tank is filled with liquid. How can i calculate the time then?

Matter of the case is really to calculate when (theoretically) the inner wall will fail, not considering any explosions or leaks.

 

[LittleInch]

In theory when all the liquid has, presumably, boiled off, but as I said before this is not the way these sorts of things are done.

Lets go back a bit here.

What sort of fire?
Where has 700C come from?
why fail at 500?
What about outer wall failure?
What is liquid boiling point?
Is the venting system good enough to handle the boiling liquid?
Is there a secondary bund?
Why is this thing double walled in the first place
What sort of fire protection do you have?

All much more important than figuring out when the inner wall fails

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[DieterVk]

What sort of fire? just A fire
Where has 700C come from? outer wall temperature when subjected to pool fire
why fail at 500? just an estimation
What about outer wall failure? I assume the vessel as lost
What is liquid boiling point? don't know
Is the venting system good enough to handle the boiling liquid? don't know
Is there a secondary bund? don't know
Why is this thing double walled in the first place? don't know
What sort of fire protection do you have? The question is, which fire protection is needed? That's the whole point of calculating time. Is there enough time to deploy a monitor, or is a fixed cooling system needed?

I don't need a precise calculation, just a rough time like is it 5 sec, 10 min, 1 hr?

 

[LittleInch]

Well your worst case is going to be when the tank is empty so use that as your start point.

Use 1m2 as your heat area which also give you your mass of steel to heat up.

Question then will be how much mass do you assume is not affected by the pool fire. - 50%?

If the tank has liquid that boils at your storage pressure a long way from 500C, then I'm pretty sure your issue is not failure of the inner tank, but the amount of vapour being generated and avoidance of the tank / vessel ( which is it - makes a big difference) exploding.

The point about type of fire is whether it is a general fire / heat or a jet fire with impingement onto the surface? The latter will be much quicker to destroy your vessel.

You might find the following guide useful

http://www.hse.gov.uk/pubns/books/hsg176.htm

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Sanikapatel]

Are there any more easy ways to calculate and get answer within sec??

Sanika Patel

http://crbtech.in/CAD-CAM-Training/

 

[kacarrol]

This is a tough one. I wonder if you could use a lumped capacitance analysis? You may have to use a numerical integration method to find the solution to the equation. Do you have a copy of Fundamentals of Heat and Mass Transfer by Incropera? Chapter 5 has some examples of calculating the time history of thermocouples where radiation is part of the equation.

 

[kacarrol]

I quickly skimmed the chapter. I think this would work for you. You have to apply a conservation of energy equation at any instant t, something like:

epsilon * area * sigma * (T^4 - Tsur^4) = rho * volume * c * dT/dt

Actually, if you only have radiation there is an exact solution to the equation, you don't have to do a numerical solution.

But you really should read the whole chapter to make sure that all the assumptions hold for your system (such as spatial effects).

 

[kacarrol]

I did a quick google search, if you don't have Incropera it looks like this slide deck is a condensed version of the chapter:

https://www.google.ca/url?sa=t&rct=...iHr9L3gsnPAhWky4MKHXy2AjoQFggpMAI&url=http://

http://www3.nd.edu/~msen/Teaching/I...zIHu2jmQ5u0Rb4wvsTYOeQ&bvm=bv.135258522,d.amc

Good luck!
K