Originally posted by kappa
If one new the dimensions of the support columns, how would you calculate the amount of heat energy required to heat them to the required temperature of plasticity? <-- right word? hehe I think you know what im saying..
Thanks
In my previous life I designed furnaces for melting metals.....I have the formulae at work.
Simply put now however....if we considered one vertical structural member which was open to the fire and extending thru the building above and below the fire.
We can then make some assumptions to simplify the calculation.
Energy (kw) is being absorbed by the structural member over its whole length that is open to the radiation source (the fire)
The rate of (thermal) energy absorbsion is a function of
surface area of the member, its black factor, the surface area of the emmitter (the stuff thats on fire all around it or the flames them selves) the emmisivity (inverse of black body factor) and the Delta T (temperature difference). There is an adjustment for the radiated distance when the size of the emmitter is small.
However we can assume that distance is irrlevant and the surface area of the emmitter is massive.
The calculation will then give you a heat transfer rate based on KW.
We then have energy loss from the system. As we are only considering radiant heat thermal losses from the rest of the environment are ignored. However as you point out there will be conductive losses along the steel member, and these will potentially limit the total energy state the member may attain locally.
To do this simply we could just take the member failure temperature and test if we can achieve an energy balance or even surplus at this point.
So we take eg 800 degrees C
We know the cross sectional area of the member lets say its an I beam 1 m x .5 x .1 thick = 0.2M^2
lets say that a sprinkler system is 2meters below and 2 meters above our inferno is cooling the steel to near ambient so delta T = 760 degrees C.
Wl=w/2= 780 *K *.2/ 2 where k is the coefficient of thermal conductivity of steel and Wl is the total conductive energy loss.
If the energy taken away is less than the energy absorbed we can safely say that the steel will exceed this temperature at a rate which is immediately calculated from the coefficient of heat capacity wattsdegreesC/kg. This coefficeint changes with temperature so as it heats up the system balance changes as all the Td's change etc.
The air temperature in there will be much hotter......but air is a lousey conductor however the environment will have ignited any combustable material that can burn including all so called fire retardent polycarbons to say nothing of the lower level plastics that abound in every environment these days. Basic organic combustables (paper and wood) will play a very small role compared to the massive thermal out put as hydro carbons start to break down at elevated temperatures. Acting like a massive candle.
The structural engineer will have been trying to achieve the maximum structural strength from the minimum weight of steel. Particularly from a very tall structure with no taper.
This would mean he is attempting to minimise cross section in his steel members. Using tubes, I beams and c sections rather than increasing basic section. if you were to take a giant razor and cut across the building horizontally you would find that the total steel cross section you have cut thru could well be less than 5% of the total. Made up of many comparatively "thin" sectioned columns.
