Cooling rate of some very hot steel

[fbeltran773]

I have a somewhat rectangle of .75 in x .75 in x 2 in. piece of low-carbon steel (1069) starting at 2250 F going to 1800 F set alone to cool off in 65 F air and need to find an approx. time between temps. I'm assuming no radiation, no forced convection, the grip holder is not affecting it and have standard conduction formula (q=hA(T-Ta). Think I can assume lumped capacitance? I'm thinking the convection coefficient will have to be up there because of how hot the steel is. Also assuming heat is leaving in all directions equally because of how small the thing is. I found a value of 45000 W/m^2*K but that seems to high and 10 seems to low. Thanks

 

[jay165]

Sounds like a homework problem, sonny

 

[fbeltran773]

NOT A HOMEWORK PROBLEM. I work at a tool forging company in Skokie, Il. 3rd day on the job and they gave me this problem. Were're more design. Think I'm going with the lumped capacitance way because the k of steel is 29 W/m*K but the L in meters is .05 so hopefully I can get away with it. Graduated back in Dec. Thanks anyways

 

[corus]

If the steel is allowed to cool off in air then the heat loss is due to radiation and natural convection. For higher temperatures radiation will be more dominant but will be less so at lower temperatures where natural convection is more significant. Using a constant heat transfer coefficient wouldn't give you the right answer though assuming a lumped mass (if that's what you mean by lumped capacitance) is a reasonable assumption to make for this case. The maths will be complicated for the non-linear radiation and natural convection but it might be made easier by assuming a heat transfer coefficient that is linear with temperature, for example. Calculate the combined heat transfer coefficient against temperature and fit a line to it. I forget how non-linear the combined relationship is. Using that will give you a differential equation that may be easier to solve. Personally I'd just use a finite element method to calculate it.
Your value of conductivity seems low as normally the value is about 50 W/m K. You'll also need density (7850 kg/m^3) and specific heat, about 500 J/kg C, if I remember.

corus

http://www.corusresearch.com

 

[fbeltran773]

Thanks corus. Your reply made me search radiation which I found H = e*theta*A*(T^4-Ta^4). Now it's all about finding the right emissivity constant for steel. The temps are so high I'm now only going to consider radiation.

 

[corus]

You should be able to find values of emissivity with temperature. To make it easy I'd use a value of 0.8, 0.95 if it were cold. Check with references though. A constant value should be good enough.
Natural convection coefficient is usually expressed in terms of A(T-Ta)^0.333 where A is a constant dependent upon the orientation of the surface. Including it in your calculation shouldn't be a problem if you're going to make the sum total of the two as a function of temperature.
You may find exact solutions have already been made for the case of radiation alone, however, as I think all you're solving is pcdT/dt=Ae(Tk^4-Tak^4). Remember Tk in the radiation part of the equation is in degrees K.

corus

http://www.corusresearch.com

 

[kmpillai]

Dear fbeltran,
I analysed the problem using the limited data provided by you, using an FE software and I found after 90 seconds the whole mass coming to below the temperature you mentioned 1800 deg F (982 deg C).In this case both radiation and natural convection is taken.
Due to very high temperature the radiation effct is a major one and natural convection heat tranfer is insignificant. In that analysis if consider only the radiation heat transfer, the time required to cool the mass below 1800 deg F is roughly 110 seconds only.

Thermal conductivity value you mentioned is alright. Mr Corus has mentioned around 50 W/m K, but that value is for around 20 deg C and at higher temperatures the value is coming down.

I am having the complete analysis results. If you want please drop mail to me.

KMP
kmpillai@hotmail.com

 

[JKEngineer]

The suggested value of e, .8-.95 seems quite hi for steel, although it may be ok at very high temperatures. Also the radiation equation needs the stefan-boltzman constant in it. Also, the temperatures in the equation can be in degrees Rankine, which is also an absolute temperature scale, like Kelvin, and is based on Fahrenheit, like the stated problem. Just be sure the units of the S-B constant are correct. Temperature dependency of the convective coefficients may be a factor. Just some random observations.

Have fun in the new job!

Jack

Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering

http://www.KleinfeldTechnical.com