Transient Heat Conduction - ABS Sheet

[WoodFE]

Hello

I've been asked to provide an estimate of the core temperature of an ABS plastic sheet heated in an oven during a thermoforming process.

I have no real knowledge of the heat input to the sheet, but I do know the skin temperature is 165-175 celsius after 180s of heating.

My question is, is it possible to calculate the core temperature of this sheet (6.5 mm) from the skin temperature, heating time and material properties alone?

The oven uses radiation as its primary heating method and I have only dealt with convective heat transfer before, so I am unsure how to proceed.

Any insight would be much appreciated.

 

[MintJulep]

If that's the only information you have then probably not.

Can you measure the surface temperature of the other side?

 

[WoodFE]

Only one side of the sheet can be measured, however the oven has elements above and below the sheet so I've assumed the surface temperature on both sides is the same.

 

[corus]

You'd have to iterate by varying the oven temperature until your calculated temperatures at 180s matched those at the surface, assuming you know the emissivity of the material and the oven is a black body. The heat transfer (k.dT/dx) would be e.Sig.(T^4-Ta^4) where temperatures are in Kelvin and e is the missivity, sig is the stefan bolzmann number, Ta is the ambient oven temperature(K). My guess though is that for such a thin sheet the core temperature would be pretty close to the skin temperature.

 

[prex]

This is a unidimensional problem of transient conduction in a slab. As you probably already know, there is a closed form solution by series summation for the case of convective heat transfer at the faces. My McAdams - Heat Transmission has charts helping in assess this problem, as the convergence of the series is slow (but that book dates from the dawn of the computer era...).
I'm afraid no closed form solution exists for your case, where the faces are much closer to a constant heat flux condition. Only a numerical solution seems too be possible: as it is a unidimensional problem it wouldn't be too difficult to write a finite difference solution in a spreadsheet.
However, if you want just an estimate, the convective solution should give you a reliable one: you could guess an equivalent heat transfer coefficient from the known surface temperature at a given time.

prex

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[WoodFE]

Thanks for your replies, I've also obtained some more information on the oven and read up on radiative heat transfer (Thorne - Technology of Thermoforming). Hopefully I am now sufficiently armed to solve the problem.

 

[IRstuff]

Seems to me that you could adapt the Biot Number to determine whether lump-model works. You can modify the Stefan-Boltzman equation into looking like the convective transfer equation and basically lump the sigma*area*episilon*(T1^3+T2^3) as a transient heat transfer coefficient.

TTFN

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[zekeman]

"Seems to me that you could adapt the Biot Number to determine whether lump-model works. You can modify the Stefan-Boltzman equation into looking like the convective transfer equation and basically lump the sigma*area*episilon*(T1^3+T2^3) as a transient heat transfer coefficient."

The problem as stated is not solvable, since the dominant heat transfer mechanism is radiation. So, we need to know the emissivity, the source temperature (at the heating elements) and the view factor.

Moreover,you cannot get a Biot number without this, since the equivalent h is

h=sigma*e*F*(Ts^4-T1^4)/(Ts4-T1)

Ts source temperature
T1 surface temperature at piece
sigma Stefan Boltzmann constant
e emissivity


Once this data is obtained, only then you can get and use the Biot number to determine if you have a "thin" slab which means the center temperature is almost equal to the surface temperature; otherwise there are a few nomographs available that can give you a good solution. I use Schneider's curves for this.
needs this I would try to upload the curves.

 

[IRstuff]

"The problem as stated is not solvable"

Not solvable as a closed-form, presumably, but there are lots of tools that could ostensibly iteratively solve the problem.

My bad on the form, though. I plead guilty to bad algebra, the residual should have been
(T_s+T₂)*(T_s²+T₂²)

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[zekeman]

You have the form right but you missed the view factor, F and what about Ts?
I only say that you need e,F and Ts to get the answer.

Do we agree?

 

[IRstuff]

It's in a radiant oven, so the view is pretty close to 2[π], but it's secondary interest.

T_s is the temperature to solve for, assuming that you can fill in the backstory of the thermal conductance, thermal capacity, and emissivity.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[zekeman]

What backstory?

You cannot do the problem without

K,rho,c and source temperature of the oven,Ts, period.

BTW, I have used Ts as the source temperature and T1 the surface temperature of the specimen.

I'm now wondering if your Ts is not the same.

 

[zekeman]

Correction, if you know the geometry and thermal properties of the specimen and e and F and the specimen surface temperature at a specific time ,t, then you can get Ts and then the temperature at the center.

 

[zekeman]

I think there is a better approach to this using the solution of Carslaw and Jaeger, "Conduction of Heat in Solids"

Assuming the source temperature is much greater than the surface temperature, (170C ) at the time of measurement, the problem is essentially constant flux into the surface at x=L and the center of the specimen by symmetry is treated as insulated.

The C& J solution is

T=FoL/k[kappa*t/L^2+.5(x/L)^2-1/6- summation of transient terms]
Fo = flux
For the problem we have
L=6.5mm=.26 "=.02 ft
k=.1BTU/HR-FT-F
Rho*c=70*.35=25
kappa=k/(rho*c)=.1/25=.004 ft^2/hr
For this problem, t=180s=1/20hr
and
kappa t/L^2=.004*1/20/.02^2=0.5
Fortuitously, for the time, 180s, the transient terms die out and we are left with a temperature distribution that is linearly rising throughout the slab with a parabolic profile

From the solution above , ignoring the transients there remains
T(0)=Fo*L/k*[kappa*t/L^2-1/6]
T(L)=Fo*L/k*[Kappa*t/L^2+1/3]
Substituting I get
T(0)/T(L)=> 0.5-.16/.8+.33=.34/.833=.408
So if T(L)=170-25=135 C, then
T(0)=.408*135=55 C
Note that Fo, the flux is not needed explicitly to get this result

For other times (kappa*t/L^2<.2) where the transient terms are significant you need to use the C&J curves to pick off T(L) and T(0) or T(x).