Temperature rise of cold side vs. time 1

[mwemag]

I need to calculate the temperature rise of the cold side in a microheat barrier (vacuum insulating) system. The constant temperature of the heat source (hot side) would be 760 deg.C, the initial cold side temperature is RT 25 deg.C and the overal thermal conductivity of the thermal barrier is as low as 0.0004 W/mK (inclusive Radiation shielding) with a vacuum-gap of 5 microns.

Since the temperature of the cold side must not exceed 400 deg.C I need to find out how much time it takes for the cold side to reach this temperature. Are there appropriate methods to calculate the temperature rise versus time?

 

[25362]

One formula, taken from the Plant Notebook section titled PREDICT HEATING AND COOLING TIMES ACCURATELY by Tosun and Aksahin, in the ChE issue of november 1993, p. 183, is:

t = (MC_p/UA) ln [(T_s-T₁)/(T_s-T₂)]​

where
t =time
T_s = 760^oC (constant)
T₁ = 25^oC (initial)
T₂ = 400^oC (final)
M = mass of fluid on the cold side
C_p = heat capacity of the cold side fluid
U = overall heat transfer coefficient
A = heat transfer area

I hope the above helps.

 

[finder1]

25362,

Looks like a very useful formula. A couple of questions:
1. Does M = mass or mass flow rate?
2. Does U = convective heat transfer coefficient?

Thank you!

 

[corus]

The formula is totally inappropriate as it refers to a fluid moving across a surface and not what is required here.

You need to solve the non linear problem of radiation to an ambient/heat source. Taking a single/constant heat transfer coefficient tto represent the radiation term is wrong. If you don't have access to a numerical solver, such as finite elements, then you can approximate the heat transfer coefficient by a single value, then solve the partial differential equation (PDE), then recalculate the heat transfer coefficient again using the calculated temperature. Continue to iterate until the terms balance. To solve the PDE you can either calculate the temperature distribution through the thickness or assume that the body is thin. For a thin cyclinder the temperature is T = Ta +exp(-h.t/p.Cp.a) where a is the thickness, h is htc, Ta is ambient or heat source in this case.

corus

 

[25362]

The quoted formula is used, for example, for steam heating (ie, constant temperature) of cold fluids. However, there is no need for these fluids to flow.
M = mass of the cold side fluid.
U is the overall heat transfer coefficient, composed by radiation, conduction and convection effects.
The colder side doesn't move, except for convection currents induced by heat, if the cold side is a fluid.

Corus is right in that while the convection coefficients are frequently not strongly dependent on temperature, radiation coefficients are indeed a very strong function of temperatures and surface emissivities, [ε].

h_r = [σ](T₁²+T₂²)(T₁+T₂) [÷] [1/[ε]₁ + (A₁/A₂)(1/[ε]₂-1)]​

I recommend to read -for example- the pertinent chapters in J.P. Holman's Heat Transfer (McGraw-Hill), on radiation and, in particular, the section concerning radiation shields.

 

[mwemag]

Thank you all for your help.
25362, I calculated the formula with the units used for an example described in the mentioned article.
M = Lbs = 2.23e-8
Cp = BTU/LB-degF = 0.215
U = BTU/hr-ft-degF = 2.31e-4
A = ft2 = 2.69e-6

Unfortunately, using these values with the formula results in a heat up time of only 5.5 seconds, which indeed would be disappointing. I doubt this result since the very low overal heat transfer coefficient and the geometry of the model matches that of microinsulation heat barriers. They have vacuum gaps of only 1-2 microns but are capable of maintaining large temperature differences provided that shielding of heat radiation is included. I can hardly imagine the convenience of such a device with an insulating capacity of only a few seconds.

http://www.osti.gov/bridge/product.biblio.jsp?osti_id=832892

Maybe I have used the wrong units. However, I can be mistaken and I am thankful for any further tips.

 

[25362]

BTW, U is in Btu/(h*ft²*^oF)

I'm afraid there is a mistake in your calculation. See, please:

t = [(2.33*e^-8)(0.215)[÷](2.31*e^-4)(2.69*e^-6)]*ln [(760-25)/(760-400)]=

[2.33*0.215*e²/(2.31x2.69)]*0.713 = 0.596*0.713 h = 25.5 min

This is assuming the basis e = 2.718. If the intention is e =10, the result becomes 5.75 hours.


Kindly correct any mistake of mine.

 

[sailoday28]

The formula provided by 25362 (Chemical) on 2 July is a good "approximation based on a "lumped mass system" which has a high thermal conductivity for the mass and low convection/radiation heat transfer coefficient--Both of which are constant. The high thermal conductivity basically relates to zero temperature gradient in the mass.
A good example is the heatup of a soldering iron with natural covection air. The temp of the metal is the same on its inside and outside surface because of the high thermal conductivity.

mwemag (Materials)- Can you justify the approximation of the metal temperature being the same on all its surfaces? If so use the formula of 25362 (Chemical) with the appropriate approximations for specific heat mass, heat trans coef.

Regards

 

[mwemag]

e = 10
First I have mistaken t = seconds instead of hours. The corrected result is then 5.5 hours (M = 2.23e-8 Lbs, not 2.33e-8).

Thanks a lot, 25362! Expecting that the formula is appropriate to estimate approximately the heat up time in my model, the resulting time is very satisfying even when introducing an uncertainty factor of 10-20 percent. I still have to substitute Btu/(h*ft*oF) with Btu/(h*ft2*oF) for U, hoping this will not dramatically reduce the heat up time. I assume the latter is the imperial equivalent to W/m2*k.

sailoday28, the cold side material in the model is mainly composed of several thin layers of pure aluminium (200 nm thick) separated by 1.5 micron thick multilayers of several dielectric materials. In the formula I used M as the mass of the overall dielectric/Al composition, not just Al, but the purpose of the heat barrier is to prevent the melting of the aluminium layers by keeping T well below the MP of 660oC, and therefore, for ease of calculation, I just used the Cp value of aluminium.

Since the Al flats have a simple square geometry of 0.5*0.5 mm I guess the metal temperature would be evenly distributed in the direction parallel to the surface and because of the layer's thinness also in the direction of the thickness in each Al layer. Sure there can be a temperature gradient perpendicular across the overall multilayer stack, and the temperature of the Al layer closest to the heat source would rise fastest, but except for not exceeding a T of around 400oC the individual layers can have different temperatures.

Are you proposing to extend the formula by introducing "specific heat mass" as a new factor or what does it stand for?

Thank you
Regards

 

[sailoday28]

ag (Materials)From your description-there will be a temp gradient, normal to the heat flux. If the dielectric has a low thermal conductivity compared to the metal, then you should treat each metal piece as a lumped mass. Otherwise your solution as described does not have an analytic basis.

Regards

 

[mwemag]

sailoday28, the dielectrics indeed have a low thermal conductivity and treating each Al layer as a single mass in the formula will result in a short up time of only 4 minutes. The cold side substrate on the other hand has a high k equal to the Al layers.

May I ask you: Could the problem be solved by introducing a high k connecting layer on both length sides of the multilayer stack, connecting all the Al layers with each other and additionally with the substrate? The path for the heat to travel from the top layer to the substrate is 1.5 microns, which is the height of the stack. I hope that in this case the masses of the Al layers and the substrate would sum up and could be treatet in the formula as one big mass, theoretically resulting in a heat up time of over 18 hours.

Is this possible?
Regards