Max thermal conductivity required for heating elements

AI Thread Summary
The discussion focuses on designing an oven with effective insulation to minimize heat loss, using thermal conductivity calculations to determine the maximum allowable value of k. The user is calculating heat generated by heating elements based on watt density and surface area, seeking confirmation on their approach and results. They aim to balance cost and efficiency, noting that more expensive insulation typically has lower thermal conductivity, which can reduce electricity costs over time. The user is reusing existing heating elements and is trying to ensure their calculations align with the specifications to avoid overspending on materials. Overall, the conversation emphasizes the importance of accurate thermal management in oven design.
Ferbs207
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I'm designing an oven and want to ensure that the insulation I specify has a low enough Thermal Conductivity (k) to resist excessive heat loss. I determine heat loss (Hout)with the following equation: Hout=A*U*(T1-T0). U is dependent on k (U=k/L). I omitted the heat transfer coefficient in calculating U to keep the calculation more conservative. My values are A=15.33m^2, T1=422K, T0=289K, L=0.041m so Hout(Watts)= 50000 * k.

I am having difficulty determining the heat generated by the heating elements. I know the watt density is 1.24*10^4 W/m^2, and the bar dimensions (L=0.914m, W=0.038m, T=0.010m) with total surface area of 8.78*10^(-2)m^2. Am I okay to multiply watt density by the total element surface area to calculate the heat generated? Along with the previous question, is it acceptable to assume that all faces of the strip heating element have the same watt density? I have (4) of these heating elements, so this calculation yields 4390 W. then kmax=50000/4390 = 0.088 W/(m*K).

I also calculated this with the Stefan Boltzmann Law. The Max Sheath Temperature of the elements is 673K, which after calculating the radiant exitance j* (j*=sigma*T^4), multiplied by element surface area and number of elements to find power yields 4085 W, or a kmax = 0.082 W/(m*K). The similar results give me confidence that I am on the right path, but am hoping someone can confirm this before I spend $2000 in building materials. Any and all help is appreciated!
 
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Sorry but your calculations are chaotic .

Let us start in a different place -

(a) what are you trying to do in principle ?
(b) what are the requirements for heating power and temperature ? Size of heating chamber ?

(c) Do you understand the relationships between input electrical power and output heating power for a heating element ?
(d) Are you using commercial heating elements ? Data sheet ?
 
(a) The goal is to buy the least expensive thermal insulation that will satisfy the design requirements. Practically speaking, the more expensive the insulation, the lower the thermal conductivity (k), so essentially I'm after the highest allowable value of k. The general principle is energy balance, the heat into the system is greater or equal to the heat out at the design temperature. Note k is proportional to temperature.

(b) The requirements are to heat a 12' long x 2-1/2' wide x 3' tall oven to 300F for an extended time (10+ Hours) using existing heating elements, controllers, and blowers (the blowers I've omitted from the analysis for now). The oven wall thickness is 1.625" on the sides and 3.625" on top. The electrical source is 240V.

(c) I have a general understanding. I think you're getting at P=V^2/R. R is an unknown at this point. The heating strips are being reused, I called the company who made them originally and they were able to find the Watt Density of 8W/in^2 and Max Sheath Temperature of 650F +/0 50F. Hence the two attempts I made to calculate the heat generated by the four strip heating elements.

(d) As stated above, the heating elements are being reused and I have not found a dedicated data sheet yet. The strip elements are similar in dimension and specs to the Omega element #CSH00313 found at this link. http://www.omega.com/pptst/CSH5_Series.html. I believe the 1000W is true for my elements as well.

Thanks for taking a look. This is out of my element so to speak.
 
Ferbs207 said:
(a) The goal is to buy the least expensive thermal insulation that will satisfy the design requirements. Practically speaking, the more expensive the insulation, the lower the thermal conductivity (k), so essentially I'm after the highest allowable value of k.
Cheaper insulation will drive up the electricity costs. If the oven is operated frequently, you should not neglect those.

If the given watt density makes sense, then I think your approaches work. It is a bit surprising, however - a lower oven temperature would lead to a higher watt density at the same surface temperature, what would happen then? Does the temperature of the heating elements go down?
 
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