Max thermal conductivity required for heating elements

In summary: 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 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
  • #1
Ferbs207
3
0
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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  • #2
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 ?
 
  • #3
(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.
 
  • #4
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?
 

What is thermal conductivity and why is it important for heating elements?

Thermal conductivity is the measure of a material's ability to conduct heat. It is important for heating elements because it determines how efficiently heat can be transferred from the element to the surrounding environment.

What is the maximum thermal conductivity required for heating elements?

The maximum thermal conductivity required for heating elements will depend on the specific application and requirements. Generally, higher thermal conductivity is preferred for faster and more efficient heat transfer.

What materials have the highest thermal conductivity for heating elements?

The materials with the highest thermal conductivity for heating elements include metals such as copper, aluminum, and silver, as well as ceramic materials such as boron nitride and diamond. These materials have high thermal conductivity due to their strong atomic bonds and ability to quickly transfer heat.

How does thermal conductivity impact the design and performance of heating elements?

The thermal conductivity of a material can greatly impact the design and performance of heating elements. Materials with higher thermal conductivity can transfer heat more efficiently, allowing for smaller and more compact designs. Additionally, materials with high thermal conductivity may be able to reach higher temperatures and provide more uniform heating.

Can the thermal conductivity of heating elements be improved?

Yes, the thermal conductivity of heating elements can be improved through the use of materials with higher thermal conductivity, optimizing the design and geometry of the element, and utilizing efficient heat transfer methods such as fins or coatings. However, it is important to also consider other factors such as cost, durability, and compatibility with the intended application when choosing materials for heating elements.

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