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| Electric Heating Elements |
This equation calculates the amount of wattage (W) needed to raise the temperature of a material a specific amount in °F (ΔF) in a given number of hours (T), you first need to know the mass (m) of the material being heated and its specific heat value (c):
m × c ×ΔF
W= -------------------
3.412 × T
The mass and specific heat of some materials may be found at www.hotwatt.com/table1.htm for metallic solids, www.hotwatt.com/table2.htm for solids other than metals, and www.hotwatt.com/table3.htm for certain liquids and gases.
The next equation calculates the amount of wattage (W) needed to maintain the temperature of a material based upon its known wattage loss per square foot (WL/SF) multiplied by the area (A) in square feet:
W = WL/SF × A
The wattage loss per square foot is found for specific materials in tables located at www.hotwatt.com/heatloss.htm. A sample table is shown (Losses from Uninsulated Metal Surfaces). Note that some materials have three different cylindrical objects as well as the top surface losses from a horizontal surface laid flat. Curve B shows the average heat loss from the top and bottom of a horizontal surface laid flat while Curve C shows the heat loss from the bottom only of a horizontal surface laid flat.The final equation determines the additional wattage necessary for melting or vaporizing a material. Whenever a material changes state, whether from a solid to liquid, or liquid to gas, it requires an additional influx of energy to initiate the change.
The heat needed to melt a solid material is known as the latent heat of fusion (Hf) while the latent heat of vaporization (Hv) determines the energy needed to change a substance from a liquid to a vapor. The equation for both is identical, with the value of Hf or Hv substituted as necessary for the value H:
m × H
W = --------------
3.412 × T
Again, m is the weight of the material in lb, while T is the heat-up time in hours. The values for the latent heat of fusion or vaporization can be found in the same look-up tables used in Equation A.
Watt Density
Watt density is the rated wattage of the heater divided by the overall area being heated. The watt density permitted for any given application depends on how well the material being heated distributes its heat throughout its volume.
Water, light oils, and metals typically have high heat distribution rates that permit the use of high watt densities. Heavy oils, syrups, hydraulic fluid, and other materials with low heat distribution need lower watt densities to prevent spot overheating. This can lead to damage of the heating element, the well, and even the heated material.
As an example, a 10 in. immersion heater is rated at 500 W. There is a 0.5 in. cold area at each end, making its total heating length (L) 9 in. The diameter (D) of the heating area is 0.75 in. To calculate the watt density of this heater, first determine the total surface area (AS) being heated:
AS = π D × L
3.1416 × 0.75 × 9 = 21.1 in2
Then divide the wattage of the heater by the surface area:
500W ÷ 21.1 in = 23.7W /in2
Note: All information, data and dimension tables in this article have been carefully compiled and thoroughly checked. However, no responsibility for possible errors or omissions can be assumed.
It is the express responsibility of the customer to determine the suitability of the product for the intended application and Hotwatt Inc. nor Belilove Company makes no claims or provides no guarantee in this respect, either written or applied.