Thermodynamics of Resistive Heating at Low Power

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SUMMARY

The discussion focuses on the thermodynamics of resistive heating in a thermally sealed chamber, particularly at low power levels. It establishes that heat loss is the primary limiting factor for achieving high temperatures, regardless of the input power. The steady state temperature is determined by the balance of heat input and heat loss, influenced by the thermal resistance of the chamber. The conversation also explores the concept of a 'thermal battery' for heat storage, emphasizing the advantages of using materials like wax for thermal energy storage due to their latent heat properties.

PREREQUISITES
  • Understanding of thermal resistance and thermal equilibrium
  • Familiarity with resistive heating principles and materials, such as Nichrome
  • Knowledge of heat transfer concepts, including conduction and convection
  • Basic principles of energy storage and phase change materials
NEXT STEPS
  • Research thermal resistance calculations for insulated chambers
  • Explore the properties and applications of phase change materials (PCMs) in thermal storage
  • Study the thermal dynamics of resistive heating elements, focusing on materials like Nichrome
  • Investigate the design and efficiency of thermal battery systems for energy storage
USEFUL FOR

Engineers, researchers, and hobbyists interested in thermal management, energy storage solutions, and the practical applications of resistive heating technologies.

  • #31
Bucket o' Magma. I wonder how much a paint-can sized Dewar would cost.

A ~9cm dia opening should radiate 1kw @ 1,000C
 
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  • #32
hmmm27 said:
Bucket o' Magma. I wonder how much a paint-can sized Dewar would cost.

A ~9cm dia opening should radiate 1kw @ 1,000C
Of course there would have to be some sort of safety screen covering the opening just like in a resistive element space heater. Good idea about the Dewar.
 
Last edited:
  • #33
Baluncore said:
To radiate 10 times the heat, will require 10 times the area, or 10 times the temperature difference, or a blend of those two. The product of element area by temperature difference will be proportional to the power.
The power radiated goes like T to the fourth power, T being.the absolute temperature. To me, this implies that you need to multiply the delta T (between the wire and its environment) by only the quartic root of 10 if you want to x10 the power emitted, i.e. 1.78, roughly. What am I missing?
 
  • #34
fluidistic said:
What am I missing?
I don't know.
I think you are assuming black body radiation from the element. But some energy radiated from the element to the sand would be reflected back to the element.
I was assuming the thermal resistance of the layer of sand that surrounds the heating element would limit heat transfer to the sand.
 

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