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Joule-Lenz law

Joule losses calculation, ohmic heating, resistive heating, electrical conductor heating

This verification example compares the current and the Joule heat generated in conductor calculated by the Ohm's law and Joule's law, and calculated in QuickField using three formulations: DC conduction, AC conduction and Transient electric.

Geometry:
All dimensions are in millimeters

Ohm law verification with FEM

Given:
V = 0.1 V - voltage applied;
f = 50 Hz - frequency in time harmonics and transient magnetics problems;
g = 1 MS/m - conductivity of conductor material.
L = 0.4 m - conductor length.
A = 0.02*0.005 m² - conductor cross-section area.

Task:
Calculate the current and Joule heat inside the conductor and compare with the value given by the Joule-Lenz law.

Solution:
To maintain the same value of Joule heat across all formulations the voltage value is adjusted:
In the time harmonics problem peak amplitude value of voltage is set √2·V.
In the transient electric voltage is set via formula V(t) = √2·V · sin(2·180·50·t).

According to the Joule-Lenz law* the power of heating generated by an electrical current I:
W = R*I², where the conductor resistance is R = (1/g) * (L/A).
The electric current value could be calculated I = V / R [A],

*Wikipedia: Joule heating.

Results:
Conductor resistance R = (1/1e6) * (0.4/0.02*0.005) = 0.004 Ohm
Current: I = 0.1 / 0.004 = 25 A
Joule heat Q = 0.004 * 25 * 25 = 2.5 W.

DC conduction:

Joule-Lenz law direct current

Time harmonics (peak current value is presented, RMS value is √2 times smaller, 35.355/1.4142 = 25 A):

Joule-Lenz law alternating current

Transient electric:

Time Current Joule heat
0.01 s 0 A 0 W
0.0125 s 25 A 2.5 W
0.015 s 35.355 A 5 W

Joule-Lenz law transient magnetic analysis

 

Heat power [W]

Joule-Lenz law

2.5

DC conduction

2.5

Time harmonics

2.5

Transient electric (time-average)

2.5

  • Video: Joule-Lenz law
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  • Download simulation files (files may be viewed using any QuickField Edition).
  • PCB design with QuickField

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