Bạn đang xem: Joule-Lenz law-QuickField FEA Software
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:
All dimensions are in millimeters
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.
Calculate the current and Joule heat inside the conductor and compare with the value given by the Joule-Lenz law.
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
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.
Time harmonics (peak current value is presented, RMS value is √2 times smaller, 35.355/1.4142 = 25 A):
|0.01 s||0 A||0 W|
|0.0125 s||25 A||2.5 W|
|0.015 s||35.355 A||5 W|
|Heat power [W]|
|Transient electric (time-average)||2.5|
Watch on YouTube
View simulation report in PDF
Download simulation files
(files may be viewed using any
QuickField Student Edition
Tera Analysis info
Chuyên mục: Kiến thức