Thermal (steady)

The Thermal physics computes the steady-state temperature from the losses of a magnetic physics (AC or magnetostatic). Air is not simulated as a flow (no CFD): it stays out of the thermal domain, and exposed surfaces exchange heat by convection. A fan enters as an air channel that renews the air and warms up with the heat it carries away.

How to use

Solve (or set up) an AC or magnetostatic magnetic physics with the working currents.
From the + of Solver, create Thermal (steady) and choose the magnetic physics under Losses from.
Set the ambient temperature, the surface convection and, in planar problems, the front/back faces.
Solve (▶). Under Results › Thermal, create a Field map: temperature surface and isotherms (contour).

Materials need the thermal conductivity k; library materials already have it (copper 400, aluminum 237, M400 steel 28, 1010 steel 50 W/m·K). Without k, a typical value of the group applies. Laminated sheets conduct \(f\cdot k\) in the drawing plane.

Cooling

Where	What	Typical values
Exposed surfaces	convection \(h\,(T - T_{amb})\) on every solid edge touching air	natural 5–10, forced 25–100 W/m²·K
Front/back faces (planar)	2D does not see the depth faces: \(2h/\text{depth}\) per volume	same as natural convection
Condition on curves	convection with another \(h\), fixed temperature or insulated, on chosen curves only	—
Air channel (fan)	flow \(Q\) (m³/h) and inlet temperature; linked curves exchange heat with the channel air	—

To apply a condition to curves: select the curves in the drawing and click Use the selected curves on the condition.

Fan (air channel)

The renewed air carries away the heat of the curves linked to the channel. With flow \(Q\), \(\dot m = \rho_{air} Q\):

\[ T_{out} = T_{in} + \frac{P}{\dot m\, c_p}, \qquad T_{air} = \frac{T_{in} + T_{out}}{2}, \]

and the convection of those curves uses \(T_{air}\). Temperature and air are solved together by iteration: low flow → the air warms up → less cooling.

Resistivity with temperature

With Resistivity with temperature on, the conductor resistivity follows \(\rho(T) = \rho_{20}\,[1 + \alpha\,(T - 20)]\) (\(\alpha\) = 0.00393/K for copper). Coil losses are recomputed with the temperature of each region, and solid conductors (eddy currents) get the AC solved again with the corrected \(\sigma\). The iteration stops when the region temperatures change by less than 0.1 K (usually 3 to 5 rounds).

Results

Variable	What
`Tmax`, `Tmin`	maximum and minimum temperature (°C)
`<region>_Tavg`, `<region>_Tmax`	mean and maximum per region
`P_perdas`, `P_dissipado`	total losses and heat leaving (they must match: balance)
`<channel>_Tar`, `<channel>_Tsaida`, `<channel>_P`	mean air, outlet and heat carried per channel

Limitations

Steady state. Adiabatic heating in a short circuit (no time for heat to leave) and the thermal transient are the next steps.
Air is not simulated: convection comes from the given \(h\). In long channels the air warms along the path; the model uses the mean of inlet and outlet.