## Incorrect joule heat dissipation in transient thermal-electric analysis

## Incorrect joule heat dissipation in transient thermal-electric analysis

(OP)

I'm trying to model a Resistance Spot Welding process using an implicit and transient thermal-electric problem on ABAQUS and came across this problem. Here's a simple test case in which i replicated the problem.

Consider the following axisymmetric analysis (no heat flows in or out of the system; any heat generated by the electric current stay within the system):

For a current of 1 A the current density is 0.318 A/m^2. For resistivity of 1 ohm-m, the resistance is 0.318 ohm. Consequently the voltage i get is also 0.318 V (IR = 1 A*0.318ohm).

The expected energy dissipated (let's call it 's') is (current_density)^2*resistivity*volume = 0.101*1*3.14 = 0.318 J

For a time-period of 1 s, I used the following fixed time increments to obtain the corresponding results:

Δt = 1 s : s = 0.105 J (one increment)

Δt = 0.5 s : s = 0.211 J (two increments)

Δt = 0.25 s : s = 0.264 J

Δt = 0.1 s : s = 0.296 J

Δt = 0.05 s : s = 0.307 J

Δt = 0.01 s : s = 0.316 J

Δt = 0.05 s - 1 s : s = 0.307 J (for automatic time control with initial time increment of 0.05 s)

Why does Δt = 1 s under predict the energy? Why is it that the first time increment always only output a third of total energy it should?

Any help is appreciated. Thanks.

Consider the following axisymmetric analysis (no heat flows in or out of the system; any heat generated by the electric current stay within the system):

For a current of 1 A the current density is 0.318 A/m^2. For resistivity of 1 ohm-m, the resistance is 0.318 ohm. Consequently the voltage i get is also 0.318 V (IR = 1 A*0.318ohm).

The expected energy dissipated (let's call it 's') is (current_density)^2*resistivity*volume = 0.101*1*3.14 = 0.318 J

For a time-period of 1 s, I used the following fixed time increments to obtain the corresponding results:

Δt = 1 s : s = 0.105 J (one increment)

Δt = 0.5 s : s = 0.211 J (two increments)

Δt = 0.25 s : s = 0.264 J

Δt = 0.1 s : s = 0.296 J

Δt = 0.05 s : s = 0.307 J

Δt = 0.01 s : s = 0.316 J

Δt = 0.05 s - 1 s : s = 0.307 J (for automatic time control with initial time increment of 0.05 s)

Why does Δt = 1 s under predict the energy? Why is it that the first time increment always only output a third of total energy it should?

Any help is appreciated. Thanks.

## RE: Incorrect joule heat dissipation in transient thermal-electric analysis

## RE: Incorrect joule heat dissipation in transient thermal-electric analysis

## RE: Incorrect joule heat dissipation in transient thermal-electric analysis

## RE: Incorrect joule heat dissipation in transient thermal-electric analysis

It computes power by integrating the electric field over the increment. In doing so, it uses the electric field at the end of the increment and the change in electric field over the increment. Consequently, it assumes zero initial voltage and thus E

^{n+1}and ΔE are equal.dP = (E

^{n+1}.σ^{E}.E^{n+1}- E^{n+1}.σ^{E}.ΔE + (1/3)ΔE.σ^{E}.ΔE) dVdP = (1/3)ΔE.σ

^{E}.ΔE dV## RE: Incorrect joule heat dissipation in transient thermal-electric analysis

## RE: Incorrect joule heat dissipation in transient thermal-electric analysis