I have an arc between a tungsten electrode and a work piece. The length of the arc remains the same. If I vary the current over a reasonable range (so as not to deform the tungsten nor to reduce the arc to instability) does the electrical resistance of the arc change?
The reason I ask is that I am attempting to quantify the heat input from a pulsed TIG arc as compared to a constant current arc.
TIA,
Ken
p.s. If the resistance does change a bonus point to anyone who can point me to a formula for quantifying the resistance
Results 1 to 9 of 9

04112013, 07:32 AM #1Senior Member
 Join Date
 Aug 2009
 Posts
 142
Electrical resistance of an arc as a function of current  an engineering question...

04112013, 08:37 AM #2
For the calculation of the heat input (Q), the relationship used for constant
current process would be Q=(VxI/s)η, where V is the voltage, I
is the current, s is the welding speed and η is the efficiency of utilization of the heat generated.
The calculation of heat input for the pulsed current
process is done by computing the mean current using the relationship Im=(Ipt+Ibtb)/ tT, where Im is the mean current, Ip is the peak current, Ib is the base current, tp is the time on peak pulse, tb is the time on base current and tT is the total time.
Does this help???JIM
Owner Operator of JNT Mobile Welding & Repair LLC
Millermatic 350P Aluma Pro
Dynasty 200DX
Maxstar 150 STL
Trailblazer 302
Suitecase 12RC
Extreme 12VS
Extreme 8VS
Spoolmatic 30A
Miller HF251D1
Passport Plus
Spoolmate 100
Hypertherm Powermax 45 and 85
Ingersoll Rand Engine Driven Compressor
Dake 75 ton HFrame Press
JD Squared Model 32 Bender
Miller Digital Elite

04112013, 12:13 PM #3Senior Member
 Join Date
 Aug 2009
 Posts
 142
Thanks jpence38,
I had not at this point considered the efficiency of utilization of the heat generated. An opportunity for further refinement. My original question related to what you refer to as the heat generated, I am afraid I have to disagree on the pulse formula you provided. As I see it...
P = I x E (power or in this case heat generated = current times potential or voltage)
E = I x R (where R is resistance) so that
P = I^2 x R (think of "I squared R" losses in a conductor)
My formula for pulsed current would be:
P = Ip^2 x Tp x Rp + Ib^2 x Tb x Rb (where the p parameters are peak and b parameters are base or background)
If I wish to determine the heat dissipated by my torch when in pulse mode and I assume that the resistance of the torch does not change I find:
Iequiv^2 x R = Ip^2 x Tp x R + Ib^2 x Tb x R
and thus I can divide through by R to get:
Iequiv = SQRT(Ip^2 x Tp + Ib^2 x Tb)
Am I in error in my thinking???
If I set my Dynasty 200DX to its default pulse parameters where background amps are 25% of peak amps and peak time is 40 percent I compute that my WP9 torch, rated at 125 amp continuous, can thus carry a peak amperage of about 190.
If the arc resistance stays the same between 190 amps and 48 amps (and the utilization efficiency is also the same) I would conclude that these pulse parameters would be adding heat to the weldment at the same as a 125 amp continuous current.
Perhaps I need to setup some experiments to gather data. The resistance I can determine by measuring the voltage drop across the arc. The efficiency will be a little more challenging. But I have some ideas.
Thanks again,
Ken

04112013, 02:27 PM #4Senior Member
 Join Date
 Sep 2009
 Location
 Colorado
 Posts
 211
Simpler approach
When I am looking at pulse welding, I generally want to use the time weighted average current.
Average current = Peak current * Time at peak + Background current * background time.
Time would be expressed as a fraction.
In the example given, with 190 amps peak, 25% background current, and 40% time high, I compute 48 amps as the background current.
Average current = 190 * 0.4 + 48 * 0.6 = 76 + 28.8= 104 amps average.
To use this formula, one would need to assume that the voltage in the arc does not vary significantly with the current. But I have not verified that this assumption is correct.
One of the problems with Ken's view is that he is looking as the arc as if it was an ideal resistive load. I doubt very much that this is the case.
The experiment is really simple to perform. Set up the torch with some kind of fixture, to insure that the arc length is constant. Vary the current, and plot the voltage. However, the high frequency voltage is likely to confuse or even damage most sophisticated voltmeters. This is a case where a really basic voltmeter might work better.
Richard

04112013, 03:16 PM #5Senior Member
 Join Date
 Aug 2009
 Posts
 142
Thanks Richard. I am not sure of the resistance characteristics of an arc, thus my original question. On the other hand, the heat lost to resistance is definitely a function of the SQUARE of the current. That is why utility power is transmitted at high voltages (low currents) and then transformed into low(er) voltages at the point of use. I maintain that the heat dissipated by a TIG torch is a time weighted average of the square of the peak and background currents.
My experimental rig as I envision it is as follows:
Starting with a piece of aluminum scrap 5" in diameter and about 6" long (I would use a block of copper but I happen to have this) I will drill a hole and insert a piece of 1/8" tungsten a couple or 3 inches into the block with 1/8" or so sticking out. This will be my work piece/heat sink. I will then make a jig to hold the TIG torch about 1/8" away from the tungsten in the work piece with a lever to allow me to bring the electrode in the torch into contact with the one in the work piece. I will then set the power source for panel amp control and lift arc start. That will prevent the HF from zapping the volt meter as you quite rightly pointed out. I should then be able to start and maintain a consistent arc while measuring the voltage at different amperages.
If I decide to get really creative I could thermally insulate the aluminum with some fiberglass or such, drill another hole in it to accommodate a thermister temperature sensor. I could then run the arc at a given current for a period of time, allow the temperature of the aluminum to reach reach uniformity and take a reading. Cool if back down, try another current etc. This would allow me to see how much of the arc heat was deposited in the weldment.
Ken

04112013, 03:45 PM #6Senior Member
 Join Date
 Sep 2009
 Location
 Colorado
 Posts
 211

04112013, 06:12 PM #7
I'm not sure I understand the point of the original question, or what use you intend to put the results to, but...
A DC arc follows Ohm's law: V = R x i
An AC arc can be approximated by using Ohm's law on the RMS (root mean square) values. Use a trueRMS digital multimeter to read them directly.
Power (in Watts) is easy to calculate using W = V x i = R x i^2
A digital meter (even a cheapo) will always be more accurate & have less impact on the circuit than an analog (even an expensive one). As long as the actual working voltage doesn't exceed what the meter is rated for, it won't fry.
But the power you calculate depends on exactly where you take your other measurements. If you measure voltage across the power supply, you'll be calculating power from the PS. For the power (heat) lost by the arc, measure voltage from the tip of the torch to the work surface. For heat due to current flow in the work, measure voltage from the arc to the ground electrode. But there will be heat lost by the arc & absorbed by the work which would be very difficult to calculate (even for NASA).
If you want to know how hot your work piece is getting, use a cheap noncontact (IR) thermometer.Last edited by Steve83; 04122013 at 11:09 AM.
Walk softly & carry a BIG SIX ! ! !
MM211 + SM100

04122013, 07:46 AM #8Junior Member
 Join Date
 Feb 2013
 Posts
 1
The arc has an effective _negative_ resistance; typically, voltage decreases with increasing current (opposite of a true resistor). This is why an arc without a welding machine to regulate it is unstable, it will consume as much current as it can take until the source blows up. Unfortunately the value depends on many variables so it is not really simple to calculate; and yes, it is not a straight line (for example, it shoots very high when voltage gets down near the ionization potential of the gas).
The total power at any instant is still V*I, so if you succeed in measuring both as you're proposing you will indeed find what you want. Interested to hear the answer.

04122013, 08:08 AM #9Senior Member
 Join Date
 Dec 2012
 Location
 Sweetwater, TX
 Posts
 201
I'd be interested in the results you obtain.
Can you please post a video showing your setup and measuring the voltage of the arc?