AWS Handbook Calculation

By - Date 10-23-2000 17:47

Hello! My name is Jeff Carney, I am an Assistant Professor in the Welding Engineering Technology Department at Ferris State University. Currently, with our juniiors, we are discussing how the cross-section of a single weld pass is roughly proportional to the energy input. We are using the AWS Handbook Volume #1, Chapter 2, pages 32-34 as the basis of our discussion.

Our end result is to calculate Aw. Invloved with calculating Aw is Q (a specific theoretical quantity of heat required to melt a given volume of metal).

My question is, "When calculating Q, what does 300 000 in the denominator represent?"

The units associated with Q are Joules/mm3.

Does a table of Q values exist for various materials?

your response to this issue would be appreciated by myself, and the students of the FSU welding programs.

Thank you,

Jeff Carney
Chairman, AWS West Michigan Section

By JohnPhillips Date 11-08-2000 08:47

The velocity of light in a vacuum is 300,000 meters per second.

John Phillips

By G.S.Crisi (****)

Date 11-09-2000 15:09

Hops! The velocity of light in vacuum is 300,000 KILOMETERS per second. Now let me ask an ignorant question: what has the velocity of light to do with the heat input in a weld?
Giovanni S. Crisi
Sao Paulo - Brazil

By JohnPhillips Date 11-10-2000 09:35

Giovanni,

Whoops! You are correct, of course...meant to say KILOmeters...slight difference.

As to what it has to do with heat input in a weld, I suspect this formula is derived from some basic law of thermodynamics that deals with heat transfer, and is related to E=MC^2, Energy (E) equals mass (M) times the speed of light (C), squared.

If this is not complicated enough for you, take a look at a paper entitled "An upwind numerical solution of nonlinear advection-diffusion problems with a moving heat source" at http://link.springer.de/link/service/journals/00231/bibs/8034004/80340287.htm.

The abstract says, "In the present work, two-dimensional temperature variations and a position of a weldpool within a workpiece during keyhole plasma arc welding are determined. The model allows to include temperature dependent thermal properties, variable welding speed, different keyhole radii and a nonlinear boundary condition of the third kind on the upper and the lower surfaces of the workpiece."

As a writer/software developer, taking up welding as a hobby, I ran across this BBS by accident, and I am amazed at the richness of the subject. I have never realized how important welding is to our civilization, or how much science is involved. I'm glad you guys are on top of this...I just hope to learn to run a nice looking bead on two pieces of mild steel with a gas torch, and maybe later learn MIG and TIG!

John Phillips

By - Date 11-11-2000 19:30

John,

In all sincerity, I wish everyone had the aspiration you stated in your last sentence, "... I just hope to learn to run a nice looking bead on two pieces of mild steel with a gas torch..." It won't solve all of our quality related problems in the industry, but I believe it would be one heck of a start in the right direction.

Robert Arranaga

By MBSims

Date 11-15-2000 05:02

Jeff,

You don't seem to be getting a rational answer on your question. It's a darn good question too! I've looked back at my thermodynamics and physical metallurgy texts and don't see an obvious answer. Q seems to be a volume heat rate since the units are in J/mm^3. The +273 converts degrees C to Kelvin. Not sure why they squared it though. My guess is that the equation is based on comparison to experimental data. Maybe they plotted melting temp. vs. volume energy for several alloys and this is the equation for the curve. The AWS handbook does say that this is an approximation.

You should be able to find a table or source for activation energies (Qm)for various metals, but they would be in J/mole. Using an approximate atomic radius, the equation for volume of a sphere and Avogadro's number (6.022 x 10^23 atoms/mole), you could convert this to J/mm^3. But this has no temperature dependance compared to the AWS handbook equation for Q.

I'll do some more digging on this next week after we get through final exams. You might try contacting A.F. Manz through AWS headquarters to see if he can answer the question directly. Tell your students to keep asking those types of questions, we need more folks like them to challenge the status quo!

Marty Sims

By MBSims

Date 11-28-2000 00:44

Without digging into it too deep, it does look like the formula gives a rough approximation of the heat needed to melt the electrode. I calculated the heat to raise 1 gram of carbon steel wire from 25 C to its melting point of about 1515 C and added the heat of fusion and came up with about 8 J/mm^3. Using the AWS handbook formula it came to about 10 J/mm^3, which is reasonably close considering there are other variables and heat losses not accounted for.

To get the straight story on where the formula came from and what significance 300,000 has, you would probably have to talk to A. F. Manz. He just had an article published in the Nov. 2000 issue of Welding Design & Fabrication and it gave his phone number as 908-687-8382 if you want to call him. He was listed as the Welding Handbook Committee Member for that chapter in the AWS Handbook and is knowledgable on welding power sources.

Hope this helps,
Marty Sims