American Welding Society Forum

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ASME gives not very good solution to calculate the Arc Energy per Pass and Area.

Def

Run is the Lenght when 1 Arc is on. The Run end when the Arc is off.

Pass is the sum of the run for the Circonference

Layer is the sum of the Pass for the relevant hight from the ID.

Area is the Sum of the Pass (Passes of Root or Support or Fill or Cap)

Example for Filling Passes of the Tube the PQR mentioned:

2 Passes is needed to be the AREA "Filling"

Pass 1 - Filling

Run 1 : 1.5kJ/mm for 10mm of Run

Run 2 : 1.3kJ/mm for 10mm of Run

Run 3 : 1.5kJ/mm for 20mm of Run

Run 4 : 1.4kJ/mm for 10mm of Run

Run 5 : 2.0kJ/mm for 2mm of Run

To calculate the Arc Energy per Pass how do we proceed?

Answer 1:

(1.5 + 1.3 +1.5 +1.4 + 2) / 5 = 1.48kJ/mm

Answer 2:

(1.5 x 10 + 1.3 x 10 + 1.5 x 20 + 1.4 x 10 + 2 x 2) / (10+10+20+10+2) =1.38kJ/mm

For me the answer is 1.38kJ/mm as this "average method" does not take into account the "non standard point" example is the 2.0kJ/mm for 2mm of lenght is proportionaly taken into account with 2mm lenght only.

Now the

Pass 2 - Filling

Run 1 : 2.5kJ/mm for 5mm of Run

Run 2 : 2.3kJ/mm for 5mm of Run

Run 3 : 2.5kJ/mm for 20mm of Run

Run 4 : 2.4kJ/mm for 10mm of Run

Run 5 : 3.0kJ/mm for 20mm of Run

Answer 1-A:

(2.5 + 2.3 +2.5 +2.4 + 3) / 5 = 2.54kJ/mm

Answer 1-B:

(2.5x5 + 2.3x5 + 2.5x20 + 2.4x10 + 3x20) / (5+5+20+10+20) =2.63kJ/mm

Now if Answer 1 is taken therefore:

(1.48 + 2.54 ) / 2 = 2.01kJ/mm for the Filling Area

Now if Answer 2 is taken therefore:

Two Cases possible: (depending how we calculate the Average)

2-A

(1.38 + 2.63 ) / 2 = 2.00kJ/mm for the Filling Area

2-B

(1.5 x 10 + 1.3 x 10 + 1.5 x 20 + 1.4 x 10 + 2 x 2 + 2.5x5 + 2.3x5 + 2.5x20 + 2.4x10 + 3x20)) / (10+10+20+10+2 + 5+5+20+10+20) = 2.11kJ/mm

The 2-B solution takes the perfect average of the Run to create the Pass and the Area. Actually I calculate like 2-B that but I dont see too much people calculate like that but more with the solution 1-A

Now another question: (supposition 1-A is taken into acount to be more easy)

When Semi Essential Variable Applies, the Arc energy Max is to be used for the WPS. What are we talking about:

Answer:

3-A

Max Run energy of the Area therfore here 3.0kJ/mm is the max of the WPS.

3-B

Max Pass Energy of the Area therefore 2.54kJ/mm is the max of the WPS.

3-C

Recorded Pass Energy of the Area therefore 2.01kJ/mm is the max of the WPS.

I ask this question because some "Run" are not relevant for the Welding procedure. You ask 1.5kJ/mm and there is one run that is 3.0kJ.

Same for the Pass you request 1.5kJ but 1 pass is 3.0kJ.

What is the pefect interpretation of the Max Arc Energy to use for our WPS and be 100% compliant with the ASME IX Code.

ASME gives not very good solution to calculate the Arc Energy per Pass and Area.

Def

Run is the Lenght when 1 Arc is on. The Run end when the Arc is off.

Pass is the sum of the run for the Circonference

Layer is the sum of the Pass for the relevant hight from the ID.

Area is the Sum of the Pass (Passes of Root or Support or Fill or Cap)

Example for Filling Passes of the Tube the PQR mentioned:

2 Passes is needed to be the AREA "Filling"

Pass 1 - Filling

Run 1 : 1.5kJ/mm for 10mm of Run

Run 2 : 1.3kJ/mm for 10mm of Run

Run 3 : 1.5kJ/mm for 20mm of Run

Run 4 : 1.4kJ/mm for 10mm of Run

Run 5 : 2.0kJ/mm for 2mm of Run

To calculate the Arc Energy per Pass how do we proceed?

Answer 1:

(1.5 + 1.3 +1.5 +1.4 + 2) / 5 = 1.48kJ/mm

Answer 2:

(1.5 x 10 + 1.3 x 10 + 1.5 x 20 + 1.4 x 10 + 2 x 2) / (10+10+20+10+2) =1.38kJ/mm

For me the answer is 1.38kJ/mm as this "average method" does not take into account the "non standard point" example is the 2.0kJ/mm for 2mm of lenght is proportionaly taken into account with 2mm lenght only.

Now the

Pass 2 - Filling

Run 1 : 2.5kJ/mm for 5mm of Run

Run 2 : 2.3kJ/mm for 5mm of Run

Run 3 : 2.5kJ/mm for 20mm of Run

Run 4 : 2.4kJ/mm for 10mm of Run

Run 5 : 3.0kJ/mm for 20mm of Run

Answer 1-A:

(2.5 + 2.3 +2.5 +2.4 + 3) / 5 = 2.54kJ/mm

Answer 1-B:

(2.5x5 + 2.3x5 + 2.5x20 + 2.4x10 + 3x20) / (5+5+20+10+20) =2.63kJ/mm

Now if Answer 1 is taken therefore:

(1.48 + 2.54 ) / 2 = 2.01kJ/mm for the Filling Area

Now if Answer 2 is taken therefore:

Two Cases possible: (depending how we calculate the Average)

2-A

(1.38 + 2.63 ) / 2 = 2.00kJ/mm for the Filling Area

2-B

(1.5 x 10 + 1.3 x 10 + 1.5 x 20 + 1.4 x 10 + 2 x 2 + 2.5x5 + 2.3x5 + 2.5x20 + 2.4x10 + 3x20)) / (10+10+20+10+2 + 5+5+20+10+20) = 2.11kJ/mm

The 2-B solution takes the perfect average of the Run to create the Pass and the Area. Actually I calculate like 2-B that but I dont see too much people calculate like that but more with the solution 1-A

Now another question: (supposition 1-A is taken into acount to be more easy)

When Semi Essential Variable Applies, the Arc energy Max is to be used for the WPS. What are we talking about:

Answer:

3-A

Max Run energy of the Area therfore here 3.0kJ/mm is the max of the WPS.

3-B

Max Pass Energy of the Area therefore 2.54kJ/mm is the max of the WPS.

3-C

Recorded Pass Energy of the Area therefore 2.01kJ/mm is the max of the WPS.

I ask this question because some "Run" are not relevant for the Welding procedure. You ask 1.5kJ/mm and there is one run that is 3.0kJ.

Same for the Pass you request 1.5kJ but 1 pass is 3.0kJ.

What is the pefect interpretation of the Max Arc Energy to use for our WPS and be 100% compliant with the ASME IX Code.

If you want to be 100% compliant with the ASME IX code forget everything about Arc Energy and all the formulas you have noted.

ASME IX is only interested in " Heat Input" !

There are huge differences between DIN/EN/ISO 15614-1 (as example) and ASME IX so if you want to work with ASME the only way is to forget what you are used to and learn something new.

ASME IX is only interested in " Heat Input" !

There are huge differences between DIN/EN/ISO 15614-1 (as example) and ASME IX so if you want to work with ASME the only way is to forget what you are used to and learn something new.

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