A Forum Newbie needing help! How do you calculate electrode consumpsion?

Ive been driving myself nuts trying to calculate how many electrodes I need to build up some forging tools that we are hoping to reclaim.
The rectangular area covered is 70mm x 470mm x 1250mm.
The material we are using is an INCO 718 which has a density of approx. 8890.We shall use manual metal arc welding method, using 5mm x 350mm electrodes.
The electrodes are purchased in 6Kg packets which contain 40 electrodes per packet. It is estimated that two thirds of the electrode weight is flux.

How many packets of electrodes do I need to purchase to cover the slab area?

Jane will be left with 4 Apples.

Sorry I can't help you, with your question, but there are some knowledgeable guys on here, so I'm sure you will get an answer or pointed in the right direction.


Welcome to the site

X x Y x Z is a volume not an area!!!
are you trying to lay down weld 70mm thick over an area 470 x 1250?
that sounds a little crazy!

you can work out the volume of steel is the rod 5mm dia rod is 19mm2 X length. 300 useable length will give 5700 mm3 = 5.7cc

once you know the area and thickness it's plain sailing

each rod will lay down approx 5700mm3 with just the filler wire but some of the electrode coating will add to the filler wire-assume a low figure of 120% so each rod could give 7000mm3 of material (sounds quite alot!) if you only use 300mm of each rod
you need to fill an area of 41125000mm3
that is a total of 5875 welding rods, or only 4570 rods if you could get 450mm long
thats how i would work it out
146 packets needed (more than a ton in total.
before you ask you will need more than a little chinese inverter box running off of 13amp socket for this onea big oilcooled 3phase unit would be better
let us all know what you decide to do and post a pic of what you need to do please.
cheers
neil

Your calcs are way off. 5700mm3 is equal to 347.8353 cubic inch

1000 mm^3 = 1 cm^3 = 1ml and these are VOLUMES not areas.

one litre is 100mm x 100mm x 100mm = 1000000mm^3 = 1000cc or ml


5cc is 5ml = 1 teaspoon therefore the volume of a rod is just over a teaspoonful!!! or .348 cu"

Shox Dr said:

Your calcs are way off. 5700mm3 is equal to 347.8353 cubic inch

Click to expand...

i hate to argue but you are the one who is way off-5700mm3 is equal to .347cubic inches

fair doos

rikrobson said:

you can work out the volume of steel is the rod 5mm dia rod is 19mm2 X length. 300 useable length will give 5700 mm3 = 5.7cc

once you know the area and thickness it's plain sailing

Click to expand...

Happy with that, plus the recovery rate which could add over 60% to the deposited weld metal (EN499)
http://www.mig-welding.co.uk/electrode-classification.htm


I use 5mmx350mm (300mm useable length) iron powder rods with a recovery rate of >160% to put down fillet welds with a 6mm leg length (a triangle). To work out number of electrodes/metre I'd use...

For the electrode;

2.5 x 2.5 x 3.142 = 19.6375mm2 (area of electrode)

19.6375 x 300 = 5891.25mm3 (volume of electrode)

Plus the recovery rate

5891.25 x 160% = 9426mm3 (volume of weld metal deposited by each iron powder rod)



For the fillet weld;

6 x 3 = 18mm2 (area of fillet = base x half the height)

18 x 1000 = 18000mm3 (volume of weld metal required for one metre)


Number of electrodes/metre;

18000 ÷ 9426 = 1.9096117 (electrodes)


So I'd price the job allowing for 2 electrodes per metre of weld (I reckon )