3 Phase welder

[socket2me]

Currently looking at a job to install power for welders. I asked for the size of the welders and customer has replied with MIG 32A 3Ph = 14KVA.
So i can work out total loadings for the installation have a i done the following sum correctly?

14kva X PF of 0.8 = 11.2kw

A= 11.2 x 1000/415 x 1.732 x PFat 0.8

I get 37.39A so 12.46 A per phase?

 

[telectrix]

you're out, IMO. i get 24.5A (19.5 @ unity pf)

 

[socket2me]

Can you show how you get there please?

 

[davesparks]

You've multiplied by the PF twice in the OP

 

[telectrix]

A = (kVA x 1000)/(1.732 x V x pf)

 

[telectrix]

14000/ (1.732 x 415 x 0.8)

14000/575

24.348A

 

[GEOFF S]

Just divide 14000/692

Cheers

 

[telectrix]

Just divide 14000/692

Cheers

and what about the PF?

 

[GEOFF S]

and what about the PF?

Not sure i follow, what about the P.F

Cheers

 

[telectrix]

OP was using a pf of 0.8. not sure if that's on the equipment spec. though.

 

[GEOFF S]

OP was using a pf of 0.8. not sure if that's on the equipment spec. though.

Okay, what is it that im doing wrong?

Cheers

 

[Darkwood]

415 x 1.732 old money converts the result to 1 phase

You got the wrong result to start with ... (see post 6 for the correct method)... disregarding thewrong result you then divided it by 3 to try get a per phase result... so you attempted to get from 415 to 240 twice as well as doing PF twice.

 

[GEOFF S]

415 x 1.732 old money converts the result to 1 phase

You got the wrong result to start with ... (see post 6 for the correct method)... disregarding thewrong result you then divided it by 3 to try get a per phase result... so you attempted to get from 415 to 240 twice as well as doing PF twice.

KVA is the total current drawn, so 14000/692 is the correct answer.

Cheers

 

[Knobhead]

It helps if you apply PF correctly.

 

[telectrix]

It helps if you apply PF correctly.

i also used your excellent spreadsheet but used 415V.

 

[GEOFF S]

It helps if you apply PF correctly.

It does, and when you have a KVA value its been adjusted. PF is just the ratio of KW/KVA.

Now S= KVA, P=Watts and Q=Var

Now KVA is the product of true and reactive power, KVA is apparent.

So we Know KVA(S) = 14, so to find our KW(P) its 14 X 0.8 = 11.2 KW.

To find KVar we need a bit of trig and we find Kvar = 8.4.

So we have all three sides to our power tringle.

S= 14kVA
P=11.2kW
Q=8.4kVAR

So the total current drawn by the welder is S, so 14000/692 is the correct answer.

Cheers

 

[Knobhead]

If you go back to the original question “what is the current loading?”

Your answer gives the current for unity PF which isn’t what was asked for. Any protection will have to account for the increase caused by 0.8PF. IE kW x the inverse PF

This develops in to an interesting situation on large motors where local PFC is installed. The O/L settings have to be lower than the motor manufacturers rating plate. RoB2 has written an excellent article about this elsewhere.

 

[Knobhead]

you're out, IMO. i get 24.5A (19.5 @ unity pf)

Can you show how you get there please?

He cheated ;-)


BTW Tel, I've chopped the spreadsheet about a bit more I'll send it when I get back home later.

 

[Bobster]

If you go back to the original question “what is the current loading?”

Your answer gives the current for unity PF which isn’t what was asked for. Any protection will have to account for the increase caused by 0.8PF. IE kW x the inverse PF

This develops in to an interesting situation on large motors where local PFC is installed. The O/L settings have to be lower than the motor manufacturers rating plate. RoB2 has written an excellent article about this elsewhere.

Not what I would call it, nothing but a pain in the ---!

 

[Knobhead]

You got the thanks for making me laugh out loud.