3 phase current

[LiamPope]

ER OK!

80,000/230 = 347.8 A total from the 3-single phases

/3 = 115.9A per phase


80,000/(400*SQRT3) = 115.4A per phase

There's only a difference at all because he uses 230 and you use 400, which isn't accurate. If the phase-neutral voltage is 230, the line voltage is really 230 * SQRT3 = 398.4. 230/400 is quoted for convenience. Do it with those figures and both answers are exactly the same.

Liam

 

[GaryMo]

ER OK!

I missed his /3 bit

 

[ColJack]

you missed this bit too...

and it's

VA / V / Sqrt 3... not VA / ( V x Sqrt 3 )

because 400 x sqrt 3 is 693V

actually, that works out the same....

 

[GaryMo]

you missed this bit too...

and it's

VA / V / Sqrt 3... not VA / ( V x Sqrt 3 )

because 400 x sqrt 3 is 693V

actually, that works out the same....

I would have done - that was an edit one minute after my post!

 

[LiamPope]

a / b / c (evaluated left to right) is the same as a / (b*c).

In the same way that a - b - c (evaluated left to right) is the same as a - (b + c)

The rule has a proper mathsy name which I've forgotten.

Liam

 

[Spark123]

Transposition of formulae?

 

[LiamPope]

Errrrrrr one of, errr associative, distributive... er... cumulatatatatatative or something. I think it's correct to say the above comes about because division (and subtraction) are not... whichever of the above rules I mean. Aaargghhh it hurts. Make it stop...

 

[Spark123]

You ought to try definitive integration, makes that stuff look easy

 

[LiamPope]

You ought to try definitive integration, makes that stuff look easy

Is that something different to a normal definite integral? Like the definitive integral - the integral to end all integrals? I can only shudder as I contemplate how awful this integral must be!

I suffered enough calculus at school/uni. Now that I'm a practicing engineer, I'm trying to forget the pure maths and just get on with things

Liam

 

[Spark123]

I'll send you some sums to try if you like

 

[ericmark]

boolean would be interesting but not sure how you would represent it? Hard enough in Word using the Equation editor but I think very quickly it would get too complex to write it down.
And the last post with maths had basic same problem as this in accuracy of results seem to remember with 170+ meters of 4mm sq cable used in ring main.
I had used excel to work it all out which I think works to something like 11 digits but some one else used a book which had only used 2 or 3 digits so answer was 2 or 3 meters out.
Many years ago when working in college with comparators well before excel I remember using 6 figure log tables instead of normal 4 figure and getting different result. And ending up with a home work to find out the plus and minus figures we were really working to.
Tolerance is quite a problem. I seem to remember in back of log tables it gave the definition of a meter as so many wave lengths of some orange light.
God knows who counted them?

 

[stunlawless]

Morning everyone,

I seem to have kicked a hornet nest here...

Blew the dust of some books last night, and found Garymo's method in a couple - no sign of mine, so for the sake of correctness I'll stick to kVA/V/sqrt3.

Thanks to Coljack for the link - saves trying to remember sqrt3 !

The intended motor is in a heat pump. Only got a skimpy brochure, saying it's a 6kW motor. I'm waiting for the manufacturer to tell me what the pf is. Just wanted to get my head around the cable sizing, so I used a round figure as an example.

Assuming a pf of 0.8, this gives 10.8 A per phase, so it looks like 2.5mm swa 4 core (heat pump controller needs a neutral) with a 16A mcb and 40A rccb will be man enough, including installation factors and volt drop.

One final question: 10 or 16 mm^2 cable back to the main earth terminal (or whatever it's called these days) ?

Thanks for looking; hope it's been fun, David

 

[bernardgreen]

Assuming a pf of 0.8, this gives 10.8 A per phase, so it looks like 2.5mm swa 4 core (heat pump controller needs a neutral) with a 16A mcb and 40A rccb will be man enough, including installation factors and volt drop.

That is probably the running current once the motor is up to speed. You should calculate voltage drop at the peak current that occurs as the motor comes up to speed from stationary.

 

[stunlawless]

Good point Bernard; I'll see what the manufacturer's spec says.

Skimpy brochure talks about "soft start"; presumably this is a device to limit the starting current ?

As you may have gathered, I'm not an expert in big motors !

Thanks, David

 

[ricicle]

That is probably the running current once the motor is up to speed. You should calculate voltage drop at the peak current that occurs as the motor comes up to speed from stationary.

You can normally ignore start - up current when calculating voltage drop, unless a specified requirement from manufacturer.