Amperage calculation using 3 phase power

[Steelhead]

I have 3 phase power (480V) supplying 6 heating elements. If the amp draw is as listed how do I calculate the total amperage for A, B, and C phases?

• AB-5.2, 10.4
• BC-10.4, 10.4
• AC-5.2, 10.4

Thanks!

 

[ptonsparky]

I a-b L1'toL2	15.6​
I b-c L2'toL3'	20.8​
I a-c L1'toL3'	15.6​
Amp: A L1	27.02​
B L2	31.63​
C L3	31.63 ​

For Delta

15.6

20.8

15.6

=SQRT(B37^2+B39^2+(B37*B39))

=SQRT(B37^2+B38^2+(B38*B37))

=SQRT(B39^2+B38^2+(B39*B38))

 

[wwhitney]

Simple conservative approximation:

When the currents A-B, B-C, and C-A are identical, say X, the currents on A, B, and C are identical and of value X * sqrt(3). The sqrt(3) factor (instead of 2) is due to the current in B from the A-B loads being 60 degrees out of phase from the current in B from the B-C loads (instead of 0 degrees out of phase).

So if you have A-B = B-C = C-A = 15.6, the currents in A, B, and C would be 15.6 * sqrt(3) = 27A. But this neglects the extra 5.2A on B-C. The simple approximation says just add that onto the B and C currents, so you get A = 27A, B = 32.2A, C = 32.2A. This is conservative, as the currents on B and C will be no more that 32.2A.

More accurate:

The 5.2A from B-C is actually at 30 degrees phase difference from the balanced currents from the balanced set of loads. So to get the B and C currents more accurately, we need to add 5.2A and 27A with a 30 degree phase difference. This can be done with the law of cosines, which says for vectors X and Y:
|X+Y|^2 = |X|^2 + |Y|^2 + 2*|X|*|Y|*cos(angle between X and Y).

That means the B and C currents are really sqrt(27^2 + 5.2^2 + 2*27*5.2*cos(30 deg)) = 31.6A. As you can see, a small difference from the conservative approximation.

Cheers, Wayne

 

[Steelhead]

Interesting, thank you. I'm going to chew on that for awhile. I'll probably be asking more questions.

 

[ptonsparky]

Interesting, thank you. I'm going to chew on that for awhile. I'll probably be asking more questions.

Put it in a spreadsheet. That's as far as my education goes and you can see the results immediately as current changes.
In my formula:
B37=15.6
B38=20.8
B39=15.6

and I didn't even know that until another thread earlier this year. Wwhitney & others drug me kicking and screaming thru the process.

 

[ptonsparky]

Load up L1-L3 and L2-L3 with nothing on L1-L2.
That's kind of an eye opener.

 

[LarryFine]

Would converting to power figures make it any easier?

 

[wwhitney]

Would converting to power figures make it any easier?

That's a good option with a caveat. It is exact for balanced loads, but non-conservative for unbalanced loading. If the loading is mostly balanced, then it's close to exact.

For example, with the loads in the OP, it would give currents of 27A, 31.5A, and 31.5A, very close to the exact 31.6A.

But if you apply it to a single L-L load of current X, it tells you the line currents are 0.866 * X, which is 13.4% too low.

Cheers, Wayne

 

[Julius Right]

Using complex number system, if we consider the same cosfi at all phases then the actual current will be:
ab=15.6*cos(0)+j*15.6*sin(0)
bc=20.8*cos(-120)+j.20.8*sin(-120)
ca=15.6*cos(-240)+j.15.6*sin(-240)
ab=15.6
bc=-10.4-j18.013
ca=-7.8+j13.51
a=ab-ca=15.6-(-7.8+j13.51)=(15.6+7.8)-j13.51= (23.4-j13.51)
b=bc-ab=-10.4-j18.013-(15.6)=(-10.4-15.6)-j18.013= (-26-j18.013)
c=ca-bc=-7.8+13.51-(-10.4-j18.013)=(-7.8+10.4)+j(13.51+18.013)= (2.6+j31.523)
a= SQRT(23.4^2+13.51^2)= 27.01999445
b= SQRT(26^2+18.013^2)= 31.63017814
c= SQRT(2.6^2+31.523^2)= 31.63004156

 

[Julius Right]

To calculate using Microsoft Excel in the complex is very simple. But translating the calculation into manual calculation is very complicated. So, I messed up the transfer a few times.