electrical three-phase supply question

bug

hrrm

a three phase mains supply the currents in L1, L2,L3, are specific quantities, does anyone know the equation to calculate the total current in the neutal?

thanks
bug

po0k

Are they balanced or unbalanced voltages?

I really should know this, but will have to consult my notes.
If no-one else has replied by tomorrow I'll put up the formula/method.

bug

its the currents Im looking at..

L1 = 227A at 0º
L2 = 218A at 120º
L3 = 222A at 240º

The actual neutral current reading is 87 A but thats way too I think big Im not very good at vector diagrams

..oh the voltage is constant and balanced accross all phases

SOL

I think the point is that it should be 0

SkepticOne

Are they just vector quantities?

Translated into X and Y coords I get


L1: (227, 0)
L2: (-109, 188.79)
L3: (-111, -192.26)

Giving a resultant of x=6.7 y=-3.464

This has a magnitude of 7.81 and a direction of 26.33.

I know nothing about electricity; this is just based on the assumption that you add up the vector quantities.

You may need to reverse the direction of the resultant vector since the current is flowing in the opposite direction in the neutral wire giving 153.67 degrees.

Mad Mike

The neutral current should be zero if the load is balanced. But it is the current magnitude and phase which determins balance not the voltages.

Skepticone did the vector sum assuming 120 degrees between the currents and got a neutral current of only 7.8A. Yet you measured 87 Amps. Two possiblities come to mind

i) Reading the meter backwards

ii) You are mixing up the phase shift of the voltage for the phase shift of the current.

Looking at your notes I suspect that you have measured the current magnitudes but do not know the phase shifts of the currents. The 0, 120 and 240 degrees you mention look to me like the voltage phase shifts. This doesn't tell us anything about the currents. It is actually very hard to measure a current's phase shift without a wattmeter.

The formula you are looking for goes like this:

Assume L1 has an angle of 0 degrees, L2: -120 degrees, and L3: -240 degrees
Assume each current has magnitude M1, M2 and M3 repectively and a phase angle phi1, phi2 and phi3 respectively with respect to its own voltage.

The the vector represenation of each current is (all angles in degrees):

I1 = (M1.cos(phi1),M1.sin(phi1))
I2=(M2.cos(phi2-120),M2.sin(phi2-120))
I3=(M3.cos(phi3-240),M3.sin(phi3-240))

Neutral Current = vector sum of I1, I2, I3 (just add the respective terms)

If you have the neutral current as a vector you can work out its magnitude by getting the square root of the sum of the squares of both terms.

ie if In =(Ix, Iy) then the magnitude of In = SQRT(Ix.Ix+Iy.Iy)


Note: Electrical Engineers usually write vectors as complex numbers using the letter j to represent the sqrt(-1) but the maths are the same as above.

SkepticOne

Originally posted by SkepticOne
You may need to reverse the direction of the resultant vector since the current is flowing in the opposite direction in the neutral wire giving 153.67 degrees.

Reconsidering this, the vector direction of the resultant neutral current is probably irrelevant since it is not a single sine wave but a sum of three sine waves.