- #1
lavalin
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I have a 45kW VSD drive with pf of .9 and FL amp of 75A, I have a 350m run from the MCC to the VSD itself and I'm trying to calculate what size cable I need so that I don't exceed the specified voltage drop of 3%. The line-line voltage where the motor is to be placed is 400V and the cable impedance is say .2ohms per km.
I was looking in the australian standards and their voltage formula is as follows:
Vdrop = (sqrt(3) * I(line) * Distance * Z(cable))/1000
V(% drop) = Vdrop/400
What I want to know is, why are they multiplying by sqrt(3), when you already have the line value Current. I thought sqrt(3) was only used to convert between phase/line values of current or voltage in delta/star configurations. The motor has a delta winding and I'm running 4C (3C+E) out to it.
From what I understand it should be:
Vdrop = I(line) * Distance * Z(cable)
So;
Vdrop = (75 * 350 * .2) / 1000
Vdrop = 5.25V
V(%) = (Vdrop/V) * 100
V(%) = (5.25 / 400) *100
V(%) = 1.31%
Thanks
I was looking in the australian standards and their voltage formula is as follows:
Vdrop = (sqrt(3) * I(line) * Distance * Z(cable))/1000
V(% drop) = Vdrop/400
What I want to know is, why are they multiplying by sqrt(3), when you already have the line value Current. I thought sqrt(3) was only used to convert between phase/line values of current or voltage in delta/star configurations. The motor has a delta winding and I'm running 4C (3C+E) out to it.
From what I understand it should be:
Vdrop = I(line) * Distance * Z(cable)
So;
Vdrop = (75 * 350 * .2) / 1000
Vdrop = 5.25V
V(%) = (Vdrop/V) * 100
V(%) = (5.25 / 400) *100
V(%) = 1.31%
Thanks