Voltage drop along cable

  • Thread starter lavalin
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  • #1
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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
 

Answers and Replies

  • #2
stewartcs
Science Advisor
2,177
3
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.

It appears they are approximating the voltage drop from line-to-line of a three phase system, not line-to-neutral.

The general approximation for a line-to-neutral voltage drop is:

VD = IRcos(phi) + IXsin(phi)

where,

VD = voltage drop in one conductor, one way
I = current flowing in one conductor
R = line resistance for one conductor in ohms
X = line reactance for one conductor in ohms
phi = angle whose cosine is the load power factor

Note that I is usually assumed to be the load carrying capacity of the conductor.

The line-to-line voltage drop for a three-phase system is then found by multiplying by sqrt(3).

CS
 
  • #3
2
0
You have given cable Z of 0.2 ohm/km which is value for 95mm2
This cable has 3ph volt drop of 0.4mV/Am
Hence VD = 0.4 *75*350/1000 = 10.5v
Volt drop usually is % of voltage at MCC which usually 415v
3% of 415V is 12.45v hence you are below this
 
  • #4
215
1
%V = ( phase voltage drop / rated phase voltage)* 100

%V (in symetrical system) = ( sqrt (3) * phase voltage drop / sqrt (3) * rated phase voltage) * 100


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