Motor starting current and fault level 3

[kvkev]

Is there any formulae available, from which you can deduce whether direct online starting of a motor will dip the mains voltage to an unnaceptable level, given the 3 phase fault current level at the motor control center.
i.e If I have a 3 phase fault level of 25kA at 415 volts at the MCC, would DOL starting of a 90kW cage motor cause a severe dip, and what % voltage dip would I get ?

KVKEV

 

[DanDel]

The voltage drop can be calculated using the impedances of the cables and the motor, and the fault current value at the source. This is typically performed in a Motor Starting Study, by software or manual calculations.

 

[bigamp]

A very quick and approximate result for voltage dip (Vd) at a bus where the available fault level (in kVA) is FL is as follows:

Vd(%) = 100*motor starting kVA/FL

The above is approximate and if it gives an answer close to your acceptable limit then you need to do more detailed calculations. Also, for FL you should use only the steady state fault level, i.e. with zero contribution from in service motors.

Another formula is this one taken from "Electricity Distribution Network Design" by Lakervi & Holmes is:

Vd(%) = 100*(sqrt(3)*I/V)*(R*cos(phi) + X*sin(phi))

I = motor starting current
R = resistive component of system impedance to bus
X = reactive component of system impedance to bus
V = system phase to phase voltage
phi = power factor of motor starting current

The above will be a little conservative. If you want an exact answer, you need to take into account that the motor starting current is reduced due to dip in voltage to the bus and voltage drop in the cable to the motor terminals.

In your example, 90kW motor starting kVA is about 680kVA, fault level to bus is 17970kVA so dip will be about 3.78%.

 

[jlazucena]

The formula is basic, of course all these are complex numbers. d)
V bus = V system – I motor starting x Z system
*****************
The main concept here is that Z system is the impedance of the system at MCC terminals, and this value can be calculated from the short circuit current (SCC):
Z system = V system / I scc
***********************************
Also, I will neglecte all other loads, and use average values for x/R ratios, motor starting currents, etc.
Analysis
V bus is the MCC voltage during motor starting
V system is the voltage at zero current or 415 V @ 0 deg (Reference voltage)

I motor starting is the starting current
Motor kVA = motor kW / (power factor x efficiency)
(Note: Take pf = .8 and eff = .9)
I motor nominal = Motor kVA / (1.732 x kV nominal)
= (90 kW /(.8*.9))/(1.732x0.415)
= 173.9 Amps
I start is 6 times nominal current at 12% power factor
I start = 6 x 173.9 @ - 83.1 deg
= (1043.4 @ -83.1 deg ) Amps

Z system = V system / I scc
Note: for this system Iscc angle may be -80 deg
Z system = (415 @ 0.0 ) / (25000 @ -80) = 0.0166 @ 80 deg Ohms

Vbus = (415 @ 0.0 ) Volts – (1043@ -83.1) x (0.0166 @ 80)
= (397.7 @ 0.1349 deg) Volts

This is a conservative value:
Real voltages may be higher because the starting current will be smaller, or
Lower if the transformer feeding this MCC is heavy loaded.

 

[jbartos]

Suggestion:
The parameters presented in the original posting do not fit the well posed problem. They require estimations of parameters.
Estimate:
1. Motor locked rotor current LRA~6xFLA=6x148=888Amps
FLA=90kW/(sqrt3 x .415kV x .94 x .9)=148Amps
2. EFF~.94
3. PF~.9
4. Feeder cable size ~ #1 AWG Copper
5. Cable impedance from NEC = .16 Ohm/1000FT = .136 + j.084, at PF=0.85 in steel conduit
6. Assume 100FT long motor feeder, Zfeeder=R+jX=.0136 + j.0084, Ohms
7. cosT=.9, sinT=.436, motor current lag
IEEE Std 141-1993 Red Book, page 98:
Actual Voltage Drop=Es + IRcosT+IXsinT-sqrt[Es**2 - (IXsinT-IRcosT)**2]
=(415/sqrt3)+888x.0136x.9+888x.0084x.436
-sqrt[(415/sqrt3)**2-(888x.0084x.9-888x.0136x.436)**2]=
=14.124Volts
or
415/sqrt3V - 14.124V=225.48Volts phase to neutral
or
225.48 x sqrt3=390.5 Volts phase to phase
at the motor terminal.
And
(390.5/415)x100%=94%
The motor terminals will see 94% of 415Volts.
(There is some small margin of error.)

 

[bigamp]

jbartos,

You have calculated the voltage drop in the motor power cable for the motor starting current and have used the motor full load power factor (cos T) of 0.9 to do so. To get a more accurate answer, you would need to use the motor starting power factor, which is typically in the 0.1 to 0.3 range.

 

[jbartos]

Suggestion: Yes, normally the motor starting power factor is used. However, it varies with the motor design. Considering many assumptions to obtain the result, my calculation above is for its form only, and it is very approximate or inaccurate.

 

[jbartos]

Corrected calculation for starting PF=.2
Suggestion:
The parameters presented in the original posting do not fit the well posed problem. They require estimations of parameters.
Estimate:
1. Motor locked rotor current LRA~6xFLA=6x148=888Amps
FLA=90kW/(sqrt3 x .415kV x .94 x .9)=148Amps
2. EFF~.94
3. PF~.9
4. Feeder cable size ~ #1 AWG Copper
5. Cable impedance from NEC = .16 Ohm/1000FT = .136 + j.084, at PF=0.85 in steel conduit
6. Assume 100FT long motor feeder, Zfeeder=R+jX=.0136 + j.0084, Ohms
7. cosT=.9, sinT=.436, motor current lag
8. cosS=.2, sinS=.98, motor starting power factor
IEEE Std 141-1993 Red Book, page 98:
Actual Voltage Drop=Es + IRcosS+IXsinS-sqrt[Es**2 - (IXsinS-IRcosS)**2]
=(415/sqrt3)+888x.0136x.2+888x.0084x.98
-sqrt[(415/sqrt3)**2-(888x.0084x.2-888x.0136x.98)**2]=
=9.725Volts
or
415/sqrt3V - 9.725=229.65Volts phase to neutral
or
22.65 x sqrt3=397.77 Volts phase to phase
at the motor terminal.
And
(397.77/415)x100%=96%
The motor terminals will see 96% of 415Volts.
(There is some small margin of error.)
The resulting 397.77V motor terminal voltage happens to be close to the 397.7V posted by jlazucena.

 

[bigamp]

jbartos,

Yourself and jlazucena have calculated two different things. The answer to both turns out to be near enough the same (as a percentage), within the context of the assumptions made regarding motor parameters and possible cable type and circuit length to the motor.

Jlazucena has calculated the voltage dip from the grid supply to the MCC to which the motor is connected. You have calculated the voltage drop from the MCC to the motor terminals.