Voltage level for MCC

[Jed057]

Dear Guru,

Our LV system is 400 V,
but there are many outgoing feeder very far from substation.
I am worry about the voltage drop.
please advise what the suitable voltage value at MCC should be by change at tap changer.
Such actual 400, 410 or 420.
If change the voltage ,is it affected to another problem because we also have overvoltage in outgoing feeder
the setting is
5% of 400 - alarm
10% of 400 - trip

thks
young eng

 

[undervoltage]

Depends upon your voltage drop particularly at the instant of motor starting and the no. of motors required to start at one time.

 

[undervoltage]

*Voltage drop at the bus. You may check this by running a motor starting study.

 

[7anoter4]

I agree with undervoltage. You have to check if the voltage drop up to each motor terminal at start will be not more than permissible to start safety.
Usually from MCC supplied maximum is 250 HP squirrel cage induction motor. This motor needs 90% of rated voltage to start in good condition [standard motor] in case of DOL [Direct on Line] start and the starting current will be at least 6 times the rated.
If the motor controller is provided with soft start or VFD the start current will be not more than 1.5*Irated.
You can take the transformer load as transformer rated and a p.f. of 0.85 average. Extract the motor rated current with the rated p.f. and add a starting current with a p.f of 0.3.
In our case -DOL start 250 hp - 350 A pf=0.85 rated at start 6.5*350= 2275 A
Let's say the transformer is 2500 kVA 6% short-circuit voltage and Xtrf/Rtrf=5
Rated xfrmr current will be 2500/.4/sqrt(3)=3608.4 A
Iactiv=3608*.85=3066.8 Ireact=3608.4*sqrt(1-0.85^2)=1900.8 A
Imotact=350*.85=297.5 Imotreact=350*sqrt(1-.85^2)=184 A
Imstact=2275*.3=682.5 Imstreact=2275*sqrt(1-0.3^2)=2170 A
Finally the current flowing through trf will be:
Iact=3066.8-297.5+682.5=3451.8 A Iract=1900.8-184+2170=3886.8 A
PF=3451.8/sqrt(3451.8^2+3886.8^2)=0.664
The voltage at transformer low voltage terminals will be [approximate]:
VL = VS-RI cos(fi)+XI sin(fi)
Let's say VS=420 [the overvoltage alarm will be activated for seconds on empty transformer!]
Uk%=sqrt(3)*Irated*Ztrf/Urated/100 Ztrf=400*6/100/3608.4/sqrt(3)=0.00384 ohm
Or the same :
Ztrf=uk%/100*Voltage^2/Strf/1000=6/100*400^2/2500/1000=0.00384 ohm
Ztrf=sqrt(Xtrf^2+Rtrf^2)=sqrt(Xtrf^2+(Xtrf/5)^2)=Xtrf*sqrt(1+1/25)
Xtrf=0.00384/sqrt(1+1/25)=0.003765 ohm
Rtrf=0.003765/5=0.000753 ohm
VL=420-(0.000753*3451.8+0.003765*3886.8)*sqrt(3)=390.15 V full load transformer at 250 HP motor DOL start.
You have to choose the cables-MCC feeder and motor feeder so that calculated voltage drop will be maximum 30 V.
Let's say the MCC rated current is 600 A and MCCpf=0.85.
IactivMCC=600*.85=510 A IreactMCC=600*sqrt(1-0.85^2)=316 A
Finally the current flowing through MCC at motor start will be:
Immcact=510-297.5+682.5=895 A Imccreact=316-184+2170=2302 A
Let's say the MCC feeder is 4*3*240 sqr.mm CU XLPE insulated 50 m from transformer.
Rcable=0.09 ohm/km at 90oC 50 Hz[including skin effect+proximity effect]
Xcable=0.0806 ohm/km
Rmcc=0.09*50/1000/4=0.001125 ohm Xmcc=0.0806*50/1000/4=0.001008 ohm
VLMCC=390-(0.001125*895+0.00108*2302)*sqrt(3)=384.3 V
The same calculation for motor feeder cable:
Let's say 3*3*240 sqr.mm CU 100m from MCC:
Rmotcable=0.09*100/1000/3=0.003 ohm Xmotcable=0.0806*100/1000/3=0.002687 ohm
VLmot=384.3-(0.003*682.5+.002687*2170)*sqrt(3)=370.7 V
As starting current is approx.direct prop. with the supply voltage you may reduce the starting current by 370.7/400 and revise the calculation.
Now, even the voltage drop in the start case at the starting motor is suitable you have to check if the voltage at terminals of other running motors at steady state is still suitable for steady state running [not less than 10% rated]
Also you have to check in the steady state of all motor if at each motor terminal the voltage will be more than 95% of rated.See NEMA MG-1 or IEC 34-1 .