Voltage Dip Estimation

[MyCleveland]

Downloaded attached formula around one year ago. Did not keep a copy or recall source. Tried finding again with no luck.

Question, is this a valid calculation method for Voltage Dip/Sag ?

If I downloaded from MH prior posts, please help me find it.

 

[Ingenieur]

that will work
I find this easier
assume 1 mva 480/3 5% puZ

ifault = 1000000/(1.732 480 0.05) = 24056
xfmr Z in Ohms = 480/24056 = 0.02

assume feeder is 0.01 Ohm (value from nec table x 1.732 x length/1000)
total Z = 0.03
motor starting i = fla x 6 or calculated from mtot code)
assume 100 x 6 = 600 A

vdrop = 0.03 x 600 = 18 vac or 3.75%

 

[MyCleveland]

Ingenieur...Thank you

Using your method on PDF problem yields 11.34%.
Assume this method is more accurate ?

Max VD at MOTOR of 10%...acceptable target value ?

Max VD at Service Equipment (for a given motor start) value ?

 

[Ingenieur]

imo theirs is wrong

should be 618.55/5454.46 x 100 = 11.34%
should not add the motor lr kva to the fault kva in the numerator

 

[publicgood]

Assume this method is more accurate ?

Max VD at MOTOR of 10%...acceptable target value ?

Max VD at Service Equipment (for a given motor start) value ?

It is just a different way of going about it.

10% depends on the target reasoning. Are you concerned about lights flickering?

Close to 25-30% could be acceptable if you were just confirming the motor controller and/or BMS wouldn’t drop out.

 

[publicgood]

imo theirs is wrong

should be 618.55/5454.46 x 100 = 11.34%
should not add the motor lr kva to the fault kva in the numerator

Faults are fed by all sources, both utility and motor contribution.

 

[Ingenieur]

Faults are fed by all sources, both utility and motor contribution.

true
but this not a fault calculation
this is a voltage drop calc due to motor inrush
the xfmr can't supply more than it's sc fault current
the motor contributes nothing, it is a load in this case

my calc uses xfmr internal Z plus conductor Z x starting current to give V at the motor terminals
very simple, like any v drop calc
matches the op formula if the motor inrush is not added to xfmr sc

 

[publicgood]

true
but this not a fault calculation
this is a voltage drop calc due to motor inrush
the xfmr can't supply more than it's sc fault current
the motor contributes nothing, it is a load in this case

my calc uses xfmr internal Z plus conductor Z x starting current to give V at the motor terminals
very simple, like any v drop calc
matches the op formula if the motor inrush is not added to xfmr sc

True I retract my statement

 

[MyCleveland]

What is the consensus on MAX target values ?

Max VD at MOTOR during start...acceptable target value ? Max %VD ?

Max VD at Service Equipment (for a given motor start) value ? Max %VD to avoid flicker?

 

[MyCleveland]

Anyone.....?
before this slips off the page.

 

[publicgood]

Anyone.....?
before this slips off the page.

The answer to the light flicker question is outlined in IEEE 241 [Gray Book]. Visible flicker is seen with as little as 0.5% voltage change. Light output for say fluorescent lighting is directly proportional to the applied voltage.

"Acceptable voltage flicker magnitude depends on the frequency of occurrence.... At 10 dips per hour, people begin to detect incandescent lamp flicker for voltage dips larger than 1% and begin to object when the magnitude exceeds 3%.... A typical motor may draw 5 kVA/hp and a transformer impedance may be 6%. The equation below estimates flicker while starting a 15 hp motor on a 150 kVA transformer: 15 hp * (5 kVA/hp) * (6%/150kVA) = 3% flicker"

It goes on to say if you are on the boarder of 10 dips at this 3% flicker you can better review via locked rotor currents. Of course you have the conductor length to consider as we have outline also.

Max VD for a motor is typically limited by the relays or contactors - subject to the CEBMA (or now ITI) curve. This value is as much as 35%, but target 30%.

Max VD for a service depends. If there is a UPS or otherwise on the service...

 

[MyCleveland]

The answer to the light flicker question is outlined in IEEE 241 [Gray Book]. Visible flicker is seen with as little as 0.5% voltage change. Light output for say fluorescent lighting is directly proportional to the applied voltage.

"Acceptable voltage flicker magnitude depends on the frequency of occurrence.... At 10 dips per hour, people begin to detect incandescent lamp flicker for voltage dips larger than 1% and begin to object when the magnitude exceeds 3%.... A typical motor may draw 5 kVA/hp and a transformer impedance may be 6%. The equation below estimates flicker while starting a 15 hp motor on a 150 kVA transformer: 15 hp * (5 kVA/hp) * (6%/150kVA) = 3% flicker"

It goes on to say if you are on the boarder of 10 dips at this 3% flicker you can better review via locked rotor currents. Of course you have the conductor length to consider as we have outline also.

Max VD for a motor is typically limited by the relays or contactors - subject to the CEBMA (or now ITI) curve. This value is as much as 35%, but target 30%.

Max VD for a service depends. If there is a UPS or otherwise on the service...

No UPS considered, just wanted to put a flag in a spreadsheet....X (max value exceeded)

 

[Julius Right]

I think it could be something like this one:
Electrical Fault Level Calculations Using MVA Method by Lee Wai Meng
http://www.jmpangseah.com/wp-content/uploads/2003/01/chapter-5.pdf

 

[Ingenieur]

I think it could be something like this one:
Electrical Fault Level Calculations Using MVA Method by Lee Wai Meng
http://www.jmpangseah.com/wp-content/uploads/2003/01/chapter-5.pdf

using that method assuming no initial load and values from op's link
Sxfmr = 0.300 mva, pu Z 5.5%, Ssc = 5.4545 mva
Smtr = 6 x 124 x 480 x sqrt3 = 0.6185 mva
% load = (0.6185 x 5.4545) / (0.6185 + 5.4545) = 0.5555
( (Smtr x Ssc)/(Smtr+Ssc) )
pu v at mtr = 5.4545/(0.5555 + 5.4545) = 0.9077 or 9.23%
( Ssc/(% load + Ssc) )

the op uses Smtr/(Smtr + Ssc) = 0.6185/(0.6185+5.454)x100=10.18%

it's very simple imo
xfmr isc = 6561 A so Z = 480/6561 = 0.0732 Ohm
vdrop thur xfmr = 6 x 124 x 0.0732 = 54.5 v or 11.34% at terminals

 

[MyCleveland]

using that method assuming no initial load and values from op's link
Sxfmr = 0.300 mva, pu Z 5.5%, Ssc = 5.4545 mva
Smtr = 6 x 124 x 480 x sqrt3 = 0.6185 mva
% load = (0.6185 x 5.4545) / (0.6185 + 5.4545) = 0.5555
( (Smtr x Ssc)/(Smtr+Ssc) )
pu v at mtr = 5.4545/(0.5555 + 5.4545) = 0.9077 or 9.23%
( Ssc/(% load + Ssc) )

the op uses Smtr/(Smtr + Ssc) = 0.6185/(0.6185+5.454)x100=10.18%

it's very simple imo
xfmr isc = 6561 A so Z = 480/6561 = 0.0732 Ohm
vdrop thur xfmr = 6 x 124 x 0.0732 = 54.5 v or 11.34% at terminals

Ingenieur
Is this not the same as my original attachment ?
Downloaded, but no time for review as of yet.
Method you showed....way back is what I used in spreadsheet I posted.

 

[topgone]

I think it's late but here's a link to an old article that deals with estimating a voltage dip when you have a large motor connected to a utility.
link

 

[Ingenieur]

I think it's late but here's a link to an old article that deals with estimating a voltage dip when you have a large motor connected to a utility.
link

lol
look at the first equation in the link
I just cranked this out

v drop = Smtrstart / Ssc x V (Smtrstart ~ LR)
v drop % = Smtrstart / Ssc x 100

from op paper
v drop % = 618.5/5454.5 x 100 = 11.34%