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1. Originally Posted by MyCleveland
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...

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Originally Posted by publicgood
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)

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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.../chapter-5.pdf

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Originally Posted by 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.../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

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Originally Posted by Ingenieur
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 ?
Method you showed....way back is what I used in spreadsheet I posted.

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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.

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Originally Posted by 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.
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%

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