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cnumartyr
03-20-2007, 06:35 PM
Currently we have a 3 phase motor that runs a chill water building supply pump.

The building has 480 volts supplied to the motor during the early spring and it pulls 19.9 Amps. The FLA on the motor is 20 while the SFA is 23.5.

If the Voltage supplied to the motor is dropped to 460 volts during the summer in "brown out" conditions, what amperage will the motor pull? I have looked everywhere to find the formula for this calculation but have been unable.

The closest thing I could think of was Ohm's Law and I=V/R and got a 19.1 Amps at 460 Volts. I know this wouldn't be right because it has to pull the same Wattage at 460 as it does at 480.

We are trying to find this because our "Energy Management" person was asking us to figure it out for him and we are unable to as we don't have the training on Electric motors to know it.

He also said something about "Corrective Amps" and finding the "safe range to run the motor in" as far as percent of FLA without burning the motor up.

Anny help would be appreciated.

lwarren
03-20-2007, 08:43 PM
Because it is three phase you have to use the square root of 3 which is 1.73 along with a couple of variables like the motor efficiency and the power factor.

The following link has some useful motor formulas:
http://www.ab.com/support/abdrives/documentation/fb/1018.pdf

Smacky
03-20-2007, 10:40 PM
I though maybe perhaps i could help you on this issue scince i just learned about it in class. I think your looking for the formula of

P (nameplate rating) = EL (voltage in the line) X IL ( amps in the line ) X 1.732 (sq root of 3 )

for a pure resistive load but seems you might be looking for VA

VA = EL X IL X 1.732

Or in your case use algebra and flop them around as you like

VA / (EL x 1.732 ) = IL

Dont forget the laws in a 3 phase load the WYE connection

IL = IP ( amps in the phase)
EL = EP x 1.732

EL = EP
IL = IP x 1.732

Power factor is your True power/apparent power

PF = P / VA

You can use ohms law to calculate anything in the phase you need

Z(inductance ) = E /I

Rememeber the power company can deviate from there voltage by 10% so in 480 system they can deviate 48 volts from as low as 432 volts to 528 usually motors are rated for that. But to me . But me personaly i think you have a wye connected motor so your amps will stay the same regardless of what the voltage states according to the laws.

hope that helps

snipe70e
03-20-2007, 10:47 PM
Regardless of the voltage the load is going to remain the same, therefore the same KW will be required.

KW=V X I X 1.72 X PF (power factor)

The power factor will change a small amount due to the change in voltage and current, but I do not think you can caculate the change, so lets assume it stays the same. So

V1 X I1 X 1.72 X PF = V2 X I2 X 1.72 X PF

V1 X I1 = V 2X I2

I2 = V1/V2 X I1

I2 = 480/460 X 19.9

I2 = 20.8 amps

Smacky
03-20-2007, 10:48 PM
I forgot to tell you that you take your voltage

460 / 1.732 = EP whic comes out to 265.5 volts

the you use ohms law

Ip = 265.5 / Z ( resistance )

THen if it is a wye connected load IP = IL

jayguy
03-21-2007, 10:30 PM
if nameplate is:
460 VAC, 20 FLA

and you are running 480 VAC then your "corrected" FLA is 19.2 FLA (@ 480 VAC).

if your SF is about 1.15, then your maximum "corrected" SFA is 22.1 amps.

when your voltage drops back to about 460 VAC, you would be drawing about 20.7 amps. well within the service factor amperage.

please keep in mind that i know very little about your motor other than what you gave me...so these are approximates.

if you gave me all of the nameplate info we could calculate things a little more accurately.

need nameplate: voltage, power factor (pf), NEMA nominal effeciency, hp, SF, etc.

RichardL
03-23-2007, 07:32 PM
:D
Snipe70e Wins this one hands down..
Whilst' others treated the motor as a set of electric strip heaters "Only" da' Snipe brought into play that the "Load" on the motor would remain constant and therefore require the same Power(Watts) to accomplish this, which will require an increase in motor amps.
Unmentioned also was the fact that as applied voltage decreased, the "Slippage" of the rotor would increase and motor heating would also increase as the motor RPM decreases.
Way to go brer' Snipe..

Smacky
03-27-2007, 12:53 PM
Really if you dont know anything about this subject you shouldnt be replying to it . AS far as snipe says you cannot calculate power factor you can calculate the difference in fact the power factor can drop as low as .6. I hope We all know the difference between power and VA. A motor is a INDUCTIVE LOAD that means voltage leads current. POwer can only be calculated IN A PURE RESISTIVE LOAD such as heaters ( power = 1 ). WE all know in change in voltage happens your current will inversly
change too and becasue it is 3 phase the square root of 3 is applied. To find power factor you take you name plate rating of kw or whatever it is he obviously gave you the volts 460 X 19.9 amps= VA
powerfactor is the difference of the apprent power / true power
WE obvioulsy didnt get enough information on this subject and there are multiple ways of getting the answer.

Smacky
03-27-2007, 01:01 PM
my correction the VA calculation for 3 phase is

Volts X amps X Sq root of 3 (1.732) = VA

big johnson
03-27-2007, 02:53 PM
[QUOTE=Smacky;1425604]Really if you dont know anything about this subject you shouldnt be replying to it .

I'm sure none of the responders to this topic realized that. Maybe we need to read the rules again. It is a forum, is it not? We are all still capable of learning.

snipe70e
03-27-2007, 10:16 PM
Really if you dont know anything about this subject you shouldnt be replying to it . .

You are just kidding? Just trying to get a reaction or what?

deanmech
03-27-2007, 11:50 PM
There is a much simpler way to go about this.

Motor Corrected FLA = (Nameplate amps X Nameplate volts)/Actual volts

Verbally it is Nameplate VA divided by actual volts.

Another related formula:

Actual HP = Rated HP X (Measured operating amps/Corrected FLA)

Make sure you always do the operation in parenthesis first in the above formulas.