# Thread: Comparing Name Plate Current to a One Phase Reading Current for a 3 phase motor

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## Comparing Name Plate Current to a One Phase Reading Current for a 3 phase motor

Everyone:

I seem to be very confused.

We were doing a routine maintenance check to measure the current draw on a 3 phase motor as it compares to its name plate FLC rating. The FLC is about 6A and we took an Amp probe and measured 3A on one phase. I've always thought that with a one phase amp probe reading you have to multiply the reading by 1.732 (so turns out to be about 5.196A) in order to convert it to the line current draw so we can compare it to the FLC rating on the motor name plate.

Is this a correct statement?

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Originally Posted by Jason T
Everyone:

I seem to be very confused.

We were doing a routine maintenance check to measure the current draw on a 3 phase motor as it compares to its name plate FLC rating. The FLC is about 6A and we took an Amp probe and measured 3A on one phase. I've always thought that with a one phase amp probe reading you have to multiply the reading by 1.732 (so turns out to be about 5.196A) in order to convert it to the line current draw so we can compare it to the FLC rating on the motor name plate.

Is this a correct statement?
No. Current you read on nameplate is the phase current measured by current clamp (ammeter) when run at full load. Your motor was probably not running at full load.

1.73 factor (3^0.5) applies when you are measuring the line to neutral voltage. in that case, you need to multiply it by 1.73 to get line-line voltage.

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Originally Posted by LMAO
No. Current you read on nameplate is the phase current measured by current clamp (ammeter) when run at full load. Your motor was probably not running at full load.

1.73 factor (3^0.5) applies when you are measuring the line to neutral voltage. in that case, you need to multiply it by 1.73 to get line-line voltage.
thanks for the response!

Is it correct to assume that if it was a Delta configured incoming power then we would use 1.732 to convert the phase current to line current?

4. Originally Posted by Jason T
thanks for the response!

Is it correct to assume that if it was a Delta configured incoming power then we would use 1.732 to convert the phase current to line current?
The motor line current is based on the voltage and does not change based on the power supply to the motor being wye or delta. You directly read the current on each of the motor conductors. At full load the current reading should be very close to the nameplate rating.

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Originally Posted by don_resqcapt19
The motor line current is based on the voltage and does not change based on the power supply to the motor being wye or delta. You directly read the current on each of the motor conductors. At full load the current reading should be very close to the nameplate rating.
Makes sense, and you are correct we were not running full load but we thought we were close. But apparently not that close...lol. Thank you for the response!

6. gar
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160619-1611 EDT

I took a look at one motor performance plot for a 35 HP motor. See figure 13-12, p 213, of "Alternating-Current Machinery", Bailey and Gualt, McGraw-Hill, 1951.

Current vs load is not quite linear but it is somewhat. What the current curve does not do is intersect at 0,0. For this motor line current is about 25% for zero load torque. Using this curve your load was possibly about 15/35 = about 43 % of full load.

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7. Originally Posted by LMAO
No. Current you read on nameplate is the phase current measured by current clamp (ammeter) when run at full load. ...
The current (ampere) value on the nameplate or read by an amp clamp is referring to line current. Phase current is the current between two lines. For example, when a motor is delta-configured the current on each of the three windings is phase current. The line current is square root of three greater than phase current.

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Originally Posted by gar
160619-1611 EDT

I took a look at one motor performance plot for a 35 HP motor. See figure 13-12, p 213, of "Alternating-Current Machinery", Bailey and Gualt, McGraw-Hill, 1951.

Current vs load is not quite linear but it is somewhat. What the current curve does not do is intersect at 0,0. For this motor line current is about 25% for zero load torque. Using this curve your load was possibly about 15/35 = about 43 % of full load.

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Load Power = sqrt3 x i x v x pf x eff
the pf and eff will be much lower at 40% fla vs fla
maybe 1/2
I would rate load at 1/2 x 15/35 = 20-25%

motor torque is ~ linear with current
P = T w (w = 2 Pi f)
for 60 Hz in rpm
T = 5252/rpm x P

5252 = (60 sec/min x 550 ft lb/HP) / (2 Pi)
a conversion factor

typ curve 45% fla pf drops to 50% from 90
eff is similar but not as dramatic
Last edited by Ingenieur; 06-19-16 at 06:35 PM.

9. gar
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160619-2215 EDT

Attached is a photo of the 3 phase induction motor curves I referenced above from Bailey nd Gault.

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Note:
Full rated load is 35 HP.
Current at zero load is 10.5 A.
At 20 A load current, 1/2 of full rated current, output power is 15 HP.

Different motor designs will have different characteristics. For any particular motor you should be able to get typical motor performance curves.

Power input to a motor is a much better way to estimate output torque and/or power.

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