Motor Amperage Question / Answer

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M achew

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How many amps does a 3-hp, 240v, 3-Phase motor draw if the motor has 90% efficiancy & a power factor of 95%?

I started with looking up the FLA = 9.6 in table 430.250

Then 9.6 / 75% = RLA (7.2)

Next 7.2 x 90% x 95% = 6.156 amps

The correct answer was 6.3

I know I'm close, but where am I off?
 
This might be better in the Education section or motors.
Full Load Amp = Running Load Amp
RLA is typically in refrigeration context

In real world I would look at the motor label to determine the FLA or RLA and that is what I would expect it to draw under full design load.
The NEC table covers a range for that HP and is for determining the conductor size.

Someone that knows should chime in.

FWIW 9.6*75%=7.2
 
ptonsparky said:
This might be better in the Education section or motors.
Full Load Amp = Running Load Amp
RLA is typically in refrigeration context

In real world I would look at the motor label to determine the FLA or RLA and that is what I would expect it to draw under full design load.
The NEC table covers a range for that HP and is for determining the conductor size.

Someone that knows should chime in.

FWIW 9.6*75%=7.2
Click to expand...
Typo in writing out the calculation...

It appears PSI Online stole many of the questions from TEST.com. I am getting some of the same questions, exactly, and with wrong answers from PSI even. Unfortunately, I am just trying to pass the JM test and the real world is not in play right now.
 
Where an ac electrical service is supplied with four (4) parallel sets of size 500 kcmil aluminum ungrounded conductors, what is the MINIMUM size copper grounding electrode conductor required when connected to the concrete-encased reinforcing building steel that does not extend on to other types of electrodes?
a. 4 AWG
b. 4/0 AWG
c. 250 kcmil
d. 2/0 AWG
In compliance with Section 250.66, the size of the grounding electrode conductor at the service at each building or structure is based on the size of the largest ungrounded service-entrance conductor or equivalent area for parallel conductors. However, if the grounding electrode conductor connected to concrete-encased reinforcing steel of a building or structure does not extend on to other types of electrodes that requires a larger size conductor, Section 250.66(C) permits the grounding electrode conductor to be not larger than 4 AWG copper.

250.66(C) - Does not give me the answer of 4AWG... I'm unclear how they came to this conclusion.
 
M achew said:
How many amps does a 3-hp, 240v, 3-Phase motor draw if the motor has 90% efficiancy & a power factor of 95%?

I started with looking up the FLA = 9.6 in table 430.250

Then 9.6 / 75% = RLA (7.2)

Next 7.2 x 90% x 95% = 6.156 amps

The correct answer was 6.3

I know I'm close, but where am I off?
Click to expand...
Here is what I get! 746*3 =2238/(240*1.732*.90*.95) =6.3 amps
 
HEYDOG said:
Sorry I repeated below!
Click to expand...
It's all good more info the better.
When he said PSI figure they were using the 746 watt for one hp method.

If it does not say use table use 746
Many of there questions on the J test use 746. At least that is what I have noticed.
Good luck on your test M achew
 
Tulsa Electrician said:
It's all good more info the better.
When he said PSI figure they were using the 746 watt for one hp method.

If it does not say use table use 746
Many of there questions on the J test use 746. At least that is what I have noticed.
Good luck on your test M achew
Click to expand...
300 questions in 2 days... My head is exploding with AC juice!
 
M achew said:
300 questions in 2 days... My head is exploding with AC juice!
Click to expand...
The first part of my PE license (Fundamentals of Engineering) was 400 questions in 8 hours. The second part was 8 questions in 8 hours.

Of course this was decades ago, the process is entirely different now.
 
M achew said:
How many amps does a 3-hp, 240v, 3-Phase motor draw if the motor has 90% efficiancy & a power factor of 95%?

I started with looking up the FLA = 9.6 in table 430.250

Then 9.6 / 75% = RLA (7.2)

Next 7.2 x 90% x 95% = 6.156 amps

The correct answer was 6.3

I know I'm close, but where am I off?
Click to expand...
NEC full load current tables in art 430 are pretty much worst case efficiency and power factor ratings one might find in the field and are intended primarily for selecting conductor size.

If you know actual efficiency, power factor and rated voltage (all based on motor being loaded to 100% output rating) then you should be able to calculate actual rated current based on 746 watts per horsepower and applying the efficiency and power factor to the mix.

If your test question says it is rated 240 volts then that likely is what you should be using. Real world that motor likely has a utilization voltage rating of 230 though.

W = Hp x .746/efficiency = VA x PF VA = V x A x 1.732

Everything you need to calculate this answer is in the line above Find W first with known figures then find VA, then calculate amps from VA.
 
kwired said:
…
If your test question says it is rated 240 volts then that likely is what you should be using. Real world that motor likely has a utilization voltage rating of 230 though. …
Click to expand...
But even if the motor nameplate says 230V, that’s just the basis of the motor design. If the actual voltage is 240, that’s always what you would use in a calculation.

You beat me to the punch on the other points though. This method is over used out in the wild because it is ASSuming that a 3HP motor is always putting out 3HP. The motor only puts out what the load requires and only draws what it needs for that, plus any internal losses. That “746” formula is ASSuming that the motor is fully loaded and the eff and PF are at those exact values. In the real world none of that is going to be so exacting. Eff. and PF change with loading so if the motor is not mechanically putting out exactly 3HP at the moment in question, the 6.3A value is not going to be exactly that.

The reason I bring that up is because I see a lot of people question the readings they take from running motors and think there is something wrong, because they use the FLC chart or the “746” formulas and come up with different values, then question why.
 
Jraef said:
But even if the motor nameplate says 230V, that’s just the basis of the motor design. If the actual voltage is 240, that’s always what you would use in a calculation.

You beat me to the punch on the other points though. This method is over used out in the wild because it is ASSuming that a 3HP motor is always putting out 3HP. The motor only puts out what the load requires and only draws what it needs for that, plus any internal losses. That “746” formula is ASSuming that the motor is fully loaded and the eff and PF are at those exact values. In the real world none of that is going to be so exacting. Eff. and PF change with loading so if the motor is not mechanically putting out exactly 3HP at the moment in question, the 6.3A value is not going to be exactly that.

The reason I bring that up is because I see a lot of people question the readings they take from running motors and think there is something wrong, because they use the FLC chart or the “746” formulas and come up with different values, then question why.
Click to expand...
Agree. Most people working with me always make that same error, regularly. I have to explain it at length about it. It would be very helpful if the actual load possible on a piece of certain equipment is known in advance so that designers can size things properly (load factors, duty, etc). But, to make quicker decisions, most will just use the FLA and start from there. These things do well if the company has deep pockets. OTOH, if you are preparing for bids, you need to be very accurate to keep the bill of materials low and win bids.
 
Jraef said:
But even if the motor nameplate says 230V, that’s just the basis of the motor design. If the actual voltage is 240, that’s always what you would use in a calculation.

You beat me to the punch on the other points though. This method is over used out in the wild because it is ASSuming that a 3HP motor is always putting out 3HP. The motor only puts out what the load requires and only draws what it needs for that, plus any internal losses. That “746” formula is ASSuming that the motor is fully loaded and the eff and PF are at those exact values. In the real world none of that is going to be so exacting. Eff. and PF change with loading so if the motor is not mechanically putting out exactly 3HP at the moment in question, the 6.3A value is not going to be exactly that.

The reason I bring that up is because I see a lot of people question the readings they take from running motors and think there is something wrong, because they use the FLC chart or the “746” formulas and come up with different values, then question why.
Click to expand...
I think there is enough exactness as a general rule, aside from maybe some cheap Chinese made motors, though that if motor nameplate states 3 hp, 230 volts, 90% efficiency 95% power factor and 6.3 amps, that all those figures will be very close to accurate if your input is 230 volts and the driven load is actually 3 HP.

What I run into in more recent years is improper selection of overload settings after adding power factor correction capacitors to a motor circuit. Not many really understand how to calculate new motor FLA at the corrected PF values.
 
kwired said:
...
What I run into in more recent years is improper selection of overload settings after adding power factor correction capacitors to a motor circuit. Not many really understand how to calculate new motor FLA at the corrected PF values.
Click to expand...
I cringe at the thought of all those motors with no FLA correction.
It wasn't so bad when I could amp clamp the motors with the customer standing there. At least they could see the difference at that moment in time.
 
M achew said:
Where an ac electrical service is supplied with four (4) parallel sets of size 500 kcmil aluminum ungrounded conductors, what is the MINIMUM size copper grounding electrode conductor required when connected to the concrete-encased reinforcing building steel that does not extend on to other types of electrodes?
a. 4 AWG
b. 4/0 AWG
c. 250 kcmil
d. 2/0 AWG
In compliance with Section 250.66, the size of the grounding electrode conductor at the service at each building or structure is based on the size of the largest ungrounded service-entrance conductor or equivalent area for parallel conductors. However, if the grounding electrode conductor connected to concrete-encased reinforcing steel of a building or structure does not extend on to other types of electrodes that requires a larger size conductor, Section 250.66(C) permits the grounding electrode conductor to be not larger than 4 AWG copper.

250.66(C) - Does not give me the answer of 4AWG... I'm unclear how they came to this conclusion.
Click to expand...
Still need an ansawer for this one?
 
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