# power of electric motor



## guitarboyled (Jun 22, 2009)

I found this for NEMA motors:

Electrical motors constructed according NEMA Design B must meet the efficiencies below:

​ 

Power (hp) Minimum Nominal Efficiency​

1 - 4 ---------------- 78.8
5 - 9 ---------------- 84.0
10 - 19 ---------------- 85.5
20 - 49 ---------------- 88.5
50 - 99 ---------------- 90.2
100 - 124 ---------------- 91.7
> 125 ---------------- 92.4​


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## raider1 (Jan 22, 2007)

The efficiency and power factor will be marked on the motor nameplate. 

When sizing motor circuit conductors, motor short circuit and ground fault protection, disconnect and starters you must use the motor ampacity values given in the motor current Tables in Part XIV of Article 430 in the NEC.

Chris


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## guitarboyled (Jun 22, 2009)

Thanks Chris,

Every answer creates more questions lolllll 

What is the ampacity value.


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## guitarboyled (Jun 22, 2009)

Also what do I need to know about torque and startup power?

http://www.dolphins-software.com/motorStartup.htm


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## raider1 (Jan 22, 2007)

guitarboyled said:


> Thanks Chris,
> 
> Every answer creates more questions lolllll
> 
> What is the ampacity value.


The tabular ampacity value given in the Tables in Part XIV of Article 430 are the worst case full load amperes for any given HP motor. These values assume a very poor power factor and effeciency. Using these values when sizing the conductors etc.. will ensure that whatever motor is installed or replaced will have sufficent sized conductors and equipment in place.

Chris


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## raider1 (Jan 22, 2007)

guitarboyled said:


> Also what do I need to know about torque and startup power?
> 
> http://www.dolphins-software.com/motorStartup.htm


During motor start up the in rushing current can be 6 to 8 times the running load current.

So for example if I have a motor that has a full load current of 20 amps, for a short duration when the motor is starting the current draw can be up to 160 amps.

This is the reason that for motors the motor branch circuit, short circuit and ground fault protection can be sized between 175% and 800% depending on the type of overcurrent protection that is being used. This information is found in Table 430.52 in the NEC.

Chris


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## guitarboyled (Jun 22, 2009)

Great explanation Chris thanks

I downloaded the NEC and this Volt software











It seems the power factor is not accounted for in induction motors. I started reading about the difference between induction motors and synchronous AC motors, pretty complexe stuff. What do I really need to know?

Values from the software seem to concur with the following table

http://www.klocknermoeller.com/dilm/hp2.htm


Power (W) = 1.73 * volts * amps

For a 1 HP motor

736W / (1.73 * 208v) = 2.04A

Why does the table and software say it’s 4.6A

Also, when I try to get a full load amperes for the equivalent synchronous motor I get a message: The specified horsepower is below the [email protected]’s range.


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## raider1 (Jan 22, 2007)

> Why does the table and software say it’s 4.6A


Because your formula does not take into account power factor and efficiency.

Chris


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## guitarboyled (Jun 22, 2009)

Power (W) = 1.73 * volts * amps * power factor * efficiency

AMPS = 746 W / (1.73 * 208v * 0.8 * 0.788) = 3.3A

Power factor = 0.8 (80%)
Efficiency = .788 (78.8%) NEMA minimum 

Still far from the 4.6A value


For a 10 HP motor the same formula results in almost the given value of 30.8A

7460W / (1.73 * 208v * 0.8 * .855) = 30.3A


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## guitarboyled (Jun 22, 2009)

Just found this, interesting reading:

Perhaps the greatest confusion arises due to the fact that early in our science educations, we were told that the formula for watts was amps times volts. This formula, watts = amps x volts, is perfectly true for direct current circuits. It also works on some AC loads such as incandescent light bulbs, quartz heaters, electric range heating elements, and other equipment of this general nature. However, when the loads involve a characteristic called inductance, the formula has to be altered to include a new term called power factor. Thus, the new formula for single phase loads becomes, watts are equal to amps x volts x power factor. The new term, power factor, is always involved in applications where AC power is used and inductive magnetic elements exist in the circuit. Inductive elements are magnetic devices such as solenoid coils, motor windings, transformer windings, fluorescent lamp ballasts, and similar equipment that have magnetic components as part of their design.
Looking at the electrical flow into this type of device, we would find that there are, in essence, two components. One portion is absorbed and utilized to do useful work. This portion is called the real power. The second portion is literally borrowed from the power company and used to magnetize the magnetic portion of the circuit. Due to the reversing nature of AC power, this borrowed power is subsequently returned to the power system when the AC cycle reverses. This borrowing and returning occurs on a continuous basis. Power factor then becomes a measurement of the amount of real power that is used, divided by the total amount of power, both borrowed and used. Values for power factor will range from zero to 1.0. If all the power is borrowed and returned with none being used, the power factor would be zero. If on the other hand, all of the power drawn from the power line is utilized and none is returned, the power factor becomes 1.0. In the case of electric heating elements, incandescent light bulbs, etc., the power factor is 1.0. In the case of electric motors, the power factor is variable and changes with the amount of load that is applied to the motor. Thus, a motor running on a work bench, with no load applied to the shaft, will have a low power factor (perhaps .1 or 10%), and a motor running at full load, connected to a pump or a fan might have a relatively high power factor (perhaps .88 or 88%). Between the no load point and the full load point, the power factor increases steadily with the horsepower loading that is applied to the motor. These trends can be seen on the typical motor performance data plots which are shown in figure 1.










Is it possible to generalize this graph to most motors? Is it for single phase? Efficiency and Power Factor seem pretty low compared to other values I came across*.*


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## guitarboyled (Jun 22, 2009)

*Found the specs for the following motors*

*MTC-001-3BD18*
*Power: 1 HP*
*Voltage: 208-230/460V*
*Current at 230V/460V : Full load 3.0A / 1.5A No Load 1.9A / 0.95A*
*FL RPM : 1760*
*FL Power Factor 0.71*
*FL Efficiency 82.5%*

*746W / (1.73 * 230v * .71 * .825) = 3.2A*

*MTC-100-3BD18*
*Power: 100 HP*
*Voltage: 208-230/460V*
*Current at 230V/460V : Full load 230A / 115A No Load 72A / 36A*
*FL RPM : 1785*
*FL Power Factor 0.87*
*FL Efficiency 94.2%*

*74 600W / (1.73 * 230v * .87 * .942) = 229A*

*Formula works out in both cases*


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## raider1 (Jan 22, 2007)

guitarboyled said:


> Power (W) = 1.73 * volts * amps * power factor * efficiency
> 
> AMPS = 746 W / (1.73 * 208v * 0.8 * 0.788) = 3.3A
> 
> ...


Correct, but the ampacity values that you get in the motor tables are the worst case scenario.

If I use a PF of 70% and an efficiency of 70% I come up with 4.2 amps

Chris


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## guitarboyled (Jun 22, 2009)

Therefore even if I have the motor specs, I’m better off using the NEC tables for sizing. The values are more restrictive therefore safer.


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## raider1 (Jan 22, 2007)

guitarboyled said:


> Therefore even if I have the motor specs, I’m better off using the NEC tables for sizing. The values are more restrictive therefore safer.


Correct, for sizing the motor circuit conductors, short circuit and ground fault protection, disconnecting means etc.. the NEC requires that you use the NEC table values instead of the actual motor nameplate values. The only thing the NEC permits you to use the nameplate value for is sizing the motor overload protection.

Chris


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