Converting Motor FLC to VA

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finhead

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Can someone explain to me why the nominal voltages listed in 220.5 (A) are used insted of the voltages specified in Tables 430.248 and 430.250 when converting motor FLC to VA?

Section 220.5 (A) specifically states that "unless other voltages are specified ...." Seems to me that in this case other voltages are specified.

Thanks
 
finhead said:
Can someone explain to me why the nominal voltages listed in 220.5 (A) are used insted of the voltages specified in Tables 430.248 and 430.250 when converting motor FLC to VA?
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They are not. For example, looking up a 1HP motor in Table 430.250, you get a current of 8.4 amps. To get the VA, you multiply 8.4 by 115 (not 120), and you get 966VA (not 1008 VA).
 
I disagree. Based on Example D3(a) in Annex D and Page 110 of the NEC handbook, voltages from the motor tables are not used. Now I know that these sources are not officially part of the NEC, but really, how much more authoritative can you get?
 
I also have to disagree, simply mlutiplying the current times voltage will give you watts, not VA.
A problem occurs because the power factor is not known, therefore properly determining VA by using current and voltage, or Hp to watts still requires some missing information to get ot VA. This is information is not provided in the NEC, nor is it explained what pf is assumed to get the values in the Table, something I have always hard a problem with.

Regardless of what pf you assume to get the VA,
 
I believe the values of current in the tables already take into account both motor efficiency and power factor. All you need to do to get VA is to multiply a tablulated value of current times a value of voltage. The original question being whether you use 115 or 120 is a matter of some debate. Anyone else want to weigh in on that question?
 
The answer is simple. Because that is assumed to be the load on the feeder or service.

Although your motor is rated at 460 volts, is is assumed that 20 volts are dropped across the feeder and branch circuit conductors. So for the 7.5HP motor, the VA's used at the motor are 460 * 11 * sqrt(3). But at the service, the VA's used are probably closer to 480 * 11 * sqrt(3).

By using the higher voltage, we are adding in the circuit losses.

Steve
 
I suppose that makes sense. But let's check the math:

480 times 11 times 1.732 is about 9145 VA (amount supplied by service).
460 times 11 times 1.732 is about 8764 VA (amount used by motor).
The difference is about 381 VA (amount lost along the conductors).

Looking at the conductors, we see 11 amps flowing, and 20 volts being dropped.

20 times 11 times 1.732 is about 381 VA.

QED.

That's one ATTABOY to Steve66!
 
kingpb said:
I also have to disagree, simply mlutiplying the current times voltage will give you watts, not VA.
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You've got that backwards. Multiplying Volts (V) times Amps (A) gives you VA. Multiplying Volts times Amps times the Power Factor gives you watts.
 
I think it was Albert Einstein (or maybe Mick Jagger) who said; "for every problem, there is a simple, logical explanation - that is wrong."

Steve's explanation and Charlie's mathematical verification depend on the assumption that the code makers have ignored voltage drop (except for FPNs) throughout the entire code, but mysteriously built it into the motor tables.

As Example D1(b) in the Annex points out, there is no hard and fast rule that VA must be used to calculate a load. In the case where are amperes are used insted of VA, the benefit of the extra VA would be lost.

Correct me if I'm wrong on this point, but I've always believed that if the voltage at the motor goes down, it will draw more current. If this is the case, it is not accurate to use the same current value for different voltage values to calculate VA.

Something else I've always thought is that the voltage values in the motor tables were purposely lower that the nominal voltages or 220.5 in order to increase the motor FLCs. This is done to ensure that the tables will accomodate the FLCs of a variety of motors.
 
finhead said:
As Example D1(b) in the Annex points out, there is no hard and fast rule that VA must be used to calculate a load. In the case where are amperes are used insted of VA, the benefit of the extra VA would be lost.
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But the amps stay the same at all points throughout the system. Our motor that draws 11 amps on its branch circuit also draws 11 amps through the service conductors. We only have to consider the losses when we are using VA's or watts.

finhead said:
Correct me if I'm wrong on this point, but I've always believed that if the voltage at the motor goes down, it will draw more current. If this is the case, it is not accurate to use the same current value for different voltage values to calculate VA.

Something else I've always thought is that the voltage values in the motor tables were purposely lower that the nominal voltages or 220.5 in order to increase the motor FLCs. This is done to ensure that the tables will accomodate the FLCs of a variety of motors.
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I agree with most of that. But I think the tables give conservative values for the currents. Most of the time you can use these currents and you don't have to worry about the exact voltage at the motor. (This may not hold if you have a really low voltage, but you will probably have other problems then. Like motors overheating and going bad, etc.)
 
sparkie001 said:
You've got that backwards. Multiplying Volts (V) times Amps (A) gives you VA. Multiplying Volts times Amps times the Power Factor gives you watts.
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Oops, thanks sparkie for pointing that out. What I was trying to say was that the Hp, cannot be converted to VA without knowing a pf.

Using the current and voltages used in the 430 tables, and working backwards, do not provide you with any realistic assumption of a pf.
 
finhead said:
Can someone explain to me why the nominal voltages listed in 220.5 (A) are used insted of the voltages specified in Tables 430.248 and 430.250 when converting motor FLC to VA?

Section 220.5 (A) specifically states that "unless other voltages are specified ...." Seems to me that in this case other voltages are specified.

Thanks
Click to expand...

I think part of the confussion may be motors are an inductive load that is rated in KW, where a resistive load ,such as a heating eliment or a light bulb is a resistive load rated in va. Inductive loads cause the current to lag the voltage "the phase shift" a resistive load doesn't cause a phase shift and is va rated.
 
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