Motor HP Equation Breakdown

[JHill161]

Gents, I saw a previous post on this exact question and a members result which was the correct answer however I am trying to understand how this was derived.
The question:
What is the efficiency and power factor of a 1HP motor, single phase, 208V motor?
A. 80.0% B. 40.7% (correct answer) C. 38.9% D. 36.8%.

The original poster was confused how to get to this answer and a response was as Follows.
"My guess would be that they took (746÷208)÷8.8=0.40756. The 8.8 comes from the full load current found in Table 430.248 for 1 hp motor at 208 volts."

My question is, how is the 746w (watts for 1HP) known to be divided by 208v then by FLC from table 430.248 8.8A? I have never seen this equation that I can recall but I have not had to study for a test in a decade.

Can someone explain that portion or show me where in the code it is that tells us we take the watts/ea HP and divide by voltage?
It just felt like this question was missing some info, which was also the response of the majority of the members, it was missing something.

I am just trying to understand if this is common knowledge I Just wasn't aware of or a wild guess that made the equation work since the one who responded had the answer of 40.7% already.
Thanks!

 

[ptonsparky]

Gents, I saw a previous post on this exact question and a members result which was the correct answer however I am trying to understand how this was derived.
The question:
What is the efficiency and power factor of a 1HP motor, single phase, 208V motor?
A. 80.0% B. 40.7% (correct answer) C. 38.9% D. 36.8%.

The original poster was confused how to get to this answer and a response was as Follows.
"My guess would be that they took (746÷208)÷8.8=0.40756. The 8.8 comes from the full load current found in Table 430.248 for 1 hp motor at 208 volts."

My question is, how is the 746w (watts for 1HP) known to be divided by 208v then by FLC from table 430.248 8.8A? I have never seen this equation that I can recall but I have not had to study for a test in a decade.

Can someone explain that portion or show me where in the code it is that tells us we take the watts/ea HP and divide by voltage?
It just felt like this question was missing some info, which was also the response of the majority of the members, it was missing something.

I am just trying to understand if this is common knowledge I Just wasn't aware of or a wild guess that made the equation work since the one who responded had the answer of 40.7% already.
Thanks!

Not in the Code book that I am aware of.
More of just education.
I was lacking.

 

[JHill161]

Ya I feel like I Should know this but we spend so many years working with plans an engineer drew up and installing never using this stuff. Now that I am doing a lot of design build and industrial I have working with them more and find this side really interesting. Seeing the responses from members on the other thread he certainly wasn't the only one stumped by this. I guess I have never really heard that formula if you care to call it that to divide watts/hp by volts and then by amps for efficiency. The main thing for me here is, it says efficiency AND power factor. Power factor has some 'typical' so the question is a little ambiguous, to me anyways.

 

[jim dungar]

The biggest thing to remember is that motors are rated in output HP, while FLA are based on the input.

You must include the efficiency of the motor when using the 746W/HP to calculate input amps. Also don't forget the power factor.

Then there is the issue with the NEC tables currents are purposely higher than anything you expect to see in the field.

 

[garbo]

For around 5 HP & higher motors and common plain Jane NEMA 2 & 4 pole motors I use 1,000 wats per HP. Only use it for quick calculation just using my head. For a 100 HP motor comes out about 4% lower then actual figure. Another quickie that I do in my head for 480 volt 3 phase motors just multiply the HP by 1.2 to get a figure that is also around 4% lower then actual FLA motor amps. Example 100 HP motor times 1.2 = 120 amps and several 100 HP 4 pole motors that I recently worked on had a FLA of 124 amps. Of course now that everybody has a calculator on thier cell phones and handy apps can get more accurate #'s within a minute. And yes I still have my slide rule that I last used in the 1960's.

 

[jim dungar]

And yes I still have my slide rule that I last used in the 1960's.

I was taught the 1.2A/HP estimate back in the 70's can't remember using it much after I started using the Square D Motor Data Calculator/slide rule.

While I still have several standard slide rules, I miss my circular one that could do polar to rectangular conversions. I also still have a CRC Math reference book on my shelves.

 

[wwhitney]

Another quickie that I do in my head for 480 volt 3 phase motors just multiply the HP by 1.2 to get a figure that is also around 4% lower then actual FLA motor amps.

Let's see,

Efficiency = Output Power / Input Power
Output Power = HP * 746 Watts / HP
Input Power = Voltage * sqrt(# phases) * Current * Power Factor

So the rule of thumb implies, at full load:

Efficiency = 746 / ( 480 * sqrt(3) * 1.2 * 1.04 * Power Factor )
Efficiency * Power Factor = 0.72

Then since both Efficiency and Power Factor are at most 1, the rule of thumb implies both typical efficiency and typical power factor (at full load) are at least 0.72, but it doesn't tell us either one specifically. If full load power factor is 0.85, then full load efficiency would also be 0.85.

Does this sounds plausible, and are there any errors above? Maybe I should use 460V instead of 480V?

Cheers, Wayne

 

[jim dungar]

Does this sounds plausible, and are there any errors above? Maybe I should use 460V instead of 480V?

The input 1.2A/HP was a guesstimate that already considered PF and Eff. It was used instead of, not in conjunction with, the 746W/HP relationship.

General calculations should be used with nominal values, like 480V.

 

[wwhitney]

The input 1.2A/HP was a guesstimate that already considered PF and Eff.

Understood, but I was back-calculating what assumed PF and Eff at full load must have been used in coming up with that guesstimate. And I couldn't pin down the PF and Eff individually, just that their product had to be 0.72.

So e.g. if assume both PF and Eff at full load are 0.85 (the square root of 0.72), then we get a 1.25A/HP rule of thumb (the 1.2A rule of thumb correctly by the 4% @garbo mentioned).

Cheers, Wayne

 

[jim dungar]

Understood, but I was back-calculating what assumed PF and Eff at full load must have been used in coming up with that guesstimate. And I couldn't pin down the PF and Eff individually, just that their product had to be 0.72.

So e.g. if assume both PF and Eff at full load are 0.85 (the square root of 0.72), then we get a 1.25A/HP rule of thumb (the 1.2A rule of thumb correctly by the 4% @garbo mentioned).

Cheers, Wayne

Actually the 1.2A was most applicable to 100 HP and larger. It looses it applicability as the HP drops, that is why it is more of a guess than an estimate.