Calculating 3 phase AC motor HP

Yes, we do know why. On the assumption that this might be a homework problem, and following the "teach a man to fish" principle, I will tell you that you did not calculate the three-phase VA properly, and you ignored one other multiplication factor. When you fix all of that, it comes out very close.

"Ben Miller" wrote in news:BLydndBRUqYVGhbVnZ2dnUVZ snipped-for-privacy@earthlink.com:

Sorry, you're asssumption of a homework problem is incorrect. The motor came from a PDF I found (through Google) last week from the company located at

The motor is their item number: 10-2363

The direct link to the PDF is:

You're welcome to reference the site and that item number and then inquire if my specifications were from a homework problem or not.

What is this thing about homework problems anyway?

Let's say for a moment it is. I performed my calculations and asked a question directly towards why my answer is incorrect.

A friend works with three phase motors and has one blowing fuses. I asked about the specs located on the plate and couldn't calculate the correct hp. I found this PDF and tried doing a cross-reference without success.

Now I've resorted to asking the question for an accurate answer and a better understanding.

TIA

Steve expressed precisely :

You enter root 3 into the equation (I*A*sqrt3). Now that is only for pure resistive power. Motors need to have the power factor entered into the equation. If you assume a PF of 0.8 then it will be very close to the posted figure.

Arlowe wrote in news: snipped-for-privacy@gmail.com:

I thought the efficency came from the PF of 0.8? The efficency comes from the phase angle between voltage and current or am I incorrect?

Steve presented the following explanation :

Power Factor is simply true power (watts) divided by apparent power (V*I). Efficency is basicaly true power in divided by true power out. The two are not really related.

The motor you have described is an inductive load on the circuit. Inductive loads cause the current to lag out of phase with the voltage. This causes the line current to increase even though the motor is in reality using less than line current. This can be proven with a wattmeter. The ratio of "line current * line voltage" to "actual current used * line voltage" gives you power factor.

So since the motor isn't using any extra power , the PF does not effect the efficency ratio of the motor. However it does effect the loading on the cables so the line current will posted as the motors running current.

Steve People do pose homework questions on this group, looking for quick answers, basically out of laziness. As originally posted, your question had all of the earmarks. Had you explained it as you just did, then there would have been no problem. And please note that I did answer your question "Does anyone know why this is?", including pointing you at the reasons.

As Arlowe explained, you multiply by 1.73 for three-phase calculations. You also needed to multiply by the power factor. Since the motor specs don't give it, you estimate it. For typical motors of that size, 0.8 is a good estimate, and gets you very close to the rated horsepower.

You are incorrect: efficiency and power factor are two different things Efficiency is the output power/input real power The input real power = output power + electrical and mechanical losses =VI*pf

Power factor is the relationship [real power(KW)]/ [apparent power VI (KVA)] or the cosine of the angle between phase voltage and phase current.

here is an example: The output of a motor is 7.5HP or 5.55KW For an efficiency of 85% the input real power is 5.55/0.85 =6.5 KW including both mechanical and electrical losses. You have an input of 230*20* root(3) =8 KVA *NOT* KW The power factor is 6.5/8 =0.82 (lagging for induction motors).

As Ben said, 0.8 pf is a good estimate.

On Sat, 26 Jul 2008 14:09:22 -0500 Steve wrote: | I am not getting the correct results for horse power. | | An online catalog for a 3-phase motor states: | | 7.5hp | 1755rpm | 230/460V | 20/10 A | Nonimal Effic: 89.5% | | | If I use the formula: hp = (V*I*Effic) / 746 | | I get 5.5hp.

Is this power 230 volt delta (L-L) 230 volt wye (L-N)? Is the current that for the individual winding leads, or for the supply circuit?

snipped-for-privacy@ipal.net wrote in news: snipped-for-privacy@news3.newsguy.com:

I'm not sure. It's a very old wiring system in an old factory. I heard it's something called a Bastard Leg but can't find much about that wiring system.

Have you ever heard of it?

That refers to a "High leg" system. It is a 240 volt delta, with one winding center tapped to ground. The high leg is 208 volts, rather than 120 volts to neutral. See

and scroll down to the last few diagrams.

Assuming the voltage and current numbers are for line voltage and current (usually are), the formula you want is:

hp = sqrt(3)*V*I*Effic*pf / 746

The term you're missing is 'power factor'. It is *not* the same thing as efficiency. If this data is accurate, then the power factor would have to be 0.78. That's not too bad for a small more such as this. For motors in this range a 'rule of thumb' power factor is between 0.8 and 0.9. For really large motors, the power factor can be as high as 0.95. For fractional single-phase motors I've seen it as low as 0.5

daestrom

---------------------- Such voltage and current ratings are always on the (external) terminal conditions because that is where one can take measurements - which are line current and line voltage rather than phase quantities. These happen to be the same for Y and delta (or a little old man in a black box measuring the voltage and output torque, with a lookup table to set the speed and current by cranking a variable impedance and pedalling madly at the same time- modern motors give him a computer rather than a lookup table).

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

On Mon, 28 Jul 2008 10:09:41 -0500 Steve wrote: | snipped-for-privacy@ipal.net wrote in news: snipped-for-privacy@news3.newsguy.com: | |> On Sat, 26 Jul 2008 14:09:22 -0500 Steve wrote: |>| I am not getting the correct results for horse power. |>| |>| An online catalog for a 3-phase motor states: |>| |>| 7.5hp |>| 1755rpm |>| 230/460V |>| 20/10 A |>| Nonimal Effic: 89.5% |>| |>| |>| If I use the formula: hp = (V*I*Effic) / 746 |>| |>| I get 5.5hp. |> |> Is this power 230 volt delta (L-L) 230 volt wye (L-N)? Is the current |> that for the individual winding leads, or for the supply circuit? |> | | I'm not sure. It's a very old wiring system in an old factory. I heard it's | something called a Bastard Leg but can't find much about that wiring | system. | | Have you ever heard of it?

Yes. It's 240V delta, or a Scott-T derived equivalent. One could also make a system equivalent enough for the purpose of this motor with 230/133 wye/star, but such systems are generally non-existant. It's definitely NOT a 400/230 or

416/240 wye/star system. That was what my question was trying to figure out.

Given that, the steps *I* use to figure this out is to convert the electrical system to its wye/star equivalent, which in this case is 230/133. I multiply the voltage of each leg relative to a (virtual) neutral (e.g. 133) times the current (20) times the phases (3) times efficiency in units (0.895) times the typical power factor (0.8), and _divide_ by 746 and get 7.66 HP.

Electrical power engineers generally do the arithmetic a little differently, multiplying just the delta or line-to-line voltage by the square root of 3 (1.732). They skip multiplying by number of phases because by multiplying by the square root of 3, that automatically includes all three phases. It's still effectively the same thing. I just worked out my steps from the science of three phase power as I learned it before I ever encountered the practiced formula (I did not study this in formal education). Engineers like shortcuts and their method is slightly shorter (less work, unless remembering the square root of 3 is considered work, which I have found is not the case) than mine.

The current ratings given for a motor is not an indication of power used, but rather of current drawn in the course of motor operation. Since the motor is partially inductive, it will store some energy in the field and "generate" it back towards the supply during part of the next voltage half-cycle. This is seen in the fact that the current phase is shifted relative to the voltage phase. The power factor is the ratio of the volts applied times current drawn divided by power used and wasted. The power factor on _this_ motor may be a tad bit lower than 0.8, or the actual current drawn may be a tad bit lower than 20A. And the usual utilization voltage listed for motors is 230/460V even though standard electrical supply in North America is 240/480V. Except in extreme cases, the accuracy of these numbers is good enough for engineering practice. It's good enough to select the proper motor and correctly rate the circuit needed to power it.

BTW, I sometimes use extreme precision of arithmetic when exploring scientific or mathematical theory related to electrical systems (or anything else I am studying). It drives the engineers nuts :-)

See Ben Miller's answer for the real circuit. It has nothing to do with a Scott-Tee, which is a way to get two phase power from a three phase input.

This is not used much anymore for power, but the equivalent is often used in servo circuits to convert "synchro" feedback into "resolver".

You need to learn how to use a slide rule for engineering calculations! That was all that was allowed when I took the Professional Engineer exam. It also prevents you from giving many-place answers derived from two-place accuracy inputs.

---------------- You also had to make estimates so that you knew whether your solution was in the ball park. ---

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

"daestrom" wrote in news: snipped-for-privacy@news4.newsguy.com:

The system in this building is 550volts and has three wires (obviously for different phases) and a ground.

If I understand correctly, the voltage was measured across two wires at a time. Two wires had 550V, another two had 550V, and the remaining two had nearly 0V.

I'm not sure why this could be but an old electrician was working on the system and informed his younger crew that it's normal for this type of system.

Does this make sense to anyone?

Steve You keep adding new information. These voltages do not make sense, and they would certainly cause the motor that you described to blow fuses (which you said was the problem). The motor is rated 230 or 460 volt, three phase, which will not run on 550 volts. Even if the voltage was normal, no three-phase motor will stay on line, as that is a "single-phasing" condition. A high leg system still has normal line-line voltage between all three pairs of wires. I can't help but think there is some information either missing or inaccurate, or the electrical system there is really messed up.

"Ben Miller" wrote in news:46-dnawzyKaOcxLVnZ2dnUVZ snipped-for-privacy@earthlink.com:

From my understanding, 550VAC three-phase motors were common in the old days. Any new motors have been rewired by a proffesional service that specializes in this. They take 460VAC three-phase motors and wire them for

550VAC three-phase.

The building has 460VAC entering and going into a step-up transformer to supply 550VAC. There is a mixture of rebuilt 460VAC motors and old standard

550VAC motors - this was easier than replacing every motor in the building.

The electricians crews measaured the conditions I mentioned above and the "old timer" said this is normal for this type of set-up. Another person I spoke to informed me it's called a 'Bastard Leg".

I apologize if I confused anyone or left out information. Some of the stuff I assumed was common knowledge for anyone familiar with this setup while the other parts were confusion at my end.

Thanks for everyones help so far.

Steve It sounds like they were measuring L-G from each phase, not L-L, and it is a

550V corner grounded delta. Then those numbers make sense. The line that is grounded measures zero, and the other two measure full L-L voltage, with respect to ground.