Calculating 3 phase AC motor HP

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.
Does anyone know why this is?

Steve wrote:
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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.

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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 www.surpluscenter.com.
The motor is their item number: 10-2363
The direct link to the PDF is:http://www.surpluscenter.com/pages/122.pdf
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 :
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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.

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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 :
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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.

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--------------- 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.

Steve wrote:
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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.

| 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 wrote:
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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?

Steve wrote:
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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 http://bmillerengineering.com/elecsys.htm and scroll down to the last few diagrams.

| snipped-for-privacy@ipal.net wrote in 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 :-)

snipped-for-privacy@ipal.net wrote:
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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".
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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.

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remove the X to answer

|> 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. | | 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.
A Scott-Tee can give you the same three phase system as you can get with a delta where one side is center tapped. But unliked closed delta, it nedes only 2 transformers. One is the same (a 120/240V center tapped secondary) with the primary wired to one phase (A-N) and the other is different (208V) with the primary wired L-L to the other phases (B-C). The secondary of the 208V transformer is wired directly to neutral and provides the high leg. There is some current rating limitation on a Scott-T.
Just because there are only 2 transformers does not mean you have to limit your view of the system as being 2 phase.
| This is not used much anymore for power, but the equivalent is often | used in servo circuits to convert "synchro" feedback into "resolver".
You can even get Scott-T in a dry-type transformer in a 3R box.
|> 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 :-) | | 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.
I know how to use a slide rule. It has no means to reduce the precision. It just has a precision limit built in.
The higher precision is used in a mathematical theory sense. It is NOT used with low precision measured values. If I measured the voltage as 122 volts L-N then I could say the volts are 244 L-L on single phase or 211 L-L on three phase. Maybe you should watch when and where and how I use higher precision and thus see how I actually apply it. Note that one way I will sometimes apply it is in humorous anecdotes. But the usual way it is applied is when mathematical comparisons are involved. I do math with high precision on all intermediate steps. That translates whatever precision and accuracy I have in the input directly to the output undistorted. One thing do not let happen is the effort of the choice of number base system influence rounding.

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wrote:
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---------------- You can use high precision in the intermediate steps but if you start with a lower precision, after the intermediate steps, in the final answer, you should round off to the original precision. We all do that as long as it doesn't need re-entry of multi-digit numbers. If you want to emphasise a higher precision mathematically then maybe you should start with 120.0000000/207.8460970 and come up with an answer to the same number of digits (less a few in multiple calculations). If you do this to bug engineers, fine- as they aren't the ones having to do the extraneous and meaningless typing.
We have been here before and likely neither of us will change :)

| You can use high precision in the intermediate steps but if you start with a | lower precision, after the intermediate steps, in the final answer, you | should round off to the original precision. We all do that as long as it | doesn't need re-entry of multi-digit numbers. If you want to emphasise a | higher precision mathematically then maybe you should start with | 120.0000000/207.8460970 and come up with an answer to the same number of | digits (less a few in multiple calculations). If you do this to bug | engineers, fine- as they aren't the ones having to do the extraneous and | meaningless typing. | | We have been here before and likely neither of us will change :)
Not that I see. I do things both ways, but I think some people just cannot see the distinction between when I do high precision, and when I do reduced precision (to match the input) and ... a third way: when I use labels that happen to be numbers. I often say "347 volts" when what I mean is a system found mostly in Canada based on 600 volts. Dividing 600 by sqrt(3) gives me 346.41016151377545870548926830117447338856105076207612561116139589038660338176 which when rounded gives me 346, not 347. So 346 is the rounded result and 347 is the label. Fortunately in most cases things are the same. But a lot of people still use the label "220" while I use 240 as the label, as well as the rounded result of a defined standard.
FYI, I did not type that high precision result in. I used copy-and-paste.

snipped-for-privacy@ipal.net wrote:
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You might be interested in a calculator called "bc", if you have not already discovered it. It was developed for Unix and has been available in Linux for some time. It has recently been made available for Windows, for those that prefer that operating system:
http://gnuwin32.sourceforge.net/packages/bc.htm
It uses arbitrary precision and can work in almost any number base. The built-in math library, in addition to the normal trig functions has bessel functions and logs:
http://www.gnu.org/software/bc/manual/bc.html
I haven't used the Windows version, but bc in Linux is excellent if somewhat specialized.

snipped-for-privacy@large.invalid says...
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Does it do fractional arithmetic in bases other than 10? Most calculators don't seem to think this is important. I have a testbench I need to verify this week (my last week ;) and placing the binary point by hand is a PITA.
<snip>

krw wrote:
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Take a look at the manual, (in several formats):
http://www.gnu.org/software/bc/manual/bc.html
It's also programmable in a "C" like language.
It is CLI only, but that makes it easy to copy and paste the results you need.