I am not getting the correct results for horse power.
An online catalog for a 3phase 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?
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 threephase 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 snippedforprivacy@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
formatting link
The motor is their item number: 102363
The direct link to the PDF is:
formatting link
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 crossreference 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: snippedforprivacy@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 threephase 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.
 I am not getting the correct results for horse power.

 An online catalog for a 3phase 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 (LL) 230 volt wye (LN)? Is the current that
for the individual winding leads, or for the supply circuit?
snippedforprivacy@ipal.net wrote in news: snippedforprivacy@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
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 singlephase 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 snippedforprivacy@shawcross.ca
remove the X to answer
 snippedforprivacy@ipal.net wrote in news: snippedforprivacy@news3.newsguy.com:

> > I am not getting the correct results for horse power.
>
> An online catalog for a 3phase 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 (LL) 230 volt wye (LN)? 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 ScottT 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 nonexistant. 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 linetoline 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 halfcycle. 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 ScottTee, 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 manyplace answers derived from
twoplace accuracy inputs.


You also had to make estimates so that you knew whether your solution was
in the ball park. 
Don Kelly snippedforprivacy@shawcross.ca
remove the X to answer
"daestrom" wrote in
news: snippedforprivacy@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
threephase motor will stay on line, as that is a "singlephasing"
condition. A high leg system still has normal lineline 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:46dnawzyKaOcxLVnZ2dnUVZ snippedforprivacy@earthlink.com:
From my understanding, 550VAC threephase motors were common in the old
days. Any new motors have been rewired by a proffesional service that
specializes in this. They take 460VAC threephase motors and wire them for
550VAC threephase.
The building has 460VAC entering and going into a stepup 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 setup. 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 LG from each phase, not LL, 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 LL
voltage, with respect to ground.
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