Replies

1. 
   foghorn | Feb 06, 2003 05:39pm | #1

   Many thanks for your knowledge and your tact.

   Regards

2. 
   dmt | Feb 06, 2003 06:19pm | #2

   An estimate of motor horsepower can be made by the formula: horsepower = (amperage x voltage) divided by 746 watts (the number of watts in 1 hp). For  example, a motor rated at 15A running on a 120V circuit would theoretically produce about 2.4 horsepower. However, in practice it doesn't work this way because a motor can't convert all of the electrical energy it uses into mechanical power. Some energy goes into magnetizing the rotor and windings. This is recognized by the use of a power factor, which is generally represented by a 20 percent loss, meaning the motor can use only 80 percent of its rating. In the example above, a closer estimate can be made by multiplying the result (2.4 hp) times 80 percent. The rating would then be 1.9 horsepower.

   Motor manufacturers test motors on a dynamometer. The horsepower is then determined with the formula: horsepower = (torque [in ft. lb.] x rpm) divided by 5,252 (constant). This then becomes the NEMA rating. NEMA standards for motors are based on the ability of a motor to deliver its nameplate rated horsepower continuously, 24 hours a day, under full load. Some motors are also time rated to represent the period of time a motor can deliver its rated horsepower without overheating.

   The service factor of an AC motor is a multiplier which, when applied to the rated horsepower, indicates a permissible horsepower loading which may be carried under the conditions specified for the service factor. Service factors are defined in NEMA (National Electrical Manufacturers Association) Standards Publication MG1 “Motors and Generators.” A condensed version of the MG1 Standard is on the NEMA website (http://www.nema.org) for free download.

   When the voltage and frequency are maintained at the value specified on the nameplate, the motor may be overloaded up to the horsepower obtained by multiplying the rated horsepower by the service factor shown on the nameplate, generally 1.0 or 1.15, which really means 100 or 115 percent).

   When the motor is operated at any service factor greater than 1, it may have efficiency, power factor, and speed different from those at rated load, but the locked-rotor torque and current and breakdown torque will remain unchanged.

   A motor operating continuously at any service factor greater than 1 will have a reduced life expectancy compared to operating at its rated nameplate horsepower. Insulation life and bearing life are reduced by the service factor load.

   Service factor is not generally an issue for non-industrial applications.

   Dan T.

1. 
   jgaiennie | Feb 06, 2003 06:39pm | #3

   DanT,

   Thanks for the informative post.  I feel like I now have a pretty clear understanding of service factor.

   >>  a motor can't convert all of the electrical energy it uses into mechanical power. Some energy goes into magnetizing the rotor and windings. This is recognized by the use of a power factor, which is generally represented by a 20 percent loss, meaning the motor can use only 80 percent of its rating.

   So is this the same as efficiency?  Or is it really called the power factor?

   jdg

1. 
   dmt | Feb 06, 2003 06:52pm | #4

   Power factor is not the same as efficiency. The NEMA standard establishes minimum and nominal efficiencies based on a 20% loss difference. The efficiency rating is based on the design of the motor and proven testing by the manufacturer.Dan T.

3. 
   TKanzler | Feb 06, 2003 06:54pm | #5

   Here's a diatribe I wrote for Breaktime (called myself TDKPE then), edited for brevity, that explains a lot of what's going on inside these dual-voltage single-phase motors typical of small WW machines.  Three-phase motors are similar, with six windings instead of two, and without the additional start winding and capacitor(s).  I just don't want to go through that whole excercise again, so I've copied and pasted.  The fourth paragraph is the beginning of the great "less filling/great taste", er, I mean, "faster starts/more efficient" debate. 

   The start winding on every motor I've ever seen that had a start winding (ie split phase, cap. start/induct. run, cap. start/cap. run) was wired in parallel with one run winding.  Wired for 115V operation, the start winding sees 115V, since all three windings are in parallel.  Wired for 230V, there's a shift from 0V at the mid-point (during starting only), since it is now a series-parallel configuration with one run winding in series with a run/start parallel winding pair.  The start/run pair will drop somewhat less than 115V due to the lower impedence.  I've measured this momentary voltage difference to ground, and I seem to remember it being in the 20V-30V range at it's peak, dropping to a volt or two after start.  In one experiment on a 1/2 hp motor starting a fairly high inertia load, the motor howled a bit during starting at 230V, but went nearly silent when I connected a neutral to the common point, thus locking each winding into 115V.  It didn't start any better, it just got quieter.  I thought that was interesting, though probably meaningless in any practical sense.

   I've attached a connection diagram and a schematic (sort of) for a 3 hp Baldor motor, but as a connection diagram it is not unique, nor is the motor itself otherwise unique.  The start portion of the circuit is in green to make it easier to visualize it not being connected after start.  A capacitor run motor has a run capacitor in parallel with the start capacitor, but wired around the start switch so that the "B" winding and run cap. remain in the circuit after starting.  These are used mostly on larger single-phase motors, with the run cap. creating a second phase, allowing more power output and higher efficiency from a smaller frame.  Sort of a 2-phase motor.

   I've also tossed in a sketch of two 60W light bulbs in series and parallel, as an analogy for how these small motors are connected.  Same voltage drop, same current, same power, just different line voltage and current.  The series case looks just like a load center and two branch circuits, if you connect the neutral.

   The core of the whole "faster starts, more power" perception of 230V operation is in how the speed/torque/efficiency/power factor curves are effected by deviations in voltage from namepate value.  The torque curve of a typical motor drops as the square of the voltage reduction from nameplate.  In other words, with a 10% drop in voltage (to 103V from 115V) at the motor terminals, locked-rotor torque, pull-up torque, and breakdown torque are reduced to 81% of full-voltage values.  At 20% voltage drop (92V, down from 115V), these values drop to 64%.  With startup current for induction motors ranging from 2 to 5 times full load current, voltage drop in branch circuit conductors can be substantial.  Doubling the voltage (and halving the current) has a huge effect on performance, all other things being equal.  Using a 1.5 hp motor as an example, with 50A locked-rotor current at 115V and 50' x 2 12 ga conductors (.162 ohm), that's an 8V drop, or about 7%, which is a drop in locked-rotor torque of about 13.5%.  Going to 230V, 25A and the same 12 ga wire, 4V dropped on 230V is about 1.7% for a locked-rotor torque reduction of only about 3.4%.  Same wire, much different performance.  Performance loss is even greater, obviously, with light extension cords and/or long wire runs.  Torque reductions are less pronounced as the motor speeds up and the current draw decreases/impedance increases (the start winding messes this up; three-phase is a smooth curve), but the perception is that the motor starts easier and "has more power".  But, of course, if a motor was wired to an infinitely stiff voltage source, 115V or 230V at the lugs, there will be no difference in performance.   My understanding of the NEMA MG-1 motor nameplate design voltage requirements of 115, 230,460V etc. as compared to the nominal supply voltages of 120,240,480V, etc. is that it is an effort to build some branch circuit voltage-drop-under-load compensation into the motor design to further reduce this effect.  An underloaded motor draws less current anyway, so a slight over-voltage is allowable and inconsequential, just as a light motor load under reduced voltage is OK.  As long as nameplate current is not exceeded. 

   It should be noted that, even though there is no performance difference when the motor leads are supplied with nameplate voltage (without sag), in actuality, you'd need 6 gauge copper at 115V to get exactly the same performance as 12 gauge at 230V, because you'd get the same percentage voltage drop that way.  But, aside from snappy starts and the ability to badly abuse a motor, when your electrician says it will work as designed when wired for the nameplate current (or the usually higher NEC requirements), well, he or she is right.  If it's bogging badly, you're abusing it, or the wiring is too long and/or light, or both.

   There is one notable difference between those Delta (and Delta is the only manufacturer I know of that does this) dual-horsepower motors and a standard motor, in that standard motors don't have the dual-hp ratings.  I'm not at all convinced they really are dual-hp, but rather they are 2 hp motors derated to 1.5 hp at 115V for reasons that may have to do with UL more than internal construction 17.2 FLA on a 15A plug maybe?)  It's telling that there's a service factor of 1.15 at 115V but only 1.0 at 230V.

   To address your questions: motors only produce to meet the load, so a 100 hp motor driving a table fan is only outputting 1/10 hp (or whatever) at the shaft.  Current draw varies with load, so an unloaded motor draws magnetizing and reactive current only (add a little for friction and air movement), but will increase current draw as load is increased.  Power factor and efficiency are poor at small loads, improve as load is increased, generally.  My 3 hp Unisaw only draws 4.4A no-load (belts, bearings, wind resistance for a load), but is nameplated 12.4A full-load.  You'd have to operate the saw for 100 hrs per month under full load, for your example.  By the way, that 4.4A is mostly reactive, so you don't pay for most of it.  Your saw spends most of it's life running under little or no load.  Production machinery is another story.

   As far as manufacturers measuring hp differently, these induction motors are built and nameplated to the NEMA MG-1 standard, and the shaft output is real.  The fudging comes from the marketing department.  Look at the motor nameplate on a big air compressor at HD; it says "sp" under hp.  The tank says "6.6 peak hp", or some such crap.  The motor manufacturer (Emerson) can't lie without risking the ability to self-certify, so they leave it blank.  The compressor manufacturer does the lying.

   Your W to hp conversion needs efficiency and power factor.  It's anyone's guess what they are, but generally they increase with an increase in hp.  Run capacitors also improve both, particularly power factor.  1/4 hp motor eff is often in the low 50's, 1000 hp high 90's.

   Finally, 2-pole 20A breakers are rated 20A on each pole, and are functionally similar to 2 20A single-pole breakers. 

   Be seeing you...

   
1. 
   jgaiennie | Feb 06, 2003 07:34pm | #6

   Tom,

   lol.  Sheesh man!  Thats a bunch of info.  I may have to take in a bit more caffine to absorb all that, and then my chances are less than even. 

   Thanks so much for the extremely informative message.

   >>  There is one notable difference between those Delta (and Delta is the only manufacturer I know of that does this) dual-horsepower motors and a standard motor, in that standard motors don't have the dual-hp ratings.  I'm not at all convinced they really are dual-hp, but rather they are 2 hp motors derated to 1.5 hp at 115V for reasons that may have to do with UL more than internal construction 17.2 FLA on a 15A plug maybe?)  It's telling that there's a service factor of 1.15 at 115V but only 1.0 at 230V.

   This obviously clears up some questions that have risen in here.  Your theory sounds good to me about them being overrated.

   So tell me this:  (laymen's terms please) 

   1.  Do you save on your electric bill if you wire 230?

   2.  Does the sawyer notice a difference at the blade - if you know?

   3.  Other advantages / disadvantages that are worthy of mention if not previously covered.

   Obviously, you've studied motors a bit.  I/we are grateful to you for sharing your knowledge with us.

   jdg

1. 
   TKanzler | Feb 06, 2003 09:10pm | #7

   1.  Do you save on your electric bill if you wire 230?

   No, not in any way you'd notice.  If you ran it 24/7 under full load, the IIR losses in the wires, and the slight reduction in sag of the efficiency curve would show, but in a home wood shop or small commercial shop it won't make any real difference.  And if you did run it 24/7 under load, it wouldn't be over the nameplate hp, at least not for very long, and that's where these effects start to snowball.

   2.  Does the sawyer notice a difference at the blade - if you know?

   Almost certainly, but that mostly has to do with the fact that folks go from 120V on 12 ga wire to 240V on the same 12 ga wire, and performance maximums (locked-rotor, pull-up, and breakdown torque) drop rapidly with drops in voltage at the motor leads.  Since people can't tell how much hp the motor is producing, they work it 'till it seems unable to work harder, which would be way beyond what it's rated for continuously (barring big voltage drops due to long extension cords and such, which makes it worse).  As the motor slows under load, it pulls even more current, causing more voltage drop, causing it to slow more (we're only talking a couple or few hundred rpm from synchronous, by the way), and so on.  The whole effect is lessened with higher voltage on the same wire, or larger wire at the same voltage.  But nobody wants to use 6 gauge wire at 120V to get exactly the same performance as they could get with 12 gauge at 240V for the same length of wire, including snappy start-ups and serious overload abuse (reread DanT's eloquent description above).  But if you never loaded the motor beyond rated output in the first place, 12 gauge would be fine for 120V operation (1.5 hp) also, with a slightly longer startup time (but then, starting up isn't cutting wood, so who cares?).  That's why folks who convert drill presses to 240V operation typically don't see any improvement; it starts easily anyway and doesn't get worked near the motor's rated output, let alone way beyond, normally.  The thermal overload devices built into most contractor saw motors does allow some heavy short-time overloads, especially if left running in between cuts (cooling time), whereas cabinet saw magnetic starters might react faster, but with bigger motors, don't often get overloaded. 

   But that motor of yours, I'm convinced, is really a 2 hp, and that's a bit big to run at 120V.  In fact, it exceeds the NEC branch circuit requirement for 125% full-load circuit ampacity for motor loads (it would be 17.2A*1.25=21.5A, more than a 20A circuit, which is why (probably partly, at least) it's dual-rated, or derated).  Nobody ever compained about a contractor saw with too much real horsepower.  I don't usually suggest converting just for the hell of it, but contractor saws generally do benefit since they're prone to being overworked, and the Delta with the 2 hp motor could benefit even more.  It basically removes the hp restriction, rather than giving you hp you really shouldn't be using, if you know what I mean. 

   3.  Other advantages / disadvantages that are worthy of mention if not previously covered.

   Disadvantages include loss of portability, need for double-pole switch, loss of panel space, expensive GFCI, if that's what you want (not code required, though), different plug/receptacle.  There was a thread recently listing these and others; I can't seem to find it now.

   Once you install a 240V receptacle or two, you can also use them for other high powered tools that can benefit, like a big shaper, wide belt sander, or big bandsaw for resawing (now that's tough work for extended periods).

   Coffee, I need more coffee........

   Be seeing you...

   Edited 2/6/2003 1:38:07 PM ET by Tom Kanzler

1. 
   jgaiennie | Feb 06, 2003 09:43pm | #8

   Tom,

   Thanks again for your response and your efforts in helping me understand this lesson.  However I still don't understand one thing - ok... many.  Just sorta dumb I guess.  I don't understand why the power being consumed isn't different when wired with the two voltages.  I'm hoping you can give me a simple answer to where my error is (see math below) - assuming I'm loading the saw enough to draw these amps and that losses in the wires are negligable.  Or is there no simple answer?  Or is it improper to make these assertions for the sake of this example?

   From my earlier post, I put:

   If I remember correctly, power = current multiplied by electromotive force (P=IE), whereas power is measured in watts, electromotive force is measured in volts and current in amps.  My power company bills me on kilowatt hours.  Correct me if I'm wrong, but this is the way I see it.  If wired at 115 volts, my saw would consume (115 x 12.8) 1472 watts.  Now if I'm paying $0.10 per kilowatt hour and I use that saw 100 hours in a month, that usage would cost me $14.72.  On the other hand, If I wired for 230 volts, consumption would be (230 x 8.6) 1978 watts and my bill would be $19.78.  So if wired for 230 volts, I'd be using less amps but consuming more power and that consumption would cost me more each month, right?

   Why doesn't the P=IE formula apply here?  Where's the problem with my math or my assumptions?

   Thanks,

   jdg

   Edited 2/6/2003 1:46:13 PM ET by jdg

1. 
   TKanzler | Feb 06, 2003 10:55pm | #9

   Simple answers are not my strong suit, but I'll try.

   For starters, you're comparing apples to oranges, i.e. power consumption for 1.5 hp at 115V against power consumption for 2.0 hp at 230V.  With 2 hp you're doing more useful work in a given time (33+% more) compared to 1.5 hp, so it should cost you more, all other things being equal.

   You also have efficiency and power factor, which is why the motor draws more current than it would in a perfect world.  [Don't worry, the simple part is coming.]  As an example, I just happen to have the certification sheet for a Marathon 3 hp cap. start, cap. run, 3600 rpm, similar to a Unisaw motor.  At full load, 3 hp, efficiency is 81.7%, and power factor is 96.1% (capacitor run, so pf is unusually high), and current is 12.4 at 230V.

   (3 hp * 746W/hp) / (0.961 * 0.817 * 230V) = 12.39A

   Power factor is the percentage of the current that goes toward magnetizing and doing work, not counting efficiency.  In other words, for that motor, 3.9% of the current is just bouncing back and forth through the wires and not contributing anything at all.  It's a little like a mass bouncing on a spring; there's a lot of energy moving around, but it's not doing anything useful, and it takes very little input to keep it going.  You also don't pay for this (the meter knows all and tells only some), unless you're a commercial/industrial customer, but that's another story.  

   Efficiency is how much of the input power goes toward making hay, er, turning the shaft.  18.3% of the remaining current is just making heat (your motor gets hot, doesn't it?).  Efficiency is usually pretty bad at low output power (61.9% at 1/4 power for this motor), so using a big motor for small loads (5 hp scroll saw?) is wasteful of both power and money (big single-phase motors ain't cheap).

   So, for your example, 12.8A * 0.90 pf * 115V = 1.325kW  At $.10/kW-hr * 100 hr, that's $13.25 per month.  I'm guessing at the .90 pf, since that's more typical of a motor this size, but if it has 2 capacitors on it, it might also be very high. 

   If this motor was reconfigured for 230V operation, with the same 1.5 hp load on it, the current is cut in half, the voltage is doubled, and the end result is the same.  Your meter adds what goes though each leg of the incoming feeders, and doesn't care how much on each is used; it gets added together.  So 10A coming out of one hot leg and returning via the neutral will turn the money counter the same speed as 5A coming out of one hot leg and returning through the other hot leg, with the process being reversed 120 times a second for 60Hz AC.  Same thing.

   So, the SIMPLE ANSWER is that you're comparing 1.5 hp output to 2.0 hp output.  You have to use the current value for one voltage, and halve or double it as required for the other voltage. 

   Clear as mud?

   Be seeing you...

   Edited 2/6/2003 3:04:37 PM ET by Tom Kanzler

2. 
   MIKEBUETTNER | Feb 07, 2003 12:33am | #10

   What an excellent thread. I love this stuff. Thanks for the great information...

   bit

3. 
   jgaiennie | Feb 07, 2003 03:18am | #12

   Tom,

   Yeps.  Clear as mud.  Hey... way clearer in some cases.  :)

   Your simple answer works for me.  BTW, that suit is strong.  Thanks.

   You're a true gentleman and a sport sir.  I/we appreciate you taking so much of your day on replys to my questions and helping with this valuable info.  I'm sure our fellow forum folks share in my gratefulness for the energy, thought and generosity that you mustered up to put all that together.  I'm going to save this thread for future reference.

   Be seeing you too...

   jdg

2. 
   foghorn | Feb 07, 2003 02:29am | #11

   Your posts have been very helpful. This one was no exception.

   Thanks

2. 
   HowardAcheson | Feb 07, 2003 07:07am | #13

   >>I'm not at all convinced they really are dual-hp, but rather they are 2 hp motors derated to 1.5 hp at 115V

   That's exactly what Delta does. They derate to keep the amps down to fit into a 20 amp 120 volt circuit.

   However, the point is that it really does produce 2.0 hp when wired for 240 volts.