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Discussion Forum

Discussion Forum

OT-electric motor question

1110d | Posted in General Discussion on October 1, 2002 08:56am

The electric motor has given up the ghost on my compressor. I need a 5hp 3450 CCW motor. This use is not commercial and is used say 3-4 times a month. I’ve done a bunch of searching for different motors and I’m frankly rather baffled by all the options. I understand there are differences between the frame types and the shaft sizes, which I’m willing to deal with. My question really has to do with motor quality and if I’ll be satisfied with a cheaper motor.

Here’s a link to the cheapest motor that I can find. The page does not specify if it’s for a compressor, but as long as it’s a capacitor start, we’re OK, right?

http://www.northerntool.com/webapp/wcs/stores/servlet/ProductDisplay?catalogId=4006970&storeId=6970&productId=461&langId=-1

Here’s a link to the Grainger page. The motor is a little more expensive (those couple of dollars was enough to get the Mrs. to balk) but I know that it’s a quality motor and was designed for a compressor.

http://www.grainger.com/Grainger/productdetail.jsp?xi=xi&ItemId=1611621890

Here’s a auction on E-bay. The ad sure looks a little hokey, but I’d use the credit card for protection.

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=2058123662&rd=1

So can you guys offer some assitance to help guide my purchase?


Edited 10/1/2002 2:11:39 PM ET by Tim

Reply

Replies

  1. JohnSprung | Oct 01, 2002 09:40pm | #1

    Ask around and find a good motor rebuilder in your area.  It may be that your existing motor can be overhauled.  And if not, the rebuilder can at leat point you in the right direction, or maybe even sell you a new or rebuilt motor.  http://www.oldwoodworkingmachines.com is a good place to ask.  That's where I found the guy who sold me the new motor for the Unisaw.

    -- J.S.

    1. 1110d | Oct 01, 2002 10:02pm | #2

      I had previously replace the pump with a larger one and the motor was a tad undersized. This new motor is a larger one that's rated for the pump. I did call a couple of rebuilders, but their rebuilt price was actually more than Grainger's new price. Thanks for the link, I'll give it a try!

      Sorry, the link is bad.

      Edited 10/1/2002 3:03:32 PM ET by Tim

  2. TKanzler | Oct 01, 2002 10:32pm | #3

    It would help hugely if you posted all of the nameplate data on the existing motor, particularly items where the data is stamped or printed into blank spaces.  Examples are HP, FLA (full load amps), SF (service factor), etc.  Unless the compressor is industrial grade, it's probably not 5 hp, despite what it says on the receiver (tank).  In fact, if it runs on 115V, it's not 5 hp (it would draw 50+ amps at 115V).  So post whatever you can.

    Edit: I didn't see your last post until after I posted this.  Take it for what it's worth. 

     

    Be seeing you...



    Edited 10/1/2002 3:41:04 PM ET by TDKPE

    1. 1110d | Oct 01, 2002 10:48pm | #4

      The compressor head has been replaced. I copied some of the technical data from the brochure for your reference. My old motor was a 3hp 3450 so I just geared the pump down and didn't have full output. I now want to reach full output. These numbers are based off a 1725rpm motor, so I figured a 3450rpm 5hp motor geared to 50% would be able to swing the larger 7.5hp pulleys.

      1. TKanzler | Oct 02, 2002 12:20am | #9

        From the catalog cut you provided, the pump speed for the model you have should be 710 rpm (or less) for a 5 hp motor when the motor is at full load speed, or 3450 in your case.  Divide the small sheave diameter by the large and multiply by 3450 for pump speed.  It sounds like you may have sheaves for a 7 1/2 hp setup, same pump, with 1725 rpm motor, so you will not be using all of the available motor power if I understand you correctly, but no harm will come to the motor or pump.  That's if I understand what you're saying.  But please do calculate pump input speed anyway and compare to the cut sheet value.  Also note that all of the motors have special shafts, in some way, based on the "z" suffix on the frame sizes.  The Grainger has a 7/8" dia shaft, the other one (Northern Tool?) is 2 1/4" dia, or something like that, and the ebay motor is 5/8" dia, which is standard for a 56 frame, but might be long or short.  The "H" in the 56HZ means it has extra holes for compatability with 143T and 145T mounting patterns.  And I agree that the full-load amp draw seems too low on the ebay unit.  The other two should be OK, just watch that shaft on the Northern Tool motor.  FWIW, those prices seem very low for a 5 hp single phase motor. 

        Be seeing you...

        1. 1110d | Oct 02, 2002 12:53am | #12

          TDKPE-Thanks for the formula. You bring up a valid point of not overspeeding the compressor. I was originally just planning on cutting the manufacturers sheave diameter in half. I'm thinking correctly right? Double the rpm, gear to 50% and the motor will have more than enough power. But what about startup? Will the motor have enough torque?

          I was figuring the Northern motor must have a typo for the shaft diameter. Either, way, the larger the shaft, the greater the rolling surface for the bearings. That would be a good thing for a high side load application like this. Right?

          1. TKanzler | Oct 02, 2002 01:59am | #13

            You bring up a valid point of not overspeeding the compressor.

            Overspeeding only in the sense that a 5 hp motor can only spin that particular head at 710 rpm at 175 psi, per their literature.  1035 rpm using a 7.5 hp motor is the next line down, but 440 rpm max with a 3 hp motor.  If you turn it faster than 710, you may fry your motor prematurely.  And rather than simply halving the motor sheave size, why not calculate the correct size for 5 hp, and have the additional air delivery available by using the appropriate sheave?  But if you're OK with 12 or 13 cfm (guestimated by eyeball interpolation), instead of the 16.5 listed for 5 hp, then it will be easier on the motor. 

            But now I'm getting confused.  You've highlighted the 7.5 hp data for that pump head.  You also stated that the pump was sized with sheaves for a 1725 rpm motor.  If you halve the motor sheave size and double the motor speed, you still need 7.5 hp to drive the pump.  So do calculate the resulting pump speed for the actual equipment you'll be using, and check that it doesn't exceed 710 rpm for 5 hp and 175 psi.  And Cairo is quite correct; frequent starts and short run times are murder on a motor, especially on compressors with relatively high inertia.   

            Be seeing you...

          2. HammerHarry | Oct 02, 2002 02:21am | #14

            Also, make sure the "unloader" valve still works.  This is a little valve that makes sure the heads are not pressurized when the compressor isn't running.  If the heads stay pressurized, aint no way your 5 hp motor will start the compressor.  But, since you've already used the compressor, unloader probably works well.  This is why it doesn't take all that much torque to start the compressor; it is actually not starting under any real load.  With a compressor, the faster you run it, the more power you will need, so as has been pointed out, to avoid overloading the 5 hp, you need to make sure you use the right size sheaves, as pointed out.

          3. 1110d | Oct 02, 2002 02:34am | #15

            Let me try this...

            If the compressor is to run at 500rpm with a 1,000rpm motor, then the motor pully would be half the size of the compressor pully.  Now if I use a 2,000 rpm motor, then the pully size would need to be 1/4 the compressor pully's diameter.  The mechanical advantage is that I just doubled the motor's power.  Therefore, if it takes a 1hp motor @ 1,000 rpms to turn the compressor, then it should take a 1/2 hp motor at 2,000 rpms to do the same work.

            ie a 5hp motor turning twice the calculated speed should have more than enough power to spin the compressor at it's max 7 1/2hp level by using a motor pully that is 1/2 the recommended diameter.

            Right?

          4. TKanzler | Oct 02, 2002 02:46am | #16

            Uh, wrong.  Rotational power is torque x speed, so a 5 hp motor at 1725 rpm produces about 15.22 lb-ft of torque.  A 5 hp motor at 3450 rpm (twice the speed) produces about 7.61 lb-ft of torque, which is one half of the previous case.  Power is power; in the end, you need to turn that pump shaft no more than 710 rpm with a 5 hp motor, however you have to gear it to make it so. 

            Be seeing you...

          5. HammerHarry | Oct 02, 2002 03:46am | #17

            That's right; power is power.  Incidentally, an often misunderstood fact is that an electric motor will deliver the power required by the machine it is driving.  If your compressor needs 5 hp, it doesn't matter if you put a 5, 7.5, or 10 hp motor on it, the power draw will be the same (well, yes, efficiency will be slightly less on the bigger motors, but you get the point). 

          6. 1110d | Oct 02, 2002 05:02am | #18

            Oh.  Well, now doesn't that just suck.  No wonder the old motor died.  I suppose you guys are also going to tell that the world isn't flat and that the tooth fairy is dead.  So is there any mechanical advantage to running the 3450 motor then?

          7. HammerHarry | Oct 02, 2002 05:15am | #19

            not if you want 700 rpm as your driven speed.  suppose the 3600 rpm motor would be smaller, though.  note here:  you will see the motors usually called 3600/1800/1200rpm etc; actual running speed will be something like 3450 rpm, but this is considered a 3600 rpm motor.

             if you are buying a new one, i would go with a 1800 rpm motor.  if there is too much of a difference in the diameters of the pulleys, you may not get enough "wrap", that is, the belt won't be in contact with enough of the pulley to transmit the power.  however, i only tend to run into this on 200 and 300 hp machines, with 5 or 7 belts;......don't think the 5 hp would be an issue.

  3. User avater
    BillHartmann | Oct 01, 2002 11:35pm | #5

    I would probably go with the North Tool one. It has a 1.15 service factor. The one on ebay the FLA appears to be too low for the that size motor.

    1. 1110d | Oct 01, 2002 11:51pm | #6

      Thanks Bill!

      I take it that the higher the service factor the better?

      1. User avater
        BillHartmann | Oct 02, 2002 12:02am | #7

        Service factor is really an "over rating". With a 1.15 SF you can run it continously with a 5.75 hp load.

        1. HammerHarry | Oct 02, 2002 12:15am | #8

          Compressor motors are notorious for their rather odd rating system.  I used to be in the industrial pump/motor/compressor bus, and all I can say is, whoever dreamed up the fantasy world of compressor motors should be hung.  Basically, all the OEM compressor motors are designed to run in the service tefactor, and are all built on the "cheapest thing possible that will do the job" theory.  I would go with the Grainger motor if you are using it all the time.  If you are only using it occaisionally, go with the cheaper one.  Beware of how often it will start/stop.  Every start of an electric motor generates heat; if the motor doesn't run or sit idle long enough between starts to cool down, you will overheat the motor.

          With regards to Service Factor, you often see the "dedicated" compressor motors have a 1.0 SF.  Again, copper equals money, so they are built as lightly as possible. 

  4. 1110d | Oct 02, 2002 12:24am | #10

    Now you guys bring up an interesting point...

    The ebay motor lists it's amps at 15 with no service factor

    The grainger motor lists it's amps at 21 with a 1.0 service factor

    The Northern motor lists it's amps at 22 with a 1.15 service factor

    Is there some creative accounting going on here?

    1. TKanzler | Oct 02, 2002 12:29am | #11

      With respect to the first item, probably.  The second, probably not; Dayton motors conform to the NEMA MG-1 standard, as far as I know.  And as to the third, the service factor is not used in the FLA, but rather shows up in SFA (service factor amps, which will be higher than the full load amps for SF>1.0), which goes into another field on the nameplate. 

      Be seeing you...

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