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  1. #301
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    Quote Originally Posted by cjbfisher View Post
    I'm in the process of doing the impeller>filter bag mod, and was wondering if the length of the transition piece will have any effect on the performance?
    Not really - its actually better that it be a little longer because this helps to smooth out some turbulence.

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  3. #302
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    Thanks Bob.

  4. #303
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    Quote Originally Posted by cjbfisher View Post
    I'm starting to mod the intake for my Carbatec 2hp dusty. I have a 16mm melamine faced MDF disc to replace the metal impeller cover and a 25mm mdf backing plate to make a total of 41mm with which to form the reverse BMH. These both have a hole machined to the id of the 6 inch DWV pipe. I then have a further 2 25mm mdf plates with holes machined to the od of the pipe so the pipe will slip in and fit to the reverse BMH.
    I have come to a standstill with regard to forming the reverse BMH. I don't have a lathe, or access to one, and a 38mm roundover bit is prohibitively expensive for what will possibly be a one time use.
    Has anyone got any other suggestions, or even a router bit that I could borrow/hire?
    @cjbfisher - If you have any photos of the unit you are making it would be good to see it, even in progress is fine. I am keen to make something similar and would like to see a photo of what you are making.

    Thanks!

  5. #304
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    Jun 2020
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    Melbourne
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    Quote Originally Posted by BobL View Post
    Next the power ratings.
    The name plate says 240V , 7.7A and 2HP (240 x 7.7 = 1848W or 2.46 HP)

    However, in stock DC configuration I find it only draws 5.0 A at 238 V.
    The naked impeller draws 5.2A at 238V,
    Whether the bags are attached makes no difference to the current draw.

    The actual power drawn is thus 5.0 x 238 = 1190 W or 1.59HP
    Compare that to my 3HP system which in stock format draws 9.4 A at 238 V or 2237 W or 2.98 HP

    It appears this 2 HP unit is not actually 2 HP but more like 1.6HP?

    I will do some air flow measurements WIGRTI.
    Hi Bob, what an informative thread this is!
    I haven't read the whole 303 posts (yet) but I couldn't find any clarification offered for the apparent inconsistency within the motor nameplate ratings.

    The simple answer is that it's an AC motor so the power factor (PF) has to be taken into account. The PF is the cosine of the phase angle between voltage and current; for a DC (= direct current) motor, PF = 1. Power = Volts x Amps x PF. For an AC motor at full load the PF is typically around 0.8 and for the motor quoted, it is (2 x 746)/(240 x 7.7) = 0.81
    At less than full load, the power factor will likely fall and could be as low as 0.2 at no load.
    Therefore, to determine the power being consumed by an AC motor it is best to use a power meter rather than measure just the current (unless you have a current vs power curve), or if you can measure the phase angle between volts and amps, you can calculate the PF.

    Apologies if this has already been explained elsewhere in the thread.
    Cheers

  6. #305
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    Quote Originally Posted by SROman View Post
    Hi Bob, what an informative thread this is!
    I haven't read the whole 303 posts (yet) but I couldn't find any clarification offered for the apparent inconsistency within the motor nameplate ratings.

    The simple answer is that it's an AC motor so the power factor (PF) has to be taken into account. The PF is the cosine of the phase angle between voltage and current; for a DC (= direct current) motor, PF = 1. Power = Volts x Amps x PF. For an AC motor at full load the PF is typically around 0.8 and for the motor quoted, it is (2 x 746)/(240 x 7.7) = 0.81
    At less than full load, the power factor will likely fall and could be as low as 0.2 at no load.
    Therefore, to determine the power being consumed by an AC motor it is best to use a power meter rather than measure just the current (unless you have a current vs power curve), or if you can measure the phase angle between volts and amps, you can calculate the PF.
    Thanks for that nice clear explanation - yes I eventually remembered this (had to drag it back from long lost memories).
    Some time after the start of this thread I made small dyno to test HP outputs of motors and used PF to calc motor efficiency.
    I have been meaning to go back and edit these issue but never got around to this.

    If its OK with you I might insert an edited version of your explanation back into the first post to save people needing to get to post #304 for an explanation.

  7. #306
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    Quote Originally Posted by BobL View Post
    I have been meaning to go back and edit these issue but never got around to this.

    If its OK with you I might insert an edited version of your explanation back into the first post to save people needing to get to post #304 for an explanation.
    Thanks Bob - yes that is perfectly okay. It's been a long time since that thread started!

  8. #307
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    Quote Originally Posted by SROman View Post
    The simple answer is that it's an AC motor so the power factor (PF) has to be taken into account. The PF is the cosine of the phase angle between voltage and current; for a DC (= direct current) motor, PF = 1. Power = Volts x Amps x PF. For an AC motor at full load the PF is typically around 0.8 and for the motor quoted, it is (2 x 746)/(240 x 7.7) = 0.81
    At less than full load, the power factor will likely fall and could be as low as 0.2 at no load.
    Therefore, to determine the power being consumed by an AC motor it is best to use a power meter rather than measure just the current (unless you have a current vs power curve), or if you can measure the phase angle between volts and amps, you can calculate the PF.
    You have no idea how happy I am that you gave the SIMPLE answer.
    I got sick of sitting around doing nothing - so I took up meditation.

  9. #308
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    Nov 2016
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    Quote Originally Posted by doug3030 View Post
    You have no idea how happy I am that you gave the SIMPLE answer.
    You should see the complex one, although it's partly imaginary.
    Dave

  10. #309
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    Mar 2011
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    I’m in the process of building a bit of a Frankenstein DC.

    Had a dead 3ph motor 3hp twin bag that i couldnt get a replacement motor for, and someone here generously donated a 2hp dc motor and blower.

    The outlet of the blower almost perfectly lines up with the inlet of a single bag chute thing, so ive made an adapter plate with an infill piece so i can bolt the 2 together without any size change in the rectangle, ending up in a similar style to this thread, but with the inlet coming horizontally



    Next step is to raise the motor up on a stand (still wip)


    Is there any way i can do the measurements at home on how it is performing? I’m trying to work out if i should just stick with a couple of metres of flexi straight to the machine, or if i should go the 150mm pvc route, transition down, and then still have 100mm flexi to the macine

  11. #310
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    [QUOTE=dcarbonetti;2304900] Is there any way i can do the measurements at home on how it is performing?[QUOTE]
    Have a read of this.
    DRAFT: FAQ - Dust Extraction (Practical Aspects)

    I’m trying to work out if i should just stick with a couple of metres of flexi straight to the machine, or if i should go the 150mm pvc route, transition down, and then still have 100mm flexi to the macine
    PVC is significantly better than flex but if you want to get the benefit of the 150mm PVC you need to open the posts up on the machine.

  12. #311
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    Hi BobL, This has been a great thread. I would like to check if my thinking here is correct, I am considering buying some form of dust collection.

    On page 3 you said that a fully mod'ed 2hp with 3M of 6" duct would in theory pull 750cfm, which relates to 5" w.c. according to the fan tables you posted. If I attach this to my table saw (sawstop pcs) with a 4" port which then extends through a short (about 12" long) 4" flex duct connected to a blade shroud under the table. Would it be correct to say this could add an additional 2" of w.c? Going by the fan curve on page 1 that would mean that with a total w.c of 7", cfm could be reduced to around 250 cfm?

    The 2hp units really seem to drop off after 5" of w.c. I have found a 2hp unit with a slightly larger 12 3/4" impeller, would that help with this? Also would running it at 60hz (in canada) then the 50hz you tested it at help?
    Last edited by Davidmr; 29th March 2023 at 11:44 AM. Reason: error

  13. #312
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    Quote Originally Posted by Davidmr View Post
    Hi BobL, This has been a great thread. I would like to check if my thinking here is correct, I am considering buying some form of dust collection.

    On page 3 you said that a fully mod'ed 2hp with 3M of 6" duct would in theory pull 750cfm, which relates to 5" w.c. according to the fan tables you posted. If I attach this to my table saw (sawstop pcs) with a 4" port which then extends through a short (about 12" long) 4" flex duct connected to a blade shroud under the table. Would it be correct to say this could add an additional 2" of w.c? Going by the fan curve on page 1 that would mean that with a total w.c of 7", cfm could be reduced to around 250 cfm?
    Unfortunately yes. 4" ducting and ports dont move a lot of air.
    This is also consistent with 6" ducting moving about 3X the amount of air compared to a 4" duct.

    The 2hp units really seem to drop off after 5" of w.c. I have found a 2hp unit with a slightly larger 12 3/4" impeller, would that help with this?
    Yes this will improve the fan curve a bit. Both in terms of Max or static pressure and all the way down the curve.
    You would need to be careful not to run the impeller wide open on 6" ducting (Inout and output) as the impeller will try and draw more air and hence more current and could damage the motor.

    Also would running it at 60hz (in canada) then the 50hz you tested it at help?
    YES about 20% more flow at 60Hz.

  14. #313
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    Thanks BobL, all of that helps.

    The dust collector im looking at has a static pressure of 11". So with that and it run at 60hz it could be possible to get 400 cfm. With that low of cfm an issue would be the velocity in the 6" duct would be quite low. It really does seem like units 2hp and smaller are intended to be near the tool.

    I have also been looking at Oneida's mini-gorilla. Looking at the fan table they provide and knowing that they tested it to provide 580 cfm with 10ft of 5" flex, once I add 2" w.c. for the tool, that equals 500 cfm in theory. I know 4" can really only flow 400cfm, but to me it seems like this option has a higher chance of providing that then the mod'ed 2hp?

  15. #314
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    Quote Originally Posted by Davidmr View Post
    Thanks BobL, all of that helps.

    The dust collector im looking at has a static pressure of 11". So with that and it run at 60hz it could be possible to get 400 cfm. With that low of cfm an issue would be the velocity in the 6" duct would be quite low.
    it should produce bout 450 cfm provided no flex is used. I don't usually worry about the flow much - its only really an issue when the suction is turned of or power fails whilst carrying sawdust in long ducting

    It really does seem like units 2hp and smaller are intended to be near the tool.
    Yes but the problem then becomes "leakage" which is why putting them outside is worthwhile and here in Oz we can employ extra (quieter) shed ventilation to remove fine dust.

    have also been looking at Oneida's mini-gorilla. Looking at the fan table they provide and knowing that they tested it to provide 580 cfm with 10ft of 5" flex, once I add 2" w.c. for the tool, that equals 500 cfm in theory. I know 4" can really only flow 400cfm, but to me it seems like this option has a higher chance of providing that than the mod'ed 2hp?
    It depends how they have done the testing - most manufactures use standardised testing methods which over estimates the air flow. It depends on the type of flex used but soft flexible flex will put a serious dent in air flow. Stuffer flex is better.

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