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Thread: NEC Table 430.250

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    NEC Table 430.250

    I have some exam prep materials that references NEC Table 430.250, and then it uses the formula: kW = 1.732 * V * I. It then uses the obtained kW in another formula, kW = hp * 0.746/efficiency, and obtains efficiency = 75%. Based on the current value it uses, the motor is an induction motor. Shouldn't the first formula include the power factor to obtain kW? If power factor is included, then the efficiency would be 94%, which does not seem very conservative.

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    show the numbers used

    the first equation should have pf
    it is the electrical power delivered to the motor, not the mechanical power delivered by the shaft
    either that or it should be va not watts
    it should also be divided by 1000

    the second is also elec power, but should be va imo
    the 0.75 is an approximate value for pf x eff

    hp x 756 = mech shaft power in watts
    mech shaft power / eff = elec input power in watts
    elec input power / pf = va = sqrt3 x v x i
    total power in va = hp x 746 / (pf x eff)

    S = total power in va
    P = real or active power in watts
    Q = reactive or imaginary power in var
    S = P + jQ
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    Quote Originally Posted by ee_username View Post
    I have some exam prep materials that references NEC Table 430.250, and then it uses the formula: kW = 1.732 * V * I. It then uses the obtained kW in another formula, kW = hp * 0.746/efficiency, and obtains efficiency = 75%. Based on the current value it uses, the motor is an induction motor. Shouldn't the first formula include the power factor to obtain kW? If power factor is included, then the efficiency would be 94%, which does not seem very conservative.
    Motor nameplate rating (usually in HP) is rated output power. 1 HP = 746 watts.

    The input power is rated output divided by efficiency. There is always some loses in the motor so input power is always greater then output power as a general rule - this can get complicated when a load has more energy and is transferring that energy back to the motor- in which it acts more like a generator but lets disregard those kind of situations for the moment. Pretty common to see efficiency ratings between 85 and 95%.

    Power factor = input kW/kVA (that is input kW and input kVA here)

    I don't know what table 430.250 has to do with anything - all that table is for is to tell us what FLA we should use in motor calculations for a certain HP rated motor. It has FLA ratings that assume the worst possible efficiency and power factor motors you would ever find. You will seldom find a motor that has a nameplate amp rating that matches what is in the table, but they want us to size circuits for say a 10Hp motor to be able to handle the current of the worst efficient 10 Hp motor ever made.
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    I think where you are confused is that if you only had kVA, you would need to use PF. If you already have kW, power factor was already used to get there. kW = kVA x PF
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    The problem states 100 hp, 460V, 3-phase motor and also states Table 430.250 gives 124 A. The question asks what efficiency does this assume. The solution uses kW = 1.732 * V * I /1000 = 1.732 * 460 * 124 /1000 (thanks Ingenieur!) = 98.8 kW. (I think the 98.8 should be kVA.). The solution then uses kW = hp * 0.746 / Eff, which results in Eff = 100 hp * 0.746 / 98.8 kW = 0.75 or 75%.

    So either the 1) efficiency = 94% (Table 430.250 appears to be not be very conservative) or 2) kW = 1.732 * V * I / 1000 (does not appear to be the correct formula), but gives a more conservative efficiency = 75%.

    Any thoughts on how to reconcile 1) and 2)?

    Thanks all for the responses!

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    imo you are correct, the first gives kva(no pf)
    kva = 98.8 kva

    kw (shaft) = kva x pf x eff so using the 0.75 as (pf x eff)
    kw (shaft) = 98.8 x 0.75 = 74.1 kw or 74100/746 = 99.3 hp(close to 100 hp from the table)

    imo the first eq is kva
    the second uses kva AND the 0.75 derived is pf x eff

    the problem is goofy
    Last edited by Ingenieur; 03-11-18 at 02:51 PM.
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    Quote Originally Posted by ee_username View Post
    The problem states 100 hp, 460V, 3-phase motor and also states Table 430.250 gives 124 A. The question asks what efficiency does this assume. The solution uses kW = 1.732 * V * I /1000 = 1.732 * 460 * 124 /1000 (thanks Ingenieur!) = 98.8 kW. (I think the 98.8 should be kVA.). The solution then uses kW = hp * 0.746 / Eff, which results in Eff = 100 hp * 0.746 / 98.8 kW = 0.75 or 75%.

    So either the 1) efficiency = 94% (Table 430.250 appears to be not be very conservative) or 2) kW = 1.732 * V * I / 1000 (does not appear to be the correct formula), but gives a more conservative efficiency = 75%.

    Any thoughts on how to reconcile 1) and 2)?

    Thanks all for the responses!
    I don't know where you are getting 94% from?

    But yes the motor FLA's listed in the tables at the end of 430 are the most inefficient motors you are likely to find. They want you to use those values for conductor selection - so that if a lesser efficient motor but still same power rating would happen to replace the original, the conductors are still sufficient.

    most 100 HP motors I have seen have a nameplate FLA of about 118 amps and nameplate efficiency close to 90%.
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    [QUOTE=kwired;1902465]I don't know where you are getting 94% from?

    I think I forgot to mention the assumed power factor of 0.8. kVA = 1.732 * 460 * 124 = 98.8 kVA. Assuming power factor = 0.8, kW = 0.8 * 98.8 = 79 kW. Eff = 100 HP * 0.746 / 79 kW = 94.4%. I guess if one used pf = 1, then the efficiency would be 75%, but I would think a power factor of 0.8 and efficiency of 94.4% seems more likely. Thanks for bearing with me as I try to explain the problem.

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    [QUOTE=ee_username;1902468]
    Quote Originally Posted by kwired View Post
    I don't know where you are getting 94% from?

    I think I forgot to mention the assumed power factor of 0.8. kVA = 1.732 * 460 * 124 = 98.8 kVA. Assuming power factor = 0.8, kW = 0.8 * 98.8 = 79 kW. Eff = 100 HP * 0.746 / 79 kW = 94.4%. I guess if one used pf = 1, then the efficiency would be 75%, but I would think a power factor of 0.8 and efficiency of 94.4% seems more likely. Thanks for bearing with me as I try to explain the problem.
    I was thinking about this some more and was about to post that you didn't have enough information. Otherwise all we have from the NEC table is the ability to calculate input KVA and the fact it is 100 HP the output power should be 74.6 kW. If we have at least one of: PF, efficiency or input kW - the others can be calculated.
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