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    Question: Research electric motor and generator efficiency and describe where and how they use energy that is not converted to power or electricity. Determine the efficiency of a 400V DC shunt wound motor that takes a current of 120A while running at 1200rpm while the field resistance is 50ohms, the armature is 0.25ohms, when the iron friction and windage losses account for 2.2kW.

    Alright so I need to find out efficiency which is Pout/Pin × 100

    Pin = VI = 400x120 = 48000

    Pout = 48000 - 2200 = 45800

    Efficiency = 45800/48000 × 100 = 95.4%

    Is it right? I'm not sure because thdy gave me resistannce and rpm, which I'm guessing should be used...
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    (Original post by rm2)
    Question: Research electric motor and generator efficiency and describe where and how they use energy that is not converted to power or electricity. Determine the efficiency of a 400V DC shunt wound motor that takes a current of 120A while running at 1200rpm while the field resistance is 50ohms, the armature is 0.25ohms, when the iron friction and windage losses account for 2.2kW.

    Alright so I need to find out efficiency which is Pout/Pin × 100

    Pin = VI = 400x120 = 48000

    Pout = 48000 - 2200 = 45800

    Efficiency = 45800/48000 × 100 = 95.4%

    Is it right? I'm not sure because thdy gave me resistannce and rpm, which I'm guessing should be used...
    All of the values given in the question are relevant except the rotational speed. I assume there is a follow on part which needs that.

    You are missing the power loss in the armature winding together with the back e.m.f. generated at 1200 r.p.m.

    E_b = E - R_aI_a

    P_a = E_bI_a

    P_{load} = P_a - P_{losses}

    \eta = \frac{P_{load}}{P_{in}} \mathrm {x \ 100 \%}
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    (Original post by uberteknik)
    All of the values given in the question are relevant except the rotational speed. I assume there is a follow on part which needs that.

    You are missing the power loss in the armature winding together with the back e.m.f. generated at 1200 r.p.m.

    E_b = E - (R_f + R_a)I_a

    P_a = E_bI_a

    P_{load} = P_a - P_{losses}

    \eta = \frac{P_{load}}{P_{in}} \mathrm {x \ 100 \%}
    The armature current is 400/0.25 = 1600A right?

    So the back voltage is 400 - 50.25×1600 = -80000V?
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    (Original post by rm2)
    The armature current is 400/0.25 = 1600A right?
    I_{supply} = I_a + I_f

    I_{field} = \frac{E}{R_f} = \frac{400}{50} = 8A

    I_a = I_{supply} - I_{field} = 120 - 8 = 112A


    E_b = E - I_aR_a

    E_b = 400 - (0.25 \mathrm {\ x \ } 112) = 372V

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    P_a = E_bI_a = 372 \mathrm {\ x \ }112 = 41664W

    P_{load} = P_a  - P_{losses} = 41664 - 2200 = 39464W

    P_{supply} = EI_{supply} = 400 \mathrm {\ x \ }120 = 48000W

    \eta = (\frac{P_{load}}{P_{supply}}) \mathrm {\ x \ }100\% = (\frac{39464}{48000}) \mathrm {\ x \ }100 = 82\% (2 s.f.)






    EDIT: apologies, I misquoted the back e.m.f. equation in an earlier post. It should not include Rf
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    (Original post by uberteknik)
    I_{supply} = I_a + I_f

    I_{field} = \frac{E}{R_f} = \frac{400}{50} = 8A

    I_a = I_{supply} - I_{field} = 120 - 8 = 112A


    E_b = E - I_aR_a

    E_b = 400 - (0.25 \mathrm {\ x \ } 112) = 372V



    EDIT: apologies, I misquoted the back e.m.f. equation in an earlier post. It should not include Rf
    Armature power = 372×112 = 41664W

    Load power = 41664-2200 = 39464W

    Efficiency = 39464÷48000 = 82.2%

    What exactly is load?
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    (Original post by rm2)
    Armature power = 372×112 = 41664W

    Load power = 41664-2200 = 39464W

    Efficiency = 39464÷48000 = 82.2%

    What exactly is load?
    The efficiency calculated is correct.

    Load in this instance means the power doing useful work after all the losses are subtracted from the input power.
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    (Original post by uberteknik)
    The efficiency calculated is correct.

    Load in this instance means the power doing useful work after all the losses are subtracted from the input power.
    Could I called it useful power output here? Not a fan of the word load and it kinda confuses me.
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    (Original post by rm2)
    Could I called it useful power output here? Not a fan of the word load and it kinda confuses me.
    There is no set rule that say's you cannot.

    However, be advised that you may get even more confused with the nomenclature when the word does crop up in later problems.
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    (Original post by uberteknik)
    There is no set rule that say's you cannot.

    However, be advised that you may get even more confused with the nomenclature when the word does crop up in later problems.
    Alright, thank you for your help.
 
 
 
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