DC Motor help

Discussion in 'Homework Help' started by dan_man555, Sep 5, 2010.

  1. dan_man555

    Thread Starter New Member

    Sep 1, 2010
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    I need help working this one out. Would really appreciate the help.

    An 80-V DC motor has a constant field flux (separately excited) and a nameplate speed of 1150 rpm with 710 W output power. The nameplate armature current is 10 Amps and the no-load current it 0.5 A. Assume constant rotational losses.

    a) Determine the armature resistance
    b) Estimate the rotational loss
    c) Find the machine constant in V/(rad/s)
    d) What is the no-load speed in rpm
    e) Find the effieciency at nameplate load, including field circuit losses of 30 W.

    I know the formula for part a) is Ra=(V-E)/Ia, but how do you find E?
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    The voltage supplied to a DC motor MUST equal the voltage dropped by armature resistance + the back EMF.
     
  3. dan_man555

    Thread Starter New Member

    Sep 1, 2010
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    Alright, so if when the motor is at rest, the induced voltage E = 0, and so the starting current is Ia= V/R
    Ra=V/Ia
    =80/10
    =0.8 ohms

    then for b)

    P-losses = P-developed - P-out

    E=V-Ia*Ra
    =80-0.5*0.8
    =79.6v

    P-developed=E * Ia
    =79.6*0.5
    =39.8W

    P-losses=39.8-710
    =-670.2W

    That doesn't seem right
     
    Last edited: Sep 5, 2010
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I would approach it as follows

    1. At rated load, the armature current is 10A at 80V DC supply.

    So P_electrical input at rated mechanical power = .....?

    2. The rated load mechanical power P_mechanical = 710W

    3. What are the total losses at rated power? P_loss_total = ....?

    4. The no load losses P_loss_no_load = I_no_load*Vsupply.

    5. The no load losses are the sum of the constant rotational losses and the no-load armature resistance losses (i.e. at Ia=0.5A).

    6. The losses at rated load would be sum of the constant rotational losses and the rated armature resistance losses (i.e. at Ia=10A).

    7. Set up two equations of the form

    P_loss_rotational+(I_no_load)^2*Ra=P_loss_no_load

    P_loss_rotational+(I_rated_load)^2*Ra=P_loss_rated_load

    8. Solve for Ra and then P_loss_rotational.

    And so on .......
     
  5. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    torque is proportional to the current through the windings,

    T = kI where T is the torque, I is the current, and k is a constant

    The wire coils have both a resistance, R, and an inductance, L. When the motor is turning, the current is switching, causing a voltage,

    V = L dI/dt

    This voltage is known as the back-emf(electromotive force), e.

    If the angular velocuty of the motor is w, then e = kw (like a generator)

    This voltage, e, is working against the voltage we apply across the terminals, and so,

    (V- kw) = IR where I = T/R

    which implies (V-kw) = (T/k) R

    The maximum or stall torque is the torque at which w = 0 or T = kV/R, and

    The stall or starting current, I = V/R

    The no load speed, w = V/k, is the maximum speed the motor can run.


    Based on the information you have,you will need to make some substitutions for the missing information you have been given. Go back and try your first calculation again, but try to get to a ohm value for the armature instead. The value for a backEMF looks right. Take that and the 80 volts given and determine what the resistance would be that allows 10 amps to flow in the loaded motor.
     
  6. dan_man555

    Thread Starter New Member

    Sep 1, 2010
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    Trying it your way t_n_k,

    P_electrical input = 10*80
    =800W

    P_mechanical
    = 710W

    P_loss_rotational+0.5^2*Ra=800

    P_loss_rotational+10^2*Ra=710

    P_loss_rotational=800.23
    Ra=-0.9023

    How does that look?
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Ra is not 0.8Ω. As you indicated earlier you don't have the motor back emf. Check my approach.

    I have Ra = 0.49875Ω.
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The no load loss will be source electrical input power under no load = 80*0.5=40W - not 800W.

    The rated load loss = rated electrical input power - P_mechanical

    Making it 800-710=90W
     
  9. dan_man555

    Thread Starter New Member

    Sep 1, 2010
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    P-rot + Ra*(0.5^2)=40
    P-rot + Ra*(10^2)=90
    That gives me Ra=0.5013Ω, and P-rot losses=39.87W

    For the machine constant I have the formula:
    T-dev= K Ia

    P-dev=Pout-P-rot losses
    =710-39.87
    =670.13W

    T-dev= Wm*P-dev
    =670.13 * (1150*2pi)/60
    =670.13*120.43
    =80,703N.m

    K=Ia/T-dev
    =0.5/80,703
    =6.2*10^-6 V/(rad/s)

    or

    E=K*Wm

    E = V - Ia*Ra
    =80-0.5*0.5013
    =79.75W

    K= 79.75 / (1150*2pi)/60
    =79.75/120.43
    =0.66 V/(rad/s)
     
    Last edited: Sep 6, 2010
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I agree with your Ra and rotational losses.

    My earlier Ra estimate was incorrect - algebraic error on my part.

    I have the machine constant as 0.62267 V/rad/sec.

    Based on this reasoning......

    At rated load the armature drop is Ra*Ia=.50125*10=5.0125V

    Hence E_rated=80-5.0125=74.987V

    Rated speed = 1150rpm=120.428 radians/sec

    Machine constant= 74.987/120.420=0.62267 V/rad/sec

    This would give (I believe) a no-load speed of 1223 rpm.
     
  11. dan_man555

    Thread Starter New Member

    Sep 1, 2010
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    Your right, i put the wrong current.
    Here's my attempt for d:

    k=0.6227
    E(no load)=79.749
    Wm=79.749/0.6227
    Nm=1223rpm

    and e:

    efficiency is usually P-out/P-in * 100%,
    so (710-30)/800 * 100
    =85%
    I assume you can just subtract the 30W field loss directly from the output power.
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    For the efficiency I would calculate the total input power as the base and the mechanical output as the power yield.

    Total rated input power = 830W (including field excitation)

    The yield is 710W mechanical.

    Efficiency = (710/830)*100=85.54% which is close to yours but slightly different.
     
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