Help me solve a DC motor task

[eng.saeed]

Hi everybody...
Can anybody help me solve this problem about DC motor manually and by using software (i.e matlab, C++,etc..)..? Please..
here it is:
A dc series motor drives an elevator load that requires a const. torque of 200 N.m. The dc supply voltage is 400 V and the combined resistance of the armature and series field winding is 0.75 Ω. Neglect rotational losses and armature reaction effect.

1) the speed of the elevator is controlled by solid-state chopper. At 50% duty cycle (i.e., α = 0.5) of the chopper, the motor current is 40 A. Determine the speed and horsepower output of the motor and the efficiency of the sys.

2) The elevator is controlled by inserting resistance in series with the armature of the series motor. For the speed of part (1), determine the values of the series resistance, horsepwer output of the motor, and efficiency of the sys.


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[VVV]

Re: DC motor problem

Here is my two cents.

1) Let’s assume that the chopper is 100% efficient.
Then the average voltage applied to the motor is:

Pin=Pmot=α*Vin*Ia=0.5*400*40=8000 W

Given that the chopper is 100% efficient and you do not have friction losses, then the only losses are caused by the motor’s internal resistance and they are:

Ploss=Ra*Ia^2=0.75*1600=1200 W

Therefore, the power available from the motor is:

Pout=Pmot-Ploss=8000-1200=6800 W

The power at the motor’s shaft is thus 6800W and it is equal to the product of the torque and angular speed:

Pout=M*ω

Hence:

ω=Pout/M=34 rad/sec,
or, in r.p.m.:
ω=34/(2*π)*60=324.67 rpm

The efficiency is:

η=Pout/Pin=6800/8000=0.85 or 85%



2) Since the motor is running under the exact same conditions, the current must still be 40A, however, it is limited by an external resistor. Since the speed does not change, the back emf is the same. And because in the first case the sum of the back emf and the drop across the internal resistance was 200V (400*0.5), equivalent applied to the motor, then in this case the rest of 200V must drop across the additional resistor. Therefore:

Rext=200/40=5 Ω

The power dissipated in it is:

Pres=Rext*Ia^2=5*1600=8000 W

And the power drawn from the source is:

Pin=Pres+Pmot=8000+8000=16000 W

Therefore the efficiency is:

η=Pout/Pin=6800/16000=0.425 or 42.5%


Incidentally, the back emf of the motor at the running speed is:

EMF=α*Vin-Ra*Ia=0.5*400-0.75*40=170V.