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DC shunt machine help

Ledwardz

Dec 21, 2010
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hey,

i am after a bit of help on the question attached.

Okay so i figured out the torque and input power quite easily.

I then figured out Total current (before it has reached field or armature) by doing total power in divided by supply voltage of 48. This gave me 13.02 Amps

right so i tried re-arranging P = Tw = Eia (where w is angular velocity)
i messed around for a while rearranging torque into k If Ia and a few other things
i changed E to V - IaRa to try and get a constant etc etc.

I think perhaps part of the problem is i dont fully understand what rated and no load mean. I assumed rated was normal operating point and no load was full speed but no torque but i don't think that can be right because then armature current would b 0? (I have used it for other questions and it gives me the right answer but i don't know why)

anyways after searching for ages i found the equation V = IwL (L is inductance)
i assume this i only for AC but woteva i tried it for the field current using the mutual inductance and no load speed and i got the right answer for field current = 1.269

can somebody explain this? surly under DC, the field winding acts as a short no? the change in current with respect to time is zero?

so when i have this i can simply do 13.02 - 1.27 to get my armature current of 11.75

forpart b, i didn't really have many problems. I would like to know the whole idea of the transient equations tho. I know how to do them and what the voltages are for

for armature i got voltage over resistor (IaRa), over armature coil (Ldi/dt) and in rotor itself ( IfwLad)
i don't get the whole rotor bit itself, is this excited emf by movement of the rotor i assume?

so basically what this is saying i have voltage lost in resistor, voltage in rotor windings (when it is first switched on before steady state) and an extra bit of voltage caused by movement of conductor in the stators magnetic field. Surely this is going to create a voltage bigger than what i supply? Have i got summit wrong here? This (If w Lad) is screwing me over.

going back to the diagram does it mean i should have a armature resistor, armature coil and the emf circle thingy?

for field i got IfRf + Lfdi/dt

and i will worry about c when i know what is going on i parts a and b.

Thanks for your help and sorry if this is a bit mumbo jumbo, :eek:

Lee.
 

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Harald Kapp

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Thanks for your help and sorry if this is a bit mumbo jumbo
Well, it is at least to me as I don't have a deep understanding of motors (forgot all about it ;) ).
Here's what I can contribute:
I assumed rated was normal operating point and no load was full speed but no torque
The first assumption is right. rated means that you have a nominal load (torque) for which a nominal power and speed are given. This could be called the "normal operating point", but "rated" is the official language. That's because in any application the "normal operating point" doesn't necessarily have to be the rated operating point. It could, for example, be perfectly normal for this motor to operate at only 80% of the rated power. This could be a design decision to leave some headroom for overload.
Similarly no load means full speed only if the input power is equal to the rated power. If you drive the motor at 1/2 input power the speed even without load is not the full speed of the motor. It is only full speed with respect to input power.
no torque but i don't think that can be right because then armature current would b 0?
That would require an ideal motor without any losses. buta the motor is real and thus has internal losses that need to be compensated to keep it rotating even without load. From the ratings you can calculate the power loss at 500W (hint efficiency = 80%). Lacking more information you can assume that this power is required even without load. And from taht you get the current under no-load conditions.

I hope that helps a bit along your way. Sorry for not having more detailed info.

Harald
 

Ledwardz

Dec 21, 2010
43
Joined
Dec 21, 2010
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That would require an ideal motor without any losses. buta the motor is real and thus has internal losses that need to be compensated to keep it rotating even without load. From the ratings you can calculate the power loss at 500W (hint efficiency = 80%). Lacking more information you can assume that this power is required even without load. And from taht you get the current under no-load conditions.

I hope that helps a bit along your way. Sorry for not having more detailed info.

Harald

Thanks for the reply and clearing up my no load and rated speed problem. I dont understand what you are implying for working out the current tho? I assume you are saying 125 W is wasted in the field and armature resistance but i don't know how much power is wasted in each one ???

If i assume the same mechanical power at 500 W i am able to work out the new torque at the no load speed and then don't have enough info to go on and wok out EMF or Ia....

Did i miss the point?
 

Ledwardz

Dec 21, 2010
43
Joined
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I found this later on in my notes..... perhaps this helps explain it
 

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