Find the brake torque in separately excited machine

[Fatima Hasan]

Homework Statement

A 300 V DC separately excited motor has the parameters Ra= 0.409, C= 1.90986 V rad/s The motor drives a constant load torque of value 286.479 Nm at speed = 1200 rpm
Part 1: If the motor needs to be braked using dynamic braking and that is by using a braking resistance of value Rb=0.6
a) The braking torque at the instant of braking
b) The braking torque when the machine speed has decreased to 800 rpm
c) The braking torque when the machine speed becomes zero.

Relevant Equations

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Here's my attempt for part 1a:

What I am not sure about is the direction, is it in the opposite direction (= - 458.37 N.m ) or it's positive?

Thanks,

 

[Tom.G]

Homework Statement:: A 300 V DC separately excited motor has the parameters Ra= 0.409, C= 1.90986 V rad/s The motor drives a constant load torque of value 286.479 Nm at speed = 1200 rpm
Part 1: If the motor needs to be braked using dynamic braking and that is by using a braking resistance of value Rb=0.6
a) The braking torque at the instant of braking
b) The braking torque when the machine speed has decreased to 800 rpm
c) The braking torque when the machine speed becomes zero.
Relevant Equations:: -

What I am not sure about is the direction, is it in the opposite direction (= - 458.37 N.m ) or it's positive?

I think that is more a linguistics question than anything else; and what your instructor uses as defaults.

The phrase "The braking torque..." could also be worded as "The torque that is applied for braking..."

Using that interpretation, it would be always be Positive.

If the value of the braking torque is used in further calculations, such as finding the speed, there will be a subtraction involved somewhere. The subtraction may be done as either the subtraction of a positive number, or the addition of a negative number. The results are identical.

My personal opinion, based only on the wording, would make the braking torque Positive. I would likely to argue the point if the instructor disagreed, unless he has stated otherwise in class.

Good Luck!
Tom

p.s. please let us know how it turns out

 

[Babadag]

As Tom already said, in a dynamic braking process, we have to disconnect the source of the armature and turn the motor into a generator producing current through Ra and Rb. The generator is consuming the kinetic energy accumulated. That means from 1200 rpm up to 0 the torque is negative [opposite to the motor torque].

If the rotor it is in a still stand position -that means rpm=0-no Emf is producing then dynamic braking it is not possible. The start current is high Istart=Vsource/Ra and the motor torque is elevated but you have to find another way-not dynamic braking-to brake it.

p.s. by the way the Ea formula has to be Ea=C*2*π*rpm/60=240

 

[Fatima Hasan]

would make the braking torque Positive

Thanks for your explanation.
I asked him about the direction of the torque during braking, is it always negative? He said "Yes".

 

[Fatima Hasan]

p.s. by the way the Ea formula has to be Ea=C*2*π*rpm/60=240

Thanks for your reply.
Could you please tell me what is the difference between them? because sometimes we use C if it's given, and mostly use Ka * Phi.

 

[Babadag]

If we calculate the mechanical- shaft- power Pm=Tq.2.π.rpm/60=36000 W [36 kW] and add the copper losses as Ra.Ia^2[Ia=150 A] -neglecting other losses [friction, ventilation, magnetic and other] we get Pinput=45000 W [divided by 300 V we get 150A].C=Tq/Ia C=286.479/150=1.90986. Then C=Ka. Φ