DC Shunt Motor

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Homework Statement


A 250V shunt dc motor runs at 1000rpm while taking a current of 25 A. The motor has armature and field resistance as 0.2W and 250W respectively. Calculate

(a) The speed when the load current is 50A.

(b) Armature torque developed at both speeds.



Homework Equations


IL = Ia + Ish

Vt = IshRsh = Ea + IaRa

Poutput = Tω = T(2πN)

The Attempt at a Solution



I already solved part a and got N2 = 979.6 rpm

I actually just have a question of solving part B. How exactly do I compute the output without the torque and angular speed.

Is Poutput = IaRa

or Poutput = IaRa - (Iron + Friction) losses
 

Answers and Replies

  • #2
cnh1995
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field resistance as 0.2W**Ω** and 250W**Ω**
The speed when the load current is 50A.
Is it the armature current or the total current supplied by the supply?
Armature torque developed at both speeds.
This means you need to find the mechanical power developed in the motor. Do you know the relation between back emf and mechanical power developed?
 
  • #3
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I dont know why it says W. It is ohms.

The load current is the current supplied by Vt.
 
  • #4
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Pinput(1) = VtIL1 = (250 V)(25 A) = 6250 A at the first speed

and it is Pinput(2) = VtIL2 = (250 V)(50 A) = 12500 A at the 2nd speed
 
  • #5
cnh1995
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Pinput(1) = VtIL1 = (250 V)(25 A) = 6250 A at the first speed

and it is Pinput(2) = VtIL2 = (250 V)(50 A) = 12500 A at the 2nd speed
Ok.
You need to find the armature current first. You know the total current and field resistance. How would you calculate the armature current?
 
  • #6
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Ia1 = IL1 - Ish = 25A - 1A = 24A

Ia2 = IL2 - Ish = 50A - 1A = 49A
 
  • #7
cnh1995
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Ia1 = IL1 - Ish = 25A - 1A = 24A

Ia2 = IL2 - Ish = 50A - 1A = 49A
Correct.
You can calculate back emf from this. What is the equation describing the relation between back emf and mechanical power developed? Back emf is the electrical feedback of mechanical energy developed.
 
  • #8
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Correct.
You can calculate back emf from this. What is the equation describing the relation between back emf and mechanical power developed? Back emf is the electrical feedback of mechanical energy developed.
That's what I'm unsure of. I know that Pinput - Poutput = the total losses (Copper, iron, friction).
And Copper Loss is I2aRa + I2shRsh
 
  • #9
cnh1995
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That's what I'm unsure of. I know that Pinput - Poutput = the total losses (Copper, iron, friction).
And Copper Loss is I2aRa + I2shRsh
Those are all electrical powers. The relation between back emf and mechanical power is Pm=EbIarmature. Can you proceed from here?
 
  • #10
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Ooohh. Yeah, its the conceptual idea I don't get. I don't get why we have to use mechanical power instead of electrical
 
  • #11
cnh1995
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Ooohh. Yeah, its the conceptual idea I don't get. I don't get why we have to use mechanical power instead of electrical
You are asked to find the torque and speed. These are associated with mechanical power. Now, how do you write mechanical power in terms of torque and speed?
 
  • #12
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You are asked to find the torque and speed. These are associated with mechanical power. Now, how do you write mechanical power in terms of torque and speed?
T = PM/ω = EaIa/(2πN)
 
  • #13
cnh1995
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  • #14
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Thanks, I see the mistake I was making! It was pretty simple, I must be tired. Appreciate the help again
 

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