Calculating Power Loss and Input for a Permanent Magnet DC Motor

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Discussion Overview

The discussion revolves around calculating power loss and input for a permanent magnet DC motor, focusing on the implications of steady-state speed, torque, and current at different rotational speeds. The context includes homework-related problem-solving and technical reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about the significance of reaching steady-state speed and how to utilize the given speed of 126 rad/s in calculations.
  • Another participant explains that at steady state, the armature current is steady, allowing the self-induction of the armature to be ignored, and proposes calculating the motor constant, Km.
  • A calculation for back EMF is presented, leading to the determination of Km as 0.749 Vs.
  • Further calculations for torque and power at 100 rad/s are suggested, although the relevance of this speed compared to 126 rad/s is questioned by participants.
  • One participant calculates torque using the derived Km and armature current, resulting in a power calculation that leads to a power loss figure.
  • Another participant confirms the correctness of the power loss calculation, but the context of using different speeds remains unclear.

Areas of Agreement / Disagreement

Participants generally agree on the calculations presented, but there is some confusion regarding the choice of rotational speed for torque and current calculations, indicating a lack of consensus on the relevance of using 100 rad/s versus 126 rad/s.

Contextual Notes

The discussion includes assumptions about the steady-state conditions and the implications for calculations, but does not resolve the choice of rotational speed for certain calculations.

Who May Find This Useful

Students and individuals interested in electrical engineering, motor dynamics, and power calculations in DC motors may find this discussion relevant.

Name15
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Homework Statement


A permanent magnet DC motor has an armature resistance of 0.4 Ohms. When a voltage of 110 V is applied to the motor it reaches a steady-state speed of rotation of 126 rad/s and draws 39 A.
(a) Find the power loss in the armature
(b) the power input to the motor

I really don't know what I'm doing. I just used the equations I already know. What is the significance of steady state speed being reached? What do i do with the 126 rad/s?

Homework Equations


P=IV (power = current x voltage)
P=(I^2)(R) (power = current squared x resistance)

The Attempt at a Solution


(a) P = (39^2) x (0.4) = 608.4 W
(b) P = 39 x 110 = 4290 W
 
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Name15 said:
What is the significance of steady state speed being reached?
The armature current will be steady, thus the self induction of the armature can be ignored.
Name15 said:
What do i do with the 126 rad/s?
The motor constant, Km, can be calculated.
Back EMF = 110V - 39A * 0.4Ω = 94.4V →
Km = 94.4V / 126 rad/s = 0.749 Vs.
You are just not asked about that, but maybe the the missing questions could be:

c) Find the steady state torque at 100 rad/s
d) Find the steady state current at 100 rad/s
 
Hi, thanks for your help!
But why would I find Torque and current at 100 rad/s and not 126 rad/s?
Continuing from your calculations: I find Torque = (0.749 Vs)x(39A) = 29.2 Nm. So using P=Tw; Power = 29.2 x 126 rad/s = 3681 W.
Power input to motor = (39 x 110) = 4290 W. So power loss is (4290 - 3681) = 609 W

Is this correct?
 
Name15 said:
But why would I find Torque and current at 100 rad/s and not 126 rad/s?
That's just an example, how the exercise could be continued.
Name15 said:
I find Torque = (0.749 Vs)x(39A) = 29.2 Nm. So using P=Tw; Power = 29.2 x 126 rad/s = 3681 W.
Power input to motor = (39 x 110) = 4290 W. So power loss is (4290 - 3681) = 609 W
Yes, that's correct.
 

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