DC Shunt Motor Starting Current

In summary: The correct answer should be 1220 rpm. In summary, the conversation discusses the calculation of efficiency, input current, and field current in a shunt motor. It also discusses how to limit the starting current and the back emf at full load. The conversation concludes with a question about the speed of the motor, with the correct answer being 1220 rpm.
  • #1
jaus tail
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Homework Statement


upload_2017-12-3_9-16-47.png


Homework Equations


Efficiency = o/p divided by i/p
I/P KW divided by input voltage gives input current
Field Current = voltage divided by field resistance
armature current = total current - field current
Back Emf = V - IaRa
T proportional to Ia
Back Emf proportional to flux and N

The Attempt at a Solution


For 1st part:
we have efficiency = 0.75 = o/p divided by input
o/p = 18KW
So we get i/p = 24KW

So Vt * (Total current) = 24kW
Vt = 200V
So Total current = 24000/200 = 120A
Field current = voltage / Rf = 200/100 = 2A --- (constant all the time)
Now to limit starting current to three times we have,
Total current must be 120 * 3 = 360A
Field current = 2A
Armature current = 358A
At start Eb = 0
Terminal Voltage = Armature Current times Ra
So 200 = 358 times Ra
This gives Ra = 0.558
Given Ra = 0.3
So we need to add extra 0.26 ohm
So for first part answer is B

But I am not able to solve second part.
At full load Torque is same irrespective of starting resistance since torque depends on load.
In shunt motor Torque proportional to Ia
So Ia is same in both cases
Back Emf is proportional to Speed times flux
Flux is same in both cases of armature resistance as Field Current is same.
So Eb1/Eb2 = 1500/(Nx)
Eb1 = 200 - 118 * 0.3 = 164.6
Eb2 = 200 - 188 * 0.56 = 133.92
So Eb1 / Eb2 = 1.23

So 1.23 = 1500 / Nx
Nx = 1220 rpm
But that is not even in the options.
Where am I wrong, or is the question wrong?
 

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  • #2
jaus tail said:
So Vt * (Total current) = 24kW
Vt = 200V
So Total current = 24000/200 = 120A
Field current = voltage / Rf = 200/100 = 2A --- (constant all the time)
Now to limit starting current to three times we have,
Total current must be 120 * 3 = 360A
Field current = 2A
Armature current = 358A
At start Eb = 0
I beileve the starting current is the armature current at the start and not the total current.
Here, the starting current would be 120-2=118A and you need to limit it at 118*3=354A and not 358A. But the answer is still B).

I agree with your calculations for the second part. I believe all the options are incorrect.
 
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Likes jaus tail

1. What is a DC shunt motor?

A DC shunt motor is a type of direct current motor that has its field winding connected in parallel with the armature winding. This allows for the motor to maintain a consistent speed even when the load changes.

2. What is starting current?

Starting current refers to the initial surge of current that a motor requires to start running. This current is typically much higher than the steady-state current and can cause damage to the motor if it is not controlled properly.

3. How is starting current different for a DC shunt motor compared to other types of motors?

DC shunt motors have a relatively low starting current compared to other types of motors. This is because the field winding is connected in parallel with the armature winding, which helps to regulate the current and prevent it from rising too high during start-up.

4. How is the starting current of a DC shunt motor controlled?

The starting current of a DC shunt motor can be controlled by using a starter or a rheostat. These devices limit the amount of current that is allowed to flow into the motor during start-up, preventing it from reaching damaging levels.

5. What are some potential issues with high starting current in a DC shunt motor?

If the starting current of a DC shunt motor is too high, it can cause damage to the motor's windings and insulation. It can also put unnecessary strain on the motor's mechanical components, leading to premature wear and tear. Additionally, high starting current can result in high energy consumption, leading to increased operating costs.

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