A four-pole DC series motor draws 60A from a 220V supply. The armature is wave wound and has a resistance of 0,25ohm. Calculate the useful flux per pole if the armature has 550conductors and rotates at 750r/min. Take the series field resistance to be 0,15ohm.

1 answer

To find the useful flux per pole in the DC series motor, we'll need to go through a few steps involving the motor's electrical parameters.

Given Data

  • Supply voltage (V) = 220 V
  • Armature current (I) = 60 A
  • Armature resistance (Ra) = 0.25 Ω
  • Field resistance (Rf) = 0.15 Ω
  • Number of conductors (Z) = 550
  • Speed (N) = 750 r/min

Step 1: Calculate the armature voltage

The voltage drop across the armature resistance can be calculated using Ohm's Law: \[ V_{a} = I \times R_a = 60 , \text{A} \times 0.25 , \Omega = 15 , \text{V} \]

The voltage across the armature (V_a) when considering the supply voltage is: \[ V_a = V - V_{a} = 220 , \text{V} - 15 , \text{V} = 205 , \text{V} \]

Step 2: Calculate the rotational speed in rad/s

Convert the speed from revolutions per minute to radians per second: \[ N = 750 , \text{r/min} \times \frac{2\pi , \text{rad}}{1 , \text{rev}} \times \frac{1 , \text{min}}{60 , \text{s}} \approx 78.96 , \text{rad/s} \]

Step 3: Calculate the back EMF (E)

The back EMF can be calculated using the formula: \[ E = V_a = 205 , \text{V} \]

Step 4: Calculate the armature reaction (induced voltage)

The relationship for back EMF is given by: \[ E = \Phi \times \frac{P \times N}{60} \times \frac{Z}{2} \] Where:

  • P = number of poles = 4
  • N = rotational speed in r/min
  • Z = total number of conductors
  • \(\Phi\) = useful flux per pole in Weber

Reorganizing for \(\Phi\): \[ \Phi = \frac{E \times 60}{P \times N \times \frac{Z}{2}} \]

Step 5: Substitute the values

Insert the values we have:

  • E = 205V
  • N = 750 r/min
  • P = 4
  • Z = 550

Substituting: \[ \Phi = \frac{205 \times 60}{4 \times 750 \times \frac{550}{2}} \]

Calculate \(\Phi\): \[ \Phi = \frac{12300}{4 \times 750 \times 275} \]

\[ = \frac{12300}{825000} \] \[ = 0.01491 , \text{Wb} \]

Conclusion

The useful flux per pole \(\Phi\) for the four-pole DC series motor is approximately 0.01491 Wb (Webers).