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Circuit analysis (with transformers and motor)

  • Engineering
  • Thread starter sandy.bridge
  • Start date
  • #1
798
1

Homework Statement


Untitled-10.jpg

Consider the motor arrangement shown, Assume ideal transformers and lossless transmission lines. The start capacitor shown in series with the motor is only used to start the motor, and is then shorted out of the circuit. The motor is loaded such that it delivers 5hp, and has a power factor of 0.6 lagging, and the motors efficiency is 90%.

A)Find the impedence of the normally running motor (ie. capacitor shorted).
[tex]P_{out}=5hp=3730W, P_{in}=3730W/0.9=4144W, F_P=0.6[/tex]
[tex]Z=V^2/S[/tex]
[tex]S=\sqrt{Q^2+P_{in}^2}=\sqrt{(tan(cos^{-1}(F_P))P_{in})^2+P_{in}^2}= \sqrt{(tan(cos^{-1}(0.6))4144W)^2+(4144W)^2}=6907VAR [/tex]
and
[tex]Z=(240V)^2/6907VAR=8.34Ω [/tex]

B) What impedence is seen by the source Es? What is the magnitude of the source current?
Firstly, I considered the motor segment to be "secondary".
[tex]Z_S=n^2Z_P,\rightarrow{Z}_P=Z_S(e_2/e_p)^2=(8.34Ω)(240V/14.4(10)^3V)^2=2.32mΩ [/tex]
 

Answers and Replies

  • #2
798
1
I don't really think I did the calculation of the impedence in part A correctly. Since the motor is 90% efficient, that indicates that 10% of its input energy is dissipated within its terminals due to internal impedence. I would really appreciate some help with this question, as I cannot move on without it! Thanks
 

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