Circuit analysis (with transformers and motor)

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SUMMARY

The discussion focuses on circuit analysis involving transformers and motors, specifically calculating the impedance of a motor delivering 5hp with a power factor of 0.6 lagging and an efficiency of 90%. The impedance of the normally running motor is determined to be 8.34Ω, derived from the input power of 4144W and the apparent power calculated as 6907VAR. Additionally, the impedance seen by the source is calculated to be 2.32mΩ, factoring in the transformation ratio of the system.

PREREQUISITES
  • Understanding of electrical power calculations, including real and apparent power.
  • Familiarity with transformer theory and impedance transformation.
  • Knowledge of motor efficiency and power factor concepts.
  • Basic skills in trigonometric functions and their application in electrical engineering.
NEXT STEPS
  • Study the principles of transformer impedance transformation in detail.
  • Learn about calculating motor efficiency and its impact on circuit performance.
  • Explore advanced power factor correction techniques for AC motors.
  • Investigate the effects of load variations on motor performance and impedance.
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Electrical engineers, students studying circuit analysis, and professionals involved in motor design and transformer applications will benefit from this discussion.

sandy.bridge
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Homework Statement


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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).
P_{out}=5hp=3730W, P_{in}=3730W/0.9=4144W, F_P=0.6
Z=V^2/S
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
and
Z=(240V)^2/6907VAR=8.34Ω

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".
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
 
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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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