How Do You Calculate Per Unit Values for a 20MVA Motor at 13.8kV?

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Discussion Overview

The discussion revolves around calculating per unit values for a 20MVA induction motor operating at 13.8kV. Participants explore how to determine the rated load current in Amps, as well as the real and reactive power drawn by the motor, all expressed in per unit form using the motor's apparent power and rated voltage as bases.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants express uncertainty about how to approach the problem, particularly regarding the application of per unit values.
  • One participant mentions confusion over multiple variations of apparent power formulas provided by their professor, indicating a lack of clarity on which formulas to use.
  • Another participant suggests that the original poster should repost the question in the homework section, adhering to specific guidelines for better assistance.
  • There is a question raised about understanding the per unit method and terminology, specifically what 1.0 p.u means.
  • A suggestion is made to select base values for power and voltage as a starting point for solving the problem.
  • One participant rephrases the question to clarify the scenario of the motor drawing full rated load current at a specified power factor.

Areas of Agreement / Disagreement

Participants generally agree that the question pertains to homework and that it should be formatted according to specific guidelines. However, there is no consensus on the best approach to solve the problem or on the understanding of the per unit method.

Contextual Notes

There are limitations in the discussion regarding the clarity of the per unit method and the specific formulas to be used, as well as the need for defined base values for power and voltage.

Who May Find This Useful

This discussion may be useful for students learning about per unit systems in electrical engineering, particularly those working on homework related to induction motors and power calculations.

DireStraits1
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1) I'm not sure how to solve this question



A 20MVA induction motor connected at 13.8kV draws a load current of 1.0 p.u, 0.85 power factor lag. Calculate the rated load current in Amps and the real power and reactive power drawn by the motor. Express these quantities in per unit form using the apparent power of the motor as the VA base, and the motor rated voltage as base voltage.
 
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I know I'm not supposed to post homework questions but my profesor told me like literally three different variations of apparent power formulas and at this point really I don't even know what formulas to use
 
There is a homework section where this should have been posted
along with a set of guidelines on how to set out your query

i suggest you repost in there using the format that is given and
that may engender some help for you :)

BTW welcome to PF :)

Dave
 
Doubtless a mod will eventually move this to homework.

Meanwhile the first question is do you understand the per unit method and terminology?

In particular what do you think 1.0 pu means?
 
DireStraits1 said:
1) I'm not sure how to solve this question



A 20MVA induction motor connected at 13.8kV draws a load current of 1.0 p.u, 0.85 power factor lag. Calculate the rated load current in Amps and the real power and reactive power drawn by the motor. Express these quantities in per unit form using the apparent power of the motor as the VA base, and the motor rated voltage as base voltage.

A good starting point would be to select base values for the power and voltage.

Once you have the rated values of power and voltage in per unit form finding the rated load current shouldn't be to difficult.
 
DireStraits1 said:
1) I'm not sure how to solve this question
Can you solve it when worded like this:

A 20MVA induction motor connected at 13.8kV draws full rated load current at 0.85 power factor lag. Calculate the rated load current in Amps and the real power and reactive power drawn by the motor.[/size][/color]
 

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