Optimizing Wire Gauge for Efficient AC Power Transmission

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Discussion Overview

The discussion centers around determining the minimum wire gauge required for efficient AC power transmission given specific parameters, including power factor, real power, voltage, and distance. Participants explore calculations related to current, resistance, and power loss in the context of AC systems.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states the parameters of the problem, seeking to find the minimum wire gauge and expresses confusion about working with AC.
  • Another participant confirms that RMS current can be used for AC calculations.
  • A participant proposes using the equation for real power (P = VI cos(theta)) to find current, questioning whether the voltage is in RMS form.
  • It is confirmed that the voltage is indeed 240 V RMS.
  • A participant calculates the current as 30 amps using the formula P = VI * power factor and finds that AWG No. 10 can handle this current.
  • The same participant inquires about calculating power losses in the wire and suggests using resistance formulas and AWG tables.
  • Another participant challenges the calculated load, noting that the power factor was not included and that the load is not purely resistive, suggesting that the current may be lower than calculated.
  • This participant highlights two limitations for wire gauge: overheating of insulation and excessive voltage drop, asking if there are guidelines for acceptable voltage drop.
  • A participant questions the understanding of power factor, asserting that it is included in the real power and seeks clarification on the nature of the load.

Areas of Agreement / Disagreement

Participants express differing views on the inclusion of power factor in calculations and the nature of the load, indicating that there is no consensus on these points. The discussion remains unresolved regarding the correct approach to calculating losses and the implications of the power factor.

Contextual Notes

Participants have not fully resolved the assumptions regarding the load type and the impact of the power factor on their calculations. There is also uncertainty about the acceptable voltage drop for the given distance.

Jason03
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The problem I am working on states that I have a power factor of .5, a real power of 3600 watts, a voltage of 240 V, and a distance between a motor and transformer of 100 feet.


Im trying to find the Minimum wire gauge that could be used...


Do I need to find the current and than go to the AWG tables to find a match?...Since I am working with AC I am a little confused.
 
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Jason03 said:
The problem I am working on states that I have a power factor of .5, a real power of 3600 watts, a voltage of 240 V, and a distance between a motor and transformer of 100 feet.


Im trying to find the Minimum wire gauge that could be used...


Do I need to find the current and than go to the AWG tables to find a match?...Since I am working with AC I am a little confused.

Even though it's AC, you can use the RMS current.
 
Ok...so I was looking at a few equations...could I use the equation for REAL POWER which is:

P = VI cos(theta)

because I have P...and I have V...and I can find cos(theta)...

my only question would be since it 240 volts...im assuming that's in RMS already so I can just plug it straight into the equation?
 
Yes, it is 240 V RMS.
 
ok so I used the formula P = VI* power factor...this allowed me to solve for I which gave me 30 amps...

So I looked up the AWG chart and found that AWG No. 10 can have a max of 30 amps...

So if that is correct...I now have to find the losses in the wire...

What is the easiest way to find the losses in the wire??

Would I need to find the resistance of the current through the wires by first using R = V^2/P...than

going to the AWG table and finding the resistance of the wire for the 100 foot length of the wire?
 
You're okay up to calculating I=30A

However, the load will be different than 16 ohms. For starters, you did not include the power factor, and secondly the load is not resistive. Notice you got a current that is only about 1/2 of the 30A you had calculated before(!)

While I am not an expert in this area, I know of two factors that limit wire gauge:
  • overheating of the wire insulation
  • too much voltage drop for long wire lengths

Given the stated 100' distance, it looks like we are dealing with the 2nd limitation. Were you given any guideline for how much voltage drop from the wiring is within tolerance?
 
I thought the power factor was included in the 3600 watts since its the real power...and if its not resistive what is it?...Im just using the methods that are shown in my text...but the example I am going off of is in DC...
 

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