Three phase total power consumed

In summary, the conversation is about a user seeking help with a math problem on a forum. The user's handwritten work is difficult to make out, so they are encouraged to type it out for better clarity. Another user points out a mistake in the conversion of arguments from negative to positive. The original user acknowledges the mistake and recalculates, but there is still an issue with the calculation of IR. The correct calculation for IR is discussed using the formula for power.
  • #1
Butterfly41398
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Summary:: I don't seem to see any problem with my work. But my final answer does not match the correct answer. (In number 3)

Here's my work
 

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  • #2
@Butterfly41398 , it's very difficult to make out the contents of your images (the handwritten workings), so I can't really help spot any errors. You'll have to clean up the images to make them useful, or better yet, type out the work so that others can quote it in replies.
 
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  • #3
Butterfly41398 said:
I don't seem to see any problem with my work.
As mentioned, it's best if you type your work into the forum to make it more legible. Please see the LaTeX Guide link below the Edit window for more information on how to format math equations at PF. Thanks.
 
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  • #4
I agree with the berkeman, of course. However, I manage to read your calculation. I still have to recalculate, but at first glance, I have a general observation: to convert the arguments from negative to positive you have to add 360 degrees and not 180.
 
  • #5
In my calculation Iy is correct, Ib is correct except the argument has to be 27.525 instead of 29.525. However Ir is quite different. Ir=28.406<-69.1127.
 
  • #6
Babadag said:
I agree with the berkeman, of course. However, I manage to read your calculation. I still have to recalculate, but at first glance, I have a general observation: to convert the arguments from negative to positive you have to add 360 degrees and not 180.
Following your entire calculation this it is not an issue. The only mistake it is wrong calculation of IR.
The power it is the real part of the sum of products: VR*conjugate(IR)+VY*conjugate(IY)+VB*conjugate(IB) where
VR=IR*ZR ;VY=IY*ZY;VB=IB*ZB
 

FAQ: Three phase total power consumed

1. What is three phase total power consumed?

Three phase total power consumed refers to the amount of electrical power used by a three phase electrical system. It is the sum of the power consumed by each phase in the system.

2. How is three phase total power consumed calculated?

To calculate three phase total power consumed, you need to measure the voltage and current of each phase in the system. Then, use the formula P = √3 x V x I x cosθ, where P is power, V is voltage, I is current, and cosθ is the power factor. Once you have calculated the power for each phase, simply add them together to get the total power consumed.

3. What are the units of measurement for three phase total power consumed?

The units of measurement for three phase total power consumed are watts (W) or kilowatts (kW), depending on the size of the system. This is the standard unit for measuring electrical power.

4. Why is it important to measure three phase total power consumed?

Measuring three phase total power consumed is important for several reasons. It allows you to monitor and manage your energy usage, which can help reduce costs and improve efficiency. It also helps to identify any imbalances or issues in the system that may be causing excess power consumption.

5. How can three phase total power consumed be reduced?

To reduce three phase total power consumed, you can implement energy-saving measures such as using energy-efficient equipment, optimizing the power factor, and minimizing idle loads. Regular maintenance and monitoring of the system can also help identify and address any issues that may be causing excessive power consumption.

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