Calculating Result Vector for Unsymmetrical Three-Phase Current

[silentcreek]

Hi all,

first I have to say, that my mother tongue is not English. So I'm not very used to special English words that belong to physics. But I try to translate as well as you can understand what I mean. Therefore it would be helpful to me, if you wouldn't use any special acronyms in your answers.

My problem belongs to three-phase current.
I want to calculate the result vector of a three-pase current system, that is used unsymetric. For example the first phase has 6A (Ampere), the second one has 5A and the third one 1A. Then I want to have the amount of current that is in the "null conductor" - the result current.
So I know that this would be a simple addition of vectors, but I have trouble finding a good relation to calculate it.
I already found a relation using trogonometric functions. That solution works fine, but it is very complicated.
My result would be for this example: 4,58A

Is there any easy relation to get the result?

I hope you understand my problem.

Tanks.

Timo

 

[eljose79]

Dear Timo,

Thank you for sharing your problem with us. To calculate the result vector of a three-phase current system, you can use the following equation:

Result vector = √(I1^2 + I2^2 + I3^2 + 2I1I2cos(θ12) + 2I2I3cos(θ23) + 2I3I1cos(θ31))

Where I1, I2, and I3 are the currents in each phase, and θ12, θ23, and θ31 are the angles between the corresponding currents.

To apply this equation to your example, we can plug in the values of 6A, 5A, and 1A for I1, I2, and I3 respectively. As the currents are not symmetrical, we can assume that the angles between them are all different. Let's say θ12 = 30°, θ23 = 60°, and θ31 = 90°.

Substituting these values into the equation, we get:

Result vector = √(6^2 + 5^2 + 1^2 + 2(6)(5)cos(30°) + 2(5)(1)cos(60°) + 2(1)(6)cos(90°))
= √(36 + 25 + 1 + 60 + 10 + 0)
= √(132)
= 11.49A

So the result vector for this example would be approximately 11.49A.

I hope this helps and is easier to understand than the trigonometric solution you found. If you have any further questions or need clarification, please let me know.

 

[Graeme Robinson]

Hello Timo,

I understand your problem and I will try to explain it in the simplest way possible. To calculate the result vector for an unsymmetrical three-phase current system, you can use the following formula:

Result vector = √[(I1² + I2² + I3²) - 2(I1I2cosθ + I2I3cosθ + I3I1cosθ)]

Where I1, I2, and I3 are the current values for each phase, and θ is the phase angle between the currents.

In your example, I1 = 6A, I2 = 5A, I3 = 1A, and θ = 0 (since the currents are in phase). Plugging these values into the formula, we get:

Result vector = √[(6² + 5² + 1²) - 2(6*5*cos0 + 5*1*cos0 + 1*6*cos0)]
= √[36 + 25 + 1 - 60] = √2

Therefore, the result current would be √2 A, which is approximately 1.41 A. This is slightly different from your calculated value of 4.58 A, so I suggest checking your calculations again.

I hope this helps. Let me know if you have any further questions.