Understanding Power Formulas in Three-Phase Motors: P= √3 Uf*If vs P= Uf*If

I have a question about three-phase motors, and more about √3 in power, e.g. active

Because sometimes I see P= √3 Uf*If

But when I solve a problem, we suddenly use Pf= Uf*If
And I don't know why that is...

I know Up is the phase-to-phase voltage and Uf is the phase-to-phase voltage
But the task said to calculate the total active power Pc AP= √3 Uf*If it's like 1 load from 3 phases but if you add all 3 it doesn't make sense ....
Pc= 3Uf*If

As if someone could explain it to me because I have no idea when P= √3 Uf*If or P= Uf*If


Thanks in advance for your help as I have not understood this for some time




Could you explain it as simply as possible? What a boy this is

markus-19 wrote:

The formulas apply to three-phase alternating current systems.

You forgot to add that the voltage and current must be a sine wave, we operate on rms values, and the load is symmetrical, i.e. the receivers in the arms of the triangle or star are identical.
If any of the waveforms (voltage, current) is not a sine wave, these patterns are g worth.
Nowadays, the voltage still holds the shape (sine wave), but the current is much worse.
Apparently the task is theoretical, but you need to be aware of the limitations for "patterns", because later flowers come out when measuring the power of, for example, LEDs.

I removed the unnecessary ones. Please speak on the subject without excursions.[Mod.TM]

markus-19 wrote:

certainly the author would like to understand the basics of calculations first,

The basis of the "calculations" is the scope of their applicability. When we go beyond this range, we count, measure ... the weather.

Thank you markus-19!

I just checked how it is with active power (probably with other powers) P= U*I*cosφ

But I saw that the power of 1 three-phase receiver is P = √3*Uf*If*cosφ and it puzzled me a lot. Forgive me for forgetting to mention this, it may be very important

I also know how it is with symmetric and unsymmetrical three-phase systems.

And thank you jack63 I didn't know that we operate with effective current but it's worth knowing but when it comes to measurements, I don't do them yet and that's why I want to know the formulas but your advice is also helpful to me! Because soon I'm starting work in connecting but I just don't know yet at what current, whether it's direct or alternating, so I repeat the direct current and alternating current to cope with the practice on adapters (knowledge tests from school) and to understand what the instructor is saying to me .

Thanks again and I look forward to your reply

Xenon02 wrote:

P= U*I*cosφ

It is on phase 1.

On 3 phases the power is only √3 higher, not 3 times because the voltages are out of phase.

Then explain to me why P= 3Uf*If*cosφ if you say it should be √3

Depends on what voltage you substitute into the formula. Phase-to-phase (typically 230VAC) or phase-to-phase (400VAC) voltage
Note that 400/230=~ 1.73, which is about the square root of 3.
Remember the assumptions! The formula applies to a symmetrical receiver!
For a system with asymmetrical reception, e.g. a house powered by 3ph, the current power must be measured or calculated knowing the receivers on individual phases. Because household receivers are usually single-phase and connected in an unbalanced star, there is no multiplier 3, only the sum of the powers of the individual phases.
He remembers that this is how we can count, measure when the receivers are linear, i.e. their current with sinusoidal excitation is also sinusoidal!
Such receivers are disappearing at a fast pace. Currents in home installations deviate more and more from the sine wave, so the above method of counting loses its sense.
You need to be aware of this in order not to make mistakes when measuring.
Here's an example of such stupidity:
https://www.elektroda.pl/rtvforum/topic3483634.html

Okay, I think I'm starting to understand about 1-phase receivers in a 3-phase network (let's assume symmetrical) P=√3 Uf*If

And if not, we add both phases

But I still have a question about the voltage distribution and the current in the triangle and in the star because it is a bit strange in my opinion, e.g. phase-to-phase voltage in the star or current in the star



Let's say they are dynamos and they create electricity

So when the star does it, the current goes through E1 and we have Current and Voltage Drop also phase-to-phase, why do we use phase-to-phase?

And the right question is probably more why we measure both circuits, since if the current has no way to return, e.g. the motor goes such a current and it reaches the resistor, then if there is no 0, e.g. in a triangle, it goes back from phase 1 to phase 2? and in the star that when the current separates, it also returns to the phase? and this voltage drop in the star that phase eg 230 and between 400? then the same current from E1 goes through e.g. E2?

In the triangle, when the current enters node 2, it spreads evenly to the receivers e12 e23.
So the current entering node 2 is higher than the current flowing through these 2 receivers together by √3.

Yes, but where does it return to its "-" because the current goes conventionally from + to -

Xenon02 wrote:

Yes, but where does it return to its "-" because the current goes conventionally from + to -

There is no permanent downside here. Minus constantly changes depending on the moment of time. And it depends on the other 2 phases.

There is no - or + constant in three-phase current.
https://www.wykop.pl/link/4162073/trojfazowa-...c-wytlumaczenia-za-pomoca-wody-i-strzykawek/#

Yes, but so that - in the phase where the current flows?

Current is water, see how it flows every moment in the video.

And if you can't see it, sponsor yourself 6 syringes and a hose.

Are you suggesting that the current goes to the receiver and then turns back?

Xenon02 wrote:

Are you suggesting that the current goes to the receiver and then turns back?

And that's 50 times a second

Mhmm, so I understood correctly ... the current goes to the receiver, there is no voltage anymore because there was a drop and it comes back because I know that + and - is variable

Power in a DC circuit
P=U*I

@ Stream of Consciousness

Strumien swiadomosc... wrote:

There is no permanent downside here. Minus constantly changes depending on the moment of time. And it depends on the other 2 phases.

There is no - or + constant in three-phase current.
https://www.wykop.pl/link/4162073/trojfazowa-...sc-wytlumacona-za-pomoca-wody-i-strzykawek/#

There is no permanent minus, but there must be a closed circuit, so where is this lower potential where the current flows for a short while?

In the video, I noticed that if the phase, say, F1 (red) is at its peak, the temporary minus is the phase F2 (yellow syringe) and F3 (blue syringe). I do not know how to put it in words so as not to offend anyone from the professionals here, but I will say that F2 consumes less power and F3 a little more and F1 only supplies current. Phase-to-phase and phase-to-phase voltages can be calculated.

Am I a bit right here?

I'll throw in my two cents.
You're right.
Regarding the main question why P=√3UpIp.
Think of the receiver as a "black" box (symmetric, you don't know how it is paired, whether Δ or γ . The general formula is 3x UfIf . If the receiver is Δ, so Uf = Up and the current Ip is greater than If by √3. If the load is γ then Uf is √3 smaller than Up and the line current is equal to the phase current.
Therefore, the general formula for active power (for R) P = √3UpIp, and therefore general because it is easiest to implement, just measure Up and Ip of one phase and you have the total load power.
Practice a bit on resistors (symmetric receiver) and you will see that no matter how you calculate the total power will always be the same.

Xenon02 wrote:

I have a question about three-phase motors, and more about √3 in power, e.g. active

Because sometimes I see P= √3 Uf*If

But when I solve a problem, we suddenly use Pf= Uf*If
And I don't know why that is...

First of all, you need to pay close attention to the indices next to the symbols of individual quantities. Because P is not the same as P f. A for example AT p is not equal AT f in a star system. Phase and conductor sizes are different.