[Solved] power of the heaters when connected in a star and in a delta?

Hello,
please help me calculate what will be the power of 3-phase heaters - each heater has a resistance of R = 8 Ohm connected in a triangle with "strength"
When connected in a star from the formula: I=U/R and P=root of 3*U*I - the power of such an object is: approx. 18 kW (3x6kW) and how to calculate it when connected in a triangle and how much will the power of the object be?
Thanks - Marcin

In the triangle, the phase-to-phase voltage will be applied to the heaters, which is greater than the phase-to-phase voltage by the square root of 3, so the power of the system is greater than that of the star.

If you were able to calculate the power at the star, then at the triangle it is the sum of your powers of each of the heaters. Regards

Added after 2 [minutes]:

You can calculate the rest, and I usually never give ready-made products, because it's not appropriate, but rather forces you to mingle a bit. thinking.

The post was reported.
Transferred from: Industrial Electronics and Power Electronics

Star connection:
I=230/8=28.75A
P=3*230*28.75=19837.5W
Threesome connection:
I=400/8=50A
P=3*400*50=60000W
I think so.

Dido230 wrote:

In the triangle, the phase-to-phase voltage will be applied to the heaters, which is greater than the phase-to-phase voltage by the square root of 3, so the power of the system is greater than that of the star.

Power increases with the square of the voltage, so 3 times.

Hello,

Dido230 wrote:

In the triangle, the phase-to-phase voltage will be applied to the heaters, which is greater than the phase-to-phase voltage o square root of 3 , so that is the greater power of the system compared to the star.

not true !
The voltage increases by √3 times, but also the current value in each load increases by that many times (linear resistor), so the power consumed by a symmetrical delta-connected load will increase three times (√3•√3=3) compared to the power consumed by star-coupled receiver.

Regards

Moderated By oldboy:

Recidivism

The power difference depending on the delta or star connection changes by √3.

And that's all if we are talking about a three-wire network with a sinusoidal alternating current with pulsation and a shape factor that is characteristic of POLISH n/n networks.

I invite my friends Quarz and pmxxyz for tutoring in mathematics and electrical engineering.

Dido230 wrote:

The power difference depending on the delta or star connection changes by √3.

NO Difference and Quotient , and
NO power and tensions .

Buddy's tutoring take yourself

Thanks for the tutoring ... but for peasant reason, if the power in the star is 3 * 6kW = 18kW, then the triangle will be 3 times more, or the first of 3 (1.73) more? ... Please give a specific formula how to count in the star a as in a triangle ... and what voltage to substitute (U = 230 or 400) when calculating resistance - I think that we substitute the star: U = 230 and the triangle: U = 400 or so?

thanks - Marcin

Do you have reading comprehension problems?
The topic is resolved.

Active power: P=U*I*cosφ
The heater is a resistor, so cosφ=1.

Ohm's Law (kids in elementary school know this): R=U/I, or I=U/R.
Substitute into the power formula and you get: P=U²/R.
Power (on the same heater) increases with the square of the voltage (or the square of the current as you transform the formula differently (P = I² * R).
Phase-to-phase voltage U=Uf*√3=230*√3≈400V.
It changes *√3, so the power changes (√3)² = 3 times.

These are the effects of the games' amnesty and the lack of compulsory math at high school.

P.S
Another example of using the book formula for three-phase current power: S=√3*U*I.
U-line voltage (in Polish = phase-to-phase) = 400V. It's 400 always in the formula.
I-conductor current (current in one wire of the installation).

In the triangle I=If*√3,
in the star I=If.

So for the furnace from the subject (R=8Ω, cosφ=1, i.e. S=P), we have:
- for a triangle If=400/8=50A, so I=86.6A.
-for star If=I=230/8=28.75A.
It is worth noting that the proportion of phase currents is √3, and of wire currents 3 (i.e. the current drawn from the network when starting motors with G/T switch is 3 times smaller).

Power for the triangle: S=P=√3*400V*86.6A≈60kW.
Star power: S=P=√3*400V*28.75A≈20kW.
Power 3 times smaller.

You insult yourself and make fun of yourself.
Where did you "learn" electrical engineering?

The power varies with the square of the voltage. Don't argue with that.
It's not like that with you (60kW does not equal 30*3)...

In practical terms, it is mainly about engines (starting).
Then, connecting the windings in a star causes a decrease in the current in the winding by √3, and the power decreases 3 times (most importantly, the torque also proportionally 3 times, because: P=M*ω, and more specifically M=c*U², where c is a constant not important at the moment).

Dido230 wrote:

The power difference depending on the delta or star connection changes by √3.

And that's all if we are talking about a three-wire network with a sinusoidal alternating current with pulsation and a shape factor that is characteristic of POLISH n/n networks.

I invite my friends Quarz and pmxxyz for tutoring in mathematics and electrical engineering.

Dude Dido230 - bastuj, don't be ashamed, I'm from Kielce too.
My advice - delete this post.

Just Quarz eats such tutors in handfuls (on an empty stomach).

sorry but I think I messed up a bit with this question ... I don't have as much knowledge as you but is there any difference with the calculations if we are talking about a resistive load - that's what I mean in a 3-phase resistance furnace
I still stick to the basic formulas: P=U*I and I=U/R

...okay 3 times more with the triangle... end of story

thanks - Marcin

In this matter, it does not matter what it is (stove or engine).

You got the answer: https://www.elektroda.pl/rtvforum/topic721377.html#3717822
TOPIC FINISHED.

P.S
Another thing:
http://wazniak.mimuw.edu.pl/index.php?title=PEE_Modu%C5%82_6

kich wrote:

sorry but I think I messed up with this question a bit ... I don't have as much knowledge as you but is there no difference with the calculations if we are talking about a resistive load - that's what I mean in a 3-phase resistance furnace?
I still stick to the basic formulas: P=U*I and I=U/R
thanks - Marcin

Substitute into the formula P=U*I this formula I=U/R
We get the formula P=U²/R, R is constant (invariant in this problem)
P1=230²/R P2=400²/R
P2/P1=3 and I don't want to leave otherwise.

LiutenetMaria - where does 2R come from in your formula?.

Aleksander_01 wrote:

[...]
LiutenetMaria - where does 2R come from in your formula?.

Because he does not notice that a third resistor is connected between the two resistors, powered from yet another phase ... Therefore, these two cannot be treated as autonomous series-connected resistors (and treated as 2R, because nonsense will come out - and they did).

It's easy for a triangle, the current of a given receiver = 400 V, divide by the resistance.
Then P1+P2+P3 = device power.
It is also easy for the star, provided that the installation is four-wire (neutral point in zero). Calculations see above with the difference that the current of a given receiver = 230 V divided by the resistance of a given phase.

A problem (but a minor one) arises when the neutral point is free.

That was 11 years ago...