Resistive Load Connected Between Two Phases

Part of the problem with blogs (this image was copied from an INTECH article) is that they are not checked for content and/or accuracy. You are right to question this, but rather than answer directly, I'm going to ask a few questions.

Shown is a single phase resistive load:

1. yet the power factor is stated as 0.86?

2. how does a resistor draw reactive current?

3. the load is unbalanced while the phasor diagram is shown as balanced?

4. without knowing anything about the source and the relative size of the load, how can you determine the power factor, power flows, degree of imbalance?

Go into a real textbook, not a blog, for the answers.

Dear Friend,

You have stated that a resistive load is connected between two phases. Since it is pure resistive load, the current drawn will be in phase with the voltage, developing pure heat and hence no reactive power.

The unbalanced current will flow through the neutral. Magnitude of the current will be 1.732 x V where V is the voltage of single phase.

DHAYANANDHAN.S

If a resistive load is connected across two phases, and the original poster has given nothing about the other connections, please explain how a current of any sort other than that induced by electromagnetism from the connected phases, flows in the neutral at all?

The current through a resistor is in phase with the voltage across the resistor. How does the phase of the voltage across the resistor compare with the R phase voltage and S phase voltage?

The voltage across the resistor doesn't "compare" with anything, it is simply the vector difference between the voltages applied at each end. If the circuit element is purely resistive then the voltage across it and the current through it are in phase.

You are in my estimation fully correct in what you say.

Some people make the error (I believe) in seeing this as a "single phase" question, when in fact it is (as you said) the vector sum of two phases at 120° to each other......

In training many years ago now, we would actually plot the voltage (and the polarity of course) for a whole 360° revolution of both of the two phases, a total of 480° in this case.

The voltage across the resistor at any point being easily read off by using a ruler (inches in my day) to see the value. The current being in phase of course with the voltage....

I personally found it far easier to understand, as a visualization is important I feel for many people. I am one of those who sincerely believe in a picture being worth a 1000 words.....

But I also had a few colleagues who could visualize such things in their heads perfectly, without any help!! (Always finishing first, time wise, in exams! But making the same errors as some others too!!)

I would not be surprised to learn that you are one of those people too who can instantly visualize such questions..... Its very useful if you can do it!

I was trying to lead the OP to figure out the answer without spelling it out, hoping he would understand about phasors.

Since G voltage it is VRS no current will flow through neutral. That means the single-phase projection it is a virtual one only. You could project on RT or TS also in the same time, but this it is a geometrical exercise only.

Let's say Zload=Rload+jXload and Xload/Rload=tan(fi).

fi=30 degrees cos(30)=sqrt(3)/2 sin(30)=0.5 tan(30)=1/sqrt(3).

Zload=Rload*(1+tan(30).j); IR=IRS=-IS.

IRS=VRS/Zload=VRS/Rload/(1+1/sqrt(3)j)=VRS/Rload*(1-1/sqrt(3)j)/[1^2+1/sqrt(3)^2].

S=VRS*VRS*(1-1/sqrt(3)j)/(1+1/3)=VRS^2/Rload*(1/4*3-3/4/sqrt(3)j)

P=VRS^2/Rload*3/4 ; Q=VRS^2/Rload*sqrt(3)/4

If Rload=1/G than SRS=3/4*VRS^2*G-sqrt(3)/4*VRS^2*G.

No meaning for SR or SS only for SRS.

Hi

When we have two load like in fig below

We took the reference, the simple voltage system (VRN, VSN, VTN)

Then

For R1

Ib1 = Ib1 Ð(phi + 0°)

For R2

Ib1 = Ib1 Ð(phi + ∏/6)

Ib2 = Ib2 Ð(phi + 7∏/6)

this is general treatment to calculate the total power consumption and the power factor.

But we can't obtain the correct power flow per phase. (No meaning for SR or SS only for SRS.)

thanks to everyone.

OK. In this case indeed Itot=VRN/R1+VRS/R2*[cos(30)+sin(30)j]

ItotRe=VRN/R1+VRS/R2*sqrt(3)/2

ItotIm=VRS/R2/2

However you have not a common voltage and then S=VRN^2/R1+VRS^2/R2 [only active power, reactive power it is still 0].

Me again, :)

Using the symmetrical components i have calculated the primary side current of Dyn11 Transformer supplying unbalanced resistive (100 A) load connected between R and S phases. (see the figure below)

the generator gives a balanced voltage system.

how obtain the power factor in each phase of primary side, using the phasor diagram?

we took simple voltage system us reference.

Thank you.

Sorry for the delay. Very busy days!

Let's take from the beginning.

In every transformer Np*Ip+Ns*Is=Np*Io

where Np=number of turns on primary winding; Ns=no.of turns in secondary and Io=no load current. Since Io is very small we may neglect it and Ip/Is=Ns/Np .

Let's label the primary terminals as R,S,T [upper case] and secondary terminals as r, s, t[lower case].

If we stated VR/Vr=1 then VRS/Vr=sqrt(3) and then Np/Ns=sqrt(3)

Then [in your case DYn11] IRS=-Ir/sqrt(3) , IST=-Is=Ir/sqrt(3) and ITR=0.[As stated by you Ir=-Is =projected on Vrn].

From the attached phasor diagram :

absolute value Ir=Irs=Is=sqrt(86.6^2+50^2)=100 [approx].

Then Ir=100*[cos(30o)+jsin(30o)=86.6+50j

Is=-Ir=-86.6-50j

On the primary side at point R,S,T -according to Kirchhoff law-sum of all currents entering the point=0.

ITR-IRS-IR=0 in point R ; ITR=0; IR=-IRS=Ir/sqrt(3)

IRS-IST-IS=0 in point S; IS=IRS-IST; IST=-Is/sqrt(3)=Ir/sqrt(3); IS=-Ir/sqrt(3)-Ir/sqrt(3)=-2.Ir/sqrt(3)

-ITR+IST-IT=0 in point T; IT=IST=Ir/sqrt(3)

If IR it will be projected on VRN [instead of Vrn]-from the attached sketch:

Absolute value IR=100/SQRT(3)=57.735

IR=57.735*COS(60)+57.735*SIN(60)i=28.868+50i

IT=IR=28.868+50i

IS=-2x[28.868+50iI]=-57.735-100i

I'll try to attach the sketch once again :

Thank you.

What about the power factor per phase?

If you will represent the voltage VRN,VSN,VTN and the currents IR,IS and IT on a phasor diagram you will get the following:

IR=28.86751+50i cos(fi) with respect to VRN cos(fi)=28.87/sqrt(28.87^2+50^2)=0.5 fi=acos(0.5)= 60o ;

In the same way IS fi=120o ; IT fi=60o.

So per phase you have IR-VRN 0.5 IS-VSN 1 IT-VTN 0.5

a power factor for resistance load.
there is a theoric transit of reactive power for unbalanced resistive load

No[total] reactive power is involved. Reactive power from phase R is positive , reactive power from phase T is the same but negative[capacitive] and reactive power in phase S is zero. Total reactive power is zero. Total active power at primary side[173.2*Voltage] it is equal with the active power in the secondary side.