Calculating Real Power

[Sahib]

Not much of a lesson. If you have a balanced resistive delta load the phase current will be in phase with the phase voltage. With a balanced resistive wye load the line current will be in phase with the line voltage. There is no way to tell the difference by any measurements on the three hot leads and the neutral.
What you cannot do is try to directly apply the angle between the line current and the phase voltage or vice versa.

Irrespective of whether it is delta or star connected balanced loads, the power factor angle to be used is the impedance angle ie the angle between phase voltage and phase current. That is the lesson in my last post.

It is really applicable for OP also.

View attachment 22390

Why don't I get the same answer using


P=3*V(phase)*I(phase)*cos(theta) or P= √3*V(line)*I(line)*cos(theta)


Since theta is always the angle between phase current and voltage, I used the angle of 46.75 degress

 

[GoldDigger]

Irrespective of whether it is delta or star connected balanced loads, the power factor angle to be used is the impedance angle ie the angle between phase voltage and phase current. That is the lesson in my last post.

It is really applicable for OP also.

And my lesson is that in all cases the angle between the phase current and the phase voltage is identical to the angle between the line current and the line voltage whenever the load is balanced. With an unbalanced load you will not necessarily have a single phase angle in either the delta or wye computation. But in all cases the power properly calculated using the delta configuration will be identical to the power properly calculated using the wye equivalent circuit.
Putting it another way, the sum of the three power numbers calculated using individual phasor phase currents and voltages will be identical to the sum of the three power numbers calculated using the three phasor line currents and line voltages.
Depending on now the load itself is characterized, one formulation may be computationally easier than the other.

 

[Sahib]

Deleted.

 

[Sahib]

View attachment 22390

Why don't I get the same answer using


P=3*V(phase)*I(phase)*cos(theta) or P= √3*V(line)*I(line)*cos(theta)


Since theta is always the angle between phase current and voltage, I used the angle of 46.75 degress

Another lesson. Whenever you encounter negative power factor angle in degrees (in impedance polar form), add 360 and use the resultant figure in the formula for real power.

 

[Sahib]

View attachment 22390

Why don't I get the same answer using


P=3*V(phase)*I(phase)*cos(theta) or P= √3*V(line)*I(line)*cos(theta)


Since theta is always the angle between phase current and voltage, I used the angle of 46.75 degress

After some struggle off and on, I found the solution. The per phase source voltage across delta connected load after its conversion to equivalent star is obtained by multiplying current (8.567A) and per phase equivalent star impedance of 7+j8 (10.63ohm). The power factor angle is impedance angle of 7+j8(48.82degrees, it is not 46.75 :happyno. Substituting the figures in formula for 3 phase power, it may be verified. As an exercise to OP, calculate total power including (1+j.5) impedance by both methods.

 

[Julius Right]

 

[BatmanisWatching1987]

After some struggle off and on, I found the solution. The per phase source voltage across delta connected load after its conversion to equivalent star is obtained by multiplying current (8.567A) and per phase equivalent star impedance of 7+j8 (10.63ohm). The power factor angle is impedance angle of 7+j8(48.82degrees, it is not 46.75 :happyno. Substituting the figures in formula for 3 phase power, it may be verified. As an exercise to OP, calculate total power including (1+j.5) impedance by both methods.

How do I use the formula S=3V_pI_p∠θ° to get the same Real Power Solution to be 1.541 kW.

I am using the values S=3*100*8.567∠48.82 = 1692.22+1934.37j VA

 

[Sahib]

How do I use the formula S=3V_pI_p∠θ° to get the same Real Power Solution to be 1.541 kW.

I am using the values S=3*100*8.567∠48.82 = 1692.22+1934.37j VA

The voltage is not 100V because of drop in (1+j0.5) impedance. To find per phase voltage drop across delta connected loads, multiply the phase current magnitude by the magnitude of per phase impedance of delta connected loads.

 

[BatmanisWatching1987]

The voltage is not 100V because of drop in (1+j0.5) impedance. To find per phase voltage drop across delta connected loads, multiply the phase current magnitude by the magnitude of per phase impedance of delta connected loads.

V_drop = I_pZ_Load = (8.567∠48.82)(7+j8) = 91.07∠2.06V

S=3VpIp∠θ° = 3*(91.07)(8.567)∠2.06--48.82 = 1476.79+j1815.89

So I get P equal to 1476.79 W.

I'm not sure if I am computing the calculation correctly with the angels?

 

[GoldDigger]

Definitely with the angels. :angel:

 

[BatmanisWatching1987]

Definitely with the angels. :angel:

I'm just not getting the same solution for Power. So I don't know where I am going wrong.

 

[Sahib]

V_drop = I_pZ_Load = (8.567∠48.82)(7+j8) = 91.07∠2.06V

S=3VpIp∠θ° = 3*(91.07)(8.567)∠2.06--48.82 = 1476.79+j1815.89

So I get P equal to 1476.79 W.

I'm not sure if I am computing the calculation correctly with the angels?

You need to take only the impedance angle of (7+j8) and current magnitude of 8.567 ie Vp and Ip only magnitudes and power factor angle is the impedance angle of (7+j8).

 

[Sahib]

V_drop = I_pZ_Load = (8.567∠48.82)(7+j8) = 91.07∠2.06V

S=3VpIp∠θ° = 3*(91.07)(8.567)∠2.06--48.82 = 1476.79+j1815.89

So I get P equal to 1476.79 W.

I'm not sure if I am computing the calculation correctly with the angels?

The angle of current=<-46.75 and not <48.82. The angle of conjugate of Ip=<46.75. So the angle of S=<2.06+46.75=<48.81.

 

[BatmanisWatching1987]

The angle of current=<-46.75 and not <48.82. The angle of conjugate of Ip=<46.75. So the angle of S=<2.06+46.75=<48.81.

I'm just a little confused by how to add the Voltage and Current for finding the Apparent Power Angle.

I understand we need to take the conjugate of the Current Angle.

So, without doing the conjugate of the Current Angle, would we add up the Voltage and Current Angle to get the Apparent Power to be <2.06+(-46.75), then we take the conjugate of the Current Angle to get <2.06+(+46.75).

 

[topgone]

I'm just a little confused by how to add the Voltage and Current for finding the Apparent Power Angle.

I understand we need to take the conjugate of the Current Angle.

So, without doing the conjugate of the Current Angle, would we add up the Voltage and Current Angle to get the Apparent Power to be <2.06+(-46.75), then we take the conjugate of the Current Angle to get <2.06+(+46.75).

The easiest way to understand vector multiplication is to plot it on paper and deduce how much angle is there between the two vectors. Easy!

 

[Sahib]

I'm just a little confused by how to add the Voltage and Current for finding the Apparent Power Angle.

I understand we need to take the conjugate of the Current Angle.

So, without doing the conjugate of the Current Angle, would we add up the Voltage and Current Angle to get the Apparent Power to be <2.06+(-46.75), then we take the conjugate of the Current Angle to get <2.06+(+46.75).

View attachment 22390
Look at your attachment above (from post#18). The angle of current phasor, as worked out, is <-46.75 degrees.

 

[Sahib]

I'm just a little confused by how to add the Voltage and Current for finding the Apparent Power Angle.

I understand we need to take the conjugate of the Current Angle.

So, with conjugate of the Current Angle, would we add up the Voltage and Current Angle to get the Apparent Power angle to be <2.06+(+46.75)? .

Yes. (I edited your quote as above in red to clarify ).

 

[BatmanisWatching1987]

I tried using the same method I just learned, but I am not getting the same solution at all with the following problem.

 

[Sahib]

View attachment 22458 View attachment 22459

I tried using the same method I just learned, but I am not getting the same solution at all with the following problem.

In this problem, they used peak value voltage and current phasors instead of RMS value phasors.

 

[Julius Right]

For the record only: in my post 26 from 02_26_2019 I did a mistake. Actually the attached note has to be:


:ashamed1::slaphead: