Calculating Real Power

[BatmanisWatching1987]

I am trying to solve Problem "B"

I got I_a = 13.995∠-90°. (Which is the Line Current = I_L)

The power formula I am also using is, P = √3V_LI_Lcosθ

What is the correct value of "θ" should I be using?

In other problems, I always used the value of θ from the current angle.

 

[mbrooke]

"Bump"

 

[LMAO]

P = 3 * V(phase) * I(phase) * cos(theta).

theta is the angle between phase current and voltage.

V(phase)=202V
I(phase)=202/25=8.1A

P=3*202*8.1*cos(60)=2448W

There are many other ways to solve the problem but there are in fact just a couple of main equations manipulated in different ways.

 

[winnie]

The I_AB equation simply says that the phase angle between the voltage V_AB and the current I_AB has a displacement of 60.

Then when I_AB 'phase current' gets converted to line current you get a -30 degree factor simply because the line to neutral voltage V_AN is 30 degrees out of phase from the line to line voltage V_AB. That -90 degree value is simply changing the reference.

If you had a single phase load connected AB, then the -90 degree value would be relevant because the L-L connection creates an apparent L-N neutral phase angle.

However the problem statement says that you have a 3 phase delta connected load, which means that the L-N phase angles will be the same as the L-L phase angles. So you would use the 60 degrees.

But one additional step, the problem statement says something about connecting 2 watt meters on lines A and B, which is confusing since watt meters must be connected to measure both voltage and current, so you would somehow need to connect the two watt meters to all 3 phases.

-Jon

 

[iceworm]

... But one additional step, the problem statement says something about connecting 2 watt meters on lines A and B, which is confusing since watt meters must be connected to measure both voltage and current, so you would somehow need to connect the two watt meters to all 3 phases.

Well, you do use all three phases. Problem statement says the the system is balanced. It doesn't say if it is Wye or Delta.

And, it doesn't matter, it is balanced. If it is Wye, there is no neutral current. So treat it like a Delta.


Connect watt meter voltage coil A - B, connect current coil around A
Connect second Wattmeter voltage coil to C - B, connect current coil around C.

They don't read Power A, or power B but rather, adding the two together gets you Summation (A + B + C)

 

[BatmanisWatching1987]

P = 3 * V(phase) * I(phase) * cos(theta).

theta is the angle between phase current and voltage.

V(phase)=202V
I(phase)=202/25=8.1A

P=3*202*8.1*cos(60)=2448W

There are many other ways to solve the problem but there are in fact just a couple of main equations manipulated in different ways.

How would I use, P = √3V_LI_Lcosθ
With, I_a = 13.995∠-90°

 

[winnie]

How would I use, P = √3V_LI_Lcosθ
With, I_a = 13.995∠-90°

You don't.

The angle value (-90 degrees) comes from the single phase A-B current, when you converted it to a line current. It doesn't tell you the line current that you would get with a 3 phase load.

-Jon

 

[BatmanisWatching1987]

I thought you could use either P = √3V_LI_Lcosθ or P = 3 * V(phase) * I(phase) * cos(theta) interchangeable.
When would I know, which formula to use.

 

[LMAO]

I thought you could use either P = √3V_LI_Lcosθ or P = 3 * V(phase) * I(phase) * cos(theta) interchangeable.
When would I know, which formula to use.

As I said in this thread these formulas are all the same and interchangeable, they are just mathematically manipulated differently. If you are not getting the same answer you are making a mistake somewhere.

In this case V(phase) is the same as V(line) because it is a delta system.

P=3*V(phase)*I(phase)*cos(theta), |I(line)|=√3|I(phase)| ⇒ P= √3*V(line)*I(line)*cos(theta)

Just remember theta is always the angle between phase current and voltage. In other words theta is your impedance angle.

 

[winnie]

I thought you could use either P = √3V_LI_Lcosθ or P = 3 * V(phase) * I(phase) * cos(theta) interchangeable.
When would I know, which formula to use.

Yes, those formulas give the same value, providing you use the correct theta. Which was your initial question: what theta should you use. Also note that those equations apply only to a balanced 3 phase load.

In the example you presented there were two angles given, 60 and 90 degrees. (Ignoring sign) The 90 degree value came from a _single phase_ calculation for one leg of a 3 phase load. But the balanced 3 phase load has a phase displacement between voltage and current of 60 degrees.

-Jon

 

[BatmanisWatching1987]

Just remember theta is always the angle between phase current and voltage. In other words theta is your impedance angle.

Thanks this will help me out a lot.

Correct me if I'm wrong,

For a Balanced 3 Phase Delta System, since Line Voltage and Phase Voltage are equal, and if the the impedance is given, we can use the phase current for theta to find the Total Power for 3 Phase Load, which was equal to 2.448 kW.

But, if we want to find the Single Phase Power we will use the line current angle, the power would be 0W? (Since the angle is 90 degrees)

 

[topgone]

Thanks this will help me out a lot.

Correct me if I'm wrong,

For a Balanced 3 Phase Delta System, since Line Voltage and Phase Voltage are equal, and if the the impedance is given, we can use the phase current for theta to find the Total Power for 3 Phase Load, which was equal to 2.448 kW.

But, if we want to find the Single Phase Power we will use the line current angle, the power would be 0W? (Since the angle is 90 degrees)

You had been told that the formula for power for every phase = phase voltage X phase current X cosine of the angle between the voltage and the current, period. That's not rocket science.

 

[winnie]

Thanks this will help me out a lot.

But, if we want to find the Single Phase Power we will use the line current angle, the power would be 0W? (Since the angle is 90 degrees)

Strange things show up when you look at the power factor of single phase loads connected to three phase sources.

Say you have a perfect resistor connected line-line on a wye source. Clearly this resistor is using power and has a 0 degree current angle.

But that 0 degree current angle is between the current flowing through the load and the l-l voltage applied to the load.

Now look at the transformer source coils supplying the load. The L-N voltages are not in phase with the L-L voltages. This means that in the transformer coils there is a non-zero current angle.

Back to your original calculation:

The load has a current angle of 60 degrees. It has a 60 degree current angle relative to the applied voltage and must be consuming some real power.

But one of the _source_ legs is seeing a 90 degree current angle, and is not supplying any real power to the load. If you do the calculation for the _other_ source leg, you should see a different result!

-Jon

 

[LMAO]

Thanks this will help me out a lot.

Correct me if I'm wrong,

For a Balanced 3 Phase Delta System, since Line Voltage and Phase Voltage are equal, and if the the impedance is given, we can use the phase current for theta to find the Total Power for 3 Phase Load, which was equal to 2.448 kW.

But, if we want to find the Single Phase Power we will use the line current angle, the power would be 0W? (Since the angle is 90 degrees)

I am not sure if I understand your question. Again, theta is you impedance angle; in this case (Zp=25<60 Ω) θ is 60 degrees. Where did you get 90 degrees? Load power always equals 3*|Iф|*|Vф|*cos(θ) where θ is the load impedance angle. Doesn't matter three phase Y, Δ or single phase. For single phase just remove the 3.

 

[winnie]

I am not sure if I understand your question. Again, theta is you impedance angle; in this case (Zp=25<60 Ω) θ is 60 degrees. Where did you get 90 degrees? Load power always equals 3*|Iф|*|Vф|*cos(θ) where θ is the load impedance angle. Doesn't matter three phase Y, Δ or single phase. For single phase just remove the 3.

Just to amplify: the power through the load always depends on the voltage and current through the _load_, exactly as LMAO says.

The original calculation converted a line-line value to a line-neutral value, an gave you the 90 degree phase angle...but this is not the current angle through the load, and can only reasonably be understood as the current angle seen by the source.

-Jon

 

[Sahib]

Deleted.

 

[Sahib]

View attachment 22295

I am trying to solve Problem "B"

I got I_a = 13.995∠-90°. (Which is the Line Current = I_L)

The power formula I am also using is, P = √3V_LI_Lcosθ

What is the correct value of "θ" should I be using?

In other problems, I always used the value of θ from the current angle.

The lesson to be learned is the power factor angle to be used in problem(B) is the impedance angle of 60 degrees ie angle between phase voltage and phase current and not the angle between line voltage and line current.

 

[BatmanisWatching1987]

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

 

[Jraef]

Cos theta = Power Factor, a number between 0 and 1 that CHANGES depending on the nature of the load. Inductive loads have lower PF, resistive loads have a PF of 1. But you cannot ACCURATELY calculate "real" power without measuring the Power Factor, all you can do is estimate it using some arbitrarily picked value.

For most "mixed" loads I use .80 PF in estimating.

 

[GoldDigger]

The lesson to be learned is the power factor angle to be used in problem(B) is the impedance angle of 60 degrees ie angle between phase voltage and phase current and not the angle between line voltage and line current.

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.