Complex question

[EEpowerOK]

Given a voltage and current with different angles (complex),

Say voltage angle is 20 degrees and current angle is 30 degrees, 20 minus 30 = -10

which is correct?

S= V * I <-10

or

S= V <20 * I <-10

and on a drop?

Vin = V<20 + I <-10 * R

or

Vin = V + I <-10 * R

 

[iahim]

EEPowerOK,

Apparent Power: S = V * I (no angles)

Complex Power: S = V * Conj(I) (with angles)

S=|S|

For your example,

S = V<20 * I<(-30)

I'm not sure about your second question. We need some circuit to understand it.

 

[EEpowerOK]

S = V<20 * I<(-30) is the same as V<0 * I<-10

On the drop, it is a generator, and the values are given

 

[iahim]

Right, they are the same. But I would just put in the calculator the voltage as given and the conjugate of the current.

For a generator: V terminal = Ef - I * X, with the angles as given in the problem.

 

[EEpowerOK]

Yes, I'm calling Vout same as V terminal.

This is CI V2 Vol2 #54. If the answer used the S=VI*, then keep the voltage reference angle of 14 degrees. But the answer used voltage angle minus current angle which he set to (14-34.9)= -20.9. That should of reset Vin (Excitation voltage) to 0 degrees. The answer continued to use 14 degrees for the Vin.

answer final was 512<14 - 47<-20.9 * (.6 + 1.3J)

He stated the current lagged so it was originally as pf of .82 lagging or -34.9 degrees, Actually the angle between the voltage and current should of been (14 - (-34.9))= 48.9 degrees not -20.9.

 

[iahim]

I think the answer from the book is correct (at least in my version, where this problem is #26).

pf=0.82 lagging, so theta - phi > 0

theta is the angle of the generated phase voltage

phi is the angle of the line current

So 14 - phi = + cos-1 (0.82)

phi = 14 - 34.92 = -20.92

KVL: 512<14 = 47<-20.92 (0.6 + j1.3) + Vout<alpha

So Vout<alpha = 455.17<9.72

 

[EEpowerOK]

Yes, this is the same problem. I guess my confusion is where you listed

14 - phi = + cos-1 (0.82)

I'm thinking

14 - phi = - cos-1 (0.82)

because it is lagging.

 

[iahim]

For lagging power factor, the current lags behind the voltage, so the angle of the current is smaller than the angle of the voltage, or phi < theta. That's where I came up with: theta - phi > 0

In general, theta - phi = +/- cos-1 (pf)

For lagging power factor, theta - phi = + cos-1 (pf) (because from above theta - phi has to be greater than zero)

For leading power factor, theta - phi = - cos-1 (pf) (theta - phi has to be less than zero)

Now, if the voltage angle is zero (theta = 0), the above equations become:

For lagging power factor, - phi = + cos-1 (pf), or phi = - cos-1 (pf)

For leading power factor, - phi = - cos-1 (pf), or phi = + cos-1 (pf)

In most problems, the voltage angle is zero, so you are probably used to the last two equations. That's probably why you're thinking the sign should be negative.

 

[msteiner]

For this particular CI problem, why is it a given that the PF describes the relationship between the current and the GENERATED voltage as opposed to the LINE voltage?

 

[EEpowerOK]

The line voltage is the generated voltage if you add back the voltage drop from the line. The generated voltage is the largest voltage point if there are no other sources.

 

[msteiner]

The line voltage is the generated voltage if you add back the voltage drop from the line. The generated voltage is the largest voltage point if there are no other sources.

I understand the concept of the generated voltage, voltage drop within the machine, and the resulting line voltage. I just don't understand why the PF angle given for the current is given with respect to the generated voltage, and not the line voltage.

 

[EEpowerOK]

You got to think in terms of the formula Vout=Voltage generated minus Current * Impedance. The current that flows out the generator is moving through the line and the impedance of the line is going to cause a voltage drop. The voltage and current originated in the generator so the PF is calculated from that point.

 

[GreenNGold]

I had a question about the solution for complex imaginary exam 2, problem 28.

Question states: A 3ph, 208V bus has 2 motors contacted to it. The first motor is a 7kVA sync motor with a leading pf of 0.78. The second motor is a 9kVA squirrel-cage motor with a lagging pf of 0.65. What is the system power factor at the bus serving these two motors?

If the first motor is is operating at 7kVA with a leading pf of 0.78, isn't the complex power 7<-38.7?

If you have a leading pf, your current angle peaks before the voltage. So if you assumed your voltage angle was 0, your current angle would be +38.7 and applying S = VI* means you would end up with a negative angle, right?

Am I missing something?

Thanks.

 

[Ship Wreck PE]

I got .98 lag. What is answer in book?

 

[GreenNGold]

That is the correct answer and what I got also. But they are saying the S1 = 7<38.7 kVA and S2 = 9<-49.5 kVA.

These are the numbers I got:

For motor 1: P1 = 5.46kW, Q1 = -4.38kVAR, and S1 = 7<-38.7 kVA

For motor 2: P2 = 5.85kW, Q2 = +6.84kVAR, and S2 = 9<+38.7 kVA

A lagging circuit is inductive so Q should be positive. Or am I making a mistake somewhere?