How am I susposed to find the active power here?

[sylent33]

Homework Statement

Multiple questions

Relevant Equations

AC Power

Hi everybody;

With a halogen transformer with the data U = 230 V, f = 50 Hz, I = 1.5 A and cos ϕ = 0.35
a reactive power compensation can be carried out with a parallel capacitor Cp. ¨
(a) Calculate the value of Cp to achieve a power factor cos ϕk = 0.9.

I did that like this;

First I calculated the active power without compensation ; $$ P = U \cdot I \cdot cos \phi $$ and the reactive power without compensation $$ P = U \cdot I \cdot sin \phi $$ and I get ;
P = 120,75 W and Q = 323,15 var

Now the active power remains the same with or without compensation,so I used that to calculated the new current that will flow when we add the capacitor to the circuit;
$$ I_{new} = \frac{P}{U \cdot cos \phi_k } $$ and it should be; I = 0,583A than I calculated Q with compensation; Qcomp = 58,44 var

And we can use this equation ; Qcomp = Qc+ Q to get the Qc value,that should be Qc = -264,71 var

$$Qc = \frac{U^2}{Xc} $$ we can find Xc out of that to be equall Xc = -200 Ohm and now;

$$ Xc = \frac{1}{\omega C} $$ we can find C to be $C = 15,3µF$

b)Represent all relevant powers with and without compensation graphically (correct angles) in the complex plane and state the scale used for thisDidnt really have any problems with this one;

c)
The capacitor Cp calculated in point (a) is wrongly connected in series with the halo gentrafo. Calculate the active power P of this arrangement.
Hint: If you cannot solve point (a), use Cp = 10 µF.

Now here is where I am having problems;

I tried calculating the Impedanz as $Z = \frac{U}{I}$ where the I is the 0,583A I calculated in A. I am assuming that the voltage remains the same because its a series connection. Z comes around to be 394,5 Ohm. Now I tried calculating R and maybe trying to find P that way but that didnt work.I calculated R like this;

R = |Z| cos phi = 355,05 Ohm

And than $P = \frac{R}{I^2}$ but that gives me the wrong result;

P should be (when we use 15mF) equall to 470,3W.

Any help would be appriciated (part a) I think is correct since that is the solution in the book)

Thanks

 

[hutchphd]

halo gentrafo? What is var? $P=\frac R {I^2}$ ?

 

[arhzz]

If I am not mistaken var should be the unit for the reactive power, (volt ampere reactive).

 

[sylent33]

As arhzz stated var is just the unit for the reactive power,thought that wasnt anytinh too out of the ordinary.halo gentrafo is the halogen transformator from a),I guess in the translation google ate some words and it came out like that.

And the quation was a latex error I wanted to type in $P = I^2 * R$ but it just came to my realization that this only works for DC.

 

[hutchphd]

If I am not mistaken var should be the unit for the reactive power, (volt ampere reactive).

Thanks. Never seen that unit (and I'm old). I believe I can live without it !

 

[Baluncore]

Never seen that unit (and I'm old). I believe I can live without it !

The real watt is the VA where Volts and Amps are in phase.
The VAR is the voltage multiplied by the Reactive current (in quadrature), which can have no real power, only circulating energy.
There is still W = I²·r heat generated in the resistance, r, of the circuit by the vector sum of the real and reactive currents.
VA and VAR should be capitalised.

 

[hutchphd]

I looked it up. It is apparently common practice, and the capitalization is optional!

from Wikipedia:

Volt-ampere reactive [edit]​

In electric power transmission and distribution, volt-ampere reactive (var) is a unit of measurement of reactive power. Reactive power exists in an AC circuit when the current and voltage are not in phase. The term var was proposed by the Romanian electrical engineer Constantin Budeanu and introduced in 1930 by the IEC in Stockholm, which has adopted it as the unit for reactive power.

Special instruments called varmeters are available to measure the reactive power in a circuit.[6]

The unit "var" is allowed by the International System of Units (SI) even though the unit var is representative of a form of power.[7] SI allows one to specify units to indicate common sense physical considerations. Per EU directive 80/181/EEC (the "metric directive"), the correct symbol is lower-case "var",[8] although the spellings "Var" and "VAr" are commonly seen, and "VAR" is widely used throughout the power industry.

Live and learn.

 

[Baluncore]

... and "VAR" is widely used throughout the power industry.

The power industry tends to use MVAR on transmission line networks.

 

[Babadag]

https://patentimages.storage.googleapis.com/58/c0/29/2cc028fa1abc66/WO1991018433A1.pdf
halogen generator [it is not electrical ,it seems to me!]

 

[Baluncore]

With a halogen transformer ...

That is an electrical transformer that generates a low voltage for use in halogen filled incandescent globes.
The expired patent covers the maintenance of the partial pressure of a halogen gas in an excimer laser.