- #1
sylent33
- 39
- 5
- 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
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