Power supplied by capacitor bank

[LagCompensator]

< Mentor Note -- thread moved to HH from the technical forums, so no HH Template is shown >[/color]

Hi,

I have the following circuit, where V2 is found to be 0.964∠-3.3 (degrees) p.u.

Then I am to find the reactive power supplied by the capacitor bank (j0.25 p.u, to the right). I got the right number, but with wrong sign, could anyone parhaps tell me if my proposal at the end here is correct?

Calculations:

Current down the capacitor branch:
$I_{c} = V_{2} \times Y_{c} = (0.964 \angle-3.3) \times j0.25 = 0.241 \angle 86.7 p.u.\\<br /> S = V \times I^{*} \rightarrow S_{c} = V_{2} \times I_{c}^{*} = 0.964 \angle(-3.3) \times 0.241 \angle(-86.7) = -j0.2323 p.u.$

So I get that Q = -0.2323, however it should have been positive since the bank is supplying the bus with power, and since injected power into a bus is defined positive.

So then I guess that current is defined as positive when entering the bus, and negative when leaving, and because of that I_c should be calculated like this instead:

$I_{c} = (0 - V_{2}) \times Y_{c}$

Thanks for any feedback.

Best regards

 

[Alpharup]

Is it really the reactance of capacitor?
I think there is mistake in question
the reactance of capacitor is always given by (-j/wC) where w is angular frequqncy...If we solve the value for C, we get a negative value...
So, I think there is mistake in question

 

[LagCompensator]

Is it really the reactance of capacitor?
I think there is mistake in question
the reactance of capacitor is always given by (-j/wC) where w is angular frequqncy...If we solve the value for C, we get a negative value...
So, I think there is mistake in question

j0.25 is the admittance not impedance, so if I solve the value for C, I get a positive value.

$Y = \frac{1}{Z} \rightarrow Z = \frac{1}{Y} = \frac{1}{j0.25} = -j4<br /> \\<br /> Z = \frac{1}{j\omega C} = -j4 \rightarrow C = \frac{1}{4 \times \omega}$

I did not write that it was the admittance, I just wrote Y instead of Z in the equations, I should have been more clear.

 

[gneill]

When you calculated the current for the capacitor bank, what direction was the assumed current?

If the the bank is supplying power, what direction should the assumed current be?

 

[LagCompensator]

When I calculated and got the wrong sign I did the following:

$(V_2-V_{Ground}) \times Y_c$
When I do it like this I am assuming the current leaves the bus, which should not be the case, therefore I should do it like this:
$(V_{Ground}-V_2) \times Y_c$
Doing it this way assumes that current flows into the bus.

Is my way of thinking OK? I'm new to "electrical stuff", and therefore I sometimes fail at stuff like this.

 

[gneill]

When I calculated and got the wrong sign I did the following:

$(V_2-V_{Ground}) \times Y_c$
When I do it like this I am assuming the current leaves the bus, which should not be the case, therefore I should do it like this:
$(V_{Ground}-V_2) \times Y_c$
Doing it this way assumes that current flows into the bus.

Is my way of thinking OK? I'm new to "electrical stuff", and therefore I sometimes fail at stuff like this.

Yes, that would be the right way to look at it.

 

[Alpharup]

I think you should look at the definition and formula of S(reactive power) for clarity.