Calculating Capacitive Reactance

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  • Thread starter JE93
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Homework Statement:

calculate the capacitive reactance/ph such that the load power factor
is increased to 0.98 lag

Relevant Equations:

1/2πfC or 1/ωt
the full question is:
The following phase schematic diagram (FIGURE 4) shows an 11 kV, 50 Hz, 3-phase, short line feeding a load. By calculation or constructing the phasor diagram (use a scale of 1 mm = 2 A) for the load current with VR as reference, determine the capacitive current and
• calculate the capacitive reactance/ph such that the load power factor is increased to 0.98 lag
• calculate the percentage reduction in line current with this value of capacitive reactance in circuit.

So far i have got to:

p.f.=1.9/3.2=0.59
Il=(3.2*10^6)/(√3*11*10^3*0.59 )=284∟-53.8
=167.7-j229.2

unsure which way to go next..
 

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  • #2
gneill
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Hi JE93,

So far i have got to:

p.f.=1.9/3.2=0.59
Il=(3.2*10^6)/(√3*11*10^3*0.59 )=284∟-53.8
=167.7-j229.2
I'm not liking the power factor that you've found. Can you justify your calculation?'

Try starting with a power triangle diagram. Which sides of the triangle represent the power values that you are given?
 
  • #3
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Hi JE93,


I'm not liking the power factor that you've found. Can you justify your calculation?'

Try starting with a power triangle diagram. Which sides of the triangle represent the power values that you are given?
Hi Gneill.

I got the power factor from doing the real divided by the reactive power which, would be 1.9mva/3.2mw.

looking at a power triangle where Z (impedance) is the hypotenuse and X (reactance) is the opposite side,
X=Zsin θ, so would this be:

167.7-j229.2 sin 53.8
presuming the p.f. is correct.

Thanks
 
  • #4
gneill
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Hi Gneill.

I got the power factor from doing the real divided by the reactive power which, would be 1.9mva/3.2mw.

looking at a power triangle where Z (impedance) is the hypotenuse and X (reactance) is the opposite side,
X=Zsin θ, so would this be:

167.7-j229.2 sin 53.8
presuming the p.f. is correct.
That is incorrect! The power factor will be the cosine of the angle subtended by the real and reactive power legs on the power triangle.

A power triangle has sides that represent powers only; no impedances. So in this case:

1603486410780.png

Edit: Corrected Power value typo in diagram.
 
Last edited:
  • #5
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That is incorrect! The power factor will be the cosine of the angle subtended by the real and reactive power legs on the power triangle.

A power triangle has sides that represent powers only; no impedances. So in this case:

View attachment 271434
I have re done this now using the power triangle above.
I realise now that I used the MVAr rather than the MVA for calculating the p.f.
Now i have:

s=√(P^2+Q^2 ) = 3.2MVA
2.3MW/3.2MVA=0.72
〖cos〗^(-1) 0.72=43.9° = θ

Would I now put the correct power factor into the previous equation, so:

Il=(3.2*10^6)/(√3*11*10^3*0.72 )=233.3∟-43.9° = 168.1-j161.8
 
  • #6
gneill
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Note that I've corrected a typo in the power triangle that I posted in #4. I had switched the digits in the real power value. Sorry about that. The new figure is repeated here:
1603488434921.png


I would suggest going first for question part (a), finding the capacitive reactance required to achieve a load power factor of 0.98 . You can work from the power triangle! Add in a line that represents the apparent power for the desired power factor. Something like:
1603488729483.png

so that ##Q_o## is the starting reactive power and ##Q_d## the desired reactive power. You should be able to calculate a value for ##Q_d##, and hence find how much reactive power the capacitor will have to carry in order to bring the reactive power down to that level.
 
  • #7
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Note that I've corrected a typo in the power triangle that I posted in #4. I had switched the digits in the real power value. Sorry about that. The new figure is repeated here:
View attachment 271443

I would suggest going first for question part (a), finding the capacitive reactance required to achieve a load power factor of 0.98 . You can work from the power triangle! Add in a line that represents the apparent power for the desired power factor. Something like:
View attachment 271444
so that ##Q_o## is the starting reactive power and ##Q_d## the desired reactive power. You should be able to calculate a value for ##Q_d##, and hence find how much reactive power the capacitor will have to carry in order to bring the reactive power down to that level.
Thanks

From re-doing the power triangle for a power factor of 0.98 I now have:

Qd= S*Sin θ = 3.2sin11.5 = 0.74MVAr

Where would I go from here to calculate the Capacitive Reactance, seem to be getting confused as the equations i know of to calculate Xc you need to know the capacitor value in farads.

Do i need to continue the way I have been going previously where:

p.f= 0.86
Il=(3.2*10^6)/(√3*11*10^3*0.86 )= 195.3∟-30.7° = 167.9-j99.7

p.f.=0.98
Il=(3.2*10^6)/(√3*11*10^3*0.98 )=171.4∟-11.5° = 167.9-j34.2

Thanks for the help
 
  • #8
gneill
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Note that the value of S will be different for the "new" power factor. Also, I think you've miscalculated the original S value. I don't get 3.2 MVA for ##\sqrt{P^2 + Q^2}##.

It's simpler to find the desired Q via the power factor angle and the constant real power P, so

##Q_{new} = P \tan{\theta}##

where ##\theta## is the power factor angle.
 
  • #9
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Note that the value of S will be different for the "new" power factor. Also, I think you've miscalculated the original S value. I don't get 3.2 MVA for ##\sqrt{P^2 + Q^2}##.

It's simpler to find the desired Q via the power factor angle and the constant real power P, so

##Q_{new} = P \tan{\theta}##

where ##\theta## is the power factor angle.
Sorry must have put it wrong, I worked it out to 3.72MVA

using P tan θ
3.2tan11.5 = 0.65
 
  • #10
gneill
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That looks better.

So if the original reactive power Q was 1.9 MVAr and the desired Q is 0.65 MVAr, how much reactive power do you need to cancel out?
 
  • #11
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That looks better.

So if the original reactive power Q was 1.9 MVAr and the desired Q is 0.65 MVAr, how much reactive power do you need to cancel out?
/
you would need 1.25MVAr to cancel out
 
  • #12
gneill
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/
you would need 1.25MVAr to cancel out
Right!

So the capacitor that's to be added across the load should provide that much reactive power. Can you think of a way to calculate the reactance (in Ohms) given that power requirement?
 
  • #13
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Right!

So the capacitor that's to be added across the load should provide that much reactive power. Can you think of a way to calculate the reactance (in Ohms) given that power requirement?
could you just use simple ohms law?
but using V^2/P
so 1.25^2/3.2=0.48kΩ

thinking this maybe wrong though
 
  • #14
gneill
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You want to consider only the reactive power here. You have a supply voltage of ##V_R## and with some reactance value ##X_c## create the necessary power value (1.25 MVAr). So that

##Q_c = \frac{V_R^2}{X_c}##

Find ##X_c##, the capacitor reactance.
 
  • #15
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You want to consider only the reactive power here. You have a supply voltage of ##V_R## and with some reactance value ##X_c## create the necessary power value (1.25 MVAr). So that

##Q_c = \frac{V_R^2}{X_c}##

Find ##X_c##, the capacitor reactance.

Qc=V^2/Xc
1.25=11^2/Xc
Xc=11^2/1.25=96.8Ω
 
  • #16
gneill
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Almost. What happened to the ##\sqrt{3}## applied to the 11 kV?
 
  • #17
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Almost. What happened to the ##\sqrt{3}## applied to the 11 kV?
sorry,
That now gives me 167.7 Ω
 
  • #18
gneill
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I don't think you're applying the ##\sqrt{3}## appropriately. Remember,

##V_R = \frac{11 kV}{\sqrt{3}}##
 
  • #19
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11*10^3/√3=6350.8
6350^2/1.25*10^6 =32.3Ω
 
  • #20
gneill
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That's it! So the required capacitive reactance will have a magnitude of 32.3 Ω . That's part (a) of your problem.

For part (b) it looks like you'll need the line currents before and after the correction is applied. What have you learned about finding the magnitude of the line current given the real power, voltage, and power factor?
 
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  • #21
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That's it! So the required capacitive reactance will have a magnitude of 32.3 Ω . That's part (a) of your problem.

For part (b) it looks like you'll need the line currents before and after the correction is applied. What have you learned about finding the magnitude of the line current given the real power, voltage, and power factor?
thanks for you help really appreciate it.

Looking back at it all, it shows using the power triangle was a massive help in working it all out. I just need to stop the silly little mistakes haha.
 
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  • #22
gneill
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Yeah, the power triangle gives you a concrete visual that's easy to comprehend and keeps the various variables and their relationships sorted out.

You're welcome for the help! Keep practicing and it'll soon become much clearer and easier.
 
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