Design of Power Factor Correction Capacitor?

  • #1

Main Question or Discussion Point

Hello...... kindly help me out in designing of power capacitor to improve the power factor as the given load is 950 KWatt and power factor is 0.7, as i want to improve the power factor to 0.92 then what formula i should use to design the capacitor as the Voltage line to line are 400Volts
 

Answers and Replies

  • #2
Baluncore
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Welcome to PF.
Is this a real world situation or a training exercise?
What is the load?
How many phases are there?

950kW is a very high power for a 400V circuit.
The capacitor bank will be very expensive because it must carry a very high current.
 
  • #3
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337
Agree with Baluncore - too much power for a casual evaluation. The other large concern is the consistency of the load. If the capacitors are fixed (constant) and the load is too low - you will then have leading PF - which causes it's own problems. This case requires a professionally engineered system.
 
  • #5
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221
Hello...... kindly help me out in designing of power capacitor to improve the power factor as the given load is 950 KWatt and power factor is 0.7, as i want to improve the power factor to 0.92 then what formula i should use to design the capacitor as the Voltage line to line are 400Volts
Sounds like HW, but I'll give you instructions.
Consider triangle of powers:
cos.png

Since given data are power, voltage, and power factors it is suitable to work with this values only.

P=P1=P2
cos φ1 → φ1=...
cos φ2 → φ2=...
Q1=P⋅tg(φ1)
Q2=P⋅tg(φ2)
Qc=Q1-Q2
Qc=V2⋅2π⋅f⋅C ⇒ C=Qc/(2π⋅f⋅V2)

Note, this is a calculation for a single phase circuit case. If you're referring to a symetrically loaded 3 phase circuit, then given power is total power of 3 phase system. This means you have to calculate with power P/3 per phase and with phase voltage is V/√3. This will give you value of cap if the caps are wye connected. If you want caps connected in delta the value is CΔ=Cwye/3.
 
  • #6
i have already gone through this calculation but couldn't found a better result... ill show my working by uploading the pics.......... Mr. Baluncore this is the real situation in the industry where i am working. as it is 3 phase 400V, and we are having 1500KVA transformer as well and on HT Panel max current set is 55 Amp.
 
  • #7
the max tripping load on acb is set to 1350Amp
 
  • #8
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the max tripping load on acb is set to 1350Amp
For P=950 kW this tripping setting is too low becouse line current is I=P/(√3⋅V⋅cosφ) and even at full compensation (cosφ=1) the current is:
I=1372 A>1350 A.
If LV instalation can't safely handle more current than that, it can't feed 950 kW load.
 
  • #9
Hello...... kindly help me out in designing of power capacitor to improve the power factor as the given load is 950 KWatt and power factor is 0.7, as i want to improve the power factor to 0.92 then what formula i should use to design the capacitor as the Voltage line to line are 400Volts
i can answer this in a very simple way and to the point. just watch out.
We have formula for it, i.e KVAR=factor*KW
whereas factor = factor K for compensation (you can get this from table-4,which i have attached also with this post)
KW = actual load
KVAR = required capacitors to improve p.f (as in market here, capacitors sell out by KVAR rating, not in FARAD. )
there are three types of capacitors available in industrial market, (i)12.5 KVAR (ii) 25 KVAR (iii) 50 KVAR, as i told you that you can purchase capacitor by KVAR ratings also.

Now come to the problem of Walli Bhatti,
KVAR = 0.59 * 930KW
=548.7KVAR
so you then need 10 capacitors of 50KVAR ratings and 2 capacitors of 25KVAR ratings, connecting these in parallel, your power factor will definitely improve upto 0.92.

thanks
 

Attachments

  • #10
as attached files.......the one with the calculation using above formulas, resulting the 75uF Capacitor which is too low value for such load. where as on the site there are two capacitors of 234uF and one 100uf are installed in parallel......as attached files..........do help me out in this regard
 

Attachments

  • #11
i think my previous post is enough for your question, i also mentioned answer of your question. according to your problem you all need only 12 capacitor. 10 of which will be 50kvar and remaining 2 of them should have been 25kvar. thats all.
 
  • #12
thnks for your guideline engr jawad........ as i have seen personally these two capacitors of 25kva and one of 15kva are installed parallel, as i switch of these capacitors due to running inductive load power factor comes to 0.75 and as i switch one this capacitor bank it turn to 0.92. along with this load of 950KW.........can u plz guide me how the three capacitors are working here to imprve upto this level
 
  • #13
1,184
221
as attached files.......the one with the calculation using above formulas, resulting the 75uF Capacitor which is too low value for such load. where as on the site there are two capacitors of 234uF and one 100uf are installed in parallel......as attached files..........do help me out in this regard
Looks to me you don't know how to multiply/divide couple of numbers or/and convert basic electrical units even when given step by step formulae...
For compensation of ≈560 000 VAr at V=400 V, f=50 Hz, the formula gives wye connected capacity value ≈11150 μF (yes, that's 11.15 mF).
Also you didn't answer question regarding dominant load in your industrial facility (motors? induction furnaces? etc) , its' variabilty, type of compensation you're dealing with...It's obvious you have very limited knowledge about this. Man, it is power equivalent of half a megawatt, and yet you want DIY and somebody to help you with it on public forum.
Friendly advice: find professional to take care of it. Good luck.
 
  • #14
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337
Switching capacitors in and out requires proper selection of devices and control, and sometimes each cap is fused. I could build this ( I have worked a number of them) and I still would't DIY it. http://www.abb.com/product/seitp329/79819934e2055d2bc1257b6a004403cc.aspx?productLanguage=us&country=US&tabKey=2 [Broken]
 
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