I am getting power factor 0.98 on plant load of 1400KVA & using the capacitor of 550 kvar, 415V.
I want to improve it up to 0.99, then what will be the value of capacitor (in KVAR) required to add in ckt?

 

Responding to Anonymous' Thu, Jan 20, 12:37pm query.

There is a trignometric relationship between PFo (original) and PFd (desired), that will yield the additional capacitive reactance. The procedure follows:

PFo = Original power factor.

PFd = Desired power factor.

kVAo = Original (Apparent) power.

kW = kVA x PFo, Real power.

kVAc = { tan[(Acos(PFo)] - (tan[(Acos(PFd)] } x kW.

You can run the numbers.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Why? The minimal cost savings will never pay for the added hardware and you are so close you don't wnat to overshoot.

Bob Peterson

 

In my opinion no calculations are necessay here.

In general there will be little to gain by improving your existing power factor from 0.98 to 0.99, the additional costs associated would never be justified.

I.m sure that your utility company would be quite happy with this and would probably wish that all its consumers were achieving this!!

You would fractionally increase your available KVA from your supply transformer but apart from that its not worth the grief.

Basically if your P.F. is above 0.95 your doing great and I would not lose any sleep over it.

 

Further to my earlier response to Anonymous' Jan 20 query:

Because I don't know the reason you asked for the calculation method provided, do not misconstrue my submittal as an endorsement to do it. Raising power factor beyond that required to fulfil an economic incentive must be done after evaluation of several technical concerns.

Those concerns are detailed as a caveat in my Sat, Jul 10, 11:58am post titled "Re: Power Factor of Induction Motors!" It explains that power factor "correction" should be treated as a double-edged sword, i.e., having a "good" and "bad" side.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Further to my earlier response of Mon, Jan 24, 6:24pm to Anonymous' Thu, Jan20, 12:37pm query:

Correction.
The formula has to many open-parentheses. The correct is,

kVARc = { tan[Acos(PFo)] - tan[Acos(PFd)] } x kW, where,

Acos = ArcCosine.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

V = 415v
KVA = 1400
KVAR = 550
PF = 0.98
PFn = 0.99

KW = 1400*098 = 1372 Constant

KVAn = 1372/0.99 = 1385.86

â = Cos invers of 0.98

KVARn = KVAn * sin â

Mr. Hakami
Electrical Engineer

 

Responding to Mr. Hakami's Mon, Jan 31 reply...

If I'm wrong please correct me. Your approach was to find the difference between the original state and new state inductive reactance values,

kVARc = kVARo - kVARn

Paraphrasing the original question, "What additional capacitive reactance will increase pf from 0.98 to 0.99?" The answer, using the short-cut formula, is 83.1!

Your calculation for total inductive reactance kVARn has an error. . . å should have been Cos invers 0.99, not Cos invers 0.98! When corrected, your approach yields,

kVARo = 278.6, kVARn = 195.5, thus kVARc = 83.1

In closing, perhaps you misread the capacitive reactance, 550 kVAR, as the original inductive reactance. I believe that it was installed to raise an earlier pf of 0.86 to 0.98.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Responding to Anonymous #62 and Anonymous #87.01-b, both of whom contacted me off-list to ask how the 0.86 power factor value was derived:

Let PFe = the earlier power factor, i.e., before the addition of the 550 kVAR capacitor. Then, substitute PFe and PFo for PFo and PFd, respectively, in the shortcut formula for KVARc. Rearrangement yields:

PFe = Cos { tan¯1 [ ( kVARc / Kw ) + Tan ( Cos¯1 (PFo) ) ] }

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

S=1400KVA
PF1=0.98
P1=1372KW
PF2=0.99
Qc=P1.Tan(Fi1-Fi2)
Fi1=ArcCos(0.98)=11.48
Fi2=ArcCos(0.99)=8.11
Qc=1372Kw.Tan(0.0588)=80.79KVar

 

<p>Responding to Kaveh's May 6, 3:40 pm, calculation:

<p>I'm assuming that you disagree with my Feb 3, 12:06 am, response to Mr. Hakami. My answer, then, was that the added capacitor reactance is Qc 83.1. So let's return to basics:
<pre>
S1 = 1,400 kVA (Given)
P1 = 1,372 kW "
Pf1 = 0.98 "
Fi1°= 11.48°
Q1 = sqrt [ (S1)^2 - (P1)^2 ] = 278.6 kVAR (inductive)

P2 = 1,372 kW (Given)
Pf2 = 0.99 "
Fi2°= 8.11°
S2 = 1,386 kVA
Q2 = sqrt [ (S2)^2 - (P2)^2 ] = 195.5 kVAR (inductive)
</pre>
<p>Then, the additional capacitive reactance, Qc = Q1 - Q2 or 83.1 KVAR.

<p>The short-cut formula solves for the additional capacitive reactance as
follows:
<pre>
Qc = P1 x [ Tan (Fi1°) - Tan (Fi2°) ]
</pre>
<p>If you agree the basic approach is correct, then, your answer for Qc 80.79 is wrong. The error is your multiplier. You used the tangent of the angle difference, that is, [Tan(Fi1°-Fi2°)]. It should have been the difference in their tangents or [Tan(Fi1°)-Tan(Fi2°)].

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

hi, i'm sanjay joshi.

my connected load is about 200 kw, and running load is about 60 KW, with 0.74 power factor and my load factor is 29.00. i have 11 KV HV supply connection. i want to improve my PF up to maximum i can. due to low pf i have to pay additional charge of Rs.18198.00 this month. i want to improve my pf with the help of automatic PF control panel.
plz give me the detail calculation. thanx.

 

Responding to Anonymous' Aug 5, 8:46pm query... formula has been provided! See the Jan 27, 2005, 2:43pm reply to this thread:

http://www.control.com/102620547/index_html

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

hi sanjay, i am shailesh.

i was working on automatic power factor callculation in my privous comp. there is 2000kw working load and about 5000 kw as connected load. for your case, selection of control scheme you should base on your major indictive load (includeing transformer) like type of load, KVA or KW, aprox duty cycle, design PF, and harmonics (if you have data). this information required for best result. send that data on control.com and my email id [email protected]

shailesh patel

 

Dear Sanjay,

From the amount you are paying as penalty, your average PF is in the range of 0.65 to 0.75. Very poor power factor. I feel that you have a single large motor load or a number of small motors which are intermittent and the load normally is lighting load. If you are serious about improving power factor, I would suggest that you purchase/rent a power factor meter and take power factor, KVA and KW readings every hour over a 24 hour period. Give the readings to any manufacturer of Power factor correction panels and he will gladly suggest a solution. Else publish the readings on the list and we will be able to suggest a solution.

In general with the GEB (Gujarat Electricity Board), if you have an average PF (calculated over the billing period of one month) above 0.95, you will get a rebate on your bills.

Regards,

Tomy Zacharia

 

May u pls assist me in this: i have a load of these ratings: P=750w,v=230V, I=5A. After calculations i found Qc=600VAr. Now how many Farads is this (C)?

e-mail: [email protected]

 

Responding to Phillip Kambonde's Aug 16, 1:26pm query:

The formula for both single and three phase capacitors is given in thread:

http://www.control.com/1026188311/index_html

Now I have a a couple of questions for the data you provided:

1) How did you derive the Qc value shown (600)?

2) What is the motor's load factor for which you seek power factor correction? Is it no-load? Or is it full-load?

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])