I have a single phase 220 with capacity 4400 VA and want to place capacitors across the line to improve the power factor. How do I calculate the value of Capacitor to get the power factor close to 1?

Is it true by adding this capacitor can save my electric bills??

Is there any formula to calculate this corresponding with electric rates?

thank you.

 

To solve this you need to have aclear idea of yor load power(kw)

 

If the oriiginal apparent load, kVAo, the original power factor, PFo, and the required power factor, PFr, are known, then the formula for capacitive reactive power, neglecting capacitor loss, is:

kVAc = K x P, where

K = tan [ Acos ( PFo ) ] - tan [ Acos ( PFr ) ], and,

P = kVAo x PFo, in kW

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

 

Power Factor correction will only save you money if your utility rate charges are calculated with Power Factor as a variable. A 220/single phase at 4400VA hardly qualifies for complex rate structures such as this.

Load correction banks usually are multi-phase, med voltage assemblies designed to bring power factor back into acceptable values.

Leave it alone, its not worth the bother.

If you electric bill is below $5K/month you are probably on a simple rate structure.

 

if you can send me more detail about your load means number of motor and transformar and their rateing than i can send distribution of capacitor and their value.
my email id is [email protected]

 

Current pf is not mentioned and let us assume it be 0.8. Then

Apparent power =4400VA or 4.4 KVA
Active power =4.4x0.8 = 3.52 KW
Reactive power =Sqr(4.4^2-3.5^2)=2.64 kvar.
Capacitor 2.64 kvar is needed to improve pf close to 1.

[email protected]

 

yep thats right. How about this one:

V = 415V
I = 10.7A
P output: 5.5kW
Eff. = 85% 0.85
Calculate the power factor?

 

<pre>p input = output/eff.
= 5500/0.85 = 6470W
S = V * I
= 415 * 10.7 = 4440.5VA which is incorrect because it is lower than 6470W...?
</pre>
<p>Does anyone know what to do?

 

Responding to Shane's Feb 24, 1:07am observation... you are correct if Anonymous wrongly observed it to be a single-phase case:

The numbers are correct for the three-phase case.

V = 415.0
A = 10.7 (Given)
kVA = sqrt(3) x A x kV = 7.69
kWin = 6.47 (Given)
PF = 6.47 / 7.69 = 0.84

Furthermore, it is not economical to provide PF correction for a single-phase load. And, if it truly is a three-phase case, one should never use unity PF as a goal. Instead, the added capacitance should just negate the motor's (if the load is a motor) magnetizing reactance.

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

 

Thanks Phil!
I asked the question above.
I should have got that question too... I've been doing a lot of Power Factor Correction just last year (Im doing Electrical Engineering in New Zealand) and we only got shown single phase then. Now I am doing 3 PHASE transformers including p.f. and the formulas have changed! I assume the question is 3-PHASE as the voltage in New Zealand is 400V 3PH & 230V 1PH 50Hz!
I'm doing my theory & regulations this year...
Cheers, Chris

 

Dear Mr. Chris,

i'm doing my study on the load flow and short circuit study, one year to go to complete my degree. do u know how if i wish to calculate the load factor and demand factor for only total connected load only know, while the max. demand is unknown?? or anyone can help???

 

If you still need definitions or parameters related to load-flow studies, contact me off-list!

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