Power factor correction problem

Thread Starter

testing12

Joined Jan 30, 2011
80
Hello ,
I had to solve the following problem:


I did this problem using admittance. (since there in parallel) I simply took the impedance above took Y= (1/Z)

and
G (susceptance) = 2(pi)fC

i got C = 312 microfarads. My question is how did they do part b in the solution manual. Im not understanding that. Please advise thank you.
 

t_n_k

Joined Mar 6, 2009
5,455
I wonder if it shouldn't be

\(C=\frac{Ptan(\theta_1-\theta_2)}{\omega V_{rms}^2}\)

where

\(\theta=(\theta_1-\theta_2)\)

is the angular difference between supply voltage and current

and

\(p.f.=cos(\theta_1-\theta_2)=cos(\theta)\)

In any case that formula should be derived or stated somewhere in the text from which the problem is quoted.

It's easy enough to show that

Motor Reactive Power, Sm

\(S_m=VIsin(\theta)=VIcos(\theta)\frac{sin(\theta)}{cos(\theta)}=P_mtan(\theta)\)

Where Pm is the motor real power.

The capacitive reactive power, Sc is

\(S_c=\frac{V^2}{X_c}=V^2\omega C\)

And assuming Sm=Sc for complete compensation.

Leads to ..

\(C=\frac{P_mtan(\theta)}{\omega V^2}\)
 
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