- #1
topcat123
- 78
- 1
The problem statement
Using the nomogram of FIGURE 10, calculate the size of
capacitor bank to give a final power factor of 0.95 for a single phase,
230 V, 50 Hz lighting installation which has a load current of 17.5 A and
operates at 0.8 pf.
The nonogram gives a value k = 0.44
Relevant equations
Qc = Pk
Xc = 1/2pifC
P = VI
The attempt at a solution
I have come up with two ways but i think the first is wrong as the current though the capasitor (If in parrallel) will not be the same as the load current.
Attempt 1
P = 230 x 17.5 = 4025W
Qc = 4.03 x 0.44 = 1.77 KVAr
P = I^2 x Xc
Xc = P/I^2 = 1.77 x 1000 x 17.5^2 = 5.783Ω
so C would come out at 550μF (as I said I think this is wrong)
Attempt 2
Power still = 4025W and Qc = 1.77 KVAr
but P = V^2/Xc so Xc = V^2/P
Xc = 230^2 x 1.77 x 1000 = 18.38Ω
so C would work out at 173μF
Just a bit of guidance would be appreciated.
Using the nomogram of FIGURE 10, calculate the size of
capacitor bank to give a final power factor of 0.95 for a single phase,
230 V, 50 Hz lighting installation which has a load current of 17.5 A and
operates at 0.8 pf.
The nonogram gives a value k = 0.44
Relevant equations
Qc = Pk
Xc = 1/2pifC
P = VI
The attempt at a solution
I have come up with two ways but i think the first is wrong as the current though the capasitor (If in parrallel) will not be the same as the load current.
Attempt 1
P = 230 x 17.5 = 4025W
Qc = 4.03 x 0.44 = 1.77 KVAr
P = I^2 x Xc
Xc = P/I^2 = 1.77 x 1000 x 17.5^2 = 5.783Ω
so C would come out at 550μF (as I said I think this is wrong)
Attempt 2
Power still = 4025W and Qc = 1.77 KVAr
but P = V^2/Xc so Xc = V^2/P
Xc = 230^2 x 1.77 x 1000 = 18.38Ω
so C would work out at 173μF
Just a bit of guidance would be appreciated.