1. 0
0
1
1
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A A
B B
C C
D D
E E
F F
G G
H H
I I
J J
V1
240 Vrms
60 Hz
0°
R1
16.224
U1
9.615 A
+ -
L1
50.31365mH
XWM1
V I
U3
9.615 A
+
-
V2
240 Vrms
60 Hz
0°
R2
16.224
U4
6.25 A
+ -
L2
50.31365mH
XWM2
V I
C2
80.761 F
U5
7.307 A
+
-
U6
9.615 A
+
-
Author: Samir Ahmadov
P = 1.5 kW
cosphi = 0.65
P = 1.5 kW
cosphi = 1
Finding of active R and reactive L parameters of single phase circuit if known active power, current, supply voltage and frequency
Compensation of power factor to make cosphi equal 1
Task: To find value of active and reactive resistance R1 and XL1, then to find a value of C in
order to compencate power factor to 1.
Given: U=240, f=60Hz, active power P=1.5kW, I=9.615A; Find: cosphi, R1, L1 and C (for compensation)
Solution: First we have to find real power factor of system, we need to do following:
S=U x I=240 x 9.615 = 2308 VA; cosphi = P/S = 1500/2308 = 0.65
Important: Due to such power factor and common industrial case we define that load has inductive nature,
then Z must be R1+jXL1 with "+" and also inductive load current lags behind from voltage.
Now we have to find full resistance Z (impedance or complex Z) of circuit, we need to do following by
using COMPLEX NUMBER method:
we have to define complex current complexI. As we know cosphi=0.65 then phi = arccos(0.65)=49.458.
Now we can be sure that complexI=9.615e^-j49.458 (we put phi with "-" as soon as we know that on
inductive load current lags behind from voltage), then reversed complexI*=9.615e^j49.458
Apparent power S= complexU x complexI*(reversed) = 240 x 9.615e^j49.458 = 2308e^j49.458
S= complexU x complexI*(reversed) = complexI x Z x complexI*(reversed) (see TOE lecture)
complexZ= S/(complexI x Z x complexI*(reversed)) = 2308e^j49.458/(9.615e^-j49.458 x 9.615e^j49.458) =
= 2308e^j49.458/(9.615 x 9.615) = 2308e^j49.458/92.448225 = 24.96e^j49.458, so now we know
that complexZ=24.965e^j49.458 , then by using Euler formula:
a x e^jphi = a x cosphi + ja x sinphi = k+jy=sqr(k^2+y^2)e^jarctg(y/k)
a x e^-jphi = a x cosphi - ja x sinphi = k-jy=sqr(k^2+y^2)e^-jarctg(y/k)
then we can find:
Z= 24.96 x cos 49.458 + j24.96 x sin 49.458 = 16.224 + j18.9678; which is means that
R1= 16.224 Om; XL1=18.9678, then we know that XL1 = w x L=2 x п x f x L; L=XL1/(2 x п x f)
L = 50.3mH
Now we need to compencate the power factor in order to make cosphi=1; for this we have to compencate Reactive power. To find reactive power we have to use formula:
Reactive power of our circuit is Q = sqr (S^2 - P^2) = sqr (2308^2 - 1500^2) = sqr (5326864 - 2250000) = 1754 VAR = 1.754 kVAR; Now we need to find a value of C which will compencate power
factor to 1. For this case reactive power of compensation unit should be defined by formula: Q=U x I (usually Q = U x I x sinphi, but when we are talking about power factor compencation unit where
we have to find a value of C to compencate power factor, then we should not use sinphi in this formula, it is because compensation unit consist of only capacitor and as we know in branch with
only capacitor absolute value of phi=90 and then sin90 =1, for better understanding using complex valued functions we can see that reactance of capacitor bank Rc=-jXc, or let's show like 0-jXc, then
Rc=sqr(0^2+Xc^2)e^-jarctg(Xc/0), as we know any number deviding by 0 is infinite and arctg of infinite is pi/2 or 90 degree, then sin(90)=1. Note that if we are using complex valued functions then
we have to show minus "-" before j90 for capacitance and plus "+" for inductance).
So Q = U^2/Xc1, then Xc1 = U^2/Q = 240^2/1754 = 32.8 Om, then Xc1 = 1/(w x C), then C=1/(w x Xc1)=1/(2 x 3.14 x 60 x 32.8) = 80.7uF (mikrofarad)