Transcript AC Power Concepts

Current Distortion
A distorted current waveform can be decomposed into a set of
orthogonal waveforms, (e.g. by Fourier analysis). The RMS
value of the composite waveform (I) may be computed as the
root-sum-squared of the RMS values of all of the orthogonal
components {Ih}.
I
2
RMS
2
1
2
2
2
2
  i  t   dt  I 0  I1  I 2  I 3 
TT
o The DC component I0 is usually (but not always) equal to zero.
o The fundamental component, I1 is the only component that
contributes to real power (and only the in-phase component).
o All the other harmonic components contribute to the RMS
harmonic distortion current, Id :
I I I 
2
d
2
2
2
3
Current Distortion
v(t )  2VRMS cos t  Fv 
Undistorted cosine, Fv = 0
i  t   2 I1, RMS cos t  F1 
Fundamental
 2 I 2, RMS cos  2t  F 2 
 2 I 3, RMS cos  3t  F3 
Harmonics

I RMS  I12  I 22  I 32 
2I1 cos t  F1   2 I1 cos  F1  cos t   I1 sin  F1  sin t 
 2I P cos t   2IQ sin t 
“In-Phase” Current
“Quadrature”
Current
… For the fundamental frequency component (from previous slide):
2I1 cos t  F1   2I P cos t   2IQ sin t 
RMS of iA (t) + iB (t)  I A2  I B2 if iA (t) and iB (t) are orthogonal
…which they are!
I1  I P2  I Q2
I12  I P2  IQ2
As always, we have two ways of looking at this . . .
IQ
I1
RS
IP
I1
P  I12 RS
Forehand
Pure
Reactance
G
RP
Pure
B
Reactance
I 2P2
P  I P RP
G
Backhand
Current Distortion
+
V
-
IQ
I1
I
ID
(Harmonics)
IP
NonLinearities
G
D  VI D
2
I PI P 
P = VI
G P  I P Q  VIQ
G
B
Fundamental
Total RMS Current
Harmonics
I I I I I 
2
2
1
2
2
I 2  I P2  IQ2  I D2
2
3
2
4
Apparent Power:
S  VIP2 V
 QI2 P2DI2Q2  I D2
Total Harmonic Distortion
Total Harmonic Distortion (THD) is defined as the ratio
of the RMS harmonic distortion current ID to the RMS
value of the fundamental component I1 :
ID
THD 
I1
thus…
(assuming zero DC)
I 2  I12  I D2
I  I1 1  THD
2
THD can be measured using a distortion analyser.
If the form of the current waveform i(t) is known . . .
Determine ID by Fourier Analysis:
2
I RMS
 I 02  I12  I D2
Given i(t) having period T,  = 2p/T:
2
I D2  I RMS
 I 02  I12
Compute IDC and IRMS
T 2
I DC
1

i  t dt  I 0

T T 2
I
2
RMS
2
1
  i  t   dt
TT
The first Fourier coefficient is:
T 2
I n 1, P
2
j1
 jnt

i
t
e
dt

I
e

1, P

T T 2
I I
2
D
2
RMS
I I
2
0
2
1
DPF = cos(
I 1)
Note:
I1 
ID
THD 
I1
1, P
2
Apparent Power
Apparent power, S, is defined as the product of RMS
voltage V, and RMS current I :

S  VI  V I1 1  THD2
I1 

I
1  THD 2
We now can express Real Power in terms of apparent
power S, DPF and THD :
P  VI1 cos F 
VI cos F
1  THD2
S
DPF
1  THD2
Power Factor
PS
DPF
1  THD
2
Power Factor is defined as the ratio of real power to
apparent power:
PF
P
DPF
cos F


2
2
S
1  THD
1  THD