Chapter 1
SINGLE PHASE CIRCUITS: POWER
DEFINITIONS AND COMPONENTS
(Lectures 1-8)
1.1
Introduction
The definitions of power and its various components are very important to understand quantitative
and qualitative power quality aspects in power system [1]–[5]. This is not only necessary from the
point of view of conceptual clarity but also very much required for practical applications such as
metering, quantification of active, reactive power, power factor and other power quality parameters
in power system. These aspects become more important when power system is not ideal i.e., it
deals with unbalance, harmonics, faults and fluctuations in frequency. We therefore, in this chapter
explore the concept and fundamentals of single phase system with some practical applications and
illustrations.
1.2
Power Terms in a Single Phase System
Let us consider a single-phase system with sinusoidal system voltage supplying a linear load as
shown in Fig. 1.1. A linear load is one which consists of ideal resistive, inductive and capacitive
elements. The voltage and current are expressed as below.
√
v(t) = 2 V sin ωt
√
i(t) = 2 I sin(ωt − φ)
The instantaneous power can be computed as,
1
(1.1)
v
i
Fig. 1.1 A single phase system
p(t) = v( t) i(t) =
=
=
=
=
V I [2 sin ωt sin(ωt − φ)]
V I[cosφ − cos(2ωt − φ)]
V I cos φ(1 − cos 2ωt) − V I sin φ sin 2ωt
P (1 − cos 2ωt) − Q sin 2ωt
pactive (t) − preactive (t)
(1.2)
R T +t
Here, P = T1 t=t1 1 p(t) dt = average value of pactive (t). This is called as average active power.
The reactive power Q is defined as,
∆
Q = max {preactive (t)}
(1.3)
It should be noted that the way Q is defined is different from P . The Q is defined as maximum
value of the second term of (1.2) and not an average value of the second term. This difference
should always kept in mind.
Equation (1.2) shows that instantaneous power can be decomposed into two parts. The first term
has an average value of V I cos φ and an alternating component of V I cos 2ωt, oscillating at twice
the line frequency. This part is never negative and therefore is called unidirectional or dc power.
The second term has an alternating component V I sin φ sin 2ωt oscillating at twice frequency with
a peak vale of V I sin φ. The second term has zero average value. The equation (1.2) can further
be written in the following form.
p(t) =
=
=
=
V I cos φ − V I cos(2ωt − φ)
p̄(t) + p̃(t)
paverage + poscillation
pusef ul + pnonusef ul
(1.4)
With the above definitions of P and Q, the instantaneous power p(t) can be re-written as following.
p(t) = P (1 − cos 2ωt) − Q sin 2ωt
2
(1.5)
√
Example 1.1 Consider a sinusoidal supply voltage v(t) = 230 2 sin ωt supplying a linear load of
impedance ZL = 12 + j13 Ω at ω = 2πf radian per second, f = 50 Hz. Express current i(t) as a
function of time. Based on v(t) and i(t) determine the following.
(a) Instantaneous power p(t), instantaneous active power pactive (t) and instantaneous reactive
power preactive (t)
(b) Compute average real power P , reactive power Q, apparent power S, and power factor pf .
(c) Repeat the above when load is ZL = 12 − j13 Ω, ZL = 12 Ω, and ZL = j13 Ω
(d) Comment upon the results.
Solution: A single phase circuit supplying linear load is shown Fig. 1.1. In general, the current in
the circuit is given as,
i(t) =
√
2 I sin(ωt − φ)
where φ = tan−1 (XL /RL ), and I = (V /|ZL |)
Case 1: When load is inductive, ZL = 12 + j13 Ω
p
√
|ZL | = RL2 + XL2 = 122 + 132 = 17.692 Ω, and I = 230/17.692 = 13 A
φ = tan−1 (X/R) = tan−1 (13/12) = 47.29o
Therefore we have,
√
v(t) = 230 2 sin ωt
√
i(t) = 13 2 sin(ωt − 47.29o )
The instantaneous power is given as,
p(t) =
=
=
=
V I cos φ(1 − cos 2ωt) − V I sin φ sin 2ωt
230 × 13 cos 47.29o (1 − cos(2 × 314t)) − 230 × 13 sin 47.29o sin(2 × 314t)
2028.23(1 − cos(2 × 314t)) − 2196.9 sin(2 × 314t)
pactive (t) − preactive (t)
3
The above implies that,
pactive (t) = 2028.23(1 − cos(2 × 314t))
preactive (t) = 2196.9 sin(2 × 314t)
Average real power (P ) is given as,
Z
1 T
P =
p(t) dt
T 0
P = V I cos φ = 230 × 13 × cos 47.29o = 2028.23 W
Reactive power (Q) is given as maximum value of preactive , and equals to V I sin φ as given below.
Q = V I sin φ = 230 × 13 × sin 47.2906◦ = 2196.9 VAr
p
Apparant power, S = V I = P 2 + Q2 = 230 × 13 = 2990 VA
2028.23
P
=
= 0.6783
Power factor =
S
2990
For this case, the voltage, current and various components of the power are shown in Fig. 1.2. As
seen from the figure the current lags the voltage due to inductive load. The pactive has an offset of
2028.23 W, which is also indicated as P in the right bottom graph. The preactive has zero average
value and its maximum value is equal to Q, which is 2196.9 VArs.
√
Case 2: When load is Capacitive, ZL = 12 − j13 that implies |ZL | = 122 + 132 = 17.692Ω,
and I = 230/17.692 = 13 A, φ = tan−1 (−13/12) = −47.2906o .
√
v(t) = 230 2 sin ωt
√
i(t) = 13 2 sin(ωt + 47.2906o )
p(t) = V I cos φ(1 − cos 2ωt) − V I sin φ sin 2ωt
= 230 × 13 cos(−47.2906o )(1 − cos(2 × 314t)) − 230 × 13 sin(−47.2906o ) sin(2 × 314t)
= 2028.23(1 − cos(2 × 314t)) + 2196.9 sin(2 × 314t)
pactive (t) = 2028.23(1 − cos(2 × 314t))
preactive (t) = −2196.9 sin(2 × 314t)
P = V I cos φ = 230 × 13 × cos 47.2906o = 2028.23 Watt
Q = V I sin φ = 230 × 13 × sin(−47.2906o ) = −2196.9 VAr
p
S = V I = P 2 + Q2 = 230 × 13 = 2990 VA
For Case 2, the voltage, current and various components of the power are shown in Fig. 1.3. The
explanation given earlier also holds true for this case.
Case 3: When load is resistive, ZL = RL = 12 Ω, I = 230/12 = 19.167 A, and φ = 0o .
4
400
6000
Voltge (V)
4000
VA, W, VAr
200
0
-200
-400
2000
0
-2000
0
0.01
0.02
-4000
0.03
p(t)
0
0.02 pact(t)
0.01
sec
sec
20
0.03
preact(t)
2400
Current (A)
0
-10
-20
Average Power (W)
Reactive Power (VAr)
2300
W, VAr
10
2200
2100
0
0.01
0.02
2000
0.03
0
0.01
sec
0.02
0.03
sec
Fig. 1.2 Case 1: Voltage, current and various power components
Therefore, we have
√
v(t) = 230 2 sin ωt
√
i(t) = 19.167 2 sin ωt
p(t) = 230 × 19.167 cos 0o {1 − cos(2 × 314t)} − 230 × 19.167 sin 0o sin(2 × 314t)
= 4408.33(1 − cos(2 × 314t))
pactive (t) = 4408.33{1 − cos(2 × 314t)}
preactive (t) = 0
P = V I cos φ = 230 × 19.167 × cos 0o = 4408.33 W
Q = V I sin φ = 230 × 19.167 × sin 0o = 0 VAr
p
S = V I = P 2 + Q2 = 230 × 19.167 = 4408.33 VA
4408.33
Power factor =
=1
4408.33
5
400
6000
Voltage (V)
4000
VA, W, VAr
200
0
-200
-400
2000
0
-2000
0
0.01
0.02
-4000
0.03
0
0.01
sec
0.02
sec
20
p(t)
p (t) 0.03
act
preact(t)
3000
Current (A)
2000
10
W, VAr
1000
0
Average Power (W)
Reactive Power (VAr)
0
-1000
-10
-2000
-20
0
0.01
0.02
-3000
0.03
0
0.01
sec
0.02
sec
0.03
0.04
Fig. 1.3 Case 2: Voltage, current and various power components
For Case 3, the voltage, current and various components of the power are shown in Fig. 1.4. Since
the load is resistive, as seen from the graph preactive is zero and p(t) is equal to pactive . The average
value of p(t) is real power (P ), which is equal to 4408.33 W.
Case 4: When the load is purely reactive, ZL = j13 Ω , |ZL | = 13 Ω, I = 230
= 17.692 A, and
13
φ = 90o . Therefore, we have
√
v(t) = 230 2 sin ωt
√
i(t) = 17.692 2 sin(ωt − 90o )
p(t) = 230 × 17.692 cos 90o (1 − cos(2 × 314t)) − 230 × 17.692 sin 90o sin(2 × 314t)
= 0 − 4069 sin(2 × 314t)
pactive (t) = 0
preactive (t) = 4069 sin(2 × 314t)
P = V I cos φ = 230 × 17.692 × cos 90o = 0 W
Q = V I sin φ = 230 × 17.692 × sin 90o = 4069 VAr
p
S = V I = P 2 + Q2 = 230 × 17.692 = 4069 VA
0
Power factor =
=0
4069
6
400
10000
Voltage (V)
8000
VA, W, VAr
200
0
6000
4000
2000
-200
0
-400
0
0.01
0.02
0.03
0
0.01
sec
0.02
sec
30
p(t)
pact(t) 0.03
preact(t)
5000
Current (A)
4000
10
3000
W, VAr
20
0
2000
-10
1000
-20
0
-30
0
0.01
0.02
-1000
0.03
Average Power (W)
Reactive Power (VAr)
0
0.01
sec
0.02
0.03
sec
Fig. 1.4 Case 3: Voltage, current and various power components
For Case 4, the voltage, current and various components of the power are shown in Fig. 1.5. The
load in this case is purely reactive, hence their is no average component of p(t). The maximum
value of p(t) is same as preactive (t) or Q, which is equal to 4069 VArs.
1.3
Sinusoidal Voltage Source Supplying Non-linear Load Current
The load current is now considered as nonlinear load. A non-linear load is one which consists
of switched elements such as diode, transistors, MOSFET, etc., in the circuit. In power circuit,
non-linear load current exists, when source supplies to power electronics based loads such as rectifier, inverter, cyclo-converters etc.. These loads cause presence of harmonics in the load current.
Assuming that all harmonics are present in the load current, the voltage and current are expressed
as following.
v(t) =
√
2 V sin ωt
∞
√ X
i(t) =
2
In sin(nωt − φn )
n=1
The instantaneous power is therefore given by,
7
(1.6)
400
5000
Voltage (V)
VA, W, VAr
200
0
0
-200
-400
0
0.01
0.02
-5000
0.03
0
0.01
0.02
sec
p(t)
pact(t) 0.03
preact(t)
30
5000
Current (A)
20
4000
W, VAr
10
0
3000
Active Power (W)
Reactive Power (VAr)
2000
-10
1000
-20
-30
0
0.01
0.02
0
0.03
0
0.01
0.02
0.03
Fig. 1.5 Case 4: Voltage, current and various power components
p(t) = v(t) i(t) =
√
∞
√ X
In sin(nωt − φn )
2 V sin ωt 2
n=1
= V
∞
X
[In 2 sin ωt sin(nωt − φn )]
n=1
= V [I1 2 sin ωt sin(ωt − φ1 )]
∞
X
+ V
[In 2 sin ωt sin(nωt − φn )]
(1.7)
n=2
Note that 2 sin A sin B = cos(A − B) − cos(A + B), using this, Eqn. (1.7) can be re-written as
the following.
8
p(t) = V I1 [cos φ1 − cos(2ωt − φ1 )] − V I1 sin φ1 sin 2ωt
∞
X
+V
In [ (cos φn − cos(2nωt − φn )) − sin φn sin 2nωt]
n=2
= V I1 cos φ1 (1 − cos 2ωt) − V I1 sin φ1 sin 2ωt
∞
X
+
V In [ cos φn (1 − cos 2nωt) − sin φn sin 2nωt]
(1.8)
n=2
=
A + B
In above equation, average active power P and reactive power Q are given by,
P = P1 = average value of p(t) = V I1 cos φ1
∆
Q = Q1 = peak value of second term in A = V I1 sin φ1
(1.9)
The apparent power S is given by
S = V Iq
S = V [I12 + I22 + I32 + .....]
(1.10)
Equation (1.10) can be re-arranged as given below.
S 2 = V 2 I12 + V 2 [I22 + I32 + I42 + ...]
= (V I1 cos φ1 )2 + (V I1 sin φ1 )2 + V 2 [I22 + I32 + I42 + .....]
= P 2 + Q2 + H 2
(1.11)
In above equation, H is known as harmonic power and represents V As corresponding to harmonics
and is equal to,
H=V
q
[I22 + I32 + I42 + .....]
(1.12)
The following points are observed from description.
1. P and Q are dependent on the fundamental current components
2. H is dependent on the current harmonic components
3. Power components −V I cos 2ωt and V I1 sin φ1 sin 2ωt are oscillating components and can
be eliminated using appropriately chosen capacitors and inductors
4. There are other terms in (1.10), which are functions of multiple integer of fundamental frequency are reflected in ’B’ terms of Eqn. (1.8). These terms can be eliminated using tuned
LC filters.
9
This is represented by power tetrahedron instead of power triangle (in case of voltage and current
of sinusidal nature of fundamental frequency). In this context, some important terms are defined
here.
Displacement Power Factor (DPF) or Fundamental Power Factor (pf1 ) is denoted by cos φ1
and is cosine angle between the fundamental voltage and current. This is equal to,
DPF = pf1 = cos φ1 =
P1
.
S1
(1.13)
The Power Factor (pf ) is defined as ratio of average active power to the total apparent power
(V I) and is expressed as,
P
S
I1
V I1 cos φ1
=
=
cos φ1
VI
I
= cos γ cos φ1
Power Factor (pf ) =
(1.14)
The equation (1.14) shows that power factor becomes less by a factor of cos γ, which is ratio of
fundamental to the total current. This is due to the presence of the harmonics in the load current.
The nonlinear load current increases the ampere rating of the conductor for same amount of active
power transfer with increased VA rating. Such kind of load is not desired in power system.
Example 1.2 Consider an ideal single-phase voltage
√ source supplying a rectifier load as given
in Fig. 1.6. Given a supply voltage, v(t) = 230 2 sin ωt and source impedance is negligible,
draw the voltage and current waveforms. Express current using Fourier series. Based on that
determine the following.
1. Plot instantaneous power p(t).
2. Plot components of p(t) i.e. pactive (t), preactive (t).
3. Compute average real power, reactive power, apparent power, power factor, displacement
factor (or fundamental power factor).
4. Comment upon the results in terms of VA rating and power output.
Solution: The above system has been simulated using MATLAB/SIMULINK. The supply voltage
and current are shown in Fig. 1.7. The current waveform is of the square type and its Fourier series
expansion is given below.
i(t) =
∞
X
4Idc
sin(nωt) where h = 0, 1, 2 . . .
nπ
n=2h+1
The instantaneous power is therefore given by,
10
i (t )
Id
v (t )
Fig. 1.6 A single phase system with non-linear load
p(t) = v(t) i(t) =
√
∞
X
4Idc
2 V sin ωt
sin(nωt).
nπ
n=2h+1
(1.15)
By expansion of the above equation, the average active power (P ) and reactive power (Q) are given
as below.
P = P1 = average value of Pactive (t) or p(t) = V I1 cos φ1
= V I1 (since, φ1 = 0, cos φ1 = 1, sin φ1 = 0)
∆
Q = Q1 = peak value of Preactive (t) = V I1 sin φ1 = 0
The rms value of fundamental and rms value of the total source current are given below.
Irms = Id = 103.5 A
√
2 2
I1 =
Id = 93.15 A
π
The real power (P ) is given by
P = V I1
√
2 2
Id = 21424.5 W .
= V ×
π
The reactive power (Q) is given by
Q = Q1 = 0.
The apparent power (S) is given by
S = V Irms
= V Id = 23805 VA .
11
300
Supply current (A)
Supply voltage (V)
200
100
0
-100
-200
-300
1
1.001
1.002
1.003
Time(Sec)
1.004
1.005
1.006
5
x 10
4
x 10
3
2.5
2
1.5
instantaneous power
Average power (W)
Reactive power (Var)
1
0.5
0
-0.5
1
1.001
1.002
1.003
Time (Sec)
1.004
1.005
1.006
5
x 10
Fig. 1.7 Supply voltage, current and instantaneous power waveforms
The displacement power factor (cos φ1 ) is,
DP F = cos φ1 = 1.
Therefore power factor is given by,
P
S1 P1
=
S
S S1
V I1
=
cos φ1
V Irms
I1
=
× DP F = 0.9 (lag)
Irms
pf =
1.4
Non-sinusoidal Voltage Source Supplying Non-linear Loads
The voltage source too may have harmonics transmitted from generation or produced due to nonlinear loads in presence of feeder impedance. In this case, we shall consider generalized case of
non-sinusoidal voltage source supplying nonlinear loads including dc components. These voltages
and currents are represented as,
∞
X
√
v(t) = Vdc +
2Vn sin(nωt − φvn )
n=1
12
(1.16)
and
i(t) = Idc +
∞
X
√
2In sin(nωt − φin )
(1.17)
n=1
Therefore, instantanteous power p(t) is given by,
∞
∞
X
X
√
√
2Vn sin(nωt − φvn )].[Idc +
2In sin(nωt − φin )]
p(t) = [Vdc +
∞
X
√
p(t) = Vdc Idc + Vdc
| {z }
I
| n=1
+
∞
X
√
(1.18)
n=1
n=1
2In sin(nωt − φin ) + Idc
2Vn sin(nωt − φvn )
n=1
{z
}
II
2Vn sin(nωt − φvn )
|n=1
∞
X
√
∞
X
√
|
{z
III
2In sin(nωt − φin )
{zn=1
}
(1.19)
}
IV
p(t) = pdc−dc + pdc−ac + pac−dc + pac−ac
(1.20)
The term I (pdc−dc ) contributes to power from dc components of voltage and current. Terms II
(pdc−ac ) and III (pac−dc ) result from the interaction of dc and ac components of voltage and current.
In case, there are no dc components all these power components are zero. In practical cases, dc
components are very less and the first three terms have negligible value compared to IV term.
Thus, we shall focus on IV (pac−ac ) term which corresponds to ac components present in power
system. The term IV can be written as,
th
IV term = pac−ac =
∞
X
√
2Vn sin(nωt − φvn )
n=1
∞
X
√
2Ih sin(hωt − φih )
(1.21)
h=1
where n = h = 1, 2, 3..., similar frequency terms will interact. When n 6= h, dissimilar
13
frequency terms will interact. This is expressed below.
√
√
2V1 sin(ωt − φv1 ) 2I1 sin(ωt − φi1 )
pac−ac (t) =
|
{z
}
A
∞
X
√
√
2Ih sin(hωt − φih )
+ 2V1 sin(ωt − φv1 )
h=2,h6=1
|
{z
B
√
√
+ 2V2 sin(2ωt − φv2 ) 2I2 sin(2ωt − φi2 )
|
{z
}
}
A
∞
X
√
√
+ 2V2 sin(2ωt − φv2 )
2Ih sin(hωt − φih ) + . . . + . . .
h=1,h6=2
{z
|
B
√
√
+ 2Vn sin(nωt − φvn ) 2In sin(nωt − φin )
|
{z
}
}
A
∞
X
√
+ 2Vn sin(nωt − φvn )
2Ih sin(hωt − φih )
√
(1.22)
h=1,h6=n
|
{z
}
B
The terms in A of above equation form similar frequency terms and terms in B form dissimilar
frequency terms, we shall denote them by pac−ac−nn and pac−ac−nh . Thus,
pac−ac−nn (t) =
∞
X
Vn In 2 sin(nωt − φvn ) sin(nωt − φin )
(1.23)
n=1
and
∞
X
√
pac−ac−nh (t) =
2Vn sin(nωt − φvn )
n=1
∞
X
√
2In sin(hωt − φih )
(1.24)
h=1,h6=n
Now, let us simplify pac−ac−nn in
pac−ac−nn (t) =
=
∞
X
n=1
∞
X
Vn In [cos(φin − φvn ) − cos(2nωt − φin − φvn )]
Vn In [cos(φn ) − cos(2nωt − (φin − φvn ) − 2φvn )]
n=1
=
=
∞
X
n=1
∞
X
Vn In [cos(φn ) − cos (2nωt − 2φvn ) − φn ]
Vn In [cos(φn ) − cos(2nωt − 2φvn ) cos φn − sin(2nωt − 2φvn ) sin φn ]
n=1
(1.25)
14
where φn = (φin − φvn ) = is phase angle between nth harmonic current and voltage. Therefore,
∞
X
pac−ac−nn (t) =
[Vn In cos φn {1 − cos(2nωt − 2φvn )}]
n=1
∞
X
−
[Vn In sin φn sin(2nωt − 2φvn )].
(1.26)
n=1
Thus, the instantaneous power is given by,
p(t) = pdc−dc + pdc−ac + pac−dc + pac−ac−nn + pac−ac−nh
| {z } | {z } | {z } | {z } | {z }
I
II
III
| IVA {z IVB }
IV
p(t) = Vdc Idc + Vdc
∞
X
√
2In sin(nωt − φin ) + Idc
2Vn sin(nωt − φvn )
n=1
n=1
+
∞
X
√
∞
X
[Vn In cos φn {1 − cos(2nωt − 2φvn )}]
n=1
∞
X
−
[Vn In sin φn . sin(2nωt − 2φin )]
(1.27)
n=1
1.4.1
Active Power
Instantaneous active power, pactive (t) is expressed as,
pactive (t) = Vdc Idc +
∞
X
[Vn In cos φn {1 − cos(2nωt − 2φvn )}]
(1.28)
n=1
It has non-negative value with some average component, giving average active power. Therefore,
Z
1 T
P =
p(t) dt
T 0
∞
X
= Vdc Idc +
Vn In cos φn .
(1.29)
n=1
The reactive component of the instantaneous power is denoted by preactive (t) and is given as following.
preactive (t) = −
∞
X
[Vn In sin φn sin(2nωt − 2φvn )]
n=1
resulting in
15
(1.30)
Q , max of (1.30) magnitude
∞
X
=
Vn In sin φn .
(1.31)
n=1
From (1.29)
P = Pdc +
∞
X
Vn In cos φn
n=1
= Pdc + V1 I1 cos φ1 + V2 I2 cos φ2 + V3 I3 cos φ3 + . . .
= Pdc + P1 + P2 + P3 + . . .
= Pdc + P1 + PH
(1.32)
In above equation,
Pdc = Average active power corresponding to the dc components
P1 = Average fundamental active power
PH = Average harmonic active power
Average fundamental active power (P1 ) can also be found from fundamentals of voltage and current i.e.,
P1
1
=
T
Z
T
v1 (t) i1 (t)dt
(1.33)
0
and harmonic active power (PH ) can be found as below.
PH =
∞
X
Vn In cos φn = P − P1 .
(1.34)
n=1
1.4.2
Reactive Power
The reactive power or Budeanu’s reactive power (Q) can be found by summing maximum value of
each term in (1.30). This is given below.
Q =
∞
X
Vn In sin φn
n=1
= V1 I1 sin φ1 + V2 I2 sin φ2 + V3 I3 sin φ3 + . . .
= Q1 + Q2 + Q3 + . . .
= Q1 + QH
16
(1.35)
Usually this reactive power is referred as Budeanu’s reactive power, and sometimes we use subscript B’ to indicate that i.e.,
QB = Q1B + QHB
(1.36)
The remaining dissimilar terms of (1.27) are accounted using prest (t). Therefore, we can write,
p(t) = pdc−dc + pactive (t) + preactive (t) +
{z
}
|
Similar frequency terms
p (t)
{z }
| rest
(1.37)
Non-similar frequency terms
where,
pdc−dc = Vdc Idc
∞
X
pactive (t) =
[Vn In cos φn {1 − cos(2nωt − 2 φvn )}]
n=1
∞
X
preactive (t) = −
[Vn In sin φn . sin(2nωt − 2 φvn )]
n=1
∞
X
prest (t) = Vdc
∞
X
√
√
2In sin(nωt − φin ) + Idc
2Vn sin(nωt − φvn )
n=1
n=1
∞
X
√
+
2Vn sin(nωt − φvn )
n=1
1.4.3
∞
X
√
2Im sin(mωt − φim )
(1.38)
m=1,m6=n
Apparent Power
The scalar apparent power which is defined as product of rms value of voltage and current, is
expressed as following.
S = V
qI
q
2
2
2
2
=
Vdc + V1 + V2 + · · · Idc
+ I12 + I22 + · · ·
q
q
2
2
=
Vdc2 + V12 + VH2 Idc
+ I12 + IH
(1.39)
Where,
VH2 = V22 + V32 + · · · =
∞
X
Vn2
n=2
2
IH
= I22 + I32 + · · · =
∞
X
In2
(1.40)
n=2
VH and IH are denoted as harmonic voltage and harmonic current respectively. Expanding (1.39)
we can write
17
S2 =
=
=
=
=
=
V 2I 2
2
2
(Vdc2 + V12 + VH2 )(Idc
+ I12 + IH
)
2 2
2 2
2 2
2 2
2
2
2
2
Vdc Idc + Vdc I1 + Vdc IH + V1 I1 + V12 Idc
+ V12 IH
+ VH2 Idc
+ VH2 I12 + VH2 IH
2
2
2
2
2
Vdc2 Idc
+ V12 I12 + VH2 IH
+ Vdc2 (I12 + IH
) + Idc
(V12 + VH2 ) + V12 IH
+ VH2 I12
2
2
2
Sdc
+ S12 + SH
+ SD
2
2
2
+ SH
+ SD
S12 + Sdc
{z
}
|
2
= S12 + SN
(1.41)
In above equation, the term SN is as following.
2
2
2
2
2
2
2 2
2
SN
= Vdc2 I12 + Vdc2 IH
+ V12 Idc
+ V12 IH
+ VH2 Idc
+ VH2 I12 + VH2 IH
+ Idc
IH + Idc
Vdc2
(1.42)
Practically in power systems dc components are negligible. Therefore neglecting the contribution
of Vdc and Idc associated terms in (1.42), the following is obtained.
2
2
2
SN
= I12 VH2 + V12 IH
+ VH2 IH
2
= DV2 + DI2 + SH
(1.43)
The terms DI and DV in (1.43) are known as apparent powers due to distortion in current and
voltage respectively. These are given below.
DV = I1 VH
DI = V1 IH
(1.44)
These are further expressed in terms of THD components of voltage and current, as given below.
T HDV
T HDI
VH
V1
IH
=
I1
=
(1.45)
From (1.45), the harmonic components of current and voltage are expressed below.
VH = T HDV V1
IH = T HDI I1
(1.46)
DV = V1 I1 T HDV = S1 T HDV
DI = V1 I1 T HDI = S1 T HDI
SH = VH IH = S1 T HDI T HDV
(1.47)
Using (1.44) and (1.46),
Therefore using (1.43) and (1.47), SN could be expressed as following.
18
2
SN
= S12 (T HDI2 + T HDV2 + T HDI2 T HDV2 )
(1.48)
Normally in power system, T HDV << T HDI , therefore,
SN ≈ S1 DI
(1.49)
The above relationship shows that as the THD content in voltage and current increases, the non fundamental apparent power SN increases for a given useful transmitted power. This means there are
more losses and hence less efficient power network.
1.4.4
Non Active Power
Non active power is denoted by N and is defined as per following equation.
S2 = P 2 + N 2
(1.50)
This power includes both fundamental as well as non fundamental components, and is usually
computed by knowing active power (P ) and apparent power (S) as given below.
√
N = S2 − P 2
(1.51)
1.4.5
Distortion Power
Due to presence of distortion, the total apparent power S can also be written in terms of active
power (P ), reactive power (Q) and distortion power (D)
S 2 = P 2 + Q2 + D2 .
(1.52)
Therefore,
D=
1.4.6
p
S 2 − P 2 − Q2 .
(1.53)
Fundamental Power Factor
Fundamental power factor is defined as ratio of fundamental real power (P1 ) to the fundamental
apparent power (S1 ). This is given below.
pf1 = cos φ1 =
P1
S1
(1.54)
The fundamental power factor as defined above is also known as displacement power factor.
19
1.4.7
Power Factor
Power factor for the single phase system considered above is the ratio of the total real power (P )
to the total apparent power (S) as given by the following equation.
P
S
P 1 + PH
= p 2
2
S1 + SN
(1 + PH /P1 ) P1
= p
1 + (SN /S1 )2 S1
pf =
(1.55)
Substituting SN from (1.48), the power factor can further be simplified to the following equation.
(1 + PH /P1 )
pf = p
pf1
1 + T HDI2 + T HDV2 + T HDI2 T HDV2
(1.56)
Thus, we observe that the power factor of a single phase system depends upon fundamental (P1 )
and harmonic active power (PH ), displacement factor (DP F = pf1 ) and THDs in voltage and
current. Further, we note following points.
1. P/S is also called as utilization factor indicator as it indicates the usage of real power.
2. The term SN /S1 is used to decide the overall degree of harmonic content in the system.
3. The flow of fundamental power can be characterized by measurement of S1 , P1 , pf1 , and Q1 .
For a practical power system P1 >> PH and T HDV << T HDI , the above expression of power
factor is further simplified as given below.
pf1
pf = p
1 + T HDI2
Example 1.3 Consider the following voltage and current in single phase system.
√
√
vs (t) = 230 2 sin(ωt) + 50 2 sin(3ωt − 30◦ )
√
√
i(t) = 2 + 10 2 sin(ωt − 30◦ ) + 5 2 sin(3ωt − 60◦ )
Determine the following.
(a) Active power, (P )
(b) Reactive power, (Q)
(c) Apparent power, (S)
(d) Power factor, (pf )
20
(1.57)
Solution: Here the source is non-sinusoidal and is feeding a non-linear load. The instantaneous
power is given by,
p(t) = v(t) i(t)
∞
∞
X
X√
√
2 Vn sin(nωt − φvn )} {Idc +
2 In sin(nωt − φin )}
p(t) = {Vdc +
n=1
n=1
(a) The active power ’P’ is given by,
Z
1 T
P =
p(t) dt
T 0
= Pdc + V1 I1 cos φ1 + V2 I2 cos φ2 + ...... + Vn In cos φn
= Pdc + P1 + PH
(1.58)
where,
φn = φin − φvn
Pdc = Vdc Idc
P1 = V1 I1 cos φ1
∞
X
Vn In cos φn
PH =
n=2
Here, Vdc = 0, V1 = 230 V, φv1 = 0, V3 = 50 V, φv3 = 30◦ , Idc = 2 A, I1 = 10 A, φi1 = 30◦ ,
I3 = 5 A, φi3 = 60◦ . Therefore, φ1 = φi1 − φv1 = 30◦ and φ3 = φi3 − φv3 = 30◦ .
Substituting these values in (1.58), the above equation gives,
P = 0 × 2 + 230 × 10 × cos 30◦ + 50 × 5 × cos 30◦ = 2208.36 W.
(b). The reactive power (Q) is given by,
∞
X
Vn In sin φn
Q =
n=1
= V1 I1 sin φ1 + V2 I2 sin φ2 + .....Vn In sin φn
= 230 × 10 × sin 30◦ + 50 × 5 × sin 30◦ = 1275 VAr.
(c). The Apparent power S is given by,
S = Vrms Irms
q
q
2
2
2
2
2
=
Vdc + V1 + V2 + ....Vn
Idc
+ I12 + I22 + .....In2
q
q
2
2
=
Vdc2 + V12 + VH2 Idc
+ I12 + IH
where,
q
V22 + V32 + ....Vn2
q
=
I22 + I32 + .....In2
VH =
IH
21
Substituting the values of voltage and current components, the apparent power S is computed as
following.
√
√
0 + 2302 + 502 22 + 102 + 52
S =
= 235.37 × 11.357 = 2673.31 VA
(d). The power factor is given by
pf =
P
2208.36
=
= 0.8261 lag
S
2673.31
Example 1.4 Consider following system with distorted supply voltages,
v(t) = Vdc +
∞
X
√
2Vn
sin(nωt − φvn )
n2
n=1
√
with Vdc = 10 V, Vn /n2 = 230 2/n2 and φvn = 0 f or n = 1, 3, 5, 7, . . .
The voltage source supplies a nonlinear current of,
i(t) = Idc +
∞
X
n=1
√
2In
sin(nωt − φin ).
n
with Idc = 2 A, In = 20/n A and φin = n × 30o for n = 1, 3, 5, 7, . . .
Compute the following.
(a) Plot instantaneous power p(t), pactive (t), preactive (t), Pdc , and prest (t).
(b) Compute P, P1 , PH (= P3 + P5 + P7 + . . .).
(c) Compute Q, Q1 , QH (= Q3 + Q5 + Q7 + . . .).
(d) Compute S, S1 , SH , N, D.
(e) Comment upon each result.
Solution: Instantaneous power is given as following.
22
!
!
√
√
∞
∞
X
X
230 2
20 2
p(t) = v(t) i(t) =
10 +
sin n(ωt − 300 )
sin(nωt)
2+
2
n
n
n=1,3,5
n=1,3,5
√
√
∞
∞
X
X
230 2
20 2
0
= |{z}
20 + 10
sin n(ωt − 30 ) + 2
sin(nωt)
n
n2
n=1,3,5
n=1,3,5
I
{z
} |
{z
}
|
II
III
!
!
√
√
∞
∞
X
X
230 2
20 2
+
sin n(ωt − 300 )
sin(nωt)
2
n
n
n=1,3,5
n=1,3,5
|
{z
}
IV
√
√
∞
∞
X
X
200 2
460
2
= |{z}
20 +
sin n(ωt − 300 ) +
sin(nωt)
2
n
n
n=1,3,5
n=1,3,5
I
|
{z
} |
{z
}
II
III
∞
X
4600
+
(cos(30o n)(1 − cos 2nωt) − sin (2nωt) sin(30o n))
3
n
n=1,3,5
|
{z
}
IV A
√
!
∞
X
230 2
+
sin nωt
2
n
n=1,3,5
|
∞
X
h=1,3,5;h6=n
20
√
h
2
!
sin h(ωt − 300 )
{z
IV B
a. Computation of p(t), pactive (t), preactive (t), Pdc ., and prest (t)
pdc−dc (t) = 20 W
∞
X
4600
pactive (t) =
cos n300 (1 − cos 2nωt)
3
n
n=1,3,5
∞
X
4600
preactive (t) = −
sin(n300 ) sin(2nωt)
3
n
n=1,3,5
√
√
∞
∞
X
X
200 2
460 2
0
prest (t) =
sin n(ωt − 30 ) +
sin(nωt)
n
n2
n=1,3,5
n=1,3,5
!
!
√
√
∞
∞
X
X
230 2
20 2
+
sin nωt
sin h(ωt − 300 )
2
n
h
n=1,3,5
h=1,3,5;h6=n
23
}
b. Computation of P, P1 , PH
1
P =
T
Z
T
p(t)dt
0
= 20 +
∞
X
4600
cos(30o n)
3
n
n=1,3,5
= 20 + 4600 cos 300 +
∞
X
4600
cos(30o n)
3
n
n=3,5,7...
= 20 + 3983.71 + (−43.4841)
= Pdc + P1 + PH
Thus,
Active power contributed by dc components of voltage and current, Pdc = 20 W.
Active power contributed by fundamental frequency components of voltage and current, P1 =
3983.71 W.
Active power contributed by harmonic frequency components of voltage and current, PH = −43.4841
W.
c. Computation of Q, Q1 , QH
∞
X
4600
sin(30o n)
Q =
3
n
n=1,3,5
0
= 4600 sin 30 +
∞
X
4600
sin(30o n)
3
n
n=3,5,7...
= 2300 + 175.7548 VArs
= Q1 + QH
The above implies that, Q1 = 4600 VArs and QH =
VArs.
P∞
3
o
n=3,5,7... (4600/n ) sin(30
d. Computation of Apparent Powers and Distortion Powers
24
n) = 175.7548
The apparent power S is expressed as following.
q
Vrms =
Vdc2 + V12 + V32 + V52 + V72 + V92 + ....
p
102 + 2302 + (230/32 )2 + (230/52 )2 + (230/72 )2 + (230/92 )2 + ....
=
= 231.87 V
(up to n = 9)
q
2
+ I12 + I32 + I52 + I72 + I92 + ....
Idc
p
=
22 + 202 + (20/3)2 + (20/5)2 + (20/7)2 + (20/9)2 + ....
= 21.85 A
(up to n = 9)
Irms =
The apparent power, S = Vrms Irms = 231.87 × 21.85 = 5066.36 VA.
Fundamental apparent power, S1 = V1 x I1 = 4600 VA.
Apparent power contributed by harmonics SH = VH × IH
VH
q
V32 + V52 + V72 + V92 + ....
=
p
(230/32 )2 + (230/52 )2 + (230/72 )2 + (230/92 )2 + ....
=
= 27.7 V
(up to n = 9)
q
I32 + I52 + I72 + I92 + ....
p
(20/3)2 + (20/5)2 + (20/7)2 + (20/9)2 + ....
=
= 8.57 A
(up to n = 9).
IH =
Therefore the harmonic apparent power, SH = VH × IH = 237.5 VA.
Non active power, N =
Distortion Power D =
VArs (up to n=9).
√
√
S 2 − P 2 = 50672 − √
3960.22 = 3160.8 VArs (up to n=9)
p
2
2
2
S − P − Q =
50672 − 3960.22 − 2475.772 = 1965.163
Displacement power factor (cos φ1 )
cos φ1 =
P1
3983.7
=
= 0.866 lagging
S1
(230)(20)
Power factor (cos φ)
P
3960.217
=
= 0.781 lagging
S
5067
The voltage, current, various powers and power factor are plotted in the Fig. 1.8, verifying
above values.
cos φ =
25
20
Current (A)
Voltage (v)
200
100
0
-100
-200
10
0
-10
-20
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0
0.005
0.01
0.015
4000
2000
0
-2000
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0
-500
-1000
0.015
0.02
0.025
0.03
0.035
0.04
Avg. active power (W)
2000
Avg. reactive power (VAr)
0.01
0.015
0.02
0.025
Distortion power (W)
0.03
0.035
0.04
Non active power (VA)
Total apparent power (VA)
2500
0.005
0.025
0.03
0.035
0.04
0.025
0.03
0.035
0.04
0.025
0.03
0.035
0.04
1000
0
0.005
0.01
0.015
0.02
2000
0
-2000
0
0.005
0.01
0.015
0.02
time (sec)
3000
1000
0
0.04
2000
time (sec)
1500
0.035
3000
0
Rest of inst. power (W)
Inst. reactive power
500
0.01
0.03
time (sec)
1000
0.005
0.025
4000
time (sec)
0
0.02
time (sec)
Inst. active power (W)
Inst. Power (W)
time (sec)
1168
1167
1166
1165
0
0.005
0.01
time (sec)
0.015
0.02
time (sec)
Fig. 1.8 System voltage, current and related powers
References
[1] IEEE Group, “IEEE trial-use standard definitions for the measurement of electric power quantities under sinusoidal, nonsinusoidal, balanced, or unbalanced conditions,” 2000.
[2] E. Watanabe, R. Stephan, and M. Aredes, “New concepts of instantaneous active and reactive
powers in electrical systems with generic loads,” IEEE Transactions on Power Delivery, vol. 8,
no. 2, pp. 697–703, 1993.
[3] T. Furuhashi, S. Okuma, and Y. Uchikawa, “A study on the theory of instantaneous reactive
power,” IEEE Transactions on Industrial Electronics, vol. 37, no. 1, pp. 86–90, 1990.
[4] A. Ferrero and G. Superti-Furga, “A new approach to the definition of power components in
three-phase systems under nonsinusoidal conditions,” IEEE Transactions on Instrumentation
and Measurement, vol. 40, no. 3, pp. 568–577, 1991.
[5] J. Willems, “A new interpretation of the akagi-nabae power components for nonsinusoidal
three-phase situations,” IEEE Transactions on Instrumentation and Measurement, vol. 41,
no. 4, pp. 523–527, 1992.
26