Review of Chapter 2, Plus Matlab Examples
2.2
Power in single-phase circuits
Let v ( t ) and i ( t ) be defined as:
v ( t ) = Vm cos (ω t + θ v )
and i ( t ) = I m cos (ω t + θ i )
then the instantaneous power is give by:
p ( t ) = v ( t ) i ( t ) = Vm I m cos (ω t + θ v ) cos (ω t + θi )
=
Vm I m
cos ( 2ω t + θ v + θ i ) + cos (θ v − θ i )
2 
{
}
= Vrms I rms cos  2 (ω t + θ v ) − (θ v − θ i ) + cos (θ v − θ i )
= V I {cos 2 (ω t + θ v ) cos (θ v − θ i ) + sin 2 (ω t + θ v ) sin (θ v − θ i ) + cos (θ v − θ i )}
p ( t ) = V I cosθ 1 + cos 2 (ω t + θ v )  + V I sin θ sin 2 (ω t + θ v )
where θ = θ v − θ i . The angle θ is called the power factor angle and is the angle between
the voltage and current at a particular 2-port. Note: the power factor is given by cos θ .
Note that the cosine is an even function thus the sign of θ is lost! For an inductive
circuit (memory helper: ELI) voltage leads the current, the carrent lags the voltage and
we call this a "lagging" power factor because the current lags the voltage. For a
capacitive circuit (memory helper: ICE) the current leads the voltage, and we call this a
"leading" power factor. Thus, we always say clearly if the power factor is lagging
(inductive) or leading (capacitive). The lag or lead is the angle by which the current lags
or leads the voltage.
V V
It is also noted that θ is the impedance angle, thus: Z ∠θ = =
∠ (θ v − θ i ) . The
I
I
instantaneous power can be expressed in two parts:
p ( t ) = pR ( t ) + p X ( t )
thus pR ( t ) = V I cos θ 1 + cos 2 (ω t + θ v )  and p X ( t ) = V I sin θ sin 2 (ω t + θ v ) .
It is common to define P and Q as follows:
1
P = avg ( p ( t ) ) = avg ( pR ( t ) ) = V I cos θ
Q = V I sin θ
Thus in terms of P and Q , the instantaneous power can be expressed as:
p ( t ) = P 1 + cos 2 (ω t + θ v )  + Q sin 2 (ω t + θ v )
Thus P is the "average" value of the cosine terms (counting DC as a zero frequency
cosine term) while Q is the "magnitude" of the sine term. It is noted that besides the DC
term, power has twice line frequency components.
Example 2.1
A supply voltage v ( t ) = 100 cos ω t is applied across a load whose impedance is given by
Z = 1.25∠60! Ω . Determine i ( t ) and p ( t ) . Use Matlab to plot i ( t ) , v ( t ) , p ( t ) , pR ( t )
and p X ( t ) over an interval from 0 to 2π .
The Matlab program is shown below:
Vm = 100; thetav = 0;
% Voltage amplitude and phase angle
Z = 1.25; gama = 60;
% Impedance magnitude and phase angle
thetai = thetav - gama;
% Current phase angle in degree
theta = (thetav - thetai)*pi/180;
% Degree to radian
Im = Vm/Z;
% Current amplitude
wt=0:.05:2*pi;
% wt from 0 to 2*pi
v=Vm*cos(wt);
% Instantaneous voltage
i=Im*cos(wt + thetai*pi/180);
% Instantaneous current
p=v.*i;
% Instantaneous power
V=Vm/sqrt(2); I=Im/sqrt(2);
% rms voltage and current
P = V*I*cos(theta);
% Average power
Q = V*I*sin(theta);
% Reactive power
S = P + j*Q
% Complex power
pr = P*(1 + cos(2*(wt + thetav)));
% Eq. (2.6)
px = Q*sin(2*(wt + thetav));
% Eq. (2.8)
PP=P*ones(1, length(wt)); % Average power with length w for plot
xline = zeros(1, length(wt));
% generates a zero vector
wt=180/pi*wt;
% converting radian to degree
subplot(2,2,1), plot(wt, v, wt, i,wt, xline), grid
title(['v(t)=Vm coswt, i(t)=Im cos(wt +', num2str(thetai), ')'])
xlabel('wt, degree')
subplot(2,2,2), plot(wt, p, wt, xline), grid
title('p(t)=v(t) i(t)'), xlabel('wt, degree')
subplot(2,2,3), plot(wt, pr, wt, PP, wt,xline), grid
title('pr(t)
Eq. 2.6'), xlabel('wt, degree')
subplot(2,2,4), plot(wt, px, wt, xline), grid
title('px(t) Eq. 2.8'), xlabel('wt, degree')
subplot(111)
2
S =
2.0000e+003 +3.4641e+003i
v(t)=Vm coswt, i(t)=Im cos(wt +-60)
100
50
4000
0
2000
-50
0
-100
0
100
200
300
wt, degree
pr(t) Eq. 2.6
400
-2000
4000
4000
3000
2000
2000
0
1000
-2000
0
2.3
p(t)=v(t) i(t)
6000
0
100
200
300
wt, degree
400
-4000
0
100
200
300
wt, degree
px(t) Eq. 2.8
0
100
200
300
wt, degree
400
400
Complex power
It is observed that the term VI * gives the result: (N.B. θ = θ v − θ i )
VI * = V I cos θ + j V I sin θ
which is identical to S = VI * = P + jQ where S is the complex power.
2
V
2
OTHER FORMS for S : S = R I + jX I = * = Z I .
Z
2
2
3
For example, for an inductive load, the current lags the voltage and the phasor diagrams
would be as shown below:
V
θ
I
S
Q
θ
P
Phasor diagrams (V, I) and the power triangle for
an inductive load
And for a capacitive load, the current would lead the voltage , and the phasor diagrams
would be as shown below:
I
V
θ
P
θ
Q
S
Phasor diagrams (V, I) and the power triangle for
a capacitive load, leading power factor angle
2.4 Complex power balance
The total value of S for a circuit is the SUM of S for the components of S .
Example 2.2:
Three impedances in parallel are suppleid by a source V = 1200∠0! V where the
impedances are given by: Z1 = 60 + j 0 , Z 2 = 6 + j12 and Z 3 = 30 − j 30 . Find the power
absorbed by each load and the total complex power. As a check, compute the power
supplied by the source.
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Matlab program follows:
V = 1200; Z1= 60; Z2 = 6 + j*12; Z3 = 30 - j*30;
I1 = V/Z1; I2 = V/Z2; I3 = V/Z3;
S1= V*conj(I1), S2= V*conj(I2), S3= V*conj(I3)
S = S1 + S2 + S3
Sv = V*conj(I1+I2+I3)
S1 =
24000
S2 =
4.8000e+004
S3 =
2.4000e+004
S =
9.6000e+004
Sv =
9.6000e+004
+9.6000e+004i
-2.4000e+004i
+7.2000e+004i
+7.2000e+004i
It is observed that the complex power is conserved.
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2.5 Power factor correction
Adding a capacitor (usually in parallel) with an inductive load can improve the power
factor.
Example 2.3
Two loads Z1 = 100 + j 0 and Z 2 = 10 + j 20 are connected across a 200-V 60 Hz source.
(a) Find the total real and reactive power, the power factor at the source, and the total
current.
(b) Find the capacitance of the capacitor connected across the loads to improve the
overall power factor to 0.8 lagging.
Q
S
QC
θd
P
Example 2.3
The Matlab program follows:
V = 200; Z1= 100; Z2 = 10 + j*20;
I1 = V/Z1; I2 = V/Z2;
S1= V*conj(I1), S2= V*conj(I2)
I = I1 + I2
S = S1 + S2, P = real(S); Q = imag(S);
PF = cos(angle(S))
thd = acos(0.8), Qd = P*tan(thd)
Sc = -j*(Q - Qd)
Zc = V^2/conj(Sc), C = 1/(2*pi*60*abs(Zc))
Sd = P + j*Qd;
Id=conj(Sd)/conj(V)
S1 =
400
S2 =
8.0000e+002 +1.6000e+003i
I =
6.0000 - 8.0000i
S =
1.2000e+003 +1.6000e+003i
PF =
6
0.6000
thd =
0.6435
Qd =
900
Sc =
0 -7.0000e+002i
Zc =
0 -57.1429i
C =
4.6420e-005
Id =
6.0000 - 4.5000i
Example 2.4
Three loads are connected in parallel across a 1400-V, 60 Hz single-phase supply. Given
that load 1 is inductive, 125 kVA at 0.28 power factor, load 2 is capacitive, 10 kW and
40 kvar, and load 3 is resistive at 15 kW.
(a) Find the total kW, kVA and the supply power factor.
(b) A capacitor is connected in parallel with the loads to improve the power factor to 0.8
lagging. Find the kvar rating of the capacitor and its capacitance in µ F.
The Matlab program follows:
V = 1400;
S1= 35000 + j*120000; S2 = 10000 - j*40000; S3 = 15000;
S = S1 + S2 + S3, P = real(S); Q = imag(S);
PF = cos(angle(S))
I = conj(S)/conj(V)
thd = acos(0.8), Qd = P*tan(thd)
Sc = -j*(Q - Qd)
Zc = V^2/conj(Sc), C = 1E6/(2*pi*60*abs(Zc))
Sd = P + j*Qd;
Id=conj(Sd)/conj(V)
S =
6.0000e+004 +8.0000e+004i
PF =
0.6000
I =
42.8571 -57.1429i
thd =
0.6435
Qd =
45000
Sc =
0 -3.5000e+004i
Zc =
0 -56.0000i
7
C =
47.3675
Id =
42.8571 -32.1429i
2.6 Complex Power Flow
This is a very important part of chapter 2. Note the diagram below. It shows four
power flows. They are:
1. S12 is the complex power at node number 1 flowing in (i.e. from node 1 in the
direction of node 2).
2. − S12 is the complex power at node number 1 flowing out (i.e. from node 1 in the
direction away from node 2).
3. S21 is the complex power at node number 2 flowing in (i.e. from node 2 in the
direction of node 1).
4. − S21 is the complex power at node number 2 flowing out (i.e. from node 2 in the
direction away from node 1).
S 12
-S 12
-S 21
S 21
Z
+
+
I12
V2
V1
-
-
Thus the power consumed by the impedance Z can be expressed as S Z = S12 + S21 .
Important Results from section 2.6: (will be shown in detail later in
the lecture)
S12 =
V1
Z
2
∠γ −
V1 V2
Z
∠ (γ + δ1 − δ 2 )
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P12 =
V1
Q12 =
V1
2
Z
cos γ −
V1 V2
sin γ −
V1 V2
Z
2
Z
Z
cos ( γ + δ1 − δ 2 )
sin ( γ + δ1 − δ 2 )
where V1 = V1 ∠δ1 and V2 = V2 ∠δ 2 and the impedance is Z = Z ∠γ . The angle δ
defined as δ = δ1 − δ 2 is often called the "power angle". N.B. This is very different from
the "power factor angle" discussed earlier.
If we assume that the circuit above represents two generators connected by a transmission
line, then the equations above still apply. In particular the impedance of a transmission
line may be assumed purely inductive, i.e. Z = 0 + jX . In this case the angle γ = 90! and
the equations above simplify to:
V V
P12 = 1 2 sin ( δ1 − δ 2 )
X
Q12 =
V1
 V1 − V2 cos ( δ1 − δ 2 ) 
X 
Since the line is "lossless" then power consumed by the line is zero, i.e. P12 + P21 = 0 . In
terms of the power angle, these equations become:
P12 =
Q12 =
V1 V2
X
sin δ
V1
 V1 − V2 cos δ 
X 
Observations (assuming R ≈ 0 ):
1. Usually δ is very small (less than 10 degrees), thus P12 ∝ sin δ , i.e. small changes in
power angle greatly change the real power flow and not the reactive power. If
δ1 > δ 2 then power flows from node one to node two. If δ1 < δ 2 , then power flows in
the opposite direction (from node 2 to node 1).
V V
2. Maximum power transfer occurs when δ = 90! and is given by: Pmax = 1 2 .
X
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3. Since δ ≈ 0 , Q ∝ V1 − V2 . Thus small changes in V1 − V2 greatly affect Q but not
P.
Thus, to control real power, we need to change the power angle δ . This is done be
increasing prime mover power (mechanical power driving the generator). To control
reactive power, we need to change the difference in voltage magnitude. This is done by
changing the DC excitation of one generator.
Example 2.5
Assume V1 = 120∠ − 5 and V2 = 100∠0 . Let Z = 1 + j 7 Ω . Determine the real and
reactive power supplied or received by each source and the power loss in the line.
The Matlab code follows:
R = 1; X = 7; Z = R +j*X;
V1 = 120*(cos(-5*pi/180) + j*sin(-5*pi/180));
V2 = 100+j*0;
I12 = (V1 - V2)/Z, I21 = -I12;
S12 = V1*conj(I12), S21 = V2*conj(I21)
SL = S12 + S21
PL = R*abs(I12)^2, QL = X*abs(I12)^2
I12 =
-1.0733 - 2.9452i
S12 =
-9.7508e+001 +3.6331e+002i
S21 =
1.0733e+002 -2.9452e+002i
SL =
9.8265 +68.7858i
PL =
9.8265
QL =
68.7858
Example 2.6:
Repeat example 2.5 using a Matlab program such that the angle of source 1 is changed
from -30 to 30 degrees in steps of 5 degrees each.
We cannot execute "chp2ex6" from within the notebook because input values are to be
entered by the user. Hence we go to a Matlab window and execute this command.
GO TO A MATLAB WINDOW AND EXECUTE "chp2ex6".
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