ECE 320
Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 1: Introduction and Three Phase Power
1.1
Review of Basic Circuit Analysis
Definitions:
Node - Electrical junction between two or more devices.
Loop - Closed path formed by tracing through an ordered sequence of nodes without passing through
any node more than once.
Element Constraints:
Ohm’s Law
v = iR
Capacitor Equation
i=C
Inductor Equation
dv
dt
di
v=L
dt
Connection Constraints:
Kirchhoff’s Current Law - The algebraic sum of currents entering a node is zero at every instant in
time.
(1.1)
∑ ik = 0
Node
Kirchhoff’s Voltage Law - The algebraic Sum of all voltages around a loop is zero at every instant in
time.
(1.2)
∑ vk = 0
Loop
Passive Sign Convention:
Whenever the reference direction of current into a two terminal device is in the direction of the
reference voltage drop across the device, then the power absorbed (or dissipated) is positive.
_
v
+
i
Figure 1. Circuit element and passive sign convention.
p (t ) = i (t ) ⋅ v(t )
(1.3)
1- 1
When the above convention is used, p(t) > 0 for absorbed power, and p(t) < 0 for delivered power.
Time Varying Signals
Although a number of exceptions can be found throughout the world, the predominance of
electric power follows 60Hz or 50Hz frequencies. North America, part of Japan, and ships at sea use
60Hz while most of the rest of the world uses 50Hz. The historical reasons for these two frequencies
stem from the differences in lighting (filaments in vacuum or filaments in a gas atmosphere). The
lower frequencies caused an annoying flicker for lights having filaments in a gas atmosphere, and
thus a higher frequency was adopted in part of the world that initially used such lighting.
While not strictly adhered to within your textbook and these notes, an attempt to use the
following conventions has been made.
Scalar time varying signals (examples):
v = vmax sin (ωt ) (V)
v(t ) = vmax cos(ωt ) (V)
v = 185 sin (377 ⋅ t ) (kV)
i (t ) = imax cos(100π ⋅ t ) (A)
Spatial vectors (bold or arrow overhead or line overhead):
G
F or F or B
If necessary, unit vectors will be used, (for example): B = Bx xˆ + By yˆ + Bz zˆ
These unit vectors should not be confused with phasors below
Phasors (phasor representation of a time varying signal):
A phasor can be represented as a complex number with real and imaginary components
such that a phasor of magnitude (or length) R that is at a phase angle of θ with respect to
the x-axis (the Real axis) can be written as Rˆ = x + jy = R ⋅ e jθ = R cos(θ ) + jR sin (θ )
where Euler’s formula e jθ = cos(θ ) + j sin (θ ) was used for the last representation. Note
that R is the magnitude and θ the phase of the phasor, and R = x 2 + y 2 and
⎛ y⎞
⎝ x⎠
θ = arctan⎜ ⎟ . The phasor can be shown graphically in Figure 2.
1- 2
Imaginary
y
R
θ
ωt
x
Real
Figure 2. Phasor of magnitude R and phase θ.
For a sinusoidal function v(t ) = V cos(ωt + θ ) , the phasor representation is
Vˆ = Ve jθ = V /θ
If the phasor is rotating counterclockwise about the origin at a rate of ω radians per
second, then we multiply the phasor by e jωt such that Vˆe jωt = Ve jθ e jωt = Ae j (ωt +θ ) and
using Euler’s formula Vˆe jωt = Ve j (ωt +θ ) = V cos(ωt + θ ) + jV sin (ωt + θ ) . Then we see that
v(t ) = V cos(ωt + θ ) is the real part of Vˆe jωt , or
{
}
v(t ) = Re Vˆe jωt = V cos(ωt + θ ) .
In the above description of the phasor, the peak value is used, however we will use the
RMS value of V as described in (1.6) instead of the peak value. This simplifies many of
the calculations, particularly those associated with power as shown below.
Phasors will be represented with a hat (or caret) above the variable. Impedance is understood to be a
complex quantity in general, and the hat (or caret ^) is left off the impedance variable
Z = R + jX where R is the resistance, and X is the reactance component respectively.
For a circuit element such as the one shown in Figure 1 representing a load in the circuit with i(t)
as the instantaneous value of current through the load and v(t) is the instantaneous value of the
voltage across the load.
In quasi-steady state conditions, the current and voltage are both sinusoidal, with corresponding
2π
amplitudes of Vmax, Imax, and initial phases, φv and φi, and the same frequency ω = 2πf =
T
v(t ) = Vmax cos(ωt + φv )
(1.4)
i(t ) = I max cos(ωt + φi )
(1.5)
The root-mean-squared (RMS) value of the voltage and current are then:
V =
1 T
[Vmax cos(ωt + φv )]2 dt = Vmax
∫
o
T
2
(1.6)
1- 3
1 T
[I max cos(ωt + φi )]2 dt = I max
∫
o
T
2
I=
(1.7)
Phasor representations for the above signals use the RMS values as Vˆ = V /φv and Iˆ = I /φi.
The instantaneous power is the product of voltage times current or:
p(t ) = v(t ) ⋅ i (t ) = Vmax cos(ωt + φv ) ⋅ I max cos(ωt + φi ) = Vmax I max cos(ωt + φv ) cos(ωt + φi )
= 2VI cos(ωt + φv ) cos(ωt + φi )
1
= 2VI [cos(φv − φi ) + cos(2ωt + φv + φi )]
(1.8)
2
The average value is found by integrating over a period of time and then dividing the result by
that same time interval. The first term in (1.8) is independent of time, while the second term varies
from -1 to +1 symmetrically about zero. Because it is symmetric about zero, integration over an
integer number of cycles (or periods) gives a value of zero for the second term in (1.8), and the
average power which we define as the real power with units of watts (W) is:
P = VI cos(φv − φi )
(1.9)
This shows the power is not only proportional to the RMS values of voltage and current, but also
proportional to cos(φv − φi ) . The cosine of this angle is defined as the displacement factor, DF. In
more general terms for periodic, but not necessarily sinusoidal signals, the power factor is defined as:
pf ≡
P
VI
(1.10)
For sinusoidal signals, the power factor equals the displacement factor, or
pf = cos(φv − φi )
(1.11)
For comparison, the voltage, current, and power for various angles between voltage and current
are shown below:
Voltage (V), Current (A), Power (W)
For leading power factors:
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt )
10
5
P = VI cos(φv − φi ) =
0
Vmax I max
cos(φv − φi ) = 5 (W)
2 2
Imaginary
-5
-10
0
0.5
1
1.5
# of Periods
2
2.5
3
I
Real
V
1- 4
Voltage (V), Current (A), Power (W)
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt + 30º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = 4.33 (W)
2 2
10
5
0
Imaginary
-5
-10
I
0
0.5
1
1.5
2
2.5
Real
3
Voltage (V), Current (A), Power (W)
# of Periods
V
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt + 60º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = 2.5 (W)
2 2
10
5
0
Imaginary
-5
I
-10
0
0.5
1
1.5
2
2.5
Real
3
Voltage (V), Current (A), Power (W)
# of Periods
V
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt + 90º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = 0 (W)
2 2
10
5
0
Imaginary
-5
-10
I
0
0.5
1
1.5
2
2.5
Real
3
Voltage (V), Current (A), Power (W)
# of Periods
V
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt + 120º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = −2.5 (W)
2 2
10
5
0
Imaginary
-5
I
-10
0
0.5
1
1.5
# of Periods
2
2.5
Real
3
V
1- 5
Voltage (V), Current (A), Power (W)
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt + 150º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = −4.33 (W)
2 2
10
5
0
Imaginary
-5
I
-10
0
0.5
1
1.5
2
2.5
Real
3
Voltage (V), Current (A), Power (W)
# of Periods
V
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt + 180º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = −5 (W)
2 2
10
5
0
Imaginary
-5
-10
I
0
0.5
1
1.5
2
2.5
Real
3
# of Periods
V
Voltage (V), Current (A), Power (W)
For lagging power factors:
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt )
10
5
P = VI cos(φv − φi ) =
0
Vmax I max
cos(φv − φi ) = 5 (W)
2 2
Imaginary
-5
-10
0
0.5
1
1.5
# of Periods
2
2.5
3
I
Real
V
1- 6
Voltage (V), Current (A), Power (W)
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt − 30º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = 4.33 (W)
2 2
10
5
0
Imaginary
-5
-10
0
0.5
1
1.5
2
2.5
Real
3
# of Periods
Voltage (V), Current (A), Power (W)
I
V
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt − 60º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = 2.5 (W)
2 2
10
5
0
Imaginary
-5
-10
0
0.5
1
1.5
2
2.5
Real
3
# of Periods
V
Voltage (V), Current (A), Power (W)
I
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt − 90º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = 0 (W)
2 2
10
5
0
Imaginary
-5
-10
0
0.5
1
1.5
2
2.5
Real
3
# of Periods
Voltage (V), Current (A), Power (W)
I
V
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt − 120º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = −2.5 (W)
2 2
10
5
0
Imaginary
-5
-10
0
0.5
1
1.5
# of Periods
2
2.5
Real
3
I
V
1- 7
Voltage (V), Current (A), Power (W)
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt − 150º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = −4.33 (W)
2 2
10
5
0
Imaginary
-5
-10
0
0.5
1
1.5
2
2.5
Real
3
# of Periods
V
Voltage (V), Current (A), Power (W)
I
v(t ) = 5 ⋅ cos(ωt )
i (t ) = 2 ⋅ cos(ωt − 180º )
V I
P = VI cos(φv − φi ) = max max cos(φv − φi ) = −5 (W)
2 2
10
5
0
Imaginary
-5
-10
0
0.5
1
1.5
2
2.5
Real
3
I
# of Periods
V
We call the power factor leading or lagging, depending on whether the current of the load leads or
lags the voltage across it. We know that current does not change instantly through an inductor, so it
is clear that for a resistive - inductive (RL) load, the power factor is lagging. Likewise the voltage
can not change instantly across a capacitor, therefore for a resistive – capacitive (RC) load, the power
factor is leading (the current changes before the voltage can). Also, for a purely inductive or
capacitive load the power factor is 0, while for a purely resistive load it is 1.
The product of the RMS values of voltage and current at a load is the apparent power, S having
units of volt-amperes (VA):
S = VI
(1.12)
The reactive power is Q with units of volt-amperes reactive (VA reactive, or VAr):
Q = VI sin (φv − φi )
(1.13)
The reactive power represents the energy oscillating in and out of an inductor or capacitor. Since
the energy oscillation in an inductor is 180º out of phase with the energy oscillating in an capacitor,
the reactive power of these two have opposite signs with the convention that it is positive for the
inductor and negative for the a capacitor.
Using phasors, the complex apparent power, Ŝ is:
Sˆ = VˆIˆ* = V/φv I/-φi
(1.14)
or
1- 8
Sˆ = P + jQ
(1.15)
As an example, consider the following voltage and current for a given load:
π⎞
⎛
v(t ) = 120 2 sin ⎜ 377t + ⎟ (V)
6⎠
⎝
(1.16)
π⎞
⎛
i (t ) = 5 2 sin⎜ 377t + ⎟ (A)
4⎠
⎝
(1.17)
⎛π π ⎞
then S = VI = 120 ⋅ 5 = 600 (W), while the power factor is pf = cos⎜ − ⎟ = 0.966 leading. Also,
⎝6 4⎠
the complex apparent power is:
Sˆ = VˆIˆ* = 120/π/6 · 5/-π/4 = 600/- π/12 = 579.6 (W) – j155.3 (VAr)
(1.18)
It should also be noted that when the angles are represented in radian, care must be taken to
assure your calculator is in radians mode. Often we represent the argument of the sine or cosine term
in mixed units. For example we might write cos(377t + 30º). The first term (377t) has units of
radians, while the second (30º) has units of degrees, and one of these must be converted before
making calculations with your calculator or computer either in radians mode or in degrees mode.
As a phasor the complex apparent power can be shown on a complex plane for the cases of
leading and lagging power factors as shown in Figure 3.
Imaginary
Imaginary
I
V
Real
Real
V
I
Imaginary
Imaginary
S
Q
P
Real
Q
P
Real
S
(a) Leading power factor
(b) Lagging power factor
Figure 3. Phasor of magnitude R and phase θ.
1- 9
Recall that the leading power factor corresponds to a resistive-capacitive load. We see from
Figure 3(a), this also corresponds to a negative value for Q. Similarly, lagging power factor has the
current lagging the voltage corresponding to an inductive-resistive load, and Figure 3(b) shows this
corresponds to a positive value for Q.
To summarize, we can use the following tables:
Equation
P = VI cos(φv − φi )
Real Power
Units
Watts
= VI ⋅ pf
Reactive Power
= S ⋅ pf
Q = VI sin (φv − φi )
Volt-Amperes-Reactive
= S ⋅ 1 − pf 2 (for lagging)
Apparent Power
S
Sˆ
Sˆ
= − S ⋅ 1 − pf 2 (for leading)
= VI
= VˆIˆ* = V/φv I/-φi
= P + jQ
Volt-Amperes
S 2 = P2 + Q2
Type of Load
Inductive
Capacitive
Resistive
Reactive Power
Q>0
Q<0
Q=0
Power Factor
lagging
leading
1
From the interdependence of the four quantities, S, P, Q, pf, if we know any two of these
quantities, the other two can be determined. For example, if S = 100 (kVA), and pf = 0.8 leading,
then:
P = S ⋅ pf = 800 (kW)
Q = − S 1 − pf 2 = −60 (kVAR), or
sin (φv − φi ) = sin[arccos(0.8)] then
Q = S sin (φv − φi )
It is important to notice that Q < 0, such that sin (φv − φi ) is a negative quantity. This can be seen
when it is understood that there are two possible answers for the arccos(0.8), that is cos(36.87º) = 0.8,
and cos(-36.87º) = 0.8 so to obtain a Q < 0 we use (φv – φi) = −36.87º.
Generally, in systems that contain more than one load (or source), the real and reactive power can
be found by adding individual contributions, but this is not the case with the apparent power. That is
Ptotal = ∑i Pi
Qtotal = ∑i Qi
(1.19)
Stotal ≠ ∑i Si
1- 10
For the above example, if the load voltage is VL = 2000 (V), then the load current would be
IL = S/VL = [100×103 (VA)]/[2×103 (V)] = 50 (A). If we use the load voltage as the reference, then:
Vˆ = 2000 /0º (V)
Iˆ = 50 /φi = 50 /+36.87º (A)
Sˆ = VˆIˆ* = [2000 /0º][50 /−36.87º] = P + jQ = 80×103 (W) – j60×103 (VAr)
1.2
Three Phase Balanced Systems
Three-phase systems offer significant advantages over single phase systems: for the same power
and voltage there is less copper in the windings, and the total power absorbed remains constant rather
than oscillating about an average value.
For a three phase system consisting of three current sources having the same amplitude and
frequency, but with phases differing by 120º as:
i1 (t ) = 2 I sin (ωt + φ )
2π ⎞
⎛
i2 (t ) = 2 I sin ⎜ ωt + φ −
⎟
3 ⎠
⎝
2π ⎞
⎛
i3 (t ) = 2 I sin ⎜ ωt + φ +
⎟
3 ⎠
⎝
(1.20)
If these are connected as shown in Figure 4, then at node n or n’, the current adds to zero, and the
neutral line n-n’ (dashed) is not needed.
⎧
2π ⎞
2π
⎛
⎛
i1 (t ) + i2 (t ) + i3 (t ) = 2 I ⎨sin (ωt + φ ) + sin ⎜ ωt + φ −
⎟ + sin ⎜ ωt + φ +
3
3 ⎠
⎝
⎝
⎩
⎞⎫
⎟⎬ = 0
⎠⎭
i1
n'
i3
n
i2
Figure 4. Balanced three phase Y-connected system with zero neutral current.
If instead we had three voltage sources Y-connected as in Figure 5 with the following values
1- 11
v1 (t ) = 2V sin (ωt + φ )
2π ⎞
⎛
(1.21)
v2 (t ) = 2V sin ⎜ ωt + φ −
⎟
3 ⎠
⎝
2π ⎞
⎛
v3 (t ) = 2V sin ⎜ ωt + φ +
⎟
3 ⎠
⎝
then, the current through each of the three loads (assuming the loads are equal), would have equal
magnitudes, but each current would have a phase that is shifted by an equal amount with respect to
the voltages, v1(t), v2(t), v3(t).
ia
+
v1
n
+
n'
+
v3
v2
ib
ic
Figure 5. Balanced voltage fed three phase Y-connected system with zero neutral current.
With equal impedances for the loads, then
ia (t ) =
2V
sin (ωt + φ + θ )
Z
ib (t ) =
2V
2π
⎛
⎞
sin ⎜ ωt + φ −
+θ ⎟
Z
3
⎝
⎠
ic (t ) =
2V
2π
⎛
⎞
sin ⎜ ωt + φ +
+θ ⎟
Z
3
⎝
⎠
(1.22)
and again the currents sum to zero at nodes n or n’, and for this balanced three phase system, the
neutral wire (dashed) is not required.
In comparison to a single phase system, where two wires are required per phase, the three phase
system delivers three times the power, and requires only three transmission wires total. This is a
significant advantage considering the hundreds of miles of wire needed for power transmission.
Y and Δ Connections
The loads in the previous two figures, as well as in Figure 6 are connected in a Y or star
configuration. If the load of Figure 6 is for a balanced Y system, then the voltages between each
phase and the neutral are: Vˆ1n = V /φ , Vˆ2 n = V /φ − 2π/3 , and Vˆ3n = V /φ + 2π/3 .
Kirchhoff’s voltage law (KVL) states that the sum of voltages around a closed loop equals zero.
This is also the case here however the voltages are complex numbers or phasors, and as such must be
1- 12
added as vectors. The phase φ can be any value, but the relative position of the phase to neutral
phasors must be 120º with respect to each other as shown in Figure 7.
3
V3n +
1
+
+
V1n
n
V12
V2n
+
2
Figure 6. Y-connected loads with voltages relative to neutral identified.
By KVL: − Vˆ12 + Vˆ1n − Vˆ2 n = 0 , or Vˆ12 = Vˆ1n − Vˆ2 n as shown in Figure 7.
-V1n
V31
V3n
V2n
V12
V1n
-V3n
-V2n
V23
Figure 7. Voltage phasors of the Y-connected loads shown in Figure 6.
We could also use the phasor representation Vˆ1n = V /φ , Vˆ2 n = V /φ − 2π/3 ,
and Vˆ = V /φ + 2π/3 to determine the line-to-line voltages as
3n
Vˆ12 = Vˆ1n − Vˆ2 n = 3V /φ + π/6
(1.23)
This shows the RMS value of the line-to-line voltage, Vl-l , at a Y load is 3 times the line-toneutral or phase voltage, Vln. In the Y connection, the phase current is equal to the line current, and
the power supplied to the system is three times the power supplied to each phase, since the voltage
and current amplitudes and phase differences between them are the same in all three phases. If the
power factor in one phase is pf = cos(φv − φi ) , then the total power to the system is:
1- 13
Sˆ3φ = P3φ + jQ3φ
= 3Vˆ1n Iˆ1*n
(1.24)
= 3Vl −l I l cos(φv − φi ) + j 3Vl −l I l sin (φv − φi )
Similarly, for a connection of the loads in the Δ configuration (as in Figure 8), the phase voltage
is equal to the line voltage however; the phase currents are not equal to the line currents for the Δ
configuration. If the phase currents are Iˆ12 = I /φ , Iˆ23 = I /φ − 2π/3 , and Iˆ31 = I /φ + 2π/3 then
using Kirchhoff’s current law (KCL) the current of line 1, as shown in Figure 8 is:
Iˆ1 = Iˆ12 − Iˆ31 = 3I /φ - π/6
Thus for the Δ configuration, the line current is
(1.25)
3 times the IΔ current.
I1
1
-I23
I31
I2
I12
I3
-I12
I31
I2
I23
I12
I1
2
-I31
I23
3
I3
Figure 8. Δ connected load, line, and phase currents.
To calculate the power in the three-phase Δ connected load:
Sˆ3φ = P3φ + jQ3φ
= 3Vˆ12 Iˆ12*
(1.26)
= 3Vl −l I l cos(φv − φi ) + j 3Vl −l I l sin (φv − φi )
which is the same value as for the Y connected load.
For a balanced system, the loads of the three phases are equal. Also, a Δ configured load, can be
replaced with a Y configured load (and visa versa) if:
1
ZY = Z Δ
3
(1.27)
1- 14
Under these conditions, the two loads are indistinguishable by the power transmission lines.
You might recall the Δ to Y transformation for resistor circuits can be remembered by overlaying
the Δ and Y configurations such as in Figure 9.
A
RB
R1
RC
R3
R2
C
RA
B
Figure 9. Δ to Y or Y to Δ resistor network transformation.
R1 =
R B RC
R A + R B + RC
(1.28)
RA =
R1 R 2 + R 2 R 3 + R 3 R1
R1
Equation (1.28) gives a similar result to that of (1.27) when RA = RB = RC and R1 = R2 = R3.
B
1.3
Calculations in Three-Phase Systems
Calculations of quantities like currents, voltages, and power in three-phase systems can be
simplified by the following procedure:
1.
2.
3.
4.
transform the Δ circuits to Y,
connect a neutral conductor,
solve one of the three 1-phase systems
convert the results back to the Δ systems
1.3.1 Example
For the 3-phase system in Figure 10 calculate the line-to-line voltage, real power, and power
factor at the load.
To solve this by the procedure outlined above, first consider only one phase as shown in Figure
11.
1- 15
j1 (Ω)
+
v1 = 120 (V)
7 + j5 (Ω)
n
+
n'
+
v3
v2
Figure 10. Three phase system with Y connected load, and line impedance.
j1 (Ω)
+
I
v1 = 120 (V)
n'
7 + j5 (Ω)
n
Figure 11. One phase of the three phase system shown in Figure 10.
For the one-phase in Figure 11,
120
= 13.02 /-40.6º (A)
j1 + (7 + j 5)
Vˆ1n = IˆZ L = 13.02 /-40.6º (7+j5) = 112/-5.1º (V)
S = Vˆ Iˆ* = (112/-5.1º)(13.02/40.6º) = 1458.3/35.5º = 1.187×103 + j0.848×103
Iˆ =
L ,1φ
L
PL ,1φ = 1.187 (kW), QL ,1φ = 0.848 (kVAr)
pf = cos(-5.1º - (-40.6º)) = 0.814 lagging
For the three-phase system of Figure 10 the load voltage (line-to-line), the real, and reactive
power are:
VL ,l −l = 3 ⋅112 = 194 (V)
PL ,3φ = 3.56 (kW)
QL ,3φ = 2.544 (kVAr)
1- 16
1.3.2 Example
For the Y to Δ three-phase system in Figure 12, calculate the power factor and the real power at
the load, as well as the phase voltage and current. The source voltage is 400 (V) line-to-line.
j1 (Ω)
+
v1
18 + j6 (Ω)
+
n'
+
v3
v2
Figure 12. Δ connected load.
First convert the load to an equivalent Y connected load, then work with one phase of the system.
400
The line to neutral voltage of the source is Vln =
= 231 (V).
3
j1 (Ω)
+
231 (V)
6 + j2 (Ω)
n
+
+
n'
Figure 13. Equivalent Y connected load.
j1 (Ω)
+
n'
IL
v1 = 231 (V)
6 + j2 (Ω)
n
Figure 14. One phase of the Y connected load.
1- 17
IˆL =
231
= 34.44 /-26.6º (A)
j1 + (6 + j 2)
VˆL = IˆL (6 + j 2) = 217.8 /-8.1º (V)
The power factor at the load is:
pf = cos(φv − φi ) = cos(− 8.1º +26.6º ) = 0.948 lagging
Converting back to a Δ connected load gives:
Iφ =
I L 34.44
=
= 19.88 (A)
3
3
Vl −l = 217.8 ⋅ 3 = 377.22 (V)
At the load the power is:
PL ,3φ = 3Vl −l I L pf = 3 ⋅ 377.22 ⋅ 34.44 ⋅ 0.948 = 21.34 (kW)
1.3.3 Example
Two three-phase loads are connected as shown in . Load 1 draws from the system PL1 = 500
(kW) at 0.8 pf lagging, while the total load is ST = 1000 (kVA) at 0.95 pf lagging. What is the power
factor of load 2?
Load 1
Power System
Load 2
Figure 15. Two three-phase loads connected to the same power source.
1- 18
For the total load we can add the real and reactive power for each of the two loads (we can not
add the apparent power).
PT = PL1 + PL 2
QT = QL1 + QL 2
ST ≠ S L1 + S L 2
From the information we have for the total load we can write the following:
PT = ST ⋅ pfT = 950 (kW)
[
]
QT = ST ⋅ sin cos −1 (0.95) = 312.25 (kVAr)
The reactive power, QT, is positive since the power factor is lagging.
For the load L1, PL1 = 500 (kW), pf1 = 0.8 lagging, thus:
S L1 =
500 × 103
= 625 (kVA)
0.8
QL1 = S L21 − PL21 = 375 (kVA)
Again, QL1 is positive since the power factor is lagging. This leads to:
PL 2 = PT − PL1 = 450 (kW)
QL 2 = QT − QL1 = -62.75 (kVAr)
and
pf L 2 =
PL 2
=
SL2
450
4502 + 62.752
= 0.99 leading.
Chapter Notes
• A sinusoidal signal can be described uniquely by:
1. Time dependent form as for example: v(t ) = 5 sin (2πft + φv ) (V)
2. by a time dependent graph of the signal
3. as a phasor along with the associated frequency of the phasor
one of these descriptions is enough to produce the other two.
•
It is the phase difference that is important in power calculations, not the phase. The phase is
arbitrary depending on the defined time (t = 0). We need the phase to solve circuit problems
after we take one quantity (some voltage or current) as a reference. For that reference quantity
we assign an arbitrary phase (often zero).
•
In both three-phase and one-phase systems the total real power is the sum of the real power
from the individual loads. Likewise the total reactive power is the sum of the reactive power of
the individual loads. This is not the case for the apparent power or the power factor.
1- 19
•
Of the four quantities: real power, reactive power, apparent power, and power factor, any two
describe a load adequately. The other two quantities can be calculated from the two given.
•
To calculate real, reactive, and apparent power when using equations (1.9), (1.12), and (1.13)
we must use absolute values, not complex values for the currents and voltages. To calculate the
complex power using equation (1.14) we do use complex currents and voltages and find
directly both the real and reactive power (as the real and imaginary components respectively).
•
When solving a circuit to calculate currents and voltages, use complex impedances, currents
and voltages.
1- 20