Notes on the analysis of AC circuits by Dr. Masoud Karimi
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Mississippi State University Department of Electrical and Computer Engineering ECE 3614: Fundamentals of Energy Systems (Spring 2013) Instructor: Dr. Masoud Karimi ([email protected]) Analysis of AC Circuits and Systems Contents 1 Concept of Phasor 2 2 Concept of Impedance and Admittance 2 3 Connection of Impedances 4 4 Real and Reactive Powers in a Single Phase ac Circuit 4 5 Powers in terms of Impedances 8 6 Examples 9 7 Per-Unit (pu) System 12 8 Three Phase Systems 14 9 Three Phase Source Connections 15 10 Phasors and Phasor Diagram of Three Phase Systems 16 11 Three Phase Load Connections 17 12 Analysis of a Three Phase Circuit 19 13 Powers in a Three Phase Circuit 19 14 Numerical Examples 21 15 Power Measurement of a Balanced Three Phase Load 23 1
Transcript
Mississippi State UniversityDepartment of Electrical and Computer Engineering
ECE 3614: Fundamentals of Energy Systems (Spring 2013)Instructor: Dr. Masoud Karimi ([email protected])
Analysis of AC Circuits and Systems
Contents
1 Concept of Phasor 2
2 Concept of Impedance and Admittance 2
3 Connection of Impedances 4
4 Real and Reactive Powers in a Single Phase ac Circuit 4
5 Powers in terms of Impedances 8
6 Examples 9
7 Per-Unit (pu) System 12
8 Three Phase Systems 14
9 Three Phase Source Connections 15
10 Phasors and Phasor Diagram of Three Phase Systems 16
11 Three Phase Load Connections 17
12 Analysis of a Three Phase Circuit 19
13 Powers in a Three Phase Circuit 19
14 Numerical Examples 21
15 Power Measurement of a Balanced Three Phase Load 23
1
1 Concept of Phasor
In an alternating current (ac) circuit with a given frequency, all the signalsare sinusoidal in the steady state and they all have the same frequency. Thefunction
u(t) =√2U cos(ωt+ θ) (1)
represents time variations of each signal in the steady state in an ac networkwhere U is the rms value and
√2U is the peak value of this signal. The root-
mean-square (rms) value theoretically means U =√
1
T
∫
Tu(t)2dt. Practically,
it means that the average power transferred by the ac voltage to a resistive loadis equal to the power that a dc voltage with the rms value delivers. It is alsocalled the “effective” value.
Given the fact that ω is normally known to good accuracy in a practical powersystem, the knowledge of doublet (U, θ) is adequate to represent u. The phasor
combines these two pieces of information into one complex number as follows.
−→U = Uejθ = U cos θ + jU sin θ (2)
The concept of phasor removes the variable t. That’s why a phasor is alsocalled the frequency-domain representation as opposed to the time-domain rep-resentation. Phasor representation implicitly assumes a reference for time orfor comparing phase angles and that is the phase angle of cos(ωt). Therefore,in phasor analysis, all angles are given with respect to a reference value. Inpractice, it is up to your decision to choose any of the signals as the reference.
2 Concept of Impedance and Admittance
In a dc network, the main circuit component is a resistor. For a resistor withresistance R, the Ohm’s law establishes the relationship between its voltage andcurrent as
v(t) = Ri(t)
for all t.
In an ac circuit, two more types of passive components exit. They are inductorand capacitor. In an inductor with inductance L, the relationship between thevoltage and current is expressed as
v(t) = Ldi
dt
and in a capacitor with capacitance C, the relationship is
i = Cdv
dt.
2
Now, let’s study what happens to the voltage/current relationship of differentcircuit components in terms of phasors.
Resistor: In a resistor of R, we have v(t) = Ri(t) for all t. Therefore, rms ofvoltage is R times rms of current and they have the same phase angle. Thismeans that −→
V = R−→I .
Capacitor: In a capacitor of C, we have i(t) = C dvdt
for all t. Therefore, if
v(t) =√2V cos(ωt+ θ), then
i = −C√2V ω sin(ωt+ θ) = Cω
√2V cos(ωt+ θ + 90).
This means that
−→I = Cω
−→V ej90
= Cω−→V (cos 90 + j sin 90) = jCω
−→V .
Inductor: In an inductor of L, we have v(t) = L didt
for all t. Therefore, if
i(t) =√2I cos(ωt+ θ), then
V = −L√2Iω sin(ωt+ θ) = Lω
√2I cos(ωt+ θ + 90).
This means that
−→V = Lω
−→I ej90
= Lω−→I (cos 90 + j sin 90) = jLω
−→I .
Definition: The Impedance−→Z of a circuit element is a phasor quantity and is
defined as−→Z =
−→V−→I
where−→V is the voltage phasor across that element and
−→I is its current phasor.
It is obvious from the above discussion that for a resistor,
−→Z = R,
for an inductor, −→Z = jLω,
and for a capacitor−→Z =
1
jCω.
3
Impedance is direct extension of resistance. Therefore, it is measured by unitof Ohms.
Definition: The Admittance−→Y is defined as reciprocal of impedance that is
−→Y =
1−→Z.
3 Connection of Impedances
When two circuit components are connected in series, they have the same cur-rent and their voltages add up. Therefore, their impedances add up:
Series connection of resistor and inductor (called series RL):−→Z = R+ jωL
Series connection of resistor and capacitor (called series RC):−→Z = R+ 1
jωC
Series connection of inductor and capacitor (called series LC):−→Z = jωL+ 1
jωC
Series connection of resistor, inductor and capacitor (called series RLC):
−→Z = R+ jωL+
1
jωC
When two circuit components are connected in parallel, they have the samevoltage and their currents add up. Therefore, their admittances add up:
Parallel connection of resistor and inductor (called parallel RL):−→Y = 1
R+ 1
jωL
Parallel connection of resistor and capacitor (called parallel RC):−→Y = 1
R+ jωC
Parallel connection of inductor and capacitor (parallel LC):−→Y = 1
jωL+ jωC
Parallel connection of resistor, inductor and capacitor (called parallel RLC):
−→Y =
1
R+
1
jωL+ jωC
4 Real and Reactive Powers in a Single Phaseac Circuit
Consider a single phase ac circuit comprising a source and a load as shown inFig. 3. Assume the voltage and current waveforms are
v(t) =√2V cos(φv), i(t) =
√2I cos(φi)
where V and I are the rms values. The phase angles φ can be written asφ = ωt+ θ where ω is the frequency (in rad/s) and θ is the initial phase angle
4
∼
i
Load
+
-
v
Figure 1: A simple source-load connection
at time zero. The instantaneous power supplied by the source (and received bythe load) is given by
p(t) = v(t)i(t)
and its dimension is watts (W).
In a dc circuit, power is constant and it flows from source to load at all times. Inan ac circuit, however, the power is not constant and is not necessarily flowingonly from source to load! With the above polarity for the voltage and the showndirection for the current,
if p(t) > 0, power flows from source to load at time instant t, andif p(t) < 0, power flows from load to source at time instant t.
The instantaneous power may be written as
p(t) =v(t)i(t) = 2V I cos(φv) cos(φi)
=V I cos(φv − φi) + V I cos(φv + φi)
=V I cos(φv − φi) + V I cos(φv − φi) cos(2φv)+
V I sin(φv − φi) sin(2φv)
=P [1 + cos(2φv)] +Q sin(2φv)
whereP = V I cos(φ), Q = V I sin(φ), φ = φv − φi.
Case (i): φv = φi (resistive load) ⇒ P > 0, Q = 0
p(t) = P [1 + cos(2φv)] > 0, for all t.
• Power flow is unidirectional from source to load.• Power has pulsation at double frequency.
Figure 4 shows that voltage, current and instantaneous power waveforms for aresistive load connected to an ac source.
This pulsation of power, which does not exist in a dc system, may be consideredas a disadvantage of an ac system. Although it does not cause visible flicker in
5
10 20 30 40
−20
−10
10
20
0
vi
f=60 Hz, V=10 V, I=4 A
t (ms)
10 20 30 40
−20
20
40
60
80
0
p
t (ms)
Figure 2: Voltage and current waveforms (top) and the instantaneous power(bottom) for a resistive load.
lighting systems (due to its high frequency) but it may cause stress on differentloads.
Case (ii): φv > φi (resistive+inductive load) ⇒ P > 0, Q > 0
Case (iii): φv<φi (resistive+capacitive load) ⇒ P > 0, Q < 0
In Case (ii) and Case (iii) where Q 6= 0, p(t) can become negative at some in-stants of time. This means that the flow of power becomes from load to sourceat those instants. Indeed, inductor and capacitor act as energy storage whichstore energy during some period of time and return it during other times. Thisexchange of energy does no work.
The term Q sin(2φv) is caused by inductive or capacitive loads. This termhas zero average over one cycle of ac signal which means an oscillation of energy
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with no net transfer over a cycle of signal. The term reactive is used to describethis component of power as opposed to the term real or active which is used forthe unidirectional component:
p(t) = P [1 + cos(2φv)] +Q sin(2φv)
instantaneouspower
=instantaneousreal power
+instantaneousreactive power
The real power is formally defined as the average of the instantaneous powerover one cycle of the signal:
Real power: 1
T
∫ T
0P [1 + cos(2φv)]dt = P
The reactive power is defined as the peak value of the reactive component ofpower:
Reactive power: Q
The dimension of reactive power is volt-ampere-reactive (VAR).
The complex power combines both quantities in one complex number:
−→S = P + jQ = Sejφ.
The apparent power is defined as the magnitude of the complex power:
S =√
P 2 +Q2 = V I.
The dimension of complex and apparent powers is volt-ampere (VA).
φ = arctan(QP) = φv − φi is called the power angle.
φ P
Power triangle:
−→S
jQ
Power Factor is defined as:
PF =P
S= cos(φ).
A lower PF at a given real power means a larger S (and a larger Q), and thusa larger current. Larger current means larger transmission capacity and largertransmission losses.
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PF is a figure of merit and its desired value is unity (corresponding to no reac-tive power).
PF does not convey the inductiveness or capacitive-ness of the load. Its valueis often accompanied with terms leading (when φ is negative) or lagging (whenφ is positive). In other words, leading means the current leads the voltage, andlagging means the current lags the voltage. Most industrial loads (includinginduction motors) are lagging loads.
5 Powers in terms of Impedances
For every circuit component with given phasor voltage and phasor current, thecomplex power is
−→S = P + jQ = V I cosφ+ jV I sinφ = V Iejφ = V ejθvIe−jθi =
−→V−→I ∗
where ∗ is the complex conjugate.
For a circuit element with impedance−→Z , and voltage and current phasors of
−→V
and−→I , the complex power is
−→S =
−→V−→I ∗ =
−→Z−→I−→I ∗ =
−→Z I2.
Or in terms of voltage phasor,
−→S =
−→V−→I ∗ =
−→V (
−→V−→Z
)∗ =V 2
−→Z ∗
=−→Y ∗V 2.
These two equations can be used to calculate powers in terms of circuit com-
ponents. For example, consider a series RL element. Its impedance is−→Z =
R+ jωL, therefore, its complex power is
−→S =
−→Z I2 = (R+ jωL)I2 = RI2 + jωLI2.
And consequently,P = RI2, Q = ωLI2.
As for a second example, consider a parallel RL element. Its admittance is−→Y = 1
R+ 1
jωL, therefore, its complex power is
−→S =
−→Y ∗V 2 = (
1
R+
1
−jωL)V 2 =
V 2
R+ j
V 2
ωL.
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And consequently,
P =V 2
R, Q =
V 2
ωL.
This method may be used to calculate real and reactive powers for any combi-nation of circuit components.
6 Examples
Example 1
Consider the 60 Hz ac circuit shown in Figure 3 comprising a source, a trans-mission line and a load. The transmission line is a series RL and its parametersare Rℓ = 0.01 Ω and Xℓ = 0.05 Ω. The load is a parallel RL and comprisesRL = 3 Ω and XL = 4 Ω. Assume that the load voltage rms value is equal to240 V.
∼
Rℓ jXℓ
RL jXL C
+
-
−→Vs
+
-
−→V
Figure 3: Circuit of Example 1
1. What is the load real power?
Solution: PL =V 2L
RL= 240
2
3= 19200 W
2. What is the load reactive power?
Solution: QL =V 2L
XL= 240
2
4= 14400 VARs
3. What is the load apparent power?
Solution: SL =√
P 2L +Q2
L = 24000 VA
4. What is the load complex power?
Solution:−→SL = PL + jQL = 19200 + j14400 VA
5. What is the load power factor?
Solution: PFL = PL
SL= 19200
24000= 0.8 lagging
6. What is the total load current rms?
Solution: I = SL
VL= 24000
240= 100 A
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7. What is the line real power?
Solution: Pℓ = RℓI2 = 0.01× 1002 = 100 W
8. What is the line reactive power?
Solution: Qℓ = XℓI2 = 0.05× 1002 = 500 VAR
9. What is the line apparent power?
Solution: Sℓ =√
P 2ℓ +Q2
ℓ = 509.9 VA
10. What is the line complex power?
Solution:−→Sℓ = Pℓ + jQℓ = 100 + j500 VA
11. What is the source real power?
Solution: Ps = PL + Pℓ = 19300 W
12. What is the source reactive power?
Solution: Qs = QL +Qℓ = 14900 VAR
13. What is the source apparent power?
Solution: Ss =√
P 2s +Q2
s = 24382 VA
14. What is the source power factor?
Solution: PFs =Ps
Ss= 19300
24382= 0.792 lagging
15. What is the source voltage rms value?
Solution: Vs =Ss
I= 24382
100= 243.82 V
16. Calculate the transmission power loss.
Solution: Pℓ = RℓI2 = 0.01× 1002 = 100 W
17. Find the capacitor C that improves the power factor to unity.
Solution: Qc = QL ⇒ V 2
XC= V 2
XL⇒ Xc = XL ⇒ 1
Cω= Lω = 4 ⇒ C =
1
4×2π×60= 6.63× 10−4 = 663 µF
18. What is the transmission power loss after compensation?
Solution: after compensation: I = VZ
= 240
3= 80 A, Pℓ = RℓI
2 =0.01× 802 = 64 W
Example 2
Consider the single phase 60 Hz ac circuit of Fig. 4 where the line parametersare Rℓ = 0.1 Ω, Xℓ = 1 Ω and the load is RL = 8 Ω, XL = 6 Ω. Consider theload voltage as the reference for phasors.
10
−→I
Rℓ jXℓ
RL
jXL
-
+
Vs = 120 V
-
+
−→VL
Figure 4: Circuit of Example 1.
1. Calculate the phasor of current.
Solution:−→ZL = 8 + j6 = 10∠36.9,
−−→Ztot = 8.1 + j7 = 10.71∠40.8,
I =120
10.71= 11.21 A ⇒ −→
I = 11.21∠(−36.9)
2. Calculate the phasor of load voltage.
Solution: VL = 11.21×√82 + 62 = 112.1 V and ∠
−→V = 0.
3. Calculate the phasor of source voltage.
Solution: Vs = 120 V and ∠−→Vs = ∠
−→I + ∠
−−→Ztot = −36.9 + 40.8 = 3.9.
4. Draw the phasor diagram showing load voltage, load current, source volt-age, and the voltages across line components.
Solution: The phasor diagram is shown in Fig. 5.
−→
VL Rℓ
−→
I
jXℓ
−→
I
−→
I
−→
Vs
Figure 5: Phasor diagram of Example 2.
5. Calculate the real power, reactive power, complex power, and apparentpower of the load.
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Solution: Complex power is
−→S =
−→VL
−→I ∗ = 112.1× 11.21× ej36.9
= 1255.5∠36.9 = 1004+ j753.8 VA,
then the apparent power is S = 1255.5 VA, real power is P = 1004 W andthe reactive power is Q = 753.8 VARs.
6. Calculate the power factor of the load?
Solution: PF = PS= 1004
1255.5= 0.8 lagging.
7. Find the value of a capacitor which when paralleled with the load increasesthe PF to unity.
Solution: The load impedance is 8 + j6 Ω. When paralleled with C, the
admittance becomes−→Y = 1
8+j6+ jCω = 0.08− j0.06 + jCω. This needs
to be a real value to increase the PF to unity. Thus, Cω = 0.06, andC = 0.06
377= 159× 10−6 F=159 µF.
Other method: the capacitor must supply all the load reactive power.Therefore, Qc = 753.8 = CωV 2 ⇒ C = 753.8
377×112.12= 159× 10−6 F.
8. Calculate the transmission loss before and after load compensation.
Solution: Pℓ = RℓI2. Before compensation: Pℓ = 0.1× 11.212 = 12.57 W.
After compensation, the load impedance is 1/0.08 = 12.5 Ω and the cur-
rent is−→I = 120
0.1+j+12.5and I = 8.89 A. Thus, Pℓ = 0.1× 8.892 = 7.9 W.
7 Per-Unit (pu) System
In the pu system of analysis, any electrical quantity is expressed as the ratio ofthe actual quantity and a base value for that quantity.
For example, if Vbase = 240 V then a voltage of 120 V is 0.5 pu and a voltageof 480 V is 2 pu.
What are the base quantities and how to choose them? Choose two vari-ables from the four main variables of (apparent) power, voltage, current andimpedance. Assign base values to those two variables. Calculate the base val-ues for the other two variables from:
Sbase = VbaseIbase, Vbase = ZbaseIbase.
Often the two variables of power and voltage are chosen and the base values forthem are selected from the rating values of a main component in the system.
Advantages of pu System
• It avoids conversion of components across a transformer (will be discussedlater).
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• Simplifies analysis of three phase systems.
• Brings the values of system variables (and components) within a reason-able range regardless of the rating of the equipment. This provides a moreconvenient framework for comparisons and also for calculations.
Notice: pu values are sometimes expressed in percentage.
Example
A generator supplies a load through a feeder whose impedance is−→Zℓ = Rℓ +
jXℓ = 1+ j2 Ω. The load impedance is−→ZL = RL + jXL = 8+ j6. The voltage
across the load is VL=120 V. Find the real and reactive power supplied by thegenerator. Solve using ordinary systm of analysis and also using pu system.
Circuit diagram is shown below:
−→I
Rℓ Xℓ
RL
XL
-
+−→Vs
-
+
−→VL
Solution Using Ordinary Approach. Take the load voltage as reference.
A balanced three phase ac source is equivalent to three single phase ac sourceswith equal amplitude and the phase angles displaced 120 apart. It is generatedby a three-phase generator whose windings are physically displaced in 120 po-sitions. A balanced three phase system supplies a real power that is three timesthat of the single phase system while it uses fewer number of conductors (3 or4 as opposed to 6). Moreover, the total supplied power by a three phase systemis smooth with no double-frequency pulsation. The bulk power generation andtransmission systems are three phase.
Two set of sequences are possible depending on the order of voltages:
Positive sequence: 0, -120, -240√
Negative sequence: 0, 120, 240
The sequence becomes very important in paralleling (inter-connecting) sourcesto a system. The positive sequence is globally accepted as the standard.
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∼
∼∼
ac
n (neutral)
b
phase-b
phase-aphase-cY (or Wye or Star) Connection:
a
∼ b
c
∆ (or Delta) Connection: ∼ ∼
Figure 6: Schematic diagram of a Y-connected and a ∆-connected source.
9 Three Phase Source Connections
Depending on how the three windings are connected, two major connectionscan be obtained: Y and Delta. In the Y-connection, the second terminals of allthree windings are connected together making a Y or Star shape. The commonconnection point of three windings is called the neutral point. In the Delta-connected, the windings are connected in series and the terminals are takenfrom the connection points making a triangle or ∆ shape.
A Y-connected source may have four or three terminal wires depending onwhether the neutral wire is used or not. A ∆-connected source has only threewires. The terminologies of three-phase-three-wire system or three-phase-four-wire system correspond to this fact.
Schematic diagrams used to show Y and Delta sources are shown in Fig. 6.
Consider a balanced Y-connection. The phase (or line-to-ground: LG) voltages
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areva = van =
√2V cos(φv)
vb = vbn =√2V cos(φv − 120)
vc = vcn =√2V cos(φv − 240)
10 20 30 40
−1
1
0
va vb vc
f=60 Hz
t (ms)
The line (or line-to-line: LL) voltages are
vab = van − vbn = · · · =√3√2V cos(φv + 30)
vbc = vbn − vcn = · · · =√3√2V cos(φv − 90)
vca = vcn − van = · · · =√3√2V cos(φv − 210)
The line voltages are√3 times larger than the phase voltages. Their phase
angles are 30 degrees advanced with respect to phase voltages.
Exercise: Verify the above equations.
The phase currents and the line currents are the same in a Y-connected source.
The phase voltages and the line voltages are the same in a ∆-connected source.
10 Phasors and Phasor Diagram of Three Phase
Systems
Figure 7 shows the phasor diagram of phase and line voltages in a Y-connectedsource. They are related according to the following equations:−→
V ab =√3−→V ae
j30 ,−→V bc =
√3−→V be
j30 ,−→V ca =
√3−→V ce
j30
Conclusion: A Y-connected voltage source can be transformed to an equiva-lent ∆-connected voltage source (and vice versa) according to:
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−→Va
−→Vb
−−→Vb
−→Vc
−→Vab
−→Vbc
−→Vca
30
Figure 7: Phasor diagram showing the relationship between phase voltages and line
voltages in a Y -connected source.
−→V ab =
√3−→V ae
j30 ,−→V bc =
√3−→V be
j30 ,−→V ca =
√3−→V ce
j30
−→V a =
1√3
−→V abe
−j30 ,−→V b =
1√3
−→V bce
−j30 ,−→V c =
1√3
−→V cae
−j30
This is a useful result in the analysis of three phase systems as will be discussedlater.
11 Three Phase Load Connections
Three phase loads are either originally three phase loads (such as three phasemotors, three-phase rectifiers etc) or are formed by a number of a number ofsingle phase loads (such as most residential loads) each of which is connectedacross two lines (or a line to neutral) of the three phase system.1
1At residential level, 208 V LL and 120 V LG are often available through a three phase
transformer. Alternatively, 240 V and 120 V may be available through a single phase trans-
former whose secondary has three taps.
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−→ZY
−→ZY
−→ZY
AC
N (neutral)
B
phase-B
phase-Aphase-C
Y (or Wye or Star) Connection:
A
−→Z∆ B
C
∆ (or Delta) Connection: −→Z∆
−→Z∆
Question: Under what condition the above two loads are equivalent?
Answer: Both loads must behave identically when connected to a voltagesource. They must draw equal currents.
Assume that the Y-connected load draws−→IA,
−→IB , and
−→IC . Therefore,
−→IA =
−−→VAN−→ZY
,−→IB =
−−→VBN−→ZY
,−→IC =
−−→VCN−→ZY
The line current of the ∆-connected load is
−→IA =
−−→IAB−
−−→ICA =
−−→VAB−→Z∆
−−−→VCA−→Z∆
=
−−→VAN −−−→
VBN −−−→VCN +
−−→VAN
−→Z∆
=3−−→VAN−→Z∆
=
−−→VAN
(−→Z∆/3)
Therefore, the condition for both circuits to have equal currents is
−→ZY =
−→Z∆
3
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Conclusion:
Both source and load entities can be transformed from one connection form toother.
12 Analysis of a Three Phase Circuit
As far as circuit analysis is concerned, a balanced three phase circuit (com-prising sources, transmission lines, loads, transformers) can always be brokeninto three equivalent single phase circuits. To understand this, convert all ∆-connected components to Y -connected with a common neutral line connection.The neutral line carries no current in a balanced system. However, the neutralline can be expanded to three lines which carry a three-phase balanced current(sum of which adds up to zero). Therefore, the three-phase circuit can be de-composed into three identical single-phase circuits. The only difference betweenthose circuits is the 120 degrees phase shift in their corresponding variables.Analysis of one of those single phase circuits provides complete information ofthe three phase circuit.
It is convenient to draw a one-line diagram of a three phase system as shownbelow (for an example where a Y-connected source supplies energy to a Y-connected load and a Delta-connected load both connected to a common bus).
∼Load 1
Load 2
G Y
Y
∆
Figure 8: One-line diagram of a three phase circuit.
13 Powers in a Three Phase Circuit
Consider the three phase source-load connection shown in Fig 9 where it is eitheroriginally in a Y form or is converted to its equivalent Y form. Let the voltagesand currents be denoted by
va = van =√2V cos(φv)
vb = vbn =√2V cos(φv − 120)
vc = vcn =√2V cos(φv − 240)
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∼
∼∼
ac
n
b
−→ZY
−→ZY
−→ZY
AC
N
B
−→Ic
−→Ia
−→Ib
Figure 9: A three phase Y-form source-load connection. Three fictitious lines are
also drawn to show how the three phase circuit can be broken into three single
phase subsystems.
ia =√2I cos(φi)
ib =√2I cos(φi − 120)
ic =√2I cos(φi − 240)
in = ia + ib + ic = 0
This circuit can be decomposed into three identical single phase circuits.The instantaneous powers corresponding to these three circuits are
pa(t) = va(t)ia(t) = · · · = P1[1 + cos 2φv] +Q1 sin 2φv
pb(t) = vb(t)ib(t) = · · · = P1[1 + cos 2(φv − 120)] +Q1 sin 2(φv − 120)
pc(t) = vc(t)ic(t) = · · · = P1[1 + cos 2(φv − 240)] +Q1 sin 2(φv − 240)
where P1 = V I cosφ and Q1 = V I sinφ and φ = φv − φi. This means thatthe circuit transfers a real power of P1 at each phase. The total real power isP = 3P1 = 3V I cosφ. Each phase also carries a reactive power of Q1 = V I sinφ.Therefore, we could say that the total reactive power is Q = 3Q1 = 3V I sinφ.
Now, assume that the load is a three-phase induction motor. The totalinstantaneous power delivered to the motor is
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p(t) =va(t)ia(t) + vb(t)ib(t) + vc(t)ic(t)
=3V I cos(φv − φi) = 3V I cos(φ)
=3× (single phase real power)
All the power pulsations cancel out! (Check as an exercise)
Smoothness of power (no double frequency pulsation) is an advantage for motorapplications because it means smooth output power transferred to the shaft ofthe motor and thus less stress.
Note: V is phase voltage. In terms of line voltage p =√3V I cos(φ).
Summary of three phase power quantities:
V : LG (or phase) voltage, I: line current, φ: phase angle difference between−→V
and−→I : φ = φv − φi
P = 3V I cos(φ): real power
Q = 3V I sin(φ): reactive power
−→S = P + jQ = 3
−→V−→I ∗: complex power
S = 3V I: apparent power
Power triangle is also valid here.
PF=PS= cos(φ): power factor
The number 3 in the above four equations becomes√3 when the LL value of
the voltage is used.
14 Numerical Examples
Example 1.
A balanced three phase Y-connected source of 208 V (LL) is connected to a
balanced Y-connected load with per-phase impedance of−→ZY = 12 + j9 Ω. The
(a) the magnitude of the line current,(b) the magnitude of the load voltage,(c) the real, reactive and apparent powers transferred to the load,
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(d) the power factor of the load,(e) the real, reactive and apparent power of the transmission line,(f) the real, reactive and apparent power supplied by the source,(g) the power factor at the source terminals.
Solution: The single phase equivalent circuit is
∼
−→
I
−→ZY
−→Zℓ
+
-
−→Vs
+
-
−→V
where Vs =208√
3= 120 V.
(a)−→Z total =
−→Zℓ +
−→ZY = 12.06 + j9.12 Ω, I = Vs
Ztotal= 120√
12.062+9.122= 7.94 A
(b) Load voltage: V = ZY I =√122 + 92 × 7.94 = 119.1 V
LL value of the load voltage: V =√3× 119.1 = 206.3 V
(c) ZL = 12 + j9 = 15∠36.9 ⇒ φ = 36.9
PL = 3V I cos(φ) = 3× 119.1× 7.94× 0.8 = 2270 W
QL = 3V I sin(φ) = 3× 119.1× 7.94× 0.6 = 1702 VAR
SL = 3V I = 3× 119.1× 7.94 = 2839 VA.
(d) PF=PS= cos(φ) = 2270
2839= 0.8 lagging
(e) Pℓ = 3RℓI2 = 3× 0.06× 7.942 = 11.3 W
Qℓ = 3XℓI2 = 3× 0.12× 7.942 = 22.7 VAR
Sℓ =√
P 2ℓ +Q2
ℓ =√11.32 + 22.72 = 25.3 VA.
(f) Ps = Pℓ + PL = 11.3 + 2270 = 2281.3 W
Qs = Qℓ +QL = 22.7 + 1702 = 1724.7 VAR
Ss =√
P 2s +Q2
s = 2860 VA.
(g) PFs =Ps
Ss= 2281.3
2860= 0.798 lagging
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Example 2.
Repeat Example 1 assuming that the load is ∆-connected with the same valueof per-phase impedance.
Solution: Based on Y-∆ transformation of load, if−→Z∆ = 12 + j9 Ω then−→
ZY = 1
3
−→Z∆ = 4 + j3 Ω. The rest is just repeat of calculations. Answers are
(a) I = 23.4 A,(b) V = 202.7 V,(c) PL = 6571 W, QL = 4928 VAR, SL = 8213 VA,(d) PFL = 0.8 lagging,(e) Pℓ = 98.6 W, Qℓ = 197 VAR, Sℓ = 220 VA,(f) Ps = 6670 W, Qs = 5125 VAR, Ss = 8411 VA,(g) PFs = 0.792 lagging.
15 Power Measurement of a Balanced Three PhaseLoad
Often the three terminals of a load are available for measurements. The realand reactive powers can be calculated from the measurements obtained fromtwo single phase Watt-meters as shown in Fig. 10. The Watt-meter measuresthe real power transferred through two terminals. The measured power satisfiesP = V I cosφ where V is the rms voltage across those two terminals, I is therms current flowing through those two terminals and cosφ is the power factorseen from the two terminals.
A
B
C Load
W1
W2
Figure 10: Circuit arrangement to measure three phase real and reactive powersusing two single phase Watt-meters
With reference to the phasor diagram of Fig. 11, the powers measured by Watt-meters are given by:
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−→VA
−→VB
−→VC
−−→VAB
−−→VBC
−−→VCA
φ− 30
−→IA
−−→VAC
Figure 11: Phasor diagram of the three phase system showing the phase andline voltages, and the line current on line a. φ is the angle between VA and IA.
