ECE 209 Network Analysis II Part 2: AC power & balanced 3-phase circuits Instructor: Dr. Ha Le Department of Electrical and Computer Engineering California State Polytechnic University, Pomona Spring 2014
What will be presented? AC Power 1. Instan Instantan taneou eous s and and averag average e powe power r 2. Maxi Maximu mum m pow power er tran transf sfer er 3. Effe Effect ctiv ive e or or RMS RMS valu value e 4. Real, Real, react reactive ive,, appare apparent, nt, comp complex lex powe powerr 5. Power Power facto factor, r, power power fact factor or corr correct ection ion 6. Power Power tria triangl ngle, e, powe powerr measu measurem rement ent
Three-Phase Circuits 1. Balanc Balanced ed 3-pha 3-phase se volta voltages ges and and curre currents nts 2. Power Power in in balan balanced ced 3-phas 3-phase e syste systems ms 3. Unba Unbala lanc nced ed 3-ph 3-phas ase e syste systems ms Reading: Chapters 11 and 12 textbook 2
Instantaneous power Instantaneously power p(t)
p (t ) v(t ) i(t ) Vm I m cos ( t
1 2
Vm I m cos (v
i )
1
2
Constant power
v )
cos ( t
i )
Vm Im cos (2 t v i ) Varying power
p(t) > 0: power is absorbed by the circuit p(t) < 0: power is absorbed by the source.
3
Average power Average power P: The average of the instantaneous power over one period.
P
1
T
T
0
p (t ) dt
P is independent of time. time.
If θv = θi , load is purely resistive.
If θv– θ i = ±900, load is purely reactive (inductive or capacitive)
P = 0 means that the circuit absorbs no average power.
1 2
V m I m cos ( v i )
4
Practice: p(t) and P Calculate the instantaneous power and the average power for an AC circuit if the voltage and the current in the circuit are:
5
Practice: Average power P Find the average power delivered to a circuit impedance Z if the current flows through the impedance is:
I 5 30 A
Z 10 25 Ω
6
Maximum average power transfer ZTH R TH j XTH ZL R L j XL The maximum average power can be transferred to the load if
ZTH = ZLOAD (=ZL) i.e. XL = -XTH and RL = RTH
Pmax
VTH
2
8 R TH
If the load is purely real:
R L
2 R 2TH X TH ZTH 7
Practice: Maximum power transfer We want to transmit the maximum power to the load Z in the circuit below. Determine Z and the maximum power.
8
Effective or RMS value The total power dissipated by R is
P
1
T
T
0
i Rdt 2
R
T
T
2 i 2 dt I rms R
0
Hence, Ieff is equal to:
I eff
1 T
T
i
2
dt I rms
0
The RMS (root-mean-square) value is a constant that depends on the shape of the function i(t).
The effective of a periodic current is the DC current that delivers the same average power to a resistor as the periodic current.
9
RMS value of sinusoids The RMS value of a sinusoidal voltage v(t) = Vmcos( t+θV) and a current i(t) = Imcos( t+θI) is given by:
Vm
VRMS
I RMS
2
Im 2
Average power written in terms of the RMS values is
P
1 T
T
0
p(t ) dt
1 2
V m I m cos ( v i )
P Vrms Irms cos (θ v
θi ) 10
Sinusoid vs. Phasor vs. RMS phasor A phasor is a complex number that represents the amplitude and phase of a sinusoid.
Phasor (frequency) domain
Time domain
i (t ) I m cos(t )
I I m
Phasor magnitude is RMS value
I RMS
I m
2
Important: 1) If you express amplitude of a phasor source in RMS, then all the answers as a result of this phasor source must also be in RMS value. 2) The same notation V and I may be used to indicate “amplitude” phasors or RMS phasors.
11
Practice: Power and RMS value The sinusoidal voltage and the current in a circuit are:
1) Express v(t) and i(t) as RMS phasors. 2) Calculate the average power P using the RMS values of v(t) and i(t).
12
Complex power (1) Complex power S is the product of the voltage and the complex conjugate of the current. S is measured in voltampere (VA).
V Vm θ v
I I m θ i
V Vm (cos V j sin V ) I I m (cos I j sin I ) S V I (Vm cos V jVm sin V ) (I m cos I jI m sin I ) *
1
Vm I m
2 P jQ
θv
θ i =
Vrms I rms θ v
θi
13
Real, reactive & apparent power S V I *
1 2
Vm I m
θv
S Vrms I rms cos (θ v
θi )
θ i =Vrms
I rms θ v
θi
j Vrms I rms sin (θ v
θi )
S P jQ P is the real part of S called real power (also average power), measured in watt (W)
Q is the imaginary part of S called reactive power , measured in volt-ampere reactive (VAR)
The magnitude of S is called apparent power , measured in volt-ampere (VA)
S
2 2 P Q
14
Power factor S Vrms I rms cos (θ v S = P + jQ = S θ v
θi ) θi
j Vrms I rms sin (θ v
θi )
S
Power factor (PF) is the cosine of the phase difference between the voltage and current. It is also the cosine of the angle of the load impedance.
PF cos(θ v
θi )
cos 15
Power triangle & Impedance triangle Power triangle represents the relation between P, Q, S, and power factor. The relation between R, X and Z is shown by impedance triangle
16
Power factor: lagging, leading, unity (1)
Lagging power factor: current lags voltage, load absorbs Q. Leading power factor: current leads voltage, load generates Q. Unity PF: current and voltage are in phase. Load draws only P. It does not generate or absorb Q. 17
Power factor vs. reactive power Q
Lagging PF:
Q > 0, inductive loads
Leading PF:
Q < 0, capacitive loads
Unity PF:
Q = 0, resistive loads 18
Summary of power relations
19
Practice: real, reactive, complex power The voltage and current through a load are:
V RMS 12060 V 0
I RMS 1025 A 0
1) Calculate complex, apparent, real and reactive power consumed by the load. 2) Find the load power factor. Is the power factor lagging or leading? 3) Calculate the load impedance. 4) Draw the power triangle and impedance triangle. 20
Power factor correction (1) PF correction is an important task in power system operation. Motivation: If a load draws lots of reactive power , the voltage where the load is connected will decrease from the nominal value. Goal: To keep voltage close to nominal value Method: To reduce the amount of reactive power that the load draws from the power system by installing capacitor banks or other devices to supply reactive power to the load. 21
Power factor correction (2) A capacitor is connected in parallel with the load to increase the load power factor (Fig. b)
22
Power factor correction (3) Qc = Q1 – Q2 = P (tan
θ1
- tan θ2)
= V2rms / ZC Q1 = S1 sin θ1 = P tan θ1
Q2 = P tan
ZC
P = S1 cos θ1
Power angle
θ2
compared to
θ1
1 C
θ2
=
2 Vrms
Qc
is smaller
PF2=cos(θ2) > PF1=cos(θ1)
C
1 Z C
=
Qc 2
Vrms
23
Practice: Power factor correction Continued from previous practice on P, Q, S: The voltage and current through a load are:
V RMS 120600 V I RMS 10250 A We want to install a capacitor in parallel with the load to raise its PF to 0.95 lagging. Determine the required capacitance C given that the circuit frequency is 260 rad/s. 24
Real power measurement: Wattmeter (1) The wattmeter is the instrument for measuring the average power . Electromagnetic wattmeter construction Consists of main two coils: the current coil and the voltage coil.
Current coil (very low impedance) is connected in series with the load.
Voltage coil (very high impedance) is connected in parallel with the load.
A wattmeter connection to the load is shown in Fig. 11.31.
25
Real power measurement: Wattmeter (2)
Smart meter
Electromagnetic wattmeter
Smart meters have many abilities: to reduce load, disconnect-reconnect remotely, interface to gas and water meters and so on.
In 2011, 493 U.S. electric utilities had installed 37,290,374 advanced ("smart") meters in the country.
Digital wattmeter
26
Reactive power measurement Reactive power is measured by an instrument called the varmeter . The varmeter is often connected to the load in the same way as the wattmeter.
A 3-phase Wattmeter/Varmeter module It consists of a wattmeter and a varmeter is designed to indicate direct measurement of active and reactive power in balanced 3-phase circuits.
27
1-phase and 3-phase circuits
A circuit that contains a source that produces sinusoidal voltages with different phases is called a polyphase system.
Most of the generation and distribution of electric power is accomplished with polyphase systems.
The most common polyphase system is the balanced three-phase system
At distribution level, many circuits are 1-phase, some are 2-phase (phase-to-phase connection). Question:
How many phases in home circuit? 28
Balanced 3-phase circuits Y-connected 3-phase source and sequence voltages
Terminals a, b and c are called line terminals or lines and n is called the neutral terminal. The source is said to be balanced if the voltages Van, Vbn and Vcn (called phase voltages), have the same magnitude and sum to zero. That is:
29
3-phase circuits: Phase sequence Balanced condition for Y-connected 3-phase source
With Vm being the amplitude of the sinusoidal phase voltage, two cases are possible to satisfy above conditions
Case 1: Van leads Vbn by 120 degrees and Vbn leads Vcn by 120 degrees. It is called positive phase sequence or abc phase sequence. Case 2: Van leads Vcn by 120 degrees and Vcn leads Vbn by 120 degrees. It is called negative phase sequence or acb phase sequence. 30
Line-line and per phase voltage
Line-line and phase voltage relationship The line to line voltage has a magnitude equal to sqrt(3) (~1.73) times the magnitude of the phase voltage and it leads the corresponding phase voltage by 30 deg. 31
Phasor diagram: Vline-line vs. Vphase Phasor diagram relating phase and line voltages
32
Possible 3-phase source & load connections Both the three-phase source and the three-phase load can be either Y- or delta-connected. We have four possible connections:
Balanced Y-connected source is most common in practice .
Delta-connected source is not common because of the circulating current that will result in the delta-mesh if the three-phase voltages are slightly unbalanced.
Balanced delta-connected load is more common than a balanced Y-connected load. 33
Balanced Y-Y connected 3-phase circuit There is no current in the neutral wire (the short circuit between terminals n and N).
Neutral wire can be removed without affecting the voltage and currents.
The line and phase currents are the same
IL = I P
34
Balanced Y-
connected 3-phase circuit (1)
Assumed that balanced abc phase sequence voltages are applied and Vab = V00. The phase currents will be
35
Balanced Y-
connected 3-phase circuit (2)
Line currents:
Line currents form a balanced set; line current Ia magnitude is sqrt(3) Iab and lags phase current Iab by 30 deg; the same applies to Ib and Ic.
Line to line voltages are identical to the phase voltages for delta connected loads i.e. VL= VP
36
Balanced Y-
load
Delta-connected load may be transformed into an equivalent Y-connected load so that the terminal behavior of the two configurations will be identical. That is corresponding line to line voltages and line currents will be the same.
For balanced load: Z Y = ZD / 3
37
3-phase versus 1-phase systems Balanced 3-phase systems have many advantages over 1-phase systems, some are: Less conductors: Each separate single-phase system requires that both the forward and return conductors have a current capacity (or ampacity) equal to or greater than the load current. A balanced 3-phase system (3 conductors) can deliver the same power as three 1-phase systems (6 conductors). Less power losses, better voltage: the total I 2R line losses in the 3-phase system are only 50% those of the three separate 1phase systems. The line-voltage drop between the source and load in the 3-phase system is 50% that of each 1-phase system. Constant instantaneous power: Total instantaneous electric power delivered by a 3-phase generator under balanced steady-state conditions is constant. Better 3-phase machines: Three-phase machines produce nearly constant torque and thereby minimize shaft vibration and noise. 38
Power in a balanced 3-phase system (1) Average power: The total average power absorbed by a three phase balanced load, or delivered by a three phase generator, is equal to the sum of the powers in each phase. The voltage and current in each phase are equal.
For a three-phase Y-connected load or generator VL= sqrt(3) VP IL = IP For a three-phase Delta-connected load or generator VL=VP IL = sqrt(3) IP 39
Power in a balanced 3-phase system (2) Average power: The total average power absorbed by a three phase balanced load, or delivered by a three phase generator, is equal to the sum of the powers in each phase. Therefore, equations for real power P, reactive power Q, and apparent power S are the same for Y- or delta-connected load or generator. Power angle theta is the phase difference between the voltage and the current.
PT 3PP 3 VP I P cos( ) 3 VL I L cos( ) QT 3QP 3 VP I P sin( ) 3 VL I L sin( ) *
*
ST 3 VP I P 3 VL I L
v i
40
Power in a balanced 3-phase system (3) Instantaneous power: For a balanced 3-phase load, the total instantaneous power is equal to the sum of the individual powers of the three phases.
pT pa p b pc vania vbnib vcnic
Vm and Im are the peak values of the phase voltages and currents respectively.
θ
is the phase angle by which the current lags the voltage in each phase.
ABC phase sequence is assumed.
41
Power in a balanced 3-phase system (4)
Using Vm = sqrt(2) VP and Im = srqt(2) IP and suitable trigonometric identities leads to
pT 3 VP I P cos 0 VP I P cos(2t ) cos(2t 1200 ) cos(2t 120 )
The second term on the right hand side of the above equation is identically zero, leading to an important property of a balanced three-phase system. Property: Total instantaneous power is time invariant and equal to total (3-phase) average power.
pT 3 VP I P cos PT
42
Power triangle, power factor Power triangle represents the relation between P, Q, S, and power factor. The relation between R, X and Z is shown by impedance triangle. For 3-phase circuit, S, P, Q are 3 times the single-phase quantities:
S3Φ = 3*S1Φ P3Φ = 3*P1Φ Q3Φ = 3*Q1Φ
43
Unbalanced 3-phase circuits
In an unbalanced 3-phase circuit, any or all of the following occur:
The source voltages are not equal in magnitude and/or are not spaced 120 deg. apart.
The load phase impedances are not equal.
The line currents are not equal.
The neutral current is not zero.
Different methods are used to analyze unbalanced 3-phase circuits. The analysis is more difficult.
44
Practice: 3-phase power, power factor Two balanced 3-phase motors, an induction motor drawing 400 kW at 0.8 power factor lagging and a synchronous motor drawing 150 kVA at 0.9 power factor leading, are connected in parallel and are supplied by a balanced, 3-phase 4160-volt source. Cable impedances between the source and load are neglected. 1)
Draw the power triangle for each motor and for the combined-motor load.
2)
Determine the power factor of the combined-motor load.
3)
Determine the magnitude of the line current delivered by the source.
4)
A delta-connected capacitor bank is now installed in parallel with the combined-motor load. What value of capacitive reactance is required in each leg of the capacitor bank to make the source power factor unity?
5)
Determine the magnitude of the line current delivered by the source with the capacitor bank installed. 45
Practice: Y and delta load
46
