EE 221 AC Circuit Power Analysis

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EE 221
AC Circuit Power Analysis
Instantaneous and average power
RMS value
Apparent power and power factor
Complex power
Instantaneous Power
ƒ Product of time-domain voltage
and time-domain current
p(t) = v(t) i(t)
ƒ Determine maximum value
ƒ Transients
ƒ Steady-state
As the transient dies out, the
circuit returns to steady-state
operation. Since the only
source remaining in the circuit
is dc, the inductor eventually
acts as a short circuit
absorbing zero power.
steady-state
Power due to Sinusoidal Excitation
ƒ v(t) = Vm cos(ωt+θ) and i(t) = Im cos(ωt +ϕ)
ƒ p(t) = Vm Im cos(ωt+θ) cos(ωt +ϕ) =
= ½ Vm Im cos(θ - ϕ) + ½ Vm Im cos(2ωt + θ + ϕ)
p(t) = ½ Vm Im cos(θ - ϕ)
+
trigonometric
identity
½ Vm Im cos(2ωt + θ + ϕ)
ƒ instantaneous
ƒ constant
ƒ periodic
ƒ two parts
ƒ independent of t
ƒ period is ½T
ƒ average
ƒ the "average"
ƒ average is zero
ƒ periodic
ƒ wanted
ƒ unwanted
ƒ active or real
Power due to Sinusoidal Excitation
p(t) = ½ Vm Im cos(θ - ϕ)
+
½ Vm Im cos(2ωt + θ + ϕ)
• voltage V = 4∠0° V, impedance Z = 2∠60° Ω, ω = π/ 6 rad/s
• I = 2∠-60° A
• p(t) = 2 + 4 cos(π / 3 - 60°) W
Power due to Sinusoidal Excitation
p(t) = ½ Vm Im cos(θ - ϕ)
½ Vm Im cos(2ωt + θ + ϕ)
+
Example 1: Average power delivered to resistor
T
2
PR = ½ Vm Im cos(θ - ϕ)
= ½ Vm Im cos(0)
1.5
= ½ Vm Im
= ½ Vm2 / R
=1W
0.5
v(t), i(t), p(t)
= ½ R Im2
1
0
-0.5
-1
V = 2∠-45° V
R = 2Ω
-1.5
-2
0
0.2
0.4
0.6
0.8
1
time (s)
1.2
1.4
1.6
1.8
2
Power due to Sinusoidal Excitation
p(t) = ½ Vm Im cos(θ - ϕ)
½ Vm Im cos(2ωt + θ + ϕ)
+
Example 2: Average power delivered to purely reactive elements
T
2
PX = ½ Vm Im cos(θ - ϕ)
= ½ Vm Im cos(±90°)
= 0
1.5
1
v(t), i(t), p(t)
0.5
0
-0.5
-1
V = 2∠-45° V
X = j2Ω
-1.5
-2
0
0.2
0.4
0.6
0.8
1
time (s)
1.2
1.4
1.6
1.8
2
Power due to Sinusoidal Excitation
p(t) = ½ Vm Im cos(θ - ϕ)
Example 3:
+
½ Vm Im cos(2ωt + θ + ϕ)
Voltage across impedance (V = 100∠25o V and Z = 50∠55o Ω)
Determine active power absorbed.
I
=V/Z
= 2 ∠25-55°
V
= 2 ∠-30°
= 1.7 - j A
25o
-30o
Z = 28.7 + j41.9 Ω
P = ½ 22 28.7
= ½ (1.72+12) 28.7
= 57.4 W
I
Power due to Sinusoidal Excitation
p(t) = ½ Vm Im cos(θ - ϕ)
Example 4:
+
½ Vm Im cos(2ωt + θ + ϕ)
(Chapter 11, Problem 8.)
In the circuit shown in Fig. 11.27, find the average power being (a) dissipated in the 3-Ω
resistor; (b) generated by the source.
ZR = 3 +
1
= 3 + 1 + j3 = 4 + j3 Ω
0.1 − j 0.3
Ignore 30° on Vs , I R = 5
(a)
2 + j5
5 29
, IR =
6 + j8
10
2
P3 Ω
1  5 29 
= 
 × 3 = 10.875 W
2  10 
(b)
(2 + j5)(4 + j3)
= 13.463∠51.94° V
6 + j8
1
= ×13.463× 5cos 51.94° = 20.75 W
2
Vs = 5∠0°
∴Ps,gen
Maximum Power Transfer
A simple loop circuit used to illustrate the derivation
of the maximum power transfer theorem as it applies
to circuits operating in the sinusoidal steady state.
Zth = Rth + j Xth
ZL = RL + j XL
ZL = Zth*
RL = Rth and XL = - Xth
Maximum Power Transfer
Example:
(Chapter 11, Problem 12.) For the circuit of Fig. 11.30;
(a) what value of ZL will absorb a maximum average power?
(b) What is the value of this maximum power?
−10 j
= 107.33∠ − 116.57° V
10 + j 5
− j10 (10 + j15)
Zth =
= 8 − j14 Ω
10 + j 5
Vth = 120
(a)
ZL = 8 + j14 Ω
(b)
IL =
107.33∠ − 116.57°
∴
16
PL ,max
1
=
2
2
 107.33 

 × 8 = 180 W
 16 
Effective Values
• Measure for sinusoidal voltages and currents
• Power outlets: 60 Hz, "voltage of 115V"
• Not the mean of T (T/2)
• Not the Amplitude ( 2 115V)
• Measure of effectiveness of a source in delivering
power to a resistive load
• Effective value of periodic current
• is equal to the DC value that
• delivers the same average power to resistor
• i(t) R → p(t) → PR and compare to IDC R
Effective Values
Mathematical expression
T
T
R 2
1 2
2
P = ∫ i Rdt = ∫ i dt =I DC
R = I eff2 R
T 0
T 0
T
I eff =
1 2
i Rdt
∫
T 0
• (Square) root of the mean of the square current
• rms value
• Defined for all periodic signals
Effective Values - Sinusoids
ƒ i(t) = Im cos(ωt +ϕ) with a period of T = 2π/ω
I eff
T
ω
1 2
2
=
I m cos (ωt + ϕ )dt = I m
∫
2π
T 0
I eff
2π / ω
∫
0
1 1

 2 + 2 cos(2ωt + 2ϕ ) dt


Im
=
2
• real quantity
• independent of phase angle
• equal to 0.707 the amplitude
• example:
2∠-30o A delivers the same as IDC = 1A
RMS value to compute average power
ƒ In general
P = ½ Vm Im cos(θ - ϕ) = Veff Ieff cos(θ - ϕ)
ƒ For resistors
P = Veff Ieff = Veff2 / R = Ieff2 R
ƒ Note:
ƒ We can use amplitude or rms value
ƒ Use V and V rms to designate voltages
RMS value to compute average power
ƒ Example: (Chapter 11, Problem 30.)
The series combination of a 1-kΩ resistor and a 2-H inductor must not dissipate more
than 250 mW of power at any instant. Assuming a sinusoidal current with ω =500 rad/s,
what is the largest rms current that can be tolerated?
The peak instantaneous power is 250 mW. The combination of elements yields
Z = 1000 + j1000 Ω = 1414 ∠45o Ω.
Vm ∠0 Vm ∠ − 45o
A and Vm = 1414 Im.
Arbitrarily designate V = Vm ∠0 , so that I =
=
1414
Z
We may write p(t) = ½ Vm Im cos φ + ½ Vm Im cos (2ωt + φ) where φ = the angle of the
current (-45o). This function has a maximum value of ½ VmIm cos φ + ½ VmIm.
Thus, 0.250 = ½ VmIm (1 + cos φ) = ½ (1414) Im2 (1.707) and Im = 14.39 mA. In terms
of rms current, the largest rms current permitted is
14.39m / 2 = 10.18 mA rms.
Apparent Power
We had
P = ½ Vm Im cos(θ - ϕ) = Veff Ieff cos(θ - ϕ)
In case of direct current we would use voltage times current:
S = Veff Ieff
• This is not the average power
• Is the "apparent" power (S or AP)
• Measured in volt-ampere or VA
• (rather than W to avoid confusion)
• Magnitude of S is always greater or equal to magnitude of P: |S| ≥ |P|
Power Factor
Defined as the ratio of average (real) power to apparent power
PF = P / S = P / (Veff Ieff)
In the sinusoidal case the power factor is
PF = cos(θ - ϕ)
• θ - ϕ is the angle the voltage leads the current: PF angle
• note:
• Purely resistive load has PF = 1
• Purely reactive load has PF = 0
• PF = 0.5 means a phase angle of ±60°
• Resolve ambiguity
• PF leading or lagging for capacitive or inductive load
Apparent Power and Power Factor
Example: (Chapter 11, Problem 30.)
(a) Find the power factor at which the source in the circuit of Fig. 11.42 is operating. (b)
Find the average power being supplied by the source. (c) What size capacitor should be
placed in parallel with the source to cause its power factor to be unity? (d) Verify your
answers with PSpice.
120
= 9.214∠ − 26.25° A rms
j192
4+
12 + j16
∴ PFs = cos 26.25 = 0.8969 lag
(a) I s =
(b) Ps = 120 × 9.214 × 0.8969 = 991.7W
j 48
1
(192 + j144)
= 4+
3 + j4
25
11.68 − j 5.76
∴ Z L = 11.68 + j 5.76 Ω, YL =
11.682 + 5.762
j 5.76
, C = 90.09 µ F
∴ j120π C =
11.682 + 5.762
ZL = 4 +
(c)
Apparent Power and Power Factor
Example: (see also chapter 10)
(Chapter 11, Problem 30.)
(d) Examine simulation output file
(AC sweep at 60Hz).
FREQ
VM($N_0003,0) VP($N_0003,0)
6.000E+01 1.200E+02
0.000E+00
FREQ
IM(V_PRINT1) IP(V_PRINT1)
6.000E+01 9.215E+00
-2.625E+01
(a) and (b) are correct
Next, add a 90.09-µF capacitor in parallel with the source:
FREQ
IM(V_PRINT1) IP(V_PRINT1)
6.000E+01 8.264E+00
-9.774E-05
(c) is correct (-9.8×10-5 degrees is
essentially zero, for unity PF).
Complex Power
• Simplifies power calculations
• We had p(t) = Veff Ieff cos(θ - ϕ) + ½ Vm Im cos(2ωt + θ + ϕ)
• Where the average (real) power P = Veff Ieff cos(θ - ϕ)
• Using complex nomenclature
P = Veff Ieff Re{ej(θ - ϕ)} = Veff Re{ej(θ)} Ieff Re{e-jϕ)}
phasor voltage
• Hence
P = Re{Veff I*eff}
• Define
S = Veff I*eff
complex conjugate of
phasor current
Complex Power
• S = Veff I*eff can be written as S = Veff Ieff ej(θ - ϕ) = P + jQ
magnitude equals
apparent power
|S| = S
PF angle
• Reactive power Q
average
(real)
power
reactive
power
• Imaginary part of complex power
• Dimensions are those of P, S, AP=S (|S|)
• Avoiding confusion by using volt-ampere-reactive or VAr
• Q = Veff Ieff sin(θ - ϕ)
• Physical interpretation:
• Time rate of energy flow back&forth between source and reactive loads
• Reactive components charge and discharge at 2ω (→ current flows)
Complex Power (Example)
p(t) = Veff Ieff cos(θ - ϕ)
+
Veff Ieff cos(2ωt + θ + ϕ)
|S|
P
• voltage V = 4∠0o V, impedance Z = 2∠60o Ω, I = 2∠-60o A, ω = π/ 6 rad/s
• p(t) = 2 + 4 cos(π / 3 - 60°) W
• voltage V = 2.83∠0o Vrms, I = 2∠-60o Arms
• S = P + jQ = V I* = 2W + j 3.46VAr = 4 ∠60o VA
Q
Power Triangle
• Commonly used graphical representation of
S = P + jQ = V I* = V (V/Z)* = V V* / Z* = |V|2 / Z* = |I|2 Z
• Useful relationships: P = |S| cos(Θ), Q = |S| sin(Θ),
Θ = power angle = tan-1(Q/P)
Q = P tan(Θ)
• Quadrant means
• 1st - power factor is lagging, inductive load
• 4th - power factor is leading, capacitive load
Im
|S|
• Need only two quantities to find third
S
Q
Re
P
(4 ∠60o VA = 2W + j 3.46VAr)
Power and Phasors
• Another interpretation of active and reactive power components
• Current components
• In phase with voltage - Ieff cos(θ - ϕ)
• 90° out of phase (quadrature component) - Ieff sin(θ - ϕ)
• Multiplied by |V| results in P and Q
Veff
Im
Ieff cos(θ - ϕ)
Ieff sin|θ - ϕ|
θ-ϕ
Ieff
Re
Power and Phasors
Example: (Chapter 11, Problem 42.)
Both sources are operating at the same frequency.
Find the complex power generated by each source
and the complex power absorbed by each passive
circuit element.
Vx − 100
V
V − j100
+ x + x
=0
6 + j4
− j10
5
 1

100
∴ Vx 
+ j 0.1 + 0.2  =
+ j 20
6
+
j
4
6
+
j
4


∴ Vx = 53.35− ∠42.66° V
100 − 53.35− ∠42.66°
∴ I1 =
= 9.806∠ − 64.44° A
6 + j4
1
∴ S1. gen = × 100 × 9.806∠64.44° = 211.5 + j 4423VA
.
2
Power and Phasors
Example: (Chapter 11, Problem 42.)
Both sources are operating at the same
frequency. Find the complex power generated by
each source and the complex power absorbed by
each passive circuit element.
S 6, abs =
1
× 6 × 9.806 2 = 288.5 + j 0 VA
2
1
( j 4) 9.806 2 = 0 + j192.3 VA
2
j100 − 53.35 − ∠ 42.66 °
= 14.99 ∠121.6 °,
I2 =
5
1
S 5 abs = × 5 × 14.99 2 = 561.5 + j 0 VA
2
1
S 2, gen = ( j100)14.99 ∠ − 121.57 ° = 638.4 − j 392.3 VA
2
2
1  53.35 
-90
0 VA
S − j10, abs = 
 ( − j10) = 0 − j142.3 VA = 142.3 ∠
2  10 
S j 4, abs =
Power Measurement
• Wattmeter(hours) measures active load
• Varmeter(hours) measures reactive load
• Average PF is used to adjust consumer's bill
(industry has to pay for unwanted losses)
• Complex power delivered to individual loads
• equals their sum
• no matter how loads are connected
• S = V I* = V (I1 + I2)* = V (I*1 + I*2) = VI*1 + VI*2
Power Factor Correction
• Large industrial consumers pay penalty when PF < 0.85
• Caused by inductive loads (motors)
• Why: Causes increased
• Device ratings
• Transmission and distribution losses
• Example:
• $0.22/kVAr above 62% of average (real) power demand
• S = P + jQ = P + j0.62P = P (1+j0.62) = P (1.77∠31.8°)
• Targets cos(31.8°) = 0.85 (pay penalty when PF < 0.85)
• Use compensation capacitors in parallel with load
Power Factor Correction
• Value of capacitance
P(tan θ old − tan θ new )
C=
2
ωVrms
corrects the PF angle from old to new at the specified frequency
and voltage
• Derived from
∆Q = V2rms / Xc = V2rms ωC
(Qold - Qnew) / V2rms = ωC
P(tan θold - tan θnew) / V2rms = ωC
Complex Power
Example: Source of 115 Vrms supplies two loads. Loads are connected in parallel:
7kW / 3 kVAr and 4kVA at 0.85 pf lagging. Find the pf of the equivalent load as seen
from the input terminal.
S1 = 7000 + j 3000
S2 = 4k [cos(θ) + j sin(θ)] = 4k [0.85 + j sin(cos-1(0.85))] = 3400 + j 2107
S = S1 + S2 = 10400 + j 5107 = 11586 ∠26.15°
θ = tan-1(Q / P) = tan-1(5107 / 10400) = 26.15°
cos(θ) = cos 26.15 ° = 0.8976
Z = |V|2 / |S| = 1152 / 11586 = 1.142Ω
Z = 1.142Ω ∠26.15° = 1.02 + j 0.5 Ω
(remember: S = |V|2 / Z*)
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