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Chapter 7
AC Power Analysis
Chapter Objectives:
 Know the difference between instantaneous power and average
power
 *Learn the AC version of maximum power transfer theorem
 Learn about the concepts of effective or rms value
 Learn about the *complex power, apparent power and power factor
 Understand the principle of conservation of AC power
 Learn about power factor correction
Instantenous AC Power
 Instantenous Power p(t) is the power at any instant of time.
v(t )  Vm cos(t  v ) i(t )  I m cos(t  i )
1
1
p(t )  v(t )i(t )  Vm I m cos(v  i )  Vm I m cos(2t  v  i )
2
2
Instantenous AC Power
 Instantenous Power p(t) is the power at any instant of time.
p (t )  v(t )i (t )
Assume a sinusoidal voltage with phase v , v(t )  Vm cos(t  v )
Assume a sinusoidal current with phase i , i(t )  I m cos(t  i )
1
1
p(t )  v(t )i (t )  Vm I m cos(v  i )  Vm I m cos(2t  v  i )
2
2
p(t )  CONSTANT POWER+SINUSOIDAL POWER (frequency 2 )
1
1
p(t )  v(t )i (t )  Vm I m cos( v  i )  Vm I m cos(2t   v  i )
2
2
 The instantaneous power is composed of two parts.
• A constant part.
• The part which is a function of time.
Trigonometri identity
Instantenous and Average Power
 The instantaneous power p(t) is composed of a constant part (DC) and a time
dependent part having frequency 2ω.
p(t )  v(t )i (t )
v(t )  Vm cos(t   v )
i (t )  I m cos(t  i )
1
1
p(t )  Vm I m cos( v  i )  Vm I m cos(2t  v  i )
2
2
Instantenous Power p(t)
Average Power
P  12 Vm I m cos(v  i )
Instantenous and Average Power
p(t )  12 Vm I m cos(v  i )  12 Vm I m cos(2t  v  i )  p1 (t )  p2 (t )
Average Power
The average power P is the average of the instantaneous power over one period .
p(t )  v(t )i (t ) Instantaneous Power
1 T
P   p(t )dt Average Power
T 0
v(t )  Vm cos(t  v ) i (t )  I m cos(t  i )
1 T
1 T1
1 T1
P   p(t )dt   2 Vm I m cos( v   i )dt   2 Vm I m cos(2 t   v   i )dt
T 0
T 0
T 0
1 T
1 T
1
P  Vm I m cos(v  i )  dt  2 Vm I m  cos(2t  v  i )dt
T 0
T 0
= 12 Vm I m cos(v  i )  0
(Integral of a Sinusoidal=0)
1
2
P  12 Vm I m cos(v  i )
1
P  Re  VI   Vm I m cos(v  i )
2
1
2

Average Power
The average power P, is the average of the instantaneous power over one period .
P  12 Vm I m cos(v  i )
1
P  Re  VI   Vm I m cos(v  i )
2

1
2
 A resistor has (θv-θi)=0º so the average power becomes:
PR  Vm I m  I m R  I R
1
2
1.
2.
3.
4.
1
2
2
1
2
2
P is not time dependent.
When θv = θi , it is a purely resistive load case.
When θv– θi = ±90o, it is a purely reactive load case.
P = 0 means that the circuit absorbs no average power.
Instantenous and Average Power
 Example 1 Calculate the instantaneous power and
average power absorbed by a passive linear network if:
v(t )  80 cos (10 t  20)
i (t )  15 sin (10 t  60)
1
1
p(t )  Vm I m cos( v  i )  Vm I m cos(2t   v  i )
2
2
=385.7  600cos(20t  10) W
P= 385.7 W is the average power flow
Average Power Problem
 Practice Problem 11.4: Calculate the average power absorbed by each of the five
elements in the circuit given.
Average Power Problem
Maximum Average Power Transfer
 Finding the maximum average power which can be transferred from
a linear circuit to a Load connected.
a) Circuit with a load
b) Thevenin Equivalent circuit
• Represent the circuit to the left of the load by its Thevenin equiv.
• Load ZL represents any element that is absorbing the power generated
by the circuit.
• Find the load ZL that will absorb the Maximum Average Power from
the circuit to which it is connected.
Maximum Average Power Transfer Condition
• Write the expression for average power associated with ZL: P(ZL).
ZTh = RTh + jXTh
ZL = RL + jXL
I
VTh
VTh

ZTh  Z L ( RTh  jX Th )  ( RL  jX L )
P
VTh
2
RL
1 2
2
I RL 
2
( RTh  RL ) 2  ( X Th  X L ) 2
Ajust R L and X L to get maximum P
VTh RL ( X Th  X L )
2
P

X L ( R  R ) 2  ( X  X ) 2  2
L
Th
L
 Th

2
2
P VTh ( RTh  RL )  ( X Th  X L )  2 RL ( RTh  RL ) 

2
2 2
RL
2 ( RTh  RL )  ( X Th  X L ) 
2
P
 0  X L   X Th
X L
P
0
RL
 RL  RTh 2  ( X Th  X L ) 2  RTh
Z L  RL  jX L  RTh  jX Th  ZTh
Maximum Average Power Transfer Condition
• Therefore: ZL = RTh - XTh = ZTh will generate the maximum power
2
transfer.
2
I L RL VTh
• Maximum power Pmax Pmax 

2
8RTh
 For Maximum average power transfer to a load impedance ZL we
must choose ZL as the complex conjugate of the Thevenin impedance
ZTh.
Z L  RL  jX L  RTh  jX Th  Z Th
Pmax 
VTh
2
8 RTh
*Maximum Average Power Transfer
 Practice Problem 11.5: Calculate the load impedance for maximum power
transfer and the maximum average power.
Maximum Average Power Transfer
Maximum Average Power for Resistive Load
 When the load is PURELY RESISTIVE, the condition for maximum power
transfer is:
XL  0
 RL  RTh 2  ( X Th  X L )2  RTh 2  X Th 2  ZTh
 Now the maximum power can not be obtained from the Pmax formula given before.
 Maximum power can be calculated by finding the power of RL when XL=0.
●
●
RESISTIVE
LOAD
Maximum Average Power for Resistive Load
 Practice Problem 11.6: Calculate the resistive load needed for maximum power
transfer and the maximum average power.
Maximum Average Power for Resistive Load
RL
 Notice the way that the maximum power is calculated using the Thevenin
Equivalent circuit.
Effective or RMS Value
 The EFFECTIVE Value or the Root Mean Square (RMS) value of a periodic
current is the DC value that delivers the same average power to a resistor as the
periodic current.
a) AC circuit
b) DC circuit
1 T
R T
2
P   i (t ) Rdt   i (t ) 2 dt  I eff 2 R  I Rms 2 R
T 0
T 0
I eff  I Rms
1 T
2

i
(
t
)
dt

0
T
Veff  VRms
1 T
2

v
(
t
)
dt

0
T
Effective or RMS Value of a Sinusoidal
 The Root Mean Square (RMS) value of a sinusoidal voltage or current is equal
to the maximum value divided by square root of 2.
I Rms
1 T 2
2

I
cos
tdt 
m

0
T
I m2
T

T
0
I
1
(1  cos 2t )dt  m
2
2
P  12 Vm I m cos(v  i )  VRms I Rms cos(v  i )
 The average power for resistive loads using the (RMS) value is:
2
V
PR  I Rms 2 R  Rms
R
Effective or RMS Value
 Practice Problem 11.7: Find the RMS value of the current waveform. Calculate
the average power if the current is applied to a 9  resistor.
4t
0  t 1
 4t
i(t )  
8  4t 1  t  2
I
2
rms
1 T 2
1
  i dt  
T 0
2 
16 
2
I rms
 
2


T 2
2

(4
t
)
dt

(8

4
t
)
dt
0
1

1
2

t 2 dt  (44t t 2 ) dt 
0
1

1
8-4t
2
16
I rms 
 2.309A
3
2
2
I rms
3
1 
 2  16
t
2
 8    4t  2t   1  
3  3
3 
PI
2
rms
 16 
R   (9)  48W
3
An Electical Power Distribution Center
Apparent Power and Power Factor
 The Average Power depends on the Rms value of voltage and current and the
phase angle between them.
P  12 Vm I m cos(v  i )  VRms I Rms cos(v  i )
 The Apparent Power is the product of the Rms value of voltage and current. It is
measured in Volt amperes (VA).
1
S  Vm I m  VRms I Rms
2
 The Power Factor (pf) is the cosine of the phase difference between voltage and
current. It is also the cosine of the angle of load impedance. The power factor may
also be regarded as the ratio of the real power dissipated to the apparent power of
the load.
P
pf   cos(v  i )
S
P  Apparent Power  Power Factor  S  pf
Apparent Power and Power Factor
 Not all the apparent power is consumed if the circuit is partly reactive.
Purely resistive
load (R)
θv– θi = 0, Pf = 1
P/S = 1, all power are
consumed
Purely reactive
load (L or C)
θv– θi = ±90o,
pf = 0
P = 0, no real power
consumption
θv– θi > 0
θv– θi < 0
• Lagging - inductive load
• Leading - capacitive load
P/S < 1, Part of the apparent
power is consumed
Resistive and
reactive load
(R and L/C)
 Power equipment are rated using their appparent power in KVA.
Apparent Power
and Power Factor
Both have same P
Apparent Powers and pf’s are different
Generator of the second load is
overloaded
Apparent Power and Power Factor
Overloading of the
generator of the
second load is
avoided by
applying power
factor correction.
Complex Power
 The COMPLEX Power S contains all the information pertaining to
the power absorbed by a given load.
2
V
1 
S  VI  VRms IRms  I 2 Rms Z  Rms
2
Z
VRms  VRms v
I Rms  I Rms i
S  VRms I Rms (v  i )
 VRms I Rms cos(v  i )  jVRms I Rms sin(v  i )
 P  jQ  Re{S}  j Im{S}  Real Power+Reactive Power
Complex Power
 The REAL Power is the only useful power delivered to the load.
The REACTIVE Power represents the energy exchange between the
source and reactive part of the load. It is being transferred back and
forth between the load and the source
The unit of Q is volt-ampere reactive (VAR)
S  P  jQ  Re{S}  j Im{S}
=Real Power+Reactive Power
S  I 2 Rms Z  I 2 Rms ( R  jX )  P  jQ
P=VRms I Rms cos(v  i )  Re{S}  I 2 Rms R
Q=VRms I Rms sin(v  i )  Im{S}  I
2
Rms
X
Resistive Circuit and Real Power
v(t )  Vm sin(t   )
i (t )  I m sin(t )
1
1
p(t )  v(t )i(t )  Vm I m cos( ) 1  cos(2t )   Vm I m sin( ) sin(2t )
2
2
 VRms I Rms cos( ) 1  cos(2t )   VRms I Rms sin( ) sin(2t )
 VRms I Rms  VRms I Rms cos(2t )
p(t ) is always Positive
  0 RESISTIVE
Inductive Circuit and Reactive Power
v(t )  Vm sin(t   )
i (t )  I m sin(t )
1
1
Vm I m cos( ) 1  cos( 2t )   Vm I m sin( ) sin(2t )
2
2
 VRms I Rms cos( ) 1  cos(2t )   VRms I Rms sin( ) sin(2t )
pL (t )  v(t )i (t ) 
 VRms I Rms sin( 2t )
  90 INDUCTIVE
pL (t ) is equally both positive and negative, power is circulating
Inductive Circuit and Reactive Power
 If the average power is zero, and the energy supplied is returned
within one cycle, why is a reactive power of any significance?
 At every instant of time along the power curve that the curve is
above the axis (positive), energy must be supplied to the inductor,
even though it will be returned during the negative portion of the
cycle. This power requirement during the positive portion of the
cycle requires that the generating plant provide this energy during
that interval, even though this power is not dissipated but simply
“borrowed.”
 The increased power demand during these intervals is a cost
factor that must that must be passed on to the industrial consumer.
 Most larger users of electrical energy pay for the apparent power
demand rather than the watts dissipated since the volt-amperes
used are sensitive to the reactive power requirement.
 The closer the power factor of an industrial consumer is to 1, the
more efficient is the plant’s operation since it is limiting its use of
“borrowed” power.
Capacitive Circuit and Reactive Power
v(t )  Vm sin(t   )
i (t )  I m sin(t )
1
1
Vm I m cos( ) 1  cos(2t )   Vm I m sin( ) sin(2t )
2
2
 VRms I Rms cos( ) 1  cos(2t )   VRms I Rms sin( ) sin(2t )
pC (t )  v(t )i (t ) 
 VRms I Rms sin(2t )
  90 CAPACITIVE
pC (t ) is equally both positive and negative, power is circulating
Complex Power
 The COMPLEX Power contains all the information pertaining to the power
absorbed by a given load.
1 
Complex Power=S  P  jQ  VI  VRms I Rms ( v  i )
2
Apparent Power=S  S  VRms I Rms  P 2  Q 2
Real Power=P  Re{S}  S cos( v  i )
Reactive Power=Q  Im{S}  S sin( v  i )
P
Power Factor= =cos( v  i )
S
• Real Power is the actual power dissipated by the load.
• Reactive Power is a measure of the energy exchange between source and reactive
part of the load.
Power Triangle
 The COMPLEX Power is represented by the POWER TRIANGLE similar to
IMPEDANCE TRIANGLE. Power triangle has four items: P, Q, S and θ.
a) Power Triangle
b) Impedance Triangle
Q0
Q0
Resistive Loads (Unity Pf )
Capacitive Loads (Leading Pf )
Q0
Inductive Loads (Lagging Pf )
Power Triangle
Power Triangle
 Finding the total COMPLEX Power of the three loads.
PT  100  200  300  600 Watt
QT  0  700  1500  800 Var
ST  600  j800  1000  53.13
Power Triangle
S  P  jQ  S1  S2  ( P1  P2 )  j (Q1  Q2 )
Real and Reactive Power Formulation
Real and Reactive Power Formulation
Real and Reactive Power Formulation
Real and Reactive Power Formulation
v(t )  Vm cos(t  v )
i (t )  I m cos(t  i )
p(t )  VRms I Rms cos(v  i ) 1  cos 2(t  v )  VRms I Rmssin(v  i ) sin 2(t  v )
=P  1  cos 2(t  v )  Q  sin 2(t  v )
=Real Power  R eactive Power
P is the REAL AVERAGE POWER
Q is the maximum value of the circulating power flowing back and forward
P  Vrms I rms cos
Q  Vrms I rms sin 
Real and Reactive Powers
REAL POWER
CIRCULATING POWER
Real and Reactive Powers
• Vrms =100 V Irms =1 A Apparent power = Vrms Irms =100 VA
• From p(t) curve, check that power flows from the supply into the load for the
entire duration of the cycle!
• Also, the average power delivered to the load is 100 W. No Reactive power.
Real and Reactive Powers
Power Flowing Back
• Vrms =100 V Irms =1 A Apparent power = Vrms Irms =100 VA
• From p(t) curve, power flows from the supply into the load for only a part of
the cycle! For a portion of the cycle, power actually flows back to the source
from the load!
• Also, the average power delivered to the load is 50 W! So, the useful power is
less than in Case 1! There is reactive power in the circuit.
 Practice Problem 11.13: The 60  resistor absorbs 240 Watt of average power.
Calculate V and the complex power of each branch. What is the total complex power?
 Practice Problem 11.13: The 60  resistor absorbs 240 Watt of average power.
Calculate V and the complex power of each branch. What is the total complex
power?
 Practice Problem 11.14: Two loads are connected in parallel. Load 1 has 2 kW,
pf=0.75 leading and Load 2 has 4 kW, pf=0.95 lagging. Calculate the pf of two loads
and the complex power supplied by the source.
LOAD 1
2 kW
Pf=0.75
Leading
LOAD 2
4 kW
Pf=0.95
Lagging
Conservation of AC Power
 The complex, real and reactive power of the sources equal the respective sum of the
complex, real and reactive power of the individual loads.
a) Loads in Parallel
b) Loads in Series
For parallel connection:
S
1
1
1
1
V I* 
V (I1*  I*2 )  V I1* 
V I*2  S1  S2
2
2
2
2
Same results can be obtained for a series connection.
Complex power is Conserved
Power Factor Correction
 The design of any power transmission system is very sensitive to the magnitude of
the current in the lines as determined by the applied loads.
 Increased currents result in increased power losses (by a squared factor since P =
I2R) in the transmission lines due to the resistance of the lines.
 Heavier currents also require larger conductors, increasing the amount of copper
needed for the system, and they require increased generating capacities by the
utility company.
 Since the line voltage of a transmission system is fixed, the apparent power is
directly related to the current level.
 In turn, the smaller the net apparent power, the smaller the current drawn from the
supply. Minimum current is therefore drawn from a supply when S = P and QT =
0.
 The process of introducing reactive elements to bring the power factor closer to
unity is called power-factor correction. Since most loads are inductive, the
process normally involves introducing elements with capacitive terminal
characteristics having the sole purpose of improving the power factor.
Power Factor Correction
Increasing the power
factor without altering
the voltage or current
to the load is called
Power Factor
Correction
Original Inductive Load
Inductive Load with improved power factor correction
Effect of capacitor on total current
Power triangle of power factor correction
Power Factor Correction
 Increasing the power factor without altering the voltage or current to the load
is called Power Factor Correction.
Qc = Q1 – Q2
= P (tan θ1 - tan θ2)
= ωCVrms2
Q1 = S1 sin θ1
= P tan θ1
P = S1 cos θ1
Q2 = P tan θ2
C 
Qc
2
ωVrms

P (tan θ1  tan θ 2 )
2
ω Vrms
Power Factor Correction
 The process of increasing the power factor without altering the voltage or current to
the original load is called power factor correction.
 Power factor correction is necessary for economic reasons.
P1  P2  P
Real power stays same
P  S1 cos 1 Q1  S1 sin 1  P tan 1
Q2  P tan  2
QC  Q1  Q2  P(tan 1  tan 2 )
C
QC
P(tan 1  tan  2 )

Vrms 2
Vrms 2
• The capacitance value needed to change the pf angle from 1 to 2 .
• Similarly the inductance value needed to change the pf angle from 1 to 2 for a
capacitive load.
Vrms 2
L
 QL
Power
Factor
Correction
Power
Factor
Correction
Power Factor Correction
 Practice Problem 11.15: Find the value of the capacitance needed to correct a load
of 140 kVAR at 0.85 lagging pf to unity pf. The load is supplied by a 110 Volt (rms),
60 Hz line.
Applications: Power Measurement
 Wattmeter is the instrument for measuring the average power. Two coils are
used, the high impedance Voltage coil and the low impedance Current coil.
Wattmeter measures the average power given by:
P  12 Vm I m cos(v  i )  VRms I Rms cos(v  i )
Wattmeter
Wattmeter connected to the load
Problem 11-74