Sinusoidal Steady-State
Power Calculation
EE3301
Kamran Kiasaleh
Learning Objectives
1.
2.
3.
4.
5.
6.
Be able to obtain the average (real) power
Be able to obtain the reactive power
Be able to obtain the complex power
Be able to obtain the apparent power
Be able to compute power and reactive factors
Be able to use Thevenin and Norton equivalent
circuits for steady-state sinusoidal inputs in order to
compute the load impedance which maximizes power
delivery
Power Delivered to a System
Definitions of Power
p(t ) = v (t )i(t )
v (t ) = Vm cos(ω 0 t + θ v )
i(t ) = Im cos(ω 0 t + θ i )
p(t ) = P + P cos(2ω 0 t ) − Qsin(2ω 0 t )
1 T
P = lim T →
p
t
dt
(
)
∫
0
∞ T
Vm Im
P=
cos(θ v − θ i )
2
Vm Im
Q=
sin(θ v − θ i )
2
Average Real
Power (kW,W)
Reactive Power
(VAR)
Power Definition
Power for purely resistive circuit
Power is
always
absorbed
by the
resistive
load
p(t ) = P + P cos(2ω 0 t )
Purely inductive circuit
Power is
exchanged
between the
pure reactive
load and the
source
Vm Im
Vm Im
Q=
sin(θ v − θ i ) =
2
2
p(t ) = −Qsin(2ω 0 t )
Purely capacitive circuit
p(t ) = −Qsin(2ω 0 t )
Power is
exchanged
between the
pure reactive
load and the
source
Vm Im
Vm Im
Q=
sin(θ v − θ i ) = −
2
2
Power Factor
pf = cos(θ v − θ i )
rf = sin(θ v − θ i )
Power factor
Reactive factor
Rf < 0 leading factor (current leads) Capacitor
Rf > 0 lagging factor (current lags) Inductor
Rf = 0 Resistive
Power delivered to a resistor
General equations
p(t ) = v (t )i(t )
1 T
P = lim T → ∞ ∫ 0 p(t )dt;Q = 0
T
2
Vrms
2
P=
= RIrms
R
1 T 2
Vrms = lim T → ∞ ∫ 0 v (t )dt
T
Irms
1
= lim T →
∞ T
∫ i (t)dt
T 2
0
Vm
= Veff
2
Im
=
= Ieff
2
v (t ) = Vm cos(ω 0 t + θ v ) ⇒ Vrms =
Vm
i(t ) =
cos(ω 0 t + θ v ) ⇒ Irms
R
Only applies for
sinusoidal signals
Simplified circuit for resistors
(any source)
Vrms
1 T 2
= lim 
v ( t )dt
T
∫
→∞
0
T
2
V
eff
2
P = Veff Ieff = RIeff
=
R
2
V
2
P = Vrmsf Irms = RIrms
= rms
R
Complex Power (sinusoidal signals)
V = Vm e
I = I me
Phasors
jθv
Complex Power
jθi
1 *
*
S = VI = Veff I eff
2
S = P + jQ
Vm I m
Vm I m
S=
cos (θ v − θi ) + j
sin (θ v − θi )
2
2
S = P2 + Q2 e
j tan −1
rf
pf
= Se
j tan −1
rf
pf
Apparent power
(kVA)
Complex Power (sinusoidal signals)
Vm I m
Vm I m
cos (θ v − θi ) + j
sin (θ v − θi )
S=
2
2
1
*
P = Re (VI * ) = Re (Veff I eff
)
2
1
*
Q = Im (VI * ) = Im (Veff I eff
)
2
Graphical Representation of S
P
pf =
S
Q
rf =
S
A phasor equivalent
equivalent impedance
Example
Veff = 250
IL,eff
250
=
= 4 − j3
40 + j30
VL,eff = (39 + j26)IL,eff = 234 − j13
SL = V
*
L,eff L,eff
I
= (4 + j3)(234 − j13)
SL = PL + jQL = 975 + j650
Some Rules
Elements in Series
(all in series)
S = S1 + S2 + ...+ Sn
(
)
*
S = V1,eff + V2,eff + ...+ Vn,eff Ieff
or
(
)
S = I1,eff + I2,eff + ...+ In,eff Veff*
Elements in parallel
(all in parallel)
Example
P1 = 8000
pf1 = 0.8 = cos (θ v − θi ) ⇒ sin (θ v − θi ) = 0.6
S1 = 8000 − j8000
0.6
= 8000 − j 6000
0.8
| S2 |= 20000
pf 2 = 0.6 = cos (θ v − θi ) ⇒ sin (θ v − θi ) = 0.8
S2 = 20000 ( 0.6 + j 0.8 ) = 12000 + j16000
S = S1 + S2 = 20000 − j10000
Q=−
Q=
| Veff |2
Reactive power
associated with a
capacitive load
Xc
| Veff |2
XL
Reactive power
associated with an
inductive load
Complex Power Levels
Equivalent Circuit Case
Thévenin equivalent
Example
Simplified circuit
Final circuit
Maximum Power Transfer
*
Vs ,eff Z L  Vs ,eff 
S L = VL ,eff I L* ,eff =


ZS + ZL  ZS + ZL 
| VL ,eff |2
| VL ,eff |2
SL =
ZL =
R + jX L )
2
2 ( L
| Z S + Z L |2
( RL + RS ) + ( X L + X S )
PL =
RL =
| VL ,eff |2
( RL + RS )
2
+ (XL + XS )
(XL + XS )
PL ,max =
| VL ,eff |2
QL ,max = −
4 RL
=
| VL ,eff |2
4 RS 2
2
2
RL ⇒
+ RS 2 ⇒ Let
∂PL
2
2
= 0 ⇒ ( RL + RS ) + ( X L + X S ) − 2 ( RL + RS ) RL = 0
∂RL
XL = −XS
| VL ,eff |2
4 RS
Xs =
| VL ,eff |2
4 RL 2
XL
ZL = ZS*
Key Result
⇒ RS = RL
Example (max power with
transformer)
Find the Thevenin equivalent circuit
Finding the Thevenin voltage
I1 = I 2 = 0;
V1
V2 = ;
4
V1 = Vi
Vi
840
VTh = − = −
∠0 = −210∠0
4
4
Finding Thevenin impedance
Short the source
0 = 60 I1 + V1 + 20 ( I1 − I 2 )
VTh = −V2 + 20 ( I1 − I 2 )
I Th
V1
I 2 = − I Th ; I 2 = −4 I1 ; I1 =
;V2 =
4
4
0 = 40 I Th + V1
V1
VTh = − + 25 I Th
4
VTh
= 35
I Th
Maximum power transfer circuit
Rs=Rl