SINUSOIDAL
STEADY-STATE
POWER
CALCULATIONS
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10.1 Instantaneous Power
10.2 Average and Reactive Power
10.3 The RMS Value and Power Calculations
10.4 Complex Power
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10.5 Power Calculations
10.6 Maximum Power Transfer
10.7 Measurement of Power
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10.1 Instantaneous Power
Every electric device or equipment has a
power rating.
Exceeding the power rating can do permanent
damage to an appliance.
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10.1 Instantaneous Power
(1) instantaneous power
瞬時功率
(2) average (real) power
平均功率,(實功率,有效功率)
(3) reactive (imaginary) power
無效功率,(虛功率)
(4) complex power
複數功率
(5) apparent power
視在功率
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10.1 Instantaneous Power
Assume passive sign convention
p (t ) = v(t ) i (t ) watt
instantaneous power
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10.1 Instantaneous Power
v '(t ) = Vm cos(ωt + θ v )
i '(t ) = I m cos(ωt + θi )
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10.1 Instantaneous Power
v(t ) = Vm cos(ωt + θ v − θi )
i (t ) = I m cos(ωt )
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10.1 Instantaneous Power
p (t ) = Vm I m cos(ωt + θ v − θi ) cos(ωt )
1
1
cos( A − B ) + cos( A + B)
2
2
V I
V I
∴ p (t ) = m m cos(θ v − θi ) + m m cos(2ωt + θ v − θi )
2
2
Q cos( A + B) = cos A cos B − sin A sin B
Q cos A cos B =
∴ p (t ) =
Vm I m
V I
cos(θ v − θi ) + m m cos(θ v − θi ) cos (2ωt )
2
2
V I
− m m sin(θ v − θi )sin(2ωt )
2
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10.2 Average and Reactive Power
p (t ) = P + P cos 2ω t − Q sin 2ω t
V I
where P = m m cos(θ v − θ i ) , in watt ( 瓦 )
2
V I
Q = m m sin(θ v − θ i ) , in var ( 乏 )
2
1 t0 + T
Q P= ∫
p (τ )dτ , called average power
T t0
Q is called the reactive power
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10.2 Average and Reactive Power
What is the corresponding relation in
phasor domain? If v ( t ) = V m cos(ω t + θ v )
i ( t ) = I m cos(ω t + θ i )
ur
then V = V m ∠ θ v
r
I = Im∠θi
It turns out that
1
V m I m cos(θ v − θ i )
2
ur r *
1
= R e[ V I ]
2
ur r *
1
Q = Im[ V I ]
2
P=
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10.3 The RMS Value and Power Calculations
The effective value, Ieff , of a periodic current
i(t) is the equivalent dc current that delivers
the same average power to a resistor as the
periodic current.
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10.3 The RMS Value and Power Calculations
P=
1 t0 + T 2
i (τ ) Rdτ
T ∫t0
P = I eff 2 R
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10.3 The RMS Value and Power Calculations
1 t0 + T 2
i (τ )dτ
T ∫t0
∴The effective value of a periodic signal
is its root mean square ( RMS ) value.
P = P ⇒ I eff =
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10.3 The RMS Value and Power Calculations
If
i (t ) = I m cos(ωt + θi )
Im2
T
then I rms =
=
∫
T
0
cos 2 (ωt + θi )dt
Im
2
Thus , we can define the rms phasor
of i (t ) as
r
I
I rms = m ∠θ
2
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10.4 Complex Power
Vm I m
cos(θ v − θi )
2
urr *
1
= Re[V I ]
2
ur r *
V I
= Re[
]
2 2
ur r *
= Re[V rms I rms ]
QP =
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10.4 Complex Power
Vm I m
V I
θ θ
sin(
θ −θ ) v − i )
sin(
Q=
21 2 uvv
= I [V I ]
*
v
1 12 Vuv urr
I
I [ V I
] ]
= =Im[
2
2 2
2= I [Vuuuvuuuv
]
urIcomplex
r *power
Then, we can define the
uv
uuuvuuuv
V
I
S = P + jQ = V
I
= Im[
]
2 2
ur r *
= Im[V rms I rms ]
Then, we can define the complex power
ur
ur r *
S = P + jQ = V rms I rms
Similarly Similarly
Q=
m m
v
i
*
m
*
m
*
m
rms rms
*
rms rms
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10.4 Complex Power
A power triangle
uv
S = P2 + Q2
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uv
S = P + jQ
uv
S : apparent power in VA
P : real power , in w
Q : reactive power , in var
(volt − ampere reactive)
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10.4 Complex Power
Power factor definition
pf = cos(θv −θi )
≤1
θv −θi is called the power factor angle
Reactive factor definition
rf = sin(θv −θi )
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10.4 Complex Power
From complex ohm ' s law
ur
r
V rms = Z I rms
r
= ( R + jX ) I rms
ur ur r *
∴ S = V rms I rms
r 2
= Z I rms
r 2
r 2
= R I rms + jX I rms
= P + jQ
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10.4 Complex Power
pf = cos θ ,
θ = θv − θi
P
pf = ur
S
Q is a measure of the energy exchange between the
source and the reactive part of the load.
Reactive power represents a lossless interchange
between the load and the source.
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10.4 Complex Power
ur
V Vm Vuv V
Z = r = Z∠
= vθ
=v −
∠θ θ
− θi
I I m = RI + jXI
uv
S cos(θ − θ )
= R +Q P ==jX
uv
uv
pf ≤ S
ur Q = 0, S resistive
load ,
Q P = S cos(
− θi ) load ,
Q >θ
0,v inductive
ur Q < 0, capacitive
ur load ,
= S pf ≤ S
m
v
i
m
v
Q = 0, resistive
i
pf = 1.0
lagging pf
leading pf
load ,
pf = 1.0
Q > 0, inductive load , lagging pf
Q < 0, capacitive load , leading pf
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10.4 Complex Power
S
jQ
θ v -θ i
Lagging pf means that
current lags voltage.
P
Leading pf means that
current leads voltage.
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10.5 Power Calculations
i (t ) = 10 cos( wt + 30°) A
ur
r
V rms = RI rms
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v(t ) = 100 cos( wt + 30°)V
ur
100
V rms =
∠30° V
2
r
10
I rms =
∠30° A
2
ur
ur
r
*
∴ S = V rm s I rm s
100
10
=
∠ 30°
∠ − 30° VA
2
2
ur
2
V rm s
r
= 500 + j0 V A =
= I rm s
R
∴ P = 5 0 0W , Q = 0 v a r
p f = c o s 0 ° = 1 .0
2
R
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10.5 Power Calculations
Time domain
p(t ) = v(t )i (t )
= 100 cos( wt + 30) ×10 cos( wt + 30)
= 500(1 + cos 2 wt )VA
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10.5 Power Calculations
i (t ) = 10
cos(t − 45°)
v(t ) = 100 cos(t + 45°)
ur ur r *
S = V rms I rms
100
10
∠45°
∠45° = 500∠90°
2
2
= 0 + j 500 VA
=
ur
r
V rms = jX I rms
∴ P = 0, Q = 500 var
pf = cos 90° = 0, lagging
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10.5 Power Calculations
Time
domain
p (t ) = v '(t )i '(t )
= 100 cos(t + 90) × 10 cos t
= 500[cos 90 + cos(2t + 90)]VA
= −500 sin 2t
= −Q sin 2t , ∴ Q = 500 var
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10.5 Power Calculations
i (t ) = 10 cos(2t + 60°) A
v(t ) = 100 cos(2t − 30°)V
ur ur r *
S = V rms I rms
100
10
=
∠ − 30°
∠ − 60° VA
2
2
= 500∠ − 90° VA = P + jQ
∴ P = 0, Q = −500 var
pf = 0, leading
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10.5 Power Calculations
Time domain i '(t ) = 10 cos 2t
v '(t ) = 100 cos(2t − 90°)
p (t ) = v '(t )i '(t )
= 1000 cos(2t − 90) cos 2t
= 500[cos(−90) + cos(4t − 90)]
= +500 sin 4t = −Qsin 4t
= −(−500) sin 4t
∴ Q = −500 var, leading
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10.5 Power Calculations
Example. Given the load voltage and current
v ( t ) = 60 cos(ω t − 10 o )V
i ( t ) = 1.5 cos(ω t + 50 o ) A
ur ur
Find (a) S & S
(b) P and Q
(c) pf and ZL
Solution ( a )
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ur
60
∠ − 10 o V
V rms =
2
r
1.5
I rms =
∠ 50 o A
2
ur ur
r*
∴ S = V rms × I rms
60
1.5
=
∠ − 10 o ⋅
∠ − 50 o
2
2
= 45∠ − 60 o
VA
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10.5 Power Calculations
r
S = 45cos(−60o ) + j 45sin(−60o )
(b)
= 22.5 − j 38.97
VA
∴ P = 22.5 W
Q = −38.97 VAR
(c )
pf = cos(−60o ) = 0.5 leading
ur
V 60∠ − 10o
ZL = r =
= 40∠ − 60o Ω
o
1.5∠50
I
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10.5 Power Calculations
Conservation of AC Power
r
I
ur
V
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ur
I1
uur
I2
Z1
Z2
Assume rms phasors
The complex power supplied by the source
ur urr * ur r r *
S = V I = V (I 1 + I 2 )
urr * urr *
= V I1 +V I 2
ur ur
= S1 + S 2
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10.5 Power Calculations
Similarly
r
I
ur
V
Z1
ur
V1
Z2
uur
V2
Assume rms phasors
The complex power supplied by the source
ur urr * ur ur r *
S = V I = (V 1 + V 2 ) I
ur r * ur r *
= V1I +V 2 I
ur ur
= S1 + S 2
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10.5 Power Calculations
The complex ( real , or reactive ) power of the
source equals the respective sum of the complex
( real , or reactive ) powers of the individual loads .
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10.5 Power Calculations
Most industrial loads are inductive and are
operated at a low lagging power factor.
Z ∠θ
r
I
S ∠θ
θ
ur
V
θ
p f = c o s θ , θ = ta n − 1
θ
ur
V
r
I
r
I =
Q
P
ur
V
Z ∠θ
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10.5 Power Calculations
This will sacrificing some generator real power output
capability to provide the reactive power.
Also, from the users’ view point, to reduce energy cost due
to the penalty of low pf, it is worth to increase the pf.
Solution: for inductive loads, QL > 0
add parallel capacitor, QC < 0
without influencing the average power P.
r
IL
uur
VL
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10.5 Power Calculations
original load
r
I
r
IL
ur
ur
S 1 = P1 + Q 1 = V
ur 2
V
=
=
Z ∠ − θ1
r
IC
ur
V
r*
× I1
ur 2
V
Z
∠ θ
1
p f1 = c o s θ 1
Suppose after pf correction
uur
pf 2 = cos θ 2
then S 2 = P2 + jQ2
uur
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uur
= S 2 × cos θ 2 + j S 2 × sin θ 2
= P2 + jQ2
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10.5 Power Calculations
ur
∴ P2 = S
uur
S1
uur
S2
θ1 θ
2
but
ur
S
∴
2
=
× cosθ
P1
co s θ
2
= P1
2
P1
then Q2 = cos θ × sin θ 2 = P1 × tan θ 2
2
ur
ur ur
∴ S C = S 2 − S1 = P1 × tan θ2 − P1 × tan θ1
ur 2
ur 2
ur
V
V ×ωc
ur
ur r *
ur V *
ur 2
SC =V × IC =V × * =
=
= − j V ×ωc
1
ZC
j
(− j
)*
ωc
ur
∴ P1 ( t a n θ 1 − t a n θ 2 ) = V
c =
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2
P1 ( t a n θ
− ta n θ
ur 2
ω × V
1
2
2
×ωc
)
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10.5 Power Calculations
Example : voltage source : Vuur = 120V ( rms )
S
ω = 60 × 2π rad / s
electrical load :absorbs 4KW at a lagging pf
of 0.8
Find C necessary to raise the pf to 0.95
Solution : pf = 0.8, θ = cos−1 0.8 = 36.870
1
1
uur
∴ S1 =
P1
4 kw
=
= 5000 VA
cos θ 1
0.8
uur
∴Q1 = S1 × sin θ1 = 5000 × sin 36.870 = 3000 VAR
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10.5 Power Calculations
when
pf 2 = 0.95
θ2 = cos−1 0.95 =18.190
uur
4 kw
4000 w
∴ S2 =
=
= 4210.5 VA
cos θ 2
0.95
uur
Q 2 = S2 × sin θ 2 = 1314.4 VAR
∴∆Q=Q2 − Q1 = 1314.4 − 3000 = −1685.6 VAR
Q
C
∴c =
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= −
ur
V
2
1
ω c
ur
= − V
2
× ω c
1685.6
= 310.5 µ F
2π × 60 × 120 2
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10.6 Maximum Power Transfer
Given a linear two-terminal ac circuit, N, with
load impedance ZL
From Thevenin’s theorem
ur
V TH
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10.6 Maximum Power Transfer
Let Z T H = R T H + j X T H
Z
= R L + jX
L
ur
ur
r
VTH
VTH
Then I =
=
ZTH + ZL (RTH + RL ) + j( XTH + XL )
L
ur 2
V
r
TH × RL
2
1
1
P = × I × RL = ( ) ×
2
2 ( RTH + RL )2 + ( X TH + X L )2
Necessary condition for maximum output P
∂P
= 0 .......(A)
∂RL
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∂P
= 0 .......(B)
∂X L
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10.6 Maximum Power Transfer
From (B) ： XL =−XTH
2
From (A) ： RL = RTH
+ ( XTH + X L )2
Hence, for maximum output power
Z
L
Pmax =
= R L + jX
ur
V TH
L
= RTH − jX
2
ur
V TH
TH
= Z T* H
amplitude phasor
2
1
×
× RTH =
2 (2 RTH ) 2
8 RTH
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10.6 Maximum Power Transfer
In case the load is purely real
Then, Z L = RL + j 0, X L = 0
∴ RL =
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2
2
RTH
+ X TH
= Z TH
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10.6 Maximum Power Transfer
Example 1： Determine ZL to get maximum average
power output PZ L
10∠0o V
amplitude phasor
Z TH = j 5Ω + 4 Ω / /(8Ω − j 6 Ω )
=2.933+j4.467 Ω
VTH =
8 − j6
× 1 0 ∠ 1 0 0 = 7 .4 5 4 ∠ − 1 0 .3 0 V
4 + (8 − j 6 )
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10.6 Maximum Power Transfer
From maximum power transfer theorem
*
Z L = Z TH
= 2.933 − j 4.467 Ω
PZ L =
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ur
V TH
8 R TH
2
= 2 .3 6 8
W
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10.6 Maximum Power Transfer
Example 2： Find RL such that it will absorb maximum
average power.
150∠30o V
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10.6 Maximum Power Transfer
ZTH = (40 − j30) / / j20 = 9.412 + j22.35 Ω
ur
V TH =
j20
×150∠300 = 72.76∠1340
j20 + 40 − j30
150∠30o V
(Amplitude phasor !!!)
From maximum average power transfer theorem
R L = Z T H = 9.412 2 + 22.35 2 = 24.25 Ω
ur
r
V TH
I =
= 1 .8 ∠ 1 0 0 .4 2 0 A
Z TH + R L
∴ Pm a x =
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=
r
1
× I
2
2
× RL
1
× (1 . 8 ) 2 × ( 2 4 . 2 5 ) = 3 9 . 2 9
2
W
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10.7 Measurement of Power
The wattmeter is the instrument for measuring the average
power.
uur
IL
uur
IL
uur
VL
uur
VL
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10.7 Measurement of Power
An electromagnetic type wattmeter consists of two coils.
the current coil (C.C) : in series with load
the voltage coil (V.C) : in parallel with load
The mechanical inertia of the moving part produces an
equilibrium deflection angle that is proportional to the
average value of the product vL (t ) iL (t ), i.e.
α
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1
T
∫
t0 +T
t0
v L (t ) × iL (t ) dt = P
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10.7 Measurement of Power
The ± sign marked on the terminals of C.C. and V.C. is
for passive sign convention to ensure upscale deflection.
Example :Find the reading of W
150∠0o V
rms
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10.7 Measurement of Power
Solution :
r
150∠0 0
150
IL =
=
A
(12 + j10) + (8 − j 6) 20 + j 4
uur r
150 × (8 − j 6)
VL = I L × (8 − j 6) =
V
20 + j 4
ur uur r * 150 × (8 − j 6)
150
S = VL × I L =
×
20 + j 4
20 − j 4
=423.7 − j 324.6 VA
∴ P = 423.7 W
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Summary
n
n
Objective 1 : Understand
r r the ac power concepts of P, Q
, S ,| S | .
Objective 2 : Know the rms phasors and be able to calculate
various ac powers and power factor.
n
Objective 3 : Understand the condition for maximum real
power transfer and be able to calculate the
maximum real power.
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Summary
n
Objective 4 : Understand the pf correction technique and be
able to find the desired compensation.
n
Objective 5 : Know how to use a wattmeter to measure the
average power of an ac load.
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Summary
Chapter Problems : 10.16
10.18
10.22
10.36
10.37
10.41
Due within one week.
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