electrical power and machines-power_1
DESCRIPTION
Electrical Power and Machines-Power_1TRANSCRIPT
Chapte
r 11
AC
Pow
er A
naly
sis
�In
sta
nta
ne
ou
s P
ow
er
�A
ve
rag
e P
ow
er
�E
ffe
ctive
or
RM
S V
alu
e
Lecture Outline
�T
he r
ate
of energ
y b
ein
g a
bsorb
ed,
rele
ased b
y o
r tr
ansfe
rred thro
ugh a
syste
m.
�T
he e
nerg
y m
ight
be in t
he form
of kin
etic,
pote
ntial,
therm
al, lig
ht,
chem
ical or
ele
ctr
ical.
�E
lectr
ical pow
er
is a
pro
duct
of curr
ent and v
oltage
Power
�E
lectr
ical pow
er
is a
pro
duct
of curr
ent and v
oltage
�E
lectr
ical pow
er:
�A
resis
tor
absorb
s p
ow
er
supplie
d to it and t
urn
s t
hem
into
heat.
�In
ducto
rs a
nd c
apacitors
absorb
, sto
re a
nd r
ele
ase p
ow
er.
�A
circuit c
onsis
ting o
f re
sis
tors
, in
ducto
rs a
nd c
apacitors
genera
lly a
bsorb
s p
ow
er
and t
urn
s it
into
heat.
�Low
pow
er
consum
ption is g
enera
lly d
esired in c
ircu
it
desig
n.
Pow
er
at any insta
nt
of tim
e
)(
)(
)(
ti
tv
tp
⋅=
Units: Watt = Volt x Ampere
Instantaneous Power
Units: Watt = Volt x Ampere
tV
tv
mω
cos
)(
=
)cos(
)(
φω
−=
tI
ti
m
If the current lags the voltage by a phase φ
)cos(
cos
)(
φω
ω−
⋅=
tI
tV
tp
mm
)cos(
cos
)(
φω
ω−
⋅=
tI
tV
tp
mm
)]2
cos(
[cos
)(
1φ
ωφ
−+
=t
IV
tp
Apply
ing 2
cos A
cos B
= c
os (A
-B) + c
os (A
+B
)
Components of Instantaneous
Power
)]2
cos(
[cos
)(
21φ
ωφ
−+
=t
IV
tp
mm
Tim
e-c
onsta
nt,
depends o
n p
hase
diffe
rence b
etw
een
voltage a
nd c
urr
ent
Tim
e-v
ari
able
,
sin
usoid
al w
ith
frequency tw
ice the
voltage a
nd c
urr
ent
time
function
Components of Instantaneous
Power (cont.)
φcos
)]2
cos(
[cos
)(
21φ
ωφ
−+
=t
IV
tp
mm
time
constant
0.40
0
0.80
0
1.20
0
1.60
0
0.4
0.8
1.2
1.6
Not
e th
at
Not
e th
at
Not
e th
at
Not
e th
at
Inst
anta
neou
s In
stan
tane
ous
Inst
anta
neou
s In
stan
tane
ous
Pow
er fr
eque
ncy
Pow
er fr
eque
ncy
Pow
er fr
eque
ncy
Pow
er fr
eque
ncy
is tw
ice
thos
e of
is
twic
e th
ose
of
is tw
ice
thos
e of
is
twic
e th
ose
of
voltage
power
Graphical Representation of
Instantaneous Power
-1.6
00
-1.2
00
-0.8
00
-0.4
00
0.00
0
010
020
030
040
050
060
070
0
Tim
e-1
.6
-1.2
-0.8
-0.4
0.0
01
00
20
03
00
40
05
00
60
07
00
Tim
e
is tw
ice
thos
e of
is
twic
e th
ose
of
is tw
ice
thos
e of
is
twic
e th
ose
of
curre
nt a
nd
curre
nt a
nd
curre
nt a
nd
curre
nt a
nd
volta
ge.
volta
ge.
volta
ge.
volta
ge.
current
�C
alc
ula
te t
he
in
sta
nta
ne
ou
s p
ow
er
if:
)20
10
cos(
80
)(
°+
=t
tv
)60
10
sin(
15
)(
°+
=t
ti
Instantaneous Power: Example
)60
10
sin(
15
)(
°+
=t
ti
)30
10
cos(
15
°−
=t
)30
10
cos(
)20
10
cos(
15
80
)(
)(
)(
°−
°+
⋅=
⋅=
tt
ti
tv
tp
)]10
20
cos(
)50
[cos(
600
°−
+°
=t
W)10
20
cos(
600
7.385
°−
+=
t
0.0
0.4
0.8
1.2
1.6
Insta
nta
neous
Pow
er
changes in
Problem with Instantaneous Power
-1.6
-1.2
-0.8
-0.4
0.0
01
00
20
03
00
40
05
00
60
07
00
Tim
e
changes in
tim
e, so it is
difficult to
measure
.
A w
ay t
o o
verc
om
e t
his
is t
o d
eal w
ith
Avera
ge P
ow
er
inste
ad.
Avera
ge o
f th
e insta
nta
neous p
ow
er
over
one p
eri
od.
∫=
T
dt
tp
P )
(1
Average Power
∫=
dt
tp
TP
0 )
(
)]2
cos(
[cos
)(
21φ
ωφ
−+
=t
IV
tp
mm
∫−
+=
T
mm
dt
tI
VT
P0
21
)]2
cos(
[cos
1φ
ωφ
Inte
gral
of a
sin
usoi
d In
tegr
al o
f a s
inus
oid
Inte
gral
of a
sin
usoi
d In
tegr
al o
f a s
inus
oid
over
its
perio
d is
zer
o.ov
er it
s pe
riod
is z
ero.
over
its
perio
d is
zer
o.ov
er it
s pe
riod
is z
ero.
φcos
21m
mI
VP=
φcos
21m
mI
VP=
If Φ ΦΦΦ
=0 =0=0=0, t
he c
urre
nt in
a c
ircui
t is
in p
hase
w
ith th
e vo
ltage
, the
circ
uit h
as a
pur
e m
mI
VP
21=
Resistive & Reactive Loads
ivθ
θφ
−=
with
the
volta
ge, t
he c
ircui
t has
a p
ure
resi
stiv
e lo
adre
sist
ive
load
resi
stiv
e lo
adre
sist
ive
load
. The
circ
uit a
bsor
bs p
ower
at
all t
imes
.If Φ ΦΦΦ
=90
=90
=90
=90o ooo
, the
cur
rent
in a
circ
uit i
s ou
t-of-
phas
e by
900
with
the
volta
ge, t
he c
ircui
t ha
s a
pure
reac
tive
load
reac
tive
load
reac
tive
load
reac
tive
load
. The
circ
uit
abso
rbs
no a
vera
ge p
ower
. It a
bsor
bs a
nd
rele
ase
equa
l am
ount
of p
ower
with
in it
s pe
riod.
mmI
VP
2=
0=
P
�C
alc
ula
te the a
vera
ge p
ow
er
if:
)20
10
cos(
80
)(
°+
=t
tv
)60
10
sin(
15
)(
°+
=t
ti
Calc
ula
ting a
vera
ge
pow
er is
much
sim
ple
r th
an
Average Power: Example 1
)30
10
cos(
15
°−
=t
φcos
21m
mI
VP=
))30
(20
cos(
15
80
21−
−⋅
= W7.
385
=
W)10
20
cos(
600
7.385
)(
°−
+=
tt
p
Note that the average
power is the constant
component of the
instantaneous power.
sim
ple
r th
an
calc
ula
ting
insta
nta
neous p
ow
er
Average Power: Example 2
Determine the power generatedby each
source and the average power absorbed
by
each passive elem
ent in the circuit.
Must evaluate the
current flow in each
mesh and the voltage
across the current
source. Can use
mesh analysis.
Average Power: Example 2 (cont.)
mesh analysis.
A 4
1=
I
030
60
5)
4(
10
22
=°
∠+
−−
II
jj
V 2.
698
.184
)1.
79
58
.10
4(10
)4(
20
1°
∠=
°∠
−+
=j
V
A 1.
79
58
.10
2°
∠=
⇒I
Power flow across the current source;
W8.67
3
)0
21
.6
cos(
498
.184
21
1−
=°
−⋅
=P
Power is supplied
by the current source.
P <
0,
pow
er
is
bein
g s
upplie
d
A 4
1=
I
V 80
)20
(4
2=
=V
W60
1 4
80
212
=⋅
=P
Average Power: Example 2 (cont.)
V 9.
11
8.105
)1.
79
58
.10
(10
L°
∠=
°−
∠⋅
=j
V
A79.1
-10.58
1.
79
58
.10
42
°∠
=°
∠−
=I-
I 1
Power absorbed by
the resistor;
2
W0
))1.
79
(9.
11
cos(
58
.10
8.105
213
=°
−−
⋅=
P Power flow across the inductor
Average Power: Example 2 (cont.)
A 1.
79
58
.10
2°
∠=
I
V 9.
11
9.52
1.79
58
.10
5C
°−
∠=°
∠⋅
−=j
V
W0
))1.79
9.11
cos(
58
.10
9.52
214
=°
−−
⋅=
P
Power flow across the capacitor;
Average Power: Example 2 (cont.)
Power flow across the voltage source;
W207.8
)1.
79
30
cos(
58
.10
60
215
=°
−°⋅
=P Power is absorbed
by the voltage source.
A 1.
79
58
.10
2°
∠=
I
P >
0,
pow
er
is
bein
g a
bsorb
ed
Average Power: Example 2(cont.)
W60
12=
P
W04=
P
W03=
P W8.
67
31=
P
W207.8
5=
P
Power
supplied
Power
absorbed
Power
absorbed
& released
Circuit Element
Average Power
V or I source
P = ½
VM IM cos(θ v- θ i)
Resistor
P = ½
VM IM = ½ IM2 R
Average Power Summary
Resistor
P = ½
VM IM = ½ IM R
Capacitor or
Inductor
P = 0
Does t
he e
xpre
ssio
n f
or
the r
esis
tor
pow
er
look
identical to
that fo
r D
C c
ircuits?
Fo
r th
e s
am
e a
mo
un
t o
f a
ve
rag
e
po
we
r a
bso
rbe
d b
y t
he
re
sis
tor,
A s
inu
so
ida
l cu
rre
nt
is f
low
ing
th
rou
gh
a r
esis
tor.
A c
ert
ain
a
mo
un
t o
f p
ow
er
is a
bso
rbe
d b
y
the
re
sis
tor.
Effective Value
po
we
r a
bso
rbe
d b
y t
he
re
sis
tor,
w
ha
t w
ou
ld b
e t
he
cu
rre
nt
va
lue
if it is
a d
c c
urr
en
t?
Th
is v
alu
e o
f th
e c
urr
en
t is
ca
lled
th
e e
ffe
ctive
va
lue
of
the
cu
rre
nt.
Th
is v
alu
e is h
igh
ly
use
ful in
an
aly
sin
g t
he
po
we
r flo
w in
a c
ircu
it. V
oltm
ete
rs a
nd
a
mm
ete
rs a
re d
esig
ne
d t
o r
ea
d
dir
ectly th
e e
ffe
ctive
va
lue
s o
f vo
lta
ge
an
d c
urr
en
t.
�R
oo
t-m
ea
n-s
qu
are
va
lue
(fo
rmu
la r
ea
ds lik
e th
e
na
me
: rm
s)
∫∫
++
==
Tt
rms
Tt
rms
dt
tv
Vand
dt
ti
I0
0
)(
1)
(1
22
Effective or RMS Values
�F
or
a s
inu
so
id:I
rms= IM/√2;
Vrm
s= V
m/√2
�F
or
exa
mp
le, A
C h
ou
se
ho
ld o
utle
ts a
re a
rou
nd
12
0 V
olts-r
ms
∫∫
==
t
rms
t
rms
dt
tv
TV
and
dt
ti
TI
00
)(
)(
Average Power
(in terms of rms values)
()
()
IV
IV
Pi
vrms
rms
iv
MM
source
cos
cos
21−
=−
=θ
θθ
θ
•The a
vera
ge p
ow
er
(P)
is:
RI
R
VI
VI
VP
rms
rms
rms
rms
MM
resistor
iv
rms
rms
iv
MM
source
2
2
212
==
==
�W
he
n w
e b
uy c
on
su
me
r e
lectr
on
ics, th
e
face
pla
te s
pe
cific
atio
ns p
rovid
e t
he
rm
s v
olta
ge
a
nd
cu
rre
nt
va
lue
s
�F
or
exa
mp
le,
wh
at
is th
e r
ms c
urr
en
t fo
r a
12
00
W
att
ha
ird
rye
r (a
lth
ou
gh
th
ere
is a
sm
all
fan
in
a
RMS in Everyday Life
Wa
tt h
air
dry
er
(alth
ou
gh
th
ere
is a
sm
all
fan
in
a
ha
ird
rye
r, m
ost
of th
e p
ow
er
go
es t
o a
re
sis
tive
h
ea
tin
g e
lem
en
t)?
�W
ha
t h
ap
pe
ns w
he
n t
wo
ha
ird
rye
rs a
re t
urn
ed
o
n a
t th
e s
am
e t
ime
in
th
e b
ath
roo
m?
�H
ow
ca
n I
de
term
ine
wh
ich
use
s m
ore
e
lectr
icity--
-a p
lasm
a o
r a
n L
CD
HD
TV
?