Power in Single Phase AC Circuits
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•
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Real Power
Reactive Power
Apparent Power
Power Factor
Complex Power
Examples
Power Factor Angle
Real Power P
Real Power: The same as the average power and
is sometimes called the active power. It has units
of watts.
• P = Pavg = Vrms Irms cos(Φv - Φi )= Vrms Irms pf
Power factor = pf = cos(Φv - Φi )
P(t) for general RLC Load
Φpf = Φv - Φi
Lagging PF: For inductive loads, the current lags
the voltage so that 0 < Φpf < 180.
• Recall that:
P(t) = v(t)i(t) = V cos(ωt +Φv) I cos (ωt +ΦI )
P(t) = ½ V I {cos(Φv - ΦI ) + cos(2 ωt +Φv + ΦI )}
Leading PF: For capacitive, the current leads the
voltage so that -180 < Φpf < 0.
• Can you show that
P(t) = VrmsIrmscos Φpf {1 + cos[2(ωt +Φv )]} +
VrmsIrms sin Φpf sin[2(ωt +Φv )] ?
ELI the ICE man
Real, Reactive, and Apparent Power
• Real Power (units = watts)
= P = VrmsIrmscos Φpf
• Reactive Power (units = VAR)
Q = VrmsIrms sin Φpf
• Apparent Power (units = VA)
|S| = VrmsIrms
Complex Power
• Given the general RLC case let
= < Φ and != < Φ"
• Then the complex power is defined
#! = !∗ = ( < Φ )( < −Φ" )
= < (Φ −Φ" ).
• In rectangular form
S = &'((Φ −Φ" ) + j ()*(Φ −Φ" ).
S = P + jQ
1
Power Triangle
S = P + jQ
Passive sign convention
The power triangle
Im
S = complex power
P = real power
Q = reactive power
With + conventional current entering the +
voltage terminal of a device, then
s
Q
Re
P
Measuring PF
P > 0 : real power absorbed by device
P < 0 : real power delivered by device
Q > 0 : reactive power absorbed by device
Q < 0 : reactive power delivered by device
Ex 1 … Derive P, Q and S for the source
• In the lab we can find PF without measuring
voltage and current phase angles since;
+
./ 012.
244 .4. .5"3
PF = |-|=00.34 012.=61/4 .4. 7 .4.
98 =
:
120?@( < 0°
"3
=
= 12C ?@( < 0°
;
10Ω
9D =
:
120?@( < 0°
"3
=
= 24C ?@( < −90°
EFD
E5Ω
:
9 9
414 = 8 + D = 26.8C ?@( < −63.4°
• EX 1 cont..
and the current into the ‘+’ side of Vin is
"3 = − 414 = 26.8 C ?@( < +116.6°
O
# = :
"3 "3 ∗= (120 < 0°)(26.8C < −116.6°)
S = 3220 VA < -116.6°
S = -1440 w – j 2880 var
• P = 1440 w delivered
• Q= 2880 var delivered
Power Factor Correction
• What is the big deal about the power factor?
Since |S| = P + QP = then for a fixed
supply voltage and a given real power P, the larger
the Q the larger the required . This mean all the
distribution equipment (lines, cables, transformers,
circuit breakers, etc.) must be sized for the larger
P ; losses will occur in
. Additionally, larger
transmission lines.
Thus we want to keep Q small so that
R =
~1
P
+ QP
2
Example 2
Calculate the power to be delivered by a capacitor
connected in parallel with the previous load in order
to increase the source power factor to 0.95 lagging.
Ex 2…
• Recall: 1/5 = 1440 T, Q1/5 = 2880 C;
And #1/5 = 3220 C
• We require VW = &'(∅3.2 = 0.95 ⇒ ∅3.2 = 18.2°
• Using the power triangle
Z[*∅3.2 =
Q
S
\]^_
+`ab
Q3.2 = 1/5 Z[*∅3.2
Q3.2 = 1440 tan(18.2°)
= 473 VAR
P
Qc0 = Q1/5 − Q3.2
= 2407 VAR
Example 3
Compare currents circuits of Ex 1 and Ex 2
• Circuit 1 has
+
deef
Rd = - = gPPf=0.44 d = 26.8C
• Circuit 2 has
RP = 0.95
P
=
6
=
deefh iejgh dkdl
= dPf
dPf
= 12.6C
Adding Complex Powers
• The total power is the sum of the component
powers regardless of their interconnection.
: !1*+ !2*}
•
#!= !*={
=(P1 +jQ1 ) +(P2 +jQ2)
=(P1 +P2 ) +j(Q1 +Q2)
= PTOT + j QTOT
•
•
•
• Clearly the higher PF leads to smaller required
current.
EX 4 Find S1 and S2
EX 4 cont.
∗
: :1 = 2220 VA <63.4 = 993 W + j 1986 VAR
#d = 1
∗
: :2 = 1790 VA <-33.7 = 1489 W - j 993 VAR
#P = 2
∗
#pqp = 9 = (120V <0)(22.3A<21.8) = 2673VA <21.8
!=
120 < 0
=
= 22.3C < −21.8
nd + nP
2 + E4 + (3 − E2)
#pqp = 2482.3W + j 993 VAR
And therefore
: = nd ! = 4.47 < 63.4 22.3 < −21.8 = 99.6V < 41.6
1
: = nP ! = 3.61 < −33.7 22.3 < −21.8 = 80.3V <-55.5
2
#pqp = pqp + j Qpqp = #d + #P = d + P + E(Qd+ QP )
3