CHAPTER EIGHT
SINUSOIDAL STEADY-STATE POWER
CALCULATIONS
8.1 RMS Value and Instantaneous Power
8.2 Average and Reactive Power
8.3 Apparent Power and Power Factor
8.4 Complex Power
8.5 Conservation of AC Power
8.6 Power Factor Correction
8.1 RMS Value and Instantaneous Power.
The effective value of a periodic current is the dc current that
delivers the same average power to a resistor as the
periodic current.
Ieff = Irms
Veff = Vrms
All current and voltage value used in power calculation are
using root mean square, rms.
Vrms = Vm/√2
Vm – peak value
Instantaneous power (in watts) is the power at any instant
of time. It is the instantaneous power being absorbed by
each electrical devices is multiplication of instantaneous
voltage and instantaneous current across it.
p(t) = v(t) x i(t)
The value is different and periodic for evey cycle. It can be
clasified into three types ;
Apperent power (kuasa ketara) (S)
Average power (kuasa purata/sebenar) (P)
Reactive power (kuasa reaktif) (Q)
Vs(t)
Load = Z/θ
Vs(t)
C
V0(t)
8.2 Average and Reactive Power
Average power or real power in watts is the average of
the instantaneous power over one period.
P = ½ x Re[VI*] = 1/2VmImcos(θv – θi)
= (VmIm)/2cosθ watt
θ – impedance, Z angle between current and voltage.
θ = V/θv / V/θi = Z/θv - /θi = Z/θ
Average power at resistor, P = (VmIm)/2cosθ watt
= (VmIm)/2cosθ = VrmsIrms
As V = IR, so P = Vrms2/Rrms = Irms2R
Average power is useful coz of resistor element. So
average power for reactive load (L and C) is zero.
Reactive Power - Power that being stored by reactive element
L or C.
Q = Vm/√2 x Im/√2 x sinθ = (VmIm)/2sinθ var
Ex : If v= 100 kos (ωt + 15º) and i= 4 sin (ωt - 15º), At network
terminal, find average and reactive power.
i= 4 kos (ωt - 105º)
P = (VmIm)/2cosθ watt
= ½ x 100 x 4 x cos(15 – (-105) = -100 watt
Average power being transfer back from load to supply
terminal.
Q = (VmIm)/2sinθ var
= ½ x 100 x 4 x sin(15 – (-105) = 173.21 var
Reactive power being absorbed by load. i
+
Network
v
-
Ex : V(t) = 120cos(377t + 450), i(t) = 10cos(377t – 100) , find
the instantaneous and average power absorb by passive
linear network
cosAcosB = ½[cos(A+B)+cos(A-B)
P = v.i = 120cos(377t + 450) x10cos(377t – 100)
P = 600[cos(754t + 350) + cos550]
Average power, P = (VmIm)/2cos(θv – θi)
= ½.120.10.cos[45 –(-10)] = 600cos550 = 344.2 w
(constant part of instantaneous power)
Ex : Find the average absorb by an empedance, Z = 30 – j70
when V = 120/00
I = V/Z = 120/00 / (30 – j70) = 1.58/66.80
Average power, P = (VmIm)/2cos(θv – θi)
= ½.120x1.58cos[0 – 66.8] = 600cos550 = 37.24 w
Ex: Sinusoid voltage maximum amplitude 625 volt being
given to terminal resistor 50Ω, find the average power
being transfer by the resistor.
Vrms = 625/√2 = 441.94 V
Average power, P = V2/R = 441.942/50 = 3906.25 W
Ex:Find the average power supplied bt the source and
average power delivered to the impedance.
I = 5/300 / ( 4- j2) = 1.12/56.570 A
Average power supplied, P =
4Ω
5/30
-j2
Maximum Average Power Transfer
For maximum average power transfer, the load impedance
ZL must equal to the complex conjugate of the Thevenin
impedance, ZTh.
ZL = RL + jXL = RTh – jXTh = Z*Th
So RL = RTh and XL = -XTh
Pmax = |VTh|2/8RTh
If load is purely real, RL = √(RTh2 + XTh2) = |ZTh|
Ex : Find, ZL for max average power and the max average
value.
j6Ω
6Ω
10Ω
ZL/θ
12/300
-j8Ω
Ex: Find the value of RL that will absorb max average
power and calculate the power.
60 Ω
120/300
-j20 Ω
j30Ω
j30Ω
RL
8.3 Apparent Power and Power Factor
Apparent power being supplied by supply to load is
multiplication of voltage and current in rms value.
S = Vm/√2 x Im/√2
= (VmIm)/2 VA
= Vrms x Irms
Z = V/I =Vrms/Irms/θv - θi
Power for ac system can be illustrate using power triangle.
|S| = √(P2 + Q2) VA
P = |S|cosθ
Q = |S|sinθ
Power factor, pf = P/S = cos(θv – θi)
Power factor is the cosine of the phase difference between
voltage and current. It is also the cosine of the angle of the
load impedance.
If current trailing voltage, power factor is trailing.
If current leading voltage, power factor is leading.
S
Q
θ
P
Extreme Case
If load is pure resistor, phase different between current
and voltage is zero.
So power factor, pf = cosθ = 1
So apparent power = average power
If load is pure C or L, phase different between current and
voltage is 900.
So power factor, pf = cosθ = 0
So apparent power = reactive power
Average power, P = VrmsIrmscosθ = Vrms2/R = Irms2R watt
Reactive power, Q = VrmsIrmssinθ VAR
Apparent power, |S| = √(P2 + Q2)
The Relation between Power Factor and Load
Trailing
θ = θv – θi = positive
Load = inductive
Q = positive
Z = R + jXL
Leading
θ = θv – θi = negative
Load = capacitive
Q = negative
Z = R + jXC
V
θi
Ref, θv= 0º
I
I
θi
V
Ref, θv= 0º
Ex : Find the power factor and the average power delivered
by the source.
Z=
p.f = cosθ
Irms = Vsrms/Z
So P = VsrmsIrmspf
8Ω
20/30
-j4 Ω
6Ω
Ex : Find the power factor and the average power supplied by
the source.
10 Ω
Pf = 0.94 lagging
P = 118 W
40/0
8Ω
j4 Ω
-j6 Ω
8.4 Complex Power, S
Product of rms voltage phasor and complex conjugate of
rms current phasor. It consist of real part is real power, P
and imaginary part, is the reactive power, Q.
S = P + jQ = ½ x V.I*
S = (VmIm)/2cosθ + j (VmIm)/2cosθ
= (VmIm)/2[cos(θv – θi) + j[sin(θv – θi)
= VrmsIrms/(θv – θi) = VrmsIrms* VA
Vrms = V/√2 = Vrms/θv Irms = I/√2 = Irms/θi
Where I* is current conjugate and it’s angle opposite to I
angle. Complex current magnitude is the apparent power.
In ac circuit, V = IZ
So S = |V2|/Z = ZI.I* = |Irms|2Z
= |Irms|2(R + jX)
= |Irms|2R + j|Irms|2X = P + jQ
Q = VC/L2/XC/L = Irms2XC/L
So P = VR2/R = Irms2R
P is average or real power and it depend on load resistance R.
Q is reactive powwr and depends on load reactance X.
P = VrmsIrmscos(θv – θi);
Q = VrmsIrmssin(θv – θi)
The real power P is the average power in watts delivered to a
load; it is the only useful power. It’s the actual power
dissipated by the load.
The reactive power, Q is a measure of the energy exchange
between the source and the reactive part of the load. The
unit is volt-ampere reactive (VAR). It’s being transferred back
and forth between the load and the source, so ;
Q = 0, for resistive load (pf = 1)
Q < 1, for capacitive load (pf leading)
Q > 1, for inductive load (pf lagging)
Complex power, S = P + jQ = ½ x V.I*
= VrmsIrms/(θv – θi)
Apparent power, S, |S| = VrmsIrms = √(P2 + Q2)
Real power, P = Re(S) = Scos(θv – θi)
Reactive power, Q = Im(S) = Ssin(θv – θi)
Power factor, pf = P/S = cos(θv – θi)
Below are power triangle and impedance triangle
S
Q
θ
|Z|
X
θ
P
R
Ex: If electrical load operate at 240 Vrms and absorb 8 KW
average power at 0.8 power factor trailing, find the complex
power at the load and load admittance.
Power factor is trailing, so inductive load. θ & Q are positive.
V
From power triangle, P = |S|cosθ
so, |S| = P/cosθ = 8kW/0.8 = 10 KVA
I
Q = |S|sinθ = 10 x 0.6 = 6 KVAR
S = P + jQ = 8 + j6 KVA
θv
Load impedance, Z = |Vrms|/|Irms| /θ
P = VrmsIrmscosθ
θi
Irms = 8kW/(240 x 0.8) = 41.76A
So, |Z| = |Vrms|/|Irms| /θ = 240/41.67 = 5.76 Ω
/Z = cos-1θ = 36.870
Z = 4.61 + j3.46
Reference, 0º
Ex: If load voltage is v(t) = 60cos(ωt - 100) V and the current is
i(t) = 1.5cos(ωt + 500) . Find (a) complex and apparent power,
(b) real and reactive power, (c) power factor and load
impedance.
Vrms = 60/√2/-100 Irms = 1.5/√2/500
a. S = VrmsI*rms = 45/-600
S = |S| = 45 VA
b. S = 45/-600 = 22.5 – j38.97 = P + jQ
c. pf = cos(-600) = 0.5 (leading)
Z = V/I = (60/√2/-100) / (1.5/√2/500) = 40/-600 capacitive
impedance.
Ex : Load Z draws 12 KVA at p.f. 0.8 lagging from 120V rms
sinusoidal source. Find (a) average and reactive power
delivered to load, (b) peak current, (c ) load impedance.
a. pf = 0.856, so θ = 31.130
P = Scosθ = 12000 x 0.856 = 10.272 kW
Q = Ssinθ = 12000 x 0.517 = 6.024 kVA
S = 10.272 + j6.204 kVA
b. I* = S/Vrms = (10.272 + j6.204)/ 120/00 = 100/31.130
Im = √2 x Irms = 141.4 A
c. Z = Vrms/Irms = 120/00 / 100/-31.130 = 1.2/31.130
8.5 Conservation of AC Power
Average power , P = VrmsIrmscosθ watt
Reactive power , Q = VrmsIrmssinθ Var
Apparent power, |S| = √(P2 + Q2) VA
Complex power, S = VrmsI*rms = P + jQ VA
Some formula from power triangle
Q = |S|sinθ
P = |S|cosθ
Q = Ptanθ = Pcosθ/sinθ
S = VrmsIrmscosθ + j VrmsIrmssinθ
= |S|cosθ + j |S|sinθ = P + jQ
The complex, real and reactive power of the source equal
the respective sum of the complex, real and reactive power
of the individual load.
e.g1. KCL : I = I1 + I2
Complex power supplied by the source,
S = ½ x VI* = ½ V (I*1 + I*2) = S1 + S2
S1 and S2 is the complex power delivered to Z1 and Z2
e.g2. KVL : V = V1 + V2
Complex power supplied by the source,
S = ½ x VI* = ½ I* (V1 + V2) = S1 + S2
I
I
I1
V
Z1
I2
Z2
V
Z1
Z2
V1
V2
Ex: Find apparent power, real power, reactive power and pf.
I1 = V/Z1 = 120/100 / 60/-300 = 2/400 A rms
I2= V/Z2 = 120/100 / 40/450 = 2/400 A rms
S1 = V2rms/Z*1 = 1202 / 60/-300 = 240/300 = 207.85 - j120 VA
S2 = V2rms/Z*2 = 1202 / 40/450 = 360/-450 = 254.6 + j254.6 VA
S = S1 + S2 = 462.4 + j134.6 VA
S = |S| =
√(P2
+
Q2)
= 481.6 VA
P = 462.4 W
Q = 134.6 VAR
pf = P/S = 462.4/481.6 = 0.96 lagging
I
I1
120/100
Z1
60/300
I2
Z2
40/450
Ex : Find :
a. Average power for the two load.
b. Supplied apparent power
c. Power factor
d. The two load reactive power
e. Complex power
(2-j1) Ω
VS rms = 60∠0º
(1+j5) Ω
KVL : Vs = IsZ1 + IsZ2 = Is(2 – j1) + Is(1 + j5)
= Is5/53.130
Is = 60/00 / 5/53.130 = 12/-53.10 A
Average power is power coz of resistor, where current across
each resistor is 12 A.
PL1 = |I2|R1 = 122 x 2 = 288 w
PL2 = |I2|R2 = 122 x 1 = 144 w
P = PL1 + PL2 = 432 w
b. Supplied apparent power, |S| = |V|.|I| = 60 x 12 = 720 VA
c. Power factor for the whole circuit, p.f. = cosθ = P/|S| =
432/720 = 0.6 trailing = cos53.10
VS = 60∠0º V and IS=720∠-53.1ºA, so I trailing V 53.1º, and
p.f. is trailing circuit.
d. VL1 = ZL1.I
= (2.24 /-26.560) x (12/-53.10)
= 26.88/-79.70 V
QL1 = V1rms.Irmssinθ = 12 x 26.88sin(-26.560) = -144 Var (Cap. ld)
VL2 = ZL2.I
= (5.1 /78.70) x (12/-53.10)
= 61.2/25.60 V
QL2 = V2rms.Irmssinθ = 12 x 61.2sin(78.70) = 720.16 Var (indt. ld)
Complex power, S = |S|/θ0
= 720/53.10 kVA
= 432.3 + j575.8 kVA
or
S = V.I*
= (60 /00) x (12/-53.10)
= 720/53.10 kVA
= 432.3 + j575.8 kVA
Ex : Load with (39+j26)Ω impedance receive supply from
250rmsV voltage source through line that the impedance are
(1+j4) Ω.
a. Find the ILoad and VLoad.
IL = Vs/ZT
= (250 /00)/(40 + j30) = (250 /00) x (50/-36.870)
= 5/-36.870)
VL = IL/ZL
= (5 /-36.870)/(39 + j26)
= (5 /-36.870) x (46.87/33.690)
= 234.36/-3.180
1Ω
j4Ω
+
39Ω
250∠0º V
(rms)
VL
IL
j26Ω
Bekalan
Talian
Beban
b. Find PL and QL
SL = P + jQ = VL.I*L
= (234.36 /-3.180) x (5/-36.870)
= 975 + j650 VA
c. Find Pline and Qline
Pline = |I2|.R = 52 x 1 = 25 watt
Qline = |I2|.X = 52 x 4 = 100 Var
Load have inductive element with positive value.
d. Find Ps and Qs
Ss = Sline + SL
= (25 + j100) + (975 + j650)
= 1000 + j750 VA
Ex: System load impedance contain -52 Ω shunt capacitance
Find :
a. VLrms and ILrms
b. PL and QL
c. Pline and Qline
d. Ps and Qs
e. Pshunt and Qshunt
j4Ω
1Ω
+
VS= 250∠0º V
(rms)
VL
39Ω
Shunt
IL
j26Ω
Source
Line
-
Load
VL = ZL1/(ZL1 + Zline) x Vs
ZL1 – load equ impedance with XCshunt
VL = VZL = VXC
ZL1 = ZL || -j52
= ((39 + j26) x (-j52))/(39 + j26) + (-j52)) = 48 – j20
= 52/-22.60
VL = (52/-22.60)/((48 -j20) + (1 + j4))) x 250/00)
= 252.18/-4.540)
IL = (252.18/-4.540)/ 46.87/-33.690)
= 5.38/38.230)
b. PL and QL
SL = VLI*L
= (252.18/-4.540) x (5.38/38.230)
= 1128.87 + j752.58 VA
c. Pline and Qline
Vline = Vs – VL
= 250 – (251.39 – j19.96)
= 20/93.980 V
Iline = Vline/Zline = 20/93.980 /(1 + j4) = 4.85/18.020 A
SL = VLI*L
= 20/93.980 x 4.85/-18.020 = 23.53 + j94.1 VA
d. Ps and Qs
ZT = Zline + ZL
= (1 + j4) + (48 – j20)
= 51.55 /-18.080
Ss = VsI*s = |Vs|2/Z*T
= 2502/51.55 /-18.080 = 1152.56 – j376.27 VA
e. Pshunt and Qshunt
Qshunt = |Vshunt|2 / XC = |VL|2 / XC = 252.182 / -52 = -1222.98 VA
8.6 Power Factor Correction
Most domestic load such as washing m/c, air-cond and
fridge, industrial load – induction motor are inductive and
operate at low lagging power factor. Although the inductive
nature of load cannot changed, we can inc it’s power factor.
The process of inc the power factor without altering the
voltage or current to original load is known as pf correction.
Pf can be corrected or improved by added a capacitor
parallel with the load, so it will reduce the phase angle, thus
inc pf. It also dec the current, so consumer pay less (P =
IL2R). So consumer and TNB try by making current at min
level and pf close to unity as possible.
Ic
I
IL
V
Inductive
load
V
IL
Ic
Inductive
load
θ1
θ2
Ic
IL
V
Pf from another perspective.
If original inductive load has apparent power S1,
Then P = S1cosθ1, Q = S1sinθ1 = Ptanθ1
If pf inc from cosθ1 to cosθ2, without altering the real power
P = S2cosθ2, then new reactive power Q2 = Ptanθ2
Dec in reactive power that coz by shunt capacitor,
Qc = Q1 – Q2 = P(tanθ1 - tanθ2)
So from previous notes, QC = V2rms/Xc = ωCV2rms.
C = QC/ ωV2rms = P(tanθ1 - tanθ2) / ωV2rms
P dissipated by load is not affected by pf
Q
correction coz Pc = 0.
S
Normally in practice load is inductive
S
By it is possible load is capacitive, so
θ1
Q
pf is leading. So inductor should be
θ2
connected across load for power correction.
c
1
2
2
Q1
So shunt inductor, can be calculated from
QL = V2rms/XL = V2rms/ωL
So L = V2rms/ωQL
Where, QL = Q1 – Q2, is the diff between new and old reactive.
Ex: Supply is 12Vrms 60 Hz power line, a load absorb 4 kW at a
lagging 0.8 pf. Find the value of capacitance necessary to
raise the pf to 0.95.
pf = 0.8, so θ1 = 36.870
Apparent power, S1 = P/cosθ1 = 4000/0.8 = 5000 VA
Reactive power, Q1 = S1sinθ1 = 5000sin36.87 = 3000 VAR
pf = 0.95, so θ2 = 18.190
Apparent power, S2 = P/cosθ2 = 4000/0.95 = 4210.5 VA
Reactive power, Q2 = S2sinθ2 = 4210.5sin18.19 = 1314.4 VAR
Qc = Q1 – Q2 = 3000 – 1314.4 = 1685.6 VAR
C = Qc/ωV2rms = 310.5 µF