SCCE
FTKEN
Chapter 7
AC Power Analysis
1
Chapter outline
Instantaneous and Average Power
Maximum Average Power Transfer
Effective or RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Correction
2
Instantaneous and Average Power (1)
• The instantaneously power, p(t)
p(t ) = v(t ) i (t ) = Vm I m cos (ω t + θ v ) cos (ω t + θ i )
1 1
= Vm I m cos (θ v − θ i ) + Vm I m cos (2ω t + θ v + θ i )
2 2
Constant power Sinusoidal power at 2ωt
p(t) > 0: power is absorbed by the circuit; p(t) < 0: power is absorbed by the source.
3
Instantaneous and Average Power (2)
• The average power, P, is the average of the instantaneous power over one period.
1 T 1
P=
T ∫ 0
p (t ) dt = Vm I m cos (θ v − θ i )
2
1. P is not time dependent.
2. When θv = θi , it is a purely
resistive load case.
3. When θv– θi = ±90o, it is a
purely reactive load case.
4. P = 0 means that the circuit
absorbs no average power.
4
Instantaneous and Average Power (3)
Example 1
Calculate the instantaneous power and average power absorbed by a
passive linear network if:
𝑣𝑣(𝑡𝑡) = 120 cos (377 𝑡𝑡 + 45°)
𝑖𝑖(𝑡𝑡) = 10 𝑐𝑐𝑐𝑐𝑐𝑐 (377 𝑡𝑡 − 10°)
Answer: P(t)=344.2 + 600cos(754t + 35°)W,
P=344.2W
5
Instantaneous and Average Power (4)
Example 2
A current I = 10 ∠ 30° flows through an impedance .
Find the average power delivered to the impedance.
Z = 20∠ − 22°Ω
6
Maximum Average Power Transfer (1)
ZTH = R TH + j X TH
ZL = R L + j X L
The maximum average power
can be transferred to the load if
XL = –XTH and RL = RTH
2
VTH
Pmax =
8 R TH
If the load is purely real, then R L = + X TH = ZTH
2 2
R TH
7
Maximum Average Power Transfer (2)
Example 2
For the circuit shown below, find the load impedance ZL that absorbs the
maximum average power. Calculate that maximum average power.
8
Maximum Average Power Transfer (2)
Excercise 2
For the circuit shown below, find the load impedance ZL that
absorbs the maximum average power. Calculate that maximum
average power.
Answer: 3.415 – j0.7317W, 1.429W
9
Effective or RMS Value (1)
The total power dissipated by R is given by:
1 T R T 2
P= ∫ i Rdt = ∫ i dt = I rms
2 2
R
T 0 T 0
T
1
Hence, Ieff is equal to: I eff =
T ∫0
i 2 dt = I rms
The rms value is a constant itself which
depending on the shape of the function i(t).
The effective of a periodic current is the dc current that delivers the
same average power to a resistor as the periodic current. 10
Effective or RMS Value (2)
The rms value of a sinusoid i(t) = Imcos(ωt)
is given by:
Im
I 2
rms =
2
The average power can be written in terms of
the rms values:
1
I eff = Vm I m cos (θ v − θ i ) = Vrms I rms cos (θ v − θ i )
2
Note: If you express amplitude of a phasor source(s) in rms, then all the
answer as a result of this phasor source(s) must also be in rms value.
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Apparent Power and Power Factor (1)
• Apparent Power, S, is the product of the r.m.s. values of voltage and current.
• It is measured in volt-amperes or VA to distinguish it from the average or real
power which is measured in watts.
P = Vrms I rms cos (θ v − θ i ) = S cos (θ v − θ i )
Apparent Power, S Power Factor, pf
• Power factor is the cosine of the phase difference between the voltage and
current. It is also the cosine of the angle of the load impedance.
12
Apparent Power and Power Factor (2)
Purely resistive θ – θ = 0, Pf = 1 P/S = 1, all power are
v i
load (R) consumed
Purely reactive θv– θi = ±90o, P = 0, no real power
load (L or C) pf = 0 consumption
Resistive and θ v– θ i > 0 • Lagging - inductive
reactive load θ v– θ i < 0 load
(R and L/C) • Leading - capacitive
load
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Complex Power (1)
Complex power S is the product of the voltage and the
complex conjugate of the current:
𝐕𝐕 = Vm ∠θv 𝐈𝐈 = Im ∠θi
1 ∗
𝑆𝑆 = V I = Vrms Irms ∠ θv − θi
2
where
𝑉𝑉 𝐼𝐼
𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = = 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 ∠θv 𝐼𝐼𝑅𝑅𝑅𝑅𝑅𝑅 = = 𝐼𝐼𝑅𝑅𝑅𝑅𝑅𝑅 ∠θi
2 2
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Complex Power (2)
1
S = V I∗ = Vrms I rms ∠ θ v − θ i
2
⇒ S = Vrms I rms cos (θ v − θ i ) + j Vrms I rms sin (θ v − θ i )
S = P + j Q
P: is the average power in watts delivered to a load and it is
the only useful power.
Q: is the reactive power exchange between the source and
the reactive part of the load. It is measured in VAR.
• Q = 0 for resistive loads (unity pf).
• Q < 0 for capacitive loads (leading pf).
• Q > 0 for inductive loads (lagging pf). 15
Complex Power (3)
⇒ S = Vrms I rms cos (θ v − θ i ) + j Vrms I rms sin (θ v − θ i )
S = P + j Q
Apparent Power, S = |S| = Vrms*Irms = P 2 + Q 2
Real power, P = Re(S) = S cos(θv – θi)
Reactive Power, Q = Im(S) = S sin(θv – θi)
Power factor, pf = P/S = cos(θv – θi)
16
Complex Power (4)
⇒ S = Vrms I rms cos (θ v − θ i ) + j Vrms I rms sin (θ v − θ i )
S = P + j Q
Power Triangle Impedance Triangle Power Factor 17
Complex Power (5)
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Example 1
Calculate the i)complex power, ii)the average power, iii)the reactive power and
iv)the power factor for the following;
(a) 𝑣𝑣(𝑡𝑡) = 160 cos 3 77𝑡𝑡 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣; 𝑖𝑖(𝑡𝑡) = 4 cos( 377𝑡𝑡 + 45°)
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Example 2
Calculate the i)complex power, ii)the average power, iii)the reactive power and
iv)the power factor for the following;
(b) 𝑉𝑉 = 80∠60° 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 ; 𝑍𝑍 = 50∠30°Ω
20
Example 2
• Calculate the power factor, the average power, the reactive power, the
apparent power and the complex power for the given circuit;
21
22
Conservation of AC Power (1)
The complex real, and reactive powers of the sources
equal the respective sums of the complex, real, and
reactive powers of the individual loads.
For parallel connection:
1 1 1 1
S= V I* = V (I1 + I*2 ) = V I1* + V I*2 = S1 + S2
*
2 2 2 2
The same results can be obtained for a series connection. 23
Example
Two loads connected in parallel are respectively 2 kW at a pf of 0.75 leading
and 4 kW at a pf of 0.95 lagging.
Calculate;
(i) the complex power supplied by the source
(ii) the pf of two loads
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Solution
25
Power Factor Correction (1)
Power factor correction is the process of increasing the
power factor without altering the voltage or current to
the original load.
Power factor correction is necessary for economic reason.
26
Power Factor Correction (2)
Qc = Q1 – Q 2
= P (tan θ1 - tan θ2)
= ωCV2rms
Q1 = S1 sin θ1 Qc P (tan θ1 − tan θ 2 )
C = =
= P tan θ1 ωVrms
2
ω Vrms
2
P = S1 cos θ1 Q2 = P tan θ2
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Application - Power measurement
• Power consumption in a AC system can be
measured using a Wattmeter.
• The meter consists of two coils; the current and
voltage coils.
• The current coil is designed with low impedance
and is connected in series with the load.
• The voltage coil is designed with very large
impedance and is connected in parallel with the
load.
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Example
• For the given circuit, calculate the wattmeter reading
29
Solution
30
Application –
Electricity consumption cost
• Utility companies divide their customers into categories;
• Residential (domestic) or small power
• Commercial or medium power
• Industrial or large power
• The amount of energy consumed in units of kilowatt-hours (kWh).
• The tariff or charge is often two part;
• First – fixed and corresponds to the cost of generation, transmission and distribution of
electricity
• Second – proportional to the energy consumed in kWh
• Bill is determined based on ;
• Total cost = Fixed Cost + Cost of Enegy
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Example
• A manufacturing industry consumes 500 MWh in a month. If the maximum
demand is 32000 kW, calculate the electricity bill based on the following rate;
• Demand charge – RM5.00 per month per kW of billing demand
• Energy charge –
• 8 cents per kWh for the first 50000 kWh
• 5 cents per kWh for the remaining energy.
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Solution
• Demand charge is RM5.00 x 32000 = RM160000
• Energy Consumed = 500 MWh = 500000 kWh
• RM0.08 per kWh for the first 50000 kWh = RM0.08 x 50000 = RM4000
• RM0.05 cents per kWh for the remaining energy
• (500000 – 50000 =450000).
• RM0.05 x 450000 = RM22500
• Total bill for the month = Demand charge + Energy consumed
= RM160000+RM4000+RM22500
= RM186500
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Example
• An 800 kW induction furnace at 0.88 power factor operates 20 hours per day
for 26 days in a month. Determine the electricity bill per month based on the
tariff;
• Energy charge: 0.06 cents per kWh
• Power factor penalty: 0.1% of energy charge for every 0.01 that power factor (pf) falls
below 0.85
• Power factor credit: 0.1% of energy charge for every 0.01 that power factor (pf) falls
exceeds 0.85
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solution
• Energy consumed = 800 kW x 20 hours/day x 26 days
= 416000 kWh
• Power factor exceed = 0.88 – 0.85 = 0.03
• Power factor credit: 0.1% of 416000 kWh x 3 every 0.01
= (0.1/100) x (416000 kWh) x 3 = 1248 kWh
• Total Energy = 416000 – 1248 = 414752 kWh
• Electricity bill = 414752 kWh x RM0.06/kWh = RM24885.12
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End of Chapter 7
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