Download 3-phase AC circuits

CA2627 Building Science
Lecture 06 Transformers, three phase AC circuits,
building power distribution
Instructor: Jiayu Chen Ph.D.
Learning Objectives
1. Analyze the ideal transformer; compute primary and secondary currents and
voltages and turns ratios. Calculate reflected sources and impedances across
ideal transformers.
2. Understand maximum power transfer.
3. Learn three-phase AC power notation; compute load currents and voltages for
balanced wye and delta loads.
4. Understand the basic principles of residential electrical wiring and of electrical
safety.
© Jiayu Chen, Ph.D.
2
Power Distribution
Electricity supply in Hong Kong
Ref.: Supply rules from CLP
https://www.clponline.com.hk/MyHome/CustomerService/OpenAccount/SupplyRule
s/Pages/Default.aspx?lang=en
Supply rules from HK Electric
http://www.heh.com/NR/rdonlyres/F98E0121-03CA-4C4F-B2DAF77774833876/0/SupplyRule_en_2008.pdf
© Jiayu Chen, Ph.D.
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Power Distribution
EMSD Code of Practice
EMSD Code of Practice for Electricity (Wiring) Regulations, 2009 Edition
(http://www.emsd.gov.hk/emsd/e_download/pps/pub/COP_E.pdf)
© Jiayu Chen, Ph.D.
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Power Distribution
Electricity is supplied by the HKE or CLP in Hong Kong. The nominal values are:
• Frequency 𝟓𝟎 𝑯𝒛 ± 𝟐%
• Phase supply
o for a single phase supply, a nominal voltage of 220 volts (root mean square value) plus or
minus 6% between a phase conductor and neutral conductor;
o for a 3 phase supply, using a 3 phase 4 wire system, a nominal voltage of 380 volts plus or
minus 6% (root mean square value) between phase conductors and 220 volts (root mean
square value) plus or minus 6% between phase and neutral conductors;
o for a 3 phase supply, a nominal voltage of 11,000 volts (root mean square value) plus 10% or
minus 2% between phase conductors; and
o for a 3 phase supply, a nominal voltage of 132,000 volts (root mean square value) plus 10%
or minus 2% between phase conductors.
© Jiayu Chen, Ph.D.
5
Power Distribution
Beyond the generation plant, an electric power network distributes energy to several
substations. This network is usually referred to as the power grid.
© Jiayu Chen, Ph.D.
6
Transformers
A transformer is a device that couples two AC circuits magnetically rather than
through any direct conductive connection and permits a “transformation” of the voltage
and current between one circuit and the other (for example, by matching a highvoltage, low-current AC output to a circuit requiring a low-voltage, high current
source).
© Jiayu Chen, Ph.D.
7
Transformers
The ideal transformer consists of two coils that are
coupled to each other by some magnetic medium.
There is no electrical connection between the coils. The
coil on the input side is termed the primary, and that on
the output side the secondary.
Let us assume the primary coil is wound so that it has 𝑛1 turns, while the secondary has
𝑛2 turns. We define the turns ratio N as
𝑛2
𝑁=
𝑛1
𝐼1
𝑉2 = 𝑁𝑉1 𝑎𝑛𝑑 𝐼2 =
𝑁
An ideal transformer multiplies a sinusoidal input voltage by a factor of N and
divides a sinusoidal input current by a factor of N.
© Jiayu Chen, Ph.D.
8
Transformers
 If N is greater than 1, the output voltage is greater than the input voltage and the
transformer is called a step-up transformer.
 If N is less than 1, then the transformer is called a step-down transformer.
 A transformer with N = 1 is called an isolation transformer and may perform a
very useful function if one needs to electrically isolate two circuits from each other.
Note:
• Any DC currents at the primary will not appear at the secondary coil.
• An important property of ideal transformers is conservation of power
• We can easily verify an ideal transformer by
𝑉2
𝑆1 = 𝐼1 𝑉1 = 𝑁𝐼2 = 𝐼2 𝑉2 = 𝑆2
𝑁
© Jiayu Chen, Ph.D.
9
Transformers
Example:
We require a transformer to deliver 500 mA at 24 V from a 120 V rms line source. How
many turns are required in the secondary? What is the primary current?
Assume we have the first coil with 3000 turns,
We will have
𝑉1 𝑉2
=
𝑛1 𝑛2
𝑛1 𝑉2
24
𝑛2 =
= 3000 ×
= 600 𝑡𝑢𝑟𝑛𝑠
𝑉1
120
Knowing the turns, we can now compute the primary current,
𝑛1 𝐼1 = 𝑛2 𝐼2
𝑛2
600
𝐼1 = 𝐼2 =
× 500 = 100 𝑚𝐴
𝑛1
3000
© Jiayu Chen, Ph.D.
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Transformers
In many practical circuits, the secondary is tapped at two different points, giving rise
to two separate output circuits. The most common configuration is the center-tapped
transformer, which splits the secondary voltage into two equal voltages.
The most common occurrence of this type of transformer is found at the entry of a
power line into a household, where a high-voltage primary is transformed to 240 V,
and split into two 120-V lines.
110 V vs 220 V ?
© Jiayu Chen, Ph.D.
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Transformers
Impedance Reflection
A very common and rather general situation is that where an AC source, represented
by its Thévenin equivalent, is connected to an equivalent load impedance by means of
a transformer.
𝑉2
𝑎𝑛𝑑 𝐼1 = 𝑁𝐼2
𝑁
At the primary connection, we can have
V1 V2 /N
1 𝑉2
1
Z′ =
=
= 2 = 2 𝑍𝐿
I1
𝑁𝐼2
𝑁 𝐼2 𝑁
𝑉1 =
© Jiayu Chen, Ph.D.
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Transformers
Impedance Reflection
We will have
𝑉𝑂𝐶 = 𝑁𝑉𝑆
𝑉𝑆 1
𝐼𝑆𝐶 =
𝑍𝑆 𝑁
And apply Thévenin equivalent
𝑍 ′′ =
𝑉𝑂𝐶 𝑁𝑉𝑆
=
= 𝑁 2 𝑍𝑆
𝑉𝑆 1
𝐼𝑆𝐶
𝑍𝑆 𝑁
Thus the load sees the source impedance multiplied by a factor of 𝑁 2, this is called
impedance reflection.
© Jiayu Chen, Ph.D.
13
Transformers
Maximum power transfer
The power transfer efficiency is defined as:
𝑃𝑙𝑜𝑎𝑑
𝜂=
𝑃𝑠𝑜𝑢𝑟𝑐𝑒
Recall that the maximum power to be transferred form the source to the load, the
impedance must be matched (Proof was omitted)
Conclusion: We know 𝑍𝑆 = 𝑅𝑆 + 𝑖𝑋𝑠
𝑍𝐿 = 𝑍𝑆∗
That is
𝑅𝐿 = 𝑅𝑆 𝑋𝐿 = −𝑋𝑆
Where 𝑍𝑆∗ is the conjugate of the source impedance
When the load impedance is equal to the complex conjugate of the source
impedance, the load and source impedances are matched and maximum power is
transferred to the load.
𝑍𝐿
= 𝑍𝑆∗
2
𝑁
𝑅𝐿 = 𝑁 2 𝑅𝑠
𝑋𝐿 = −𝑁 2 𝑋𝑆
© Jiayu Chen, Ph.D.
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Transformers
Example:
Find the transformer turns ratio and load reactance that results
in maximum power transfer in the circuit.
𝑉𝑠 = 240∠0 𝑉; 𝑅𝑆 = 10 Ω; 𝐿𝑆 = 0.1 𝐻; 𝑅𝐿 = 400 Ω;𝜔 = 377
For the maximum power transfer, we will require
𝑅𝐿 = 𝑁 2 𝑅𝑆
Thus,
𝑅𝐿 400
𝑁2 =
=
𝑁 = 40 = 6.325
𝑅𝑆
10
Further, to cancel the reactive power we require that 𝑋𝐿 = −𝑁 2 𝑋𝑆
𝑋𝑆 = 𝜔 × 0.1 = 37.7
And
𝑋𝐿 = −40 × 37.7 = −1508
Thus the load reactance should be a capacitor with the value of
1
1
𝐶=−
=
= 1.76 𝜇𝐹
𝑋𝐿 𝜔 −1508 × 377
© Jiayu Chen, Ph.D.
15
Transformers
A substantial amount, 20% of the power from the source, is lost in the line if no
transformers are used.
Since power goes into heating up the line (line has resistance), and P ~ I2R, the loss in
the line can be reduced by reducing the current in the line, by using transformers.
Transformers at the source step up the voltage and step down the current, keeping the
power constant, before transmission. At the load, we use a step-down transformer to step
down the voltage and step up the current.
© Jiayu Chen, Ph.D.
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Transformers
Example:
If the transformer delivers 50 A at 110 V rms with a certain resistive load, what is the
power transfer efficiency between the source and the load?
𝑉𝑊 = 𝑛𝑉𝑠𝑒𝑐 = 220 𝑉 𝑟𝑚𝑠
The transformer primary voltage
The transformer primary current
The current through the inductance
𝐼𝑊 =
𝐼𝐿 =
𝑉𝑊 220
=
∠ − 90° = 11∠ − 90° 𝐴 𝑟𝑚𝑠
𝑗20
20
The source current is (add phasor currents)
The source power is
1
𝐼 = 25 𝐴 𝑟𝑚𝑠
𝑛 𝑠𝑒𝑐
𝐼𝑆 =
2
𝐼𝐿2 + 𝐼𝑊
= 27.3 𝐴 𝑟𝑚𝑠
𝑃𝑆 = 1Ω ∙ 𝐼𝑆2 + 𝑃𝑠𝑒𝑐 = 745 + 5.5 𝑘 = 6.245 𝑘𝑊
Power transfer efficiency
𝜂=
© Jiayu Chen, Ph.D.
𝑃𝑙𝑜𝑎𝑑
𝑃𝑠𝑒𝑐
5.5
=
=
= 0.88
𝑃𝑠𝑜𝑢𝑟𝑐𝑒 𝑃𝑠𝑜𝑢𝑟𝑐𝑒 6.245
17
3-phase AC circuits
Generation of Three Phase AC Electricity
The generation of electrical power is more efficient in 3phase systems.
A 3-phase system employs three balanced voltages, equal in
magnitude and differing in phase by 360°/3 = 120°.
http://www.fam-oud.nl/~plugsocket/Generator_3phase.html
http://ele.aut.ac.ir/~wind/en/tour/wtrb/syncgen.htm
http://www.launc.tased.edu.au/online/sciences/physics/3pha
se/threeph.htm
© Jiayu Chen, Ph.D.
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3-phase AC circuits
As the rotor rotates counterclockwise at 50 r/sec, its magnetic field cuts the armature
windings, thereby inducing in them the sinusoidal voltages. These voltages have
peaks at one-third of a period apart, or 120° apart, because the armature windings are
displaced by 120° in space. As a result, the alternator produces three voltages of the
same rms value, but with phase differences of 120°.
𝑉𝑎𝑎′ = 𝑉𝑚 𝑐𝑜𝑠𝜔𝑡
𝑉𝑏𝑏′ = 𝑉𝑚 cos(𝜔𝑡 − 120°)
𝑉𝑐𝑐′ = 𝑉𝑚 cos(𝜔𝑡 − 240°)
In HK: phase voltage = 220 V
line voltage = 380 V
© Jiayu Chen, Ph.D.
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3-phase AC circuits
A three-phase circuit generates, distributes, and uses energy in the form of three
voltages equal in magnitude and symmetric in phase. The three similar portions of a
three-phase system are called phases. .
Line voltage, abbreviated from line-to-line voltage
Because the voltage in phase aa' reaches its maximum first, followed by that in phase bb'
and then by that in phase cc', we say the phase rotation is abc.
The three voltages are said to be balanced voltages because they have identical
amplitude, Vm, and frequency, ω, and are out of phase with each other by exactly 120°.
𝒗𝑎𝑎′ + 𝒗𝑏𝑏′ + 𝒗𝑐𝑐′ = 0
The positive phase sequence is abc; the sequence acb is called the negative phase
sequence.
© Jiayu Chen, Ph.D.
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3-phase AC circuits
3-phase circuits and machines possess some unique advantages:
• The power transmitted in a three-phase circuit is constant or independent of time
rather than pulsating, as it is in a single-phase circuit.
• Three-phase motors start and run much better than do single-phase motors.
• The transmission of electrical power is more efficient in 3-phase systems
employing three sinusoidal voltages. For example, for the addition of one more
conductor, a three-phase supply provides 73% more power than a single-phase
supply. With a three-phase supply the voltage between two line or phase cables is
√3 = 1.73 times that between the neutral and any one of the line cables, i.e. 220 volts
x 1.73 =380 volts.
© Jiayu Chen, Ph.D.
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3-phase AC circuits
Y and Δ connections
Referring to the generator, there are six terminals and three voltages, va, vb and vc. We use
phasor notation and assume that each phase winding provides a source voltage in series
with a negligible impedance. Under these assumptions, there are two ways of
interconnecting the three sources.
Y connection
The Y connection selects terminals a', b', and c'
and connects them together as neutral. The
common terminal is called the neutral
terminal and is labeled n. The neutral terminal
may or may not be available for connection.
Balanced loads result in no current in a neutral
wire, and thus it is often not needed.
© Jiayu Chen, Ph.D.
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3-phase AC circuits
Δ connection
The Δ source connection is seldom used in
practice because any slight imbalance in
magnitude or phase of the three-phase voltages
will not result in a zero sum. The result will be
a large circulating current in the generator coils
that will heat the generator and depreciate the
efficiency of the generator.
© Jiayu Chen, Ph.D.
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3-phase AC circuits
Y-connected source
The line-to-line voltage, 𝑉𝑎𝑏 , of the Y-connected sources is
𝐕ab = 𝐕a − 𝐕b
𝑽𝒂 = 𝑉𝑚 ∠0° 𝑎𝑛𝑑 𝑽𝒃 = 𝑉𝑚 ∠ − 120°
𝑽𝒂𝒃 = 𝑉𝑚 − 𝑉𝑚 −0.5 − 𝑖0.866 = Vm 1.5 + i0.866
𝑽𝒂𝒃 = 3𝑉𝑚 ∠30°
And
𝑽𝒃𝒄 = 3𝑉𝑚 ∠ − 90°
𝑽𝒄𝒂 = 3𝑉𝑚 ∠ − 210°
Therefore, in a Y connection, the line-to-line voltage is 3 times the phase voltage and is
displaced 30° in phase. The line current is equal to the phase current.
The relation can also be obtained from the sine law,
𝑽𝒂𝒃
𝑽𝒂
=
𝑠𝑖𝑛120° 𝑠𝑖𝑛30°
© Jiayu Chen, Ph.D.
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3-phase AC circuits
Example:
The Y-connected three-phase voltage source 𝑽𝒄 = 120∠ − 240° V rms. Find the line-toline voltage 𝑽𝒃𝒄
Answer: 207.8∠ − 90° V rms
© Jiayu Chen, Ph.D.
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3-phase AC circuits
Y-to-Y circuit
This three-phase circuit consists of three parts: a three-phase source, a three-phase load,
and a transmission line.
4-wire Y-to-Y
The transmission line used to connect the
source to the load consists of four wires,
including a wire connecting the neutral node of
the source to the neutral node of the load.
3-wire Y-to-Y
The three-phase source is connected to the load
using three wires, without a wire connecting
the neutral node of the source to the neutral
node of the load.
© Jiayu Chen, Ph.D.
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3-phase AC circuits
4-wire Y-to-Y
Each impedance of the three-phase load is connected directly across a voltage source of the
three-phase source. Therefore, the voltage across the impedance is known, and the line
currents are easily calculated as:
𝐕𝐚
𝐕𝐛
𝐕𝐜
𝐈𝐚𝐀 =
, 𝐈𝐛𝐁 =
, and 𝐈𝐜𝐂 =
𝐙𝐀
𝐙𝐁
𝐙𝐂
The current in the wire connecting the neutral node of the source to the neutral node of the
load (the neutral wire) is:
𝐕𝐚 𝐕𝐛 𝐕𝐜
𝐈𝐍𝐧 = 𝐈𝐚𝐀 + 𝐈𝐛𝐁 + 𝐈𝐜𝐂 =
+
+
𝐙𝐀 𝐙𝐁 𝐙𝐂
The total average power delivered by the three-phase source to the three-phase load is
calculated by adding up the average power delivered to each impedance of the load.
P = PA + PB + PC
Where, eg.PA is the power absorbed by ZA. Since we can calculate IaA , we can then
calculate PA .
Denote the phase voltages of the Y-connected source as
𝐕𝐚 = Vp ∠0° V rms, 𝐕𝐛 = Vp ∠ − 120° V rms, and 𝐕𝐜 = Vp ∠120° V rms,
Notice that we are using effective values because the units of Vp are V rms
© Jiayu Chen, Ph.D.
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3-phase AC circuits
4-wire Y-to-Y
For a balanced load, 𝑍𝐴 = 𝑍𝐵 = 𝑍𝐶 = 𝑍 = 𝑍∠𝜃, the load is said to be a balanced load.
The currents are:
𝐕𝐚 Vp ∠0°
𝐕𝐛 Vp ∠ − 120°
𝐕𝐜 Vp ∠120°
𝐈𝐚𝐀 =
=
,𝐈 =
=
, and 𝐈𝐜𝐂 =
=
𝐙𝐀
𝑍∠𝜃 𝐛𝐁 𝐙𝐁
𝑍∠𝜃
𝐙𝐂
𝑍∠𝜃
Vp
Vp
Vp
𝐈𝐚𝐀 = ∠ − θ, 𝐈𝐛𝐁 = ∠ − θ − 120°, and 𝐈𝐜𝐂 = ∠ − θ + 120°
Z
Z
Z
The line currents have equal magnitudes and differ in phase by 120°.
The current in the neutral wire (which connects the neutral node of the source to the
neutral node of the load) is:
Vp
𝐈𝐍𝐧 = 𝐈𝐚𝐀 + 𝐈𝐛𝐁 + 𝐈𝐜𝐂 =
1∠0° + 1∠ − 120° + 1∠120° = 0
Z
Vp
𝐈𝐍𝐧 = 𝐈𝐚𝐀 + 𝐈𝐛𝐁 + 𝐈𝐜𝐂 =
1∠ − 𝜃 + 1∠ − 𝜃 − 120° + 1∠ − 𝜃 − 120° = 0
Z
This shows that the neutral wire has no current.
Remember that we are using eff or rms values. The average power delivered to the load is
V
V
V
V2
P = PA + PB + PC = V cos −𝜃 + V cos −𝜃 + V cos −𝜃 = 3 cos(𝜃)
Z
Z
Z
Z
The average power delivered to the load can be determined by calculating PA and
multiplying by 3
© Jiayu Chen, Ph.D.
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3-phase AC circuits
3-wire Y-to-Y
A 3-wire Y-Y circuit would not necessarily have the neutral current InN = 0, and the node
equation at node N needs to be solved. However, it can be shown that for a balanced 3wire Y-Y circuit, the neutral current InN = 0. The balanced three-wire Y-to-Y circuit acts
like the balanced four-wire Y-to-Y circuit. The line currents and the average power
delivered to the load are as given above.
The per-phase equivalent circuit
For balanced circuits, all 3 phases are similar, and we need only calculate one line current,
say IaA. The per-phase equivalent circuit consists of the voltage source and impedances in
the one of the phases.
The neutral nodes, n and N, are connected by a short circuit in the per-phase equivalent
circuit to indicate that VnN=0 in a balanced Y-to-Y circuit. The per-phase equivalent circuit
can be used to analyze either 3-wire or 4-wire balanced Y-to-Y circuits, but it can only be
used for balanced circuits.
© Jiayu Chen, Ph.D.
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3-phase AC circuits
The per-phase equivalent circuit
The Balanced Y-to-Y Circuit
© Jiayu Chen, Ph.D.
30
3-phase AC circuits
Example:
Determine the complex power delivered to the three-phase load of a four-wire Y-to-Y
circuit. The phase voltages of the Y-connected source are 𝑽𝒂 = 110∠0° V rms, 𝑽𝒃 =
110∠120° V rms, and 𝑽𝒄 = 110∠120° V rms. The load impedances are 𝒁𝑨 = 𝒁𝑩 =
𝒁𝑪 = 50 + 𝑖80 Ω
We need to calculate only one line current, 𝐼𝑎𝐴 , and the complex power 𝑆𝐴 , delivered to
only one of the load impedances, 𝑍𝐴 . The power delivered to the three-phase load is 3𝑆𝐴 .
We begin by calculating 𝐼𝑎𝐴 as
𝑽𝒂
110∠0°
𝑰𝒂𝑨 =
=
= 1.16∠ − 58° 𝐴 𝑟𝑚𝑠
𝒁𝑨 50 + 𝑖80
The complex power delivered to 𝒁𝑨 is
∗
𝑺𝑨 = 𝑰𝒂𝑨
𝐕𝐚 = 1.16∠ − 58° ∗ 110∠0° = 68 + 𝑖109 𝑉𝐴
The complex power delivered to the three-phase load is
3𝑺𝑨 = 204 + 𝑖325 𝑉𝐴
© Jiayu Chen, Ph.D.
31
3-phase AC circuits
Example:
Determine the complex power delivered to the three-phase load of a four-wire Y-to-Y
circuit. The phase voltages of the Y-connected source are 𝑽𝒂 = 110∠0° V rms, 𝑽𝒃 =
110∠120° V rms, and 𝑽𝒄 = 110∠120° V rms. The load impedances are 𝒁𝑨 = 50 +
𝑖80 Ω, 𝒁𝑩 = 𝑖50 Ω, and 𝒁𝑪 = 100 + 𝑖25 Ω
The line currents of an unbalanced four-wire Y-to-Y circuit are calculated using,
𝑽𝒂
110∠0°
𝑽𝒃 110∠ − 120°
𝑽𝒄 110∠120°
𝑰𝒂𝑨 =
=
, 𝑰𝒃𝑩 =
=
, 𝑎𝑛𝑑 𝑰𝒄𝑪 =
=
𝒁𝑨 50 + 𝑖80
𝒁𝑩
𝑖50
𝒁𝑪 100 + 𝑖25
So
𝑰𝒂𝑨 = 1.16∠ − 58° 𝐴 𝑟𝑚𝑠, 𝑰𝒃𝑩 = 1.2∠150° 𝐴 𝑟𝑚𝑠, 𝑰𝒄𝑪 = 1.07∠106° 𝐴 𝑟𝑚𝑠
The complex power delivered to 𝒁𝑨 is
∗
𝑺𝑨 = 𝑰𝒂𝑨
𝐕𝐚 = 1.16∠ − 58° ∗ 110∠0° = 68 + 𝑖109 𝑉𝐴
Similarly we can find 𝑺𝑩 and 𝑺𝑪
𝑺𝑩 = 2.2∠150° ∗ 110∠ − 120° = 𝑖242 𝑉𝐴
𝑺𝑪 = 1.07∠106° ∗ 110∠120° = 114 + 𝑖28 𝑉𝐴
The total complex power is
𝑺𝑨 + 𝑺𝑩 + 𝑺𝑪 = 182 + 𝑖379 𝑉𝐴
© Jiayu Chen, Ph.D.
32
3-phase AC circuits
Y-Δ transformations
The Δ-to-Y and Y-to-Δ transformations convert Δ-connected loads to equivalent Yconnected loads and vice versa.
© Jiayu Chen, Ph.D.
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3-phase AC circuits
Y-Δ Circuit
We use the phasor diagram. In a Δ connection, the line current is √3 times the phase current
and is displaced -30° in phase. The line-to-line voltage is equal to the phase voltage.
IA
IAB
Using sine law, sin120°
= sin30°
We will have
𝐈𝐚𝐀 = 3 𝐈
Or
IL = 3Ip
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3-phase AC circuits
Balanced 3-phase circuit
We have only two possible practical configurations for three-phase circuits, Y-to-Y and
Y-to-Δ and we can convert the latter to a Y-to-Y form. Thus, a practical three-phase
circuit can always be converted to the Y-to-Y circuit.
Y-to-Δ circuit
Y-to-Y circuit
𝑍Δ
𝑍𝑌 =
3
© Jiayu Chen, Ph.D.
The per-phase
equivalent circuit
35
3-phase AC circuits
Instantaneous and Average Power in a Balanced 3-Phase Load
One advantage of three-phase power is the smooth flow of energy to the load. Consider a
balanced load with resistance R. Then the instantaneous power is
𝑣𝑎𝑏 𝑡 2 𝑣𝑏𝑐 𝑡 2 𝑣𝑐𝑎 𝑡 2
p t =
+
+
R
R
R
Where vab (𝑡) = Vcosωt , and the other two-phase voltage have a phase of ±120°.
Furthermore,
cos2 αt = (1 + cos2α)/2
Therefore,
V2
p t =
1 + cos2ωt + 1 + cos2 ωt − 120° + 1 + cos2 ωt − 240°
2R
3V 2 V 2
=
+
[cos2ωt + cos 2ωt − 240° + cos(2ωt − 480°)]
2R 2R
The bracketed term is equal to zero for all time. Hence,
3V 2
p t =
2R
The instantaneous power delivered to a balanced three-phase load is a constant.
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Earthing System - Additional
Earthing System:
An electric connection to the general mass of earth, whose dimensions are
very large in comparison to the electrical system being considered.
The terms ‘Ground’ and ‘Grounding’ are synonymous with ‘Earth’ and
‘Earthing’ and are more prevalent in some countries like North America.
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37
Earthing System - Additional
Earth Electrode 1 :
• Electrical connect to earth
• Independent earth electrode for
each
system/equipment
• Maintain sufficient distance, 3.5m or twice of
driven length, from on another without
significant affect the potential of the others
Main Earthing Terminal 6 :
• Connect group of protective conductors to the
earth electrode 1 through earth conductor 2
Protective / bonding Conductor 3 :
• Exposed conductive parts
• Conductive part of electrical installation - not a
live part
• Extraneous conductive parts 4
• Conductive part not forming part of the electrical
installation
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38
Earthing System - Additional
Residual Current Devices (RCDs)
The RCD is a circuit breaker which continuously compares the current in the phase with that in
the neutral. The difference between the two will flow to earth, because it has left the supply
through the phase and has not returned in the neutral.
2
1
3
The purpose of the residual current device is to monitor the residual current and to
switch off the circuit quickly if it rises to a preset level.
When the amount of residual current, and hence of tripping current, reaches a predetermined level, the circuit breaker trips, opening the main contacts and interrupting
the circuit.
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Lightning System - Additional
Thunder Cloud
•
•
•
Unstable upper atmosphere before lightning generation
Positive charge and negative charge of thunder cloud separated
Light weight positive charge at higher altitude, heavier negative charge at the base of the
cloud
Lightning Discharge
•
•
•
•
May be cloud to cloud or cloud to ground
Faintly luminescent leader channel from
cloud toward to ground into many fingers
Rapid escalating electric field at ground
cause streamer move upward
Discharge commenced the ionized channel
completed at the junction of the leader and
streamer
Useful lighting projection system wiki:
http://www.electrical-installation.org/enwiki/General_rules_of_lightning_protection
© Jiayu Chen, Ph.D.
40
Lightning System - Additional
The Surge Protection Device (SPD) is a component of the electrical installation
protection system.
This device is connected in parallel on the power supply circuit of the loads that it has
to protect It can also be used at all levels of the power supply network. This is the
most commonly used and most efficient type of overvoltage protection.
Note: Surge Protection Devices (SPD) are used for electric power supply networks, telephone
networks, and communication and automatic control buses.
© Jiayu Chen, Ph.D.
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Thank You!
© Jiayu Chen, Ph.D.
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