1 The Transformer 1. 2 X XX X X B field into page L L No voltage on terminals.
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Transcript of 1 The Transformer 1. 2 X XX X X B field into page L L No voltage on terminals.
1
The Transformer
ELEC 3105 BASIC E&M AND POWER ENGINEERING
1
2
Review: Faraday’s law of induction
X
XXX X B field into page
L
𝒗 (𝒕 ) L
𝒗 (𝒕 )=− 𝝏𝚽𝝏 𝒕
=−𝝏 ∙𝑳𝟐
𝝏𝒕
Case A: If B and L and orientation, , constant
𝒗 (𝒕 )=− 𝝏 ∙𝑳𝟐
𝝏𝒕=𝟎 No voltage on terminals
3
Review: Faraday’s law of induction
X
XXX X B field into page
L
𝒗 (𝒕 ) L
𝒗 (𝒕 )=− 𝝏𝚽𝝏 𝒕
=−𝝏 ∙𝑳𝟐
𝝏𝒕
Case B: If B changes as a function of time with L and orientation, , constant
𝒗 (𝒕 )=𝝎𝑳𝟐𝑩𝒎𝒂𝒙𝐜𝐨𝐬 (𝝎𝒕)Voltage on terminals due to a changing magnetic field
=−𝐵𝑚𝑎𝑥sin (𝜔𝑡)
4
Review: Faraday’s law of induction
X
XXX
X B field into page
L
𝒗 (𝒕 ) L𝒗 (𝒕 )=− 𝝏𝚽
𝝏 𝒕=−
𝝏 ∙𝑳𝟐𝝏𝒕
Case C: If B and orientation, , constantwith L increasing as a function of time and
𝒗 (𝒕 )=𝐋𝐁 𝝏𝑴𝝏𝒕
Voltage on terminals “Linear motor”
“Rail Gun”
X
XXX
M
𝒗 (𝒕 ) L𝑣
XX
5
Review: Faraday’s law of induction
X
XXX
X B field into page
L
𝒗 (𝒕 ) L𝒗 (𝒕 )=− 𝝏𝚽
𝝏 𝒕=−
𝝏 ∙𝑳𝟐𝝏𝒕
Case D: If B and L constant with orientation, , changing as a function of time and
𝒗 (𝒕 )=𝝎𝑳𝟐𝐁𝒔𝒊𝒏 (𝝎𝒕 )
Voltage on terminals “Rotational motor”
“Generator”
𝜔 𝑡
∙=𝐵𝑐𝑜𝑠 (𝜔𝑡 )
Transformer
𝒗 (𝒕 )
Zg
I(t)
𝒗 ′ (𝒕 ) Zload
Loop on source side “Primary”
Loop on load side “Secondary”
(𝑡)
Transformer
N1 N2
corePrimary Secondary
Transformer optimize coupling,
perform transformation
Transformer N1 N2
corePrimary Secondary
Core: Designed such that as much produced in the primary passes to the secondary
Only a few field lines pass through
secondary
Iron core
Magnetic circuit guides field lines from primary to secondary
8
TRANSFORMER VOLTAGE RELATION
XArea A
One loop
=BAX
Area 2ATwo loops
=2BA
XArea 3A
Three loops
=3BAX
Area N1AN1 loops
=N1BA
ON PRIMARY SIDE
9
TRANSFORMER VOLTAGE RELATION
X
ON PRIMARY SIDE
𝒗 (𝒕 )=− 𝝏𝚽𝝏 𝒕
=𝑵𝟏[− 𝝏𝑩𝑨𝝏𝒕 ]Voltage produced for one loop
Voltage produced for N1 loops
1=BA Flux for each loop on primary
𝒗 (𝒕 )
N1 loops
=N1BA
Area N1A
𝒗𝟏 (𝒕 )=𝑵𝟏[− 𝝏 𝑩𝑨𝝏𝒕 ]
TRANSFORMER VOLTAGE RELATION
10
XArea A
One loop
’=BAX
Area 2A
Two loops
’=2BA
XArea 3A
Three loops
’=3BA
XArea N2A
N2 loops
’=N2BA
ON SECONDARY SIDE
TRANSFORMER VOLTAGE RELATION
11
XArea N2A
N2 loops
’=N2BA
1’=BA Flux for each loop on secondary
𝒗 ′ (𝒕 )=− 𝝏𝚽 ′𝝏𝒕
=𝑵𝟐[− 𝝏 𝑩𝑨𝝏𝒕 ]
𝒗 ′ (𝒕 )
Voltage produced for one loop
Voltage produced for N2 loops
ON SECONDARY SIDE
𝒗𝟐 (𝒕 )=𝑵𝟐[− 𝝏 𝑩𝑨𝝏𝒕 ]
12
TRANSFORMER VOLTAGE RELATION
X𝒗𝟏 (𝒕 )
=N1BA
Area N1A
12
XArea N2A
’=N2BA
𝒗𝟐 (𝒕 )
𝒗𝟐 (𝒕 )=𝑵𝟐[− 𝝏 𝑩𝑨𝝏𝒕 ]
𝒗𝟏 (𝒕 )=𝑵𝟏[− 𝝏 𝑩𝑨𝝏𝒕 ]combine
𝒗𝟏 (𝒕 )𝑵𝟏
=𝒗𝟐 (𝒕 )𝑵𝟐 Voltage transformation
Prefect flux coupling
TRANSFORMER CURRENT RELATION
X𝒗𝟏 (𝒕 )N1
13
XN2
𝒗𝟐 (𝒕 )𝒗𝟏 (𝒕 )𝑵𝟏
=𝒗𝟐 (𝒕 )𝑵𝟐
IDEAL TRANSFORMERNo power loss
Voltage transformation
Power (IN)
𝒊𝟏(𝒕 )
𝒊𝟐(𝒕 )
𝑷 𝒊𝒏=𝒗𝟏 (𝒕 )𝒊𝟏 (𝒕 ) Power (OUT)
𝑷 𝒐𝒖𝒕=𝒗𝟐 (𝒕 ) 𝒊𝟐 (𝒕 )combine
𝒊𝟐 (𝒕 )𝑵𝟏
=𝒊𝟏 (𝒕 )𝑵𝟐 Current transformation
TRANSFORMER IMPEDANCE RELATION
X𝒗𝟏 (𝒕 )N1
14
XN2
𝒗𝟏 (𝒕 )𝑵𝟏
=𝒗𝟐 (𝒕 )𝑵𝟐
IDEAL TRANSFORMERNo power loss
𝒊𝟏(𝒕 )
𝒊𝟐(𝒕 )
𝒊𝟐 (𝒕 )𝑵𝟏
=𝒊𝟏 (𝒕 )𝑵𝟐
Zload𝒗𝟐 (𝒕 )
Zeq looking into transformer
com
bin
e
𝒁 𝒆𝒒=(𝑵𝟏
𝑵𝟐)𝟐
𝒁 𝒍𝒐𝒂𝒅
TRANSFORMER IMPEDANCE RELATION
X𝒗𝟏 (𝒕 )N1
15
XN2
IDEAL TRANSFORMERNo power loss
𝒊𝟏(𝒕 )
𝒊𝟐(𝒕 )
Zload𝒗𝟐 (𝒕 )
𝒗𝟏 (𝒕 )N1
𝒊𝟏(𝒕 )
Zeq
Remove transformer𝒁 𝒆𝒒=(𝑵𝟏
𝑵𝟐)𝟐
𝒁 𝒍𝒐𝒂𝒅
16
Transformer Calculation Example
A 10-kVA; 6600/220 V/V; 50 Hz transformer is rated at 2.5 V/Turn of the winding coils. Assume that the transformer is ideal. Calculate the following: A) step up transformer ratio B) step down transformer ratio C) total number of turns in the high voltage
and low voltage coils D) Primary current as a step up transformer E) Secondary current as a step down
transformer.SOLUTION PROVIDED IN CLASS
17
Transformer Calculation Example
Find N1/N2 ratio such that maximum power transfer to the load is observed.
SOLUTION PROVIDED IN CLASS
𝒗 (𝒕 )
2Ω
I(t)
𝒗 ′ (𝒕 ) 32Ω
N1 “Primary” N2 “Secondary”
18
Maximum Power to Load
𝒗 (𝒕 )
Zs
I(t)
Zload ~𝐼=~𝑉~𝑍
=~𝑉
~𝑍𝑠+~𝑍 𝑙𝑜𝑎𝑑
Work in phasor domain
In general: ~𝑧 𝑠=𝑅𝑠+ 𝑗 𝑋 𝑠~𝑧 𝑙𝑜𝑎𝑑=𝑅 𝑙𝑜𝑎𝑑+ 𝑗 𝑋 𝑙𝑜𝑎𝑑
~𝐼=~𝑉
𝑅𝑠+𝑅 𝑙𝑜𝑎𝑑+ 𝑗 (𝑋 𝑠+𝑋 𝑙𝑜𝑎𝑑 )Then
Time average power to load 𝑃=|~𝐼|2
𝑅 𝑙𝑜𝑎𝑑
2Find
maximum
19
Maximum Power to Load
𝑃=|~𝑉|2
|√ (𝑅𝑠+𝑅 𝑙𝑜𝑎𝑑)2+(𝑋 𝑠+ 𝑋 𝑙𝑜𝑎𝑑)2|2𝑅𝑙𝑜𝑎𝑑
2Find
maximum
Step 1: Make as small as possible with respect to the complex “reactance” part 𝑋 𝑙𝑜𝑎𝑑=− 𝑋 𝑠
𝑃=|~𝑉|2
|√ (𝑅𝑠+𝑅 𝑙𝑜𝑎𝑑)2|2𝑅𝑙𝑜𝑎𝑑
2Find
maximum
20
Find maximum
Rs= 50Ω
|𝑉 𝑠|2=200
Rload= 50Ω
Maximum Power to Load~𝑧 𝑠=𝑅𝑠+ 𝑗 𝑋 𝑠
~𝑧 𝑙𝑜𝑎𝑑=𝑅𝑠− 𝑗 𝑋 𝑠
𝑃=|~𝑉|2
|√ (𝑅𝑠+𝑅 𝑙𝑜𝑎𝑑)2|2𝑅𝑙𝑜𝑎𝑑
2
Transformer Types
N1N2
corePrimary Secondary
Autotransformer
Ip Is
Starting motors ELEC 4602
Transformer Types
N1 N2
corePrimary Secondary
Single phase
Ip Is
Transformer Types
Single phase
Transformer Types
Maximum power transfer to the load
Transformer Types
N1
Ns1
corePrimary Secondary
Center tapped
Ip Is1
Ns2Is2
Vs1
Vs2 ground
With ground Vs1 180o out of phase with Vs2. Two phase household
Transformer Types
Transformer Types
N1
Ns1
corePrimary Secondary
Center tapped
Ip Is1
Ns2Is2
Vs1
Vs2 ground
With ground Vs1 180o out of phase with Vs2. Two phase household
Transformer TypesCenter tapped
Transformer Types
NA1 NA’2
Three phase
A A’
NB1 NB’2B B’
NC1 NC’2C C’
Interconnection
Transformer Types
Transformer Types
Transformer Types
Transformer Types n turns ratio
35
Three phase AC to DC converter
Topic of ELEC 3508: Power Electronics
36
Pow
er d
istrib
utio
n
37
Three phase power transformers
LabVolt Module used in ELEC 3508