lesson 8: transformer theory and operation notes/lesson 8_et332b.pdf2/4/2016 2 ideal transformer...
TRANSCRIPT
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LESSON 8: IDEAL
TRANSFORMER THEORY AND
OPERATION
ET 332b Ac Motors, Generators and Power Systems
1 Lesson 8_et332b.pptx
Learning Objectives
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After this presentation you will be able to:
Explain how an ideal transformer operates
Find the voltages and currents on both sides of an ideal
transformer using the turns ration
Reflect impedances through a transformer
Identify and compute the no-load currents that flow in a
non-ideal transformer
Draw the no-load circuit model of a non-ideal transformer.
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Ideal Transformer Action
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Principle: Stationary coils, time varying flux due to ac current flow. Flux
produced by one coil must link to other coil to induce voltage
Lentz's Law
dt
dNe m
11
dt
dNe m
22
Induced voltage
has opposite
polarity from
source
Ideal Transformer Action
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For sinusoidal sources
maxpp fN44.4E
maxss fN44.4E
maxs
maxp
s
p
fN44.4
fN44.4
E
E
s
p
s
p
N
N
E
E
Dividing the above equations gives:
Voltage relationship
for Ideal transformer
Voltage ratio equals
the turns ratio Where: E’p = voltage induced in the primary (V)
E’s = voltage induced in the secondary (V)
Np = turns in the primary coil
Ns = turns in the secondary coil
Primary Secondary
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Assumptions for Ideal Transformer
Operation
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1) All flux produced in the primary coil links to the secondary coil
2) no core losses due to hysteresis or eddy currents
3) no power losses
4) permeability is infinite (no saturation no magnetizing )
5) windings have zero resistance
6) no current required to magnetize the iron core
For ideal transformer
s
p
s
p
s
p
V
V
N
N
E
Ea
Where: a = turns ratio
Vp = nameplate rated primary voltage (higher V)
Vs = nameplate rated secondary voltage (lower V)
E’p = induced primary voltage
E’s = induced secondary voltage
Ideal Transformer Equations
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sp
s
p
s
p
EaE
N
N
E
Ea
Voltage Ratio
The turns ratio is a
scalar. Introduces no
phase shift
Apparent Power balance
sp
sspp
SS
IEIE
No power losses in
idea transformer
Current Ratio Current ratio is the
inverse of the voltage
ratio a
1
I
I
s
p
sp Ia
1I
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Ideal Transformer Equations-
Impedance Transforms
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Impedances Reflected Through Ideal Transformers
Load impedance as
seen from primary
side of transformer Zin
p
p
inI
EZ
By Ohm’s Law
Zload
s
sload
I
EZ
Write Es and Is in terms of primary values ps
p
s IaI a
EE
2
p
p
p
p
p
p
s
s
a
1
I
E
Ia
1
a
E
Ia
a
E
I
E
in
2
load2inload ZaZ a
1ZZ
Load impedance is
increased when
viewed from primary
side
Ideal Transformer Equations-
Impedance Transforms
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Derive equation when impedances are connected to the primary side and
viewed from the secondary side.
p
p
loadI
EZ
Zin Zload a
II EaE s
psp
s
sin
I
EZ
Write primary values in terms of
secondary and substitute in the
Zload equation.
s
s2
s
ss
s
p
p
loadI
Ea
I
aEa
a
I
Ea
I
EZ
in2
load2
inload Za
Z aZZ
Generally : Moving impedance from secondary
to primary multiply by a2. Moving from
primary to secondary, divide by a2.
s2
p2
sp Za
Z aZZ
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Ideal Transformer Calculations
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Example 8-1: A 25 kVA, 7200 - 240/120 center-tap single phase
transformer operates at rated voltage. It supplies a single phase load that has
an equivalent impedance of 7.2 +36.9o ohms. Assume Ideal operation and
find:
a.) turns ratio
b.) secondary current
c.) primary current
d.) load Z as seen from primary side
e.) PT, ST, QT, and Fp
240 V
120 V
120 V
7200 V
Example 8-1 Solution (1)
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a) For ideal transformers
Ans
b) Secondary current Use Ohm’s law to find Is
Ans
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Example 8-1 Solution (2)
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c) Find the primary current
Ans
d) Find the input impedance as seen from the primary side
Ans
Example 8-1 Solution (3)
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e) Find the power and the power factor of the load
Using secondary side quantities
Ans
Using primary side quantities
Ans
Power equal on both sides of ideal transformer
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Example 8-1 Solution (4)
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Now find the power factor and the active and reactive powers
Ans Ans
Ans
Ideal Transformer Calculations
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Example 8-2: 300 kVA 2400-120, 60 Hz single phase
transformer operates at 2300 volts on the primary side. It supplies
115 kVA to a load that has a power factor of 0.723 lagging.
Assume idea operation and find:
a.) secondary voltage at operating voltage
b.) secondary current
c.) impedance of the load as seen on the secondary side
d.) impedance of the load as seen on the primary side
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Example 8-2 Solution (1)
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a) Find secondary voltage at operating voltage
Use rated values to find turns ratio
Ans
b) Find secondary current at operating voltage
Power is equal on both
sides of ideal transformers Ans
Example 8-2 Solution (2)
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c) Find load impedance seen on secondary side
Angle between V and I.
Change sign for
impedance angle
Ans
Next find impedance angle
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Example 8-2 Solution (3)
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d) Find load impedance seen on primary side of transformer
Reflecting impedance from secondary to primary-multiply by a2.
Ans
Non-Ideal Operation-No-load
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Practical transformers draw current with no load connected to secondary winding.
Current caused by two non-ideal conditions: power losses and core magnetization
Control hysteresis losses - use alloy steels designed
for magnetic circuits
Control eddy current losses - laminate core,
insulate laminates
Active power losses Hysteresis losses - power losses due to
repeated change in magnetic polarity. It takes
more mmf (NI) to demagnetize core in one
direction than the other.
Eddy currents - ac currents induced in iron core
due to changing magnetic field
Active power loss
Control
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Non-Ideal Operation-No-load
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A finite amount of current is necessary to drive mutual flux between coils.
Permeability is finite so reluctance is finite Some NI = F needed.
Al
R m = mutual flux R = reluctance R
INm
R
2NL In terms of inductance so core has inductance with associated
inductive reactance
Define above as the magnetizing inductance with associated magnetizing
reactance Xm
No-Load Circuit Model
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VT = the primary voltage I0 = exciting current
Ife = core-loss component IM = magnetizing component
Rfe = resistance that represents the core losses
Xm = inductive reactance that represents the core magnetizing L
VT
Io
Rfe XM
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No-Load Circuit Model
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Model equation using phasors
fe
Tfe
R
VI
M
TM
jX
VI
Mfeo III
Mfeo IjII Add current magnitudes at 90 degrees
No-load apparent power oTM IVS
fe
2
Tfe
fe
2
Tfe
P
VR
R
VP
M
TM
I
VX Model parameter formulas
Core loss resistance Magnetizing reactance
No-Load Transformer Example
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Example 8-3: Computing the values of magnetizing reactance
and core loss resistance. A 50 kVA 7200-240 V, 60 Hz single
phase transformer is operating with no load. With the primary
connected to a 7200 V system, it draws 248 W and has a
power factor of 0.187 lagging. Find:
a) the exciting current and its components
b) the magnitudes of magnetizing reactance, XM and core loss R
c) Repeat parts a and b if the transformer is energized from
the secondary (low voltage) side.
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Example 8-3 Solution (1)
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a) Find current Io
SM = magnetizing apparent power
Ans
Example 8-3 Solution (2)
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b) Find the value of core loss resistance and magnetizing reactance
Ans
Ans
Ans
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Example 8-3 Solution (3)
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c) Find same parameters on secondary side
Power constant through transformer
Ans
Example 8-3 Solution (4)
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q
Compute Rfe and XM