Transcript EMT 113 Chapter 1

ELECTRICAL ENGINEERING
TECHNOLOGY
EMT 113/4
CHAPTER 1:
TRANSFORMER
Electrical Machines
Device that can convert either mechanical energy
to electrical energy or vice versa.
TRANSFORMER: Device that changes ac electric energy at
one voltage level to ac electric energy at another voltage level
through the action of magnetic field.
WHY NEED TRANSFORMER ??
Power efficiency over a long distance
• Power at high voltage is necessary to decrease
the lines losses.
• Power at low voltage is necessary to be used at
safe level in home appliances and most
equipments.
Principal purpose: convert ac power at one voltage
level to ac power of the same frequency at another
voltage level.
Construction
1.
2.
3.
The primary winding -the input winding, connected to
an ac power source
The secondary winding is the output winding.
The windings are not directly connected. Magnetic flux
present in ferromagnetic core connects the windings.
CORE
Windings wrapped
around the two sides of a
rectangular laminated
piece of steel.
TYPES
TYPES OF
OF
TRANSFORMER
TRANSFORMER
SHELL
Windings wrapped
around the center leg of a
three-legged laminated
core.
** both core is built up of thin laminations, which are electrically
isolated from each other to minimize eddy currents
Transformer : Operation
• AC voltage is applied to the primary winding (N1), result an AC
current.
• The AC primary current i1 sets up a time-varying magnetic flux φ in
the core. The flux links the secondary winding (N2) of the
transformer.
• Flux in the core induce electromagnetic forces (EMF) in N1 and N2
due to a time rate of change of φM (mutual flux), as stated by the
Faraday’s Law.
Faraday’s Law:
- Show that voltage,e is generated by a coil of wire when
magnetic flux in circuit changes for any reason
d
d
e 
dt
dt
…………………………(1.2)
Where,
e : instantaneous voltage induced by magnetic field (emf)
 : number of flux linkages between the magnetic field and
the electric circuit.
 : effective flux
• The voltage induced in the primary is nearly equal to the applied
voltage, and the voltage at the secondary winding also differs by only a
few percent from the voltage induced into that winding.
• Thus, the primary-to-secondary voltage ratio is essentially equal to the
ratio of the number of turn in the two windings. Turn ratio, a:
N1 V 1
a

N2 V2
•
………………………………(1.3)
According to Faraday’s law, the voltage induced is proportional to
the number of the turn in the windings, thus
e1 
d
N1
dt
and
e2 
d
N2
dt
………......……..(1.4)
• If the resistance is neglected, equation (1.2) becomes.
N 1 e1

N 2 e2
v1  e1  N
d
1(
);
dt
d
v 2  e 2  N 2( )
dt
…........…..(1.5)
• By neglecting the power losses,
Power in primary winding = Power in secondary winding
Substituting equation 1.1 into equation 1.4
d
d
eN
 N ( max sin 2ft )…………......…….(1.7)
dt
dt
Solve for this equation …then the rms value of the induced voltage is
given as
E
N max
2
 4.44 fN max
………………….(1.8)
f = frequency in hertz ; also known as the emf equation
Combine equations 1.3; 1.4;1.5 and 1.6,
N 1 v1 e1 i 2
a
  
N 2 v 2 e2 i1
……………………………..............................(1.9)
Turn ratio, a will classify the transformer :
If,
a > 1  Step down transformer
a < 1  Step up transformer
a = 1  Isolation Transformer
TRANSFORMER CLASSIFICATIONS:
Step-up transformers
– Permit higher voltage at secondary windings
– connected between the generator and transmission line.
– Turn ratio, a < 1
Step-down transformers
– connected between the transmission line and various electrical loads.
– Permit low voltage at secondary windings
– Turn ratio, a > 1
TRANSFORMER POLARITY - DOT CONVENTION:
• The dot convention appearing at one end of each winding tell the
polarity of the voltage and current on the secondary side of the
transformer.
• If the primary voltage is positive at the dotted end of the winding
with respect to the undotted end, then the secondary voltage will
be positive at the dotted end also. Voltage polarities are the same
with respect to the doted on each side of the core.
• If the primary current of the transformer flow into the dotted end
of the primary winding, the secondary current will flow out of the
dotted end of the secondary winding.
Theory of Transformer : Operation
Ideal Transformer
• Characteristics of an ideal transformer
- Windings with zero impedance
- Lossless
v1=e1
- Infinite permeability core
- 100% efficiency
•
Zero resistance result in zero voltage drops between the terminal
voltages and induced voltages
N 1 v1 e1 i 2
a
  
N 2 v 2 e2 i1
• In term of phasor quantities (or rms value), these quantities are
N 1 V 1 E1 I 2
a



N 2 V 2 E2 I1
v2=e2
•The power supplied to the transformer by the primary winding:
Where:
Pin = V1I1 cos 1
cos  = power factor
1
= the angle between the primary voltage and the primary current
•The power supplied by the transformer secondary winding:
Pout = V2I2 cos 2
Where:
2 = the angle between the secondary voltage and the secondary current
For an ideal transformer,
1=2 ;same power factor, then
Pout = Pin
Same with Q and S:
The reactive power (Q)
Qout = Qin = V1I1 sin  = V2I2 sin  (VAR)
The apparent (complex) power (S)
Sout = Sin = V1I1 = V2I2 (VA)
S = VI = P ± jQ
S = Apparent power, unit=VA.
P = Average power (also known as real power) , unit = Watt
Q = Reactive power, unit=VAR
Power factor also = ratio between real power and complex power
= P/S
Ideal Transformer : Impedance
Impedance - the ratio of the phasor voltage
across it to the phasor current flowing
through it.
VL
ZL 
IL
V V
Z  
I
I
s
2
s
2
L
Impedance of primary circuit
Vp aVS
2 VS
Z'L  
a
Ip IS / a
IS
Z'  a Z
Figure: a) Definition of Impedance;
b) Impedance through transformer
2
L
L
Non Ideal Transformer
• Two components of flux exist:
leakage flux - flux links only the primary or
secondary winding.
mutual flux - links both primary and
secondary windings
•
For a non-ideal/practical transformer, the
output power is less than the input power
because of losses.
• Practical transformer has :
a) Hysteresis losses
b) Eddy-current (or core) losses
c) Leakage flux losses
d) Resistive losses
In primary and secondary windings
Non Ideal Transformer
LOSSES IN TRANSFORMER:
• Copper losses – The resistive heating losses in the primary and secondary
windings
• Eddy Current Losses - The resistive heating losses in the core of the
transformer
• Hysteresis losses - Associated with the re-arrangement of the magnetic
domains in the core during each half cycle. They are complex, nonlinear function
of the voltage applied to the transformer.
• Leakage flux – the fluxes at primary and secondary which escape the core and
pass through only one of the transformer windings.
These losses that occurred in real transformers are modeled in the
transformer model
Exact Equivalent model
Approximate model
Non Ideal Transformer
NO LOAD
NoLoad
Power out = 0 (no load at secondary )
Power in = power out + power loss
Power loss = core loss + Cu loss
Cu = 0 (no load)
Power in = core loss
=Ic2Rc Watt
Non Ideal Transformer
UNDER LOAD
Symbol
Description
a
Turns ratio
E1
E2
Primary and secondary induced voltages
V1
V2
Primary and secondary terminal voltages
I1
I2
Primary and secondary currents
I
I0
No load current
R1
x1
Primary winding resistance and reactance
R2
x2
Secondary winding resistance and reactance
Im
Xm
Magnetizing current and reactance
Ic
Rc
Core loss current and resistance
The previous figures are accurate model of a transformer, but to
analyze practical circuits containing transformer, it is necessary to
refer to its primary side or to its secondary side because it is
necessary to convert the entire circuit to the equivalent circuit at a
single voltage level.
Referred to primary
EQUIVALENT CIRCUIT
Referred to
secondary
Non Ideal Transformer
APPROXIMATE EQUIVALENT CIRCUIT
Referred to primary
I2/a
Req_1 = R1 + a2R2
jXeq_1 = X1 + a2X2
Non Ideal Transformer
APPROXIMATE EQUIVALENT CIRCUIT
Referred to secondary
Req_2 = R1/a2 + R2
jXeq_2 = j(X1/a2 + X2)
Non Ideal Transformer
APPROXIMATE EQUIVALENT CIRCUIT
In some applications, the excitation branch may be neglected entirely without
causing error. In this cases, the equivalent circuits in the previous slides reduces to
these simples circuits:
Referred to primary
Referred to secondary
Transformer Characteristics
Transformer characteristics can be defined by:•
Efficiency
•
Voltage regulation.
Good transformers has high efficiency and low voltage
regulation.
Through short circuit and open circuit test, parameter, power loss,
efficiency and voltage regulation can be determined
Transformer Efficiency
Efficiency of a transformer is defined as :
P

X 100%
P
out
in
In practice, the efficiency of a transformer is about 97% or better
For a non-ideal transformer, the output power is less than the input
power because of losses.
2 types of losses – Copper losses (winding or I2R losses)
- Core losses (Hysteresis & eddy-current losses )
Pout

X 100%
Pout  Ploss
Ideally,
Pout  Pin
For non-ideal transformer, losses are considered, therefore
Pout  Pin  Plosses
Pin  Pout  Plosses
Then,
Pin  Plosses

X 100%
Pin
Pout

X 100%
Pout  Plosses
Pout

X 100%
Pout  Plosses

Pout
Pout  PCopper  PCore
X 100%
Transformer Voltage Regulation
•
•
Voltage regulation - a measure of the change in the terminal voltage of
the transformer with respect to loading.
Defined as
V.R

V2 noload  V2 fullload
V2 fullload
X 100%
In calculation of voltage regulation, the equivalent circuit can be referred to
primary and secondary side.
Good practice to have a small voltage regulation as possible.
For an ideal transformer, V.R = 0 %
Open Circuit Test
In open Circuit Test:
• one winding of the transformer is open while the other is excited by applying rated
voltage
•Test is conducted on the low voltage, LV side of the transformer (power, voltage
and current are measured from LV side)
•It means that voltage is applied at low voltage, LV side while the high voltage, HV
side is opened
Open Circuit Test
Open circuit test : Need to determine RcLV & XmLV
Poc  Voc/ RcLv
2
RcLV  Voc / Poc
2
XmLV  Voc / Im
Im  Ioc sin oc
Cosoc  Poc / VocIoc
2
Short Circuit Test
This test is designed to determine the winding resistances and leakage resistance
In short circuit test:
• one winding of the transformer is short while another winding is excited by
applying rated voltage
• test is conducted on high voltage, HV side with short circuit on the low voltage, LV
side
Short Circuit Test
Short circuit test : Need to determine ReHV & XeHV
Psc  Isc Re
Re HV  Psc / Isc
2
2
XeHV  Zscsin sc
Zsc  Vsc / Isc
Psc  VscIsc cos sc
cos sc  Psc / VscIsc
Simplified Circuit
Simplified Circuit referred to HV
(primary)
Simplified Circuit referred to LV
(secondary)
Transformer Phasor Diagram
•
•
•
•
What is phasor diagram?
– A sketch of phasor voltages and currents in the transformer.
Why need it?
– Easiest way to determine the effect of the impedances and the current phase
angles on the transformer voltage regulation.
Vs is assumed to be at an angle of 0 degree, and all other voltages and currents are
compared to that references.
A transformer phasor diagram is presented by applying Kirchhoff's Voltage law to
the transformer equivalent circuit and an equation will be as follows.
Transformer Phasor Diagram
Lagging Power Factor
Unity Power Factor
Leading Power Factor
Transformer Application
1.
2.
3.
4.
Voltage level adjustment (step-up and step-down transformers).
Voltage and current measurement.
Isolation for safety (isolation transformers)
Impedance matching (for maximum power transfer from the source to
the load)
Review:
• Unit transformer – Connected the output of a generator and used to step the voltage up
to transmission levels (110kV)
• Substation transformer – Connected at the other end of the transmission line which
steps the voltage down from transmission level to distribution levels (2.3 to 34.5 kV).
• Distribution transformer – Takes the distribution voltage and steps it down to the final
voltage (110V, 208V,220V,etc)
• Special-purpose transformers :
» Potential transformer
» Current transformer
Transformer Application
Elements of Power Transmission and Distribution System
Three Phase Transformer
Almost all the major power generation and distribution systems in the world
today are three-phase ac system.
Two ways of constructing transformer of three-phase circuit;
(i) Three single phase transformers are connected in three-phase bank.
Three Phase Transformer
(ii) Make a three-phased transformer consisting of three sets of windings
wrapped on a common core.
The three-phased transformer on a common core is preferred because it is
lighter, smaller, cheaper and slightly more efficient.
Three Phase Transformer
Advantages three phase transformer
Less material for the same three phase power and voltage ratings
Smaller/lighter because all connection are made internally
Less cost to manufacture
Less external wiring
It has slightly better efficiency
Disadvantages three phase transformer
Failure of one phase puts the entire transformer out of service.
Three Phase Transformer
The primary and secondary windings of the three phase transformer may be
independently connected in either a WYE (Y) or DELTA () connection. As a result,
four types of three phase transformers are commonly use.
Winding Connections
Applications
Wye-Wye (Y-Y) Connection
Seldom used because of possible voltage
unbalances and problems with third harmonic
voltages
Wye-Delta (Y-Δ) Connection
Used in stepping down or as in distribution
transformer. Frequently used to interconnect high
voltage networks (240 kV/345 kV). The delta
winding filters the 3rd harmonic, equalizes the
unbalanced current and provides a path for ground
current
Delta-Wye (Δ-Y) Connections
Step up transformer in generation station
Delta-Delta (Δ-Δ) Connection
Used for medium voltage (15kV)
Example 1
1. A 250 kVA, 11000V/400V, 50Hz single phase transformer has
80 turns on the secondary. Assume the transformer is ideal.
Calculate:
a) The values of the primary and secondary currents
b) The number of primary turns
c) The maximum value of flux, фm.
Solution Example 1
SOLUTION:
a) I1
= S1/V1
= 250kVA/11kV = 22.73A
I2
= V1I1/V2
= (11kV)(22.73A) / 400V = 625 A
b)
a = N1/N2 = I2/I1 = V1/V2
a = V1/V2 = 11000V/400V = 27.5
N1 = aN2 = (27.5)(80) = 2200 turns
c)
E1
E1
Фmax
= 4.44 fN1 фmax
= V1
= V1/4.44 fN1
= 11000V / (4.44)(50Hz)(2200) = 22.5mWb
Example 2
2. A single phase power system consists of a 480-60Hz generator
supplying a load Zload= 4 + j3Ω through a transmission line of impedance
Zline = 0.18 + j0.24Ω. Answer the following question about the system. If
the power is exactly as described in the Figure1, what will the voltage at
the load be? What will the transmission line losses be?
Solution Example 2
Solution:
IL = V/ (Zline + Zload)
Zline + Zload
IL
= (0.18 + j0.24Ω) + (4 + j3Ω)
= 4.18 + j3.24 = 5.29 < 37.8o
= 480 < 0o / (5.29< 37.8o) = 90.74 < -37.8o A
VLoad
= IL x ZLoad
= (90.74 < -37.8o) (5 < 36.9o) = 4537 < - 0.9o V
Line Losses = ILine2 x RLine Power in Watt only consider Resistance
= (90.74) 2 (0.18) = 1482 W
Example 3
3. Test are performed on a 1ф, 10 kVA, 2200/220V, 60 Hz transformer
and the following results are obtained.
Open Circuit test (HV
side open)
Short Circuit test (LV side
shorted
VOC = 220V
VSC = 150V
IOC = 2.5A
ISC = 4.55A
POC = 100W
PSC = 215 W
a)Derive the parameters for the approximate equivalent circuits
referred to the low voltage and high voltage side.
b)Express the excitation current as a percentage of the rated
current
c)Determine the power factor for the no load and short circuit test
Solution Example 3
Solution a):
OPEN CIRCUIT TEST: HV side opened
(means calculate at LV side)
RCLV
= VOC2 / POC
= (220)2 / 100 = 484 Ω
XMLV
Im
POC
Cos θOC
θOC
= VOC / Im
= IOC sin θOC
= VOC IOC cos θOC
= POC / VOC IOC
= 100 / (220 x 2.5)
= 0.182
= 79.52o
Im
= IOC sin θOC
= 2.5 sin 79.52o = 2.458 A
XMLV
= VOC / Im
= 220 / 2.458 = 89.5 Ω
SHORT CIRCUIT TEST : LV side shorted (calculate at HV side)
ISC2
PSC
REHV
=
REHV
= PSC / ISC2
= 215/ (4.55) 2 = 10.38 Ω
ZEHV
= VSC / ISC
= 150 / 4.55 = 32. 97 Ω
XEHV
Ω
= √ (ZEHV) 2 – (REHV) 2
= √ (32.97) 2 – (10.38) 2 = 31.29
From the calculation above, we get these calculated values:
RCLV = 484 Ω
XMLV = 89.5 Ω
REHV = 10.38 Ω
XEHV = 31.29 Ω
Equivalent circuit refer to HV (primary side)
Value for REHV and XEHV remain the same.
RCLV and XMLV need to convert to HV (primary side)
Refer to primary side : multiply a2
RCHV
= a2 RCLV
= (V1/V2)2 RCLV
= (2200/220) 2 x 484 = 48.4 kΩ
XMHV
= a2 XMLV
= (V1/V2)2 XMLV
= (2200/220) 2 x 89.5 = 8.95 kΩ
Equivalent Circuit (refer to primary side)
Equivalent circuit refer to LV (secondary side)
Value RCLV and XMLV remain the same.
REHV and XEHV need to convert to LV (secondary side)
Refer to primary side : multiply 1/a2
RELV
= (1/a2) REHV
= (V2/V1)2 RELV
= (220/2200) 2 x 10.38
= 0.1038 Ω
XELV = (1/a2) XELV
= (V2/V1)2 XEHV
= (220/2200) 2 x 31.29
= 0.3129 Ω
Equivalent Circuit (refer to secondary side)
b) Excitation current as percentage of the rated current
IOC = IO + I1
I1 =0 because primary opened
Irated
= S/V2
= 10k / 220 = 45.45 A
IOC = 2.5 A
Percentage
= ( IOC / Irated ) x 100%
= (2.5 / 45.45) x 100%
= 5.50 %
c) Power factor for the no load and short circuit test
Power factor at no load : Power factor at no load (refer to open circuit test!!!)
Cos θOC = POC / VOC IOC
= 100 / 220 (2.5) = 0.182
Power factor short circuit :
Cos θSC = PSC / VSC ISC
= 215 / 150 (4.55) = 0.315
END OF CHAPTER 1