Transcript SMJE 2103 Electiral Power System

SMJE 2103
Electrical Power System
3- Ph Power Apparatus
Power Apparatus
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Transmission Lines
Transformers
Circuit Breaker
Switchgear
Rotating Machine (Generator – Motor)
Transformer
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Introduction to power TX
Connection Types and symbol
Functionality
Application
Structure
Material
Equivalent Circuit
Testing
3-Phase Tx
Transformer
1.
2.
3.
4.
Step up
Step down
Distribution Tx
Special purpose (PT, CT)
Transformer
Transformer
-Connection Types and Symbol
Transformer
-Connection Types and Symbol
Transformer
-Connection Types and Symbol
Transformer
-Connection Types and Symbol
Transformer
-Connection Types and Symbol
Transformer
-Operation (Ideal x – Turn Ratio)-
v p t 
vs t 

Np
Ns
a
i p t 
1

is t  a
Transformer
-Operation (Ideal Tx – Power)Pin = Vp Ip cos θp
Pout = Vs Is cos θ
Pout = Vs Is cos θs
Also, Vs = Vp/a and Is = a Ip
θp - θ s = θ
Pout 
Vp
a
aI cos
p
Pout = Vp Ip cos θ = Pin
Also same for power Q and power S
Transformer
-Operation (Ideal x – Impedance)VL
ZL 
IL
VP
ZL '
IP
VS
ZL 
IS
Transformer
-Operation (Ideal x – Impedance)aVS
VP
2 VS
ZL ' 

a
IP IS / a
IS
ZL’ = a2 ZL
Transformer
-Operation (Ideal Tx – Analysis)-
(a) If the power system is exactly as described above in Figure (a), what will the voltage at the
load be? What will the transmission line losses be?
(b) Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line
and a 10:1 step-down transformer is placed at the load end of the line (Figure (b)). What will
the load voltage be now? What will the transmission line losses be now?
Transformer
-Operation (Real Tx - flux)-
- Flux generated
- The flux from each coil
- Average flux level
eind
d

dt
 

N
N
   i
i 1
eind
d
N
dt
Transformer
-Operation (Real Tx – Voltage Across)-
 
1
vP (t ) dt

NP
Primary Side:
Hence,
vP (t )  N P
Or rewritten,
P  M  LP
dP
d
d
 N P M  N P LP
dt
dt
dt
vP (t )  eP (t )  eLP (t )
Pri or sec voltage due
to mutual flux
d M
eP (t )  N P
dt
Relationship, Vp and Vs
eP (t ) dM eS (t )


NP
dt
NS
eP (t ) N P

a
eS (t ) N S
Transformer
-Operation (Real Tx – Magnetization)Magnetization current, iM
– current required to produce flux in the core.
Core-loss current, ih+e
– current required to compensate
hysteresis and eddy current losses.
iex  im  ihe
Transformer
-Application-
Transformer
-Application-
Transformer
-Application-
Transformer
-Structure-
Transformer
-Structure (Shell Form)-
Transformer
-Structure (Core Form)-
Transformer
-Structure-
Transformer
-Structure-
Transformer
-Material• Copper conductors
• Silicon
iron/laminated piece
of steel – eddy
current
• Oil
• Insulation materials
Transformer
-Equivalent Circuit-
Transformer
-Equivalent Circuit (simplified)-
Transformer
-Equivalent Circuit (simplified)-
Example/Tutorial
• The secondary winding of a transformer has a
terminal voltage of vs(t) + 282.8 sin 377t V. The turns
ratio of the transformer is 100:200 (a = 0.5). If the
secondary current of the transformer is is(t) =7.07 sin
(377t -36.87o) A, what is the primary current of this
transformer? What are its voltage regulation and
efficiency? The impedances of this transformer
referred to the primary side are
Transformer
-Testing-
To know the value of the inductance and resistance
Transformer
-Short circuit test-
Z SE
VSC

I SC
Power factor, PF = cos θ = PSC / VSC ISC
Z SE 
The series impedance ZSE
(lagging)
VSC 0 VSC

 
I SC    I SC
= Req + jXeq
= (RP + a2 RS) + j(XP + a2 XS)
Transformer
-Open circuit test-
Z SE
VSC

I SC
Power factor, PF = cos θ = PSC / VSC ISC
Z SE
The series impedance ZSE
(lagging)
VSC 0 VSC


 
I SC    I SC
= Req + jXeq
= (RP + a2 RS) + j(XP + a2 XS)
Transformer
-Equivalent Circuit (Example)The equivalent circuit impedances of a 20kVA, 8000/240V,
60Hz transformer are to be determined. The open circuit
test and the short circuit test were performed on the
primary side of the transformer, and the following data
were taken:
Open circuit test (primary)
VOC = 8000 V
IOC = 0.214 A
POC = 400 W
Short circuit test
VSC = 489 V
ISC = 2.5 A
PSC = 240 W
Find the impedance of the approximate equivalent
circuit referred to the primary side, and sketch the
circuit.
Example/Tutorial
A 1000 VA 230/115 V transformer has been tested to determine its
equivalent circuit. The results of the tests are shown below
All data given taken from the primary side of the transformer.
(a) Find the equivalent circuit of this transformer referred to lowvoltage side of the transformer.
(b) Find the transformer’s voltage regulation at rated conditions
and (1) 0.8 PF lagging, (2) 1.0 PF, (3) 0.8 PF leading.
(c) Determine the transformer’s efficiency at rated conditions and
0.8 PF lagging.
Transformer
-Impulse voltage test-
Transformer
- Voltage Regulation-
VR 
V S ,nlVS , fl
VS , fl
x 100%
Transformer
- EfficiencyPout

x100%
Pin
Pout

x100%
Pout  Ploss
Types of losses incurred in a
transformer:
Copper I2R losses
Hysteresis losses
Eddy current losses
VS I S cos

x100%
PCu  Pcore  VS I S cos
3-Phase Transformer
Transformers for 3-phase circuits can be constructed in two
ways:
connect 3 single phase transformers
Three sets of windings wrapped around a common core
The primaries and secondaries of any three-phase transformer
can be independently connected in either a wye (Y) or a delta (∆)
The important point to note when analyzing any 3-phase
transformer is to look at a single transformer.
The impedance, voltage regulation, efficiency, and similar
calculations for three phase transformers are done on a perphase basis.