LESSON GUIDE: 15 HOURS: 2 TITLE

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Transcript LESSON GUIDE: 15 HOURS: 2 TITLE 

Transformer
Agus Purwadi, Qamaruzzaman
& Nana Heryana
Laboratorium Penelitian Konversi Energi Elektrik
Institut Teknologi Bandung
Transformer
A transformer is a device that
convert ac electric energy at one
voltage level at another voltage
level.
Transformer are important to
modern life
1.
• Thomas A. Edison (1882)
Power Distribution System : New York-USA ; 120 Vdc
Problems : Losses & Voltage Drop = inefficient
Example :
1000 A
Rs=0,05 ohm/10km
50 V
120 V
120 kVA
70V
50 kVA
70 kVA
70 kVA
  58%
High Voltage
1A
Rs=0,05 ohm/10km
0,05 V
70 kV
~
70 kVA
70kV
0,05 VA
70 kVA
70 kVA
Transformer
are important to modern power system
Low voltage
High Voltage
Real Application on Power System
PM
G
380 V/20 kV
TR
TR
M
20 kV/380 V
Transformer
1. Power Frequency , 50/60 Hz
2. Radio frequency, > 30 kHz
1. Power Transformer
2. Instrument Transformer
2. Type and Construction Of Transformer
3. The Ideal Transformer
IP(t)
IS(t)
+
VP(t)
+
NP
NS
-
vP (t ) N P

a
vS (t ) N S
VS(t)
-
Turn ratio a
a
NP
NS
The Ideal Transformer
IP(t)
IS(t)
+
VP(t)
+
NP
NS
-
N P .iP (t )  N S .I S (t )
VS(t)
-
Turn ratio a
iP (t ) 1

iS (t ) a
The Ideal Transformer
IP(t)
IS(t)
+
+
VP(t)
NP
NS
VS(t)
-
-
In term of phasor quantities
VP
a
VS
IP 1

IS a
Power in ideal Transformer
Power
input
ideal
PIN  VP I P cos P
Power
input
POUT  VS I S cos
P  S  
Turn ratio
equation
POUT  VS I S cos S
VS  VP / a and I S  aIP
POUT
VP

aI P cos  PIN
a
Power in ideal Transformer
Active
power
POUT  VP I P cos  PIN
QOUT  VS I S sin  VP I P sin  QIN
reactive
power
Apparent
power
SOUT  VS I S  VP I P  S IN
Evaluasi
• Untuk memahami konsep trafo ideal berikan ke
mahasiswa, contoh sederhana misalnya :
– Cara mendapatkan tegangan sekunder y volt
jika tegangan primernya x volt.
– Arus sekunder b ampere pada tegangan y
volt, cari arus primer a ampere pada
tegangan X volt.
– Ambil nilai x dan y yang bedanya sangat
besar
– Soal pada sistem 1 fasa dulu saja.
Impedance Transformer
Impedance =
ratio of the phasor voltage across it ti the phasor current
IL
VL
ZL
VL
ZL 
IL
Impedance Transformer
IP
VP
Z 'L 
IP
+
+
VP
VS
-
if
VP  aVS
IS
VS
ZL 
IS
ZL
-
IP  IS / a
aVS
VP
2 VS
Z 'L 

a
IP IS / a
IS
Z 'L  a2 Z L
Theory of operation of real singlephase transformers

IP(t)
IS(t)
+
+
VP(t)
NP
NS
VS(t)
-
-
The basis transformer operation can be derived
from Faraday’s law

Flux
linkage
N
   i  
i 1

N
eind
d

dt
eind
d
N
dt
EMF equation of a transformer
Faraday law of
electromagnetic induction
  max cost
e  Nmax sin t
RMS Value
E
Nm
2
 4,44 fNm
d
e  N
dt
Transformer Losses
• Copper (I2R) losses
• Eddy current losses
• Hysteresis losses
• Leakage flux
The equivalent circuit of a transformer
IP
RP
XP
XS
+
VP
RS
IS
+
Rc
jXM
NP
NS
-
VS
Ideal
transformer
the model of a real transformer
The equivalent circuit of a transformer
IP
IS /a
+
+
RP
j
j XP
VP
Rc
a2 XS
a2
RS
aVS
jXM
-
Referred to primary
a IP
j XP /a2
IS
j XS
+
+
RP
VP / a
/a2
RS
RC /a2
VS
j XM /a2
-
Referred to secondary
Phasor diagram corresponding – referred to
primary
V1
j I1X1
E
E 1=a 2
I1R1
Im
I0

I2/a
IX
aI2R2
I1
2
ja
2
aV2
2
IC
The equivalent circuit of a transformer
IP
Req
+
j Xeq
IS /a
+
Ih+e
Im
aV
VP
Rc
jXM
-
Approximate transformer models
S
Test on transformers
• Open-Circuit Test
wattmeter
Ip (t)
A
+
v(t)
~
V
Vp (t)
-
transformer
• Result : VOC IOC POC
Open-Circuit Test Result
Conductance of the core-loss
resistor
1
GC 
RC
1
XM
Susceptance of the magnetizing inductor
BM 
Admittance
1
1
YE 
j
RC
XC
YE  GC  jBM
Magnitude admittance
I OC
YE 
VOC
Open-Circuit Test Result
Power Factor
Power Factor angle
Admittance
POC
PF  cos 
VOC .I OC
POC
  cos
VOC .I OC
1
I OC
YE 
 
V OC
I OC
YE 
  cos1 PF
V OC
Test on transformers
Short-Circuit Test
wattmeter
Ip (t)
IS (t)
A
+
v(t)
~
V
Vp (t)
-
transformer
Result : VSC ,ISC ,POC
Short-Circuit Test Result
Z SE
Series impedance
VSC

I SC
Power Factor of the current
PSC
PF  cos 
VSC .I SC
Lagging - Current angle (-),Impedance angle (+)
Therefore,
Z SE
VSC 0 0 VSC



I SC    I SC
PSC
  cos
VSC .I SC
1
Short-Circuit Test Result
Series impedance
Z SE  Req  jX eq
Admittance
Z SE  (RP  a 2 RS )  j( X P  a 2 X S )
2.2
The equivalent circuit impedances of a 20 kVA, 8000 / 240 V, 60 Hz
transformer are to be determined. The open circuit test were performed
on the primary side of the transformer, and the following data were
taken :
Open-circuit test
(on primary)
VOC = 8000 V
Shots-circuit test
(on primary)
VOC = 489 V
IOC = 0,214 A
IOC = 2,5 A
POC = 400 W
POC = 240 W
Ip
+
Vp
-
Req
38.4 
I h e
Im
Rc
159 k
j Xm
j 38.4 k
j X eq
Is
a
j 192 
+
a Vs
-
The per-unit system of measurement
actual quantity
Quantity per unit 
base valueof quantity
Pbase , Qbase or Sbase  Vbase I base
In single-phase system :
Z base
Vbase

I base
Ybase
I base

Vbase
Z base
(Vbase ) 2

S base
Voltage Regulation & Efficiency
Voltage Regulation
VR 
VS ,no load  VS , full load
VS , full load
x100%
Efficiency
Pout

x100%
Pin
Pout

x100%
Pout  Ploss
Autotransformer
IH
Step-down autotransformer
ISE
VC N C

VSE N SE
NSE
IL
VH
IC
NC
VL
NC

VH N SE  N C
VL
N C I C  N SE I SE
VL  VC
VH  VC  VSE
I H  I SE
I L  I SE  I C
I L N SE  N C

IH
NC
Power rating -autotransformer
S IO N SE  NC

SW
N SE
SIO =Input and Output apparent powers
SW
= apparent power in the transformer windings
For example, a 5000 kVA autotransformer connecting a 110 kV
system to a 138 kV system would have an NC/NSE turn of ratio of
110 : 28. Such an autotransformer would actually have
windings rated at :
SW  S IO
N SE
28

5000kVA  1015kVA
NC  N SE 28  110
Three-phase transformer
NP1
NS1
NS2
NP2
NP3
NS3
A three-phase transformer bank composed of independent transformer
NP1
NP2
NP3
NS1
NS2
NS3
A three-phase transformer wound on a single three-legged core
Three-phase transformer connections
•
•
•
•
Wye-wye (Yy)
Wye-delta (Yd)
Delta-wye (Dy)
Delta-delta (Dd)
Wye-wye (Yy) Connection
V P
V S
a
VLP

VLS
3 V P
3 V S
a
Wye-delta (Yd) Connection
VLP

VLS
3 V P
V S
VLP
 3a
VLS
Delta-wye (Dy) Connection
V P
VLP

VLS
3 V S
VLP
a

VLS
3
Delta-delta (Dd) Connection
VLP V P

a
VLS V S
A 50 kVA 13.800 / 208 V, Dy distribution transformer
has a resistance of 1 percent and reactance of 7
percent per unit.
(a) What is the transformer’s phase impedance
referred to the high-voltage side?
(b) Calculate this transformer’s voltage regulation at
full load and 0,8 PF lagging, using the calculated
high-side impedance.
(c) Calculate this transformer’s voltage regulation
under the same conditions, using the per-unit
system.
Instrument transformers
• Potential transformer (PT)
• Current transformers (CT)
CT
PT
S
V
W
A