l7 course notes
TRANSCRIPT
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8/3/2019 L7 Course Notes
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1Sharif University of
Technology
Electric Machines
52331
Energy conversion I
Lecture 7:
Topic 2: Transformers & its performance (S. Chapman, ch. 2)
- Introduction
- Types and Construction of Transformers.
- Ideal Transformer.- Theory of operation of real single-phase transformers.
- The Equivalent Circuit of a Transformer.
- The Per-Unit System of Measurement.
- Transformer voltage regulation and efficiency.- Autotransformers.
- Three phase transformers.
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8/3/2019 L7 Course Notes
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2Sharif University of
Technology
Electric Machines
52331
Voltage regulation:
Series transformer impedance is the origin ofoutput voltage variations even if theinput voltage is constant (load regulation)
VR : Voltage regulation (percent)Vs,nl : No-load Secondary voltageVs,fl : full-load Secondary voltage (usually rated or 1 p.u)
Using Simple equivalent circuit and p.u values:
Transformer voltage regulation
%100,
,,x
V
VVVR
flS
flSnlS
%100,,
,,,x
V
VVVR
puflS
puflSpup = Series equivalent impedance
limits short circuit current
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8/3/2019 L7 Course Notes
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3Sharif University of
Technology
Electric Machines
52331
Transformer voltage regulation
Load effect :
Using phasor diagram:
Vp,pu
= Vs,pu
+ Req
Is,pu
+jXeq
Is,pu
considering Vs as the reference phasor:
Vs
Vp
IsReqsIs
jXeqsIs
Vs
Vp
Is ReqsIs
jXeqsIs
Vs
Vp
Is
ReqsIs
jXeqs
Is
Lagging power factor
Leading power factor
Unity power factor
Note: Voltage regulation can benegative for capacitive loads.
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8/3/2019 L7 Course Notes
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4Sharif University of
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Electric Machines
52331
Transformer Tap changer
Transformer Tap changer is a device to change the voltage ratio forregulation ofthe output voltage:
On load tap changer: can change the voltage ratio while transformer is loaded
Off load tap Changer: changes the voltage ratio of a no-load transformer
Example: having 4 taps of 2.5% in HV for a 13200/480 V transformer means:
+5.0% tap : 13860/480 V+2.5% tap : 13530/480 V
Nominal : 13200/480 V
-2.5% tap : 12870/480 V-5.0% tap : 12540/480 V
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Electric Machines
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Transformer Efficiency
Transformer efficiency:
Considering power Losses:
Pcu: copper (winding RI2) Loss (proportional to load current).
Pcore: Core (Eddy current and Hystersis) loss (proportional to voltage).
Show that for a given output voltage and power factor, maximum efficiencyhappens if Pcu = Pcore
%100xP
P
in
out
%100cos
cos%100 x
IVPP
IVx
PP
P
SScoreCu
SS
lossout
out
++=
+=
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Electric Machines
52331
Transformer voltage regulation and efficiency
Example:
The equivalent Circuit parameters of a 15kVA, 2300/230 V transformer referred to low
voltage side are:
Req = 0.0445
, Xeq = 0.0645
, Rc = 1050
, XM = 110
A- Calculate the full load voltage regulation @ 0.8 lagging power factor, 1.0 power
factor and 0.8 leading power factor.
B- What is the efficiency at full load with 0.8 lagging power factor.
Solution:
The full-load current in the LV side is: Is,n = Sn / Vs,n = 15000/230 = 65.2 A
For 0.8 lagging power factor: Is
Vp = Vs + Is(Req+jXeq)=
o9.362.65 =
oj 4.085.234)0645.00445.0(9.362.650230 =++
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8/3/2019 L7 Course Notes
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Electric Machines
52331
Therefore: VR=
For unity power factor: Is
Vp = Vs + Is(Req+jXeq) =
Therefore: VR=
For 0.8 leading power factor: Is
Vp = Vs + Is(Req+jXeq) =
Therefore: VR=
%1.2%100230
23085.234=
o02.65 =
oj 04.194.232)0645.00445.0(2.650230 =++
%28.1%100230
23094.232=
o9.362.65 =
oj 27.185.229)0645.00445.0(9.362.650230 =++
%062.0%100
230
23085.229=
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Electric Machines
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B- efficiency at full load with 0.8 lagging power factor.
Losse can be calculated as:
Pcu = Req Is2 = 0.0445(65.2)2 = 189 W
Pcore
= (Vp
)2 / Rc = (234.85)2 / 1050 = 52.5 W
Output power is : Pout = VsIscos = 230 65.2 cos (36.9o) = 12000 W
= Pout
/ (Pout
+ Ploss
) 100%= 12000 / (12000+189+52.5) 100%= 98.03%
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Electric Machines
52331
Autotransformers
For voltage ratios close to 1 it is usual to use autotransformer instead ofTransformer.
For voltage ratios close to one, autotransformer is much smaller in weightand volume compared to transformer.
For voltage ratios close to one, kVA rating of autotransformer is muchhigher than transformer used to make it.
Problem: no electrical isolation!
c
SE
SE
C
SE
c
SE
c
N
N
I
I
N
N
V
V
=
=
C: Common coilSE: Series coil
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Electric Machines
52331
Step-up Autotransformers
VL = VC
VH = VC + VSE
IL = IC + ISE
IH = ISE
SEC
C
SEC
C
H
L
NN
N
VV
V
V
V
+=
+=
STR = ISEVSE
SATR = IH VH = ISE VH
SE
SEC
SE
H
N
NN
V
V +==
TR
ATR
S
S>>1 If voltage
ratio is close to 1
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Electric Machines
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Autotransformer rating
C
SEC
L
H
SE
SEC
SE
H
N
NN
V
V
NNN
VV
+=
+==
TR
ATR
SS
STR : Transformer or each winding kVA (SW)
SATR : Autotransformer kVA (SIO)
Example: For a 5000 kVA, 110 kV / 138 kV we can use a 110 / 28 transformer:
NC: 110 , NSE = 28
STR = SW = 28 / (28 + 138) * 5000 = 1015 kVA
Think about Series impedance
of Autotransformer comparedto Transformer
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Electric Machines
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Variable Autotransformer (VARIAC)
Used as Variable AC voltage Source
I2No
V1
NC
IC
IL
Iload
V2
NSE
Z
Y
Sliding
terminal
Output voltage varies between 0 to (NSE+NC)/NCV1