power 6
TRANSCRIPT
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Electrical Power Engineering (2)
Faculty ofEngineering Tanta University
Dr. Ahmed Mohamed Azmy
Department of Electrical Power and Machine EngineeringTanta University - Egypt
Code: EP2207
Lecture: 4 Tutorial: 4 Total: 8
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valuebase valueactualvalue unitPer
base
basebase V
SI
base
2base
base
basebase S
VIVZ
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Impact of TransformersI1
V1
-
+V2
-
+ I2
base11
2base2 V
NNV
basebase2base1 SSS
base12
1base2 I
NNI
pu2pu1 VV
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ExampleA transformer is rated at 13.8kV/138kV/4.157kV with rated power of 50MVA/40MVA/10MVA. Terminations are as followings:13.8 kV winding: 13.8 kV Source138 kV winding: 35 MVA load, pf = 0.866 lag.4.157 kV winding: 5 MVA load, pf = 0.866 leading
Using Sbase = 10 MVA, and voltage ratings as bases, (a) Solve for the primary current, power, and power factor (b) Draw the pu equivalent circuit
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Example
1
3
2
---
+
++
V1
V2
V3
+
-
I1
I2
I3
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Solution
...... up 30 53s up 531035s 22
...... up 30 50s up 50105s 33
......... up 00 01VVV up 01813813V 3211
...*
up 30 53VSI
2
22
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Solution
.. 30 5.0*
3
33 up
VSI
..... up4233.775 30 5030 53III 321
....* up423 7753IVS 111
lagging 0.9177pf MVA7537107753S1 ,..
AI 27360138.010775.31
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Solution
1
3
2
--
+
+V1
V2
V3
+
-
I1
I2 =
I3 =
3053.
3050.
001.
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Steps of converting actual single-phase system to the
per unit quantities:
• Choose MVA base MVAbase for the whole system
• Choose a kilo voltage base kVbase for one section
• Calculate the voltage base in other sections in the
network using the transformation ratio of transformers
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• Calculate the pu value using the general equation
valuebase valueactual valueunitPer
• Calculate the impedance base and current base using MVAbase and kVbase in each section
base
2base
base
basebase S
VIVZ
base
basebase V
SI
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To convert the actual three-phase system to the per unit quantities, perform the following steps:
• Choose an apparent power base "VA3-base" for the
three phase for all parts in the system
• Choose a line-to-line voltage base VL-base for one
section in the circuit
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• Calculate the impedance base and current base in
terms of VAbase and Vbase in each section of the power
system using the following equations:
baseL
basebase V
SI
3
3
• Calculate the voltage base in other sections in thenetwork using transformation ratio of transformersconsidering the method of winding connection
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For star connection:
base3
2baseL
base SVZ
For delta connection:
base
baseLbase S
VZ
3
2
3
• Calculate the pu value using the general equation
valuebase valueactualvalue unitPer
ZZY 31
YZZ 3
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Changing the base
If the values of the parameters are given for certainbases and it is required to calculate the values forother bases, a certain relation can be usedThis relation modifies the old value according to theration between new and old bases according to thefollowing relation:
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Example:For the network shown, the data are:
Draw the impedance diagram for the networkassuming bases of 20 MVA and 11 kV in the G1 area.
G1 G2
G3
J 10 WTr1 Tr2
Tr3
Tr1: 20 MVA, 11/110 kV, x=J0.12 p.u.G1: 20 MVA, 11 kV, x=J0.4 p.u.Tr2: 10 MVA, 6/110 kV, x=J0.15 p.u.G2: 10 MVA, 6 kV, x=J0.35 p.u.Tr3: 5 MVA, 3/110 kV, x=J0.1 p.u.G3: 5 MVA, 3 kV, x=J0.3 p.u.
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SolutionMVAb = 20 MVA
kVb-1 = 11 kV
kVb-II = 110 kV
kVb-III = 6 kV
kVb-IV = 3 kV
XL = 10 * 20 / (110)2 = j 0.017
Xg1 = 0.4 * 20 / 20 (11/11)2 = j 0.4
Xg2 = 0.35 * 20 / 10 (6/6)2 = j 0.7
G1 G2
G3
J 10 WTr1 Tr2
Tr3
III
IV
III
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Solution
B
J 0.017
J 0.3
C
J 0.4
A
J 0.12
J 0.017 J 0.017
J 0.4 J 1.2 J 0.7
Xtr2 = 0.15 * 20 / 10 (110/110)2 = j 0.3
Xtr3 = 0.1 * 20 / 5 (110/110)2 = j 0.4
2.1j33
5203.0X
2
3g
12.0j110110
202012.0X
2
1tr
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Generator G1 200 MVA, 20 kV, Xd = 15%Generator G2 300 MVA, 18 kV, Xd = 20%Generator G3 300 MVA, 20 kV, Xd = 20%Transformer T1 300 MVA, 220Y/22 kV, Xd = 10%
Transformer T2Three single-phase units each rated 100 MVA, 130 Y / 25 D kV, X = 10%
Transformer T3 300 MVA, 220/22 kV, X = 10%
Example: The power system shown in the figure has the following ratings:
G1 G2
G3
J 75 WT1
T2
T3
Y Y
J 75 W
Y YJ 50 W
Y YY
The transmission linereactances are asindicated in thefigure. Draw thereactance diagram inper unit choosing thegenerator 3 circuit asthe base.
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MVAbase = 300 MVAH.V.S. of transformer T2 is Y connectedIts rated line to line voltage is √3 x 130 = 225 kVL.V.S. is connected in Its line voltage is 25 kVVb, III = 20 kV, Vb, IV = 20 x 220/22= 200 kV
G1 G2
G3
J 75 WT1
T2
T3
Y Y
J 75 W
Y Y
J 50 W
Y YY
I II
III
IV
22 220
22 220 225 25
Vb, I = 200 x 22/220= 20 kV
Vb, II = 200 x 25/225= 22.22 kV
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Vb, I = 200 x 22/220= 20 kV
Vb, II = 200 x 25/225= 22.22 kV
Generator G1
Generator G2
Generator G3
Transformer T 1
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Transformer T 2
Transformer T 3
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