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Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi 1
Per-unit analysis
 The per unit value of any quantity is defined as the ratio of the
actual value of the quantity to the base value expressed as
decimal.
 Per unit value=
𝐴𝑐𝑡𝑢𝑎𝑙 𝑉𝑎𝑙𝑢𝑒
𝐵𝑎𝑠𝑒 𝑉𝑎𝑙𝑢𝑒
 Base Impedance 𝑍 𝑏=
𝑘𝑉𝑏
2
𝑀𝑉𝐴 𝑏
 𝑋 𝑝.𝑢 𝑛𝑒𝑤=𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛*
𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤
𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛
*
𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛
𝑘𝑉 𝑏 𝑛𝑒𝑤
2
2
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
Per phase representation of various components
of a power system.
3
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
Impedance Diagram
 The impedance diagram on single phase basis under balanced
operating conditions can be drawn form the one line diagram.
Assumptions:
 Single phase transformers equivalents are shown as ideal
transformers with transformer impedance indicated on
appropriate side.
 Magnetizing reactance’s of the transformer have been neglected.
 Generators are represented as voltage sources and with series
resistance and inductive reactance.
 The shunt capacitances are also neglected.
 Loads are represented by resistance and inductive reactance.
 Neutral grounding impedances are neglected.
4
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
Reactance Diagram
The reactance diagram can be obtained from impedance diagram
if we omit all static loads, all resistances, shunt branches of
transformer and capacitance of transmission lines in the
impedance diagram.
Sample system:
5
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
Impendence & Reactance Diagram
6
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
ExamplesObtain the per unit impedance and reactance diagram of the power system shown in fig.(13)
G1: 30MVA, 10.5kV, X”=1.6 ohm
G2: 15MVA, 6.6kV, X”=1.2 ohm
G3: 25MVA, 6.6kV, X’=0.56 ohm
T1: 15MVA, 33/11kV, X=15.2Ω/ph on H.T side.
T2: 15mva, 33/6.2kV, X=16Ω/ph on H.T side.
Transmission Line:20.5Ω/ph
Load A: 15MW, 11kV, 0.9p.f lagging
Load B: 6.2kV, 34MW+j20.07MVAR
7
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
SOLUTION:
Base MVAnew=30MVA Base KVnew=33kV
Transmission Line:
Actual impedance=j20.5Ω/ph.
P.u impedance=
𝑍 𝑎𝑐𝑡
𝑘𝑉𝑏
2 *MVAb =
𝑗20.5
332 *30=j0.565p.u
Zp.u=j0.565p.u
Transformer T1(H.T Side)
Zp.u =
𝑍 𝑎𝑐𝑡
𝑘𝑉𝑏
2 *MVAb =
𝑗15.2
332 *30= j0.418p.u
KVbnew(LT Side)= KVb(HT Side)×
𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
Generator 1:(L.T Side) = 33×
11
33
= 11kV
Zp.u =
𝑍 𝑎𝑐𝑡
𝑘𝑉𝑏
2 *MVAb =
𝑗1.6
112 × 30=j0.397 p.u
8
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
Load A:
P=15MW, 0.9 p.f lagging
SL=15 𝑐𝑜𝑠−1
0.9 = 13.5+j6.54MVA
Actual impedance = ZL =
𝑘𝑉 𝑏
2
𝑆 𝐿
∗ =
112
13.5−𝑗6.54
= 7.259+j3.52Ω
Zp.u =
𝑍 𝑎𝑐𝑡
𝑘𝑉 𝑏
2 *MVAb =
7.259+𝑗3.52
112 × 30 = 1.8+j0.87 p.u
Transformer T2(H.T Side):
Zp.u =
𝑍 𝑎𝑐𝑡
𝑘𝑉 𝑏
2 *MVAb =
𝑗16
332 × 30 = j0.44 p.u
Generator 2:(L.T Side)
KVbnew(LT Side)= KVb(HT Side)×
𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
= 33×
6.2
33
= 6.2kV
Zp.u =
𝑍 𝑎𝑐𝑡
𝑘𝑉 𝑏
2 *MVAb =
𝑗1.2
6.22 × 30=j0.936
9
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
Generator 3
Zp.u =
𝑍 𝑎𝑐𝑡
𝑘𝑉 𝑏
2 *MVAb =
𝑗0.56
6.22 × 30 = j0.437 p.u
Load B:
Actual impedance = ZL =
𝑘𝑉 𝑏
2
𝑆 𝐿
∗ =
𝑘𝑉 𝑏
2
34−𝑗20.07
= 0.838+j0.495
Zp.u =
𝑍 𝑎𝑐𝑡
𝑘𝑉 𝑏
2 *MVAb = =
0.838+𝑗0.495
6.62 × 30 =0.654+j0.386
10
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
The single line diagram of an unloaded power system shown in fig. along with components data
determine the new perunit values and draw the reactance diagram. Assume 50MVA and
13.8kV as new base on generator1.
11
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
Solution:
Base MVAnew=50MVA Base KVnew=13.8kV
Reactance of G1:
𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 *
𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤
𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛
*
𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛
𝑘𝑉 𝑏 𝑛𝑒𝑤
2
= j0.2×
50
20
×
13.8
13.8
2
=j0.5p.u
Reactance of T1:(L.T Side)
𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 *
𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤
𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛
*
𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛
𝑘𝑉 𝑏 𝑛𝑒𝑤
2
= j0.1×
50
25
×
13.8
13.8
2
=j0.2p.u
Reactance of Transmission Line1:
KVbnew(HT Side)= KVb(LT Side)×
𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
=13.8×
220
13.8
= 220kV
Zp.u =
𝑍 𝑎𝑐𝑡
𝑘𝑉𝑏
2 *MVAb =
𝑗80
2202 ×50 = j0.0826p.u
Reactance of Transmission Line2:
Zp.u =
𝑍 𝑎𝑐𝑡
𝑘𝑉 𝑏
2 *MVAb =
𝑗100
2202 ×50 = j0.1033p.u
12
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
Reactance of Transformer T3(H.T Side)
𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 *
𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤
𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛
*
𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛
𝑘𝑉 𝑏 𝑛𝑒𝑤
2
= j0.1×
50
35
×
220
220
2
=j0.1429 p.u
Reactance of Generator G3:
Generator is connected on LT side of transformer.
KVbnew(LT Side)= KVb(HT Side)×
𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
= 220×
22
220
= 22kV
𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 *
𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤
𝑀𝑉𝐴 𝑏 ,𝑔𝑖𝑣𝑒𝑛
*
𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛
𝑘𝑉 𝑏 𝑛𝑒𝑤
2
= j0.2×
50
30
×
20
22
2
=j0.2755 p.u
Reactance of Transformer T2(H.T Side)
T2 is a three phase transformer bank formed using three single phase transformer with
voltage rating 127/18kV. In this the HT side is star connected and LT side is delta connected.
∴ Voltage ratio of line voltage of 3-phase transformer =
𝟑×𝟏𝟐𝟕
𝟏𝟖
=
𝟐𝟐𝟎
𝟏𝟖
kV
𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 *
𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤
𝑀𝑉𝐴 𝑏 ,𝑔𝑖𝑣𝑒𝑛
*
𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛
𝑘𝑉 𝑏 𝑛𝑒𝑤
2
=j0.2×
50
30
×
220
220
2
= j0.2755 p.u
13
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
Reactance of Generator G3:
Generator is connected on LT side of transformer.
KVbnew(LT Side)= KVb(HT Side)×
𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1
= 220×
18
220
= 18 kV
𝑋𝑝.𝑢 𝑛𝑒𝑤 =𝑋𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 *
𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤
𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛
*
𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛
𝑘𝑉 𝑏 𝑛𝑒𝑤
2
= j0.2×
50
30
×
18
18
2
= j0.333 p.u
14
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi
References
1. Hadi Saadat, ‘Power System Analysis’, Tata McGraw Hill Education
Pvt. Ltd., New Delhi, 21st reprint, 2010.
2. Kundur P., ‘Power System Stability and Control, Tata McGraw Hill
Education Pvt. Ltd., New Delhi, 10th reprint, 2010.
3. Pai M A, ‘Computer Techniques in Power System Analysis’, Tata Mc
Graw-Hill Publishing
Company Ltd., New Delhi, Second Edition, 2007.
4. J. Duncan Glover, Mulukutla S. Sarma, Thomas J. Overbye, ‘ Power
System Analysis & Design’, Cengage Learning, Fifth Edition, 2012.
5. Olle. I. Elgerd, ‘Electric Energy Systems Theory – An Introduction’, Tata
McGraw Hill Publishing Company Limited, New Delhi, Second Edition,
2012.
6. C.A.Gross, “Power System Analysis,” Wiley India, 2011.
7. M.Jeraldin Ahila “POWER SYSTEM ANALYSIS”,Lakshmi Publications,
Chennai, Tenth Edition 2016.
15
Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by,
Mrs.s.Revathi

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Per unit analysis

  • 1. Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi 1
  • 2. Per-unit analysis  The per unit value of any quantity is defined as the ratio of the actual value of the quantity to the base value expressed as decimal.  Per unit value= 𝐴𝑐𝑡𝑢𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 𝐵𝑎𝑠𝑒 𝑉𝑎𝑙𝑢𝑒  Base Impedance 𝑍 𝑏= 𝑘𝑉𝑏 2 𝑀𝑉𝐴 𝑏  𝑋 𝑝.𝑢 𝑛𝑒𝑤=𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛* 𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤 𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛 * 𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛 𝑘𝑉 𝑏 𝑛𝑒𝑤 2 2 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 3. Per phase representation of various components of a power system. 3 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 4. Impedance Diagram  The impedance diagram on single phase basis under balanced operating conditions can be drawn form the one line diagram. Assumptions:  Single phase transformers equivalents are shown as ideal transformers with transformer impedance indicated on appropriate side.  Magnetizing reactance’s of the transformer have been neglected.  Generators are represented as voltage sources and with series resistance and inductive reactance.  The shunt capacitances are also neglected.  Loads are represented by resistance and inductive reactance.  Neutral grounding impedances are neglected. 4 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 5. Reactance Diagram The reactance diagram can be obtained from impedance diagram if we omit all static loads, all resistances, shunt branches of transformer and capacitance of transmission lines in the impedance diagram. Sample system: 5 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 6. Impendence & Reactance Diagram 6 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 7. ExamplesObtain the per unit impedance and reactance diagram of the power system shown in fig.(13) G1: 30MVA, 10.5kV, X”=1.6 ohm G2: 15MVA, 6.6kV, X”=1.2 ohm G3: 25MVA, 6.6kV, X’=0.56 ohm T1: 15MVA, 33/11kV, X=15.2Ω/ph on H.T side. T2: 15mva, 33/6.2kV, X=16Ω/ph on H.T side. Transmission Line:20.5Ω/ph Load A: 15MW, 11kV, 0.9p.f lagging Load B: 6.2kV, 34MW+j20.07MVAR 7 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 8. SOLUTION: Base MVAnew=30MVA Base KVnew=33kV Transmission Line: Actual impedance=j20.5Ω/ph. P.u impedance= 𝑍 𝑎𝑐𝑡 𝑘𝑉𝑏 2 *MVAb = 𝑗20.5 332 *30=j0.565p.u Zp.u=j0.565p.u Transformer T1(H.T Side) Zp.u = 𝑍 𝑎𝑐𝑡 𝑘𝑉𝑏 2 *MVAb = 𝑗15.2 332 *30= j0.418p.u KVbnew(LT Side)= KVb(HT Side)× 𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 Generator 1:(L.T Side) = 33× 11 33 = 11kV Zp.u = 𝑍 𝑎𝑐𝑡 𝑘𝑉𝑏 2 *MVAb = 𝑗1.6 112 × 30=j0.397 p.u 8 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 9. Load A: P=15MW, 0.9 p.f lagging SL=15 𝑐𝑜𝑠−1 0.9 = 13.5+j6.54MVA Actual impedance = ZL = 𝑘𝑉 𝑏 2 𝑆 𝐿 ∗ = 112 13.5−𝑗6.54 = 7.259+j3.52Ω Zp.u = 𝑍 𝑎𝑐𝑡 𝑘𝑉 𝑏 2 *MVAb = 7.259+𝑗3.52 112 × 30 = 1.8+j0.87 p.u Transformer T2(H.T Side): Zp.u = 𝑍 𝑎𝑐𝑡 𝑘𝑉 𝑏 2 *MVAb = 𝑗16 332 × 30 = j0.44 p.u Generator 2:(L.T Side) KVbnew(LT Side)= KVb(HT Side)× 𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 = 33× 6.2 33 = 6.2kV Zp.u = 𝑍 𝑎𝑐𝑡 𝑘𝑉 𝑏 2 *MVAb = 𝑗1.2 6.22 × 30=j0.936 9 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 10. Generator 3 Zp.u = 𝑍 𝑎𝑐𝑡 𝑘𝑉 𝑏 2 *MVAb = 𝑗0.56 6.22 × 30 = j0.437 p.u Load B: Actual impedance = ZL = 𝑘𝑉 𝑏 2 𝑆 𝐿 ∗ = 𝑘𝑉 𝑏 2 34−𝑗20.07 = 0.838+j0.495 Zp.u = 𝑍 𝑎𝑐𝑡 𝑘𝑉 𝑏 2 *MVAb = = 0.838+𝑗0.495 6.62 × 30 =0.654+j0.386 10 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 11. The single line diagram of an unloaded power system shown in fig. along with components data determine the new perunit values and draw the reactance diagram. Assume 50MVA and 13.8kV as new base on generator1. 11 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 12. Solution: Base MVAnew=50MVA Base KVnew=13.8kV Reactance of G1: 𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 * 𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤 𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛 * 𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛 𝑘𝑉 𝑏 𝑛𝑒𝑤 2 = j0.2× 50 20 × 13.8 13.8 2 =j0.5p.u Reactance of T1:(L.T Side) 𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 * 𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤 𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛 * 𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛 𝑘𝑉 𝑏 𝑛𝑒𝑤 2 = j0.1× 50 25 × 13.8 13.8 2 =j0.2p.u Reactance of Transmission Line1: KVbnew(HT Side)= KVb(LT Side)× 𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 =13.8× 220 13.8 = 220kV Zp.u = 𝑍 𝑎𝑐𝑡 𝑘𝑉𝑏 2 *MVAb = 𝑗80 2202 ×50 = j0.0826p.u Reactance of Transmission Line2: Zp.u = 𝑍 𝑎𝑐𝑡 𝑘𝑉 𝑏 2 *MVAb = 𝑗100 2202 ×50 = j0.1033p.u 12 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 13. Reactance of Transformer T3(H.T Side) 𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 * 𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤 𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛 * 𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛 𝑘𝑉 𝑏 𝑛𝑒𝑤 2 = j0.1× 50 35 × 220 220 2 =j0.1429 p.u Reactance of Generator G3: Generator is connected on LT side of transformer. KVbnew(LT Side)= KVb(HT Side)× 𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 = 220× 22 220 = 22kV 𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 * 𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤 𝑀𝑉𝐴 𝑏 ,𝑔𝑖𝑣𝑒𝑛 * 𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛 𝑘𝑉 𝑏 𝑛𝑒𝑤 2 = j0.2× 50 30 × 20 22 2 =j0.2755 p.u Reactance of Transformer T2(H.T Side) T2 is a three phase transformer bank formed using three single phase transformer with voltage rating 127/18kV. In this the HT side is star connected and LT side is delta connected. ∴ Voltage ratio of line voltage of 3-phase transformer = 𝟑×𝟏𝟐𝟕 𝟏𝟖 = 𝟐𝟐𝟎 𝟏𝟖 kV 𝑋 𝑝.𝑢 𝑛𝑒𝑤 =𝑋 𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 * 𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤 𝑀𝑉𝐴 𝑏 ,𝑔𝑖𝑣𝑒𝑛 * 𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛 𝑘𝑉 𝑏 𝑛𝑒𝑤 2 =j0.2× 50 30 × 220 220 2 = j0.2755 p.u 13 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 14. Reactance of Generator G3: Generator is connected on LT side of transformer. KVbnew(LT Side)= KVb(HT Side)× 𝐿𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 𝐻𝑇 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑇1 = 220× 18 220 = 18 kV 𝑋𝑝.𝑢 𝑛𝑒𝑤 =𝑋𝑝.𝑢 𝑔𝑖𝑣𝑒𝑛 * 𝑀𝑉𝐴 𝑏 𝑛𝑒𝑤 𝑀𝑉𝐴 𝑏,𝑔𝑖𝑣𝑒𝑛 * 𝑘𝑉 𝑏𝑔𝑖𝑣𝑒𝑛 𝑘𝑉 𝑏 𝑛𝑒𝑤 2 = j0.2× 50 30 × 18 18 2 = j0.333 p.u 14 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi
  • 15. References 1. Hadi Saadat, ‘Power System Analysis’, Tata McGraw Hill Education Pvt. Ltd., New Delhi, 21st reprint, 2010. 2. Kundur P., ‘Power System Stability and Control, Tata McGraw Hill Education Pvt. Ltd., New Delhi, 10th reprint, 2010. 3. Pai M A, ‘Computer Techniques in Power System Analysis’, Tata Mc Graw-Hill Publishing Company Ltd., New Delhi, Second Edition, 2007. 4. J. Duncan Glover, Mulukutla S. Sarma, Thomas J. Overbye, ‘ Power System Analysis & Design’, Cengage Learning, Fifth Edition, 2012. 5. Olle. I. Elgerd, ‘Electric Energy Systems Theory – An Introduction’, Tata McGraw Hill Publishing Company Limited, New Delhi, Second Edition, 2012. 6. C.A.Gross, “Power System Analysis,” Wiley India, 2011. 7. M.Jeraldin Ahila “POWER SYSTEM ANALYSIS”,Lakshmi Publications, Chennai, Tenth Edition 2016. 15 Kongunadu College of Engineering and Technology Per Phase & Per-Unit Analysis Prepared by, Mrs.s.Revathi