
reading / British practice / Vol D - 1990 (ocr) ELECTRICAL SYSTEM & EQUIPMENT
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Lou+. 2.18
Transformer design characteristics and equivalent taps
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i mpedance |
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I mpedance |
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Tap |
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Equivalent tap value in 'Yu |
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based on System voltages |
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Transformer |
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of |
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on rating |
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on 100 MVA |
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Principal |
Minimum |
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432/ |
Generator |
800 |
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+6.67 |
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1.2 |
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15.20 |
8.0 |
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23.5 |
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to |
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-13.35 |
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400/ |
Supergrid |
240 |
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1.43 |
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15.01 |
-0.72 |
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132 kV |
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0.0832 |
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-5 |
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132/ |
Station |
90/60/ |
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0.037 H-N |
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1.33 |
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2.53 |
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11.8/ |
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25 L-L |
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0.239 L- N |
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11.8 kV |
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23.5/ |
Unit |
60 |
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14.5 |
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0.278 |
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±7.5 |
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6 |
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2.33 |
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0.21 |
-6.78 |
- 1337 |
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11.8 kV |
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11/ |
Station and |
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2.39 |
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5.22 |
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-13.90 |
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3.45 kV |
unit auxiliary |
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11/ |
Essential |
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a |
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0.874 |
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8 |
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2.39 |
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5.22 |
-4.350 |
-13.90 |
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3.45 kV |
auxiliary |
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I ll |
Chlorination |
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a |
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2.39 |
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5.22 |
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-13.90 |
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3.45 kV |
plant |
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3300/ |
TH H & V services |
2 |
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7.5 |
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4.082 |
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2.4 |
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0.64 |
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-8.96 |
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433V |
Reactor services |
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IS) |
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Turbine services |
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Essential services |
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('X' and 'Y') |
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3300/ |
Chlorination |
1.6 |
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4.082 |
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2.4 |
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0.64 |
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- 8.96 |
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433V |
plant services |
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Admin and |
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workshop services |
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Reactor lighting |
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Reactor services |
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3300/ |
Turbine lighting |
1.0 |
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5.171 |
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0.64 |
-- 4.16 |
- 8.96 |
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433V |
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Noie: The system voltages are 400, 132, 23.5, II, 3.3 and 0.415 kV.
analysis performance system Power

Electrical system analysis |
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Chapter 2 |
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I LOAD FLOW RESULTS. BUSBAR PU VOLTS & LINE MVA OADING |
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235VG |
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1.10, 08 , |
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KV031 |
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11091 |
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3 3.501 |
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3 3F01 |
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1 043 |
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I 044 |
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053l |
0319 |
UTS5 |
SS1A |
C |
0T34 |
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CWPH TO |
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BHA |
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• 059 |
046 |
1 DSO |
1 044 |
1 044 |
1 Ns |
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1 044 |
1 04? |
HvWS |
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JETTY |
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' 034 |
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1 053 |
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11, 04065 29 Jun "504 09 34 14
FIG, 2.54 Voltage profile of an oil-fired power station electrical system at minimum load
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LOAD FLOW RESULTSBUSBAR PU VOLTS & LINE MV |
OADING |
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HYYYS |
FPH |
OHS , |
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CHI. |
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JLTTY |
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015 |
0 994 |
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FA5344045 29 Jun ?maze 29 35
Fro. 2.55 Voltage profile of an oil-fired power station electrical system at maximum load
1 54

Power system performance analysis
LOAD FLOW RESULTS - BUSBAR PU VOLTS & LINE MVA LOADING
2600 ¶ 320
Flo. 2.56 Voltage profile of a network with a 3.3 kV diesel generator supplying an isolated part of a SES
LOAD FLOW RESULTS - BUSBAR FU VOLTS & LINE MVA LOADING
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GI 781 |
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E um. E |
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86CY |
E |
eficx. |
Noy |
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8600 |
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1 014 |
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0.2 |
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A3 |
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.1600' |
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I 014 |
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I 1_014 |
0.2 • • 011 |
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1:170? 87071 |
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0 2 |
0 .2 |
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1P$0060 29 Jun 1959 ■ 616
Flo. 2.57 Voltage profile with turbine and reactor off-load
165

Electrical system analysis |
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Chapter 2 |
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LOAD FLOW RESULTS |
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BUSBAR PU VOLTS |
& LINE MVA LOADING |
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1896.868 |
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29 ,2,,n '998 12734:04 |
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FIG. 2.59 Voltage profile with the first II kV starting/standby feedpump being started
156

Power system performance analysis
LOAD FLOw RESULTS - BuSBAR PU VOLTS & LINE MVA LOADING
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AM |
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1 012 |
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2 5 •I '°6 |
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1 013 |
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15. 1 |
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0 975 |
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1 005 |
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IPSA9914 29.10 198a 17.41 24
Etc. 2.60 Voltage profile with the first 11 kV starting/standby feedpump at speed and full load
, LOAD FLOW RESULTS. BUSBAR pu VOLTS & LINE MVA LOADING
1 M 91
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IP008615 01 Jul 191111 5551.11
Flu. 2,61 Voltage profile with main generator at 40% full load and gas circulators at full load
157

Electrical system analysis |
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Fla. 2.62 Voltage profile with second 11 kV starting/standby feedpump being started
LOAD FLOW RESULTSBUSBAR PU VOLTS & LINE MVA LOADING
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31286908 01 Jo 1946 1593 19 |
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FIG. 2.63 Voltage profile with second 11 kV starting/standby feedpump at speed and full load
158

Power system performance analysis
|
1 |
024 |
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14634468 01 44 1511411610.50
FIG. 2.64 Voltage profile with the unit at full load
NETWORK DIAGRAM DRAWING AND MODIFYING
TYPE '/I 1 IF YOU NEED ANY HELP
J7
Flo. 2.65 Example of computer generated draughting
159

Electrical system analysis |
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Chapter 2 |
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LINE DATA 1 |
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PAGE 1 OF 3 |
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335-52'07 136-21:2 |
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AES:57-73 24527-31: |
535C-2U 252-3-23 |
232-2-211 |
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4:147 |
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3-12 |
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3:177 |
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1 |
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3.243:3 |
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1:331 |
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1:332 |
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1.33323 |
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11332 |
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3.04959 |
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3.3111311 |
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710:3 |
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4.09126 |
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3.73313 |
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73513 |
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3.33313 |
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4.23695 |
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3.3531 |
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2324: |
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1.14329 |
5,34695 |
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3.1231 |
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4,X31.1 |
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3.31302 |
0.34:75 |
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5.3531 |
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2,511 |
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1.21756 |
4.11111 |
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OPTIONS: |
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REDISPLAY |
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SYSTEM BASE MVA= |
100.0 |
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NEXT PAGE |
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HELP |
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RETURN |
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1206860 01 :ul 1485 :6:22:24 |
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FIG, 2.66 Example of line data entry |
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INDUCTION MOTOR DATA 1 |
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PAGE |
1 OF |
1 |
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535250/JD |
2251211-143 31.12-22 |
5166-0 |
|
37-3-211 |
52-X-PU ROT-R-24.1 001-6-PU 5-OUT-T1 514-IN-21 3-2121-12 361-15- 122 |
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11331 |
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5.401 |
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92.934 |
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1.0330 |
1.3146 |
0.1556 |
1.3146 |
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1133: |
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5.370 |
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92.994 |
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0.3330 |
1.8146 |
1,1558 |
1.3146 |
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1:351 |
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1 |
.120 |
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79.628 |
|
0.3546 |
1.8504 |
0.1095 |
1.0512 |
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||||||||
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11351 |
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7.100 |
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19.628 |
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0.0596 |
1.0504 |
0.1045 |
1.0512 |
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1:311 |
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1.493 |
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182.220 |
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3.1155 |
7.7041 |
0.5014 |
7.7043 |
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1511 |
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1.480 |
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79.628 |
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3.0596 |
1.8504 |
0.1095 |
1.0512 |
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|||||
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11531 |
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162.220 |
|
3.3155 |
7.7011 |
0.5014 |
7.7047 |
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||||||||
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11332 |
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5.910 |
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92.494 |
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0.0910 |
1.8146 |
0.1558 |
1.3146 |
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||||||
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11312 |
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, ,300 |
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71.628 |
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0.0596 |
1.8504 |
0.0095 |
1.0502 |
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:5 |
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11460 |
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182.220 |
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0.3155 |
2.1047 |
0.5014 |
1.1041 |
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06 |
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7,001 |
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13.628 |
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0.0546 |
1.8504 |
0.1395 |
1.1512 |
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5.511 |
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92.994 |
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0.0930 |
1.11146 |
1.1558 |
1.3146 |
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3.10313 |
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11750 |
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131.214 |
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0.3195 |
5.5015 |
0.2520 |
3.8280 |
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|||||||
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3.12315 |
|
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1.000 |
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229.520 |
|
0.5542 |
1.6271 |
0.4413 |
10.9916 |
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|||||||
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3 |
133:3 |
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0.350 |
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241.710 |
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0,5886 |
11.1344 |
0.4640 |
11.5700 |
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||||||
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33113 |
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1.120 |
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229.620 |
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0.5542 |
9,6217 |
0.4413 |
11.4916 |
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|||||||
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1,13323 |
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0.950 |
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241.110 |
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0.5686 |
10.1344 |
1.4640 |
11.5100 |
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|||||||
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3.33324 |
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1.100 |
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229.620 |
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2.5592 |
4.6277 |
0.4413 |
10.4916 |
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|||||||
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:9 |
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1.750 |
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131.214 |
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0.1195 |
5.5015 |
0.2520 |
3.8280 |
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|||||||
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3.322 |
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1.000 |
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229.620 |
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0.9592 |
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9.5271 |
0.4413 |
10.4916 |
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1.050 |
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252.525 |
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0.8499 |
12.3045 |
1.3836 |
1.6053 |
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3.322 |
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1.390 |
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214.110 |
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0.1258 |
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1.6275 |
3.6239 |
6.2429 |
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3.121 |
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1,050 |
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252.525 |
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0.4999 |
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12.3049 |
1.1836 |
7.8053 |
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1.1E1 |
2.950 |
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96.281 |
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0.3486 |
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3.4515 |
0.3661 |
3.0505 |
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3.3E4 |
2,150 |
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252.525 |
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0.9999 |
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12.3049 |
1.0636 |
7.8053 |
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5.5E4 |
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86.297 |
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5.1416 |
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1.4515 |
0.3661 |
3.0505 |
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:10 |
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1.330 |
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214.700 |
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0.8259 |
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7.9275 |
1.8258 |
6.0409 |
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.111 |
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1.150 |
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252.5251 |
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0.8990 |
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12.3049 |
1.0836 |
7.8051 |
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J14 |
1.391 |
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9.529 |
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0.0536 |
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1.0504 |
3.1095 |
1.2512 |
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J:3 |
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102.220 |
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0.3155 |
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7,7047 |
3.5014 |
7,7147 |
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OPTIONS: |
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SYSTEM RASE MVA2 |
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100.0 |
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REDISPLAY |
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HELP
RETURN
1P51860 01 Jul 1980 16;22,44
FIG. 2.67 Example of induction motor data entry
160

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Power system performance analysis |
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TRANSFORMER DATA 1 |
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PAGE 1 OF 2 |
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31.75 - 5353 1175 - .53C |
TA2-3TA2: |
MIN-TAP TAP-STEP |
MAE - TAP |
V-SPEC ? - s2:r7 |
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: 71022 |
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1.13 |
-6.43 |
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15.20 |
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8.01 |
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1.20 |
15.20 |
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4:1 , 2 |
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3 15 , 3 . |
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-4.4: |
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2.33 |
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4:34'2 |
1531 |
-2.62 |
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1.04 |
2.51 |
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' 3 |
-2.52 |
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1.04 |
2.58 |
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-13.71 |
2.33 |
3.21 |
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-4,35 |
1.11 |
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-4.35 |
-0.32 |
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33: |
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-4.35 |
-4.35 |
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-4.36 |
-4.35 |
:).01 |
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_1531 |
1.353: |
-4.35 |
-4.35 |
.3.11 |
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11532 |
3.3532 |
-4.35 |
-4.35 |
0.01 |
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11332 |
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-4.35 |
-4.35 |
0.21 |
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-4.35 |
-4.35 |
5.01 |
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1 _572 |
3.3'2323 |
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1.3T32A |
-4.35 |
-4.35 |
0.01 |
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111:32 |
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1 1532 |
22 |
-4.36 |
-4.35 |
5.01 |
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3.51331A |
14310 |
-4.15 |
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0.11 |
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3.13313 |
74310 |
-4.15 |
-4.15 |
0.01 |
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3.3631B |
064231 |
-4.16 |
-4.16 |
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3.3531 |
FUEL: |
-4.16 |
-4.16 |
3.01 |
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3.3531 |
AX3L1 |
-4.15 |
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0.31 |
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1.3531 |
3A01 |
-4.16 |
-4.16 |
0.01 |
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3.3532 |
?HEAT |
-4.16 |
-4.15 |
0.01 |
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1.3532 |
RAN |
-4.16 |
-4.16 |
1.01 |
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3.3532 |
325602 |
-4.16 |
-4.15 |
0.7: |
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3.3532 |
FUEL; |
-4.16 |
-4.16 |
0.01 |
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3.113125 |
04012 |
-4.16 |
-4.16 |
0.01 |
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1.312324 |
114320 |
-4.16 |
-4.16 |
0.01 |
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3.112720 |
161326 |
-4.16 |
-4.16 |
0.01 |
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3.3E3 |
E3 |
-4.16 |
-4.16 |
0.01 |
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OPTIONS: |
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(RED ISP LAY |
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SYSTEM BASE MVAm |
100.0 |
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NEXT PAGE |
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HELP |
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RETURN |
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12556514 01 261 |
1489 |
6122.56 |
|
FIG, 2.68 Example of transformer data entry
ANALYSIS PARAMETERS |
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PAGE 1 OF 1 |
5.85E WA |
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100.0 |
14VA |
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0153 FION CONL,ERSENCE ACCURACY |
0.010 |
MVA |
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HAX:NUM LOAD FLOW ITERATIONS |
25 |
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SLACK RUSOAR |
400KV |
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FAULT TIME |
0.100 |
SEC |
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SYSTEM FRE.TUENCy |
50.0 |
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REFERENCE MACHINE 6V535.0 NAME |
40061 |
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RETERENCE HACHINE IDENTIFIER |
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5102Y DTRATIDH TIME |
1.00 |
SEC |
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5124311.174 51E0 LENGTH |
0.010 |
SEC |
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56153 ANSLE / POLE 53:5' 0:625 |
540 |
DECREES OR POLE PAIRS |
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3103 555123. MACHINE SPEEDS? |
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R3.114TO311 SYSTEM DATA1 |
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6T5.310:71 20I510U5 1117E01563 |
0.040 |
SEC |
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21 500121 15052628 VOLTAGES? |
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RAINT.= AVR DATA? |
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3R0513131 GOVERNOR DATA? |
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OPTIONS: |
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' |
REDISPLAY |
SYSTEM BASE MVAm |
100.0 |
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I |
HELP |
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11RETURN |
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1255660 01 Jul 1904 16:23:04 , |
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FIG. 2.69 Example of analysis parameters
161

Electrical system analysis |
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Chapter 2 |
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SUB TRANSIENT
PERIOD
SLOWLY DECAx;NG TRANSIENT pERioo
STEADY STATE
PERIOD
Xd
7
s
Fin. 2.70 Traosient short-circuit of a synchronous rnachlne — only one phase envelope shown
\, oltage and current in the other two phases. However, when a network has unbalanced loading between phases, or is subject to an electrical fault involving one or two phases only, then phase symmetry is destroyed.
Method of symmetrical components
The method of symmetrical components [10] was developed to reduce the complexity of unbalanced fault analysis and is widely used in computer programs. This method represents an unbalanced system of voltages and currents by the superposition of two symmetrical three-phase systems of opposite phase sequence and a zero phase sequence system, i.e., a single-phase alternating system.
Three components are defined:
V a |
|
Vo |
+ |
V1 + |
V2 |
Vb |
= |
VO |
|
a 2 V1 |
+ aV2 |
V c |
= |
Vo |
+ |
aVi |
+ a 2 V2 |
Or |
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|
Vo |
= 1/3 (V, |
Vb |
Vc) |
V I = 1/3 (V, |
aVb + cx 2 V c ) |
||
V2 |
– 1/3 (Va |
+ a 2 Vb + Vc) |
Where
V a , Vb, V, are network voltages in phases a, b, c
V i |
positive sequence component voltage |
V2 = negative sequence component voltage
Vo – zero sequence component voltage
a is a phase rotation of 120 0 = exp j 2r/3 a 2 is a phase rotation of 240 ° = exp j 47r/3
(a 3 is a phase rotation of 360 0 = exp j 27r = 1) or 0°
Similar equations can be written for network currents. The component V1 is the normal value of the phase voltage for a balanced system with normal phase sequence (a, b, c) and is called the positive phase sequence component of voltage. Component V2 has the opposite
phase sequence (a, c, b) — this is indicated by the phase rotation coefficients a and a 2 — and is called the negative phase sequence component of voltage. Component Vo has no phase displacement between phases a, b and c — there are no phase rotation coefficients present — and is called the zero phase sequence component of voltage.
The total power in an unbalanced system is the sum of the symmetrical component powers.
The reason for the symmetrical component transformation being so useful in analysis is that, for most types of equipment used in power systems, their positive, negative and zero sequence components are independent of each other, or 'decoupled'. This means the associated matrices used during analysis are diagonal and, in consequence, are easily manipulated mathematically.
The values of the various sequence component impedances for generators, overhead lines, cables, transformers, and other electrical apparatus can be obtained by test or from analysis. Positive and negative phase sequence component impedances are identical for static apparatus. Zero sequence impedance is associated with earth return paths and may be greater or less than the positive sequence impedance.
Driving voltage is only associated with the positive phase sequence component network. It follows that negative and zero sequence component currents can flow only when their networks are connected to the positive phase sequence component network. The connections between the various networks under fault conditions and the associated equations are given on Fig 2.71,
This assumes negligible impedance at the point of fault but this is not always so. In practice, fault impedance is sometimes considered in single line to earth faults and this requires an impedance of 3 x fault impedance to be inserted in the zero sequence network. However, design analysis is usually concentrated on worst conditions and this means setting fault impedances to zero. If fault impedance (Zr) is to be taken into account, for single-line to earth faults, 3Zf is inserted in the zero sequence network. For a doubleline to earth fault, the impedance between phases (Zr) is inserted in each phase network and 3Z f in the zero sequence network. For a three-phase to earth fault Z p is inserted in each phase network, 3Z 8 in the zero sequence network (Z 8 is the impedance to ground) and an additional parallel path is introduced having an impedance of (Zf – Z p )/3 (Fig 2,72).
Generator representation
Programs are written to calculate generator fault current contribution according to the amount of machine data entered by the analyst. For complete representation of fault current decrement with time, synchronous, transient and subtransient data must be entered. If only one value of machine reactance is entered, the
162