marine and offshore power system

TET4200 MARINE AND OFFSHORE POWER SYSTEM

PROJECT WORK SPRING 2012

Mamta Maharjan, Linda Rekosuo

iii

SUMMARY In this project an electrical power system for a though offshore platform is studied. Short

circuit analysis are done for single-phase, two-phase and three-phase to ground faults.

Especially fault currents are under examination. Also power flow and transient analysis are

done and critical clearing times for induction machines are solved. Motor starting analysis are

done in two different ways. First start-up is analyzed when motor is connected to system.

Later start-up is examined when motor is connected to long subsea cable and star-up is

controlled by converter.

iv

Table of content

1. Introduction ......................................................................................................................... 1

1.1. System description ....................................................................................................... 1

2. SIMPOW Simulations ........................................................................................................ 2

1.1. Power flow ................................................................................................................... 2

1.2. Characteristics for induction machines ........................................................................ 3

1.3. Short-circuit analysis ................................................................................................... 4

1.1.1. Single phase to ground fault ................................................................................. 4

1.1.2. Two-phase fault .................................................................................................... 5

1.1.3. Three phase fault .................................................................................................. 5

1.4. Motor starting analysis ................................................................................................ 9

1.5. Transient analysis ...................................................................................................... 10

1.1.4. Contingency analysis .......................................................................................... 10

1.1.5. Critical clearing time .......................................................................................... 11

3. ATPDraw simulations ....................................................................................................... 12

4. Conclusions ....................................................................................................................... 14

1

1. Introduction

In this project an electrical power system for a though offshore platform is studied. Four

analyses are done: power flow, short-circuits, motor starting and contingencies. Also start-up

of induction motor from frequency converter via long subsea cable is examined more detail.

First four analyses are done by using SIMPOW. Start-up with converter is analyzed with

ATPDraw.

1.1. System description Single-line-diagram for system is shown in Figure 1. It consists of three turbine/generator

sets, eight motors, three passive loads and three transformers. The number of generators

running in each simulation is varied. Voltage levels are 13.8 kV, 6.0 kV and 0.44 kV.

Generators are two pole synchronous machines with brushless excitation system (type

IEEEX2). Turbines are twin-shaft aero-derivative gas turbine engines. The speed is controlled

with regulator. Motors are induction machines with squirrel-cage designed rotor.

All transformers have two windings and isolated neutral. Passive loads act as constant

impedances and they consist of active and reactive part. Cables are three-phase types and they

are modeled with resistance, inductance and susceptance. Detailed data for each component

can be seen in Appendix 1.

2

2. SIMPOW Simulations

1.1. Power flow Power flow analyses are done in two different situations. In Figure 1 can be seen power flow

results when two of the three generators are working. Generator G3 is not connected to

system.

Figure 1 Single-line-diagram and power flow results for system with two generator sets

working.

00

0

0

0 0

BUS1

U = 1 p.u.

FI = -3.02274E-022

degrees

P = -4.16264 MW

Q = -2.4933 Mvar

P = -4.16032 MW

Q = -2.49386 Mvar

P = 15.3636 MW

Q = 9.08899 MvarP = 15.36 MW

Q = 9.1 Mvar

P = 0 MW

Q = -0 Mvar

P = -4 MW

Q = -1.9373 Mvar

P = -4.86918 MW

Q = -3.21012 Mvar

P = -1.04447

MW

Q = -0.562251

Mvar

M1

P = -6.237 MW

Q = -3.74218 Mvar

M2

P = -6.25 MW

Q = -3.74998 Mvar

BUSG3

U = 1 p.u.

FI = -3.02274E-022

degrees

P = 0 MW

Q = -0 Mvar

G3

P = 0 MW

Q = -0 Mvar

BUSG1

U = 1 p.u.

FI = 0 degrees

P = -15.3636 MW

Q = -9.08899 Mvar

G1

P = 15.3636 MW

Q = 9.08899 Mvar

BUS2

U = 0.998661 p.u.

FI = -0.0214586

degrees

P = 4.158 MW

Q = 2.49442 Mvar

M3

P = -4.158 MW

Q = -2.49442 Mvar

BUSG2

U = 1 p.u.

FI = -3.02274E-022

degrees

P = -15.36 MW

Q = -9.1 Mvar

G2

P = 15.36 MW

Q = 9.1 Mvar

BUS7

U = 1 p.u.

FI = -3.02274E-022

degrees

P = 4 MW

Q = 1.9373 Mvar

0

P = -4 MW

Q = -1.9373 Mvar

BUS3

U = 0.968803 p.u.

FI = -2.76627

degrees

P = 1.037

MW

Q = 0.49524

Mvar

0

M4

P = -1.037 MW

Q = -0.49524 Mvar

BUS4

U = 0.94939 p.u.

FI = -4.19837

degrees

P = 4.83346 MW

Q = 2.70104 Mvar

0

P = -0.95078 MW

Q = -0.579376 Mvar

0

P = -1.80268 MW

Q = -0.87304 Mvar

M5

P = -1.04 MW

Q = -0.624314 Mvar

M6

P = -1.04 MW

Q = -0.624314 Mvar

BUS5

U = 0.923247 p.u.

FI = -6.36011

degrees

P = 0.945192 MW

Q = 0.528144 Mvar

0

0

P = -0.426192 MW

Q = -0.206448 Mvar

M7

P = -0.519 MW

Q = -0.321697 Mvar

BUS6

U = 0.999331 p.u.

FI = -0.0107136

degrees

P = 4.158 MW

Q = 2.49443 Mvar

M8

P = -4.158 MW

Q = -2.49443 Mvar

3

Table 1 shows power flow when all the three generators are running. Single-line diagram for

this situation is in Appendix 2. In both cases voltage levels on several buses are lower than

they should be. Generators can be a little overloaded and they cannot keep voltage levels high

enough.

Table 1 Power system results when all three generators are working

1.2. Characteristics for induction machines Torque-speed and current-speed characteristics for induction motors are shown in Figure 2. It

can be seen that torque is maximum when speed reaches its nominal value.

Figure 2 Torque-speed and torque-current charasteristics for induction motor

30

0

0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900

ASYNC M1 SPEED p.u.

0.0

2.0

4.0

6.0

8.0

10.0

TET 4200 "Marine and Offshore power systems". Spring 2012.

DATE 15 MAR 2012 TIME 13:49:18 JOB mini_project_2012_Part4 Simpow 11.0.008 Diagram:3STRI Software

ASYNC M1 ME TORQUE MNm/ W0

30

0

0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900

ASYNC M1 SPEED p.u.

0.00

0.20

0.40

0.60

0.80

1.00

TET 4200 "Marine and Offshore power systems". Spring 2012.

DATE 15 MAR 2012 TIME 13:49:18 JOB mini_project_2012_Part4 Simpow 11.0.008 Diagram:4STRI Software

ASYNC M1 I POS. kA

4

Current stays 1.2 kA first and starts to decrease when speed is reaching a nominal value.

1.3. Short-circuit analysis Three different short-circuit analysis are done: during three-phase, two-phase and one-phase

to ground faults. In every case fault impedance is assumed to be zero.

1.1.1. Single phase to ground fault Single-phase to ground fault is examined. Two generator/turbine sets are running and system

is under normal load. First fault occurs on Bus 1, between phase A and ground, at t=1.0 s.

Phase fault currents (RMS) for phase A is shown in Figure 3. As can be seen a maximum fault

current in phase A is 4.10 kA. Also generator currents in phase A during the fault are shown

in Figure 3. Before the fault current is 750 A and just after fault it increases to 2.28 kA.

Figure 4 shows current when fault occurs on Bus 3.Now fault current is 51.5 A. Generator

current just after fault is 747 A. In this case currents are significantly smaller than when fault

occurs on Bus 1.

Figure 3 Phase A fault current and generator G1 current. Fault occurs on Bus 1.

Figure 4 Phase A fault current and generator G1 current. Fault occurs on Bus 3.

0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100

TIME SECONDS

0,00

0,50

1,00

1,50

2,00

2,50

3,00

3,50

4,00

TET 4200 "Marine and Offshore power plants". Spring 2012.

Dynpow file for Mini-project. 2 T/G sets in operation. Normal load.

Short-circuit on BUS 1 with regulator (one-phase fault)

DATE 26 APR 2012 TIME 21:24:05 JOB mini_project_2012_Part5-2-a-iii Simpow 11.0.009 Diagram:1STRI Software

1,020, 4,10

FAULT FEL1 I PHASE A kA

0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100

TIME SECONDS

0,80

1,00

1,20

1,40

1,60

1,80

2,00

2,20

0,80

1,00

1,20

1,40

1,60

1,80

2,00

2,20





1,019, 2,28

SYNC G1 I PHASE A kA


0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100

TIME SECONDS

0,0000

0,0100

0,0200

0,0300

0,0400

0,0500





1,012, 0,0515


0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100

TIME SECONDS

0,75400

0,75500

0,75600

0,75700

0,75800

0,75900

0,76000

0,74700

0,74800

0,74900

0,75000

0,75100

0,75200

0,75300

0,75400





1,020, 0,74707



5

1.1.2. Two-phase fault Next two-phase, phases A and B, line-to-ground fault is simulated. During the fault system is

working without load. Generator G3 is disconnected from the system. Fault current and

generator G1 current in phase A are shown in Figure 5 . Fault current is 7.21 kA which is also

the sum of generator currents of the system.

Figure 5 Fault current in phase A and generator current during the fault

1.1.3. Three phase fault

System working without load

Three-phase fault occurs on Bus 1 at t=1.0. System is working with one generator/turbine set

which is running without mechanical load. Initial maximum short-circuit current and

instantaneous fault current can be calculated following way:

√

√

Details of calculations can be seen in Appendix 3. Figure 6 shows simulation result of this

case. From the simulation fault current is approximately 4.8 kA which is quite close to

calculated value.

When two generator sets are running fault currents are twice as big as in previous case:

Using these values κ-value can be examined

0,9900 0,9950 1,0000 1,0050 1,0100 1,0150 1,0200 1,0250 1,0300 1,0350 1,0400

TIME SECONDS

0,00

1,00

2,00

3,00

4,00

5,00

6,00

7,00


Dynpow file for Mini-project. 2 T/G sets in operation. No load.

Short-circuit on BUS1 (two-phase symmetric fault)

DATE 15 APR 2012 TIME 12:04:10 JOB mini_project_2012_Part5-2-a-ii Simpow 11.0.009 Diagram:1STRI Software

1,0000, 7,21


0,9900 0,9950 1,0000 1,0050 1,0100 1,0150 1,0200 1,0250 1,0300 1,0350 1,0400

TIME SECONDS

0,00

0,50

1,00

1,50

2,00

2,50

3,00

3,50



Short-circuit on BUS1 (two-phase symmetric fault)


1,0001, 3,61


6

√

√

This situation is also shown in Figure 6. From the figure fault current is about 9 kA and it is

almost the same as calculated value.

Figure 6 Initial maximum currents when system is running with 1/3 (on left) and 2/3

generator sets

System under load

Now two generators are working and system is under

normal load. Balanced three-phase fault occurs on

Bus 1. Calculations of generator current during the

fault give

and for the maximum instantaneous current

Details of calculations can be seen in Appendix 3. Figure 7

shows simulation results of this current. DC-offset of

generator current can be calculated as

Fault currents are shown in Figure 8. As can be seen current is approximately 12.8 kA and a peak

value of instantaneous current is about 35 kA.

0,800 0,850 0,900 0,950 1,000 1,050 1,100 1,150 1,200 1,250 1,300

TIME SECONDS

0,00

0,50

1,00

1,50

2,00

2,50

3,00

3,50

4,00


Dynpow file for Mini-project. 1 T/G set in operation. No load.

Short-circuit on BUS1 (three-phase symmetric fault)

DATE 5 APR 2012 TIME 11:12:00 JOB mini_project_2012_Part5-2-a-i Simpow 11.0.009 Diagram:1STRI Software

FAULT FEL1 I POS. kA

0,800 0,850 0,900 0,950 1,000 1,050 1,100 1,150 1,200 1,250 1,300

TIME SECONDS

0,00

1,00

2,00

3,00

4,00

5,00

6,00

7,00

8,00



Short-circuit on BUS1 (three-phase symmetric fault)



0,990 1,000 1,010 1,020 1,030 1,040 1,050 1,060 1,070 1,080 1,090

TIME SECONDS

1,00

1,50

2,00

2,50

3,00

3,50

4,00

4,50



Short-circuit on BUS 1 w ithout regulator (three-phase symmetric fault)


1,000, 4,59


Figure 7 Generator current during fault

7

Figure 8 Initial short-circuit current and instantenous fault current. Fault on Bus 1.

Using these simulation results, κ-value can be calculated as

√

√

This is almost the same value than in case that generator was working without load.

Also situations where fault occurs on Bus 4 and 5 are simulated. Figure 9 shows fault current

when fault is on Bus 4. Maximum RMS value of fault current is 6 kA and peak value of

instantaneous current is 14.5 kA.


When fault occurs on Bus 5 fault current is 25.5 kA and peak value of instantaneous current is

about 64 kA as can be seen in Figure 11.


0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100 1,120 1,140 1,160 1,180 1,200

TIME SECONDS

0,0

2,0

4,0

6,0

8,0

10,0

12,0



Short-circuit on BUS 1, 4 and 5 (three-phase symmetric fault)



0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100

TIME SECONDS

-30,0

-25,0

-20,0

-15,0

-10,0

-5,0

0,0

TET 4200 "Marine and Offshore power plants". Spring 2012. (MASTA)


Short-circuit on BUS 1 (three-phase symmetric fault)



0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100 1,120 1,140 1,160 1,180 1,200

TIME SECONDS

0,00

1,00

2,00

3,00

4,00

5,00

6,00






0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100

TIME SECONDS

-14,0

-12,0

-10,0

-8,0

-6,0

-4,0

-2,0

0,0






0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100 1,120 1,140 1,160 1,180 1,200

TIME SECONDS

0,0

5,0

10,0

15,0

20,0

25,0






0,900 0,920 0,940 0,960 0,980 1,000 1,020 1,040 1,060 1,080 1,100

TIME SECONDS

-60,0

-50,0

-40,0

-30,0

-20,0

-10,0

0,0






8

Contribution of induction machines and currents of outgoing feeders

RMS values of induction machine currents during the fault are examined. Results are seen on

Table 2. Curves of currents can be found on Appendixes 4, 5 and 6. It can be seen motor M7

has the highest current in every case. This is because motor M7 has the highest rated current.

Fault bus Peak RMS current during the fault [kA]

M1 M2 M3 M4 M5 M6 M7 M8

Bus 1 0.972 0.975 0.367 0.353 0.328 0.328 1.788 0.639

Bus 4 0.355 0.355 0.236 0.131 0.384 0.385 2.108 0.236

Bus 5 0.318 0.318 0.212 0.117 0.134 0.134 2.275 0.212

Table 2 Induction motor currents during the fault

If fault occurs on Bus 1 all currents coming from generators are almost the same. The highest

current of outgoing feeders from Bus 1 is between Bus 1 and transformer T1.

If fault occurs on Bus 4 or 5 current coming from generator G2 is the highest. When fault

occurs on Bus 4 motor, M5 and M6, currents are the highest of outgoing currents. Also when

fault occurs on Bus 5 motor (M7) is the highest current leaving from Bus 5.

If one generator is disconnected from the system and fault occurs on Bus 1 generetor currents

are almost the same than in case that every generator were working. If fault occurs on Bus 4

or 5 currents from generators become higher. Overcurrent breakers can be used to prevent

generators and othercomponents to be damaged by too high currents.

Influence of Automatic Voltage Regulator during three phase fault

Three-phase fault is examined with and without generator voltage regulators. Two

generator/turbine sets are working. Fault occurs on Bus 1.

Figure 11 Generator speed without (on the left) and with regulator

Figure 11 shows generator G1 speed with and without regulator during the fault. Graphs are

similar for generator G2. Voltages are seen on Figure 12. With regulator generator speed does

not get so high value and it becomes stable faster. Although speed also goes under nominal

value before it reaches 1 p.u.

Without regulator generator voltage drops about 8 kV and does not reach original value

anymore. With regulator voltage drops first but reaches original value in 2 seconds.

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0 11,0 12,0

TIME SECONDS

1,000

1,020

1,040

1,060

1,080

1,100





SYNC G1 SPEED p.u.

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0

TIME SECONDS

0,99800

1,00000

1,00200

1,00400

1,00600



Short-circuit on BUS 1 w ith regulator (three-phase symmetric fault)


SYNC G1 SPEED p.u.

9

Figure 12 Generator voltages without (on the left) and with regulator

1.4. Motor starting analysis Motor starting analysis for induction motor M1 without mechanical load is done. System is

running with two generator/turbine sets. All of the other motors are working under normal

load. Speed and current (RMS) curves of motor M1 during start-up are seen in Figure 13.

Figure 13 Voltage (RMS) and speed curves for motor during start-up. Motor without load

Motor speed reaches its nominal value in 9 seconds and its rising is almost linear. Voltage

drops almost to 6 kV. Figure 15 shows generator G1 and motor M2 speed changes when

motor is started.

1,00 2,00 3,00 4,00 5,00 6,00 7,00 8,00 9,00

TIME SECONDS

0,00

1,00

2,00

3,00

4,00

5,00

6,00

7,00





SYNC G1 U PHASE A kV

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,00

1,00

2,00

3,00

4,00

5,00

6,00

7,00

8,00



Short-circuit on BUS 1 w ith regulator (three-phase symmetric fault)


SYNC G1 U PHASE A kV

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

7,00

7,50

8,00

8,50



Motor starting analysis


ASYNC M1 U POS. kV

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,00

0,20

0,40

0,60

0,80

1,00





ASYNC M1 SPEED p.u.

10

Figure 14 Generator speed and motor M2 speed when motor M1 is started

Now motor is running with full load. Start-up voltage and speed curves are seen in Figure 15.

Speed is not rising linearly and it takes almost 40 s to reach nominal value. Minimum voltage

is 6.59 kV.

Figure 15 Start-up voltage and speed curves when motor under full load

1.5. Transient analysis

1.1.4. Contingency analysis System is working with two generator settings and under normal load. Motor M1 is

disconnected for a short time. Figure 16 shows speed and voltage changes of motor M2

during disconnection. Curves of other motors can be seen in Appendix 7. Speed of every

motor has become stable in 9 second after disconnection.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,9800

0,9850

0,9900

0,9950

1,0000

1,0050





SYNC G1 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,9700

0,9750

0,9800

0,9850

0,9900

0,9950

1,0000





ASYNC M2 SPEED p.u.

0,0 5,0 10,0 15,0 20,0 25,0 30,0 35,0 40,0 45,0 50,0 55,0 60,0

TIME SECONDS

7,00

7,50

8,00

8,50





1,2, 6,59

ASYNC M1 U POS. kV

0,0 5,0 10,0 15,0 20,0 25,0 30,0 35,0 40,0 45,0 50,0 55,0 60,0

TIME SECONDS

0,000

0,200

0,400

0,600

0,800





ASYNC M1 SPEED p.u.

11

Figure 16 Voltage and speed curves of motor M2 during disconnection

1.1.5. Critical clearing time Critical clearing time for system is examined. System is working with two generators. Three-

phase fault occurs on Bus 1. Clearing times for induction motors are seen on Table 1. Curves

are shown in Appendix 8. As can be seen from table the critical clearing time is 8.5 s.

Clearing times [s]

M1 M2 M3 M4 M5 M6 M7 M8

8.3 8.3 8.5 8.4 8.4 8.4 8.5 8.3

Table 3 Clearing times when two generators are running.

Same analysis is done when all three generators are working. Results are seen in Table 4.

Now critical clearing time is 10.1 s.

Clearing times [s]

M1 M2 M3 M4 M5 M6 M7 M8

10.1 10.1 9.9 9.9 9.9 9.9 9.9 10

Table 4 Clearing times when three generators are running.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,991100

0,991200

0,991300

0,991400

0,991500

0,991600

0,991700





ASYNC M2 SPEED p.u.

0,80 0,90 1,00 1,10 1,20 1,30 1,40 1,50 1,60 1,70 1,80 1,90 2,00 2,10 2,20 2,30

TIME SECONDS

7,960

7,980

8,000

8,020

8,040

8,060





ASYNC M2 U PHASE A kV

12

3. ATPDraw simulations

Starting of induction motor with frequency converter via long subsea cable is examined.

Virhe. Viitteen lähdettä ei löytynyt. shows a picture of the system. First starting currents of

motor are calculated. Cable capacitances and motor and transformer magnetizations are

neglected. Boost is assumed to be 65 %. Cable has rated 12 kV and 288 A. Starting currents

and frequencies needed to cause the correct starting torque are seen on Table 5Table 5. It can

be seen starting current rises with increasing cable length and break-away torque. In case of

30 km, 40 km and 50 km length cable start-up currents are quite high. Also start-up

frequencies in some cases reach too high values.

Calculations are seen in Appendix 9.

Same situation is examined with simulations using

ATPDraw. Figure 18 shows current and speed when cable

length is 10 km and break-away torque is 0.1 p.u. Starting

current with converter is almost one third smaller than

current without converter. Appendix 10 shows other

simulations. Simulations and calculations are not equal. This

is because neglected parameters of calculations.

Break

away

torque

[pu]

Start-up current /Start-up frequency

Cable length

10 km 20 km 30 km 40 km 50 km

0.1 191 A 7.0

Hz

221

A

9.4

Hz

255 A 12.5

Hz

290 A 16.2

Hz

327A 20.6

Hz

0.2 381 A 14.9

Hz

442

A

18.9

Hz

509 A 25.0

Hz

580 A 32.4

Hz

654A 41.1

Hz

0.3 572 A 21.0

Hz

664

A

28.3

Hz

764 A 37.5

Hz

870 A 48.6

Hz

980A 61.7

Hz

0.4 762 A 28.0

Hz

885

A

37.7

Hz

1019A 50.0

Hz

1161A 64.8

Hz

1307A 82.3

Hz

Table 5 Starting currents and frequencies of induction motor with converter controlled start-

up. Boost is 65 %.

Also simulations show that increasing cable length increases starting current. For cable

lengths 10 km and 20 km motor can be started with previous cable. When length is higher

than this starting is not possible anymore. To able motor starting cable resistance have to be

increased. For 30 km and 40 km a cable with rated 12 kV and 464 A is used. But this cable

cannot start 50 km length cable. Cable data is seen in Appendix 11.

Figure 17 Induction motor connected

to long cable and converter

13

Figure 18 Motor current and speed during start-up when breakaway torque is 0.1 pu and

cable legth 10 km. Green and pink curves are for system with converter. Other are without

converter

Figure 19 shows starting current and speed with cable length 30 km and break-away torque

0.1 p.u. Initial frequencies are varied. It can be seen the starting current with initial frequency

20 Hz is almost 500 A higher than current with 10 Hz initial frequency. Starting current can

be minimized by decreasing initial frequency.

Figure 19 Starting current and speed when cable lenght is 30 km, break-away torque is 0.1

p.u. and boost is 65 %. Initial frequencies are 10 Hz (on the left) and 20 Hz (on the right)

(file startup1(1).pl4; x-var t) m:I1 m:I2 u1:OMEGM u2:OMEGM m:I1 0 5 10 15 20 25[s]

0

150

300

450

600

750

900

(file startup1(1).pl4; x-var t) m:I2 u1:OMEGM

0 5 10 15 20 25[s]0

100

200

300

400

500

600

700

800


0 5 10 15 20 25[s]0

200

400

600

800

1000

1200

14

4. Conclusions

In this report an electrical system for offshore platform was examined. Analyses were done by

using programs SIMPOW and ATPDraw. Simulations shower that under normal conditions

system cannot keep voltage levels high enough. This can mean generators are working

overloaded. From short circuit analysis it can be seen three-phase-to-ground fault causes the

highest fault currents. If fault occurs on Bus1 and system is running with two generator sets it

takes 8.5 seconds for induction motors to cover. If all three generator are working the clearing

time increases to value 10.1 s. Also if one motor is disconnected from the system the

recovering time is about 9 seconds. For generators automatic voltage regulators can be used to

control speed and voltage and make the generator stable faster after fault.

Motor starting analyses were done in two different situations. First one induction were started

while other system were running normally. If motor is running without mechanical load start-

up time is about 9 s and almost linear. When motor is running full load start-up lasts almost

40 s. In other situation motor start-up where examined when motor was connected to long

subsea cable and converter. Analyses showed the cable length increased starting current and

finally made start-up impossible. If cable was changed to other cable with higher rated current

start-up were possible also cable lengths 30 km and 40 km. Although when cable got length of

50 km start-up were not able anymore. Other thing which affected to start-up was initial

frequency. Increased initial frequency also increased starting current.

15

Appendixes

Appendix 1

Component data

16

Appendix 2

Power Flow

First table shows power flow when two of the three generators are working. Second table is

for the system of three generators.

17

00

0

0

0 0

BUS1

U = 1 p.u.

FI = 3.16042E-021

degrees

P = -4.16264 MW

Q = -2.4933 Mvar

P = -4.16032 MW

Q = -2.49386 Mvar

P = 0.363609 MW

Q = 2.08899 MvarP = 15.36 MW

Q = 9.1 Mvar

P = 15 MW

Q = 7 Mvar

P = -4 MW

Q = -1.9373 Mvar

P = -4.86918 MW

Q = -3.21012 Mvar

P = -1.04447

MW

Q = -0.562251

Mvar

M1

P = -6.237 MW

Q = -3.74218 Mvar

M2

P = -6.25 MW

Q = -3.74998 Mvar

BUSG3

U = 1 p.u.

FI = 3.16042E-021

degrees

P = -15 MW

Q = -7 Mvar

G3

P = 15 MW

Q = 7 Mvar

BUSG1

U = 1 p.u.

FI = 1.1416E-033

degrees

P = -0.363609 MW

Q = -2.08899 Mvar

G1

P = 0.363609 MW

Q = 2.08899 Mvar

BUS2

U = 0.998661 p.u.

FI = -0.0214586

degrees

P = 4.158 MW

Q = 2.49442 Mvar

M3

P = -4.158 MW

Q = -2.49442 Mvar

BUSG2

U = 1 p.u.

FI = 3.16042E-021

degrees

P = -15.36 MW

Q = -9.1 Mvar

G2

P = 15.36 MW

Q = 9.1 Mvar

BUS7

U = 1 p.u.

FI = 3.16042E-021

degrees

P = 4 MW

Q = 1.9373 Mvar

0

P = -4 MW

Q = -1.9373 Mvar

BUS3

U = 0.968803 p.u.

FI = -2.76627

degrees

P = 1.037

MW

Q = 0.49524

Mvar

0

M4

P = -1.037 MW

Q = -0.49524 Mvar

BUS4

U = 0.94939 p.u.

FI = -4.19837

degrees

P = 4.83346 MW

Q = 2.70104 Mvar

0

P = -0.95078 MW

Q = -0.579376 Mvar

0

P = -1.80268 MW

Q = -0.87304 Mvar

M5

P = -1.04 MW

Q = -0.624314 Mvar

M6

P = -1.04 MW

Q = -0.624314 Mvar

BUS5

U = 0.923247 p.u.

FI = -6.36011

degrees

P = 0.945192 MW

Q = 0.528144 Mvar

0

0

P = -0.426192 MW

Q = -0.206448 Mvar

M7

P = -0.519 MW

Q = -0.321697 Mvar

BUS6

U = 0.999331 p.u.

FI = -0.0107136

degrees

P = 4.158 MW

Q = 2.49443 Mvar

M8

P = -4.158 MW

Q = -2.49443 Mvar

18

Appendix 3

Short circuit calculations

Generator without load:

Nominal current is

√

√

Initial maximum value of short-circuit current with one generator working is

Total short circuit current is

DC-offset:

√

It gets the maximum value when .

First peak is when t=0.0083 s

√ √

With load

Load of generator:

( )

( )

( )

( )

19

√

DC-offset:

√

It gets the maximum value when .

First peak is when t=0.0083 s

√

√

20

Appendix 4

Current contribution from induction machines during three-phase fault on Bus 1

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,300

0,400

0,500

0,600

0,700

0,800

0,900





ASYNC M1 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,200

0,300

0,400

0,500

0,600





ASYNC M3 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,300

0,400

0,500

0,600

0,700

0,800

0,900





ASYNC M2 I POS. kA

21

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,100

0,150

0,200

0,250

0,300

0,350





ASYNC M4 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,150

0,200

0,250

0,300





ASYNC M5 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,80

1,00

1,20

1,40

1,60





ASYNC M7 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,150

0,200

0,250

0,300





ASYNC M6 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,200

0,300

0,400

0,500

0,600





ASYNC M8 I POS. kA

22

Appendix 5


0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,2900

0,3000

0,3100

0,3200

0,3300

0,3400

0,3500





ASYNC M1 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,2800

0,2900

0,3000

0,3100

0,3200

0,3300

0,3400

0,3500





ASYNC M2 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,1050

0,1100

0,1150

0,1200

0,1250

0,1300





ASYNC M4 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,1900

0,2000

0,2100

0,2200

0,2300





ASYNC M3 I POS. kA

23

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,80

1,00

1,20

1,40

1,60

1,80

2,00





ASYNC M7 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,150

0,200

0,250

0,300

0,350





ASYNC M6 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,150

0,200

0,250

0,300

0,350





ASYNC M5 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,1900

0,2000

0,2100

0,2200

0,2300





ASYNC M8 I POS. kA

24

Appendix 6


0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,2850

0,2900

0,2950

0,3000

0,3050

0,3100

0,3150





ASYNC M1 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,2900

0,2950

0,3000

0,3050

0,3100

0,3150





ASYNC M2 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,1950

0,2000

0,2050

0,2100





ASYNC M3 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,1060

0,1080

0,1100

0,1120

0,1140

0,1160





ASYNC M4 I POS. kA

25

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,1200

0,1220

0,1240

0,1260

0,1280

0,1300

0,1320





ASYNC M5 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,1200

0,1220

0,1240

0,1260

0,1280

0,1300

0,1320





ASYNC M6 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

1,00

1,20

1,40

1,60

1,80

2,00

2,20





ASYNC M7 I POS. kA

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00

TIME SECONDS

0,1950

0,2000

0,2050

0,2100





ASYNC M8 I POS. kA

26

Appendix 7

Voltage variation of induction motors when motor M1 is disconnect

0,80 0,90 1,00 1,10 1,20 1,30 1,40 1,50 1,60 1,70 1,80 1,90 2,00 2,10 2,20 2,30

TIME SECONDS

7,960

7,980

8,000

8,020

8,040

8,060






0,80 0,90 1,00 1,10 1,20 1,30 1,40 1,50 1,60 1,70 1,80 1,90 2,00 2,10 2,20 2,30

TIME SECONDS

7,960

7,980

8,000

8,020

8,040

8,060






0,80 0,90 1,00 1,10 1,20 1,30 1,40 1,50 1,60 1,70 1,80 1,90 2,00 2,10 2,20 2,30

TIME SECONDS

7,940

7,960

7,980

8,000

8,020

8,040






0,80 0,90 1,00 1,10 1,20 1,30 1,40 1,50 1,60 1,70 1,80 1,90 2,00 2,10 2,20 2,30

TIME SECONDS

3,4350

3,4400

3,4450

3,4500

3,4550

3,4600

3,4650

3,4700






27

0,80 0,90 1,00 1,10 1,20 1,30 1,40 1,50 1,60 1,70 1,80 1,90 2,00 2,10 2,20 2,30

TIME SECONDS

3,3600

3,3650

3,3700

3,3750

3,3800

3,3850

3,3900

3,3950

3,4000






0,80 0,90 1,00 1,10 1,20 1,30 1,40 1,50 1,60 1,70 1,80 1,90 2,00 2,10 2,20 2,30

TIME SECONDS

3,3600

3,3650

3,3700

3,3750

3,3800

3,3850

3,3900

3,3950

3,4000






0,80 0,90 1,00 1,10 1,20 1,30 1,40 1,50 1,60 1,70 1,80 1,90 2,00 2,10 2,20 2,30

TIME SECONDS

0,24600

0,24650

0,24700

0,24750

0,24800

0,24850






0,80 0,90 1,00 1,10 1,20 1,30 1,40 1,50 1,60 1,70 1,80 1,90 2,00 2,10 2,20 2,30

TIME SECONDS

7,940

7,960

7,980

8,000

8,020

8,040

8,060






28

Frequency variations when motor M1 is disconnect

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,991200

0,991400

0,991600

0,991800





ASYNC M1 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,991100

0,991200

0,991300

0,991400

0,991500

0,991600

0,991700





ASYNC M2 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,990900

0,991000

0,991100

0,991200

0,991300

0,991400

0,991500

0,991600





ASYNC M3 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,988900

0,989000

0,989100

0,989200

0,989300

0,989400

0,989500

0,989600





ASYNC M4 SPEED p.u.

29

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,991100

0,991200

0,991300

0,991400

0,991500

0,991600

0,991700

0,991800





ASYNC M5 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,991100

0,991200

0,991300

0,991400

0,991500

0,991600

0,991700

0,991800





ASYNC M6 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,989600

0,989700

0,989800

0,989900

0,990000

0,990100

0,990200

0,990300





ASYNC M7 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,991000

0,991100

0,991200

0,991300

0,991400

0,991500

0,991600





ASYNC M8 SPEED p.u.

30

Appendix 8

Induction motor curves for critical clearing analysis, Mechanical load k=0.85

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0 11,0 12,0 13,0 14,0 15,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





9,3, 0,99121

ASYNC M1 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





9,5, 0,99107

ASYNC M3 SPEED p.u.

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0 11,0 12,0 13,0 14,0 15,0

TIME SECONDS

0,98400

0,98500

0,98600

0,98700

0,98800

0,98900

0,99000





9,4, 0,98904

ASYNC M4 SPEED p.u.

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0 11,0 12,0 13,0 14,0 15,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





9,3, 0,99121

ASYNC M2 SPEED p.u.

31

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0 11,0 12,0 13,0 14,0 15,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





9,4, 0,99123

ASYNC M5 SPEED p.u.

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0 11,0 12,0 13,0 14,0 15,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





9,4, 0,99123

ASYNC M6 SPEED p.u.

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0 11,0 12,0 13,0 14,0 15,0

TIME SECONDS

0,98400

0,98500

0,98600

0,98700

0,98800

0,98900

0,99000





9,5, 0,98975

ASYNC M7 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





9,3, 0,99108

ASYNC M8 SPEED p.u.

32

Three generator/turbine sets working

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





11,1, 0,99121

ASYNC M1 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





11,1, 0,99121

ASYNC M2 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





10,9, 0,99107

ASYNC M3 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98400

0,98500

0,98600

0,98700

0,98800

0,98900

0,99000





10,9, 0,98904

ASYNC M4 SPEED p.u.

33

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





10,9, 0,99123

ASYNC M5 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





10,9, 0,99123

ASYNC M6 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98400

0,98500

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100





10,9, 0,98975

ASYNC M7 SPEED p.u.

0,0 2,0 4,0 6,0 8,0 10,0 12,0 14,0 16,0 18,0 20,0

TIME SECONDS

0,98600

0,98700

0,98800

0,98900

0,99000

0,99100

0,99200





11,0, 0,99108

ASYNC M8 SPEED p.u.

34

Appendix 9

Matlab code for starting currents and frequencies

function [] = torque_marine( )

clear all hold off %control values B=0.65;

Tba1=0.1; Tba2=0.2; Tba3=0.3; Tba4=0.4;

%system

fn=50; l=50; %motor:

Tn=9900; Un=6.6*10^3; Unpu=1.0; Sn=1.2*10^6; In=Sn/(sqrt(3)*Un); Zbase=Un^2/Sn; sn=(2*fn*pi-996*2*pi/60)/(2*pi*fn);

Rs=0.0215; Rr=0.011; X1s=0.0673; Xs2=0.0673; Rr1=Zbase*Rr;

%cable: %In=241;

Rac=0.25; Racpu=Rac/Zbase; Xc=0.37*10^(-3)*2*pi*50; Xcpu=Xc/Zbase;

%transformer St=5*10^6; Zt=St*0.0403/Sn; Rt=Zt/sqrt(8.9^2+1); Xt=8.9*Rt;

Rsys=Rs+Rr+Racpu*l+Rt; Xsys=X1s+Xs2+Xt+Xcpu*l;

Zpu=sqrt(Rsys^2+Xsys^2) Z=Zpu*Zbase f=0.0;

while f <= 50

35

Tpu=(3*(Unpu*(1+B)*f/(fn*Zpu))^2*Rr/(f/fn));

plot(f,Tpu);

hold on f=f+0.1; end

xlabel('Frequency [Hz]') ylabel('Torque [pu]') title(['Start-up torque,B=',num2str(B), ', l=',num2str(l), ' km']); grid;

%current fba1=Tba1/(3*(Unpu*(1+B)/(fn*Zpu))^2*Rr*fn) Ist1=sqrt(Tba1*Tn*2*pi*fba1/(3*Rr1))

fba2=Tba2/(3*(Unpu*(1+B)/(fn*Zpu))^2*Rr*fn) Ist2=sqrt(Tba2*Tn*2*pi*fba2/(Rr1*3))

fba3=Tba3/(3*(Unpu*(1+B)/(fn*Zpu))^2*Rr*fn) Ist3=sqrt(Tba3*Tn*2*pi*fba3/(3*Rr1))

fba4=Tba4/(3*(Unpu*(1+B)/(fn*Zpu))^2*Rr*fn) Ist4=sqrt(Tba4*Tn*2*pi*fba4/(Rr1*3)) end

36

Appendix 10

ATPDraw simulations of starting currents and speeds

Cable: 2XS(FL)2YRAA 6/10(12) kV, In=288 A

B=0.65, l=10 km, Tba=0.2, finit=14 Hz:

B=0.65, l=10 km, Tba=0.3, finit=21 HZ:

B=0.65, l=10 km, Tba=0.4, f=28 Hz


0

150

300

450

600

750

900


0

150

300

450

600

750

900

37

B=0.65, l=20 km, Tba=0.1, finit=9.4 Hz

B=0.65, l=20 km, Tba=0.2, f=19 Hz

B=0.65, l=20 km, Tba=0.3, f=29 Hz


-2500

-2000

-1500

-1000

-500

0

500

1000

(file startup1(1).pl4; x-var t) m:I1 m:I2 u1:OMEGM u2:OMEGM 0 5 10 15 20 25[s]

0

100

200

300

400

500

600

700


-2200

-1700

-1200

-700

-200

300

800

38

B=0.65, l=20 km, Tba=0.4, f=38 Hz

B=0.65, l=30 km, Tba=0.1, f=12.5 Hz, cannot start

Cable changed to In=327 A

B=0.65, l=30 km, Tba=0.1, f=10 Hz


-2200

-1700

-1200

-700

-200

300

800


-2500

-2000

-1500

-1000

-500

0

500

1000


-2000

-1000

0

1000

2000

3000

4000

5000

39

but when Tba=0.2 stalling again.

Cable changed to In=464 A

B=0.65, l=30 km, Tba=0.1, f=10 Hz:

B=0.65, l=30 km, Tba=0.2, f=10 Hz

B=0.65, l=30 km, Tba=0.3, f=10 Hz


0

100

200

300

400

500

600

(file startup1(1).pl4; x-var t) m:I2 u1:OMEGM 0 5 10 15 20 25[s]

0

100

200

300

400

500


0

100

200

300

400

500

40

B=0.65, l=30 km, Tba=0.4, f=10 Hz

Tba=0.1, l= 50 km, f=10 Hz, when Tba=0.2 stalling


0

100

200

300

400

500


0 5 10 15 20 25[s]0

100

200

300

400

500


0

100

200

300

400

500

600

41

Tba=0.1, l= 40 km, finit=10 Hz

Tba=0.3, l= 40 km, f=20 Hz

Tba=0.4, l= 40 km, f=20 Hz


0

100

200

300

400

500

600


0

100

200

300

400

500

600

700


0

100

200

300

400

500

600

700

42

Appendix 11

Cable data

marine and offshore power system

Documents

norm al load

offs hore

offshore power

critical clearing

long subsea

async m6 speed

async m3 speed

async m2 speed