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EE321Power Systems Analysis

Experiment 6Semester I , 2016

School of Engineering & Physics Faculty of Science & Technology

Electrical & Electronics Engineering Discipline

Team Members

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Asesela Sivo

Edward Chan

Elia !nior Ma"i

Mohammed M!#a""ar $han

S11111%01

S0000&033

S110''(31

S110))'()

*ate + 2(th April 2016

ab Session+ Monda-s )am.12am

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2

Asesela SIVO

Edward CHAN

Elia Junior MAFI

Mohammed Muzaffar

S1111141

S!""

S11##$"1

S11%%#$%

Week9-Lab6

Al&erna&or 'ower

/ECTIES

• To observe the effect of dc excitation on the power delivered by an alternator.

• To observe the effect of power delivered by an alternator on the torque of the primemover.

E.A

Problem 1

Three physically identical synchronous generators are operating in parallel. They are all rated

for a full load of 3 MW at 0. !" lagging. The no#load frequency of generator $ is %& '() and

its speed droop is 3.* percent. The no#load frequency of generator + is %&., '() and its speed

droop is 3 percent. The noload frequency of generator - is %0., '() and its speed droop is 2.%

 percent.

• 3 synchronous generators operating in parallel• $ll rated for a full load of 3MW 0. !" lagging

•  / 1$ frequency of generator $ is %&'() speed droop 3.*4

•  / 1$ frequency of generator + is %&.,'() speed droop 34

•  / 1$ frequency of generator - is %0.,'() speed droop 2.%4

a5 6f a total load consisting of 7 MW is being supplied by this power system) what will the systemfrequency be and how will the power be shared among the three generators8

�� =�� −

����

� 100%

=��� −

������

���

� 100%

� ���� ��������� =

 ��100

+ 1

• Therefore we can calculate the full load frequencies of the three generators.

��������� � ∶ � =

 

���

�� ��100

+ 1

��������� � ∶ � =

 �

��

�� ��100

+ 1

�����

����

�

=

 ��

��100

+ 1

61��=3.4

100+ 1

61.5��=3

100+ 1

60.5 ��=2.6

100+ 1

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3

Asesela SIVO

Edward CHAN

Elia Junior MAFI

Mohammed Muzaffar

S1111141

S!""

S11##$"1

S11%%#$%

Week9-Lab6

Al&erna&or 'ower

= ��. � � ~� �� � = ��.

� �~�

� �� = ��. � � ~� �� �

• We must now determine the slopes of the power frequency curves for each generator.�3 

��� == �. ��/

2��

�3 

��� = 1.79��= �. � ��/

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�3 

��� = 1.53��= �. � ��/

• "rom the question we 9now that the load of the system is 7MW) from which we can

now determine our system frequency by the ���� equation provided in notes.

= ��� (���� − �) : ��� (���� − �   ) : ��� (���� − �   )����

7 � =1  (1.5)(61 − �) + (1.67)(61.5 − �) + (1.96)(60.5 − �   )

• ;olving for �   gives us

5.137 ∗ �   = 306.2

     � = ��. � �  

There"ore the "re4!en5- o" the s-stemis ()60 7#

• We can now determine the power shared among each generator 

 � = ��  (���   −   � = ��  (���  − � = ��  (���  −

= (1.5)(61 − 59.61)   � = (1.67)(61.5 −  � = (1.96)(60.5 −

� = �. ��� � = �. ���� � = �. ��

Therefore we have< ower s!pplied b- 8enerator A 9 20'(M:

ower s!pplied b- 8enerator 9 31(63M:

ower s!pplied b- 8enerator C 9 1&%M:

 b5 -reate a plot showing the power supplied by each generator as a function of the total power supplied to all loads =you may use M$T1$+ to create this plot5. $t what load does one of the

generators exceed its ratings8 Which generator exceeds its ratings first8

• 6n order to draw the plot we need to represent the system frequency =�   5 as a functionof load

 ���� = (1.5)(61 − �) + (1.67)(61.5 − �   ) + (1.96)(60.5 − �   )

313.2 − ���� = 5.137 ∗ �

∴ �   =

 

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313.2−

 ����5.137

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Matlab Code+

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/!tp!t+

Graph 1: Power Sharing Vs Total Load Power 

"rom the plots on M$T1$+) we can view the power sharing problems for both high and low loads. "romthe plots) we can see that >enerator + exceeds its load power ratings at about approximately %.,MW.

c5 6s this power sharing in =a5 acceptable8 Why or why not8

This type of power sharing is generally not acceptable because the >enerator#2 exceeds its load power rating at about %.,MW and we have an unbalanced situation which is not acceptable in

 power system applications.

d5 What actions could an operator ta9e to improve the real power sharing among these

generators8

• To improve the real power sharing among these generators) the operator could decreasethe set points on generator + while simultaneously increasing the set points on generator $ and generator -.

• The operator can maintain the system frequency of the generators close to %0'( bysynchroni(ing the speeds of each generator by some means.

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roblem 2

$ 20#M?$) &2.2#9@) 0.#!"#6agging) @#connected synchronous generator has a negligible

armature r esistance and a synchronous r eactance of &.& per unit. The gener ator is connected in

 parallel with a ,0#'() &2.2#9? infinite  bus that is ca pable of supplying or consuming any

amount of r eal or r eactive  power with no change in frequency or terminal voltage.

a5 What is the synchronous reactance of the generator in ohms8

��� =

3(�∅,��)2

=���

3(7044)2

= 7.44Ω20000000�

��� = 0Ω, �   = (1.1)(7.44) = 8.18

 b5 What is the internal generated voltage  E $ of this generator under rated conditions8

��=

�� = 

=√3  

20 �� = 946� , power factor is 0.8 lagging 

√3(12.3 )� 

. => �� = 946∠   − 36.87°

�� = �∅ + ���� + ���� = 7044+j (8.18) (4! ∠  "#!.87°) = 1#$#0∠   $7.% 

c5 What is the armature current  I A in this machine at rated conditions8

�� = 946 ∠   − 36.87°

d5 ;uppose that the generator is initially operating at rated conditions. 6f the internalgenerated voltage A$ is decreased by , percent) what will the new armature current  I A be8

 =   3 ∅ �� ��� =

��������, ���� = �

  ��� 

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&' ecrease ��2 =12570���1

�1 1 �2 2

13230�

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���−1

(��2

) ��� 2 =

���−1  (12570

)��� 27.9° = 29.5°

 *ence new aratre crrent ��=

��2−  ∅� 

= 1$&70-

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e) !lot the magnitude of the armature current   $ as a function of E $.

MATA Code+

/!tp!t+

Graph 2: Armature Current Vs E  A

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I;STenerator 

M/*E

2&&

;ynchronous MotorB>enerator 2*&

- ?oltmeterB$mmeter *&2$- $mmeter *2,

$- ?oltmeter *2%Three#!hase Wattmeter **&

;ynchroni(ing Module %2&!ower ;upply 2&'and Tachometer C20

-onnection 1eads C*&Timing +elt C*2

/CE*enerator) - MotorB>enerator) ;ynchroni(ing Module)

!ower ;upply) $- $mmeter and $- ?oltmeter) connect the circuit shown in "igure ,#&.

 /ote that the output of the alternator is connected through the ;ynchroni(ing Module to

the fixed *&, ?) 3#phase output of the !ower ;upply) terminals &)2 and 3. The rotor of

the synchronous motorBgenerator is connected to the variable 0#2*0 ? dc output of the!ower ;upply) terminals 7 and /. The dc shunt motor =dc motorBgenerator5 is connected

to the fixed 2*0 ? dc output of the !ower ;upply) terminals and /.

:arnin=+ 7i=h volta=es are present in this aborator- Experiment> *o not ma?e an-5onne5tions with the power on> The power sho!ld be t!rned o"" a"ter 5ompletin= ea5h

individ!al meas!rement>

;/TE+ A ad@!stments i" an- !sin= the dial on an- mod!leB while the s-stem is

powered on m!st be done slowl- steadil- to avoid an- transients> ail!re to 5ompl-

will res!lt in the s-stem o!ta=e d!e to 5ir5!it brea?er operation or ma- 5a!se "ire>>

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Figure6.1

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&.

a5 -ouple the - MotorB>enerator to the alternator =;ynchronous MotorB>enerator5

with the Timing +elt.

 b5 ;et the field rheostat of the - MotorB>enerator at its full cw position =forminimum resistance5.

c5 !lace the synchroni(ing switch in its open position.

2.

a5 Turn on the !ower ;upply. Dsing your 'and Tachometer) adEust the rheostat ofthe - MotorB>enerator for a motor speed of &,00 rBmin.

 b5 Turn on the switch ; and adEust the dc excitation of the alternator until the output

voltage A) *& , ? ac.

c5 ;ynchroni(e the alternator with the power line and turn on the synchroni(ingswitch.

d5 -arefully adEust the dc excitation of the alternator as well as the speed of the

motor until both watteter and /areter indicate (ero. Measure A&) 6&) and 62.

A& %2() ? ac 6&

003 m$ ac 62

011 m$ dc

The alternator is now FfloatingF on the power line. 6t is neither receiving power 

from the line nor delivering power to the line.

T7E EDEIME;T C/

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*ISC

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*ISC

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And of Axperiment#% Geport =2, B 0* B 20&%5