42
1— يهق ا ا ه ر ؟ت ا وت ا ت ي انه اق ا ا ا) و ( ان دا ر ا ا) ور ا ) 2 -- م DIRECTIONAL RELAY ا ا ه ؟ و ؟. directional relay ال م bus bar-generators -transformer ا وه م over current relay ه ال ع ا اrelay ء وث اى (fault) ا ا ان وث اى وا دى ا ا ر ا د ا ا(fault) ال ا و ر ان ا relay م الrelay الC.B . ال وdirectional relay : 1 -- م ىع ا اover current relay رس ا ا عا ا وهر ا د وز اInstantaneous directional relay--2 عا ا وه م و رس ا اد و ات ا ال bus bar 3 -- ن ا ا ا اه BUSBAR DIFFRENTIAL RELAY ؟؟. اف ا ا voltage supervision relay 4 -- ذا اHVHRC ؟ ج: ــ ر إا ا هHigh Voltage High Rupturing Capacity fuse تت ات و ا ات ا م وهتت واآ وا ا ا. 5 -- ل ذا ؟ م ا ر اار ا د رل ا ) ر ا أو ا ا ن ا أو ا... إ( ر د ووف ا ي ا) ل ا .( ر أي أنلوف ا ا ا اوف دا . ار ا م اج ا ة دا م يل ا ا ت آ ي ا. ر أي أن ا ا) أ ( ر ا . ر ا م ) ي ا ( ج إ و ا ا ا ا آ اي ا ا . ية ال دام ا و ر ل ا ا د ا ا مم، و ير ا ن ا ذي اار ا ا) ا أو ا.( ق و اي ا) ا( ا إ ا ةت آ إي ا ة آ ي ة ا أو ا ل أو ا ورةل أو ا ا. ا آ ا ه ل ي ا ا ا ا ت وآ ا ارة ا ا واوارات ا ا ا. ام أ ولت اآ ى إ ء ا أ - ية ا دا

اسئلة واجوبة في الحماية الكهربائية

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Page 1: اسئلة واجوبة في الحماية الكهربائية

��� ����ت ا����� و ����ت ا�� ر ؟ � ه� ا��ق ا��ه�ي —1�� ا��ه�ي ان ����ت��� ا��ق ا�� � ��� و� !���� (ا����� -,(+ &*() '&% $�# �"!

�� �� ��"�� دا2) /� ��1 ان) �'&% /���� �� ���� ا�� ر -,(+ &*() '&% /�# �"!89 �7 -� ر ا�() و� !���� '&% 456( ا�4�4 !�: ��"��) ��<4م -- 2: +�/ DIRECTIONAL RELAYJ'ا�Kه� ا ؟ ؟ و�

. directional relay ال J: ��<4م /% 1(- bus bar-generators -transformer _ J`:�aا bcوه ��<4م �*5: d over current relay

ء relayا(!J`a ا(��gع '&�5 ال ه% -�(%a2 )�� 41وث اىfault( b�� /% ا4ا (fault) ا� J ا* د:J :(� ا�� ر /% /% ا�-� b ا* دى &4ا��b و�k '!4 41وث اى /!�4 ان /%

% :(� /% ا-� b ال- ن ا-� b ا�� ر :!*lk و�/ relay �`�/م الrelay (,�� الC.B : directional relayو:��K 46'�� �� ال.

1 -- J*� 4م>�� س ا�� ر over current relay ا!�ع اcى :k*K!�ع :�,) �*4 اا اcوه ا(`!� وز: د-J '� ا�� ر

Instantaneous directional relay--2 4*� �q� (,�: !�عا اcوه س ا�� ر /�J و:��<4م �(k*Kو�6د ا J� :J ا(�4ات /% 1

��bus barl�K %&' �:4 ال3 --��&g ن ا��ds`ا �: )� �d1 ه� ا�( :� ا(,� BUSBAR DIFFRENTIAL

RELAY؟؟ :� ا��xاف ا����.)1 voltage supervision relay

؟ HVHRC ا��{ � ذا :*!% } -- 4 High Voltage High Rupturing Capacity fuse هcا ا��{ إ�2, ر ــ: ج

ت��k)ت ا :� ا(���ت و�1)� % �ت وه� :��<4م /% �1 ت ا�45 ا*� kت وا وا(��آ .ا�% -*() '&% 456 ا��"�)

ا�� ر '!4 '4م ا-, 5+ ��()؟ ( ذا :�8 $,� �/� ���ل -- 5 �� �) أو ا<� ا( ر(/� ���ل ا�� ر -��4د $�(� -� ر ا���4اk*(����/� ان ا ds`...أو ا�إ (

1�8 ��وف ا��kd و� د2) �� ر J�)�$ �/ ي�K أي أن -� ر ). '&% 'lk ���ل ا�45(ا���`) '� ��وف ا(��ل� ��5 ��وف دا��-J ا� �K:� ا���4ا�/ )� . �� ر ا���4ا�- +q*� م�`:

��� ج ا��# ا("!�K�� /4 $�ة دا�ي :`�م ���c(��ل ات /� $&8 ا *� آ����5 /� �&�) /� ا� �Kي(:`�م -� ر ا�() . -� ر ا("!�a) /� أ_&Jd(ا���4ا�� :(�) أي أن -� ر. ا� �Kي

ج�K5(� إ)� �� � � :�4 �� ا��# ا(�,) و�)� �� آl ��# ا���4ا*� ��� !"� #�/��ل -� ر ��() �7 و/� 1 � '4م ا-, ل دا��ة ا� �Kي (. �/� ا(&� ا� �Kي ا�45 '&%

5� `� ��dkا ��� د &��# ا("!s)ا �����ا J*� ي :!*4م، و:!*4م�K �����1 /�ن -� ر ا� �� ر ا���4ا�� ذي ا`�(� ا*�- �� �- ا�45 ��� �/� وc�!�1 :�-�7 /�ق). أو ا* �� 46ا(ا!

ت آ��dة 46ا $4 -,) إ% ا�) ا(�����1(ا� �Kي :��� � آ��dة k) �� 4 اcيإ% �>� 8d�:آ( :���� ا`&8 . ا(��ل أو ا(� ورة J ا(��ل أو &�<� ا(�* �) �*J أو &(*4ة ا�� -���ي

�)�` � � �� ا�4:4ي &(��ل /� هbc ا�� �Jg�*- �� Jdd ا* �� 46ا &��# ا("!- )� ت:���� {c4:4 وآ�7 اd��& !ارة ا��ا �� �� -�� �� ا�� رات ا4وا��� وا�<&� '

��� .ا("!Jq1�� J�&)'

دا��ة ا� �Kي ����- '� _�� $,4–أ�! ء ا*() /� إ41ى �xآ ت ا��dول $ م أ41 ا(5!��94

Page 2: اسئلة واجوبة في الحماية الكهربائية

�. (��ل -� ر أ�! ء /�,J 4ا��ة -�"�) و-�k+ �*`4ة �41 ا`�ا 7K ا!���� إ� �� ا{��) وآ و�) ����K b4: �/ ز���ق را -<�ج �� 56K Jd�: JK�� J� م �&.

��`� �� ا��� ا��4 &(���ت �7 ا`�ا 7 -- 6� )� و_��ه ؟ ( ذا :��% دا1 � ا��� _�� ا���k�: 4ن /�اغ ه�ا�% /% هbc ا(!�k: J`aن %/ JK� ��!: )� �: �& (� $

Ja`!ا bcل /% ه *�xاى :�4ث ا �)>� :*�ف �� J!' 7 -- ر ا4dء /% ا(��آ ت؟ه ��- (�&`� J�4>�� % ا�aق ا(

(�&`� J�4>�� ر ا4dء &(��آ ت ا�% -*() '&% �56د �"��b �� ا�aق ا(�- 1-soft starter

2- )�K J&���- د� /% ا(��آ ت ا�% -*() '&% 456 ����9

��ا% �`&�) ا�� ر ا(��ك و-�kن ��,&J �7 ا*�s ا4وار � ا�9<4ام �` و�J '!4 �4ء -�"�)�- ا! وآ+ ا�$�. :�4ث (��� /����'!4 -( س ا41 ا� �ار ا���� و�`�� /� ا*() � ذا-- 8

ا(��"�ق &�4ث ؟ و( ذا؟1 � �4ء ا�( س وذ} �ن) د$�`�50( -!��� 2�ل �� J�/ ي 41ثcر ا�a-!�`) �� ا �����ا

£: )� �)�&� ا% ا� �ار اgر دي ا%ا*d:�`- 5�)�$ �*g %ا �)�&� .ار-� ع ا� �ار ا) � ه� أه(�� -- 9Tertiary Winding(���ت؟ (ا �/

Tertiary Winding ال%/ ¦ /� إ% /% ا(���ت ه% �&� �g§ ا(��ل �

��<4م (�ور ��آ�d ا�� ر ا,��: ا(&� ت ا§��4ا��� و ا� �K:� و :��) '&% ه��� د� و:� ج 456 � ¦ &(��ل و :<�&� $�(� /� 1 � '4م إ-{ان ا�1( ل '&% ا(��ل و�K§ 4م>��:

��� و /� آ��� �� ا��1 ن -�kن $4ر-5 �&¦ $4رة ا`4رة '&% هcا ا(&� '� ا(&�����ا� ,ا(&� ت ا��2ى

ج ا`4رة و�k (�ور ��آ�d و /� أ�1 ن�K§ �&)ا اc,��:�أ�2ى � :�+ إ�9<4ام هر ا ا���`/.

�4ة ا<{ان ا���1 � /� ����ت ا`4رة ؟ -- 10 / � -- 1 (�&`- �:}& l�� �1 ا����a &�5اء ا(� . ا(-- 2 � 1 �/ ���� ( look out) ا!�sح -*�:# ا<{ان ا� ت ا��:� 3 --d&`�ا �� �:} � +k��4د.ا)�: � ء �1¦ ان ا{:�x �&`�:و ���

ت -- 11kdx (2-�4ث دا ��ا �������� ا� ��5kء ا a2ه� ا� ? ا!`) � .ز: دة ا�() -- 1 �x ��� ا� �ار او( دوا�� ا`,� -- 2d� س )- 4: 5 و��� ا(�!��) ($7a /� ا41 ا� �ار(ا4وا�� ا(����1 -- 3 س ��� ا� �ار وا�رض(ا* ز�� -- 4)-)

؟.ا�Kاع ����ت ا��ر:# -- 12 ا{6{اج ���ل 1 -- open delta ���ل أ�ــ -- 2

��<4م /� ����ت ا4� 2و1ا(���ت - 3 -- �� و-��<4م /� -�ر:# ا(�4ات.. ���ل ا(* و$� ا* ذا :��<4م-- 13) l�و J6او ��� م ا�qK ����5k`4رة اا ���q!� �/4؟؟6ال5او

مqK ا�9<4ام (s�: d9� J6او ��� :ب ا�-��ا�1 -- J6او ��� م ا�qK �/ ت `&� ا��ا/`� .$&� ا�k&�� ا�$�, د:� -- 24 �� -� ر -- 3�& �K �&���- ��K(�ود�k4ء ا�dا.

Page 3: اسئلة واجوبة في الحماية الكهربائية

ه� و���� -���&� ال ـ-- 14� ?? Open delta ا�45 -�<4م /% ���ل

J641وث $,� ��� ا41 ا�و � ��9<4ا�� و ا�رض&�,�ل '&% 456 ا(�آ�d ا,��:� /% 1 /�� او ا�$ :� �� ز: دة� ر ا`,� /% ا�$ :� ا(�- b ا�� ر /ى��4:4 ا-�

�4ة ر�� ا<� ا�ا�7 -- 15 / ر '!4 ر�� ا(&� ت ��a:`�) ا�ر��(��� ؟(Y)ا()� � �K(�از4ة �6اء '4م ا� وذ} �(�:� ا�� رات ا{ا

16 -- ATS?? ه� � م :��<4م /� -�qK ����5ه�k`4رة ا4 ا�) ac( 8 رة /� ا"d' وه� J� (,�: ك د:{ل��� �' alternator �$ aل ا��: bي �4ورcوا ��k�K ا% آ����5k�)اac ة رات ا* د:� �7 ��ا'�� ��K -�آ��d '&% اk��� م q!ا اcز ه و:(�

زآ( :(� ac to dc وذ} ���:) matching ودا��ة rectifier و�6د دا��ة alternator %&' J-4ر`� lk' {(��ك وذ} /`� �(��د -�"�) اوذ ���&a)`4رة اج ا �Kإ

رات ا(�4ات ا* د:� /��� � -�-�7 $4رة ) د:!(�(ا� واcي :�a&8 ر/7 $4رة ا(��ك و� ج ا4:!(� '&%�K§ا وا® أ'&+.اcه

'4ة� ت أآ�� '� هcا ا(��gع ��6 ء ا(��&*� J:4 �� و +k��ت ا 1� �' �� >��)qK!�ع �� ا�ا اc��4 /� -�"�) ه>�� .ا(

ذا -- 17) �`d ر�6 او �ت :�4x 8ه �� ا"�ف ا<� k4 8�9 ا!' Armory ؟ �ت� k4 8�9 ا!' �`d ر�6 او و� :�+ 4xه �� 4x 8�: Armoryه �� ا"�ف ا< ا(��) او ا* زل

�) و�c} :`)ان -+ 4xه �� ا(��) �9ف :�k7 اa`� �1 � � &(��) و-`) �a�9ا + ()�-(� kا

�� ر و2��&Is.c �ت �� ا* زل /c5ا :£دي� k4 اx +- ان و$,� '(�isolating b ا� $&� ا�

18 --What is the IP standard "Protection degrees of enclosures :�kن ��kن �� ��ث ار$ م :� bc5 ا(*4ة �1¦وه� در�6 ا�( IP آ) 56 ز او �*4ة :�kن 5

ex. IP 543 dust �� وه� در�6 1( :� ا�5 ز" 5"ا�$+ ا�ول �K water or "liquid" وه� در�6 1( :� ا�5 ز ��" 4"ا�$+ ا�

mechanical impact وه� در�6 1( :� ا�5 ز ��" 3"ا� ¦ ا�$+�: �<�&�� k) ر$+ وه! ك در6 ت 1(

Kانو �g�: {8 وذ�k: � ¦ ن ا�$+ ا�K ر ان ا�1d�'ا� �/ c2 هbc ا( آ�!� ��``� mechanical impact

ذا -*�ف '� ال -- 19� ANSI DEVICE NUMBERS ؟ % � ت آ: رb '� ار$ م � ���d*- J'� ا�Kاع �<�&�J �� ا�(d' %ه:

NO.2 MEANS TIME DELAY NO.21 MEANS DISTANCE

NO.25 MEANS SYNCHRONISM-CHECK NO.27 MEANS UNDERVOLTAGE NO.30 MEANS ANNUNCIATOR

NO.32 MEANS DIRECTIONAL POWER NO.37 MEANS UNDERCURRENT OR UNDERPOWER

NO.38 MEANS BEARING NO.40 MEANS FIELD

NO.46 MEANS REVERSE-PHASE

Page 4: اسئلة واجوبة في الحماية الكهربائية

NO.47 MEANS PHASE-SEQUENCE VOLTAGE NO.49 MEANS THERMAL

NO.50 MEANS INSTANTANEOUS OVERCURRENT NO.51 MEANS AC TIME OVER CURRENT

NO.59 MEANS OVER VOLTAGE NO.60 MEANS VOLTAGE BALANCE

MEANS PRESSURE (MECHANICAL PROTECTION) NO.63 NO.64 MEANS APPARATUS GROUND

NO.67 MEANS AC DIRECTIONAL OVER CURRENT NO.68 MEANS BLOCKING

NO.69 MEANS PERMISSIVE NO.74 MEANS ALARM

NO.76 MEANS DC OVER CURRENT NO.78 MEANS OUT-OF-STEP

NO.79 MEANS AC RECLOSING NO.81 MEANS FREQUENCY

NO.85 MEANS CARRIER OR PILOT-WIRE NO.86 MEANS LOCK OUT

NO.87 MEANS DIFFERENTIAL NO.94 MEANS TRIPPING

� ا����� ؟ � ه� 8d9 ار-� ع 9*� ا�,) /� -- 20s�>Kا $�ا 7 ا4ا��ة آ&(� ا����� $&� ا��ارةs�>Kا .ا��)4ة '!4 '(&�� ا�,) و�c} -�-�7 9*� ا�,( آ&(

ن ؟ ا�� ر ا(��دد او ا�� ر ا(��(�؟ ا:5( اآ�� �a2ا '&% -- 21�Kا� ��(� وذ} �JK �*&�م ان ا�� ر ا(��دد :(� 2�ل ا�� ر ا �k- 4$ن ا�6 �� �ول وه&�: ج)

�(� &���ر 2�ل ا,*`� ا ����5k�k؟ �a`!� J��d�6 ا,�� �(: ن ���K§ا %&' �a2(��دد أر ا ا�� ر ا(��(� ، هcا &��ددات ا(!<��s ا��

2000Hz �s�>!)رات ا 5-�3 ه��-{ أ60a2-50ا�� ر ا(��دد و:�kن -���� 30mAوا���)�� ف ا�� ر -���� ا�� ر ا(*gأ.

8 �!�!% 2 ص، و-,) أ'&% $�(� -� `�(� ا��دد وه� 1*d- (��ددر ا -<�&� �a2رة ا�� . ه��-{50 – 60�a>&رة '!4 ا��دد

ا�� ر ا(�-��d�*: 7 أ$) �a2ا �` رl�K 7� �K ا`�(� ا�� ر ا(��دد ذو ا��ددات ا(�-�*� وذو - ر&�� ر�- ��� ،�)�� �� ا�� ر ��1000&� أ���d '!4 -�دد ���40دد ا(K ه��-{ :*��d أآ�� أ�

�)�� �� ا�� ر ا(��دد 30ا(K :*�l�K �d ا�� ر ا(��(� أآ�� أ�)!�� ، ��d��13&� أ��&�� . ه��-{500أ���d '!4 -�دد

- �)�� ن '!4 $�+ �<�&�� و:<�&� '&% :<�&� أ�� ا�� ر ا(��دد '� أ�� ا�� ر ا(�K§ا +�6 1�8 ا��دد، /`�k: 4ن أ�� -� ر ذو b��أ s:أ K .$�(� -�دد ' � :!�,� '&% ا��وق أ�1

ن أو !`) در�6 -�Kº �� /�ق ا�45، �` و�� : ا<�aرة -*�4) '&% ��9 '�ا�4x (ة ا§� +� ر(ا���& �&kر ا � ر ا�� ر /� ا��+، وا{�� اcي 4xة ا�� ر، �Kع ا�� ر، ،(أو ا(��

�/ J *�ض�,*`� :�+ اا. ن ه� /� ا�� ر و�l /� ا����� -�K§ا %&' %)q*رة ا�a>ا. ��� ه� � ه�ة إ�2�ج ا`&8 �� - ��5k,*`� اب /� ا :�*�ض J ا(,� �a2أ

fibrillation �$�- � � م د$ ت ا`&8 و�q�Kاب إ�agه�ة إ �g ا4م أو g*� ، وه� ��dوأآ ،�s`&8 اا '&% ا�a2 (k�: �5/ {cدد، و��)ر ا ه�ة ه� ا��qا bc5� 8d��.

b وا41 أو ا(��دد ذو -�,) -�-�� �)�� ا��+ ����K (�ور ا�� ر ا(:� ا<�aرة '&% 2

Page 5: اسئلة واجوبة في الحماية الكهربائية

d��: )� :� و� ��� ا<:��) ا§��kو��� /� ا< �2�ل ا �� �s�>!� 8-�ددات ��:���(� � ت ا(���:5 اK�:ف -�ز:7 ا���2�� 8d��) وا� � .ا

5 ��b4 ���ة �:&� و���¦ :(� �� ا����� -�ا«�!� ه� ا����� ا�� :��7�a ا�<� (J�2 ،J� *�ر�ا �k): � ر �-

ر ا���:� -�- Let Go Current b4!' يcر ا �J ه� ا���K �:��- ن �K§7 ا�a��: J��!� .وه� و/` &(!�!% ا(cآ�ر

ب ا����� -� ن، /�+ -�4:4 ا����� ا��!� -+ 1�K§ا +�ا«�!� �! ءا '&% أ$) 41 (` و�� 6�، أ� ا����� ا«�!� &�� ر ا(��(� /�5 65 ه��-{ �`�(� 50ا(��دد &�� ر�/ 110،��/

م ا� /� و¼56{ة /� ��وف '4م و�6د ر� ���، أ� إن و46ت ا� ��� /�ن ا�4هcا ¼6�، و&�45 ا(��(� 30ا«�� &�45 ا(��دد ه� �/ 60��/ .

- �/ gرة -*�(4 إ�a>ا ،J *�ض�ي :�+ اc{�� ار '&% ا وا(!�!% إ% $�(� ا�� ر و�Kع ا��l� &{�� ه�(��دد وا �)�� .�a2 و:<�&� � ��� ا�� ر ا(

- d�- �����رة ا�a2�/ �`/ ر�& ��- �5 ر ' ز�� ا��+ وا��وق ا!Kإ ����. 22 -- d�-�- ��-»رات ا ر-8 ا��:4' ,-

Ir �!`)ر ا ر ا�()IL - ا���- - Io.cر ر ا�,) Ioff - ز: دة ا���- - Is.c �,`ر ا �- -InL ()1� ر ا�- J� Ioff , InL,IL,Ir,Io.c,IS.C : ا�6 � ا�$ :� وا�( :� ؟؟� ه�ا��ق �� -- 23

�: م ��) K`�ل أ56{ة ا�$ :� �� ا��:� : ان ا�$q!6{ء �� ا l� ز -a&� '&% ان ا�5fire fitting

�: م ��) أ56{ة ا�( :�: ا� ا�(q!ز ه� 6{ء �� ا �� ز: دة ا�� ر -a&� '&% ان ا�5 ض وز: دة ا�45�>Kدة وا وز:

/� ا% ا��ق ��� ا�(gأ�: �: : وا�$�: 56 ز ا�$ (�� (a*& ��ر أ�و ( ا2�d&{( ا" ز:� 56 ز ا�$ :� :*() '!4 41وث ا«�

J�و:*() '&% ازا (a*4 41وث ا*� ()*: �: 56 ز ا�( ن ا�k4ر ا��/ �'��� ه� ا��ق ��� ا`��ز وا(�(( ت ؟ -- 24�

- ���k- ز���(�(+ : اا �k *� J}*-و (a*ا J,�>)ت ا ���k ا*a) و:,4ر أ�� &(5( ذ} ا*a) �*{ل) ا`�ا 7( ج ا%) ا� �Kى(ا���ز �-�� ج ا% �,4ر &�"c:� أ� ا(�+ -��: �:c"�& ر (�,4ر ����ت ا��

) ����ت ا�45 ا�-� ه�� وا(�(( ت ا"�� ا-� ه��؟ � ه� ا��ق ��� ا(�(( ت -- 25

Jه� ض -/`� ��) ��(+ 4g ز: دة ا�� ر ه% ا�% -*() � `�(�ا(�(( ت ا"��أ-��>Kدة وا ز: ��) ��(+ 4g ا���ب ا�رg% ا(�(( ت-ا�45 *� b ا�-� ه�� ه% ا�% -*() � `�(� وا�-�

س ا`4رةk*Kب-ا�� ا�� g&%-ا�رg% ا�-� ه% 4g ا� demand load ?و��� ال connected load ا��ق ��� ال -- 26

connect load �K ءة ه� ��(�ع ا�1( ل ا(��&J /% ا4ا��ة اJ���5k �9اء آgل '!4ك ا ��<4م او � ���

½� �Kوا �-��d)و-&��{:�ن وآ J ���<4م آ) دb /% و$� واl� 41 ال و-��k و��J6 و_� connect load J6 6 ت دى او اى 1 � ه� ��(�ع آ) ا�K&% ام ا q!8 ا�1 J&����

. ر4���9 demand load ا�

J /% و$� �*�� وه% ���ن ه� ا�() ا(a&�ب /%"x ن�k�� %&ت ا 6 و$� ��4د :*!% ا� رb '� الd' connect load %/ وب�s� factor ال J)9ا d:�`- diffarcity factor

J ن�k�� d _ b&�) '&% ا ودم وا��ا %/ %K � �9 س ان آ�/ ت '&% ا9 س ا�K ��<�&� �� و$

Page 6: اسئلة واجوبة في الحماية الكهربائية

bت زرو .واو$ ت ا�5�9�ك ���kن /�5 ا$) /% او$27 -- J�d!64ول ا�4م ا>�� ذا -) recepticals bye 3 wire not 2wire as Egypt؟

زل ��!) � ��� �4 :�4ث �k�- hot and neutral onlyن ا���J ا�!!*/ short )5!����5 ء ا%kا ¾:��- +�: �gا�ر.

5�)&: ن ا�<� :*() آ�{ء �� ا4ا��ة و:`�م �4ور ا(` و�� و:(�/*!�4�Kر /� ا� ا��bر آ��� a2ة وا .و�� ا((�k ان :£دي هcا ا�� ر ا% ا�/

�kو %)�: �2¿ {&9 :(� ا�{ء ا�آshort �d /*!4 41وث ground ا��½ ا����� :�kن �5�� {&� ن ا%�ن �` و��J ا$) �ground `� �� ���k ا�� ر ��K� ن و� :(� ��Kو�� ا� ر �-

ن�Kا� %&' ��� - J l� 46ا ��"�. ا� �� ا�xح آ) �� ا�`�ات

: Recovery voltage: 7a$ 4*� 4ورة7 ا $ %&' �5q- ��ا �����ارة ه� ا��ا.

Resitriking voltage: ء 7 ا4ورة أ�!$ %&' �5q- ��ا �����ء ه� ا '(&�� إ � .�ارةا�

Transient over voltage: �/ ��q�&ت ا ' ا����� ا�� -�4ث /� ا��kd ه� ا�ر-�أ6{اء ا��kd أو /� 1 � 41وث -�� -���� '4ة '�ا�) ��) /,) أو -���) أي 6{ء ��

.ا,�ا'�First phase to clear the fault: 7 ه�a$ %&' 4ورة7 ا $ ��&� $)3Phase

short circuit) �(d$ 7a` 7 7 $� م ا41 ا� �ار �a$ ��&' ن�k: ¦�1 �:�2»ر:� ا�aا�,Kر:� ا«�2:� �(�ة و�aأ'&% �� ا ����/ d:�`- . ¦�1 ارة��ا J� À�a!- أي أن أول �ر

J� 7�-�-,�� و% ار إ أي اJK :�8 أن . �bc5 ا!`Resitriking voltage �a :,) ا�� .ا«�2:� :� أآ�� �� ا�aر1.5:��() /����

Symmetrical fault: ��� ل ا�� -�4ث '&% ا� �ار ا�a'ا وه� ا�cه �/ ���q!)وا1 � ا-{ان �/ ���q!)ا %`d- ل a'!�ع �� ا�ا.

bcن ه�k-و ل '&% �K'�� ه(a'ا�: 3phase fault

3phase to ground fault As Symmetrical fault: -�4ث '&% ا ��ل ا a'و-�4ث /� وه� ا� ��� ��5kا ���q!)

� 1 �' ���q!)ل -<�ج ا a'!�ع �� ا�ا اc41وث $,� '&% �ر وا41 أو �ر:� و/� ه� .ا�-{ان 1

ل '&% ���� أ�Kاع ه�a'ا� bcن ه�k-و: Single phase to ground fault Two phase to ground fault

Phase to phase fault Zero sequence impedance: �a>ر ا وه� $�(� ا(` و�� ا�� -*��ض ��ور -�

(Fault current If) �d' �a>ا �a`K �� ا�رض وه� (Fault) د إ% �a`K ا��(Neutral point) ���)ع ا�رض /� ا�K %&' .(ا�...... ,ر�&��, �<�:�(و-*�(5�)�$ 4

Source impedance وه� �k): ��45 وا�4ر ا,� �*K 5 ا(( K*� ا4ا2&�� �(�)�K أن�5q- ��ا �*K .'!4 41وث دا��ة $,� '&% �,4ر ا�45 وه� ا((

Inductive type voltage impedance: 414م ا>��أ�Kاع ����ت ا����� ا�� -ا!�ع �� ا(���ت '� �4dأ '() ا(��� �_�اض ا`� س وا�( :� و� :<�&� �4dأ '() هcا

و-*���� �d� 110V$�+ ا�45 ا* � إ% $�(� ��<4م �<��# $�(� ا�45 ��و-, ا�'�� د:��d�! ر c} -��� ��� ا�aر وا�رض ا�45 �` و�� ' �� 46ا ���&.

Page 7: اسئلة واجوبة في الحماية الكهربائية

Out off phase switching: 7 ه� '4م -{ا�� '(&�� ا�,) وا����) � �ار $ 5!�� أي :�8 '&% ا` 7 -�() ��) هbc ا� � , ا� �ار أي و�6د /�ق ز�!� ���, ا4ورة /�(�d,: JKأي آ� Short circuit %&' 2phase ة�x d� 7 .و'&% �/� ا`

Over current ??ز: دة ا�� ر و Over load ا��ق ��� ز: دة ا�() --- 29 ز: دة ا�()

%��5kر ا �ت ���ة &�() '� ا`�(� ا(`!!J و -��(&5 ا(* ه% $�(� ا{: دة /% ا��� k4ة او ا .% 25ا% % 10-��اوح � ��� ز�!�� دون ان -�&� و

ل�� رة '� ���كd' %��51() آ K4!' ن �8 و �,(+ أذا آ آ��5% :`�م ���"�) ��9 !`) ا�`�' 4:}:� J�&' 8� 8 -� ر 1000 '&% ان :�kن وزن ا����آ�&��6ام و '!4 هcا ا�() :

�8 أ���d /�ذا �200`4ارة آ�&� �6ام /c5ا �*! b ان ا(��ك '&� ن 1200ا% زاد وزن ا�`8 -� ر آ%��5�� أ���d '� ا(,,+ '&�J و �c} 40ز: دة /�(`!: J�) هcا ا�() �9ف :

%&kر ا -��a2 46رةا{: دة /% ا�� ر '� ا`�(� ا(`!!J �9ف و ه! . أ��d,:240��d ا�� % �9ف :£دى هcا ا% -&� ا( دة ا* ز� ا�� :£دى ا% ار-� ع در�6 �1ارة� و . �ت و�

�: )1 7g�: {c ()�دة ا .&(��آ ت 4g ز: -,(+ ا��ت ا����5k ان -��() ز: دة /% ا�()d ���ة ز�!�� % 25 - 10-��اوح ��� _

.&(*4ة (*�/� ا�� ��) و :�8 ��ا6*� ا��آ� ا(,!*�. $,��ة دون ان -�&� رز: دة ا�� Over current

ا�� ر ا��5k% '� ا�� ر ا(`!� ا�% -£دى ا% إ-�ف ا(*4ة ا����5k دون ه% $�(� ا{: دة /%��2�-

�� �dن اآ�k- � d .�� $�(� ا�� ر ا(�!� % 50ز�!% و _����&� 5 -��() ز: دة -� ر -,(+K����5 اk(*4ات ار $,�( ا ان -�&� و(4ة ��ث ��اK% دون) -�

.:�8 ان -*() ا56{ة ا�$ :� $d) هcا ا{�� ا{:�% و ا(��غ �� ا�5اء ؟ � ا��ق ��� ا` 7 ا" زى و --- 30

�% ��� ا`�ا 7 ه� �Kع ا( دة ا* ز���ا(��<��4 /% إ � ء ا��ارة ا����5k ا�! ء ا��ق ا�7 `& ����� ./,) K`� ا���l ا�

Vacuum Circuit Breaker �� ا�5اءا` 7 ا(��غ - 1 �� ا�5اء �4ر�6 ' �� 46ا 46ا هcا ا!�ع �� ا`�ا �k- 7ن� _�/� أ � ء ا��ارة ���_� -(

-,) ا%1000000000 / 1 Torr ى��ا �"sا ��-

�K دا2&���� ()' �k): � {c !�ع وا اcب ه��' �� �d�*: اc7 و ه `& ����� ت ا����)& رات '&% هcا ا!�ع �� ا`�ا 7 و $� س ا(` و�� ا4ا2&�� � ا`�ا 7 و'!4 ا�6اء�d�2ا�

ت و و46 ان���)& �� 4:}: )� (� k 5 _�� 9&�(� :�+ ا4d�9ال _�/� أ � ء ا��ارة ��)�$�� k- %�1 �5د�4م /% ا>�� . آ�&� /��36ا,� �K و هcا ا!�ع : Oil Circuit Breaker ا` 7 ا{:�% - 2

-�kن _�/� أ � ء ا��ارة �(&£ة هcا ا!�ع �� ا$4م ا�Kاع ا`�ا 7 و � زال :��<4م �1% ا«ن و ت��� '4 '&% أ � ء ا��ارة ��� ا(� زل :' �:}� 8�: JKا �q1�� 8�: �k و �����ا�

رات دور:� &{:� �*4 '4ة '(&� ت /,)d�2{م ا� '() ا&`,� و :�+ -"��ة اذا 4م >���� و :'���� ان �1(� آ��d 46ا /% 1 � ا�9<4ا�� /% ا�45 /% ا��5د ا(!<��s و ا(���a9 و ��

% .ا* SF6 Circuit Breaker ا` 7 ا" زى - 3

ر /% ا�و�K ا���2ة +��Kا� %/ c2`�ا 7 ا!�ع �� اا اcآ���ة و ��*4دة و ه J �� �{ا: ت ا�5:��� .آ�&�/�� �1100د ا(<�&�� %�1 :��<4م /% 7�)6 �

Page 8: اسئلة واجوبة في الحماية الكهربائية

�:�dkدس /&�ر:4 ا _�/� أ � ء آ��9 ' زل داSF6 (2 و /% هcا ا!�ع :��<4م _ ز 9 .ا��ارة

ت : 31a�)� �ت وا���9 ت ا�% :�8 -�ا/�ه�� ا(���ت وإدارات ا�$ :� ؟ � ه% ا

�� �ت ا��� :�8 أن -��ا/� ا1 - %��� % single line diagram�a�)&أو ا���4ا� ا�9+ ا�- 2 �: wiring diagram وا�� ر ا(��(� و-�ز:7 ا�� ر ا(�"�� ر9+ ا4وا�� ا� �K:� وا�$3 - �: �9) ا�$ رات 4--d�2وا� �K �9) ا,�5 - (�"�� � �� �9) ا�*&�( ت ا< ا�!�� ا�*&�( ت 6 -- 7 �: آ�وت أ56{ة ا�$5 دات ا��2 8 -x�K ر و��ا�� ا,�d :�ة ا*4ادات وأ56{ة ا`� س 9 -*� (�9

ر دوا�� ا���K�ك 10 -d�2�9) ا - 11 (�9 �a�)& ��!�ت ا K �dا 32 – ������ أو ا���4ا�� ت ا4وا�� ا�K�k� %ه ؟ � آ�� 1 -k� ا(� -�� وا ن - 2ds`ا ����ت ا�� ر- 3 ا�45 ����ت - 4���ا(���ت 5 -�� ا� '4ة -6� ا(���ت ا(

56 ز ا�$ :� ا�&�+ ؟ � ه% ا��وط – 33 %/ ا�ا86 -�ا/�ه- 1 ��9 � ر $,� أو -�دد :�8 أن :�kن 56 ز: ا��- ($�� bر�*x �64رس 46ا �1 �: ا�$

5�)�$ %a>�- 5 ا�5 ز أو _�� ذ} '!�4�&' �: ا`�(� ا(*- 2 ��� `�Kء:�8 أن :`�: ا� `�K � �: م ا�$qK م �kd�ا �� b��_ دون J}'`,� ون ا k� - 3 �'�� ن: اk� ف و'{ل م :�8 أن :�+ اآ��qK ن ��'� ا(a&��� وآ&( آ ا`,� �

ذ} -�aرت أ56{ة ا�$ :� �� ا(�1�ت ا�$ :� أآ�� �9'� آ&( آ ن ذ} أ/s) و'&% ��k�K إ% ا��9 -���k إ%k�5و��k$ ا�ا��)

در '&% ا*(): ا(���$�� 4 -$ �: م ا�$qK `� /% أن�وه% -*!% ا �`a!)ا %/ (a*ء ا أ�!��_ � !� %/ (a' ء و�6د ا(! � ا(���ل '!5 وأJK ا�% :�(�5 /`� و:�8 أ� :*() أ�!

J� '&% أداd&9 ��£: )� �: رات و/% ا�56{ة ا�4:�� :�46 � :�4ث أى 2&) /% 56 ز ا�$xإ إcKار '!4 و�6د 'a) دا5� %&2

��<4م ؟ � ا(`,�د : 34- �(���ت ا`� س و/�( ـ: ����ت ا`� س أو ����ت ا�56{ة -�() '&%

����ت ا�245 -����ت ا�� ر 1 - 89 أ56{ة و-��<4م هbc ا(���ت ���:) ا�� رات!� �s�>!� +�$ %46ا إ �� وا��5د ا*

س وا��k+ و'4اداتا�$ :� وا�` � $� ا�* � وا"�� /*aا . 5�$�� 8�إ% ا�56{ة ا�% -+ ذآ�ه وذ} �!�8 وهbc ا(���ت -`�م �!`) 1 � ا��kd و1

� آ( /% ����ت ا�� ر � 1�8 ا�� ر ا)`!� ¼56{ة و/% 5/ 400 أو(:��-400/ 1 � 8� 11000 أو 100 /11000أو 110/ 66000 أو 100 /����66000ت ا��k- 45ن ا!

/110��4>��1�8 ا�45 ا(`!� ¼56{ة ا( ��� . /�ا�4 ����ت ا`� س ؟ : 35�

Page 9: اسئلة واجوبة في الحماية الكهربائية

م ا`4رة إ% $�+ �"��ة 1 -qK رات ��� أ56{ة ا`� س -��:) �56د و-�� �d9 -�kن �!�: وا��k+ وا(�ا$�d وأ56{ة ا�$

5 ذات -� ر و456دوا�� ا�56{ة '� ا4وا�� ا '{ل 2 -- �و�� !q+ ا`4رة وا�% -�kن �&� م ا�$ :� أو ا`� سqK ر 89 �45 و-�!� ��_ % '

c"- ��9ى �5 ا�56{ة /(�� :�kن ا�� ر ا� �Kى ا(`!� /% -�4�1 $�+ ا�� ر وا�45 3 -�$ +�`� أو �100ت ا�45 أ���d وا�45 ا� �Kى ا(`!� /% ���5أ���d أو ����1ت ا�� ر �/

110 ��/ 36 – ��)�& �d-ا�ا +�` ا(`,�د �� rated burden ؟

5 '&% ����ت ا�� ر أو ا�45 �,�� دا�(� '&% أن ه% ا`4رة&�)�- �k): %�ا ��dأ� ��� �(q- �$4ى ا���� 8�1 &(���ت $�(� ا<�a /% ا�� ر وزاو:� ا�J6 /% ا�4ود ا(�(�ح �5

37 – % �5 ا(`,�د � ��{ ا�p20 , 30 VA ؟ �1� %&' 8�k- �:�*�ا Name plate }��وا p ىc$+ ا�وا �: :*!% أJK ���ل -� ر &�$ ر�: %&' �5q: ف��ا p +$4$�5وه� رى ا���� %!*: Accuracy class ىc$+ ا�وا �) أ$,% 41ود اp �$4'&% :(�� ا��ف*� (�): ALF !*: ر وه� 20% أ�k): JK ��ور -� ر ا�ا-8��& �*g ن�k- أن �a:�x ��d ا<�a /% ا�4ود ا(`�رة 5K ء (��ل ا�� ر �7 �`

� أ�30��dا(��&� '&�J ا�1( ل�/ . ��&kا �a>ا �d�K %!*- أ�2ى �`:�a� (`!� و/% آ) ��20ور '!4 % 5أور ا �� ا���*g

%&' �&a: 30ا��1ال VA ا�d-ا�ا +�` ��)�& 8�k:5وP أو X 45�ر و/% ����ت ا 5 ر�{ 6pأو 3p'&% ���ل ا��`d�: CL أو KL

أو class 2أو class 0.5 و/% أ56{ة ا`� س -�kن class :*!% ا4$� أو ����ى ا�$45 20_�� ذ} �� ا`�+ وا��2ى `d�: n 8-ا�ر ا ت ا���' s� 4د' .

38 : � d�2ر ؟ه% ا� رات ا�ز�� &��آ4 �� ����1 ���ل ا�� س ا��9(�ار:� 1 -�$ �:�K ت ا��&)& )continuity( ر ا*{ل ��ا�a9 ا(��� 2 -d�2و�1000ا ��/ 7� �:�K و�� ا*{ل &(&� ت ا�`� (`-

�' %g*{ل10ا�رو�� ا أوم و� -`) �`��� �' %g7 ا�ر� ��� أوم ��20&(&� ت ا���4ا�

- 3 �a9ا�� ��da`ر ا d�2ر:� ا adا ��d ا���:) وذ} � ��ار -� ر �56 ا���4ا�% و$� س ا�� ر 4 -K ر d�2ى ا�K ا� ر ا��7d وذ} ���&�� 456 '&% ا(&� ا� �Kى و$� س $�(� ا�� ر 5 -d�2% ا�1% -,) إ

��d ��1&� ا��7d ا�% -4dأ �� ا!`�a ا�% إذا زاد ا�45 /�5!� /�ن -� ر ا(&� ا� �Kى % 10 �d� ن أى ز: دة �"��ة /% % 4:}:50 �!/ �a`!ا bcة 46ا و�*4 ه��dدة آ ا�45 -£دى إ% ز:

7d��(��ل ��1&� ا24) ا: {cر و� . /% ا��6- �k): �)�� س ا(` و�� ا4ا2&�� &(&� ا� �Kى وذ} '� �:� -���) �,4ر 456 ��$

�' J�/ +k��45ا�ر وا و�� ��"��ة و:�+ ر/7 ا�45 -4ر:�� و$� س $�+ ا��`� �:� �)��ا(�$� ا(` و�� * �Kن أوم �$ 8� ����9 $�(� ا�� ر ÷ ����9 $�(� ا�45= و-��8 ا(` و�� 1

س 1(�� ����ت ا�� ر ؟ – 39�$ �k): آ�� �� أ$�ب روز-� و:�+ -���) K , L أو S2 , S1 ا� �اف ا� �K:� (���ت ا�� ر :�+ /,)(���ت &�45 ا(��دد :(�k ا��k+ /% $�(�J إ% K`� ا����) ا(` �&� ¼ �اف ا� �K:� �,4ر

�È1 $�+ ا�� رKو K %�1,) إ% $�(� ا�� ر وا�% -"cى أ56{ة ا�$ :� و:�+ ر/7 ا�45 -4ر:��5 و:�+ ا�45 ا(! ��و:�+ -���) $�(� Inا�� ر ا�ا-�d (���ت ا�� ر

ب ا�(�� � ا�ا-8 &(��ل ا�� ر× ) '!4 ��ور ا�� ر ا�ا-8 ($�(� ا�45 ا(` س = 1J�&' �K(4ور وا و-` رن � `�(� ا�ا-�d �(�� ���ل ا��

ر ا(��(�؟– 40�� � �a�)ا �:c"- م qK ح�xا

Page 10: اسئلة واجوبة في الحماية الكهربائية

ت ه%a�)ا %/ �)�� ر: ت ا�% �,4ر ا�� ر ا(ad% ا�ا ���أ56{ة ا �a9ا�� 5!�x +�: ر: ت -`�م ���4�1 ا�� ر ا(��ددadا %/ JKا�2{ا �k): �)�� و-��:&J إ% -� ر ��: _ %/ �)�� آ�� وا�� ر ا(k� `�ا 7 وا� �� ا�ه(�� إذ :��<4م /% دوا�� اk!��ول ا<

�)qKأ �:c"- %/:� وا و��م ا qKار وcK§ت وا� ءة ا��agار:� وأ��d �� ا�$ :� وا�-,g§ �1 ����� ا�ه(��x �� ر:� وأآ�� a� �� �9<4ام أآ�� ا�% :�8 -�/�� �, در �4:&� وذ} �

و�6د ا�� ر ا(��(� /% أ�9أ ا�qوف �1ارة ا{:� و���d در�6 �1ارة ا(&� ت ؟ � ه� ا��ق ��� ��d� ()' �:�qK در�6 - 41

ز �1ارة:*() ا�5 ز ا< ص �`� س در�6_ J� � خ �"(�ر /% ا{:��Kن �� ا�k�: وه� �ا{: �) -(4د آ��d وا" ز :,) ��*� J +k��:و ���dKأ �a9ا�� l�� خ إ% ا(£�x وK`� ا���Kا�

�`K �� ���'�)�� %/ ز ا§cKار 65ا���l ا{�d`�� و:(Jadg �k '&% در�6 �1ارة ا�5 ��K ذ} &�,) أو _��95وا�

�� إ� إنأ� ا� خ :���� �5 ز ا< ص �`� س در�6 �1ارة ا(&� ت /�5 ��) ا�5 ز ا��Kا� ر�- و�� :(� �5`� �� ��- /� إ% ا��ارة ا!g§ � � 89 �7 ا�� ر ا( ر ���ارة ا{:!�:

(��ل وهcا ا�� ر اcى :c2£ �� ���ل -� ر دا2) ا(��ل� ��و�c�k- {ن ا��ارة ا(£��ة �J�&'ت و �1ارة ا(&�)Kوإ ت ا�و% -�ds �1ارة ا{:� و41ه��� ت �� ا�'�)�� ����

%&'50��K در�6 و/% أ�Kاع 100 در�6 وا� �� &�,) 70cKºار در�6 ��"�) ا(�اوح وا� ت ��"�) ��(�'�% ا(�اوح وا§cKار وا�,) أ�2ى أر�*� ��(�' ت ����� ا�

؟؟) ا�$ :� ا! -�� �� -�&) ا" زات( ��2&{ '() 56 ز-�k/ �' +&kة – 42 رة '� و' ء �*K4% :���ى '&%d' �4!' �'�ا���� �� ا���!��م -a��ان '&% �a9 ا{:

� وآ):} � ء �(�&�K§ن ا�k: ن�k-و ��`d���l ز- �`K %/ +k��-ك �1ل ���ر و���- 5!� ) �1���� l��� ا*�ا�� $�&� K`� ا���l و:�آ8 هcا ا*�ا�� /�� / K`� ا�adذا ه�

���dKز '&% أ �J و-�آ8 ا�5�K (��لو��� ا % ا{:� ا�% -,) ��� 2{ان ا{:� ا���1 در�6 �7 ا�/`% �k: %�1ن ��ور ا{:� �� ا<{ان ا*&�ى إ% ا(��ل ��2) ا����dK �{او:�

2�ل �� �/ Jق و�� ء ��k+ ا§_K§ز ��2&{ وا 56 �a9وا41ة �� ا'&% -��� و-"&� ��ا �� م)� �� أو �*�k- 4:� ا{::} و-��<4م §�2اج ا" زات �*4 �¼ ا(��ل �

/�`�) وا�7g ا* دى)5�&' �"s- زات ن إ� /% 1 � K`� ا{:� أو و�6د _��/ ا*�ا�� ن 1&� ن إcKار و/,)�� ت '&% ا�� �7 وه(��� ا�

��dآ�� :(�k ا�,�ل –43K %&' ر ؟ d�2ر �4ون ا -��:) ����ت ا�45 و����ت ا���J أو �� ا*4اد ا(�آ�k): %' 8 ا�,�ل '&% �*&�� ت ���ل�K (��ل45 �� '&% ا�ا'&% ا(��ل أو '&% 2&�� ا`� س و:���ط أن -�kن هbc ا("cى أو �� ا��� ��� ا(�آ8

� ا* ا�56{ة ��ردة �7 ا&��1 أو� � �)�`� �� ��d ����ت ا�� ر إذا آ ن ا�K %&' 4اد�d�K ب�g (� ��d ����ت ا�� ر× ����ت ا�45 ����ب ��K

�k): ر �J أو �� ا����� أو و&�,�ل '&% �*&�� ت ����ت ا���K (��لا �� ا�,�ل '&�5��d ���� �� ا*4اد ا(�آ8 '&% ا<&�� أو ��K %&' ���d)ر ا ت ا�� ر أو 56 ز ز: دة ا��

�d�K %&' *4ادا �� � �)���ب �`�� �� ����ت ا�45 إذا آ ن ا�44 : (d$ ء ا�% :�8 ا��آ5!� 4�xه% ا� و7g ا�رg% ا(�&% ؟ � ف :�kن. ك 11-� ر 456 ا��آ4 �� /,) ا(5(� ا(�اد ا*() '&�5 و��k ����ت1 -

ا(�� ح ��,�ل و2 رج ا<&�� ل-���� '4م 2 -d`�9ا %g7 أرgو %/ �d_� 456 وذ} ����غ ا��5 ا("cاة � ر ا��آ4 �� '4م و�6د 456 3 -d�2ا� ,' �:� �' (� kا76 �� ا� ���d ا�45( �,' ) - 4 %g7 و�&� ا�رgو �K �% أو� �+ �� �56 ا(5(� ��� �� �56 ا�رg% ا�

ت : 45:c")� (`,�ده� ا 5 ؟ا��ازى وآ���� �*� (� ا�* ت ا�% -"cى �1 ت:c")رة -4ل '&% ا d' %ه �a�� %/ ازى��ز:7 و���&� '&% ا��ا

ن ا��ز:7 ds$ �� � ن ا��ز:7 و/% �1�1ا(���ت ��ds$ %&' 7:ز��ن 1ا ds$ %&' أو

Page 11: اسئلة واجوبة في الحماية الكهربائية

ت /% �!�5% ا<�aرة �ن '!4 / وا�ا�� ���) �&��1 ا��ز:27 ,1ا��ز:7 :c")و-&} ا (, ت �� ا(��k: �aن ه! ك:c")ح �� أ41 -&} ا �) إذا + :�+ /,) ا(��kا76 '&% ا� � 456

7� (� �) أو ����ت ا�� ر إ� �*4 ا��آ4 �� '4م و�6د 456 ��1 ا��ز:7 و� :�+ ا�*kا ��آ4 �� ف ا��ازى /� :�8 ا*() '&% �24) ا4ا��ة إ� �*4 ا��. ك66و:!�da ذ} '&% دواا��2ى '� �:� ا��k+ و���q1 و�6د ا�45 '� �:� ���ل ا�45 /,) ا4ا��ة �� ا��5

. ف. ك 66ا4ا��ة أو ', ���d ا�45 وا< �� ��45 ا(��6د '!4 �24)46 : ()* ن ا��ز:7 456 �� ا(,�ح 5+ �ds$ %&'11 ف ؟. ك

�K ن (%94!5 ,�ds`ا {&- %&' ()* �K وه+ / :,�ح ��, ر ا�!��� �dوآ �����`� آ( نds`& 45�ا �:c"- در - : وا*() � :�+ إ� � ��وط ا� �� �d2 +5:4ة �(,

ن 1 -ds`& ىc")ا %��� �� ا����5 وإ�2اج '��� ا(�� ح 2 رج ا<&�� /,) ا(��ل ا� ت2 -:c"<�وج و�ت ا :c"� 7�)6 (,/ ت ا��2ى و���)ا �� ���رج ا 2 ��- إ�2اج ا(�

و'4م:� ا< ت ا��� إذا + :�+ ا��آ4 �� ا�,) �� ا�$��اب �� ا�{ء:c"� ازى أو��ت ا :c") %&��ا

ا��2ى ا��5 ( 2&�� ا`� س( إ�2اج '��� ����ت ا�45 2 رج ا<&�� 3 - ن 456 ا��آ4 4 -ds`<&��. ك ��11 /,) را�� ارج ا 2 J1 ��� 7gف وو ر و�6د 456 5 -d�2ر 456 ا d�2ا� ,' �a9ن ��ا ds`ف. ك11'&% ا ����� ا*() و-�kن أو% �a2ات ا����) و7g ارg% ��&% وه�a2 �2¿ bcة 6 - ر ا*{ل -"a% آ) 7 -d�2وا �/ q!ن �*4 ا ds`& �&��)ت ا ا���

رة �7g ا��"�) ؟ا(! و ف �*4. ك �11 :�J&)' 8 '!4 د�2ل 2&�� �2وج 456 – 47- 1 %&' J!� 7�$��ز:7 وا��ل ا���5 ء ا(! ورة �7 �K) إ����7 -*�ق اK4م و�6د أى ��ا' -

�) إن 2 -kع '&% رأس ا�g�)ا %gر/7 ا�ر J*/آ4 �� ر��و46 وا - 3 5 و-"��هK�{:��� وا ��- إذا {م ا��� ��ا6*� ����ى ا{:� /% ا(� ر '��� ا(�� ح �1%ا��آ4 �� '4م و 4-�� �"��ة �6د أى �*4ات 2&� أو /% �K و� آ- 5 �g�' ن�k- � %�1 �:}!d � 5�/ qK 4*� ��- ل �7 أى �xارة '4م د�2ل ا(�*�x� ا��آ4 �� '4م و�6د آ5!� '&% ا(�� ح أو أى �*4ات 6 - ن ا*() 7 -k� �' ل *�x� �&� إ�* د ا(�اد ا` :� وا��"�) ��) �� -��ا��آ4 �� -� 8-$�& �:c")ا �)�� ا�� ر ا(- 9 7gح /% و ر وا����) وا�,) ا��5k% -���) روز-� ا(��d�2ا�

��k�K �ى �, '8 د�2ل ا(�� ح /% ���ى ا�24ل ا�:(� 10 -k�)آ� ا��ا �q1�وا�:�� و� ح -�اJ6 ا�24ل وا�*��� ا,�����)&

$+ ا�رد:� و'!4 ا����) :�kن ��ا�a9'4م ا���xاك 11 -a *() إ�رد:� /% ا�ا l��ر §�6اءات*� ن ا����) و��k� �' ن �*�4ا�k: �2»رد:� ا�) و/�د ا����ا

ع ا��ز:7 '&12%a$ 7� ���-���) ا("cى �� إذا آ ن ا("cى :"cى ��1 ا��ز:7 :�+ ا�!��{ا�� �7 ا����) &��1 وذ} �2/ �� 41وث 'a)ا(��a أو� و�*4ه ا����) �� د�2ل ا

ا��� '� 8d9 ا�,) '&% �"cى �2وج �� ��1 ا��ز:7 /��,) ا("cى � (��a و:<�&�JK kو� - 13 ��&>& ��� :*� Jg����K ��4,& أى �a2 '!4 ا����) � :�8 أن :�kن ا�!% �)��� 9 48 – 4*� %��� ا,� �K ؟ � :�J&)' 8 '!4 -���( ���ل ر

5 ء أ�� ا�")1 -Kآ4 �� إ��ا أ/�اد ا,� �K ا��آ4 �� 2&� ا(�$7 �� آ) 2 - -�� ا(��ل �� ا����5 ا��آ4 �� '4م و�6د أى �5( ت '&% ا(��ل أو � ���:7 أو '&% 3-��

Page 12: اسئلة واجوبة في الحماية الكهربائية

�a`K %&' (��6دة(��ل ار:# ا�- �!�k9 (���-دل و ا�*4 - �� (� ا("cى &(5( تا��آ4 �� -���) ا�� ر ا(- 5 (d$ 4آ��ح ا آ�� �� /,) 6(�7 أوJ6 ا(��k� -���) ا5 ء ا*() - 6�K � +k��& رة xل إ وا�,�ل '&% ا(�ا/`� '&% ا����) إر9 آ�� - 7k� ع -*&�( ت ا��k+ �����) اd-(��ل �� �56 اح ا ف . ك 66أو� :�7d ذ} ���

�56 �� ف. ك 11و�*4ه ن 456 را��/,) - 8ds`45 '&% . ك11ا�وى ا �ف �*4 ا��آ4 �� -�(�) ا(��ل و-

ا(���� ن 9 -ds`45 '&% ا�ا �dg 4!' È1�:1, 2JK� �k45 و�ات �"�� ا�a2 وى � � :�kن ��

ف وا("��:� �<�&��� /% '4د ا<�aات. ك 66ر�( ا4ا��-�� �<�&���� /% ا�45 �� �56 وا!�ع49 – :�8 '(&J '!4 -���) دا��ة �*4 ا,� �K ؟�

5 ء أ�� ا�") �� دا2) ا(��a أو �� 2 رج ا(��a � :�+ ا4dء /1% -Kأى '() إ� �*4 إ ا��k+ وإ��غ

ا4ا��ة و��ا/`�J '&% ا����) &��آ4 �� '4م و�6د 56 ت أ�2ى -*() '&% ا��آ4 �� -���) ا�� ر ا(��(� ا("cى &(5( ت 2 -- 3 %�!�k9 (���- +k��ت ا )�&*- 8� ن 1ds`<� واا ا��2ى أو� وإ��غ ا��k+ '!4 9`�ط أ41 :�+ ��ا6*� ا�45 إذا -+ ا����) �� ا��5 4 -

J6ا�و - 5 �*� ا���d� %&' 45 456 ا4ا��ة إذا �4ء ا��k+ ا����) �� -���) ��� ح ا4ا��ة و��{�56 �`� :�+ '() ا<�aات - 6�� إ�xاف �5!4س ا��k+ا�-

50 – �a�) ؟ � ه% 41ود '() /�د ا��"�) � دى وإذا ا4��9 /,) ����ت أو' 7g�ا ) ت ا��"�) )�&*� `d ()*ا +�: �� ك 66دوا

ا��"�) ا* دى ف -�kن -*&�( ت ا*() '� �:� ا��k+ و� :�8 -"��� و7g. ك 220ف أو . +k��ا �:� �' ���d)*() �1¦ أنا(,&�� ان ا�� ظ '&% ا��kd دون '&+ ا��k+ و� آ

����� ا��k+ و� :�+ /,) �k9!� ا<� :4و: إ� �*4 ا��آ4 ��� �a�)4م و�6د 456 �� ا' ح د�2ل ا4ا��ة و:(!7 د�2ل��� J6و�� /,) 6(�7 أو �&� ا���:7 را/* :4ك و:�s) ا(`

ا��ادث ذآ� ا�� ��) و'4م ا�"��� � دة �� �d2ة ا{��ء و'!4ا��وى /% أc2 ا`�ارات وا��9�8d9 (�&�- (5 ا� د�� و'4م: %�1 �`�`�& -�kاره

51 – %`- ��"�) وا,� �K ا�% ر�(� ��&� ت ا,"��ة &*d6ا�ه% ا � آ��dة ؟ �>� �� 1 - ��k��*() و1,� ارج ا ا�`�q ا� �� و-�ك آ) ا(� آ) 2)�/ ��s- � %�1 J� �&k-

�dأآ �&k�� {��! ت - 2:c")ل ا 4 ���q1 $�اءة أ1(*� �: ت �� أ56{ة ا�$��و'4د ��ات ا�,) ور/7 ا*

ت��' (`!- � %�1 س �( -8�k و�Kا1% -4و:!5��ت 2 �� &��ادث وا§1�&�- 5*d�: �):4$ $� �ن k) -"��� ا�1( ل أو در6 ت ا��ارة أو ���� ا�"��� /% $�+aأو '4ادات ا �ى ا{:

8�: 8d9 ح �4 �� ا�ر� ظ '&% ا(5( ت و-�`�� أ'&% '��& J�/�*� و����ى ا{:� و1(&J وو7g -�"�) ا(�واح ���q1 �1ارة ا(��ل - 3ا(5( ت و� ر/7 :4ك � ���:7 و� ا*() /% أآ�� �� � :�8 -�ك ��اد ���و�� � `�ب �� 4 -

1 ()'{��K و-*�ض ���- � %� �a>& و_��ك دى دون �2ف 5 -' ��_ bا�- )� �K �� ا�5�9{اء /��( إ��غ ر��l ا(��a أو ����ل ا,�

�1 ة أ�2} و-(!7 1 د�� آ��dة c`!- - 6 {� ع �!�g4م را76 آ) ا�و!- ��k (9 k�- و_��ك و� {��K �' ل���� ا(Kأ ) ر:� 7 -adا È1� 5�/ qKا$% و�4رع ار:� ه% ا adن ا� ���ا ��d(�&�ل وأ�ا �/ وآ�

Page 13: اسئلة واجوبة في الحماية الكهربائية

�a�)& آ�� 8-k� ع ا(� -�� واg(�ا6*� أون �dت ا d) È1�

52 - �:��dا�� �*�/� �a2ط ا!`) ا�5ا��� اآ�d �� ا�*� ا����d:� /� ا<�aط ( ذا ا ا�ر��g؟ ع*x§ارى ا��ك (ا� ل ا��ارb �� ا§9`�KإJ�gت ا�ر�� kأ�9ع �� ا J�� /% ا<�aط ا�5ا

89 �7 ا(* �) ا��ارى!�- Jو� و�� ا(*�وف أن ا(` ه� ا" :� �� و�6د - 53� � ت ا�{:���q!� �/ �)���s<� -4ار �(��ك :*() '&% ا�� ر ا(

4K ا(*4ات� ت ا) ا(�4 وا(*4ات ا4وارة ا��2ى, ا��ر��� ( ا4وارة (a�� �/4��� ���؟ ��5kا

عa`K(*4ات �ن ا�ة اc5 وري 46ا�g ت >s)ة اcوب ان و�6د ��) ه ا(� �6ء &�� ر ا(�!%&' 5- ا�� ر ا(�! وب /� ا��"�) ا�'�� دي :£دي ا% -�$� �!���q ا�{:�� ا�� -*() ���آ

4K ��� ا�1 �( :£دي ا% 1,�ل ا�gار $4 -��4 ا(�� و�1% ا(��ر ا�K . ت /� ا(!���q وا��>s)!�ع �� اا اcو�6د ه �kو �q� ��k�- -24) ا*() �,�رة او-��

4K وا(��ر� ع ا�� ر ا(�! وب :�(� ا(a`Kاي -&� ا �� ����cا :�+ ا� آ�4 '&�5 .ا� .و/�,5 �,�رة دور:�

؟؟ Frequency load shedding ا�1( ل � ��دد � �*!% ر�� -- 54�'�$���� $� ا�() � ��دد 'k'

When the load increase ------------------the frequency decrease when the load decrease ------------------the frequency Increase

ت ا��دد /*!4 ز: دة ا�())k��� 7�a��- � ��dآ (k�� Frequency Governor #:�*- ن ا-�دد -`�م,`K م ا�� ظ '&%ا�(q!& %!� م %�1 :�q!ل ا�$) اه(�� /� ا ت ��,) ا�1(:

ا��9`�ار اي :(�k ا`�ل

��� ��5kل ا �4ر:� '!�4 :`) ا��دد ��8d ا{: دة :*!� هcا ا(,/ %&' �&a,) ا�1(�'&% ا���9 �� ا��:*� speed governor ا(� �6ة '&% ا�() و '4م $4رة 1 آ+ ا��دد

ن ا��دد ا(� À6 و :�+ /,)/ دة{:,`K ء او /� ا4dا:� و ان 9*d ��)`&�&� ا�هل ا ا�1( مq!ا �`���� و /`4 -�از�K ا�7g و + : ��5kا stability frailer %م ا q! � (,K !K /

�مqا � 1 l4ا�ا : How the voltage affect by Reactive Power??55

Answer: The voltage affects directly with the reactive power in the power systems so if the reactive power increased the voltage increase and vise versa when the reactive power

decrease the voltage decrease. ه� ا��ق ��� ا�ر�g وا!��-�ال ؟؟ -- 56� ا� ر:#

�� �� ا�56{ة او ا(*4اته� دا��ة 1( ��5kر ا ��ب ا��- �� �: ��� ��5kا ��� ب �,*`� آ��5,: � ر ��Kر '� ا� ر ا���� ���"- اذا و�) -� ر ا���ب وا"�ض �!5

�/ 5)�),- �/ �� 2 �`:� J #:ر��وا ��dا� �,K 41ود ��� ب آ��5a$و-�46 ا�ن ا

و�� ا�رض �:�+ _�59 /� ا�رض '!4 ����ى`� (*�: �' �5)3-{:4� اوم وذ} � �رض� J�_��-ر و ��ور ا��

ا!��-�ال ر ا�� ر و'�د-J ا% ا(,4ر و:�kن ا!��-�ال /�5�) ()k)وه� ا �)qKا�7 /� ا�ا {&� ا

-*�a 456:� واذا آ ن 4:} �,4ر ذو ار�*� ا9�ك و�5( آ ن b456 /� ا(, در ا��

Page 14: اسئلة واجوبة في الحماية الكهربائية

456 ا�J6 و:*c63{�aر )/ ا<�(456 ا�&}=4 ا(��2ذ �� ا41 ا�9�ك وا!��-�ال / ن ا�5 آ( ان

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Page 15: اسئلة واجوبة في الحماية الكهربائية

J�: 5K %/ !`) و4ا:� �2 ا� %/ 5*g,�ا'� :�+ وا �*K � )4*� (d$ و �a�)ا �� J6�2و -24) '&% د�2 J�a�)& ا� �� و -*() � K*� ا,�ا'� '&% -��:¾ ا��!�� (d$ % ا�رضإ

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�x 41وث"� ا�و% /� ا�d� ��_ أو �x d� (k�� �g<� وا�را ��� l��- " 7a`!- ء�Kء '� آ) أ ��5kا �����4 ELCB ا(!{ل ، ���K� *() ا` 7 ا�!' d ، و:�4ث ذ} _

��- ./� و7g ا��"�) -�kن ا(�

Page 16: اسئلة واجوبة في الحماية الكهربائية

��K �x أو"/� ا� � ا�d� (k�� 4: x 41وث -��l ��� ا<� وا(�d� ��_� " �/ {cوآ����K ،(!{لا �� �����5 ء '� 6{ء �k7 اa`!- ،�*ا��ا � MCB *() ا` 7 ا�s:وأ ،

7gو �/ ��- �4 -�kن ا(�!' d .ا��"�) :�4ث ذ} _ �� �x أو _��"/� ا� � ا�d� (k�� �g4 وا�ر: �x 41وث -��l ��� ا(�d� " 7a`!-

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ا��ق ��� ال -- 66� electric field ,وال magnetic field؟ ت 9 آ!� أو آ!���� ��6د (electric field) ا��ق ه� ان!�x و�6د ����K ��!:

(magnetic field) 7� ��"�� ��}ا ال)!)��magnetic fieldت ����آ� أو ( !�� ����K ��!: electric field) ��"�� /% ا(��k ا�� ر ��`4م -- 67)!�� � '� ا�45؟ �J /% ا(&� ا�� ر �� �2 '� ا��

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V=Vmcos(wt) � �&� ا��k� 4*� ر /��kن �* د� ا��

I(t)=imsin(wt) � :�kن ا�� ر ����2 ب � در90�6و�

� 1 �/ ا(�k�/ ��kن ا�� ر ه� -� g) ا�45 و-�kن �* د�J آ «-� أ�I(t)=-imsin(wt)

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ه� ا��ق ��� ���ل ا�� ر ا(��<4م /� ا`� س '� -- 69� ���� �/ J�4>��ت ا�� ر ا( ا�( :J ؟؟

ا`� س ه� $� س ا�� ر /� 1 �ت ا�() ا�da*� و���� ����ت ا�� ر ا< �� ��56{ة}56�� �� ل أ� ����ت ا�� ر ا<a'ل ا�� س ا�� ر 2�$ 5�����/ J: ا�(bc5 �a!")ن �!�!� ا�/ J�&'س -�(�{ �(!�!� و ا(���ت -*�(4 '&% و����5 /(���ت ا`�

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�� bc5 ا(���ت ��-�*J أ:�JK � (KNEE POINT) و-�kن '�وة ا(!�!% ا�� رات ا* ع &`&8dxل :�4ث ا a'ا� � .ا�4:4ي bc5 ا(���ت /� 1

ذا :��<4م -- 70) fuse&CB %/ ؟ ��)' J: 1 ت ا��ز:7 او /% ا�(� �: )1 &�() وا(,4ر ا�( :J ه� '() 1( :Jا45ف ��

CB �6د ا6{اء�ا �qK {وذ 5&�/ � :J او� و/% 1)� 5 -`�م ال -`�م �� J�k�K k�� FUSE ك ز�� �*�� ��� '() ال 5 و-�,) ا4ا��b وه!&)*� FUSE &CB 4,`: � bوه�

J� ( PROTECTION COORDINATION)

ه� أ�Kاع ال -- 71� Faults �؟؟/� ا ��� ��5kت ا kd Fault In Electrical Power System:

Page 17: اسئلة واجوبة في الحماية الكهربائية

1- Multi-phase Short Circuit "High Current will flow a- Balanced three phase Fault

b- Single Phase - to - ground Fault c- Double phase to ground Fault

d- Phase to phase Fault ه� Single Phase - to - ground Fault أ�5x أ�Kاع ا`,� ه�d:�`- J�41و �d�Kو

80 % ��� ل و :�kن /� ا<�aط ا�5اa'ت و �� ا� و:)�kأو ا ���-� �8d -&�ث ا*�ازل ��و آc} :�4ث '!�x 4خ أو آ�� أ41 ا*�ازل . ا`,� '!4 ز: دة ا� ��� أو K{ول ا(�a :�4ث

�K أوd)<� /�ق ا8 ��ور اd� ت ا(�-�*� -�� ا<���ور أو �!1 ا��`K �&- 8d��ت ا�ر��g :�4ث �d�k& �d�! . أو ��8d أ'( ل ا��� joints ا&� م �

2- Open Circuit/Phases "No current" 3- Contacts with Ground "Isolated system or High Impedance

grounded system

Page 18: اسئلة واجوبة في الحماية الكهربائية

1 -- Why cables are not used for long distance transmission?

Cables are not used for long distance transmissions due to their large

charging currents.

2 -- Mention the 3 main parts of the cable?

Conductor, dielectric, sheath

3 -- What is the function of conductor?

Conductor provides the conducting path for the current.

4 - .What is the function of sheath in cables?

The sheath does not allow the moisture to enter and protects the cable

from all external influences like chemical or electrochemical attack, fire

etc.

5 -- Mention the conductor materials in cables?

Copper, Aluminium All occupies greater space than copper for a given conductance.

Both are high electrical conductivity.

6 -- What is the purpose of stranding of conductors?

Stranding increases the cable. It has flexibility. Resistance of the

7 -- Define the segmental conductors.

The stranded wires which are compacted by the rollers to minimize the air

spaces between the individual wires are called segmented conductors.

Here the conductor size is reduced for a given conductance.

8. State the properties of insulating materials.

Page 19: اسئلة واجوبة في الحماية الكهربائية

It should have high insulation resistance, high dielectric strength, and

good mechanical properties, non-hygroscopic, capable of being operated

at high temperatures, low thermal resistance and low power factor.

9. Mention the commonly used power cables.

Impregnated paper, Polyvinyl chloride, polyethylene

10. Mention the advantages of PVC over paper insulated cables.

Reduced cost and weight, Insulation is resistant to water, Simplified

jointing, increased flexibility No plumbing required.

11. State the merits of paper insulated cables.

High current carrying capacity, long life and greater reliability

12. Where polythene cables are used?

-being non-hygroscopic used in cables for submarines and damp soil. - being lighter used as aerial cables for vertical installations.

13. State the advantages of polythene insulators.

They are non-hygroscopic, light in weight, low dielectric constant, low loss

factor and low thermal resistance.

14. By what materials cable sheaths are made?

Lead sheaths and Aluminums sheaths.

15. In what way Al sheaths are superior to lead sheaths?

Al sheaths are smaller in weight, high mechanical strength, greater

conducyivity, cheap, easy to manufacture and install, withstands the

required gas pressure without reinforcement.

16. Where CSA sheath is used in cables?

Corrugated seamless aluminums sheath is used in high voltage oil filled

cables and telephone lines.

Page 20: اسئلة واجوبة في الحماية الكهربائية

17. Why it is used?

It is used because it is very flexible and easily by repeated bending the

sheath is not distorted and it is not damaged. It has lesser weight and

reduced thickness.

18. Why protective covering is done in cables?

To protect the cables from mechanical damage, corrosion and electrolytic

action when laid direct in the ground the protective covering is made.

19. By what material protective covering is made?

Bitumen &Bituminized materials, PVC and layers of fibrous materials.

20. What is meant by serving of a cable?

Layers of fibrous material permitted with waterproof compound applied to

the exterior of the cable is called serving of a cable.

21. Why armoring is done in the cables?

To protect the sheath from mechanical damage.

22. Why armoring is not done in single core cables?

The presence of magnetic material within the alternating magnetic field of

a single core cable produces excessive losses. Hence single core cables are

left unarmored with non-magnetic materials like tin-bronze or silicon-

bronze tapes or wires.

23. Why Al is used as an armour material?

It has non-magnetic properties, high conductivity and mechanical

strength.

24. What is meant by grading of cables?

The method of equalizing the stress in the dielectric of the cable is called

the grading of cables.

Page 21: اسئلة واجوبة في الحماية الكهربائية

25.Mention the methods of cables. Capacitance grading and intersheath grading.

26.Why the capacitance of the cable is very high than the capacitance of the overhead lines?

The distance between the conductors are small. The distance between the cores and the earthed sheath is also small. The permittivity of the cable insulation is 3 to 5 times greater than

that of air insulation. 27.Write the expression of the capacitance of a single core

cable.

( (2πε) ----------- = C=q/v

( ln(R/r)

28.What is meant by charging current of a cable? The capacitance of a cable determines the charging current. The charging current restricts the use of cables on EHV lines. The current carrying capacity of an a.c cable is also reduced by the

charging current.

29.Why power loss occurs in the dielectric of a cable? -due to conductivity of insulation

-Dielectric hysteresis or dielectric absorption. -Ionization or corona

30. Why all transmission and distribution systems are 3 phase systems?

A 3 phase a.c circuit using the same size conductors as the single phase circuit can carry three times the power which can be carried by a 1 phase circuit and uses 3 conductors for the 2 phases and one conductor for the neutral. Thus a 3 phase circuit is more economical than a 1 phase circuit in terms of

initial cost as well as the losses. Therefore all transmission and distribution systems are 3 phase systems.

31. Why the transmission systems are mostly overhead systems?

Because of the cost consideration, the transmission systems are mostly overhead systems.

32.Why all overhead lines use ACSR conductors? ACSR conductors comprises of hard drawn aluminium wires

Page 22: اسئلة واجوبة في الحماية الكهربائية

stranded around a core of single or multiple strand galvanized steel wire. They provides the necessary conductivity while the steel provides the necessary mechanical strength. Has less corona loss. The breaking load is high and has less weight. 33.Why transmission lines are 3 phase 3 wire circuits while

distribution lines are 3 phase 4 wire circuits? A Balanced 3 phase circuit does not require the neutral

conductor, as the instantaneous sum of the 3 line currents are zero. Therefore the transmission lines and feeders are 3 phase 3

wire circuits. The distributors are 3 phase 4 wire circuits because a neutral wire is necessary to supply the 1 phase loads

of domestic and commercial consumers. 34.Why overhead line conductors are invariably stranded? They are stranded to make them flexible during erection and

while in service.

35 . Stat the advantages of interconnected systems. Any area fed from one generating station during overload hours

can be fed from another power station and thus reserved capacity required is reduced, reliability of supply is increased

and efficiency is increased.

36. What is a ring distributor? A ring distributor is a distributor which is arranged to form a

closed circuit and is fed at one or more than one point. 37. State any two advantages of ring main system.

Less voltage fluctuations at consumer’s terminals.Less copper is required as each part of the ring carries less current than in

radial system. 38. Mention the disadvantages of a 3 wire system

In 3 wire system a third wire is required.The safety is partially reduced.A balancer is required and therefore cost is increased. 39. What are the advantages of a 3 wire dc distribution system

over a 2 wire dc distribution system? If 3 wire system is used to transmit the same amount of power

over the same distance with same efficiency with same consumer voltage we require 0.3125 times copper as required in

2 wire system.

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40. Mention the differences between 3 wire and 3 phase 4 wire distribution system?

3 phase 3 wire is employed for balanced loads, but 3 phase 4 wire is employed for unbalanced loads.

3 phase 3 wire is used for transmission but 3 phase 4 wire is used for distribution of power to consumers.

41. State kelvin’s law. The annual expenditure on the variable part of the transmission system should be equal to the annual cost of energy wasted in

the conductor used in that system. 42. State any two limitations of kelvin’s law.

It is difficult to estimate accurately the annual charge on the capital outlay. It does not give the exact economical size of the

conductor. 43. Define resistance of the transmission line.

It is defined as the loop resistance per unit length of the line in a single phase system. In 3 phase system it is defined as the

resistance per phase. 44. Define inductance of a line.

It is defined as the loop inductance per unit length of the line.Its unit is henrys per meter.

45. Define capacitance of a line. It is defined as shunt capacitance between the two wires per

unit line length. Its unit is farads per meter. 46. What is skin effect?

The steady current when flowing through the conductor ,does not distribute uniformly, rather it has the tendency to concentrate near the surface of the conductor. This

phenomenon is called skin effect. 47. Why skin effect is absent in dc system?

The steady current when flowing through a conductor distributes itself uniformly over the whole cross section of the

conductor.That is why skin effect is absent in dc system. 48. What is the effect of skin effect on the resistance of the line?

Due to skin effect the effective area of cross section of the conductor through which current flow is reduced. Hence the resistance of the line is increased when ac current is flowing.

49. On what factors the skin effect depend? Nature of the material, Diameter of the wire , Frequency and

shape of the wire.

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50. Define symmetrical spacing. In 3 phase system when the line conductors are equidistant

from each other then it is called symmetrical spacing. 51. What is the necessity for a double circuit line?

To reduce the inductance per phase and to increase the efficiency.

52. Mention the factors governing the inductance of a line. Radius of the conductor and the spacing between the

conductors. 53. Define a neutral plane.

It is a plane where electric field intensity and potential is zero. 54. Define proximity effect.

The alternating magnetic flux in a conductor caused by the current flowing in a neighbouring conductor gives rise to a circulating current which cause an apparent increase in the resistance of the conductor.This phenomenon is called as

proximity effect.

55. What is the effect of proximity effect? It results in the non uniform distribution of current in the cross

section, and the increase of resistance. 56. What is a composite conductor?

A conductor which operates at high voltages and composed of 2 or more subconductors and run electrically in parallel are called

composite conductors. 57. What is a bundle conductor?

It is a conductor made up of 2 or more sub conductors and is used as one phase conductors.

58. Mention the advantages of using bundled conductors. Reduced reactance, reduced voltage gradient , reduced corona

loss .reduced interference 59. What is corona?

The phenomenon of violet glow, hissing noise and production of ozone gas in an overhead line is called corona.

60. Mention the factors affecting corona. Atmosphere,conductor size, spacing between conductors and

line voltage. 61.Define critical disruptive voltage.

It is defined as the minimum phase voltage at which corona

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occrus. 62. Define visual critical voltage.

It is the minimum phase voltage at which corona appears all along the line conductors.

63. State any two merits of corona. Reduces the effects of transients produced by surges.System

performance is improved. 64. Give the two demerits of corona.

The transmission efficiency is affected. Corrosion occurs. 65. Mention the methods of reducing corona effect.

By increasing the conductor size and conductor spacing.

. Why ACSR conductors are used in lines? If the size of the conductor is larger corona effects are reduced and reduces the proximity effect.Hence they are used in lines.

67. Define medium lines. Lines having length between 60 and 150 km and line voltages

between 20 and 100kv are called medium lines. 68.Define short lines.

Lines having lengths below 60km and voltages below 20kv are called short lines.

69.Mention the limitations of end condenser method. This over estimates the effects of line capacitance.It is assumed

to be lumped or concentrated. 70. Mention the demerits of HVDC transmission.

Electric power cannot be generated at high dc voltages. The dc voltages cannot be stepped up for transmission of power

at high voltages. The dc switches and circuit breakers have their own limitations.

71. What are the advantages of high voltage ac transmission. The power can be generated at high voltages.

The maintenance of ac substation is easy and cheaper. 72. Mention the disadvantages of high voltage ac transmission.

An ac line requires more copper than a dc line. The construction of an ac line is more complicated than a dc

transmission line. Due to skin effect in the ac system the effective resistance of the

line is increased. 73. Mention the limitations of using very high transmission

voltage. The increased cost of insulating the conductor.

The increased cost of transformers ,switch gears and other

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terminal apparatus. 74. Mention the terminal equipments necessary in HVDC

system. Converters, mercury arc valves and thyristors.

75. Define sag of a line. The difference in level between the points of supports and the

lowest point of the conductor is called as sag.

76. Mention the factors that affect sag in the transmission line. Weight of the conductor,length of the span , working tensile

strength and the temperature. 77. What is the reason for the sag in the transmission line? While erecting the line , if the conductors are stretched too much between supports then there prevails an excessive

tension on the line which may break the conductor. In order to have safe tension in the conductor a sag in the line is allowed.

78. Mention the methods of laying the cables. Direct laying , draw in system and solid system

79. Mention the advantages of direct laying of cables. It is simple and cheaper method.It gives the best conditions for

dissipating the heat generated in the cables. 80. State any 2 disadvantages of direct laying method.

Localisation of fault is difficult . It cannot be used in congested areas where excavation is inconvenient.

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1. What are the functions of protective relays

To detect the fault and initiate the operation of the circuit breaker to isolate

the defective element from the rest of the system, thereby protecting the system from

damages consequent to the fault. 2. Give the consequences of short circuit.

Whenever a short-circuit occurs, the current flowing through the coil increases to an

enormous value. If protective relays are present , a heavy current also flows through the

relay coil, causing it to operate by closing its contacts.The trip circuit is then closed , the

circuit breaker opens and the fault is isolated from the rest of the system. Also, a low

voltage may be created which may damage systems connected to the supply.

3. Define protected zone. Are those which are directly protected by a protective system

such as relays, fuses or switchgears.If a fault occurring in a zone can be

immediately detected and or isolated by a protection scheme dedicated to that particular

zone. 4. What are unit system and non unit system?

A unit protective system is one in which only faults occurring within its

protected zone are isolated.Faults occurring elsewhere in the system have no influence

on the operation of a unit system.A non unit system is a protective system which is

activated even when the faults are external to its protected zone. 5. What is primary protection?

Is the protection in which the fault occurring in a line will be cleared by its

own relay and circuit breaker.It serves as the first line of defence.

6. What is back up protection?

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Is the second line of defence , which operates if the primary protection

fails to activate within a definite time delay. 7. Name the different kinds of over current relays.

Induction type non-directional over current relay,Induction type directional

over current relay & current differential relay. 8. Define energizing quantity.

It refers to the current or voltage which is used to activate the relay into operation.

9. Define operating time of a relay. It is defined as the time period extendind from the occurrence of

the fault through the relay detecting the fault to the operation of the

relay. 10. Define resetting time of a relay.

It is defined as the time taken by the relay from the instant of isolating

the fault to the moment when the fault is removed and the relay can be reset.

11. What are over and under current relays? Overcurrent relays are those that operate when the current in a

line exceeds a predetermined value. (eg: Induction type non-

directional/directional overcurrent relay, differential overcurrent relay)whereas undercurrent relays are

those which operate whenever the current in a circuit/line drops below a

predetermined value.(eg: differential over-voltage relay)

12. Mention any two applications of differential relay. Protection of generator & generator transformer unit; protection

of large motors and busbars .

13. What is biased differential bus zone reduction? The biased beam relay is designed to respond to the differential

current in terms of its fractional relation to the current flowing through the

protected zone. It is essentially an over-current balanced beam relay type with an additional

restraining coil. The restraining coil produces a bias force in the opposite direction

to the operating force.

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14. What is the need of relay coordination? The operation of a relay should be fast and selective, ie, it

should isolate the fault in the shortest possible time causing minimum disturbance to

the system. Also, if a relay fails to operate, there should be sufficiently quick backup

protection so that the rest of the system is protected. By coordinating relays, faults can always

be isolated quickly without serious disturbance to the rest of the system.

15. Mention the short comings of Merz Price scheme of protection applied to a power transformer.

In a power transformer, currents in the primary and secondary are to be compared.

As these two currents are usually different, the use of identical transformers will give

differential current, and operate the relay under no-load condition. Also, there is usually a

phase difference between the primary and secondary currents of three phase transformers.

Even CT’s of proper turn-ratio are used, the differential current may flow through the

relay under normal condition. 16. What are the various faults to which a turbo alternator is

likely to be subjected? Failure of steam supply; failure of speed; overcurrent; over

voltage; unbalanced loading; stator winding fault .

17. What is an under frequency relay? An under frequency relay is one which operates when the

frequency of the system (usually an alternator or transformer) falls below a certain value. 18. Define the term pilot with reference to power line protection. Pilot wires refers to the wires that connect the CT’s placed at the

ends of a power transmission line as part of its protection scheme. The

resistance of the pilot wires is usually less than 500 ohms.

19. Mention any two disadvantage of carrier current scheme for transmission line only.

The program time (ie, the time taken by the carrier to reach the other end-upto

.1% mile); the response time of band pass filter; capacitance phase-shift of the

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transmission line . 20. What are the features of directional relay?

High speed operation; high sensitivity; ability to operate at low voltages; adequate

short-time thermal ratio; burden must not be excessive. 21. What are the causes of over speed and how alternators are

protected from it? Sudden loss of all or major part of the load causes over-

speeding in alternators. Modern alternators are provided with mechanical centrifugal

devices mounted on their driving shafts to trip the main valve of the prime mover when a

dangerous over-speed occurs.

22. What are the main types of stator winding faults? Fault between phase and ground; fault between phases and

inter-turn fault involving turns of the same phase winding.

23. Give the limitations of Merz Price protection. Since neutral earthing resistances are often used to protect

circuit from earth-fault currents, it becomes impossible to protect the whole of a star-

connected alternator. If an earth-fault occurs near the neutral point, the voltage may be

insufficient to operate the relay. Also it is extremely difficult to find two identical CT’s. In

addition to this, there always an inherent phase difference between the primary and

the secondary quantities and a possibility of current through the relay even when there is

no fault. 24. What are the uses of Buchholz’s relay?

Bucholz relay is used to give an alarm in case of incipient( slow-developing)

faults in the transformer and to connect the transformer from the supply in the event of

severe internal faults. It is usually used in oil immersion transformers with a rating over

750KVA. 25. What are the types of graded used in line of radial relay

feeder? Definite time relay and inverse-definite time relay.

26. What are the various faults that would affect an alternator? (a) Stator faults

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1, Phase to phase faults 2, Phase to earth faults

3, Inter turn faults (b)

1, Earth faults 2, Fault between turns

3, Loss of excitation due to fuel failure (c) 1, Over speed 2, Loss of drive

3, Vacuum failure resulting in condenser pressure rise, resulting in

shattering of the turbine low pressure casing (d) 1, Fault on lines 2, Fault on busbars

27. Why neutral resistor is added between neutral and earth of an alternator?

In order to limit the flow of current through neutral and earth a resistor is

introduced between them. 28. What is the backup protection available for an alternator?

Overcurrent and earth fault protection is the backup protections. 29. What are faults associated with an alternator?

(a) External fault or through fault (b) Internal fault

1, Short circuit in transformer winding and connection 2, Incipient or slow developing faults

30. What are the main safety devices available with transformer? Oil level guage, sudden pressure delay, oil temperature

indicator, winding temperature indicator .

31. What are the limitations of Buchholz relay? (a) Only fault below the oil level are detected.

(b) Mercury switch setting should be very accurate, otherwise even for

vibration, there can be a false operation. (c) The relay is of slow operating type, which is unsatisfactory. 32. What are the problems arising in differential protection in

power transformer and how are they overcome? 1. Difference in lengths of pilot wires on either sides of the relay.

This is overcome by connecting adjustable resistors to pilot wires to

get equipotential points on the pilot wires.

2. Difference in CT ratio error difference at high values of short

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circuit currents that makes the relay to operate even for external or through

faults. This is overcome by introducing bias coil.

3. Tap changing alters the ratio of voltage and currents between HV and LV sides

and the relay will sense this and act. Bias coil will solve this. 4. Magnetizing inrush current appears wherever a transformer is

energized on its primary side producing harmonics. No current will be seen by

the secondary. CT’s as there is no load in the circuit. This difference in current

will actuate the differential relay. A harmonic restraining unit is added to the

relay which will block it when the transformer is energized.

33. What is REF relay? It is restricted earth fault relay. When the fault occurs very near

to the neutral point of the transformer, the voltage available to drive the earth

circuit is very small, which may not be sufficient to activate the relay, unless the

relay is set for a very low current. Hence the zone of protection in the winding of the

transformer is restricted to cover only around 85%. Hence the relay is called REF relay.

34. What is over fluxing protection in transformer? If the turns ratio of the transformer is more than 1:1, there will

be higher core loss and the capability of the transformer to withstand this is limited

to a few minutes only. This phenomenon is called over fluxing. 35. Why busbar protection is needed?

(a) Fault level at busbar is high (b) The stability of the system is affected by the faults in the bus

zone. (c) A fault in the bus bar causes interruption of supply to a large

portion of the system network.

36. What are the merits of carrier current protection? Fast operation, auto re-closing possible, easy discrimination of

simultaneous faults .

37. What are the errors in CT?

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(a) Ratio error Percentage ratio error = [(Nominal ratio – Actual ratio)/Actual

ratio] x 100 The value of transformation ratio is not equal to the turns ratio.

(b) Phase angle error: Phase angle _ =180/_[(ImCos _-I1Sin _)/nIs]

38. What is field suppression? When a fault occurs in an alternator winding even though the

generator circuit breaker is tripped, the fault continues to fed because EMF is

induced in the generator itself. Hence the field circuit breaker is opened and stored

energy in the field winding is discharged through another resistor. This method is known as

field suppression. 39. What are the causes of bus zone faults?

_ Failure of support insulator resulting in earth fault _ Flashover across support insulator during over voltage

_ Heavily polluted insulator causing flashover _ Earthquake, mechanical damage etc.

40. What are the problems in bus zone differential protection? _ Large number of circuits, different current levels for different

circuits for external faults.

_ Saturation of CT cores due to dc component and ac component in short

circuit currents. The saturation introduces ratio error. _ Sectionalizing of the bus makes circuit complicated.

_ Setting of relays need a change with large load changes. 41. What is static relay?

It is a relay in which measurement or comparison of electrical quantities is made

in a static network which is designed to give an output signal when a threshold condition

is passed which operates a tripping device. 42. What is power swing?

During switching of lines or wrong synchronization surges of real and reactive

power flowing in transmission line causes severe oscillations in the voltage and current

vectors. It is represented by curves originating in load regions and traveling towards relay

characteristics. 43. What is a programmable relay?

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A static relay may have one or more programmable units such as microprocessors

or microcomputers in its circuit. 44. What is CPMC?

It is combined protection, monitoring and control system incorporated in the static

system. 45. What are the advantages of static relay over electromagnetic

relay? o Low power consumption as low as 1mW

o No moving contacts; hence associated problems of arcing, contact bounce,

erosion, replacement of contacts o No gravity effect on operation of static relays. Hence can be

used in vessels ie, ships, aircrafts etc.

o A single relay can perform several functions like over current, under

voltage, single phasing protection by incorporating respective functional

blocks. This is not possible in electromagnetic relays o Static relay is compact

o Superior operating characteristics and accuracy o Static relay can think , programmable operation is possible

with static relay

o Effect of vibration is nil, hence can be used in earthquake-prone areas

o Simplified testing and servicing. Can convert even non-electrical

quantities to electrical in conjunction with transducers. 46. What is resistance switching?

It is the method of connecting a resistance in parallel with the contact space(arc).

The resistance reduces the restriking voltage frequency and it diverts part of the arc

current. It assists the circuit breaker in interrupting the magnetizing current and capacity

current. 47. What do you mean by current chopping?

When interrupting low inductive currents such as magnetizing currents of the

transformer, shunt reactor, the rapid deionization of the contact space and blast effect

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may cause the current to be interrupted before the natural current zero. This phenomenon

of interruption of the current before its natural zero is called current chopping.

48. What are the methods of capacitive switching? • Opening of single capacitor bank

• Closing of one capacitor bank against another 49. What is an arc?

Arc is a phenomenon occurring when the two contacts of a circuit breaker

separate under heavy load or fault or short circuit condition. 50. Give the two methods of arc interruption?

_ High resistance interruption:-the arc resistance is increased by elongating, and

splitting the arc so that the arc is fully extinguished _ Current zero method:-The arc is interrupted at current zero

position that occurs100 times a second in case of 50Hz power system

frequency in ac. 51. What is restriking voltage?

It is the transient voltage appearing across the breaker contacts at the instant of arc being extinguished.

52. What is meant by recovery voltage? The power frequency rms voltage appearing across the breaker

contacts after the arc is extinguished and transient oscillations die out is called

recovery voltage. 53. What is RRRV?

It is the rate of rise of restriking voltage, expressed in volts per microsecond. It is

closely associated with natural frequency of oscillation. 54. What is circuit breaker?

It is a piece of equipment used to break a circuit automatically under fault

conditions. It breaks a circuit either manually or by remote control under normal

conditions and under fault conditions. 55. Write the classification of circuit breakers based on the

medium used for arc extinction? _ Air break circuit breaker

_ Oil circuit breaker _ Minimum oil circuit breaker

_ Air blast circuit breaker

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_ SF6 circuit breaker _ Vacuum circuit breaker

56. What is the main problem of the circuit breaker? When the contacts of the breaker are separated, an arc is struck

between them. This arc delays the current interruption process and also

generates enormous heat which may cause damage to the system or to the breaker itself. This is

the main problem. 57. What are demerits of MOCB?

_ Short contact life _ Frequent maintenance _ Possibility of explosion

_ Larger arcing time for small currents _ Prone to restricts

58. What are the advantages of oil as arc quenching medium? • It absorbs the arc energy to decompose the oil into gases,

which have excellent cooling properties

• It acts as an insulator and permits smaller clearance between line conductors

and earthed components 59. What are the hazards imposed by oil when it is used as an

arc quenching medium? There is a risk of fire since it is inflammable. It may form an

explosive mixture with arc. So oil is preferred as an arc quenching medium.

60. What are the advantages of MOCB over a bulk oil circuit breaker?

• It requires lesser quantity of oil • It requires smaller space

• There is a reduced risk of fire • Maintenance problem are reduced

61. What are the disadvantages of MOCB over a bulk oil circuit breaker?

o The degree of carbonization is increased due to smaller quantity of oil

o There is difficulty of removing the gases from the contact space in time

o The dielectric strength of the oil deteriorates rapidly due to high degree of carbonization.

62. What are the types of air blast circuit breaker? _ Arial-blast type

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_ Cross blast _ Radial-blast

63. What are the advantages of air blast circuit breaker over oil circuit breaker?

o The risk of fire is diminished o The arcing time is very small due to rapid buildup of dielectric

strength between contacts

o The arcing products are completely removed by the blast whereas oil

deteriorates with successive operations 64. What are the demerits of using oil as an arc quenching

medium? • The air has relatively inferior arc quenching properties

• The air blast circuit breakers are very sensitive to variations in the rate of rise

of restriking voltage • Maintenance is required for the compression plant which

supplies the air blast 65. What is meant by electro negativity of SF6 gas?

SF6 has high affinity for electrons. When a free electron comes and collides with

a neutral gas molecule, the electron is absorbed by the neutral gas molecule and negative

ion is formed. This is called as electro negativity of SF6 gas. 66. What are the characteristic of SF6 gas?

It has good dielectric strength and excellent arc quenching property. It is inert,

non-toxic, noninflammable and heavy. At atmospheric pressure, its dielectric strength is

2.5 times that of air. At three times atmospheric pressure, its dielectric strength is equal to

that of the transformer oil. 67. Write the classifications of test conducted on circuit

breakers. _ Type test

_ Routine test _ Reliability test

_ Commissioning test 68. What are the indirect methods of circuit breaker testing?

o Unit test o Synthetic test

o Substitution testing o Compensation testing

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o Capacitance testing 69. What are the advantages of synthetic testing methods? • The breaker can be tested for desired transient recovery

voltage and RRRV. • Both test current and test voltage can be independently varied.

This gives flexibility to the test

• The method is simple • With this method a breaker capacity (MVA) of five time of that

of the capacity of the test plant can be tested.

70. How does the over voltage surge affect the power system? The over voltage of the power system leads to insulation

breakdown of the equipments. It causes the line insulation to flash over and may

also damage the nearby transformer, generators and the other equipment connected to

the line. 71. What is pick up value?

It is the minimum current in the relay coil at which the relay starts to operate.

72. Define target. It is the indicator used for showing the operation of the relay.

73. Define reach. It is the distance upto which the relay will cover for protection.

74. Define blocking. It means preventing the relay from tripping due to its own

characteristics or due to additional relays. 75. Define a over current relay.

Relay which operates when the current ia a line exceeds a predetermined value.

76. Define an under current relay? Relays which operates whenever the current in a circuit drops

below a predetermined value.

77. Mention any 2 applications of differential relays. Protection of generator and generator-transformer unit:

protection of large motors and bus bars

78.Mention the various tests carried out in a circuit breaker at HV labs.

Short circuit tests, Synthetic tests& direct tests. 78. Mention the advantages of field tests.

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The circuit breaker is tested under actual conditions like those that occur in the

network. Special occasions like breaking of charging currents of long

lines ,very short line faults ,interruption of small inductive currents etc… can be

tested by direct testing only.

79. State the disadvantages of field tests. The circuit breaker can be tested at only a given rated voltage

and network capacity.

The necessity to interrupt the normal services and to test only at light load

conditions. Extra inconvenience and expenses in installation of controlling

and measuring equipment in the field.

80. Define composite testing of a circuit breaker. In this method the breaker is first tested for its rated breaking

capacity at a reduced voltage and afterwards for rated voltage at a low

current.This method does not give a proper estimate of the breaker performance.

81. State the various types of earthing. Solid earthing, resistance earthing , reactance earthing , voltage

transformer earthing and zig-zag transformer earthing.

82. What are arcing grounds? The presence of inductive and capacitive currents in the

isolated neutral system leads to formation of arcs called as arcing grounds.

83. What is arc suppression coil? A method of reactance grounding used to suppress the arc due

to arcing grounds.

84. State the significance of single line to ground fault. In single line to ground fault all the sequence networks are

connected in series. All the sequence currents are equal and the fault

current magnitude is three times its sequence currents.

85. What are symmetrical components? It is a mathematical tool to resolve unbalanced components into

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balanced components.

86. State the three sequence components. Positive sequence components, negative sequence components

and zero sequence components.

87. Define positive sequence component. -has 3 vectors equal in magnitude and displaced from each

other by an angle 120 degrees and having the phase sequence as original vectors.

88. Define zero sequence component. They has 3 vectors having equal magnitudes and displaced from

each other by an angle zero degees.

89. State the significance of double line fault. It has no zero sequence component and the positive and

negative sequence networks are connected in parallel.

90. Define negative sequence component. It has 3 vectors equal in magnitude and displaced from each

other by an angle 120 degrees and has the phase sequence in opposite to its

original phasors. 91. State the different types of faults.

Symmetrical faults and unsymmetrical faults and open conductor faults.

92. State the various types of unsymmetrical faults. Line to ground ,line to line and double line to ground faults 93. Mention the withstanding current in our human body.

9mA 94. State the different types of circuit breakers.

Air ,oil,vacuum circuit breakers. 95. Define per unit value.

It is defined as the ratio of actual value to its base value. 96. Mention the inductance value of the peterson’s coil.

L=1/3wc2 97. Define single line diagram.

Representation of various power system components in a single line is

defined as single line diagram. 98. Differentiate between a fuse and a circuit breaker.

Fuse is a low current interrupting device. It is a copper or an aluminium wire.Circuit breaker is a high current interrupting

device and it act as a switch

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under normal operating conditions. 99. How direct tests are conducted in circuit breakers?

Using a short circuit generator as the source. Using the power utility system or network as the source.

100. What is dielectric test of a circuit breaker? It consists of overvoltage withstand test of power frequency

lightning and impulse voltages.Testa are done for both internal and external insulation

with switch in both open and closed conditions.

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