Page 1 Page 1Fault calculation Fault perhitungan by JHNaylor oleh JHNaylor Chapter 3 Bab 3 3.1 Introduction 3,1 Pendahuluan 3.1.I Purpose of fault calculation Tujuan 3.1.I kesalahan perhitungan Fault calculation is the analysis of power system electrical behaviour under fault perhitungan Fault adalah analisis sistem tenaga listrik di bawah salah perilaku conditions, with particular reference to the effects of these conditions on the power kondisi, dengan rujukan khusus pada dampak dari kondisi daya system current and voltage values. sistem nilai arus dan tegangan. Together with other aspects of system analysis, Bersama dengan aspek lain dari analisis sistem, fault calculation forms an indispensable part of the whole function and process of kesalahan perhitungan merupakan bagian tak terpisahkan dari keseluruhan fungsi dan proses power system design. daya sistem desain. Correct design depends essentially on a full knowledge and desain yang benar pada dasarnya tergantung pada pengetahuan yang penuh dan understanding of system behaviour and on the ability to predict this behaviour for memahami perilaku sistem dan pada kemampuan untuk memprediksi perilaku ini untuk the complete range of possible system conditions. kisaran lengkap kondisi sistem mungkin. Accurate and comprehensive Akurat dan komprehensif analysis, and the means and methods of achieving it, are therefore of essential analisis, dan sarana dan metode untuk mencapainya, adalah karena itu penting importance in obtaining satisfactory power system performance and in ensuring the penting dalam mendapatkan kinerja daya sistem yang memuaskan dan dalam memastikan continued improvement in performance which results from the development and lanjutan perbaikan kinerja yang hasil dari pengembangan dan application of new methods and techniques. penerapan metode baru dan teknik. The applications of power system analysis cover the full range of possible system Penerapan sistem analisis daya mencakup berbagai sistem mungkin conditions, these being divisible into two main classes, namely conditions in which kondisi, ini yang dibagi menjadi dua kelas utama, yaitu kondisi di mana the power system is operating in a normal healthy state, and others in which it is sistem tenaga beroperasi dalam keadaan normal dan sehat, dan lain-lain di mana ia subjected to one or more of a wide variety of possible fault conditions. dikenakan untuk satu atau lebih dari berbagai macam kondisi kesalahan mungkin. The analysis Analisis of these conditions and their effects on the power system is of particular relevance kondisi ini dan efeknya pada sistem daya adalah relevansi khusus R) such considerations as: R) pertimbangan seperti: (a) the choice of a suitable power system arrangement, with particular reference (A) pilihan dari pengaturan sistem kekuasaan yang cocok, dengan referensi khusus to the configuration of the transmission or distribution network untuk konfigurasi jaringan transmisi atau distribusi (b) the determination of the required load and short-circuit ratings of the power (B) penentuan beban-sirkuit yang dibutuhkan dan peringkat singkat kekuasaan system plant sistem tanaman

(c) the determination of the breaking capacity required of the power system (C) penentuan kapasitas melanggar dibutuhkan dari sistem tenaga switchgear and fusegear switchgear dan fusegear (d) the design and application of equipment for the control and protection of the (D) dan aplikasi desain peralatan untuk kontrol dan perlindungan power system daya sistem (e) the operation of the system, with particular reference to security of supply (E) sistem operasi, dengan referensi khusus terhadap keamanan pasokan and economic considerations dan pertimbangan ekonomi

Page 2 Page 254 54 Fault calculation Fault perhitungan (f) the investigation of unsatisfactory performance of the power system or of (F) tidak memuaskan investigasi kinerja sistem kekuasaan atau individual items of power system plant. item individual sistem pembangkit listrik. The present chapter is concerned principally with the analysis of system fault Bab ini berkaitan terutama dengan analisis sistem sesar conditions, these conditions being of direct and particular relevance to the design kondisi, kondisi kesejahteraan dan khusus relevansi langsung untuk merancang and application of power system protection. dan penerapan sistem proteksi tenaga. The methods of analysis employed, Metode analisis yang digunakan, however, are essentially applications of general analysis and, as such have equal Namun, pada dasarnya aplikasi dari analisis umum dan, karena itu telah sama application to a wide range of other problems whose solution is dependent on aplikasi untuk berbagai solusi masalah-masalah lain yang tergantung pada electrical network analysis. Analisis jaringan listrik. 3.1.2 Types of fault 3.1.2 Jenis-jenis kesalahan In the context of electrical fault calculation, a power system fault may be defined Dalam konteks perhitungan arus listrik, kesalahan sistem tenaga listrik dapat didefinisikan as any condition or abnormality of the system which involves the electrical failure sebagai keadaan atau kelainan sistem yang melibatkan kegagalan listrik of primary equipment, the reference to primary (as opposed to ancillary) equip- peralatan utama, referensi untuk primer (sebagai lawan tambahan) peralatan ment implying equipment such as generators, transformers, busbars, overhead lines an menyiratkan peralatan seperti generator, transformator, busbars, saluran udara and cables and all other items of plant which operate at power system voltage. dan kabel dan semua item lainnya dari pabrik yang beroperasi pada sistem listrik tegangan. Electrical failure generally implies one or the other (or both) of two types of Kegagalan Listrik umumnya menyiratkan satu atau yang lain (atau keduanya) dari dua jenis failure, namely insulation failure resulting in a short-circuit condition or conducting- kegagalan, yaitu kegagalan isolasi menghasilkan kondisi sirkuit pendek atau melakukan- path failure resulting in an open-circuit condition, the former being by far the more jalan yang mengakibatkan kegagalan dalam kondisi sirkuit-terbuka, yang pertama yang jauh lebih common type of failure. umum jenis kegagalan.

The principal types of fault are listed and classified in Table 3.1.2A, and are Jenis utama dari kesalahan yang terdaftar dan diklasifikasikan dalam Tabel 3.1.2A, dan discussed in greater detail below. dibahas secara lebih rinci di bawah ini. Table 3.1.2A Types of fault Tabel 3.1.2A Jenis kesalahan Short-circuited Hubung pendek phases fase Three-phase fault clear of earth Tiga-fase kesalahan yang jelas bumi Three-phase-to-earth fault Tiga fase-ke-bumi kesalahan Phase-to-phase fault Tahap-ke-fase kesalahan Single-phase-to-earth fault Single-fase-ke-bumi kesalahan Two-phase-to-earth fault Dua-fasa-ke-bumi kesalahan Phase-to-phase plus single-phase-to-earth fault Tahap-ke-satu fasa-fasa-ke-bumi ditambah kesalahan Open-circuited Buka-hubung phases fase Single-phase open-circuit Satu-fasa terbuka sirkuit Two-phase open-circuit Dua-fasa terbuka sirkuit Three.phase open-circuit Three.phase sirkuit terbuka Simultaneous Serentak faults kesalahan A combination of two or more faults at the same time, the faults Kombinasi dari dua atau lebih kesalahan pada saat yang sama, kesalahan being of similar or dissimilar type and occurring at the same or keberadaan atau sejenis berbeda dan terjadi pada saat yang sama atau different locations. lokasi yang berbeda. Typical examples are the cross-country earth- Contoh-contoh umum adalah cross-negara bumi fault and the open-circuit-with-earth-fault condition kesalahan dan sirkuit-terbuka-dengan-bumi-kondisi kesalahan Winding faults Berliku kesalahan Winding-to-earth short-circuit Berliku-ke-bumi-sirkuit pendek Winding.to-winding short-circuit Winding.to berliku-sirkuit pendek Short-circuited turns Hubung pendek ternyata Open-circuited winding Buka-hubung berliku

Page 3 Page 3Fault calculation Fault perhitungan 55 55 Short-circuited phases: Hubung singkat-fase: Faults of this type are caused by insulation failure Kesalahan jenis ini disebabkan oleh kegagalan isolasi between phase conductors or between phase conductors and earth, or both, the antara konduktor fasa atau antara konduktor fasa dan bumi, atau keduanya, result being the short-circuiting of one or more phases to earth or to one another, Hasil menjadi hubungan arus-pendek dari satu atau lebih fase ke bumi atau satu sama lain, or both. atau keduanya. The full range of possible fault conditions of this type is illustrated in Kisaran penuh kondisi kesalahan yang mungkin dari tipe ini digambarkan dalam Fig. Gambar. 3.1.2A. 3.1.2A. The three-phase fault, which may be to earth or clear of earth, is the The-fase tiga kesalahan, yang mungkin ke bumi atau bumi yang jelas, adalah

only balanced or symmetrical short-circuit condition, the presence or absence of hanya seimbang atau simetris-kondisi sirkit pendek, ada atau tidak adanya the earth connection being normally of little or no significance unless the fault sambungan bumi adalah normal dari sedikit atau tidak ada arti kecuali kesalahan occurs simultaneously with a second unbalanced fault involving earth. terjadi secara bersamaan dengan kesalahan kedua yang melibatkan bumi tidak seimbang. The three- Ketiga- phase short-circuit is commonly used as a standard fault condition as, for example, fase sirkuit pendek biasanya digunakan sebagai suatu kondisi kesalahan standar, misalnya, in the determination of system fault-levels, these levels being normally quoted as dalam penentuan tingkat-kesalahan sistem, tingkat ini adalah normal dikutip sebagai three-phase short-circuit values. tiga fase-sirkuit pendek nilai. Three-phase clear 1" earth Tiga-tahap yang jelas 1 "bumi "l'hree-phase-tt-earth "L'hree-fase-tt-bumi - - - - - - t t I Aku i aku Phase-to-phase Tahap-ke-fase Single+phase-to-earth Single-+ fase ke-bumi ii ii Tw, o-phase-to-earth Tw, o-fase-ke-bumi Phase-to-phase plus Tahap-ke-fase plus single-phase-to-earth satu-fase-ke-bumi 4 4 . . . . . . . . I Aku - - --- --- i aku Fig. Gambar. 3.1.2A 3.1.2A Shorpcircuited-phase faults Shorpcircuited-fase kesalahan Open-circuited phases: Buka-hubung fase: This type of fault, illustrated in Fig. Jenis kesalahan, diilustrasikan pada Gambar. 3.1.2B, is the failure 3.1.2B, adalah kegagalan of one or more phases to conduct. dari satu atau lebih tahap untuk melakukan. The more common causes of this type of fault Semakin umum penyebab jenis kesalahan are joint failures on overhead lines and cables, and the failure of one or more adalah kegagalan bersama pada saluran udara dan kabel, dan kegagalan dari satu atau lebih phases of a circuit-breaker or isolator to open or close. fase sebuah pemutus-rangkaian atau isolator untuk membuka atau menutup. The single-phase and two- Fase-tunggal dan dua-

Page 4 Page 456 56 Fault calculation Fault perhitungan phase conditions are of particular interest because they both tend to produce un- kondisi fase yang menarik khususnya karena keduanya cenderung menghasilkan un-

balance of the power system currents and voltages with consequent risk of damage keseimbangan sistem arus listrik dan tegangan dengan risiko akibat kerusakan to rotating plant. untuk memutar tanaman. Simultaneous faults: A Simultan kesalahan: A simultaneous fault condition, sometimes termed a kondisi kesalahan simultan, kadang-kadang disebut sebuah multiple fault condition, is defined as the simultaneous presence of two or more beberapa kondisi kesalahan, yang didefinisikan sebagai kehadiran simultan dari dua atau lebih faults of similar or dissimilar types at the same or different points on the power kesalahan atau berbeda jenis yang sama di titik yang sama atau berbeda daya system. sistem. Such conditions may result from a common cause, from different but Kondisi tersebut dapat hasil dari penyebab umum, dari berbeda tetapi consequential causes or, extremely rarely, from quite separate and independent menyebabkan konsekuensial atau, sangat jarang, dari cukup terpisah dan independen causes. penyebab. The commonest simultaneous fault condition is undoubtedly the double- Kesalahan paling umum adalah kondisi simultan diragukan lagi dua kali circuit overhead-line fault in which a common cause (for example lightning or sirkuit overhead-line kesalahan di mana penyebab umum (misalnya petir atau accidental contact) results in a fault on each of the two circuits concerned. kebetulan kontak) menghasilkan kesalahan di masing-masing dari dua sirkuit yang bersangkutan. These Ini Single-phase Fase tunggal ()pen-circuit () Pena-sirkuit ,pcn-circui! , PCN-circui! Three-phast. Tiga-phast. tlt.n-circuil tlt.n-circuil Fig. Gambar. 3.1.2B 3.1.2B Open-circuited-phase faults Buka-hubung-fase kesalahan two faults, although possibly geographically coincident, will be electrically separate dua kesalahan, meskipun mungkin secara geografis bertepatan, akan elektrik terpisah to an extent determined by the point of fault and the particular power system sampai batas yang ditentukan oleh titik kesalahan dan sistem kekuasaan tertentu configuration. konfigurasi. A simultaneous fault condition of particular interest is that Sebuah kondisi kesalahan simultan kepentingan tertentu adalah known as the cross-country earth-fault, in which a single-phase-to-earth fault at dikenal sebagai negara-bumi-kesalahan silang, di mana-fase-ke-bumi kesalahan tunggal pada one point in the power system occurs coincidentally with a second single-phase-to- satu titik dalam sistem kekuasaan terjadi secara kebetulan dengan single kedua fase-ke- earth fault on another phase and at some other point in the system. bumi kesalahan pada tahap yang lain dan pada beberapa titik lain di dalam sistem. This condition Kondisi ini is most commonly experienced on impedance-earthed systems where the second ini paling sering dialami pada sistem dibumikan-impedansi dimana kedua earth-fault may be initiated by the increased healthy-phase voltage resulting from bumi-kesalahan yang dapat diawali oleh fase tegangan yang sehat meningkat akibat

the neutral displacement produced by the first. perpindahan netral yang dihasilkan oleh yang pertama. As already stated, a simultaneous Seperti telah dinyatakan, sebuah simultan fault condition may consist of two different types of fault at the same point, and kondisi kesalahan dapat terdiri dari dua jenis kesalahan pada titik yang sama, dan one example of this is the open-circuit-with-earth-fault condition in which two salah satu contoh dari ini adalah rangkaian-terbuka-dengan tanah-kesalahan-kondisi di mana dua faults, namely a kesalahan, yaitu single-phase fase tunggal open-circuit and a single-phase-to earth fault, occur rangkaian-terbuka dan satu-fase-untuk kesalahan bumi, terjadi coincidentally on the same phase and at the same point in the power system. kebetulan pada tahap yang sama dan pada titik yang sama dalam sistem kekuasaan. Such Seperti itu a condition can occur on an overhead line for example, due to a phase conductor Kondisi ini dapat terjadi pada saluran udara misalnya, karena konduktor fasa breaking at a point near to a tower, the conductor on the tower side of the break melanggar pada titik dekat sebuah menara, konduktor di sisi menara break being held by the suspension insulator and that on the other side falling to ground. ditahan oleh isolator suspensi dan bahwa di sisi lain jatuh ke tanah. The fault conditions described are divisible into two distinct classes, namely kondisi gangguan yang dijelaskan terbagi menjadi dua kelas yang berbeda, yaitu

Page 5 Halaman 5Fault calculation Fault perhitungan 57 57 balanced or symmetrical fault conditions and unbalanced or unsymmetrical fault seimbang atau kondisi gangguan simetris dan tidak seimbang atau simetris kesalahan conditions, the former class comprising all conditions which are symmetrical with kondisi, mantan kelas yang terdiri dari semua kondisi yang simetris dengan respect to the three phases and the latter class the remainder. hormat kepada tiga fase dan kelas kedua sisanya. Of the faults listed. Dari kesalahan yang terdaftar. only the three-phase short circuit (to earth or clear of earth) and the three-phase hanya sirkuit-pendek fase tiga (ke bumi atau menghapus bumi) dan tiga fasa open-circuit are balanced fault conditions, the normal balanced three-phase load sirkuit terbuka kondisi kesalahan seimbang, fase beban seimbang tiga-normal condition being a further example of the balanced or symmetrical condition. kondisi menjadi contoh lebih lanjut dari kondisi seimbang atau simetris. Winding faults: The Berliku kesalahan: The types of fault which can occur on machine and transformer jenis kesalahan yang dapat terjadi pada mesin dan transformator windings are illustrated in Fig. gulungan diilustrasikan pada Gambar. 3.1.2C and consist mainly of short-circuits, from one 3.1.2C dan terutama terdiri dari pendek-sirkuit, dari satu phase winding to earth, from one phase winding to another or from one point to fase berliku-liku ke bumi, dari satu tahap berliku-liku ke lain atau dari satu titik ke titik another on the same phase winding. lain pada tahap yang sama berliku. The last mentioned condition is known as a Kondisi yang disebutkan terakhir dikenal sebagai

short-circuited turns fault, and is of particular interest from the protection stand- hubung singkat-kesalahan berubah, dan kepentingan tertentu dari perlindungan stand- point in that the fault current in the short-circuited turns may be very large and titik dalam kesalahan saat di-hubung singkat dapat berubah sangat besar dan that in the remainder of the winding very small. bahwa dalam sisa gulungan sangat kecil. The open-circuited winding condi- Open-hubung berliku Condi- tion is quite rare in practice and is usually the result of damage to the winding as a SI cukup langka dalam praktek dan biasanya merupakan akibat dari kerusakan gulungan sebagai consequence of a preceding winding short-circuit at or near the point of fault. konsekuensi dari berliku-sirkuit pendek sebelumnya pada atau dekat titik kesalahan. Open Buka circuits in transformers may also occur as a result of failure of the tap-change equip- sirkuit transformator juga dapat terjadi sebagai akibat dari kegagalan dari perubahan-tap peralatan ment. an. Phase-to-earth fault Tahap-ke-bumi kesalahan Phase-to-phase fault Tahap-ke-fase kesalahan __ __ t'Y'V'V'g'v"r t'Y'V'V'g'v "r . . . . . . . . ¢'V"WY% 'V ¢ WY% " _. _. Short-circuited turns Hubung pendek ternyata Open-circuited winding Buka-hubung berliku Fig, 3.1.2C Gambar, 3.1.2C Winding faults Berliku kesalahan Changing-fault conditions: The Mengubah-kesalahan kondisi: The types of fault which have been referred to can all jenis kesalahan yang telah disebut dapat semua be regarded as fixed fault conditions, in that the type of fault remains unchanged dianggap sebagai kondisi kesalahan tetap, dalam jenis kesalahan tetap tidak berubah for the duration of the fault. selama kesalahan. The great majority of fault conditions are of this type Sebagian besar kondisi kesalahan adalah dari jenis ini but there are bthers, known as changing-fault conditions, in which the type of fault tetapi ada bthers, dikenal sebagai salah-kondisi yang berubah, di mana jenis kesalahan changes during the course of the fault. perubahan sepanjang kesalahan. Such changing-fault conditions can result Tersebut berubah-kesalahan kondisi dapat mengakibatkan

Page 6 Page 658 58 Fault calculation Fault perhitungan from a number of causes, the most common being the spreading of a fault arc, or dari beberapa penyebab, yang paling umum adalah penyebaran dari busur kesalahan, atau

of the ionised gases from a fault arc, to other phases and even to other circuits. gas terionisasi dari busur kesalahan, dengan tahapan lain dan bahkan untuk sirkuit lainnya. A A typical example is a single-phase-to-earth fault which develops into a two-phase-to- contoh khas adalah fase-ke-bumi kesalahan tunggal yang berkembang menjadi dua-fase-ke- earth fault and possibly, later, into a three-phase fault. bumi dan kemungkinan kesalahan, kemudian, menjadi tiga fase kesalahan. The analysis of a changing Analisis suatu perubahan fault condition presents no particular difficulty, since the condition can be menyajikan kondisi kesalahan ada kesulitan khusus, karena kondisi dapat considered as a succession of fixed fault conditions, each of which can be analysed dianggap sebagai suksesi kondisi kesalahan tetap, masing-masing yang dapat dianalisis individually. individual. 3.1.3 Factors affecting fault severity 3.1.3 Faktor-faktor yang mempengaruhi tingkat keparahan kesalahan The severity of a power system fault condition may be assessed in terms of the Tingkat keparahan kesalahan kondisi daya sistem dapat dinilai dari segi disturbance produced and the fault damage caused, the magnitude of the fault gangguan diproduksi dan kerusakan yang disebabkan kesalahan, besarnya kesalahan current and its duration being of particular interest, especially in relation to the saat ini dan durasi yang menjadi kepentingan tertentu, terutama dalam kaitannya dengan design and application of power system protection. desain dan penerapan sistem proteksi tenaga. The factors which affect fault Faktor-faktor yang mempengaruhi kesalahan severity must therefore be given due consideration in all aspects of power system keparahan karena itu harus dipertimbangkan dalam semua aspek sistem tenaga analysis in order to ensure results which are truly representative of the conditions analisis untuk memastikan hasil yang benar-benar mewakili kondisi which can occur in practice. yang dapat terjadi dalam praktek. The factors which normally require to be considered Faktor-faktor yang biasanya perlu dipertimbangkan are: adalah: (a) Source conditions: These (A kondisi Sumber): ini relate to the amount and disposition of all connected berhubungan dengan jumlah dan disposisi semua terhubung generation (including all other power sources such as interconnections with generasi (termasuk semua sumber daya lainnya seperti interkoneksi dengan other systems), the two extremes of minimum and maximum connected plant lain sistem), dua ekstrim maksimum dan minimum terhubung tanaman being of particular interest. menjadi kepentingan tertentu. The minimum and maximum plant conditions are Dan kondisi tanaman maksimum minimum normally those corresponding to the conditions of minimum and maximum biasanya yang sesuai dengan kondisi minimum dan maksimum connected load. beban tersambung. (b) Power system configuration: (B) Power konfigurasi sistem: This is determined by the items of plant, namely Hal ini ditentukan oleh item tanaman, yaitu generators, transformers, overhead-line and cable circuits etc., assumed to be Generator, transformator, overhead-line dan kabel dll sirkuit, diasumsikan

in service for the particular condition being investigated and by such other dalam pelayanan untuk kondisi tertentu yang sedang diselidiki dan lain seperti itu factors as have a bearing on the topology of the equivalent system network. faktor seperti memiliki bantalan pada topologi jaringan sistem setara. The system configuration may change during the course of a fault with conse- Konfigurasi sistem dapat berubah selama berlangsungnya kesalahan dengan konsekuensi- quent changes in the magnitude and distribution of the fault current, typical quent perubahan besar dan distribusi kesalahan saat ini, khas causes being the sequential tripping of the circuit.breakers at the two ends of menyebabkan menjadi tersandung berurutan dari circuit.breakers pada kedua ujung a faulted transmission line and the sequential clearance of multiple fault garis transmisi menyalahkan dan clearance berurutan beberapa kesalahan conditions. kondisi. (c) Neutral earthing:Faults (C) dibumi Netral: Kesalahan which involve the flow of earth current (for example, yang melibatkan aliran bumi saat ini (misalnya, a singie-phase or two-phase fault to earth, a single-phase or two-phase open sebuah singie-fase-fase atau kesalahan dua ke bumi, fase-tunggal atau dua-fasa terbuka circuit, etc.) may be influenced considerably by the system neutral-earthing sirkuit, dll) sangat mungkin dipengaruhi oleh sistem pentanahan netral- arrangements, particularly by the number of neutral earthing points and the pengaturan, terutama oleh jumlah titik pentanahan netral dan presence or absence of neutral earthing impedances. ada atau tidak adanya impedansi pentanahan netral. Power systems may be Power sistem mungkin single-point or multiple-point earthed and such earthing may be direct (that tunggal atau beberapa titik-titik dibumikan dan pentanahan tersebut dapat langsung (yang is, solid earthing) or via impedance, typical examples being the direct multiple- adalah, dibumi padat) atau melalui impedansi, contoh-contoh yang khas langsung ganda earthing employed on the British 132 kV, 275 kV and 400 kV systems and dibumi yang dipekerjakan pada sistem Inggris 132 kV, 275 kV dan 400 kV dan the single-point and sometimes multiple-point resistance-earthing commonly titik-tunggal dan kadang-kadang beberapa point-pentanahan umumnya resistensi employed at 66 kV and below. bekerja pada 66 kV dan di bawah ini. Earthing impedance can be used to limit the Impedansi pentanahan dapat digunakan untuk membatasi earth-fauit current to a very low and even negligibly small value, as in the case bumi-fauit saat ini untuk yang sangat rendah dan bahkan diabaikan nilai kecil, seperti dalam kasus of a system earthed through a Petersen coil or a generator earthed through a suatu sistem dibumikan melalui kumparan Petersen atau generator dibumikan melalui

Page 7 Page 7Fault calculation 59 Fault perhitungan 59 voltage transformer. transformator tegangan. (d) Nature and type of fault: (D) Sifat dan jenis kesalahan: From what has already been said, it will be evident Dari apa yang telah dikatakan, maka akan jelas

that the type of fault and its position in the power system may have a con- bahwa jenis kesalahan dan posisinya dalam sistem kekuasaan mungkin memiliki con- siderable effect on the magnitude and distribution of the system fault current, siderable berpengaruh pada besarnya dan distribusi kesalahan sistem saat ini, this being particularly the case in respect of earth-faults as compared with ini khususnya menjadi kasus dalam hal bumi-kesalahan dibandingkan dengan phase faults, open-circuits as compared with short-circuits and faults within fase kesalahan, terbuka dibandingkan dengan sirkuit-sirkuit pendek dan kesalahan dalam machine and transformer windings as compared with similar faults at the mesin dan gulungan transformator dibandingkan dengan kesalahan yang sama di winding phase-terminals. belitan fase-terminal. Similarly, the effects of a given fault condition may Demikian pula, efek dari suatu kondisi kesalahan mungkin diberikan be considerably modified by the simultaneous presence of one or more other akan sangat dimodifikasi oleh kehadiran simultan dari satu atau lebih lainnya fault conditions as, for example, in the combination of a short-circuit and an kondisi sebagai kesalahan, misalnya, dalam kombinasi dari sirkuit-pendek dan open-circuit phase condition. buka-fase kondisi sirkuit. A further factor which may require considera- Faktor selanjutnya yang mungkin memerlukan pertimbangan- tion is the possible effect of fault impedance (for example, fault-arc resistance SI adalah efek yang mungkin timbul dari impedansi kesalahan (misalnya,-busur resistensi kesalahan and the ohmic resistance of any metallic or non-metallic fault path, etc.), this dan ketahanan ohmik dari setiap non-logam kesalahan path atau logam, dll), ini being of particular importance in matters relating to the design and application yang penting dalam hal yang berkaitan dengan desain dan aplikasi of distance protection. perlindungan jarak. The wide range of possible system fault conditions and the many factors Beragam kondisi gangguan sistem mungkin dan banyak faktor which influence them result in a wide range of possible levels of fault severity, yang mempengaruhi mereka menghasilkan berbagai kemungkinan tingkat keparahan kesalahan, ranging from extremely low levels up to the maximum levels possible for the mulai dari tingkat yang sangat rendah sampai tingkat maksimum yang mungkin untuk system being considered. sistem yang dipertimbangkan. It is therefore of value, in referring to fault severity Oleh karena itu nilai, dengan mengacu kesalahan keparahan generally, to be able to refer to a standard fault condition, namely the three- umumnya, untuk dapat merujuk pada suatu kondisi kesalahan standar, yaitu tiga phase short-circuit, and to the level of fault severity produced by this fase-sirkuit pendek, dan tingkat keparahan kesalahan yang dihasilkan oleh particular fault condition, namely the three-phase fault level. kesalahan kondisi tertentu, yaitu fase tingkat kesalahan tiga. This level may Tingkat ini mungkin be expressed in amperes or, as is more usual, in three-phase MVA correspond- diekspresikan dalam satuan ampere atau, seperti yang lebih biasa, dalam tiga-tahap MVA sesuai- ing to the rated system voltage and the symmetrical value of the three-phase ing dengan tegangan sistem nilai dan fase simetris nilai tiga-the fault current. arus gangguan. The three-phase short-circuit can normally be regarded as the Tiga-fasa sirkuit pendek biasanya dapat dianggap sebagai most severe condition from the point of view of fault severity, and it is kondisi paling parah dari sudut pandang keparahan kesalahan, dan itu

accordingly the maximum possible value of the three-phase fault level which sesuai dengan nilai maksimum yang mungkin fasa tiga tingkat kesalahan yang normally determines the required short-circuit rating of the power-system biasanya menentukan peringkat sirkuit pendek yang diperlukan dari sistem-daya switchgear. switchgear. A factor which may also have to be taken into account is the Faktor yang juga mungkin harus dipertimbangkan adalah maximum value of the single-phase-to-earth fault current which, in a solidly- nilai maksimum fasa-ke-bumi arus gangguan-tunggal yang, dalam sebuah solid- earthed system, may exceed the maximum three-phase fault current. sistem dibumikan, dapat melebihi kesalahan tiga fase maksimum saat ini. The three-phase fault levels experienced in this country range up to 35 Sampai dengan 35 tiga fase tingkat kesalahan di negeri ini mengalami berbagai MVA at the lowest distribution voltage of 415 V up to some 35 000 MVA MVA pada tegangan distribusi terendah dari 415 V sampai dengan sekitar 35 000 MVA at the highest supergrid transmission voltage of 400 kV, the maximum fault pada tegangan transmisi supergrid tertinggi 400 kV, kesalahan maksimum current in the latter case being of the order of 50 000 A for a three-phase fault saat ini dalam kasus terakhir menjadi dari urutan 50 000 A untuk sebuah kesalahan fasa-tiga and 60 000A for a single-phase-to-earth fault. dan 60 000A untuk fase-ke-bumi kesalahan tunggal. Fault clearance times range Fault clearance kali rentang from less than a tenth of a second to one second or more depending on the dari kurang dari sepersepuluh dari satu detik detik atau lebih tergantung pada protective arrangements employed, low clearance times being of particular pelindung pengaturan bekerja, kali clearance rendah makhluk tertentu importance at the higher fault levels. penting di tingkat kesalahan yang lebih tinggi. 3.1.4 Methods of fault calculation 3.1.4 Metode kesalahan perhitungan The information normally required from a fault calculation is that which gives the Informasi yang biasanya dibutuhkan dari seorang perhitungan kesalahan adalah bahwa yang memberikan values of the currents and voltages at stated points in the power system when the nilai arus dan tegangan pada titik-titik dinyatakan dalam sistem tenaga ketika latter is subjected to a given fault condition, the fault location and system operating terakhir ini mengalami kondisi kesalahan tertentu, lokasi kesalahan dan sistem operasi

Page 8 Page 860 60 Fault calculation Fault perhitungan conditions being specified. kondisi yang ditentukan. Fault calculation is therefore essentially a matter of Oleh karena itu perhitungan Fault dasarnya masalah network analysis and can be achieved by a number of alternative methods, namely: analisis jaringan dan dapat dicapai dengan beberapa metode alternatif, yaitu: (a) direct solution of the network equations obtained from the mesh-current or (Solusi langsung) dari persamaan yang diperoleh dari jaringan-mesh saat ini atau nodal-voltage methods, tegangan nodal-metode, (b) solution by network reduction and back-substitution and, (B) solusi dengan pengurangan jaringan dan back-substitusi dan, (c) solution by simulation using a fault calculator or network analyser. (C) larutan dengan simulasi menggunakan kalkulator kesalahan atau network analyzer. The choice of method will normally depend on the size and complexity of Pilihan metode biasanya akan tergantung pada ukuran dan kompleksitas

network and on the amount of information required from the analysis, a further jaringan dan pada jumlah informasi yang diperlukan dari analisis, lebih jauh important factor being the availability of suitable computing facilities. menjadi faktor penting ketersediaan fasilitas komputer yang sesuai. Direct Langsung solution of the network equations is now commonly employed using suitable solusi dari persamaan jaringan umum digunakan sekarang menggunakan cocok digital-computer facilities and appropriate computer programs, such use of the digital-fasilitas komputer dan program komputer yang tepat, penggunaan tersebut dari computer making it possible to study a wide range of system and fault conditions komputer sehingga memungkinkan untuk mempelajari berbagai sistem dan kondisi gangguan speedily and economically, particularly in the case of the larger networks. cepat dan ekonomis, terutama dalam hal jaringan yang lebih besar. Solution by network reduction using manual (that is slide rule or desk-calculator Solusi dengan pengurangan jaringan menggunakan manual (yang slide aturan atau meja-kalkulator computation) is widely used for such problems as involve, or can be represented by, komputasi) secara luas digunakan untuk masalah seperti melibatkan, atau dapat diwakili oleh, a network of limited size and complexity, there being a large number of fault calcu- jaringan ukuran dan kompleksitas yang terbatas, ada menjadi sejumlah besar kesalahan calcu- lations which fall into this class. lations yang jatuh ke dalam kelas ini. Its use in more complicated cases, however, is Penggunaannya dalam kasus-kasus yang lebih rumit, bagaimanapun, adalah limited only by the amount of time required to obtain a solution, the cost of such hanya dibatasi oleh jumlah waktu yang dibutuhkan untuk mendapatkan solusi, biaya tersebut time being a not altogether unimportant consideration. waktu menjadi pertimbangan penting sama sekali tidak. Solution by simulation Solusi dengan simulasi using a fault calculator or network analyser has the advantage of simplicity of menggunakan kalkulator kesalahan atau penganalisa jaringan memiliki keuntungan dari kesederhanaan application due to the one-to-one correspondence between the real and the simu- antara aplikasi karena satu-ke-satu korespondensi yang nyata dan Simu- lated system, but although widely used in the past, prior to the advent of computer lated sistem, tetapi walaupun banyak digunakan di masa lalu, sebelum munculnya komputer methods, its use is now generally limited to smaller networks and to situations metode, penggunaannya sekarang pada umumnya terbatas pada jaringan yang lebih kecil dan untuk situasi where computer facilities may not be readily available. dimana fasilitas komputer mungkin tidak tersedia. An essential part of power system analysis and fault calculation is that which Sebuah bagian penting dari sistem kekuasaan dan perhitungan analisis kesalahan adalah yang concerns the determination of the equivalent system network for the system menyangkut penentuan sistem jaringan yang setara untuk sistem operating conditions and fault conditions under consideration. kondisi operasi dan kondisi gangguan dalam pertimbangan. As already seen, the Seperti yang telah dilihat,

fault conditions to be analysed normally fall into one or other of two classes, kondisi kesalahan yang akan dianalisis biasanya jatuh ke dalam satu atau lain dari dua kelas, namely balanced or symmetrical fault conditions (for example the three-phase yaitu seimbang atau kondisi gangguan simetris (misalnya tiga-fase short-circuit) and unbalanced or unsymmetrical fault conditions, the latter class sirkuit pendek) dan tidak seimbang atau kondisi gangguan tidak simetris, kelas terakhir being normally analysed by the symmetrical-component method. yang biasanya dianalisa dengan metode-komponen simetris. As will be seen Seperti akan kita lihat later, both classes of fault are analysed by reducing the power system, with its kemudian, kedua kelas kesalahan dianalisis dengan mengurangi sistem kekuasaan, dengan perusahaan fault condition, to an equivalent single-phase network. kondisi kesalahan, ke jaringan tunggal-fase setara. 3.2 Basic principles of network analysis 3,2 Prinsip dasar analisis jaringan 3.2.1 Fundamental network laws 3.2.1 hukum jaringan Fundamental The great majority of fault calculations are concerned with the behaviour of the Sebagian besar kesalahan perhitungan prihatin dengan perilaku power system under steady-state conditions or conditions which, from the point of daya sistem di bawah kondisi negara stabil atau kondisi yang, dari titik view of analysis, may be regarded as steady-state conditions. melihat analisis, dapat dianggap sebagai kondisi negara stabil. It can also usually be Hal ini juga dapat biasanya menjadi assumed that all the power system currents and voltages vary sinusoidally with time berasumsi bahwa semua aliran sistem tenaga dan tegangan sinusoidal dengan waktu bervariasi at a common constant frequency and can therefore be treated as vector quantities pada frekuensi konstan umum dan karenanya dapat dianggap sebagai besaran vektor and be expressed, together with the power system impedances and admittances, in dan diekspresikan, bersama-sama dengan impedansi sistem kekuasaan dan admittances, di complex-number form. kompleks-nomor formulir.

Page 9 Page 9Fault calculation Fault perhitungan 61 61 The relationship between the currents, voltages and impedances in any linear Hubungan antara arus, tegangan dan impedansi dalam setiap linier network is governed by the three basic network laws, namely Ohm's Law and the jaringan diatur oleh tiga jaringan dasar hukum, yaitu Hukum Ohm dan two laws of Kirchhoff, a formal statement of these laws in terms of vector quantities dua hukum Kirchhoff, pernyataan formal dari hukum dalam hal jumlah vektor being given below. yang diberikan di bawah ini. Ohm's Law: Ohm's Law states that the vector voltage drop V produced by a Ohm Hukum: Hukum Ohm menyatakan's bahwa vektor drop tegangan V yang dihasilkan oleh vector current I flowing through a complex impedance Z is given by the vector vektor arus I mengalir melalui suatu impedansi kompleks Z diberikan oleh vektor equation persamaan V = IZ V = IZ 3.2.1.1 3.2.1.1 An alternative form is Bentuk alternatif

I= VY I = V 3.2.1.2 3.2.1.2 where Y is the reciprocal of Z and is the complex admittance. di mana Y adalah kebalikan dari Z dan pengakuan kompleks. The law is illustrated Hukum digambarkan in Fig. pada Gambar. 3.2.1A from which it will be noted that the sense of the voltage-drop Vis in 3.2.1A dari yang akan mencatat bahwa perasaan-drop tegangan Vis di opposition to that of the current I. oposisi dengan yang saat ini I. Kirchhoff's First Law: Kirchhoff's First Law states that the vector sum of all the Hukum Pertama Kirchhoff: Hukum Pertama Kirchhoff menyatakan bahwa jumlah vektor semua currents entering any junction or node in a network is zero or, stated in equation arus memasukkan persimpangan atau node dalam sebuah jaringan adalah nol atau, dinyatakan dalam persamaan form bentuk Z Ii Z Ii = = 0 0 3.2.1.3 3.2.1.3 i aku where I i is the vector current flowing into the node from branch i, the summation di mana saya i adalah vektor saat ini mengalir ke node dari cabang i, penjumlahan yang extending over all the branches connected to the node. memperpanjang atas semua cabang terhubung ke node. Outflowing currents with Arus mengalir keluar dengan respect to the node are simply treated as negative inflowing currents. terhadap node hanya diperlakukan sebagai arus pemasukan negatif. The law, also known as the Junction Law is illustrated in Fig. Hukum, juga dikenal sebagai Hukum Junction diilustrasikan pada Gambar. 3.2.1B. 3.2.1B. Kirchhoff's Second Law." Kirchhoff's Second Law states that the vector sum of Hukum Kedua Kirchhoff negara.'s Kedua Kirchhoff Hukum "bahwa jumlah vektor all the driving voltages (that is source voltages) acting round any closed path or semua mengemudi tegangan (yang adalah sumber tegangan) putaran bertindak setiap lintasan tertutup atau mesh in a network is equal to the vector sum of the voltage drops in the impedances mesh dalam jaringan adalah sama dengan jumlah vektor dari tegangan turun di impedansi of the component branches of the path. komponen cabang jalan. Thus, in equation form Jadi, dalam bentuk persamaan L Ei = . L EI =. llZl llZl 3.2.1.4 3.2.1.4 i aku i aku Fig. Gambar. 3.2.1A 3.2.1A Ohrn'$ Law Ohrn '$ UU ! ! z z ,. ,. z z I Aku ; ; o o

- - v v v =lZ v = LZ where Ei is the vector driving mana EI adalah vektor mengemudi voltage voltase in branch i, Ii the vector current in the di cabang i, Ii vektor arus di branch and Z i the complex impedance of the branch, the summation extending cabang dan Z i impedansi kompleks cabang, penjumlahan memperluas over all the component branches of the path or mesh. atas semua cabang komponen path atau mesh. The driving voltages and Menyetir tegangan dan currents must all be measured in the same direction round the path. arus semua harus diukur di putaran jalan arah yang sama. Expressed in Disajikan dalam another way, the law simply states that the vector sum of all the voltages (that is cara lain, hukum hanya menyatakan bahwa jumlah vektor dari semua tegangan (yang driving voltages and voltage drops) acting round a closed path or mesh is zero. tegangan dan tegangan) tetes bertindak bulat mengendarai lintasan tertutup atau mesh adalah nol.

Page 10 Page 1062 62 Fault calculation Fault perhitungan The law, illustrated in Fig. Hukum, diilustrasikan pada Gambar. 3.2.1 C, is also known as the Mesh Law. 3.2.1 C, juga dikenal sebagai Hukum Mesh. These three basic network laws applied to the branches, nodes and meshes of Ketiga undang-undang dasar jaringan diterapkan pada cabang, node dan jerat dari any linear network (that is, a network in which the impedances are constant and setiap jaringan linear (yaitu, jaringan di mana impedansi yang konstan dan independent of the currents through them), enable the branch currents to be found independen dari arus melalui mereka), memungkinkan arus cabang dapat ditemukan if the branch driving voltages and branch impedances are known. jika cabang yang menggerakkan tegangan dan impedansi cabang diketahui. Fig. Gambar. 3.2.1B 3.2.1B I h Saya h I d Aku d Kirchhoff'$ First Law Kirchhoff '$ Pertama Hukum \ \ I a + I b + I c + I d : 0 Aku a + b + I aku c + aku d: 0 I." 2 I. "2 j'/ j '/ E l E l E 3 E 3 Z s Z s Z 3 Z 3 E 5 E 5 I Aku i aku ! ! E l + E 2 + E 3 + E 4 + E 5 = IIZ I + 12Z 2 + 13Z 3 + 14Z 4 + 15Z $ E l + E 2 + E 3 + 4 + E E 5 = IIZ I + 13Z 12Z 2 + 3 + 14z 4 + $ 15Z Fig. Gambar.

3.2.1C 3.2.1C Kirchhoff'$ Second Law Kirchhoff '$ Kedua Hukum As a simple example consider the elementary two-machine system shown in Sebagai contoh sederhana mempertimbangkan sistem dua-mesin dasar yang ditunjukkan pada Fig. Gambar. 3.2.1D and comprising a generator and a synchronous motor, the former 3.2.1D dan terdiri dari generator dan motor sinkron, mantan represented by a driving voltage of 118 +j24 V behind an impedance of 1 +]4 [2 diwakili oleh tegangan mengemudi dari 118 + j24 V balik impedansi 1 +] 4 [2 and the latter by a driving voltage of 100 +/0 V behind an impedance of 2 +/'5 f2. dan yang terakhir oleh tegangan mengemudi 100 + / 0 V balik impedansi 2 + / '5 f2. The current flowing from the generator to the motor is 3 - j 1 A, and this satisfies Mengalir arus dari generator untuk motor adalah 3 - j 1 A, dan ini memenuhi Kirchhoff's Second Law because the total voltage drop in the path is Hukum Kedua Kirchhoff karena jatuh tegangan total di jalan adalah

Page 11 Page 11Fault calculation Fault perhitungan 63 63 (3 - / 1 X3 +/'9) v, that is 18 + j 24 V, this being equal to the total driving voltage (3 - / 1 + X3 / '9) v, yaitu 18 + j 24 V, hal ini menjadi sama dengan tegangan total mengemudi acting round the path, namely 118 ÷/24 V minus 100 +/0 V. bertindak putaran jalan, yaitu 118 ÷ / minus 24 V 100 + / 0 V. 118 +j24V 118 + j24V 100 +j0V 100 + j0V 3 -jlA 3-JLA I11 +jI3V I11 + jI3V N N Fig. Gambar. 3.2.1D 3.2.1D Assume now that a short-circuit occurs at the generator terminals, this short- Asumsikan sekarang bahwa arus pendek listrik terjadi pada terminal generator, ini pendek circuit path having an impedance of 1 + ]0 I. The conditions are then as shown in sirkuit jalan memiliki impedansi 1 +] 0 I. Kondisi ini kemudian seperti ditunjukkan pada Fig. Gambar. 3.2.1E where 11 is the current from the generator, 12 the current from the 3.2.1E di mana 11 adalah arus dari generator, 12 arus dari motor (now acting as a generator) and motor (sekarang bertindak sebagai generator) dan 13 13 the current through the impedance of the arus melalui impedansi dari short-circuit path. pendek sirkuit jalan. It is required to determine 11,12 and 13, and hence applying Kirchhoff's First Hal ini diperlukan untuk menentukan 11,12 dan 13, dan karenanya menerapkan Pertama Kirchhoff Law we obtain the equation Hukum kita memperoleh persamaan 11 +I2 - 11 + I2 - 13 = 0 13 = 0 3.2,1.5 3.2,1.5

while application of the Second Law to each of the two meshes in turn gives the sedangkan penerapan UU Kedua untuk masing-masing dari dua jerat pada gilirannya memberikan equations persamaan (1 +j4)11 + (1 +j0)I3 = 118 +j24 (1 + j4) 11 + (1 + j0) I3 = 118 + j24 3.2.1.6 3.2.1.6 (2 +j5)I2 + (1 (2 + j5) I2 + (1 +/0)13 + / 0) 13 = 110 +j0 = 110 + j0 3.2.1.7 3.2.1.7 the first of these equations relating to the generator mesh and the second to the pertama dari persamaan yang berhubungan dengan mesh generator dan yang kedua ke motor mesh. motor mesh. 118 +j24V 118 + j24V I00 +jOV I00 + jOV ! ! 3 3 I + jO I + JO Fig. Gambar. 3.2.1E 3.2.1E

Page 12 Page 1264 64 Fault calculation Fault perhitungan Eliminating 13 from eqns. Menghilangkan 13 dari eqns. 3.2.1.6 and 3.2.1.7 by use ofeqn. 3.2.1.6 dan 3.2.1.7 oleh ofeqn digunakan. 3.2.1.5, we obtain 3.2.1.5, kita memperoleh the two equations kedua persamaan (2 +/4)11 +(I (2 + / 4) 11 + (I +/0)12 + / 0) 12 = 118 +/24 = 118 + / 24 (I (Aku +/0)11 + / 0) 11 +(3 +jSY2 = II0+/0 + (3 + jSY2 = II0 + / 0 3.2.1.8 3.2.1.8 3.2.1.9 3.2.1.9 these being a pair of simultaneous equations in the two unknowns 11 and 12. pasangan ini menjadi persamaan simultan dalam dua diketahui 11 dan 12. Eliminating/2 from these equations so as to solve for 11 we obtain Menghilangkan / 2 dari persamaan tersebut sehingga untuk memecahkan 11 kita memperoleh [(3 +jSX2 +]'4)- (1 +/oX1 +/0)] I, [(3 + jSX2 +] '4) - (1 + / oX1 + / 0)] I, = [(3 +j5XI18 +j24)- (1 +j0)(ll0+j0)] = [(3 + j5XI18 + j24) - (1 + j0) (ll0 + j0)] which reduces to yang mengurangi ke (-15 +/22)I = 134 +/662 (-15 + / 22) I = 134 + / 662 giving pemberian I1 = 17.7-/18.1 A I1 = 17.7-/18.1 A The value of I1 can now be used in eqn. Nilai I1 sekarang dapat digunakan dalam eqn. 3.2.1.9 to obtain 12, thus 3.2.1.9 untuk mendapatkan 12, sehingga (100 +j0)- (1 +/0X17-7 -i18.1) (100 + j0) - (1 + / 0X17-7-i18.1) I2= I2 = 3 +/5 3 + / 5 82.3 +/18.1 82,3 + / 18,1

3 +/5 3 + / 5 giving pemberian la = 9.9-/10.5A la = 9.9-/10.5A Finally, substituting the values Ofll and 12 in eqn. Akhirnya, menggantikan nilai-nilai Ofll dan 12 di eqn. 32..1.5 gives 32 .. 1,5 memberikan Is = 27.6 -/28.6 A Apakah = 27,6 -/28.6 A The voltage at the point of fault, namely the product of I3 and Za, is seen to be Tegangan pada titik kesalahan, yaitu produk I3 dan Za, ini terlihat giving pemberian /:3 = (27.6 -/28.6X1 +/0) /: 3 = (27,6 -/28.6X1 + / 0) v3 = 27.6 -/28.6 V v3 = 27,6 V -/28.6 This simple example has been solved by treating the network branch currents as the Contoh sederhana ini telah diselesaikan dengan memperlakukan arus jaringan cabang sebagai unknowns in the network equations, and while this is obviously the most direct tidak diketahui dalam persamaan jaringan, dan sementara ini jelas yang paling langsung approach, it nevertheless suffers from certain disadvantages if the number of meshes pendekatan, itu tetap menderita kerugian tertentu jika jumlah jerat in the network is at all large. dalam jaringan sama sekali besar. These disadvantages stem from the fact that the Kelemahan ini berasal dari kenyataan bahwa number of unknowns, namely the branch currents, will generally be larger than is jumlah yang tidak diketahui, yaitu arus cabang, umumnya akan lebih besar daripada yang necessary for the solution of the problem and that the equations containing these diperlukan untuk pemecahan masalah dan bahwa persamaan yang mengandung unknowns will not generally be amenable to a systematic method of solution. umumnya tidak diketahui tidak akan dapat menerima untuk suatu metode sistematis solusi. To Untuk avoid these difficulties, network analysis is better carried out using mesh-current menghindari kesulitan-kesulitan ini, analisis jaringan yang lebih baik dilakukan dengan menggunakan mesh-saat ini analysis and nodal-voltage analysis. analisis dan analisis nodal tegangan. These methods are briefly described in the Metode ini secara singkat dijelaskan dalam following two Sections. Bagian berikut dua.

Page 13 Page 13Fault calculation Fault perhitungan 65 65 3.2.2 Mesh-current analysis 3.2.2 Mesh-saat analisis This method of analysis can be understood by considering the simple network Metode analisis ini dapat dipahami dengan mempertimbangkan jaringan sederhana shown in Fig. ditunjukkan pada Gambar. 3.2.2A. 3.2.2A. It will be noted that each mesh of the network is assumed to Ini akan dicatat bahwa setiap mesh jaringan diasumsikan carry a circulating current, and it is these so-called mesh currents which are treated membawa beredar saat ini, dan inilah disebut mesh arus-begitu yang diperlakukan as the unknowns in the problem, the current in any given branch being readily sebagai tidak diketahui dalam masalah ini, arus dalam setiap cabang tertentu yang mudah obtained once the mesh currents associated with that branch have been determined. diperoleh setelah arus mesh terkait dengan cabang telah ditentukan.

Because the mesh currents associated with any given node flow through that node it Karena arus mesh yang terkait dengan aliran node diberikan melalui node yang can easily be seen that they satisfy Kirchhoff's First Law. dengan mudah dapat dilihat bahwa mereka memenuhi Hukum Pertama Kirchhoff. We can therefore proceed Oleh karena itu kita dapat melanjutkan to write down the equations which result from application of the Second Law, menuliskan persamaan yang hasil dari penerapan UU Kedua, there being one such equation for each mesh of the network. ada yang satu persamaan tersebut untuk setiap lubang jaringan. Za f Za f Ea() Ea () ZC ZC Z b Z b ( ) Eb () Eb Fig. Gambar. 3.2.2A 3.2.2A Mesh-current notation Mesh-saat notasi Thus, remembering that the current in any branch is the vector sum (taking due Dengan demikian, mengingat bahwa saat ini di cabang manapun adalah jumlah vektor (mengambil karena account of direction) of the mesh currents flowing in that branch, we obtain the rekening arah) dari arus mesh yang mengalir di cabang itu, kami memperoleh following equations for meshes 1, 2 and 3, respectively: persamaan berikut untuk jerat 1, 2 dan 3, masing-masing: zJ, + za(,q - I2) + zN, - I3) --, zJ, + za (, q - I2) Zn +, - I3) -, Za(I2 - 11) + ZcI2 + Ze(I2 - la) =-Ec Za (I2 - 11) + ZcI2 + Ze (I2 - la) =- Ec Zf(13 - 11 ) + Ze (I3 - 12 ) + Zb ZF (13 - 11) + Ze (I3 - 12) + ZB 13 = Ec - El, 13 = Ec - El, 3.2.2.1 3.2.2.1 Rearranging these Menata ulang ini where dimana equations in a more systematic form, we obtain persamaan dalam bentuk yang lebih sistematis, kita memperoleh Z111 +Z1212 Z1212 Z111 + +Z1313 =El ] + Z1313 = El] Z211 + Z212 + Z2ala = E2 Z211 Z212 + + = E2 Z2ala Za111 + Za]2 + Z3ala = Ea Za111 + Za] 2 Z3ala + = Ea ZI =Z, +Z, +Z r ZI = Z, Z +, + Z r Z22 = Zc + Za + Ze Zc + Z22 = Za + Ze Z33 = Za + Z + ZI Z33 = Za + Z + ZI 3.2.2.2 3.2.2.2 3.2.2.3 3.2.2.3

Page 14 Page 1466 66 Fault calculation Fault perhitungan ZI2 =Z21 ='-Zd / ZI2 = Z21 = '-ZD / Z23 = Z32 =--Ze Z23 = Z32 =-- Ze Z31 = Zl3 = -Z: Z31 = Zl3 =-Z: 3.2.2.4 3.2.2.4

E 2 -- -Ee E 2 --Ee Es --Ec- El, Es - Ec-El, 3.2.2.5 3.2.2.5 It will be noted that Zl i is the sum of the impedances of the branches forming Ini akan dicatat bahwa Zl i adalah jumlah dari impedansi cabang membentuk mesh 1, Z22 the sum of those forming mesh 2, and Z33 the sum of those forming mesh 1, Z22 jumlah dari pembentukan 2 mesh, dan Z33 jumlah dari pembentukan mesh 3. mesh 3. It will also be noted that Z2, equal to Z21, is equal to minus the branch Hal ini juga akan mencatat bahwa Z2, sama dengan Z21, sama dengan minus cabang impedance common to meshes 1 and 2, that Z23, equal to Z32, is equal to minus impedansi umum untuk jerat 1 dan 2, bahwa Z23, sama dengan Z32, sama dengan minus the branch impedance common to meshes 2 and 3, and Z3, equal to Z13, is equal impedansi cabang umum untuk jerat 2 dan 3, dan Z3, sama dengan Z13, sama to minus the branch impedance common to meshes 3 and 1. untuk dikurangi impedansi cabang umum untuk jerat 3 dan 1. Finally it will be noted Akhirnya akan dicatat that El is the sum of the driving voltages in the branches forming mesh 1, E2 the bahwa El adalah jumlah tegangan mengemudi di cabang-cabang membentuk mesh 1, E2 sum of those in the branches forming mesh 2, and E3 the sum of those in the jumlah orang-orang di cabang-cabang membentuk mesh 2, dan E3 jumlah mereka yang berada di branches forming mesh 3. cabang membentuk mesh 3. It should be noted that the branch driving voltages are Perlu dicatat bahwa cabang mengemudi tegangan adalah treated as positive when they act in the same direction as the mesh current in the diperlakukan sebagai positif ketika mereka bertindak dalam arah yang sama dengan mesh lancar pada mesh to which the equation refers, namely, in a clockwise direction, and are treated mesh yang mengacu persamaan, yaitu, searah jarum jam, dan diperlakukan as negative when they act in the opposite, anti-clockwise direction. sebagai negatif ketika mereka bertindak bertentangan anti-arah searah jarum jam,. Eqns. Eqns. 3.2.2.2 are a set of three simultaneous equations in the three unknown 3.2.2.2 adalah seperangkat tiga persamaan simultan di tiga tidak diketahui mesh currents lt,/2 and/3 and it will be evident that the general case of a network arus mesh lt, / 2 dan / 3 dan itu akan menjadi jelas bahwa kasus umum jaringan of n meshes will result in a similar set of n simultaneous equations in terms of the dari jerat n akan menghasilkan serangkaian serupa n persamaan simultan dalam hal n unknown mesh currents. n arus mesh diketahui. Knowing the branch driving-voltages and impedances, Mengetahui mengemudi cabang-tegangan dan impedansi, this set of equations can be virtually written down by inspection from the rules himpunan persamaan ini dapat hampir ditulis oleh inspeksi dari aturan already given. sudah diberikan. The solution of these equations to obtain the mesh currents can be Solusi persamaan ini untuk mendapatkan arus mesh dapat achieved by a number of different methods, each branch current being then dicapai dengan beberapa metode yang berbeda, masing-masing cabang yang kemudian saat ini obtained from its component mesh currents. diperoleh dari komponen arus mesh tersebut.

3.2.3. 3.2.3. Nodal-voltage analysis Tegangan nodal analisis In this method of analysis one of the network nodes is chosen as the reference node Dalam metode analisis salah satu node jaringan dipilih sebagai simpul referensi and the voltages of the remaining nodes, measured with respect to the reference dan tegangan dari node yang tersisa, diukur sehubungan dengan referensi node, are treated as the unknowns in the problem. node, diperlakukan sebagai tidak diketahui dalam masalah. The voltage across any branch of Tegangan di setiap cabang a given mesh is equal to the difference between the node voltages at the two ends of jaring yang diberikan adalah sama dengan perbedaan antara tegangan simpul di kedua ujung the branch in question. cabang yang bersangkutan. It is seen, therefore, that the summation of the component Hal ini terlihat, oleh karena itu, bahwa penjumlahan komponen branch voltages round the mesh must be zero because the node voltages, whose tegangan mesh cabang bulat harus nol karena tegangan node, yang difference constitutes any given branch voltage, will be cancelled out in the summa- Perbedaan merupakan cabang diberikan tegangan apapun, akan dibatalkan dalam summa- tion by the contributions which the same node voltages, but reversed in sign, make SI oleh kontribusi yang tegangan simpul yang sama, tapi dibalik tanda, membuat to the voltages across the adjacent branches of the mesh. ke tegangan di cabang yang bersebelahan mesh. The node voltages thus Tegangan node sehingga satisfy Kirchhoff's Second Law and we can therefore proceed to apply the First memenuhi Hukum Kedua Kirchhoff dan karena itu kita dapat melanjutkan untuk menerapkan Pertama Law to each node of the network, in turn, with the single excepton of the reference Hukum untuk setiap node jaringan, pada gilirannya, dengan excepton referensi tunggal node. node. Thus, for the simple network shown in Fig. Dengan demikian, untuk jaringan sederhana ditunjukkan pada Gambar. 3.2.3A, we obtain the following 3.2.3A, kita mendapatkan berikut equations for nodes 1, 2 and 3, respectively: persamaan untuk node 1, 2 dan 3, masing-masing:

Page 15 Page 15Fault calculation Fault perhitungan 67 67 Fig, 3.2.3A Gambar, 3.2.3A Ya Ya 1 1 Yc Yc 2 2 v 3f3 v 3f3 ) ) I Aku ( ( o o Nodal-voltage notation Notasi tegangan nodal- Yb Yb

E b E b (O+Ea- V,)Y. (O + Ea-V,) Y. +(V2- V,)Yc+(V3- + (V2-V,) + YC (V3- V,)Ya =0 / V,) Ya = 0 / (0 + Eb - V2 ) Yt, + ( V, - V2) Ye + (V3 + Ec - V) Ye = 0 (0 + Eb - V2) Yt, + (V, - V2) + Ye (V3 + Ec - V) Ye = 0 (0- V3)Yf+(V 1 - V3)Ya (0 - V3) YF + (V 1 - V3) Ya +(V2- + (V2- Ec- V3)Ye Ec-V3) Ye =0 = 0 3.2.3.1 3.2.3.1 It will be noted that these equations are written in terms of the branch admittances Ini akan dicatat bahwa persamaan ini ditulis dalam hal admittances cabang as compared with the branch impedances used in the mesh-current method of dibandingkan dengan impedansi cabang yang digunakan dalam metode-arus mesh analysis. analisis. Rearranging the equations in a more systematic form, we obtain Menata ulang persamaan dalam bentuk yang lebih sistematis, kita memperoleh where dimana YllVt + YIiV2 + YI3Va =11 YllVt + YIiV2 + YI3Va = 11 Y21 gl + Y22 V2 Gl + Y22 Y21 V2 + Y23 V3 + Y23 V3 =12 = 12 Y31 VI + Y32V2 + YaaVa =Ia Y31 VI + Y32V2 YaaVa + = IA YIl = Ya + Ye + Yd ] YIl = Ya + + Ye yd] Y22 = Yb + Ye + }re Y22 = Yb + Ye +) re Y33 = Yd + Ye + Yf Y33 = yd + Ye + YF 3.2.3.2 3.2.3.2 3.2.3.3 3.2.3.3 YI2 = Y21 = - YI2 = Y21 = - Y¢ Y ¢ | | Y23 = Y32 = - Y23 = Y32 = - Ye Kamu Y31 = Y13 =- Y31 = Y13 =- Yd Yd 3.2.3.4 3.2.3.4 I1 = Ea Ya I1 = Ea Ya ] ] I2 = Eb Yb + Ee ]re I2 = Eb Yb + Ee] kembali I Aku 3.2.3.5 3.2.3.5 13 = -Ec }re 13 =-Ec) kembali It will be noted that Yt i is the sum of the admittances of the branches connected Ini akan dicatat bahwa Yt i adalah jumlah dari admittances cabang terhubung to node 1, Y22 the sum of those connected to node 2, and Y33 the sum of those untuk node 1, Y22 jumlah dari node terhubung ke 2, dan Y33 jumlah dari connected to node 3. terhubung ke node 3. It will also be noted that Y12, equal to Y2, is equal to minus Hal ini juga akan mencatat bahwa Y12, sama dengan Y2, sama dengan minus

Page 16 Page 1668 68 Fault calculation Fault perhitungan the admittance of the branch connecting nodes 1 and 2, that Y2a, equal to Yaz, is pengakuan dari cabang yang menghubungkan node 1 dan 2, bahwa Y2a, sama dengan Yaz, adalah equal to minus the admittance of the branch connecting nodes 2 and 3, and that sama dengan minus pengakuan dari cabang yang menghubungkan node 2 dan 3, dan bahwa Fax, equal to Yla, is equal to minus the admittance of the branch connecting Fax, sama dengan Yla, adalah sama dengan minus pengakuan dari cabang yang menghubungkan nodes 3 and 1. node 3 dan 1. Finally it will be noted that I is the sum of the products of driving Akhirnya akan dicatat bahwa saya adalah jumlah dari produk mengemudi voltage and admittance for each of the branches connected to node 1,!z the sum tegangan dan penerimaan untuk masing-masing cabang terhubung ke node 1, jumlah! z of these products for each of the branches connected to node 2, and la the sum of produk ini untuk masing-masing cabang terhubung ke node 2, dan la jumlah these products for each of the branches connected to node 3, the branch driving produk ini untuk masing-masing cabang yang terhubung ke node 3, cabang mengemudi voltages being treated as positive when they act towards the node in question and tegangan diperlakukan sebagai positif ketika mereka bertindak terhadap node tersebut dan negative when they act in the opposite direction, namely away from the node. negatif bila mereka bertindak dalam arah yang berlawanan, yaitu jauh dari node. Eqns. Eqns. 3.2.3.2 are a set of three simultaneous equations in the three unknown 3.2.3.2 adalah seperangkat tiga persamaan simultan di tiga tidak diketahui node voltages VI, V2 and Va and it will be evident that the general case of a tegangan node VI, V2 dan Va dan akan menjadi jelas bahwa kasus yang umum dari network of n nodes, excluding the reference node, will result in a similar set of n n node jaringan, termasuk node acuan, akan menghasilkan satu set serupa n simultaneous equations in terms of the n unknown node voltages. persamaan simultan dalam hal tegangan node n tidak diketahui. Knowing the Mengetahui branch driving voltages and admittances, this set of equations can be virtually tegangan cabang mengemudi dan admittances, himpunan persamaan ini dapat hampir written down by inspection from the rules already given. ditulis oleh inspeksi dari aturan yang sudah diberikan. The set of simultaneous linear equations obtained is identical in general form Himpunan persamaan linier simultan diperoleh identik dalam bentuk umum with that obtained using the mesh-current method of analysis and is amenable to dengan yang diperoleh menggunakan metode-arus mesh analisis dan setuju untuk the same methods of solution. sama metode solusi. Having solved the equations to obtain the node Setelah memecahkan persamaan untuk mendapatkan node voltages, each branch current is then readily obtained from the node voltages at its tegangan, setiap cabang saat ini maka dengan mudah diperoleh dari tegangan node pada level daya two ends, the branch current being given by the product of the branch admittance dua ujung, cabang saat ini yang diberikan oleh produk pengakuan cabang

and the voltage across this admittance. dan tegangan di seluruh penerimaan ini. In obtaining this latter mentioned voltage Dalam memperoleh tegangan yang disebutkan terakhir ini from the node voltages, due account must, of course, be taken of the driving dari tegangan node, account karena harus, tentu saja, diambil dari mendorong voltage, if any, in the branch in question. tegangan, jika ada, di cabang yang bersangkutan. 3.2.4 Application of mesh-current and nodal-voltage analysis 3.2.4 Aplikasi-arus dan tegangan nodal analisis mesh The mesh-current and nodal-voltage methods of analysis have both been shown to The-saat ini dan tegangan nodal-metode analisis mesh memiliki keduanya telah terbukti result in a set of simultaneous linear equations, the solution of which leads to a menghasilkan satu set persamaan linear simultan, solusi yang mengarah ke complete determination of all the network branch currents. penentuan lengkap semua cabang arus jaringan. As already stated, the Seperti telah disebutkan di atas rules for the formulation of the equations are such that, with either of the two aturan untuk perumusan persamaan seperti itu, dengan salah satu dari dua methods, the resulting set of equations can be virtuaUy written down by inspection metode, yang dihasilkan set virtuaUy persamaan dapat ditulis oleh inspeksi once the network meshes or nodes have been numbered. sekali jerat jaringan atau node telah bernomor. The solution of a set of simultaneous linear equations is a standard computa- Solusi dari persamaan linier simultan adalah standar computa- tional procedure and may be achieved by a number of methods including elimina- prosedur nasional dan dapat dicapai dengan beberapa metode termasuk elimina- tion, determinant and matrix methods and iterative methods. SI, penentu dan metode matriks dan metode iteratif. Details of these Rincian ini methods will be found in a number of the references given in the bibliography. metode akan ditemukan dalam beberapa referensi diberikan dalam bibliografi. With Dengan all methods of solution, however, the use of manual computation, even with the semua metode larutan, Namun, penggunaan perhitungan manual, bahkan dengan aid of a desk computer, is not normally a practical proposition if the number of bantuan komputer meja, biasanya tidak proposisi praktis jika jumlah equations is at all large, since the computation time tends to be considerable and persamaan sama sekali besar, sejak waktu komputasi cenderung cukup besar dan increases rapidly with the number of equations. meningkat pesat dengan jumlah persamaan. Solution by high-speed digital Solusi oleh digital berkecepatan tinggi computer is however, very much a practical proposition, such computers being komputer Namun, sangat banyak proposisi praktis, komputer tersebut menjadi capable of handling very large sets of equations (that is, simultaneous equations mampu menangani set sangat besar persamaan (yaitu, persamaan simultan representing a large number of meshes or nodes) and of providing a rapid and mewakili sejumlah besar jerat atau node) dan penyediaan yang cepat dan accurate solution. solusi akurat.

Page 17 Page 17Fault calculation Fault perhitungan 69 69

The choice between the mesh-current and nodal-voltage methods will normally Pilihan antara arus dan tegangan nodal-metode mesh biasanya akan be determined by the need to reduce computing time to a minimum. ditentukan oleh kebutuhan untuk mengurangi waktu komputasi untuk minimum. It is of Ini adalah interest to note, therefore, that the great majority of networks have fewer nodes menarik untuk dicatat, karena itu, bahwa sebagian besar jaringan telah node lebih sedikit than meshes and are therefore more suited to solution by the nodal-voltage method. dari jerat dan karenanya lebih cocok untuk solusi yang dilakukan oleh tegangan-metode nodal. This is readily demonstrated by the simple two-machine network of Fig. Hal ini mudah dibuktikan dengan jaringan dua-mesin yang sederhana pada Gambar. 3.2.1E 3.2.1E which, it will be noted, has two meshes but only one node, apart from the reference yang akan dicatat, memiliki dua jerat tetapi hanya satu node, selain dari referensi node. node. Thus, applying the mesh-current method we obtain the two equations Dengan demikian, menerapkan metode-saat mesh kita memperoleh dua persamaan (2 +j4)/, - (1 +/0)/2 = 118 +/24 (2 + j4) /, - (1 + / 0) / 2 = 118 + / 24 / / =(I +]O)/l +(3 +/5)/2 = (I +] O) / l + (3 + / 5) / 2 =- =- I00-]0 I00-] 0 I Aku 3.2.4.1 3.2.4.1 where I and/2 are the left-hand and right-hand mesh currents, respectively and are di mana saya dan / 2 adalah tangan kiri dan kanan arus mesh, masing-masing dan assumed to act in a clockwise direction. diasumsikan untuk bertindak dalam arah searah jarum jam. Applying the nodal-voltage method, on the other hand, gives the single equation Menerapkan metode-tegangan nodal, di sisi lain, memberikan persamaan tunggal [ [ 1 1 1 1 l ] l] +j24 100+/0 l + J24 100 + / 0 l , , l+j4 1+/0 2+/5 l + j4 1 + / 0 2 + / 5 +/4 + / 4 2 2 where Vx is the voltage at the point of short-circuit, and is denoted by V3 in mana VX adalah tegangan pada titik-sirkuit pendek, dan dilambangkan oleh V3 di Section 3.2.1. Bagian 3.2.1. It will be left to the reader to verify that Equation 3.2.4.1 and Equation 3.2.4.2 Ini akan diserahkan kepada pembaca untuk memverifikasi bahwa Persamaan dan Persamaan 3.2.4.1 3.2.4.2 both give the same results, these having already been obtained in Section 3.2.1. baik memberikan hasil yang sama, karena ini sudah diperoleh dalam Bagian 3.2.1. 3.2.5. 3.2.5. Network theorems and reduction formulas Jaringan teorema dan formula pengurangan

The mesh-current and nodal-voltage methods of analysis are both a means of ob- The-saat ini dan tegangan nodal-metode analisis mesh keduanya alat ob- taining the complete analytical solution to any linear network problem. taining solusi analisis lengkap untuk setiap masalah jaringan linier. They are Mereka being used to an increasing extent, their use being limited only by the availability digunakan sampai batas meningkat, penggunaannya hanya dibatasi oleh ketersediaan of suitable computing facilities and the fact that an analytical solution can also be fasilitas komputer yang sesuai dan fakta bahwa solusi analisis juga dapat achieved by other means, which in many cases may be simpler and more direct. dicapai dengan cara lain, yang dalam banyak kasus mungkin lebih sederhana dan lebih langsung. This alternative means of analysis, known as the network-reduction method, Alternatif ini berarti analisis, yang dikenal sebagai metode pengurangan jaringan, depends on the reduction of the network to a simpler equivalent form, the method tergantung pada pengurangan jaringan ke bentuk yang ekuivalen sederhana, metode being particularly useful where the solution is required to give the values of only a yang sangat berguna di mana larutan yang diperlukan untuk memberikan nilai-nilai hanya limited number of branch currents. terbatas jumlah arus cabang. Network reduction is the process of combining network branches so as to reduce Jaringan reduksi adalah proses penggabungan jaringan cabang sehingga mengurangi the given network to an equivalent network with fewer branches, this equivalent jaringan yang diberikan ke jaringan setara dengan cabang-cabang yang lebih sedikit, setara ini network being amenable to direct and simple solution. jaringan yang setuju dengan solusi yang langsung dan sederhana. Thus, a complete and Jadi, yang lengkap dan complex network, as viewed from any given pair of nodes, may be reduced in this jaringan yang kompleks, seperti dilihat dari setiap pasangan node tertentu, dapat dikurangi dalam way to its simplest equivalent form, namely a single equivalent branch. cara untuk membentuk setara sederhana, yaitu setara cabang tunggal. The network Jaringan theorems and reduction formulae most commonly required are described below. teorema dan formula pengurangan diperlukan paling sering dijelaskan di bawah ini. Combination of series branches: Kombinasi seri cabang: Taking first the general case of any number of Mengambil pertama kasus umum sejumlah branches connected in series, any given branch i comprising a driving voltage E t and cabang yang terhubung secara seri, setiap cabang yang diberikan terdiri dari i t tegangan E mengemudi dan series impedance Zl, the equivalent single branch comprises a driving voltage impedansi seri Zl, cabang tunggal setara terdiri dari tegangan mengemudi E r in E r di

Page 18 Page 1870 70 Fault calculation Fault perhitungan series with an impedance Zr where seri dengan suatu impedansi Zr mana Er= . Er =. EI= Ex + E2 + ... EI = Ex + E2 + ... + En + En i aku and dan

Zr = . Zr =. Zt =ZI +Z2 + ... Zt = ZI + Z2 + ... +Zn + Zn i aku 3.2.5.1 3.2.5.1 3.2.5.2 3.2.5.2 and n is the number of branches. dan n adalah jumlah cabang. The driving voltages are all measured in the same Tegangan mengemudi semua diukur dalam konteks yang sama direction with respect to the end nodes of the series combination. arah sehubungan dengan node akhir seri kombinasi. These rules, Aturan-aturan ini, applied to the combination of three series-connected branches, are illustrated in diterapkan pada kombinasi dari tiga cabang yang terhubung seri, yang diilustrasikan dalam Fig. Gambar. 3.2.5A. 3.2.5A. Z Z Z Z Z 3 Z 3 p p o o q q _ _ p p Zr")---oq Zr ")--- OQ Fig. Gambar. 3.2.5A 3.2.5A Combination of series branches Kombinasi seri cabang Combination of parallel branches: Taking, again, the general case of any number Kombinasi cabang paralel: Mengambil, sekali lagi, kasus umum nomor apapun of branches but now connected in parallel, any given branch i comprising a driving cabang tapi sekarang terhubung secara paralel, setiap cabang yang diberikan terdiri mengendarai i voltage E i and series admittance Yi (that is series impedance l[Yi), the equivalent tegangan seri E i dan penerimaan Yi (yaitu impedansi seri l [Yi), setara single branch comprises a driving voltage Er in series with an admittance Yr (that is, cabang tunggal terdiri dari mengemudi tegangan Er secara seri dengan sebuah yr pengakuan (yaitu, an impedance 1/Yr), where impedansi 1/Yr), dimana Er=(l/Yr) EiYt=(I[Yr)(EIY +EY2 + ... +EnYn) Eh = (l / yr) EiYt = (I [yr) (EIY + EY2 + ... + EnYn) 3.2.5.3 3.2.5.3 i aku and dan Yr = Z Yi = Y1 + Y2 +... Yr = Z Yi = Y1 + Y2 + ... + Yn + Yn 3.2.5.4 3.2.5.4 i aku and n is the number of branches. dan n adalah jumlah cabang. The driving voltages are all measured in the same Tegangan mengemudi semua diukur dalam konteks yang sama direction with respect to the common nodes of the parallel combination. arah sehubungan dengan node umum dari kombinasi paralel. These Ini

rules, applied to the combination of three parallel-connected branches, are illustrated aturan, diterapkan pada kombinasi dari tiga cabang yang terhubung paralel, diilustrasikan in Fig. pada Gambar. 3.2.5B. 3.2.5B. Star-to-delta transformation: The star-to-delta transformation permits any set of Delta-untuk-transformasi Star: The-untuk-transformasi delta bintang izin setiap himpunan three star-connected branches with isolated star-point to be replaced by an equiva- tiga cabang terhubung dengan bintang-bintang titik terisolasi untuk diganti oleh equiva- lent set of three delta-connected branches. meminjamkan set tiga cabang terhubung delta. Thus, let a, b and c denote the three Jadi, mari a, b dan c menunjukkan tiga terminals of the star and its equivalent delta, and let the star-connected branches terminal bintang dan delta yang setara, dan membiarkan cabang-terhubung bintang comprise driving voltages Ea, E and E c and impedances Z a, Z b and Z c respect- terdiri mengemudi tegangan Ea, E dan c E dan impedansi Z a, b, Z dan Z c-menghormati ively, the voltages being measured in the direction away from the star point. ively, tegangan yang diukur dalam arah yang jauh dari titik bintang. Then Kemudian denoting the driving voltages and impedances of the equivalent delta-connected yang menunjukkan tegangan mengemudi dan impedansi dari setara delta-terhubung

Page 19 Page 19Fault calculation Fault perhitungan 71 71 P o P o y3l_. y3l_. <) q <) Q Fig. Gambar. 3.2.5B 3.2.5B Combination of parallel branches Kombinasi cabang paralel Yr Yr "---p "--- P I Aku I,oq Aku, OQ branches by cabang dengan Eab , Et, c Eab, Et, c and Eta and dan Eta dan Zab, Zbc Zab, Zbc and Zca respectively, the latter values dan Zca masing, yang kedua nilai-nilai are given in terms of the former by the equations diberikan dalam hal yang pertama oleh persamaan Eab = E a Eab = E a - - E b E b + IZab + IZab Ebc = .E b - E c + .[Zbc EBC = b. E - E c +. [Zbc Eca= E e - E a + 1Zea ECA = E e - E a + 1Zea 3.2.5.5 3.2.5.5 ( ( co / co /

Z a Z a Fig. Gambar. 3.2.5C 3.2.5C Star and delta circuits Rangkaian bintang dan delta and dan ZaZb + ZoZc + ZcZa ZaZb + ZoZc + ZcZa Zat - Zat - Zc Zc ZaZ b -I- ZbZ c ..I- ZcZ a ZaZ b-aku-ZbZ c .. I-ZcZ sebuah Zbc - Zbc - Za Za ZaZb + ZbZc + ZcZa ZaZb + ZbZc + ZcZa Z ea -- Z ea - Zb ZB a sebuah 3.2.5.6 3.2.5.6

Page 20 Page 2072 72 Fault calculation Fault perhitungan where I, equal to di mana saya, sama dengan (Eab + Ebc + Eca)/(Zab + Zbc + (Eab + EBC + ECA) / (Zab + + Zbc Zca), is any arbitrarily chosen Zca), adalah setiap sewenang-wenang dipilih current assumed to be circulating in an anticlockwise direction through the saat ini diasumsikan beredar dalam arah berlawanan arah jarum jam melalui branches of the delta. cabang delta. It is normally convenient to assume this current to be zero. Hal ini biasanya mudah untuk menganggap ini saat ini menjadi nol. The driving voltages Menyetir tegangan Eab, Ebe Eab, EBE and Eta act from b to a, c to b and a to c, respect- dan Eta bertindak dari b ke c ke b dan c untuk, menghormati- ively, in the branches concerned. ively, di cabang-cabang yang bersangkutan. These rules are illustrated in Fig. Peraturan-peraturan ini diilustrasikan pada Gambar. 3.2.5C. 3.2.5C. Delta-to-star transformation: The Delta-untuk-bintang transformasi: The delta-to-star transformation permits any set of Bintang-untuk-transformasi delta izin setiap himpunan three delta-connected branches to be replaced by an equivalent set of three star- tiga cabang terhubung delta untuk diganti oleh setara set tiga bintang- connected branches with isolated star-point, the relationships between the star and cabang terhubung dengan titik bintang-terisolasi, hubungan antara bintang dan delta quantities, using the nomenclature of the previous section, being given by the delta jumlah, menggunakan nomenklatur bagian sebelumnya, yang diberikan oleh equations persamaan and dan E a - E b = Eab - [Zab E a - E = [- b Eab Zab E# - E c = Ebc - 1Zoc E # - E c = EBC - 1Zoc E c -E a = Eta - 1Zca E c-E a = Eta - 1Zca ZatZca ZatZca Z a = Z a = Zab + Zbc + Zca Zab + Zbc + Zca

Z bcab Z bcab Z# = Z # = Zab + Zb + Z Zab + ZB + Z Z caZ bc Z Caz bc Zc = Zc = Zab + Zbc + Z Zab + + Zbc Z 3.2.5.7 3.2.5.7 3.2.5.8 3.2.5.8 where I, equal to di mana saya, sama dengan (Eab + Ebc + Eca)/(Zab + Zbc + Zca), (Eab + EBC + ECA) / (Zab + Zbc + Zca), is the current circulating in adalah beredar saat ini di an anticlockwise direction through the branches of the delta. arah berlawanan arah jarum jam melalui cabang delta. It should be noted that any one of the three driving voltages Perlu dicatat bahwa salah satu dari tiga tegangan mengemudi E a, E b E a, b E or atau E c E c can be dapat chosen quite arbitrarily, this value being then used to determine the remaining two dipilih cukup sewenang-wenang, nilai ini yang kemudian digunakan untuk menentukan dua sisa driving voltages. tegangan mengemudi. It is normally convenient to arrange that the sum of the three Hal ini biasanya mudah untuk mengatur bahwa jumlah tiga driving voltages is equal to zero, in which case mengemudi tegangan sama dengan nol, dalam hal Eo -- '/3 [Eb - E, - I(Z,b - Eo - '/ 3 [Eb - E, - I (Z, b - Z)] Z)] Eb = t/3[Ebc - Eab - I(Zbc - Eb = t / 3 [EBC - Eab - I (Zbc - Z,,t,)l Z,, t,) l E, = '/ [E - Eb, - S(Z, - Zbc)] E, = '/ [- Eb E, - S (Z, - Zbc)] 3.2.5.9 3.2.5.9 where I, as already stated, is given by di mana saya, sebagaimana telah dinyatakan, diberikan oleh a +Eb+E, a + Eb + E, I= I = Zab + Zbc + Zca Zab + Zbc + Zca 3.2.5.10 3.2.5.10 For the special case in which Untuk kasus khusus di mana Eab + Ebc Eab + EBC + Eta = 0, giving I = 0, the star driving + Eta = 0, memberikan saya = 0, bintang mengemudi voltages then become tegangan kemudian menjadi

Page 21 Page 21Fault calculation Fault perhitungan 73 73 Ea = ]/3(Eat- Eca) Ea =] / 3 (Makan-ECA) Eb = t/3 (Eric- Eb = t / 3 (Eric- Eat,) Makan,)

Ec = R/3 (Eca - Ebe) Ec = R / 3 (ECA - EBE) 3.2.5.11 3.2.5.11 3 3 I Aku * iOZ * IOZ --K i3V / ¢ 6Z - I3V K / ¢ 6Z 2- lOW 2 - RENDAH Fig. Gambar. 3.2.5D 3.2.5D Example of equivalent star and delta circuits Contoh bintang rangkaian delta setara dan The transformation rules are illustrated by the circuit shown in Fig. Aturan-aturan transformasi diilustrasikan oleh rangkaian ditunjukkan pada Gambar. 3.2.5D. 3.2.5D. Combination of equal driving-voltages: If two or more branches have a common Kombinasi tegangan mengemudi-sama: Jika dua atau lebih cabang memiliki kesamaan node and identical driving-voltages with respect to this node, then the individual node dan identik mengemudi-tegangan sehubungan dengan node ini, maka individu branch driving-voltages can all be removed and replaced by a single external driving cabang mengemudi-tegangan semua dapat dihapus dan diganti dengan mengemudi eksternal tunggal l'J q J l'q "'% r "'% R Fig. Gambar. 3.2.gE 3.2.gE q q /a / A P P f f Combination of equal driving voltages Kombinasi mengemudi sama tegangan voltage connected to the common node, the latter driving voltage being equal to the tegangan terhubung dengan node umum, mengemudi tegangan terakhir menjadi sama dengan common value of the original branch driving-voltages. umum nilai mengemudi cabang asli-tegangan. The theorem is illustrated in teorema ini digambarkan dalam Fig. Gambar. 3.2.5E and follows readily from the fact that equipotential points can be 3.2.5E dan mengikuti mudah dari kenyataan bahwa poin ekuipotensial dapat joined together without any resultant change in the electrical state of the network. bergabung bersama-sama tanpa ada perubahan yang dihasilkan di negara jaringan listrik. The theorem is particularly useful in power system analysis in the reduction of a Teorema ini sangat berguna dalam analisis sistem kekuasaan di suatu pengurangan multiple-source network to its single-source equivalent. beberapa sumber-jaringan tunggal setara sumbernya. Superposition Theorem:The Superposition Theorem, applicable to any linear Teorema superposisi: The Teorema Superposisi, berlaku untuk setiap linier network, states that the current which flows in any branch of a network as a result jaringan, menyatakan bahwa arus yang mengalir dalam cabang jaringan sebagai hasilnya of the simultaneous action of several driving voltages is equal to the vector sum of dari tindakan simultan dari beberapa mengemudi tegangan adalah sama dengan jumlah vektor

Page 22 Page 2274 74 Fault calculation Fault perhitungan 118 , i24V 118, i24V © © I " j4, Aku "j4, 15.6- j2 1.2A 15,6-j2 1.2A I' Aku ' 2 + j5Z 2 + j5Z A A . . ] ] N N 1.7 - j4.2A 1,7 - j4.2A 17.3 17,3 jl7.0A jl7.0A I + jO I + JO v v I " j4,:' Aku "j4,: ' --2.1 - j3.1A - 2.1 - j3.1A I' Aku ' 2 + jSl2 2 + jSl2 -- - "- "- ) ) 8.2 - j14.7A 8,2 - j14.7A 10.3 -.il 1.6A -. Il 10,3 1.6A I + ,iOSZ I +, iOSZ I00 + jOV I00 + jOV @ @ 118 i24V 118 i24V I00 +jOV I00 + jOV 17.7 17,7 ilt.lA ilt.lA 9.9 -- jlO.SA 9,9 - jlO.SA "-- 27.6 "- 27,6 j28.6A j28.6A I + jO I + JO N N Fig. Gambar. 3.2.5F 3.2.5F Illustration of the superposition theorem Ilustrasi dari teorema superposisi the currents which would flow in the branch in question with each driving voltage arus yang akan mengalir di cabang yang bersangkutan dengan masing-masing mengemudi tegangan acting individually and all the remaining driving voltages equal to zero, that is, bertindak secara individu dan semua sisa mengemudi tegangan sama dengan nol, yaitu,

short-circuited. hubung pendek. The theorem is illustrated in Fig. teorema ini diilustrasikan pada Gambar. 3.2.5F using the simple two- 3.2.5F sederhana dengan menggunakan dua machine problem of Section 3.2.1. masalah mesin Bagian 3.2.1. Thevenin's Theorem: Thevenin's Theorem, similarly applicable to any linear Teorema Thevenin: Teorema Thevenin, juga berlaku untuk setiap linier network, states that any such network containing driving voltages, as viewed from jaringan, menyatakan bahwa setiap jaringan seperti mengandung tegangan mengemudi, dilihat dari any two terminals, can be replaced by a single driving voltage acting in series with a setiap dua terminal, dapat digantikan oleh tegangan mengemudi tunggal bertindak secara seri dengan single impedance. tunggal impedansi. The value of this driving voltage is equal to the open-circuit Nilai tegangan ini mengemudi adalah sama dengan sirkuit-terbuka voltage between the two terminals, and the series impedance is the impedance of tegangan antara dua terminal, dan impedansi seri adalah impedansi the network as viewed from the two terminals with all the driving voltages equal to jaringan ditinjau dari dua terminal dengan semua tegangan mengemudi sebesar zero, that is short-circuited. nol, yaitu hubung pendek. The theorem is illustrated in Fig. teorema ini diilustrasikan pada Gambar. 3.2.5G and its appli- 3.2.5G dan yang Appli- cation to the simple two-machine circuit of Section 3.2.1 is shown in Fig. kation dengan rangkaian dua-mesin sederhana dari Bagian 3.2.1 ditunjukkan pada Gambar. 3.2.5H. 3.2.5H.

Page 23 Page 23Fault calculation Fault perhitungan 75 75 Active network Aktif jaringan ,,0 ,, 0 q q TVoc TVoc Same network with Sama jaringan dengan all semua source voltages sumber tegangan removed and replaced dihapus dan diganti by short circuits oleh arus pendek P P o o 0 0 q q n°- n ° - t t i aku P P I Aku ! ! O O V°c V ° C i aku

0 0 I Aku .j . J q q Fig. Gambar. 3.2.5G 3.2.5G Thevenin's Theorem Teorema Thevenin II1 II1 +jl3V + Jl3V I +j412 I + j412 2 2 + jSs2 + JSs2 ! ! J J I Aku I Aku j j P P J J 0 0 ! ! ; ; ! ! I Aku ' ' I + j012 I + j012 I Aku i aku i aku o o i aku l l N N Fig. Gambar. 3.2.5H 3.2.5H Thevenin's Theorem applied to the example of Fig. Teorema Thevenin diterapkan pada contoh Gambar. 3.2.1E 3.2.1E Norton's Theorem: Norton's Theorem is the dual of Thevenin's Theorem and Teorema Norton: Norton Teorema adalah ganda Teorema Thevenin dan states that any linear network containing driving voltages, as viewed from any two menyatakan bahwa setiap jaringan linear yang mengandung tegangan mengemudi, seperti yang dilihat dari setiap dua terminals, can be replaced by a single driving current shunted by an impedance. terminal, dapat diganti dengan mengemudi tunggal saat ini didorong oleh sebuah impedansi. The Itu value of this driving current is equal to the short-circuit current which will flow nilai ini saat mengemudi adalah sama dengan arus-pendek yang akan mengalir between the two terminals when connected together, and the shunt impedance is antara dua terminal ketika terhubung bersama-sama, dan impedansi shunt adalah the impedance of the network as viewed from the two terminals with all the driving impedansi jaringan dilihat dari dua terminal dengan semua mengemudi

voltages equal to zero, that is short-circuited. tegangan sama dengan nol, yang pendek-hubung. The theorem is illustrated in Fig. teorema ini diilustrasikan pada Gambar.

Page 24 Page 2476 76 Fault calculation Fault perhitungan Active network Aktif jaringan P P q q lsc LSC Same network with Sama jaringan dengan all source voltages semua sumber tegangan removed and replaced dihapus dan diganti by short circuits oleh arus pendek P P 0 0 • 4-'-" • 4-'-" Z Z , , q q r r I Aku i aku p p I Aku ! ! o o 0 i 0 i i aku lsc LSC z z i aku i aku i aku o o Fig. Gambar. 3.2.51 3.2.51 Norton's Theorem Teorema Norton 3.2.5I and its application to the simple two-machine circuit of Section 3.2.1 is 3.2.5I dan aplikasinya pada rangkaian dua-mesin sederhana dari Bagian 3.2.1 adalah shown in Fig. ditunjukkan pada Gambar. 3.2.5J. 3.2.5J. It is important to note that the driving current is constant in Penting untuk dicatat bahwa saat mengemudi adalah konstan dalam value irrespective of the voltage across it, just as a given driving voltage is constant terlepas dari nilai tegangan di atasnya, seperti mengemudi diberikan tegangan konstan in value irrespective of the current through it. nilai terlepas dari arus melewatinya. A constant driving current can be Sebuah mengemudi konstan arus dapat P P i aku I + j4F I + j4F

2 + j51"I i 2 + j51 "Aku i I + j0D, I + j0D, i aku io io .; N N.; F" F " [ [ 19.S - j43.6A 19.S - j43.6A [ [ i aku "" "" i aku i aku -@ - @ i aku i aku i aku i aku i aku i aku L.. L.. Fig. Gambar. 3.2.5,) 3.2.5,) Norton's Theorem applied to the example of Fig. Teorema Norton diterapkan pada contoh Gambar. 3.2. 3.2. IE IE regarded in physical terms, as the current produced by an infinite voltage acting in dianggap dalam hal fisik, sebagai arus yang dihasilkan oleh tegangan tak terbatas bertindak series with an infinite impedance, the ratio of the voltage to the impedance being seri dengan suatu impedansi yang tak terbatas, rasio tegangan ke impedansi yang the required driving current. menyetir yang dibutuhkan saat ini.

Page 25 Page 25Fault calculation Fault perhitungan 77 77 A number of other useful theorems and reduction formulae will be found in the Sejumlah teorema berguna lainnya dan formula pengurangan akan ditemukan dalam references given in the bibliography. referensi diberikan dalam bibliografi. 33 Calculation of balanced fault conditions 33 Perhitungan kondisi gangguan seimbang 3.3.1 Single-phase representation 3.3.1 Single-tahap representasi The component items of electrical plant which together form a three-phase power Item komponen tanaman listrik yang bersama-sama membentuk tiga fase daya system, namely generators, transformers, overhead-line and cable circuits, etc., can sistem, yaitu generator, transformator, overhead-garis dan rangkaian kabel, dll, dapat all be regarded, for most practical purposes, as having electrical characteristics semua dianggap, untuk sebagian besar tujuan praktis, memiliki karakteristik listrik which are balanced or symmetrical with respect to the three phases. yang seimbang atau simetris terhadap tiga fase. Thus, the plant Dengan demikian, tanaman

impedance characteristics are usually such that the phase self-impedance values can karakteristik impedansi biasanya seperti yang fase impedansi diri-nilai dapat be regarded as the same for all three phases and the phase-to-phase mutual- dianggap sebagai hal yang sama untuk ketiga fase dan fase-ke-fase saling impedance values regarded as symmetrical with respect to the three phases. dianggap sebagai nilai-nilai impedansi simetris terhadap tiga fase. Simi- Simi- larly, the driving voltages produced by the generators and any other synchronous larly, yang menggerakkan tegangan yang dihasilkan oleh generator dan lain sinkron machines are balanced in that the three phase-to-neutral emfs produced by any mesin seimbang dalam tiga fase-ke-netral emfs diproduksi oleh given machine are all equal in magnitude and symmetrically spaced, at 120 ° intervals, mesin yang diberikan semuanya sama besar dan simetris spasi, pada 120 ° interval, in time-phase, the phase order beingthe same for all the machines. dalam waktu-fasa, maka urutan beingthe fase yang sama untuk semua mesin. This phase order, Perintah fase, namely the order in which the instantaneous phase voltages attain their maximum yaitu urutan fasa sesaat tegangan mencapai maksimum values in the cycle, is termed the positive-sequence phase-order. nilai dalam siklus, yang disebut-urutan-urutan fase positif. It readily follows Ini mudah berikut from this system phase-symmetry that conditions which are themselves balanced dari fase-simetri sistem yang kondisi yang seimbang sendiri with respect to the three phases (for example, balanced4oad and three-phase short- sehubungan dengan tiga fase (misalnya, balanced4oad dan tiga fase pendek circuit conditions) will preserve this phase symmetry in that the resulting phase kondisi sirkuit) akan mempertahankan simetri ini fase dalam fase yang dihasilkan currents and similarly, the resulting phase voltages at any point in the system will arus dan sama, fase tegangan yang dihasilkan pada setiap titik dalam sistem akan also be balanced with respect to the three phases. juga diimbangi dengan hormat kepada tiga fase. Thus, for the conditions referred Dengan demikian, untuk kondisi dimaksud to, the three phase currents and, similarly, the three phase-to-neutral voltages at untuk, arus fase tiga, sama, tiga fasa-ke-netral tegangan di any given point in the system will be equal in magnitude and equally spaced at suatu titik tertentu dalam sistem akan sama besarnya dan sama spasi di 120 ° intervals in time-phase, the phase order being the same as that of the generated 120 ° interval waktu-fase, urutan fase menjadi sama dengan yang dihasilkan phase-to-neutral voltages, namely the positive-sequence phase-order. fase-ke-netral tegangan, yaitu urutan-urutan fase positif. These currents Arus ini and voltages are termed positive-sequence currents and voltages and are represented dan tegangan yang disebut-urutan arus positif dan tegangan dan diwakili in vector-diagram form in the manner shown in Fig. dalam bentuk diagram vektor dengan cara yang ditunjukkan dalam Gambar. 3.3.1A. 3.3.1A. The mathematical relationships between these balanced (positive-sequence) Hubungan matematis antara seimbang (positif-urutan) phase values at any given point in the power system are the same for both current tahap nilai pada suatu titik tertentu dalam sistem kekuasaan yang sama untuk kedua saat ini .and voltage, namely yaitu. dan tegangan, and dan

i = cao i = Cao V b =aVa V b = Ava v =aVo v = avo 3.3.1.1 3.3.1.1 where the subscripts a, b and c denote the three phases in positive-sequence phase- dimana subskrip a, b dan c menunjukkan tiga fase dalam tahap positif-urutan- order and the symbol a (not to be confused with the phase reference) is a constant ketertiban dan simbol yang (tidak harus bingung dengan referensi fasa) adalah konstan

Page 26 Page 2678 78 given by diberikan oleh Fault calculation Fault perhitungan 1 .V3 1. V3 a . sebuah. . . . . . . 1 [120 ° 1 [120 ° 2+/"--2 "- 2 +/"-- 2 "- and in known as the 120°operator. dan dikenal sebagai operator sebesar 120 °. V c V c V V h • h • a sebuah IC IC 1 h 1 jam 3.3.1.2 3.3.1.2 Fig. Gambar. 3.3.1A 3.3.1A Phasor-diagrarn representation of positive-sequence voltages and currents Fasor-diagrarn representasi-urutan tegangan positif dan arus It will be noted that multiplication of any given vector by a gives a resultant Ini akan dicatat bahwa multiplikasi dari setiap vektor yang diberikan oleh memberikan resultan vector equal in magnitude to the given vector but advanced in phase (that is rotated vektor besarnya sama dengan vektor yang diberikan tetapi maju dalam fase (yang diputar in an anticlockwise direction) by 120 ° with respect to the given vector. dalam arah berlawanan arah jarum jam) dengan 120 ° terhadap vektor diberikan. Similarly Demikian pula multiplication by aa give a phase-advance of 240 °, because perkalian oleh aa memberikan fase-muka 240 °, karena ( 1 _.3)( 1 ./3) (1 _.3) (1 3. /) i aku x/3 x / 3 32 = 2"+1 32 = 2 "1 -7 +I 7 Aku -7 + 7 =-'2- j 2 = 1[240° =- '2 - J 2 = 1 [240 ° 3.3.1.3 3.3.1.3 J J Further multiplication gives perkalian lebih lanjut memberikan aS = - 7 +'i "2-; 2-1 '-= 1/360° = 1[0° AS = - 7 + '2 "i -; 2-1' -= 1 / 360 ° = 1 [0 °

3.3.1.4 3.3.1.4 and so on. dan sebagainya. It should be noted that Perlu dicatat bahwa a = 1 [120 ° = 1/-240 ° a = 1 [120 ° = 1/-240 ° a 2=1[240 °=l[- 120 ° 2 = 1 [240 ° = [l - 120 ° 1 +a +a 2 =0 1 + + a 2 = 0 3.3.1.5 3.3.1.5 bearing in mind that a vector rotation forward through any given angle is the same mengingat bahwa rotasi vektor maju melalui sudut yang diberikan adalah sama as a rotation backwards through an angle equal to 360* minus the given angle. sebagai rotasi mundur melalui sudut sebesar 360 * minus sudut diberikan. The balanced conditions referred to are illustrated by Fig. Kondisi seimbang dimaksud diilustrasikan oleh Gambar. 3.3.1B which shows 3.3.1B yang menunjukkan an 11 kV generator with an impedance of 0 +]1 I2 per phase supplying a balanced sebuah kV generator 11 dengan impedansi 0 +] 1 I2 per fase memasok seimbang load of 20 + ]5 2 per phase through a transmission line of 2 +/3 I2 per phase, the beban 20 +] 5 2 per fase melalui jalur transmisi dari 2 + / 3 I2 per fasa, maka generator neutral point being earthed, as shown, and the transmission line having a generator titik netral yang dibumikan, seperti yang ditunjukkan, dan jalur transmisi memiliki three-phase fault clear of earth at 0ae end F remote from the generator. tiga-tahap yang jelas kesalahan bumi pada akhir F 0ae jauh dari generator. Denoting Yang menunjukkan

Page 27 Page 27Fault calculation Fault perhitungan 79 79 --0 - 0 ,--4m,, , - 4m,, @ @ © © 0 I jl$ 0 Saya jl $ L....-J ...- L. J 0 + 0 + illl illl I Aku ; ; I Aku i aku C C (; (; 2 2 j3l j3l I Aku ] ] 2 2 j3l j3l ; ; ] ] 2 2 + .i3$Z + $. I3 Z

20 20 j55l j55l ; ; ") ") 20 4 i55 20 4 i55 20 20 j5-l j5-l I c Aku c Ec Ec c Ea c Ea z,-"" VI z, - "" VI V b V b I a Aku seorang Fig. Gambar. 3.3.1B 3.3.1B Illustration of a balanced three-phase fault condition Ilustrasi kondisi kesalahan tiga fase seimbang the three phases, in positive-sequence phase order, by the subscripts a, b and c, the tiga fase, di urutan-urutan fase positif, oleh subskrip a, b dan c, generator phase-to-neutral driving-voltages are given by generator fasa-ke-netral mengemudi-tegangan yang diberikan oleh E a E sebuah = 6350 +/0 = 6.350 + / 0 = 6350/0 ° V = 6.350 / 0 ° V E b = - 3175 -/5500 = 6350/- 120 ° V E b = - 3175 -/5500 = 6.350 / - 120 ° V E c E c = - 3175 +/5500 = 6350/120°V = - 3175 + / 5500 = 6350/120 V ° the magnitude of the phase to neutral voltage namely 6350 V, being the line-to-line besarnya tegangan fase ke netral yaitu 6.350 V, menjadi garis-untuk-line voltage of 11 000 V divided by /3. tegangan 000 V 11 dibagi dengan / 3. It will be noted that emf Ini akan dicatat bahwa ggl E a E sebuah has been telah arbitrarily chosen to have zero phase-angle. sewenang-wenang memilih untuk memiliki fase-sudut nol. Using the rule given in Section 3.2.5 for the combination of parallel branches it Menggunakan aturan yang diberikan dalam Bagian 3.2.5 untuk kombinasi cabang paralel itu will be seen that the voltage to earth, at F, of the three short-circuited phases is akan terlihat bahwa tegangan ke bumi, pada F, dari hubung pendek tiga fase adalah

Page 28 Page 2880 80 Fault calculation Fault perhitungan given by diberikan oleh 2+i4[E. 2 + i4 [E. +eb + Eb o o IrF=- IRF =- 3 3 2+14 .j= 2 14. J =

because the three phase driving-voltages sum to zero and neutral current being karena ketiga fase mengemudi-tegangan jumlah ke nol dan netral saat ini sedang absent, the neutral point N and the earth are equipotential points. absen, titik netral N dan bumi adalah poin ekuipotensial. The impedance impedansi yang voltage-drop in each phase must therefore be equal to the phase-to-neutral driving- tegangan-drop di setiap fase sehingga harus sama dengan fase-ke-netral mengemudi- voltage in the phase in question and, hence, the three phase currents in the tegangan dalam fase tersebut dan, karenanya, arus fase tiga pada generator and transmission line are given by generator dan saluran transmisi yang diberikan oleh 6350 +10 10 6350 Ia IA = = = 635 - 11270 = 1420/-63.4°A = 635-11270 = 1420/-63.4 a ° 2+14 2 14 -3175 -/5500 = - 1418 +i85 = 1420[176.6 ° A -3.175 -/5500 = - 1.418 + i85 = 1420 [a ° 176,6 /'b = / 'B = 2+/4 2 + / 4 -3175 +/5500 -3.175 + / 5500 I c Aku c = = = 783 +11185 = 1420[56"6 ° A = 783 11185 = 1420 [56 "6 ° A 2+/4 2 + / 4 The phase-to-earth (and phase-to-neutral) voltages at the generator terminals G are Fase-ke-bumi (dan fase-ke-netral) tegangan pada terminal generator G equal to the phase voltage drops in the transmission line and are therefore given by sama dengan tegangan fase tetes di jalur transmisi dan karenanya diberikan oleh Ir a Ir a = (635 -/1270) (2 +i3) = 5080 -/635 = (635 -/1270) (2 + I3) = 5.080 -/635 = 5119/-7.1"v = 5119/-7.1 "v vb = (- 1418 +i85) (2 +/'3) = - 3091 - 14084 = 5119[ - 127.1°v vb = (- 1418 + i85) (2 + / '3) = - 3.091-14.084 = 5.119 [- 127,1 ° v vc vc = (783 +11185) (2 +13) = - 1989 +/4179 = 5119[112.9°v = (783 11185) (2 13) = - 1989 + / 4.179 = 5.119 [112,9 ° v The three-phase short-circuit at F has been assumed to be clear of earth in this Tiga-fase pendek sirkuit F telah dianggap jelas bumi ini example but it will be evident that, because contoh tetapi ini akan menjadi jelas bahwa, karena VF VF has been shown to be zero, a three- telah terbukti menjadi nol, tiga- phase-to-earth fault at F would result in zero current in the earth connection and -Ke-bumi di F kesalahan fase akan menghasilkan nol saat ini di bumi dan sambungan would thus give precisely the same results as those already obtained for the fault dengan demikian akan memberikan hasil yang sama persis dengan yang sudah diperoleh untuk kesalahan clear of earth. yang jelas bumi. The three-phase vector diagrams of the currents and voltages are Vektor diagram fase-tiga dari arus dan tegangan adalah shown in Fig. ditunjukkan pada Gambar. 3.3.1B. 3.3.1B.

Now, it will be evident from Eqns. Sekarang, itu akan terbukti dari Eqns. 3.3.1.1 that a knowledge of the phase 3.3.1.1 bahwa pengetahuan tentang fase currents and voltages relating to any one chosen reference phase will enable the arus dan tegangan yang berhubungan dengan salah satu fasa referensi dipilih akan mengaktifkan corresponding values of current and voltage in the other two phases to be deter- sesuai nilai-nilai arus dan tegangan pada fase dua lainnya akan menghalangi- mined. ditambang. Thus, if phaso-a is chosen as the reference phase then the phase-b values of Jadi, jika sebuah phaso-dipilih sebagai referensi maka fase-fase nilai b current and voltage are simply the corresponding phase-a values multiplied by a 2 arus dan tegangan secara sederhana adalah sebuah fase yang sesuai-nilai dikalikan dengan 2 and the phase-c values the corresponding phase-a values multiplied by a, where a is dan fase-fase nilai c-suatu nilai yang sesuai dikalikan dengan suatu, di mana a adalah the 120 ° operator. operator sebesar 120 °. For the purposes of analysis, therefore, the power system of Untuk keperluan analisis, oleh karena itu, sistem kekuasaan Fig, 3.3.1B can be represented by the single-phase network shown in Fig. Gambar, 3.3.1B dapat diwakili oleh fase jaringan tunggal ditunjukkan pada Gambar. 3.3.1C, 3.3.1C, this single-phase network and the associated single-phase vector diagram both ini-fase jaringan tunggal dan vektor diagram fase tunggal terkait baik relating to the chosen reference phase, namely phase a. berkaitan dengan fase referensi dipilih, yaitu fase a. The single.phase representation, although demonstrated for the extremely simple Representasi single.phase, meskipun menunjukkan untuk yang sangat sederhana system of Fig. Gambar sistem. 3.3.1B, is equally applicable to any balanced three-phase network 3.3.1B, sama-sama berlaku untuk setiap jaringan tiga fase seimbang

Page 29 Page 29Fault calculation Fault perhitungan 81 81 operating under balanced conditions, for example balanced4oad or three-phase beroperasi dalam kondisi seimbang, untuk balanced4oad misalnya atau tiga-fasa short-circuit conditions. kondisi sirkuit pendek. The single-phase network obtained is termed the system The-fase jaringan tunggal yang diperoleh disebut sistem positive-sequence network, and the impedances represented in the network are the Jaringan urutan-positif, dan impedansi terwakili dalam jaringan adalah impedances appropriate to positive sequence conditions, termed the positive- impedansi urutan sesuai dengan kondisi positif, disebut positif- sequence impedances. impedansi urutan. It will be noted from Fig. Ini akan dicatat dari Gambar. 3.3.1C that the three-phase short- 3.3.1C bahwa tiga fase pendek circuit is represented in the positive-sequence network by the short-circuiting sirkuit diwakili dalam jaringan urutan-positif oleh arus pendek connection from the point of fault F to the zero-potential (neutral) bar represented sambungan dari titik F kesalahan ke netral) bar-potensi (nol diwakili by the thick horizontal fine. oleh baik horisontal tebal. With this short-circuiting connection absent, the Dengan hubungan arus pendek koneksi tidak ada, yang conditions represented are those corresponding to the balanced-load condition kondisi diwakili adalah mereka yang sesuai dengan kondisi beban seimbang

existing before the application of the three-phase short circuit. yang ada sebelum penerapan fase sirkuit pendek-tiga. _•• _ • • O O .il . Il F F 21) jS 21) JS f f ; ; r-'-I r-'-aku G G 2 i3fl 2 i3fl | | I Aku Fig. Gambar. 3.3.1C 3.3.1C V V Single-phase representation of Fig. Single-tahap representasi dari Gambar. 3.3.1B 3.3.1B 3.3.2 Use of a common voltage base 3.3.2 Penggunaan dasar tegangan umum The simple example of the previous Section was solved by direct analysis of the Contoh sederhana dari Bagian sebelumnya diselesaikan dengan analisis langsung dari given network using actual values of plant impedance together with actual values of diberikan untuk jaringan dengan nilai yang sebenarnya dari impedansi tanaman bersama-sama dengan nilai aktual the power-system currents and voltages. dengan sistem arus listrik dan tegangan. This direct application of the network laws Aplikasi ini langsung dari jaringan hukum was possible because the network interconnections were direct (that is, not through mungkin karena jaringan interkoneksi adalah langsung (yaitu, tidak melalui transformers) or, put in another way, because the whole system had the same transformer) atau, menaruh dengan cara lain, karena seluruh sistem telah sama nominal rated voltage for all its component items of plant. tegangan nominal untuk semua item komponen tanaman. The power systems Sistem listrik which have to be considered in practice, however, almost invariably consist of a yang harus dipertimbangkan dalam prakteknya, hampir selalu terdiri dari number of sections operating at different nominal rated voltages, the interconnec- jumlah bagian yang beroperasi pada voltase berbeda nominal, yang interconnec- tions between sections being by means of transformers. tions antara bagian yang melalui transformator. The analysis of such Analisis tersebut multiple-section systems is achieved by representing the given network by an -Bagian sistem beberapa dicapai dengan mewakili jaringan yang diberikan oleh equivalent network whose several sections all have the same nominal rated voltage. jaringan yang setara beberapa bagian semua memiliki tegangan nominal yang sama. This procedure allows the transformer interconnections to be replaced by equiva- Prosedur ini memungkinkan interkoneksi transformator akan digantikan oleh equiva- lent direct interconneetions, the resulting network being then amenable to direct interconneetions langsung dipinjamkan, sehingga jaringan yang kemudian setuju untuk langsung solution using the normal methods of network analysis. solusi menggunakan metode analisis jaringan normal.

Page 30 Page 3082 82 Fault calculation Fault perhitungan The derivation of this so-called common-voltage equivalent network can be Penurunan ini biasa disebut jaringan tegangan setara sehingga dapat readily understood by considering the two network sections, P and Q, shown in mudah dipahami dengan mempertimbangkan dua bagian jaringan, P dan Q, yang ditunjukkan pada Fig. Gambar. 3.3.2A, the two sections being interconnected by an ideal transformer of turns 3.3.2A, dua bagian yang saling berhubungan oleh ternyata trafo ideal ratio M (= rasio M (= Np/Nq) Np / Nq) where dimana nominal rated voltage of network-section P nominal tegangan jaringan-bagian P M= M = 3.3.2.1 3.3.2.1 nominal rated voltage of network-section Q tegangan nominal jaringan-bagian Q The two network sections are each represented, for simplicity, by a single branch Kedua bagian jaringan masing-masing diwakili, untuk kesederhanaan, dengan satu cabang consisting of a driving voltage and an impedance in series, their values being terdiri dari tegangan mengemudi dan impedansi seri, maka nilai yang Ep Ep and dan Zp Zp for network-section P and untuk jaringan-bagian P dan Eq Persamaan and dan Zq Zq for network-section Q. The ideal- untuk jaringan-bagian Q. ideal transformer terminal-voltages and currents are similarly denoted by terminal transformator tegangan dan arus sama-sama dilambangkan dengan Vp Vp and dan lp lp and dan Vq Vq and dan 1q, 1Q, where dimana I t, Aku t, is the current flowing out of network-section P and adalah arus yang mengalir keluar dari jaringan-bagian P dan lq LQ that bahwa flowing into network-section Q. Now, because the transformer is ideal (that is has mengalir ke bagian jaringan T. Sekarang, karena trafo ideal (yang memiliki windings of zero leakage.impedance and takes zero exciting current), it follows that gulungan dari nol leakage.impedance dan mengambil nol saat ini menarik), maka yang

Vp = MVq = V; Vp = MVq = V; say I kataku = Iq/M Iq = / M - - ' ' - - I say Kataku 3.3.2.2 3.3.2.2 I Aku and it is easily seen that these same equations are satisfied by the equivalent system- dan mudah dilihat bahwa persamaan yang sama dipenuhi oleh sistem setara network of Fig. Gambar jaringan. 3.3.2B in which the ideal transformer is represented by a direct 3.3.2B di mana transformator ideal adalah diwakili oleh langsung zero-impedance interconnection. impedansi nol-interkoneksi. This equivalent network has the same nominal Jaringan ini setara memiliki nominal yang sama rated voltage throughout, namely that of network-section P, and consists of the tegangan di seluruh, yaitu bahwa bagian jaringan-P, dan terdiri dari given network-section P and the equivalent or referred network-section Q'. diberikan bagian jaringan P dan setara atau disebut jaringan-bagian Q '. It will Ini akan be noted that the referred network-section Q' contains the referred values dicatat bahwa bagian jaringan disebut-Q 'berisi nilai-nilai dimaksud E = MEq E = mEq = = lq'= Iq/M LQ '= IQ / M Zq Zq = M 2 Zq M = 2 Zq P P F . F. . . . . . . i aku ! ! Fig. Gambar. 3.3.2A 3.3.2A 3. 3. :2.3 : 2.3 . . . . . . . . . . "7 "7 i aku Ip Ip M : I M: Aku Iq Zq Iq Zq [ [ '; tl i '; Tl i i aku Vp I Vp Saya Vq Vq ! ! _1 _1 J J

Simplified representation of two networks with ideal-transformer interconnection Sederhana representasi dari dua jaringan interkoneksi dengan-transformator ideal in place of the actual network values, the expression for the referred impedance di tempat nilai-nilai jaringan yang nyata, ekspresi untuk impedansi dimaksud following from the fact that the voltages in the referred network are all M times the berikut dari kenyataan bahwa tegangan di jaringan dimaksud adalah kali M semua corresponding voltages in the actual network. sesuai tegangan di jaringan yang sebenarnya. Thus, the voltage-drop across Jadi, tegangan drop di Zq in Zq di the actual network is jaringan sebenarnya adalah lqZq lqZq and the corresponding voltage-drop in the referred net- dan penurunan tegangan yang sesuai-sebagaimana dimaksud dalam net-

Page 31 Page 31Fault calculation Fault perhitungan 83 83 f f i aku i aku i aku i aku i aku ! ! k.. k.. 1 1 r r /p Ii, [ / P Ii, [ i aku r--1 ; I r - 1; saya i aku li li ' ' i aku i aku o o i aku i aku Q, Q, Itt/M M2Zq Itt / M M2Zq I Aku v v f f i aku i aku I MVq Aku MVq 0 0 Fig. Gambar. 3.3.2B 3.3.2B

Common-voltage equivalent of Fig. Common-tegangan setara Gambar. 3.3.2.4 3.3.2.4 work is kerja IqZq, IqZq, so that sehingga MIqZq MIqZq zi, = zi, = = MZq = MZq Now, eqns. Sekarang, eqns. 3.3.2.3 will apply to all the voltages, currents and impedances in 3.3.2.3 akan berlaku untuk semua tegangan, arus dan impedansi di network-section Q when the latter is considered as a general network rather than as jaringan-bagian Q saat terakhir dianggap sebagai jaringan umum daripada sebagai a single branch. cabang tunggal. For this general case, therefore, the voltages, currents and Untuk kasus yang umum, karena itu, tegangan, arus dan impedances given in eqns. impedansi diberikan dalam eqns. 3.3.2.3 can now be regarded as general network values, 3.3.2.3 sekarang dapat dianggap sebagai nilai jaringan umum, the equations given representing the relationships between actual values and their diberikan persamaan yang mewakili hubungan antara nilai yang sebenarnya dan mereka referred equivalents. disebut setara. This general case is represented in block-diagram form in Ini kasus yang umum direpresentasikan dalam bentuk blok-Diagram dalam Fig. Gambar. 3.3.2C. 3.3.2C. I' Aku ' QI QI r r i aku F F ] ] i aku i aku io io ,, ,, ,o , O i aku i aku % % '" '" °i ° i I Aku < < i aku i aku i aku i aku i aku I Aku I Aku I Aku i aku

i aku i aku i aku i aku v, v, ,, ,, , , i aku V V p p Z Z il il t I Aku t "t I "Saya t o I o Aku ,o , O I Aku i aku J J t t J J Fig. Gambar. 3.3.2C 3.3.2C Generalised representation of Fig. Generalised representasi Gambar. 3.3.28 3.3.28 Consider, now, an ideal-transformer interconnection between network-section Q Pertimbangkan, sekarang, sebuah transformator-interkoneksi yang ideal antara jaringan-bagian Q and a third network-section S, the turns ration N (-- dan bagian ketiga jaringan S, ternyata ransum N (- Nq/Ns) Nq / Ns) of this transformer transformator ini being given by yang diberikan oleh nominal rated voltage of network section Q tegangan nominal bagian jaringan Q N = N = 3.3.2.4 3.3.2.4 nominal rated voltage of network section S nominal tegangan bagian jaringan S By similar reasoning to that already described, the two network sections Q and S Dengan alasan serupa dengan yang sudah dijelaskan, kedua jaringan bagian Q dan S can be represented by an equivalent network consisting of the given network- dapat diwakili oleh jaringan setara yang diberikan terdiri dari jaringan- section Q and an equivalent or referred network-section S', the referred network- bagian Q dan setara atau disebut jaringan-bagian S ', yang disebut jaringan

Page 32 Page 3284 84 Fault calculation Fault perhitungan section S' containing the referred values bagian S 'yang memiliki nilai-nilai dimaksud E; = NEs E; = Nes v; =NG v; = NG z; = NZ z; = NZ 3.3.2.5 3.3.2.5

in place of the actual network values. di tempat nilai-nilai jaringan yang nyata. The ideal-transformer interconnection The-trafo ideal interkoneksi between network-sections Q and S is now represented as previously by a direct antara jaringan-bagian Q dan S sekarang diwakili seperti sebelumnya dengan langsung zero-impedance interconnection in the equivalent system-network. impedansi nol-interkoneksi di jaringan-sistem setara. The equivalent system-network is shown in block-diagram form in Fig. Jaringan setara sistem ditampilkan dalam bentuk diagram blok pada Gambar. 3.3.2D. 3.3.2D. Suppose, now, that all three network sections are interconnected through the Misalnya, sekarang, bahwa ketiga bagian jaringan saling berhubungan melalui given ideal transformers. diberikan transformator ideal. Then it will be evident that the equivalent network con- Maka akan tampak jelas bahwa jaringan setara con- sisting of Q and S' can itself be replaced by another equivalent network consisting sisting Q dan S 'sendiri bisa diganti dengan jaringan lain setara yang terdiri of Q' and S", this latter equivalent network being directly connected through zero- Q 'dan S ", jaringan ini setara terakhir secara langsung terhubung melalui nol impedance to the network-section P. Now the values in the network-section Q' are impedansi ke bagian-jaringan P. Sekarang nilai-nilai di bagian-jaringan Q 'adalah (.) (.) S S r r I Aku r r 1 1 ' ' I Aku I Aku I Aku , , i aku i aku i I i saya I Aku e e j j I:,t I:, t Il Il o . o. [ [ o o I Aku I Aku ! ! i aku i aku i aku I Aku ; ; i aku i aku

s s ' ' i aku i aku v v ' ' ; ; I Aku I Aku i aku I Aku v v ,. ,. i aku io io ! ! o o , , -._ -. _ J J t t ,, ,, Fig. Gambar. 3.3.20 3.3.20 those already given by eqns. mereka yang sudah diberikan oleh eqns. 3.3.2.3 while those in S", are given by applying these 3.3.2.3 sementara mereka yang di S ", yang diberikan dengan menerapkan ini same equations to the values in S'. persamaan yang sama dengan nilai dalam S '. Hence, it follows that Oleh karena itu, berikut bahwa E' = M.NE, E '= M. NE, G' =MVG G '= MVG q' q ' = ;,/Mlv =;, / Mlv z;' = (.v)z z; '= (. v) z 3.3.2.6 3.3.2.6 C C l'.p l '. p I p Saya p Vp Vp /p / P Q i T i S" S " | | r r I Aku ii ii if jika l l , i , I oi oi i aku

i aku i aku ':, ':, ':, ':, ii ii ': ': '" i '"I ii ii ii ii i aku o o o o o o I Aku o o I Aku vlq vlq ' ' " " ' ' ,' , ' ! ! I Aku v, v, " " I Aku o o o o ii ii I ! Aku! I Aku j L ° L ° j °j ° j t t i aku Fig. Gambar. 3.3.2E 3.3.2E

Page 33 Page 33Fault calculation Fault perhitungan 85 85 the resulting equivalent network for the full three-section system being represented jaringan setara dengan yang dihasilkan untuk sistem tiga-bagian diwakili penuh in block-diagram form by Fig. dalam bentuk diagram-blok oleh Gambar. 3.3.2E. 3.3.2E. From eqns. Dari eqns. 3.3.2.1 and 3.3.2.4 it is seen that 3.3.2.1 dan 3.3.2.4 terlihat bahwa nominal rated voltage of network section P nominal tegangan bagian jaringan P MN = MN = nominal rated voltage of network section S nominal tegangan bagian jaringan S

and it thus follows that the rules for referring any given network section to a given dan dengan demikian berarti bahwa aturan untuk merujuk semua bagian yang diberikan kepada suatu jaringan yang diberikan common nominal rated voltage, termed the common voltage base, are tegangan nominal yang sama, disebut tegangan dasar umum, adalah where dimana E' = KE E '= EK V'=KV V '= KV I' = I/K Aku '= I / K Z' = IZ Z '= IZ 3.3.2.7 3.3.2.7 common voltage base umum tegangan basis K = K = 3.3.2.8 3.3.2.8 nominal rated voltage of given network section nominal tegangan dari bagian jaringan yang diberikan and the primed and unprimed symbols represent referred and actual values dan prima dan unprimed simbol mewakili nilai-nilai dimaksud dan aktual respectively. masing. It will be noted that these equations are simply a more general state- Ini akan dicatat bahwa persamaan ini hanyalah negara yang lebih umum ment of eqns. bangan eqns. 3.3.2.3 and 3.3.2.5. 3.3.2.3 dan 3.3.2.5. Although these conversion equations have been derived by considering network- Meskipun persamaan konversi telah diturunkan dengan mempertimbangkan jaringan sections interconnected as a single point it is not difficult to see that they are equally bagian yang saling berhubungan sebagai titik tunggal maka tidak sulit untuk melihat bahwa mereka sama valid for any number of interconnection points. berlaku untuk jumlah titik interkoneksi. The process of network analysis by the use of a common voltage base can now Proses analisis jaringan dengan menggunakan dasar tegangan umum sekarang dapat be seen to consist of three stages: terlihat terdiri dari tiga tahap: (a) the conversion of the given system network to an equivalent network having (A) konversi jaringan sistem diberikan ke jaringan setara memiliki a single common nominal rated voltage, namely the chosen base voltage, for yang umum tegangan nominal tunggal, yaitu tegangan dasar dipilih, untuk all its sections semua nya bagian (b) the solution of this equivalent network in terms of its common-base values of (B) solusi jaringan ini setara dalam hal bersama-dasar nilai-nilai dari voltage, current and impedance tegangan, arus dan impedansi (c) the reconversion of the equivalent network, with its now determined (C) reconversion jaringan setara, dengan kini ditentukan common-base values of current and voltage, to its original actual form con- umum-nilai dasar arus dan tegangan, dengan bentuk aslinya sebenarnya con- taining actual values of current, voltage and impedance. taining nilai yang sebenarnya arus, tegangan dan impedansi. Eqns. Eqns. 3.3.2.7, used to determine the referred values of driving voltage and 3.3.2.7, digunakan untuk menentukan nilai-nilai dimaksud berkendara tegangan dan impedance for the equivalent system network, are, of course, equally valid for the impedansi untuk jaringan sistem setara, tentu saja, sama-sama berlaku untuk

final stage of reconversion from referred values to actual values. tahap akhir reconversion dari nilai-nilai dimaksud nilai yang sebenarnya. These equations, Ini persamaan, rewritten in the form appropriate to this reconversion, are ditulis ulang dalam bentuk yang sesuai dengan reconversion ini, adalah E= Et/K E = Et / K v = V'/K v = V '/ K I =KI' I = KI ' z = z'/x z = z '/ x 3.3.2.9 3.3.2.9

Page 34 Page 3486 86 Fault calculation Fault perhitungan where the primed symbols represent common-base values and the unprimed dimana simbol-dasar prima merupakan nilai-nilai bersama dan unprimed symbols actual values. simbol nilai-nilai aktual. The symbol K is the ratio of common base voltage to actual simbol K adalah rasio dari tegangan dasar umum untuk aktual nominal rated voltage, as already defined. tegangan nominal, sebagaimana telah didefinisikan. The use of the common voltage base is illustrated by the simple example Penggunaan dasar tegangan umum digambarkan oleh contoh sederhana shown in Fig. ditunjukkan pada Gambar. 3.3.2F. 3.3.2F. It is important to note that the use of the common voltage base, as so far des- Penting untuk dicatat bahwa penggunaan tegangan dasar umum, sejauh des- cribed, has assumed ideal-transformer interconnections between the different cribed, telah diasumsikan-transformator interkoneksi yang ideal antara yang berbeda network sections, the voltage-transformation ratios of these transformers being jaringan bagian, transformasi rasio tegangan dari transformator yang 11/33 kV 11/33 kV 3316.6 kV 3.316,6 kV 3 + j6S i' 3 + j6S i ' 4 jS.L 4 jS.L /.-.... /.-.... [ [ 12 kV 12 kV phase-to-phase fase-ke-fase I Aku l/6.6 kV l/6.6 kV I -i jill Aku-i Jill 4160 4.160 +.i0v +. I0v 0.12 + 0,24Sg SG 0,12 + 0,24 o o o o 1.44 + jl,80, 1,44 + jl, 80, ! ! ; ; o o I Aku

I + jlS I + jlS 4160 + jOY 4.160 + JOY j0.72 S. j0.72 S. 0.11 + .iO.21 0,11 +. IO.21 + jl. + Jl. 4 160 + 4 160 + jOY JOY I.II , j1.'43S I. II, J1. '43S 31 31 j1620A j1620A Fig. Gambar. 3.3.2F 3.3.2F Application of the common voltage base for a system with ideal-transformer Aplikasi dasar tegangan umum untuk sistem dengan ideal-transformator interconnections (common base voltage 6.6 k V) interkoneksi (tegangan umum dasar 6,6 k V)

Page 35 Page 35Fault calculation Fault perhitungan 87 87 identical to the ratios of the nominal rated voltages of their associated network identik dengan rasio dari voltase nominal yang terkait jaringan mereka sections. bagian. As already shown, such ideal-transformer intercormections are represented Seperti telah ditunjukkan, intercormections ideal-trafo tersebut diwakili in the common-voltage equivalent circuit by direct intereonnections of zero di sirkuit tegangan setara-umum oleh intereonnections langsung dari nol impedance. impedansi. The representation necessary ha the case of practical transformers is Representasi ha diperlukan kasus transformator praktis discussed in the following section. dibahas pada bagian berikut. 3.3.3 Representation of nominal-ratio transformer circuits 3.3.3 Representasi rasio transformator sirkuit-nominal Practical transformer circuits differ from the ideal transformers of the previous rangkaian transformator Praktis berbeda dengan transformator ideal sebelumnya section in that they have non-hat'mite values of exciting impedance and non-zero values of leakage impedance. From the point of view of their representation in a common voltage equivalent circuit they may ha addition, have transformation ratios which differ from the nominal ratios so far considered. Two-winding transformers: The common-voltage equivalent circuit of two- winding transformer can be derived by considering the single-phase transformer shown ha Fig. 3.3.3A, where H and L denote the high-voltage and low-voltage windings, respectively. The transformer is assumed to have a high-voltage-to-low- voltage turns-ratio N (= NH/NL) equal to the ratio of the nominal rated voltages of the associated network sections. As shown in the figure, the given transformer can be represented completely by an ideal transformer of this same turns-ratio N together with the lumped impedances ZH, ZL and dan Ze. The Itu

impedances Zt4 and ZL are the leakage impedances of the windings H and L, respectively, and Ze the itu exciting impedance, Ze + Ztt being the impedance which would be measured by an open-circuit test on the high-voltage winding. N:I I"- "" "" i" i " I,,,,. H H L L .< ,. ,. IH IH ZH ZH N : I ZL ]L c "; " :- : - t t ;'; o o TI' IIITr VH VH Z e Z e EH EH EL EL VL VL C ..... , , O O Fig. Gambar. 3.3.3A Equivalent circuit of a two-winding transformer

Page 36 Page 3688 88 Fault calculation Fault perhitungan Now, the voltage EL EL across the low-voltage winding of the ideal transformer is given by diberikan oleh EL EL = = Vz,

+ ILZL 3.3.3.1 where dimana VL VL is the voltage and/i, the outflowing current at the terminal of winding L. The voltage across the high-voltage winding of the ideal transformer is therefore EH = NEL = IV(VL + IL ZL ) 3.3.3.2 which can be rearranged to read EH = NVL + (IL/N)N2 ZL 3.3.3.3 3.3.3.3 It will be noted that NVL NVL and dan IL/N are the equivalent values of V'L and IL when ketika referred to the nominal rated voltage of the high.voltage winding, so that EH EH = = VL + fLZ 3.3.3.4 where Z (= N:ZL)is the referred value ofZL. From Eqn. Dari Eqn. 3.3.3.4 it will be evident that the equivalent T circuit of Fig. 3.3.3B is the required common-voltage equivalent circuit of the transformer referred to the nominal rated voltage of the high-voltage winding H. The exciting impedance Ze can be ignored in the great majority of calculations because the exciting current le can normally be considered to be negligibly small. i H i H ZH ZH c c . . ; ; ; ; - - V HI Z e Z e z (=N2Z L ) =[(=L/N) EH(=EL) V'L(=NV L) o o ! ! H H ZHL(=ZH+N2ZL) I Aku > >

r-'-I > > O O % % Fig. Gambar. 3.3,3B Common-base equivalent circuit of a two-winding transformer In such cases, therefore the common-voltage equivalent circuit reduces to the single series impedance ZH + NaZL = ZHL , say mengatakan 3.3.3.5 shown in Fig. ditunjukkan pada Gambar. 3.3.3B. It will be noted that ZHL is the short-circuit impedance of the transformer as measured from the high-voltage side with the low-voltage winding short-circuited. The equivalent impedance of the transformer referred to the nominal rated voltage of the low-voltage winding, namely ZLH is readily

Page 37 Page 37Fault calculation Fault perhitungan 89 89 obtained from the fact that ZHL = ZLH Ving Ving ZLH = = ZL + + (I/N). ZH ZH 3.3.3.6 The impedance ZLH is, of course, the short-circuit impedance of the transformer as measured from the low-voltage side with the high-voltage winding short-circuited. Auto-transformers: The auto-transformer differs from the two.winding trans- former in that a single tapped winding is used in place of two separate windings, the arrangement being as shown in Fig. 3.3.3C. The full winding, which constitutes the high.voltage winding, is assumed to have 2V H turns and the common portion, which constitutes the low-voltage winding, is assumed to have NL NL turns. berubah. As before, the Seperti sebelumnya, transformer is assumed to have a high-voltage-to-low-voltage turns-ratio N (= NH/NL) equal to the ratio of the nominal rated voltages of the associated network sections. As shown in the figure, the given transformer can be represented completely by an ideal auto-transformer of this same turns-ratio N together with N:I

L L VH VH IH IH ZS 1 L 1 L f f zel It L ZC .,I 0 0 0 0 VL VL Fig. Gambar. 3.3.3C Equivalent circuit of an auto-transformer lumped impedances Zs, Z e and Ze. The impedance Z s Z s is the leakage impedance of that portion of the winding which lies between the high-voltage terminal and the tapping point, the series portion, this portion having NH - Nt. turns. berubah. Similarly, Demikian pula, Zc is the leakage impedance of the remaining portion of the winding, namely that between the tapping point and the common terminal. The impedance Z¢ is the

Page 38 Page 3890 90 Fault calculation Fault perhitungan exciting impedance, Zs + Ze + Ze being the impedance which would be measured by an open-circuit test on the high-voltage winding. The tapping point, it will be noted, constitutes the low-voltage terminal of the transformer. Now, the voltage EL across the low-voltage (common) winding of the ideal transformer is given by Ez, = VL + (1 L - 1H)Zc 3.3.3.7 where VL is the voltage and IL the outflowing current at the low-voltage terminal of the winding L and 1H the inflowing current at the high-voltage terminal of winding H. The terminal voltage VH of the high-voltage winding H is given by VH = VL +(N- 1)EL +IHZs 3.3.3.8 Substituting for EL in this latter equation usin eqn. 3.3.3.7 gives VH =NVL +IH[Zs- (N- 1)Zc] +IL (N- 1)Ze 3.3.3.9 which, rearranged, gives VH =NVL +IH[Zs- (N- 1)Zc] + (IL/N)N(N- 1)Z e 3.3.3.10 It will be noted, as before, that NVL and IL/N are the equivalent values of VL and I£ when referred to the nominal rated voltage of the high-voltage winding, so that sehingga VH VH = =

If L +IH[Z s- (N- 1)Zc] +N(N- 1)Z c 3.3.3.11 Now the open-circut impedance of the auto-transformer as measured from the high-voltage side is Zs + Ze + Zc 3.3.3.12 and that measured from the low-voltage side, but referred to the high-voltage side, is adalah N 2 Ze + Ze IH IH ZS-(NZ)Z c N(NI)Z c l l q q ; ; i aku ; ; ; ; > > Z e Z e + + NZ c 3.3.3.13 Fig. Gambar. 3.3.3D IH IH ZHL [= Zs+(N- I)2Zc] I. I. c > I--"l > > VH VH Common-base equivalent circuits of an auto-transformer

Page 39 Page 39Fault calculation Fault perhitungan 91 91 The required common-voltage equivalent-circuit for the auto-transformer can now be obtained using eqns. 3.3.3.11, 3.3.3.12 and 3.3.3.13 and is as shown in Fig. Gambar. 3.3.3D, this equivalent circuit being that referred to the nominal rated voltage of the high-voltage winding. If the magnetising impedance Ze can be ignored, as is usually the case, the common-voltage equivalent circuit reduces to the single series impedance Z s Z s + + (N- 1)2Ze = ZHL

3.3.3.14 shown in Fig. 3.3.3D. It will be noted that ZHL is the short-circuit impedance of the transformer as measured from the high-voltage side with the low-voltago wind- ing short-circuited. The equivalent impedance of the transformer referred to the low-voltage side is obtained by dividing the above value by N a , as before, the result being menjadi ZLH = (1]N)2 Zs + (N- 1)2 Zc/N 2 3.3.3,15 As before, the impedance ZLH is the short-circuit impedance of the transformer as measured from the low-voltage side with the high.voltage winding short-circuited. Three-winding transformers: The common-voltage equivalent-circuit of a single. phase three-winding transformer can be obtained by similar reasoning to that already employed, the result being the equivalent-star circuit shown in Fig. 3,3.3E, It is assumed, as before, that the turns ratio of each pair of windings is equal to the ratio of the nominal rated voltages of the associated network sections. The letters H, L and T denote the high-voltage, low-voltage and tertiary-voltage windings, respectively, and the component branch impedances ZH, Z'L and Z'T are all assumed to be values referred to the high voltage side. (Note that primed symbols denote referred values.) These component branch impedances are obtained by similar o o o o T T H - z, L I Aku ! ! | U ZH ZH ] ] I Aku ; ; j j z'T T' Fig. Gambar. 3.3.3E Common-base equivalent circuit of a Chree-winding ransforrnet

Page 40 Page 4092 92 Fault calculation Fault perhitungan short-circuit impedance measurements to those used for two-winding transformers and autotransformers. Thus, let ZH/

= = effective impedance of winding winding T open-circuited Z, = effective impedance of winding winding H open-circuited ZH ZH = = effective impedance of winding winding L open-circuited H with winding L short-circuited and L with winding T short-circuited and T with winding H short-circuited and all three values being referred to the high-voltage side. Then, from the equivalent circuit it is seen that ZHL = ZH + Z'L Z'Lr = = Z'L + Zr Zr t t grH = = Zr Zr + + ZH ZH 3.3.3.16 from which dari yang zH = = I(ZHL + ZH - - ZT) Z' L = I (zL + Z'Lr - Z'rH) t t 1 t ZT = 2(Zr +ZH - ZL) 3.3.3.17 Fig, 3.3.3F ZH ZH HC HC ."'1----- z,,. LI LI | | "1 "1 0 T' Z e Z e Common.base equivalent circuit of a three-winding transformer with exciting

impedance impedansi included. disertakan. The equivalent-star representation of Fig. 3.3.3E is equally valid whether the three windings are electrically separate or not and is therefore also applicable to the quite common case in which two of the windings form an auto-transformer, the third winding only being electrically separate. If it is required to represent exciting impedance in the equivalent circuit this can be done by adding an impedance of appropriate value between the star-point of the

Page 41 Page 41Fault calculation 93 93 12 12 kV kV phase-to-phase fase-ke-fase 4160 4.160 + j0V .-n.lm- I Aku 1/33 kV ZHL=J1.512, 3 3 + j6,$I I Aku I Aku 4 4 + jS [. jO.06l 0.12 + jO.241Z 1.44 1,44 +jI.8012 33/6.6 kV ZHL=J2, l l i +jl, ZHL=JI ,I 1 1 I/6.6 kV jO.O8l 4160 4.160 +jOV j0.7212 O.12 + j0.33 1 +jill II II E::::3- 4160 + jOV 1.12 + j2.OSlZ r'-"l 855 -jlS6SA

Fig. Gambar. 3.3.3G Example of Fig. 3.3.2F but with practical transformer Interconnectlona. (common base voltage 6.6 k W impedances impedansi ZH, Z'L and Z- and neutral, as shown in Fig. 3.3.3F. This exciting impedance must, of course, be to the same common voltage base as th¢ other impedances. The representation of transformer impedance is illustrated by the simplo example shown in Fig. 3.3.3G, this being the same as that given in Fig. 3,3,2F except that the transformers, previously assumed ideal, now have the given values of winding impedance.

Page 42 Page 4294 94 Fault calculation Fault perhitungan It might be expected that transformers having more than three windings would be represented by an equivalent-star circuit similar to that shown in Fig. 33.3E but with the number of branches in the star (exciting impedance ignored) equal to the number of windings. This form of representation is not normally valid, however, for transformers having more than three windings, and such transformers normally require to be represented by an equivalent-mesh circuit. This mesh circuit o o c c o o c c 0 0 (, (, 1) I o o BI BI (,e 0 0 A A c c o o E E ¢ ¢ Fig. Gambar. 3.3.3H B B 0 0 0 0 IJl IJl A A EI EI DI DI (,i Common-base equivalent circuits of four-winding and five-winding transformers with exciting impedance ignored

has one terminal per winding with every pair of terminals interconnected by an impedance. impedansi. Thus, a four-winding transformer would be represented by four terminals interconnected by six impedances and a five-winding transformer by five terminals interconnected by ten impedances. This form of representation is shown in Fig. pada Gambar. 3.3.3H. 3.3,4 Representation of off-nominal-ratio transformer circuits It has so far been assumed that the transformers are all nominal-ratio transformers, that is that the turns ratio of each transformer is equal to the ratio of the nominal rated voltages of its associated network sections. Put in another way, it has been

Page 43 Page 43Fault calculation Fault perhitungan 95 95 assumed that the nominal rated voltage of each transformer winding is identical to that of the network section to which it is connected. This assumption is commonly made in fault calculations but it is nevertheless sometimes necessary to consider cases in which the transformer ratio is different from this nominal value, the most common cause of such a difference being the ratio change produced by transformer tap-change equipment. A treatment of off-nominal-ratio transformers is given in Appendix 3.8.1. 3.3.5 Transformer phase-shifts The single-phase (reference-phase) representation of three-phase transformers, as required for the system positive-sequence network, has so far proceeded on the tacit assumption that all the transformer windings are star connected. It is necessary, Itu perlu, R (a) v v (b) (B) a(c) Fig. Gambar. 3.3.5A r (a) r"y"y'"y y(h) r*'T*'TY b (c) therefore, to examine the validity of this single-phase representation for those cases where other winding connections (for example delta and interconnected-star connections) are employed. The point at issue arises from the fact that, for a power system without transformers, the three system phases have an identity which extends throughout the network. With transformers present, however, the identity of a given phase conductor is limited to the particular network section concerned, this identity terminating at each and every transformer interconnection between that network section and the rest of the system. The single-phase representation is valid if all the transformers are star connected since the phase references a, b and c can then apply to the corresponding phases of all the network sections, the reference phase (phase a) then having an identity which extends throughout the network. These conditions are shown in Fig. 3.3.5A. Consider, however, the star-delta transformer shown in Fig. 3.3.5B in which the star-side and delta-side phase-terminals are denoted by the phase letters R, Y, B and

r, y, b, respectively, allocated as shown. The relationship between the star-side and the delta-side currents and voltages can be obtained by considering each limb of the transformer core and its associated two windings to constitute a single-phase transformer, and combining the delta-side phase-winding voltages and currents to

Page 44 Page 4496 Fault calculation obtain the phase-terminal values. The vector diagrams show the resulting relation- ship between the star-side and delta-side positive-sequence voltages under no-load conditions and that between the star-side and delta-side positive-sequence currents, exciting current ignored. The delta-side currents and no-load voltages, it wig be noted, all lag their corresponding star-side values by 30 ° , this phase shift being caused by the particular star-delta connection of the windings and the given alloca- tion of the phase references. It readily follows, therefore, that with the transformer providing the interconnection between two network sections, the effect of the star-delta connection as compared with the star-star connection is that the currents and voltages at any point in the delta-side network-section are all retarded in phase by 30 ° with respect to any reference vector in the star-side network. Now, all the R R Y Y B B rE1 V b V b No-Load Voltages 1 B 1 B Iy 0 I b illl illl l\x [ Currents Arus Fig. Gambar. 3.3.6B Phase-shifts introduced by a star/delta transformer

Page 45 Page 45Fault calculation Fault perhitungan 97 97 transformers in an interconnected power system must have compatible phase-shifts. In other words, the phase-shift introduced between any two network sections as a result of transformer winding.connections must be the same for all the inter- connecting paths between the two sections. This means, in effect, that the phase- shifts introduced by the transformer winding-connections have no effect on the magnitudes of the current and voltage values, these magnitudes being identical to those which would be obtained if all the transformer windings were star-connected. The only effect of the transformer winding-connections, therefore, is to shift the vector phase.positions of the currents and voltages in any given network section with respect to those in any other network section or with respect, for example, to the chosen common reference vector for the whole power system. The amount of Jumlah this phase-shift is known if the winding connections of the interconnecting trans- formers are known, the total phase-shift introduced by a number of transformers

in series being the algebraic sum of their individual phase-shifts. The effect of transformer winding-connections can now be seen to be readily obtained by simply analysing the system network with all the transformer windings treated as if they were star-connected (that is by using the single-phase equivalent circuits already described) and using the phase reference a, b and c to denote the phases of this hypothetical system, these phase references denoting actual phase conductors in one, or possibly more, of the network sections. The phase currents and voltages so obtained can then be corrected in phase, as necessary, to allow for the transformer phase-shifts, the resulting values then being correct in both magni- tude and phase. Thus, considering the particular case of a two-section system interconnected by the star-delta transformer of Fig. 3.3.5B, let it be assumed that the reference phase (phase a) has been chosen to correspond to phase R on the star side of the trans. former. mantan. Then, at any given point in the star-side network-section, the phase voltages and currents are given by vs = Vo zs = I° V y = a 2 V a I r = a2 Ia VB =aV° 18 =a/o where Va and Ia are the values of the reference-phase positive-sequence voltage and current at the point in question. The symbol a is the 120 ° operator already defined. Now the reference phase on the delta side of the transformer is a hypothetical con- tinuation of that on the star side based on the assumed star-star transformer-inter- connection and hence, for any point in the delta-side network-section, the phase voltages and currents are given by v, = 1/- 30°1I, Vy = 1[- 300a 2 Va Vb = 1/- 30*aV a i,= l/- 30% ly = 1/- 30°aZ la /b = 1/-3O°a/.

Page 46 Page 4698 98 Fault calculation Fault perhitungan where dimana V a and dan Ia IA are now the values of the reference-phase positive-sequence voltage and current at the point in question in the delta-side network section. The symbol a is the 120 ° operator, as before, the vector multiplier l/s_._* representing the necessary phase-shift correction. The phase correction required in any given network-section is determined from a knowledge of the winding connections of the transformers present in any inter-

connecting path between that particular network-section and the network-section containing the chosen system reference vector. The phase-shifts are normally known from the system phasing diagram but can otherwise be obtained from a knowledge of the vector-symbol reference (for example, Ydl, Dyl 1, Yy0, etc.)of the transformers concerned and their phase markings. The vector-symbol references are defined in British Standard BS 171:1978, the first (capital)letter denoting the connection of the high-voltage winding (Y for star, D for delta and Z for inter- connected-star), the second (small) letter the connection of the low-voltage winding using similar small-letter symbols (y, d and z) and the remaining figure (0, I, 6 or 11) being the clock-dial reference representing the hour position of the low-voltage phase-vector in relation to that of the corresponding (similarly lettered) high-voltage phase-vector, the latter being assumed to occupy the 0 (that is 12 o'clock) position. In determining the network-section phase-shifts due account must, of course, be taken of the actual phase references employed, usually red, yellow and blue, and their allocation with respect to the BS transformer phase-terminal references. The Itu latter are A, B and C for the high-voltage winding and a, b and c for the low-voltage winding and should not be confused with the general phase references a, b and c employed in the network analysis. In so far as balanced-load and three-phase short-circuit conditions are concerned, it is of interest to note that correction for transformer phase-shifts is not normally of any practical interest or value, since such correction merely ensures the correct relative phase displacements of the currents and voltages in different network- sections. bagian. The currents and voltages in any given network-section have correct phase angles with respect to each other whether phase correction is applied or not, since the effect of phase correction is merely to shift all the voltage and current vectors in any network-section by the same fixed amount. In analysing balanced-load or three-phase short-circuit conditions, therefore, phase correction is usually ignored and the network analysed by assuming star-connected transformer windings. As will be seen in later sections, however, the effect of transformer phase-shifts must be considered in obtaining the correct analysis of unbalanced-fault conditions. 3.3.6 Representation of synchronous machines The representation of a synchronous machine in the positive-sequence network is by means of the single-phase (reference-phase) circuit shown in Fig. 3.3.6A, the

Page 47 Page 47Fault calculation Fault perhitungan 99 99 Z Z Fig. Gambar. 3.3.6A Positive-sequence equivalent circuit ofa synchronous machine o o ru ru o o .... .... q.%%. JVVVVVUVUVVI --A -- Stead y-state va|ue • • L___ JV'J",

VVVUvvuuvv I h Saya h o o Steady-state value -----'u uvv Fig. Gambar. 3.3.6B Instantaneous phase-currents of a synchronous machine during a three-phase short-circuit at its terminals

Page 48 Page 48100 100 Fault calculation Fault perhitungan circuit comprising a positive-sequence driving.voltage in series with the machine positive-sequence impedance. The values of voltage and impedance to be employed in the representation, however, are dependent on the machine operating conditions at the instant in question and on whether the condition to be analysed is a steady- state condition or a condition following a sudden change in electrical conditions, for example a system short-circuit. The factors which determine the required representation can be understood by considering the effect of a three-phase terminal short-circuit on an unloaded machine as shown by the osciUograms of Fig. 3.3.6B, the field excitation voltage being assumed constant at the value required to produce rated voltage at the machine terminals prior to the application of the fault. The variation with time of Stead y-state value 0:2 0:2 oY3 0:, Time in seconds Fig. Gambar. 3.3.6C Variation with time of the rms, ac component of short=circuit current of Fig. Gambar. 3.3.6B the rmsac, component of the phase current is shown in Fig. 3.3.6C, this value being the same in each of the three phases. It will be noted that the reduction in the short-circuit current magnitude with time is exponential in form, and this reduction is caused by the effects of mutual coupling between the stator and rotor windings, as represented by armature reaction, and of similar coupling between each of these windings and another effective winding representing induced-current paths in the body and poles of the rotor and in any damping windings which might be present. hadir. The rms, current magnitude at any time t after the instant of short-circuit is given by diberikan oleh t t t t Est Et Et Tst Et Es Tt E s

I = I = e e + + e e +-- + - 3.3.6.1 Zst Zt Zt Zs Z s Z s and is seen to consist of a constant component and two exponentially-decaying components. komponen. The first exponentially-decaying component is a rapidly-decaying component of time-constant Tst and the second a more slowly decaying component of time-constant T t. The machine impedances Zs, Zt and Z s are termed the sub- transient, transient and synchronous impedances, respectively, and can be regarded

Page 49 Page 49Fault calculation 101 for most practical purposes as purely reactive, that is as/Xst,/Xt and/Xs. The Itu voltages Est, Et and E s are the voltages behind these impedances, their values being the vector sum of the machine terminal voltage and the appropriate impedance voltage-drop in the machine immediately prior to the short-circuit. The time- constants Tst and T t are termed the sub-transient and transient short-circuit time- constant, respectively, the former usually being of the order of twenty or thirty milli-seconds and the latter of the order of one or two seconds. For the case of a Untuk kasus short-circuit on an unloaded machine, the voltages Est, Et and E s would, of course, all be equal to the pre-fault value of the open-circuit phase-to-neutral terminal voltage of the machine. E s I"'.., p / "'.7 Fig. Gambar. 3.3.6D Phasor diagram for a round-rotor synchronous machine It will be noted from eqn. 3.3.6.1 that the current at the instant of short-circuit (t = 0) is equal to Est/Zst and is69thus determined by the sub-transient impedance only. saja. Similarly, the current under steady-state conditions (t' much greater than Tst and Tt) is given by Es/Z s and is determined by the synchronous impedancE only. If Jika the initial, rapidly-decaying, transient component is ignored, the Initial current is effectively given by Et/Z t and is thus determined by the transient impedance only. The vector diagram for a synchronous machine (cyliNdrical-rotoR type assumed) iS shown in Fig. ditunJukkan pada Gambar. 3.3.6D where V represents the phase-to-neutral terminal-voltage of the machine and I the current flowing out of the machine. In addition to the ac components of short-circuit current, two at least of the three phases will have an exponentially-decaying dc component of current as

shown by the off-set of the current waves with respect to the zero axis in Fig. 3.3.6B. The initial value of this dc component in any given phase depends on the instantaneous value of the phase voltage at the instant of short-circuit and is a maxi- mum (equal to the peak value of the initial ac component of the phase-current) when the instant of short-circuit (t -- O) coincides with an instant at which the voltage behind the sub-transient impedance is instantaneously zero for the phase in

Page 50 Page 50102 102 Fault calculation Fault perhitungan question. pertanyaan. Conversely, the dc component of current will be completely absent when the instant of short-circuit coincides with an instant at which the voltage behind the subtransient impedance is at its maximum (peak) value. The time- constant of the dc component of current is practically the same as the sub-transient short-circuit time-constant Tst. Because the dc currents in the rotor field-winding and damping-circuits are associated with ac machine-frequency currents in the stator windings, it follows that the transient dc component of current in the stator phases will be associated with a corresponding ac machine-frequency component of current in the rotor field winding and damping-circuits. The variation in the field-current of a short-circuited machine operating with constant field voltage is shown in Fig. ditunjukkan pada Gambar. 3.3.6E. If Jika Stead y-sigHt, value 0 0 0 I 0 2 0 2 0.3 0,3 0.4 0,4 I'Jlllt" ill sakit eCcbllds Fig. Gambar. 3.3.6E Field current of a synchronous machine during a three-phase short-circuit at its terminals terminal Effect of automatic voltage regulators: The consideration of synchronous- machine performance presented so far has assumed constant field voltage. In Di practice, however, machines are commonly equipped with automatic voltage regulators whose function is to control the excitation and thereby attempt to main- tain the machine terminal-voltage at a constant pre-set value. Under short-circuit conditions, therefore, the regulator will increase the field voltage in an attempt to restore the terminal voltage to its normal pre-fault value and will thus increase the short-circuit current to a higher value corresponding to the higher field current. The Itu effect of a typical regulator for a terminal three-phase short-circuit on an unloaded generator is shown in Fig. 3.3.6F, the variation in the short-circuit current with time being a function of the response characteristics of the regulator, the field- circuit time-constant and the maximum available field-supply voltage.

Saturation effects." The synchronous reactance of a synchronous machine is normally defined, in per.unit value (see Section 3.3.7), as the ratio of the field current required to produce rated armature current on sustained three-phase terminal short.circuit to that which would be required to produce rated armature voltage on no-load if no saturation were present, that is with the open-circuit- voltage/field-current characteristic assumed to be an extrapolation of its initial straight-line form. (See Fig. 3.3.6G.) The synchronous reactance is thus, by defini- tion, an unsaturated value and is therefore applicable to those steady-state condi- tions in which the main flux paths of the machine can be assumed to be unsaturated.

Page 51 Page 51Fault calculation 103 C C cJ Cj .m . M E E Fig. Gambar. 3.3.6F 6 6 Without AVR . . . . . . . . With AVR Generator excited to gnve rated voltage on No-Load ! ! 2 2 3 3 4 4 Time in seconds In the case of the sub-transient and transient reactance, the flux paths concerned are leakage-flux paths which are partly in iron and partly in air. These reactances are therefore not strictly constant but are dependent on the degree of saturation of the iron paths, it being necessary, for precision, to know the conditions for which the stated values apply. The values obtained from a terminal short-circuit test with rated pre-fault terminal voltage are known as the rated-voltage values of sub- transient and transient reactance, the values being saturated values because of the high value of armature current for this condition. Unsaturated values are obtained by performing the short-circuit test at reduced pre-fault terminal voltage, for example, at 50% of rated voltage. Quoted values of sub-transient and transient reactance can usually be assumed to be unsaturated values, unless otherwise stated, these values usually being some 10 or 15% higher than the saturated values. Salient-pole machines: The equivalent circuit of Fig. rangkaian setara Gambar. 3.3.6A is, strictly speaking, only valid for the case of round-rotor machines, there being no equivalent represen- tation for salient-pole machines, the analysis of which requires the use of the two-

Page 52 Page 52104 Fault calculation 1.5 1,5

m m := .o ¢-, E E .o.5 Open-circuit test • • .(o.. -y -Y If I Jika saya If 2 Field Lapangan current arus io io ._. . _. E E - - 0.5 0,5 r- r- P P Short-circuit test If3 Field Lapangan current arus Per unit synchronous impedance = lf3 ifl Short-circuit ratio = lf.-.2 l f3 Fig. Gambar. 3.3.6G Open-circuit and short-circuit characteristics of a synchronous machine axis theory. The given representation is, however, sufficiently valid for most practical cases of fault calculation provided the impedances used are the direct-axis values. nilai-nilai. 3.3.7 Use of per-unit and per-cent values The network analysis described so far has all been in terms of actual or referred values of voltage, current and impedance. In the great majority of cases, however, the plant impedance values are given in per-unit or per-cent value, and it is accord- ingly normally preferable to employ such values directly in the analysis rather than convert them to their actual or referred ohmic equivalents. Per-unit values: The basis of the per-unit method can be understood by consider- ing the application of Ohm's Law to a single impedance of Z ohms, the voltage-drop V in volts produced by a current of Iamps flowing through the impedance being given by diberikan oleh V=IZ 3.3.7.1 where V and I are vector values and Z is the complex impedance, all values being expressed by complex numbers. Now, letting Vbase and lt, ase be real numbers representing stated values of voltage in volts and current in amps, the above equa-

Page 53 Page 53

Fault calculation Fault perhitungan 105 105 tion may be written "I ' 1I'-1 = = 7= 7 = = = which, in per-unit notation is written 3.3.7.2 V pu = (I puX Z pu) 3.3.7.3 The per-unit values of voltage, current and impedance are thus defined by V V V pu = 3.3.7.4 Vse I Aku I pu = 3.3.7.5 Zp.u. =m = M -- - Z Z Zbase 3.3.7.6 where the base impedance is given by Vt, ase Zt, a = 3.3.7.7 It should be noted that the base values of voltage, current and impedance are all real numbers and it therefore follows that the per-unit values V p.tt, I pu and Z pu are complex numbers with the same arguments as V, I and Z. The per-unit values, being ratios of similar quantities (that is volts, amps or ohms), are dimen- sionless. The base values of voltage and current, which must of course be stated, can be any convenient values. In the analysis of three-phase systems the base voltage is always chosen to be equal to the rated phase-to-neutral voltaGe of thE73plant item or network section conCerned anD the base current, the current correspoNding to a73stated three-phase power-level (usually expressed in MVA), at this same rated voLtage. tegangan. In referring to individual plant items it is usual for the stated powerqevel to be chosen to Be equal to the load rating in MVA of the plant item concerned. Thus Demikian considering a 132 kV transmission line with a positive-sequence impedance of 0-076 +/0.379 I2 per km and a load rating of 175 MVA, it follows that 132000 Vt, ase = - - = 76 300 V 43 43 so that sehingga 175 x lO s

Itse = = 765.4 A /3 x 132000

Page 54 Page 54106 106 Fault calculation Fault perhitungan from which dari yang 76 300 76 300 Zte = = 99.7 I2 765.4 Hence, the per-unit impedance of the transmission line is given by 0-076 +]0-379 Zp.u. =' = ' 99.7 99,7 per km per km from which Zt,.u ' = 0.000762 +]00380 per km to the stated rating of 175 MVA. The base current for a given base voltage is directly proportional to the chosen MVA base and hence the per-unit impedance of any given item of plant is also directly proportional to the MVA base. Thus the per-unit impedance of the above- mentioned transmission line to a base of 100 MVA is given by 100(0-000762 +/0.00380) Zp.u. = = per km per km 175 175 giving pemberian Zp.u. = 0.000435 +/0.00217 per km to the new base of 100 MVA. Per-cent values: It is common practice in power system analysis to express plant impedances in per-cent value rather than per-unit value, the former being simply the latter value multiplied by 100. Expressed in per-cent values, Ohm's Law becomes I%Z V% = 3.3.7.8 100 100 where the per-cent values of voltage and current are similarly the corresponding per- unit values multiplied by 100. Thus, the transmission line referred to in the previous example has an impedance of 0.0435 + ]0.217 per-cent per km to a base of I00 /VIVA. Use of common-base per-unit and per-cent values: The use of per-unit or per- cent values of voltage, current and impedance in the analysis of a complete network requires that all such values should be referred to a common MVA base,just as the use of values expressed in volts, amps and ohms requires that these latter-mentioned values, should be referred to a common voltage base. Thus, consider any item of plant, say a transmission line, forming a part of a complete three-phase power system and assume it to have a rated line-to-line voltage of V, and VA rating of S

and ohmic impedance of Z 2 per phase. Then, as already shown in Section 3.3.2, this item of plant would be represented in the system positive-sequence network by the referred impedance value Z' = Z Z 3.3.7.9 where Vtzse is the line-to-line value of the chosen common base voltage. Expressing Mengekspresikan this referred impedance in per-unit value on its own three-phase VA load-rating S,

Page 55 Page 55Fault calculation Fault perhitungan 107 107 and remembering that this referred impedance value now represents an equivalent plant item with a rated line-to-line voltage of l/b, it is seen from eqn. 3,3,7,6 that where dimana Z pu = 3.3.7.10 v'b / 43 S S =/base 43Vba and dan Vbase[ x/3 = phase-to-neutral value of the base voltage Eqn. Eqn. 3.3.7.10 reduces to Z pu = Z Z 3,3,?.11 Eqn. Eqn. 3.3.7.11, it will be noted, is the per-unit value of the actual plant Impedance Z, thus showing that reference to a common voltage base is unnecessary when using per-unit or per-cent values. Consider, now, the representation of the given plant item in the system positive, sequence network. The positive-sequence network expressed in terms of per-unit or per-cent values of impedance must be identical except for a common multiplier, to the same network expressed in terms of referred ohmic impedances to a common voltage base Vbase. In other words, the impedances of corresponding branches in the two networks will be in the same constant ratio for all the branches. Now, from eqns. eqns. 3.3.7.9 and 3.3.7.11 it is seen that Zp.u _ S S 3.3.7,12 Z' Z ' V V which is seen to be a constant ratio if all the plant items are assumed to have the same VA load rating S. Conversion of all the per-unit or per-cent impedance values to a common MVA base is readily achieved bearing in mind the direct proportion, ality between MVA rating and per-unit or per-cent impedance. Thus, the required conversion of all the impedance values to a stated common/VIVA base, which carl be arbitrarily chosen, is given by

[ Z pu (or Z%) to ] =[CommonMVAbase][Zp.u.(or. Z)to 13,3.7.13 common MVA base | Plant MVA rating 1 [ plant MVA rating A frequently used value for the common MVA base is the MVA load rating of the

Page 56 Page 56108 Fault ca/cu/at/on largest individual plant item concerned or some standard value such as 100 MVA, 500 MVA or 1000 MVA appropriate to the system under examinaffon. 3,3,8 Fault-calculation procedure The analysis of a three-phase balanced-fault condition consists, in general, of three parts, the first being the representation of the given power system with its fault condition by the equivalent positive-sequence network, the second the solution of this network in terms of its common-base values of voltage, current and impedance, and the third the conversion of the resulting common-base values to actual values. The system positive.sequence network as already stated, is the equivalent single- phase (reference-phase) representation of the complete power system. In this Dalam hal ini network each component item of plant is represented by its equivalent positive- sequence circuit using common-base values of voltage, current and impedance, namely values to a stated common voltage base or, as is more usual, per-unit or pot-cent values to a stated common MVA base. Overhead.line and cable circuits: Overhead-line and cable circuits are represented by their equivalent-lt circuits, it being usually sufficient to employ the nominal-It circuit in which the series arm represents the total series impedance of the circuit (k + j,.)L)I c .......... c .......... t t ; ; i aku -2j I wC! . . . . . . . . ii ii (a) Nominal - representation perwakilan A A -2J (.'1 (R + jL)I ¢ ....... I Aku I Aku 0 0 III III (h) Simple series-circult representation Fig, 3,a,SA Equivalent circuits of an overhead-line or cable circuit of length I and parsrnete R, L and C per unit length

Page 57 Page 57Fault calculation 109 concerned and each of the two shunt arms the impedance corresponding to one half of the total phase-to-neutral capacitance. The shunt-arm impedances are always large in comparison with the series-arm impedance, and representation by the series arm alone is usually sufficiently exact for most practical purposes, particularly for overhead.line circuits. Transformers and synchronous machines: Transformer and synchronous machine impedances are predominantly reactive with X/R X / R ratios typically between ten and twenty, or sometimes more. It is, therefore, usually sufficiently exact to ignore the resistive component of the impedances and to assume all the impedances to be purely reactive. Loads: Load impedances are always large in value in compariaon with the series impedances of the power system plant and they therefore have only a small effect on the value of the total fault current under short-circuit conditions. Load can therefore be ignored in the majority of short-circuit calculations unless the analysis is particularly concerned with the combined effect of load conditions and short, circuit conditions. Transformer tap-position: For the great majority of fault calculations it is usually sufficient to ignore actual tap positions and to assume all the transformers to be operating on the nominal-ratio tap-position, the error introduced by this assumption being normally quite small insofar as values of total short-circuit current are concerned. The use of the exact off-nominal-ratio representation, referred to in Section 3.3.4, has many applications in practice, however, where it is required to determine the precise division of load or fault current between transformers operating on different taps. Equivalent sources: The representation of a complex power,system network can often be simplified considerably by the use of an equivalent generator to represent the whole or certain parts of a given network. Thus, a complete network, as seen from any given point, may be represented, using Thevenin's Theorem, as a single driving voltage in series with a single impedance. This equivalent-generator circuit can often be obtained with sufficient accuracy from an estimated knowledge of the system three-phase fault-level at the point in question, the pre-fault value of the voltage at this point being assumed equal to the nominal rated value. Treatment of complex impedances: The impedances in the positive,sequence network are all complex impedances and must, therefore, be represented in R +iX form. formulir. In many cases, however, the resistance components of the impedances are small compared with the reactance components and in such cases it is often sufficient to treat the impedances as pure reactances, thus ensuring a considerable simplification in computation. The use of such a pure-reactance form of representa, tion, it should be noted, results in a short-circuit current slightly greater than the

true value. nilai benar. Plant-impedance values: The impedance values employed in any particular fault calculation should, wherever possible, be the known values appropriate to the particular items of plant concerned. Where precise actual values are not known, however, it may be permissible to use typical values appropriate to similar plant of

Page 58 Page 581 I0 Fault calculation Fault perhitungan similar load and voltage rating. Tables of typical plant impedance values are given in Section 3.4.3. Neutral earthlng: Neutral earthing arrangements have no effect on balanced three-phase conditions, whether load conditions or three-phase short-circuit condi- tions, and are therefore disregarded in the derivation of the system positive-sequence network. jaringan. 3.3.9 Example 1 The power system shown in Fig. 3.3.9A develops a three-phase short-circuit on the 132 kV busbars at Station C. Determine the resulting value of the three-phase fault current and its distribution in the 132 kV system, given that the pre-fault value of Zg = j!8.0% Z t = ji2.S% Z t = jlO.0% Z s = j17.5% II kV I00 MVA AJ 132 kV |B 60 MVA II kV I00 MVA J[ J [ l l 60 MVA 4.6 + j26.0. 2.3 + jl 3.0 FJgJ3.3.gA the line-to.line wltage at the point of fault is 140 kV. The system can be assumed to be operating at no4oad prior to the fault. Solution: Solusi: Using a common voltage base of 132 kV, the resulting positive-sequence network is as shown in Fig. 3.3.9B, the generator and transformer impedances having been converted to equivalent 132 kV 2. Thus, considering the 100 MVA generator, its rated current is given by 100 100 x x 106 106 I = I = = 5250 A × 11000

and its rated phase-to-neutral voltage is given by 11 11 000 000 V = V = = 6350 V 45 45 from which the corresponding base impedance is seen to be 6350 6.350 Z Z = = = 1.21 5250 5.250

Page 59 Page 59Fault calculation 111 80.8 kV @ @ j3! .41Z r---1 80,8 kV kV j21.8 6.9 + j39.0 j29.1 l j50,912 l " A I/B II ' Illl' '' 4.6 +j26,012 C" 2.3 +jl3.02 I Aku Fig. Gambar. 3.3.9B The ohmic impedance of the generator is therefore given by Z =/'0.180 x 1.21 =/0.218 and the value referred to the common baSe voltage of 132 kV is given by = '-/ j0"218 =J31"42 Because the pre-fault conditions have been assumed to be no-load79eondhions, the generator driving-voltages, referred to the coMmon baSe79voltage of 132 kV, are both given by diberikan oleh 140 000 140 000 E = E = = 80 800 V B B j53.22 4 . 6 6 +j26,0 Fig. GambaR. 3.3.9C

Page 60 Page 0112 Fault calculAtion

this voltage being used as the vector base. Because the two drIving Voltages are equal they can be connected in parallel, 4hus giving the equivalent circuit sHown in Fig. Gambar. 3.3.9C. B B j80.Oll.I 5 + J6.SOl j53.21 , , 2.30 + jl 3.O80A A Fig, 3.3.9D ThE delta-connected line-impedances 2.3 +/13.0, 4.6 +/26.0 and 6.9 +/39.0 can now be replaced by the equiValent Star-connected impedances shown in Fig. 3.3.9D. Thus, the impedance conneCting StAtion C to the star point is given by ,(2.3 +/13.0X4.6 +/26.0) = 0.7780+/4.33 12 (2,3 +/13.0) + (µ.6 +j26.) + 6.9 +/39.0) udng eqn, 3,2,5.8, the two reMaining impedances connecting Stations A and B to the star point being sImilarly obtained. The80remaiNing steps in the reduction process are shown in Fig. 3.3.9E From whicH the resultant three-phase fault current is seen to80be given by 80 800 If" = 79. -/1930 A 1.72 +/41.8 ving Ving80left 1932/-- 87 ° 40' A The80distribution of the fault current iN the power system can now be obtained by the Process of back-substitUtion. This cOnsists of retracing the steps in the Reduc- tion pRocess starting with the total fault cur2ent and dividinG this faulT currEnt among the dIfferent branches by straightforward application of the network laws. Thus, the curren4 fed Into the 132 kV system by the generatOr/transformer unit at

Page 61 Page 61FAult calculation Fault peRhitungan 113 113 Station A i380given by (1.15 +]86.5X79.7 -/1930) (1.15 +/86.§) (2.30 /66.2) = = 55.3 55,3 -80- 11094 A = 1095 [- 87 ° 7' A80and hence the phase-to-neutral voltage of the reference phase at Station A Is given by oleh 80 800 -/53.2(55.3 -/1094) = 22 550 -/2944 V

= 22 720 [ - 7 ° 26' V 1.15 + j86.5 80.8kV 0.77 + j4,3311 2.30 2,30 + + j66.21Z 80.8 80,8 kV kV 0.954 + j37.512 0.77 + j4,33 f----] 80.8 80,8 kV kV 1.72 1,72 + j41.82 f-"l C Jplf Fig. Gambar. 3.3.9E Similarly, the current fed into the 132 kV system from Station B is given by (79.7 -/1930) - (55.3 -/1094) = 24.4 - 1836 A = 836 [ - 88* 19' A

Page 62 Page 62114 Fault calculation giving a voltage at Station B of 80 800 - ]80.0(24-4 -/836) = 13 920 - ]1952 V = 14050 [- 7 ° 58'V Now, since the voltage at Station C is zero because of the fault, the current flowing from Station A to Station C is 22 550 -/2944 - - 38.9 -/.860 A 4.6 +/26.0 = 861 [-87 ° 25' A Also, the current flowing from Station B to Station C, by subtraction, is (79"7 -/1930) - (38.9 - j860) -- 40.8 -/1070 A = 1071 [- 87 ° 49'A Fig. Gambar. 3.3,9F 109 [ 7-87071A 235 [ -85°SgPA 836 [ -88°19 / A "-- '-!A •I ] ( ---a 861 [ -87°25 t A • C • C t 1071 / -870491A If= 1932 [-87°40 tA Finally, the current flowing from A to B is (55.3 - ]1094) - (38.9 -/'860 = 16.4 -/.234 A =235[-85"59'A

The required fault-current distribution is thus as shown in Fig. 3.3.9F. The voltage at the generator terminal at Station A, to the common base voltage of 132 kV, is 80 800 - ]31-4(55.3 - ]1094) = 46 430 - j1737 V the actual value, transformer phase-shift ignored, being 11(460430 - ]1737) 132 132 - 3870- j145 V = 3870 [- 2 ° 8'V

Page 63 Page 63Fault calculation 115 Similarly, the actual generator current at Station A, transformer phase-shift ignored, is adalah 132(55.3 -/1094) 11 11 = 663.5 - ]13 130 A = = 13 150/- 87 ° 7' A The corresponding values of voltage and current for the generator at Station B can be similarly shown to be 3190 -/103-6V (=3190 [ - 1 ° 51' V) and 293 -/10030 A (= 10 030 L- 88 ° 19' A), respectively. 3.3.10 Example 2 The power system shown in Fig. 3.3.10A develops a three-phase short-circuit at the point F on the hv terminals of one of the two auto-transformers at Station B, the pre-fault voltage at the 132 kV busbars of Station B being 138 kV. The load supplied by the 132 kV busbars can be assumed to be passive (that is, to contain no 120 MVA 275/132 kV ji so% 210 MVA 200 MW 16.5/275 kV 5.72 + j24.M 150 MVA Zst -- j18.2% l 5.72 +j24.8 Z s = j143% , , 138 kV 120 MVA Fig. Gambar. 3.3.10A 275/132 kV jis .0% driving voltages) and to be representable by a constant impedance. Determine the Tentukan value of the short-circuit current and the network branch currents at the instant of fault. kesalahan. Determine also the final steady-state value of the fault current and the net- work branch currents assuming the fault to remain on the system and there to be

no voltage-regulator action. Solution: Solusi: Using the per-unit method and a common base of 200 /VIVA, the system positive-sequence network for the pre-fault load condition is as shown in Fig. Gambar. 3,3.10B, the branch impedances having all been converted to common-base per-unit values. The generator driving-voltage and impedance depend on the fault condition being investigated, as already mentioned, and the values are therefore denoted by per-unit values Eg Misalnya and Zg as shown, the symbol Vg denoting the generator terminal voltage.

Page 64 Page 64116 Fault calculation The conversion to common-base per-unit values has already been described in Section 3.3.7. Thus, the impedance of the generator step-up transformer, namely /14.0 per-cent on its own rating of 210 MVA, becomes 200 x/0.140 =/0.133 pu on 200 MVA 210 210 Similarly, the impedance of each auto-transformer, namely/15.0 per-cent on their rating of 120 MVA, becomes 200 x/0.150 =/0.250 pu on 200 MVA 120 120 The 275 kV transmission-line impedances, each of which corresponds to 76.5 km of a sebuah twin-conductor-bundle line design, are given in ohmic value in Fig. 3.3.10A and these have been convERted by83the use of eqn. 3.3.7.11, the resulting per-unit impeDanCe of each line bEin' given by 200 200 x x l0 s x X 5-72 +i2µ.8) (2750 =8301§ i0.05§ pu to the CoMmon83base of8320083MVA. NOw,83the currEnt83tAKEn by tHe83load oF 1§ MVA83at tHe StaTed83prefaULt83voLTaGE83Of 13883kV83iS 5 x83106 628 A 1

732 x84000 And the cUrrent coRrESpONdinG84To the base MVA of 200 at a rateD voLtage oF 12 kV is 00 x 106 =84875 A 1.32 x 32000 HEncE, the scalar value84of the load currenT in per-unit value iS84628842 0.18 pU on 200 MVA 875 875 The scalar value of the vOltage across the load

namely 38 kV, is8513 138 =851

0µ5 pu 132 132 and hence the scalar value86of the86load86impedance is givEn by 1.045 1,045 = 1.455 pu on 200 MVA 0-8

Page 586PagE 6586FaulT calculation8611 NOW the power faCtOr86of86the load is 0.850, lAgging, correspoNding86to a loaD- imPedance phase-angle of 31.8, the sine aNd co3ine of this phase86aNgle beiNg 0.52 and .85 respectively. The cOmplex value86of tHE lOad86impedance iS therefore given by oleh 1.455(0.850 +]0.527) = 1.236 +]0.767 pu AdoptinG86the pre-fault refeRence-phase voltage at the 32 KV busBar aS the86vector bAse, its compLex value thus being 1.045 + ]0 pu, the current supplied to the load is given by diberikan oleh 104§ 145 = 018/- 31"8 ° 1.455 [ 31.8 ° = 0.610 -/0.379 pu This is also the current supplied by tHe Generator through its step-up transform%R, the86curreNt dividing equally between the TWo transformer-feeder Circuits. O.OlS + jO.O6SS jO.SO AZ g j0.3 B !'236 + j0.76 0.015 + j0.06SS86jo.2so Vg = 1.160 +j0.175 0.610 - j

379 F)g. Gambar. 3

3.10B The geNerator terminal voLtage is given by88Vg = (1.0µ5 +/0) [(0-015 /0.31§5) +/0.133]88(0.610 - /0.379) GiVinG pemberian Vg = 1.160 +/0.175 pu The pre-fault condi4ions are Thus as shown in88FIg. 3..0B, and the Fault aNalysis cAn now proceed by appLying the fault connection between the pOint of fault F AnD the neutral bAr wiTh appropriate values of Eg and Zg for the conditioN under investigation. penyelidikan. Considering, first, the cOnditions88at88the instant of short-circuit, The required generatOr impedance is the sub-transient value, nameLy/18.2 Per-cent on the machine ratinG of 222 MVA, the required common-base per-unit vaLue being 200 xj0.182 222 22 =]0.164 pu on 200 MVA

PAge 66 Page 66118 118 Fault calculation Fault perhitungan THe generator driviNg-voltage is the pre-fault valuE88of the generator termInal-voltage88plus the pre-Fault impedance droP in the machine sub-transient impedance, giving Eg = Est = (1

60 +/0.175) +/0.164(0.610 -/0.379) from which dari yang Eg = 1.222 +]0.275 pu89The Positive-sequence network for the conditions at the instant of fault is shown in Fig. Gambar. 3.3.10C and this can bE SOlved foR the89network branch currents By tHe process of netwoRk Reduction And back-substitution, the reduction procedure89being showN in Fig. pada Gambar. 3.3.10D. 0.15 + j0.0655 j.250 1.222 1,222 +89+ o.275 !"-"-! o o / / jO.164 jO.133891.236 + j.767 C"-Iof A 8 8 ,""'} 0.015 j0.0655 j0.250 I89Aku ; ; ; ; I Aku89F I If FIg. GAMBar. 3.3.10C From Fig. DaRi GambAr

3.3

10D, the given netWOrk is SEen to reduce To the91generAtoR drivinG- Voltage Est in series WiTh the single imPedanCe 0.0124 +]0.35691pu and, Hence, the generatOr current is given91by 1.222 +/'0

75 =920.892 -¤3.405 pu 0.0124 +/0.356 ApplYing the ProceSs oF back-substitution to obtain the remaining branch currents, the voltage92at The point A is92obtained from VA = (0.892 - ]3.405) (0.0124 +/0.0590) giving pemberian V.4 = 0.2105 +/0.0983 pu The Current fLowing frOm A to F through the 275 kV Line associated With the faUlted transforMer can now be ObtaIned and is 0.210§ +/0.00983 0.015 +/0.655 = 0.842 - 3.20 pU The currenT flowing FrOm A to B tHroUGh the othER 275 kV line is the generAtoR Current mInus the current jUst obtaiNed and is given by

Page926792Page 67Fault calculation 11 1.22 + j0.75 j0.297 i aku I Aku ; ; 0.05 + j.5§ 0.05 0,015 + j0.3155 ["-'3 j0.250 B i.236 + j0.767 1.222 + jO.275 0.015 + jO.0655 0.OI5 + j0.3155 i""'1 0.0302 + j0.2252 B B 1.222 + j0.275 jO.297 "i t t ; ; 0.015 + jO.0655 0.0452 + j0.5407 | | I Aku 1.222 + j0.275 __ __ j0.297 i aku f f

; ; 0.0124 + j0.0590 1.222 + j0.275 0.0124 + jO.356 - - ; ; 1 1 lg lg Fig. Gambar. 3.3.10D (0.892 - j3.405) - (0.842 - j3.020) = 0.050 - j0.385 pu The voltage at B can now be obtained and is given by (0.2105 +/0.00983)- (0.050- ]0.385) (0.015 +]0.3155) = 0.0882 - ]0-00027 pu The current flowing to the fault through the transformer on the faulted circuit is

Page 68 Page 68120 120 Fault Kesalahan calculation perhitungan given by diberikan oleh 0.0882 0,0882 -/0.00027 -- - 0.001 - 0-353 pu; i0-250 and the total fault current at F is thus: /f = (0.842 - ./3.020) + (- 0.001 -/0.353) from which dari yang I/= 0.841 - i3.373 pu The base current corresponding to 200 MVA at 275 kV is given by 200 200 xlO 6 -= 420 A A 1.732 x 275 000 and hence the fault current in amps is given by I! Aku! = (0.841 -/3.373)420 giving pemberian I/= 353 -/1416 A = 1460 / - 76.0°A The fault current and the branch currents are shown in per-unit value in Fig. Gambar. 3.3.10F. Considering now the determination of the final steady-state values of the fault 1.222 + j0.275 -@ - @ [----U'--1 ;.-..J I Aku • • ] ] o.oso - j0.385

/ / - - / oo , - jo.o, '> "> ; 2t--rg--f-'--. - - 0.842 -j3.0 0 ' [ -0.001 -j0.353 , If = 0.841 - j3.373 Fig. Gambar. 3.3.10E 163 / -82.6 ° ° ll.9/J2.7°kV • > A A s2"s / 32'1° [- [- 148F / -9B. " " If = 1460[--76.0 ° Fig. Gambar. 3.3.10F

Page 69 Page 69Fault calculation Fault perhitungan 121 121 current and the network branch-currents, the generator impedance applicable to this condition is the synchronous value, namely/143 per-cent on the machine rating of 222 MVA. The required common-base per-unit value is thus 200 x/1.43 222 222 =/1.288 pu on 200 MVA The required generator driving-voltage is the pre-fault voltage behind the synchronous impedance (the excitation being assumed to remain unaltered) and this is given by Eg = E s = (1.160 +/0.175) +/1.288(0.610 -/0.379) from which dari yang Eg Misalnya = 1.648 +/0.961 pu The positive-sequence network for the final steady-state condition, therefore, is as shown in Fig. 3.3.10G and this can be solved for the network branch currents, as before, by the process of network reduction and back-substitution. Because the Karena 0.015 + j0.0655 j0.250 1.648 + j0.961 } ) l l jO.13 1.236 ÷ j0.767 ' ' A A B, B,

f--'l lO.OJ s jo.o6ss jo.2so Fig. Gambar. 3.3.10G positive-sequence network external to the generator terminal is the same as for the previous case, its equivalent impedance as seen from the generator terminal is the same as before, namely /0.133 + (0.0124 +/0.0590) -- 0.0124 +/0.192 pu The equivalent positive-sequence network, as seen from the generator, is therefore as shown in Fig. 3.3.10H and comprises a driving voltage of 1-648 +/0.961 pu in series with an impedance of /1.288 + (0-0124 +/0.192) = 0.0124 +/I480 pu

Page 70 Page 70122 122 Fault calculation Fault perhitungan 1.648 + j0.961 0.0124 +j1.480 lg lg Fig. Gambar. 3.3.10H The current supplied by the generator, therefore, is given by 1.648 +/0.951 = 0.659 -/1-108 pu 0.0124 +/1.480. and, using the process of back-substitution, the network branch currents are found as shown in Fig. seperti ditunjukkan pada Gambar. 3.3.101, the fault current at F now being given by If = 0.637 -/1.101 pu ,6.+,090, : : > > : : , , , , 0,057 --j0.132 / / t t 1 I 1 saya 1 1 "" "" 0.659 -- j1.108 10.022 -- jO.O07 • O ; ; <" <" oo -j 35 - jo.l:5 I Aku ll = 0.637 - jl.lOl

- - i aku Fig. Gambar. 3.3.101 The fault current value in amps, therefore, is given by If = (0.637 - i1.101)420 giving pemberian If = 268 - ]463 A = 535 [ - 60.0°A The values of the network branch currents in amps are shown in polar form, in Fig. Gambar. 3.3.10J. It should be noted that the results obtained by the method of solution employed above could have been obtained equally well by the use of Thevenin's Theorem and the Superposition Theorem. Thus, the impedance of the network of Fig. 3.3.10C, as measured between point F and the neutral bar with the fault connection removed and the driving voltage Est short-circuited, is 0.0549 + ]0-310 pu, as can be seen from the network reduction shown in Fig. 3.3.10K. Also the pre-fault

Page 71 Page 71Fault calculation 123 () () 18.2 [ 30.2__ kV Q Q ol s I Aku ...s.='_' > > Fig. Gambar. 3.3.10J I'Z ! ! -,6 6' ,'" I " - k"A 1' ' '-' 6.'" I-"--"1 LLJ-t" ..... ..... ; ; I Aku If = 535 [ --60.0 ° voltage Ef at the point of fault F is given by Ef = (1.045 +i0) + 0.500(0.610 -/'0.379)(/0.250) giving pemberian Ef = 1.092 +/0.076 pu Hence, applying Thevenin's Theorem, the fault current for a short-circuit at F is given by diberikan oleh 1-092 +/0.076 0.0549 +/'0.310 giving pemberian If = 0.841 -/3.373 pu This is seen to be identical to the value obtained by direct solution for the network branch currents.

Knowing the total fault current and the pre-fault branch currents, the network branch currents with the fault condition present are readily obtained by use of the Superposition Theorem as shown in Fig. 3.3.10L. Thus considering the pre-fault load condition represented by diagram (a) of this figure let E¢ denote the pre-fault voltage at the point of fault, the value of Ef being 1.091 + j0.076 pu as already computed. dihitung. It will be evident that the network branch currents in diagram (a) will be undisturbed if a fictitious driving voltage of value Ef Ef is assumed to be connected between the neutral bar and the point F, the direction of El being from the neutral bar to the point F. The pre-fault load condition indicated by diagram (a) can there. fore be regarded as caused by the joint action of the driving voltages Est and El. By Oleh Thevenin's Theorem, however, the fault current at F is obtained by applying a driving voltage Ef between the neutral bar and the point of fault F, a shown in diagram (b), with the driving voltage Est removed (that is short-circuited), It should be noted in this diagram that the voltage Ef acts from the point F to the neutral bar, as is necessary to obtain the correct direction for the fault current, this being the same as an applied driving voltage of -El acting from the neutral bar to the point of fault. Diagram (c) represents the required fault-plus-load condition for wtdeh it is required to determine the values of the network branch currents and it will be noted that this corresponds to diagram (a) with the assumed fictitious driving

Page 72 Page 72124 Fault calculation 0.01S + j0.0655 j0.2S0 jo.3131 c"-! j0,164 : /- -:' .A B B 1 1 0.015 0,015 + j0.0655 j0.250 I Aku I Aku o--1 "'1 F F 1.236 + j0.767 ! ! ! ! j0,297 0.0075 + j0.0345 j0.125 1.236 + j0.767 A A

S S B B F F iii II ]]I IIII 0.0047 + j0.02617 |,236 + j0.892 o.oo,, • j . j. 98s t I Aku II II III III III III I Aku I Aku Fig. Gambar. 3,3,10K 0.0047 + j0.02617 0.0549 +jO.310 t t I Aku o o f'7"'l F' F ' voltage E/short-circuited. Let the current in any given branch in diagrams (a), (b) and (c) be denoted by Ikd, licit and I/oadarault, respectively. Then by the Superposition Theorem it readily follows that: lJoad = IIoad-pfault - [fault

Page 73 Page 73(a) (A) 1.222 + jO.27S .610-j0.379 Fault calculation Fault perhitungan 0.305 -- jO., 89 B 3 , 9 9 0.305 0,305 j0.189 FIE fA = 1,092 + jO,O'/6 / / 125 125 (b) (B) [ [ : > , , ) ) 0.282 - j3.026

[--'1 > ; I --0.255 -- j0,196 I Aku ^ ^ B < :'I F" F " ,. ,. . . - / °'$59-j°'347 0.S37 - j2.831 I Aku 1,222 +j0.275 ] ] ; J [ - - " / / O.OSO - jO.38S / / 89 - j3.4os| /o.os,- jo,o32 I -->' ;<" I Aku (c) (C) °"'-'"°'° I-°°°' -'°"" I Aku Fig. Gambar. 3.3.10L from which the required value of branch current for the load.plus-fault condition is given by: diberikan oleh: load.ofauu = Ik, ad + Ifau# This relationship is true for each branch of the network as is easily soon from Fig. Gambar. 3.3.10L. The use of the Superposition Theorem in this manner is particularly useful in cases where fault analysis is required to take due account of a known pre. fault load condition. Next Page Next Page