{"id":1696,"date":"2018-01-11T14:15:17","date_gmt":"2018-01-11T11:15:17","guid":{"rendered":"http:\/\/mekanikelektriktesisat.com\/?p=1696"},"modified":"2018-01-11T14:15:17","modified_gmt":"2018-01-11T11:15:17","slug":"gerilim-dusumu","status":"publish","type":"post","link":"https:\/\/benga.pro\/index.php\/2018\/01\/11\/gerilim-dusumu\/","title":{"rendered":"Gerilim D\u00fc\u015f\u00fcm\u00fc"},"content":{"rendered":"

Gerilimde bir d\u00fc\u015fme olmasa ne g\u00fczel olurdu. \u0130ncecik kablolarla kayna\u011f\u0131n gerilimini istedi\u011fimiz kadar uza\u011fa ta\u015f\u0131r, en ufak bir enerji kayb\u0131 da ya\u015famazd\u0131k.\u00a0 Ama maalesef b\u00f6yle bir d\u00fcnya yok. En az\u0131ndan s\u00fcper iletken teknolojisi ilerleyene kadar bu gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131n\u0131 yapmaya devam etmek zorunday\u0131z.<\/p>\n

Bildi\u011finiz gibi bir gerilim kayna\u011f\u0131n\u0131, bir t\u00fcketece iletkenle ba\u011flar\u0131z. \u0130letkenler de neticede bir \u00f6z empedans i\u00e7erirler ve t\u00fcketecin ihtiyac\u0131 olan ak\u0131m iletkenden ge\u00e7erken \u00f6z empedanslar\u0131 oran\u0131nda enerjiyi kullan\u0131rlar. Is\u0131 ve manyetik alan olu\u015fturmak i\u00e7in kulland\u0131klar\u0131 bu enerji\u00a0 kimsenin i\u015fine yaramad\u0131\u011f\u0131 i\u00e7in kay\u0131p enerjidir. Durum b\u00f6yle olunca kablo tip ve kesit\u00a0 se\u00e7iminde bu kaybolan bu enerjinin minimum olmas\u0131 kriteri g\u00f6z \u00f6n\u00fcne al\u0131n\u0131r.<\/p>\n

Mevzuat\u0131m\u0131zda Gerilim D\u00fc\u015f\u00fcm\u00fc<\/h4>\n

Elektrik \u0130\u00e7 Tesisleri Y\u00f6netmeli\u011fi Madde 57’de “\u0130letken ve kablolar\u0131 boyutland\u0131r\u0131lmas\u0131” ba\u015fl\u0131kl\u0131 bir k\u0131s\u0131m vard\u0131r. Anala\u015f\u0131laca\u011f\u0131 \u00fczer bu k\u0131s\u0131mda kablolar\u0131n nas\u0131l boyutland\u0131r\u0131laca\u011f\u0131 madde madde belirtilmi\u015ftir. Bu maddelerin 3.s\u00fcnde gerilim d\u00fc\u015f\u00fcm\u00fc anlat\u0131lmaktad\u0131r.<\/p>\n

Y\u00f6netmelikte \u015fu ifadeler vard\u0131r:<\/p>\n

i) \u0130\u00e7 tesis hatlar\u0131nda s\u00fcrekli en b\u00fcy\u00fck i\u015fletme ak\u0131m\u0131 ile i\u015fletme gerilimine g\u00f6re y\u00fczde gerilim d\u00fc\u015f\u00fcm\u00fc,<\/em><\/p>\n

Yap\u0131 ba\u011flant\u0131 kutusu ile t\u00fcketim ara\u00e7lar\u0131 aras\u0131nda, a<\/em>yd\u0131nlatma ve priz devreleri i\u00e7in % l,5’i, m<\/em>otor devreler i\u00e7in % 3 ‘\u00fc, <\/em>ge\u00e7memelidir.<\/em><\/p>\n

ii) Yap\u0131n\u0131n yada yap\u0131 k\u00fcmesinin beslenmesi i\u00e7in bir transformat\u00f6r\u00a0<\/em>kullan\u0131lm\u0131\u015fsa, bu transformat\u00f6r\u00fc \u00e7\u0131k\u0131\u015f u\u00e7lar\u0131 ile yap\u0131 ba\u011flant\u0131 kutusu aras\u0131ndaki gerilim d\u00fc\u015f\u00fcm\u00fc % 5’i ge\u00e7memelidir. (<\/em>A\u00e7\u0131klama : Gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131, gerekli g\u00f6r\u00fcld\u00fc\u011f\u00fcnde g\u00f6r\u00fcnen g\u00fc\u00e7 g\u00f6z \u00f6n\u00fcne al\u0131narak yap\u0131lmal\u0131d\u0131r.)<\/em><\/p>\n

Bu durumda, kendi trafosu bulunan tesislerde toplam gerilim d\u00fc\u015f\u00fcm\u00fc, ayd\u0131nlatma ve priz devreleri i\u00e7in %6,5, motor devreleri i\u00e7in %8 olabilir, sonucu \u00e7\u0131kmaktad\u0131r.<\/p>\n

\u0130kinci maddeye bir a\u00e7\u0131klama ifadesi eklenip, gerekli g\u00f6r\u00fcld\u00fc\u011f\u00fcnde g\u00f6r\u00fcnen g\u00fc\u00e7 g\u00f6z \u00f6n\u00fcne al\u0131narak hesap yap\u0131lmas\u0131 gerekti\u011fi vurgulanm\u0131\u015ft\u0131r ama o gerekli haller anlat\u0131lmam\u0131\u015ft\u0131r. G\u00f6r\u00fcn\u00fcr g\u00fcce at\u0131fta bulunuldu\u011funa g\u00f6re X, yani reaktans de\u011feri akl\u0131m\u0131za gelmelidir. Yani devremizdeki\u00a0 reaktans de\u011feri y\u00fcksekse (ki bu daha \u00e7ok b\u00fcy\u00fck kesitli ana besleme kablolar\u0131nda ve orta gerilim \u015febekelerinde dikkate al\u0131nmas\u0131 gereken noktalara ula\u015f\u0131r) gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131nda reaktans\u0131 da ihmal etmeden i\u015flemler sokmal\u0131y\u0131z.<\/p>\n

iii) Elektrik i\u00e7 tesislerinde gerilim d\u00fc\u015f\u00fcmlerini hesaplanmas\u0131nda a\u015fa\u011f\u0131daki form\u00fcller kullan\u0131labilir:<\/em><\/p>\n

-Bir fazl\u0131 alternatif ak\u0131m tesislerinde:<\/em>
\nAk\u0131m biliniyorsa : u = 2 L. I.cosp \/\u00a0X.S<\/em>
\nG\u00fc\u00e7 biliniyorsa : u = 2 L. N \/ X. S. U.<\/em>
\nYada y\u00fczde gerilim d\u00fc\u015f\u00fcm\u00fc olarak : % e =2 . 100 .L . N \/\u00a0X . S .U2<\/em><\/p>\n

-\u00dc\u00e7 fazl\u0131 dengeli y\u00fckl\u00fc alternatif ak\u0131m tesislerinde :<\/em>
\nAk\u0131m \u015fiddeti biliniyorsa : e =\u00a0l,73 L. I.cos p \/ X.S<\/em>
\nG\u00fc\u00e7 biliniyorsa : e =\u00a0L.N \/ X .S. U yada % e =\u00a0100. L. N\/ X .S U<\/em><\/p>\n

Yukar\u0131daki form\u00fcllerde <\/em><\/p>\n

e: Gerilim d\u00fc\u015f\u00fcm\u00fc (Volt)<\/em>
\n(Bir fazl\u0131 hatlarda gidi\u015fi ve d\u00f6n\u00fc\u015f iletkenleri \u00fczerindeki; \u00fc\u00e7 fazl\u0131 hatlarda ise yaln\u0131zca faz iletkeninin \u00fczerinde gerilim d\u00fc\u015f\u00fcm\u00fc hesaplanacakt\u0131r.)<\/em>
\nL: Hat uzunlu\u011fu (metre)<\/em>
\nI: Ak\u0131m \u015fiddeti (amper)<\/em>
\nU: \u0130\u015fletme gerilimi (\u00fc\u00e7 fazl\u0131 \u015febekelerde faz aras\u0131 gerilimi) (volt)<\/em>
\nCosp : g\u00fc\u00e7 katsay\u0131s\u0131 (omik y\u00fcklenmede ve do\u011fru ak\u0131mda cosp=1 al\u0131n\u0131r)<\/em>
\nN: G\u00fc\u00e7 (Watt)<\/em>
\nX:\u00d6zg\u00fcl iletkenlik katsay\u0131s\u0131 (m\/ohm.mm2)-bak\u0131r i\u00e7in = 50 m\/ohm.mm2 al\u0131n\u0131r)<\/em>
\nS : \u0130letken kesiti (mm2)<\/em><\/p>\n

A\u00e7\u0131k\u00e7as\u0131, gerilim d\u00fc\u015f\u00fcm\u00fc ile ilgili y\u00f6netmeli\u011fimizdeki bu form\u00fcllerin olduk\u00e7a s\u0131\u011f oldu\u011funu d\u00fc\u015f\u00fcn\u00fcyorum.Belki bilgisayar destekli proje haz\u0131rlama imkanlar\u0131n\u0131n olmad\u0131\u011f\u0131 eski d\u00f6nemlerde baz\u0131 hesaplar\u0131 ampirikle\u015ftirmek i\u015fleri kolayla\u015ft\u0131rabilirdi ama g\u00fcn\u00fcm\u00fczde hala eski metotlarda \u0131srar etmek anla\u015f\u0131l\u0131r gibi de\u011fil. Bug\u00fcn, t\u00fcm d\u00fcnyan\u0131n otorite kabul etti\u011fi IEEE’nin yay\u0131nlar\u0131n\u0131 inceledi\u011fimizde art\u0131k gerilim d\u00fc\u015f\u00fcm\u00fcn\u00fcn tam de\u011ferini veren form\u00fcllerin kullan\u0131ld\u0131\u011f\u0131 g\u00f6r\u00fclmektedir.<\/p>\n

Bu noktada, “tam de\u011fer mi?” diye sorabilirsiniz. A\u00e7\u0131k\u00e7as\u0131, bug\u00fcne kadar yap\u0131lan gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131 belli bir hata pay\u0131 i\u00e7inde kalmak suretiyle yakla\u015f\u0131k olarak hesaplanmaktayd\u0131.\u00a0 Hatta baz\u0131 hesaplarda iletken reaktans de\u011ferlerini ihmal etmekle bu hata pay\u0131 daha da artmaktayd\u0131.<\/p>\n

Gerilim D\u00fc\u015f\u00fcm\u00fc Teknik Tan\u0131m\u0131<\/h4>\n

Burada linki <\/a>verilen EMO yay\u0131n\u0131nda, gerilim d\u00fc\u015f\u00fcm\u00fc i\u00e7in \u015fu tan\u0131m yap\u0131lmaktad\u0131r.<\/p>\n

Bir hatt\u0131n ba\u015f\u0131ndaki gerilim faz\u00f6r\u00fc ile sonundaki gerilim faz\u00f6r\u00fc aras\u0131ndaki farka gerilim d\u00fc\u015f\u00fcm\u00fc ad\u0131\u00a0verilmektedir. Gerilim d\u00fc\u015f\u00fcm\u00fc boyuna ve enine gerilim d\u00fc\u015f\u00fcmleri olarak iki bile\u015fenden olu\u015fur. Al\u00e7ak gerilim \u015febekelerinde boyuna gerilim d\u00fc\u015f\u00fcm\u00fc etkin olup, enine gerilim d\u00fc\u015f\u00fcm\u00fc dikkate al\u0131nmaz. Orta gerilim \u015febekelerinde gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131nda her iki bile\u015fen de hesaplanmal\u0131d\u0131r. Boyuna gerilim d\u00fc\u015f\u00fcm\u00fc ufak bir hata ile hatt\u0131n ba\u015f\u0131ndaki gerilim faz\u00f6r\u00fcn\u00fcn hat sonu gerilim faz\u00f6r\u00fc \u00fczerindeki izd\u00fc\u015f\u00fcm\u00fc ile hat sonu faz\u00f6r\u00fc aras\u0131ndaki fark gerilim olarak hesaplan\u0131r. <\/em><\/p>\n

\"resim1\"<\/p>\n

Yukar\u0131da verilen tan\u0131m ve \u00e7izim gayet net olarak Gerilim D\u00fc\u015f\u00fcm\u00fc olay\u0131n\u0131 anlat\u0131yor ama konuyu biraz detayland\u0131rmakta fayda var.<\/p>\n

A\u015fa\u011f\u0131da tekhat \u015femas\u0131 verilmi\u015f basit bir devre var. Bu devre bir panodan beslenen bir fazl\u0131 y\u00fck ve faz+n\u00f6tr iletkeni bulunan bir besleme kablosundan olu\u015fan basit bir devre.<\/p>\n

\"res3\"<\/p>\n

Bu tekhat \u015femas\u0131 verilen devreyi a\u015fa\u011f\u0131daki \u015fekilde geni\u015fletebiliriz.<\/p>\n

\"res1\"<\/p>\n

Burada verilen Z Faz, faz iletkenin empedans\u0131n\u0131, ZN y\u00fck\u00fcn empedans\u0131n\u0131, Zn\u00f6tr ise n\u00f6tr iletkeninin empedans\u0131n\u0131 simgeliyor. Burada Zfaz ve Zn\u00f6tr birbirine e\u015fittir ve toplanarak tek empedans( Z H) olarak ifade edilebilir. Bu durumda devremiz \u015fu hale gelir.\"res2\"<\/p>\n

Burada, hat empedans\u0131n\u0131n ak\u0131m\u0131n \u015fiddetine katk\u0131s\u0131 ihmal edilir. \u00c7\u00fcnk\u00fc devre ak\u0131m\u0131n\u0131 y\u00fck\u00fcn ak\u0131m\u0131 olarak kabul etmek yeterlidir, bu anlamda hat empedans\u0131 \u00e7ok d\u00fc\u015f\u00fckt\u00fcr. Elbette bu durum bir \u00e7eli\u015fki olarak alg\u0131lanmamal\u0131d\u0131r. Yani hem k\u00fc\u00e7\u00fck deyip ihmal ediyoruz hem de gerilim d\u00fc\u015f\u00fcm\u00fcne sebep olarak bu empedans\u0131 i\u015faret ediyoruz. Unutmayal\u0131m ki,\u00a0 hattan y\u00fck ak\u0131m\u0131 ge\u00e7ti\u011finde, ihmal etti\u011fimiz o k\u00fc\u00e7\u00fck empedans gerilimi d\u00fc\u015f\u00fcrmeye yetmektedir. Hatt\u0131n empedans\u0131 ak\u0131m\u0131 belirlerken ihmal edilir, gerilimi belirlerken dikkate al\u0131n\u0131r.<\/p>\n

Dilenirse elbette hassas hesaplamalarla devre ak\u0131m\u0131, hat ve y\u00fck empedanslar\u0131 eklenip, kaynak gerilimine b\u00f6l\u00fcnerek tespit edilir ama \u00e7\u0131kacak sonu\u00e7 \u00e7ok k\u00fc\u00e7\u00fck bir farkla ihmal etti\u011fimiz durumun ak\u0131m de\u011ferinden a\u015fa\u011f\u0131da olacakt\u0131r. Haliyle hat empedans\u0131n\u0131 ihmal etmek hesaplamalara emniyetli tarafta kalmak anlam\u0131nda katk\u0131 sa\u011flamaktad\u0131r.<\/p>\n

 <\/p>\n

Yukar\u0131da ula\u015ft\u0131\u011f\u0131m\u0131z denklemleri faz\u00f6r b\u00f6lgesinde ifade edersek, i\u015fimiz birka\u00e7 geometrik problem \u00e7\u00f6zmeye kal\u0131r.<\/p>\n

\"res4\"<\/p>\n

Bu resimdeki gerilim d\u00fc\u015f\u00fcm\u00fc asl\u0131nda p ve t noktalar\u0131 aras\u0131ndaki mesafedir. Ancak s ve t noktalar\u0131 aras\u0131 ihmal edilir ve gerilim d\u00fc\u015f\u00fcm\u00fc olarak p ve s noktalar\u0131 aras\u0131 kabul edilir.<\/p>\n

Yani p ve s noktalar\u0131 aras\u0131ndaki mesafe kadar gerilim kaynaktan y\u00fcke ula\u015f\u0131rken hat \u00fczerinde kaybolmu\u015ftur. Geometrik bir tak\u0131m \u00e7\u00f6z\u00fcmlemelerle p ve s noktalar\u0131 aras\u0131ndaki mesafeyi yani gerilim d\u00fc\u015f\u00fcm\u00fcn\u00fc \u015fu \u015fekilde bulabiliriz.<\/p>\n

\"res5\"<\/p>\n

Ancak tekrar vurgulamak gerekirse, bu ifadede s ve t noktalar\u0131 aras\u0131ndaki mesafeyi ihmal ettik. Asl\u0131nda bu mesafeyi hesaplamak zor say\u0131lmaz. Sadece yukar\u0131daki gerilim d\u00fc\u015f\u00fcm\u00fc form\u00fcl\u00fcm\u00fcze s ve t noktalar\u0131 aras\u0131ndaki mesafeyi verecek bir ifade daha eklenecek o kadar.<\/p>\n

s ve t noktalar\u0131 aras\u0131ndaki mesafeyi hesaplamak i\u00e7in biraz geometri \u00e7al\u0131\u015fal\u0131m:<\/p>\n

\"res6\"<\/p>\n

Yukar\u0131daki \u015fekilde s\u00a0 ve v noktalar\u0131 aras\u0131ndaki mesafeyi, w ve v aras\u0131ndaki mesafeden s ve w noktalar\u0131 aras\u0131ndaki mesafeyi \u00e7\u0131kararak bulabiliriz. Yani \u00a6sv\u00a6=\u00a6wv\u00a6-\u00a6sw\u00a6<\/p>\n

B\u00f6ylece o, s ve v noktalar\u0131 aras\u0131nda bir dik \u00fc\u00e7gen olu\u015ftu. Bu dik \u00fc\u00e7genin, o ve s noktalar\u0131 aras\u0131ndaki kenar uzunlu\u011funu bulabiliriz.<\/p>\n

\"res7\"<\/p>\n

B\u00f6ylece s ve t noktalar\u0131 aras\u0131ndaki mesafeyi de bulmu\u015f olduk. Daha yukar\u0131da buldu\u011fumuz gerilim d\u00fc\u015f\u00fcm\u00fc form\u00fcl\u00fcm\u00fcze bu de\u011feri de ekleyerek hi\u00e7 hata pay\u0131 olmayan gerilim d\u00fc\u015f\u00fcm\u00fc form\u00fcl\u00fcm\u00fcz\u00fc elde etmi\u015f oluruz.<\/p>\n

\"res8\"<\/p>\n

IEEE’nin yay\u0131nlar\u0131nda (IEEE Standard 141) gerilim d\u00fc\u015f\u00fcm\u00fc form\u00fcl\u00fc olarak bu buldu\u011fumuz ifade verilmektedir. Fakat a\u00e7\u0131k\u00e7as\u0131, hata pay\u0131 dedi\u011fimiz k\u0131s\u0131m \u00e7ok k\u00fc\u00e7\u00fckt\u00fcr. Bu nedenle ifademizi \u015fu \u015fekilde yazal\u0131m:<\/p>\n

\"res10\"<\/p>\n

Gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131nda ileri derece hassasiyet aranm\u0131yorsa yani hesaplamalar\u0131m\u0131z kritik seviyelerde de\u011filse veya bilgisayarda hesaplama program\u0131 yap\u0131lm\u0131yorsa form\u00fcl\u00fc karma\u015f\u0131kla\u015ft\u0131rmaya hi\u00e7 gerek yoktur. hata=0 kabul ederek ilk ula\u015ft\u0131\u011f\u0131m\u0131z form\u00fcl g\u00f6n\u00fcl rahatl\u0131\u011f\u0131yla kullan\u0131labilir.<\/p>\n

\"res5\"<\/p>\n

Hat empedans\u0131<\/h4>\n

Bir i\u015fi yapmak i\u00e7in belirli bir \u015fiddette ak\u0131m \u00e7ekmek zorunday\u0131z, haliyle gerilim d\u00fc\u015f\u00fcm\u00fcn\u00fc azaltaca\u011f\u0131z diye y\u00fck ak\u0131m\u0131n\u0131 azaltamay\u0131z. Yapaca\u011f\u0131m\u0131z tek \u015fey y\u00fck ak\u0131m\u0131n\u0131 ta\u015f\u0131d\u0131\u011f\u0131m\u0131z iletkenin empedans\u0131n\u0131 do\u011fru se\u00e7mektir.<\/p>\n

Empedans\u0131n iki bile\u015feni vard\u0131r: Diren\u00e7(R) ve reaktans (X)<\/p>\n

Diren\u00e7 (R)<\/em><\/h5>\n

SI \u00f6l\u00e7\u00fclerine g\u00f6re, 1m uzunlu\u011funda ve 1mm2 kesitindeki bir ilekenin, elektrik ak\u0131m\u0131na kar\u015f\u0131 g\u00f6sterdi\u011fi dirence \u00f6zdiren\u00e7 (ro)\u00a0denir. \u00d6zdiren\u00e7 de\u011feri, malzemenin yap\u0131s\u0131ndan kaynaklanan ve o malzemeye \u00f6zg\u00fc bir de\u011ferdir.<\/p>\n

Al\u00fcminyum i\u00e7in\u00a0 ro\u00a0Al<\/sub>\u00a0= 0,028264 Ohm<\/b>\u00a0mm2<\/sup>\/m ,<\/b>bak\u0131r i\u00e7in\u00a0 ro\u00a0Cu<\/sub>\u00a0= 0,017857 Ohm<\/b>\u00a0mm2<\/sup>\/m\u00a0<\/b>dir.<\/p>\n

\u00d6zdirecin tersine\u00a0iletkenlik<\/strong>\u00a0denir. \u0130letkenlik, Al\u00fcminyum i\u00e7in SAl<\/sub>\u00a0= 35,38 m\/Ohm<\/b>\u00a0mm2<\/sup><\/b>\u00a0 bak\u0131r i\u00e7in\u00a0 SCu<\/sub>\u00a0= 56 m\/Ohm<\/b>\u00a0mm2\u00a0<\/sup><\/b>dir.<\/p>\n

\u00d6zdirenci, iletkenin L uzunlu\u011fu ile \u00e7arp\u0131p S kesit de\u011ferine b\u00f6lersek R direncini buluruz.<\/p>\n

R= ro * L \/ S<\/p>\n

Diren\u00e7 ve S\u0131cakl\u0131k<\/em><\/p>\n

Ancak vurgulamak gerekir ki diren\u00e7 de\u011feri s\u0131cakl\u0131kla de\u011fi\u015fir. Yukar\u0131da verilen de\u011ferler 20\u00baC ortam s\u0131cakl\u0131\u011f\u0131ndaki de\u011ferlerdir. \u0130letken \u00fczerinden ak\u0131m ge\u00e7ip iletken \u0131s\u0131nd\u0131k\u00e7a diren\u00e7 de\u011feri artacakt\u0131r. Haliyle gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131nda kablonun i\u015fletme s\u0131cakl\u0131\u011f\u0131ndaki diren\u00e7 de\u011ferlerini hesaba almak do\u011fru olur.<\/p>\n

Bir iletkenin T1<\/sub>\u00a0s\u0131cakl\u0131\u011f\u0131ndaki direnci R1<\/sub>\u00a0olsun, s\u0131cakl\u0131k biraz daha art\u0131p T2\u00a0<\/sub>s\u0131cakl\u0131\u011f\u0131na geldi\u011finde iletkenin direnci R2<\/sub>\u00a0olacakt\u0131r.<\/p>\n

Buna g\u00f6re direncin s\u0131cakl\u0131kla de\u011fi\u015fimi a\u015fa\u011f\u0131daki denkleme g\u00f6re ger\u00e7ekle\u015fir<\/p>\n

(R2<\/sub>\u00a0\u2013 R1<\/sub>) \/ R1<\/sub>\u00a0= \u03b1 (T2<\/sub>\u00a0\u2013 T1<\/sub>) \u00a0veya<\/p>\n

R2<\/sub>\u00a0= R1\u00a0<\/sub>x [1 + \u03b1 (T2<\/sub>\u00a0\u2013 T1<\/sub>) ]<\/p>\n

\u03b1 iletkenin direncinin s\u0131cakl\u0131\u011fa g\u00f6re de\u011fi\u015fim katsay\u0131s\u0131d\u0131r. Her maddenin kendine \u00f6zg\u00fc bir de\u011fi\u015fim katsay\u0131s\u0131 bulunur. Bak\u0131r iletkenlerde\u00a0 \u03b1 =0,00393, al\u00fcminyum iletkenlerde \u03b1=0,00403 al\u0131n\u0131r.<\/p>\n

Diren\u00e7 ve Deri Etkisi ile Yak\u0131nl\u0131k Etkisi<\/em><\/p>\n

Bir alternatif ak\u0131m sisteminde gerilim d\u00fc\u015f\u00fcm\u00fc hesab\u0131 yap\u0131yorsak, hesaplaraa DC diren\u00e7 de\u011fil AC direnci esas almak durumunday\u0131z. Kablo kataloglar\u0131nda 20\u00baC s\u0131cakl\u0131kta DC diren\u00e7 de\u011ferleri verilir. Halbuki i\u015fletme \u015fartlar\u0131nda bu direncin de\u011ferini artt\u0131ran ba\u015fka fakt\u00f6rler olu\u015fur. Alternatif ak\u0131m sistemlerinde ortaya \u00e7\u0131kan, deri etkisi ve yakla\u015f\u0131m etkisi denilen iki fenomen vard\u0131r. Bu iki etkinin sonu\u00e7lar\u0131n\u0131n nas\u0131l hesapland\u0131\u011f\u0131 IEC 60287-1-1 Part1-1 ‘de detayl\u0131 olarak anlat\u0131lmaktad\u0131r. Bu yaz\u0131da fazlaca detaya girilmeyecektir ancak bu iki etkiyi hesaba alarak kablo kesiti baz\u0131nda 90\u00baC s\u0131cakl\u0131kta AC diren\u00e7 kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bulmak i\u00e7in a\u015fa\u011f\u0131daki tabloyu kullanabiliriz.<\/p>\n

Tablodan da g\u00f6r\u00fclece\u011fi gibi AC diren\u00e7 de\u011ferini kullanmak, k\u00fc\u00e7\u00fck kesitli kablolarda pek \u00f6nem arz etmese de \u00f6zellikle b\u00fcy\u00fck kesitli kablolarda \u00f6nemli de\u011fi\u015fikliklere yol a\u00e7maktad\u0131r.<\/p>\n

70\u00baC i\u015fletme s\u0131cakl\u0131\u011f\u0131na sahip kablolar i\u00e7in AC Diren\u00e7 tablosu a\u015fa\u011f\u0131dad\u0131r. (Tablo-1)<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Kesit<\/td>\nDC Diren\u00e7 20\u00baC<\/td>\nDC Diren\u00e7 70\u00baC<\/td>\nDeri Etkisi Katsay\u0131s\u0131<\/td>\nYakla\u015f\u0131m Etkisi Katsay\u0131s\u0131<\/td>\nAC Diren\u00e7 Katsay\u0131s\u0131<\/td>\nAC Diren\u00e7 70\u00baC<\/td>\n<\/tr>\n
mm2<\/td>\nOhm\/km<\/td>\nOhm\/km<\/td>\n<\/td>\n<\/td>\n<\/td>\nOhm\/km<\/td>\n<\/tr>\n
Bak\u0131r \u0130letkenler (Yuvarlak Kesitli)<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
1,5<\/td>\n12,1<\/td>\n14,5<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n14,5<\/td>\n<\/tr>\n
2,5<\/td>\n7,41<\/td>\n8,87<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n8,87<\/td>\n<\/tr>\n
4<\/td>\n4,61<\/td>\n5,52<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n5,52<\/td>\n<\/tr>\n
6<\/td>\n3,08<\/td>\n3,69<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n3,69<\/td>\n<\/tr>\n
10<\/td>\n1,83<\/td>\n2,19<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n2,19<\/td>\n<\/tr>\n
16<\/td>\n1,15<\/td>\n1,38<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n1,38<\/td>\n<\/tr>\n
25<\/td>\n0,724<\/td>\n0,866<\/td>\n0,0001<\/td>\n0,0000<\/td>\n1,0001<\/td>\n0,866<\/td>\n<\/tr>\n
35<\/td>\n0,524<\/td>\n0,627<\/td>\n0,0002<\/td>\n0,0000<\/td>\n1,0002<\/td>\n0,627<\/td>\n<\/tr>\n
50<\/td>\n0,387<\/td>\n0,463<\/td>\n0,0004<\/td>\n0,0001<\/td>\n1,0005<\/td>\n0,463<\/td>\n<\/tr>\n
70<\/td>\n0,268<\/td>\n0,321<\/td>\n0,0008<\/td>\n0,0006<\/td>\n1,0014<\/td>\n0,321<\/td>\n<\/tr>\n
95<\/td>\n0,193<\/td>\n0,231<\/td>\n0,0015<\/td>\n0,0021<\/td>\n1,0037<\/td>\n0,232<\/td>\n<\/tr>\n
120<\/td>\n0,153<\/td>\n0,183<\/td>\n0,0024<\/td>\n0,0054<\/td>\n1,0078<\/td>\n0,184<\/td>\n<\/tr>\n
150<\/td>\n0,124<\/td>\n0,148<\/td>\n0,0037<\/td>\n0,0124<\/td>\n1,0161<\/td>\n0,151<\/td>\n<\/tr>\n
185<\/td>\n0,0991<\/td>\n0,1186<\/td>\n0,0058<\/td>\n0,0299<\/td>\n1,0357<\/td>\n0,1228<\/td>\n<\/tr>\n
240<\/td>\n0,0754<\/td>\n0,0902<\/td>\n0,0100<\/td>\n0,0847<\/td>\n1,0948<\/td>\n0,0988<\/td>\n<\/tr>\n
300<\/td>\n0,0601<\/td>\n0,0719<\/td>\n0,0157<\/td>\n0,1887<\/td>\n1,2045<\/td>\n0,0866<\/td>\n<\/tr>\n
400<\/td>\n0,0470<\/td>\n0,0562<\/td>\n0,0255<\/td>\n0,3948<\/td>\n1,4203<\/td>\n0,0799<\/td>\n<\/tr>\n
500<\/td>\n0,0366<\/td>\n0,0438<\/td>\n0,0415<\/td>\n0,6817<\/td>\n1,7231<\/td>\n0,0755<\/td>\n<\/tr>\n
600<\/td>\n0,0283<\/td>\n0,0339<\/td>\n0,0678<\/td>\n0,9645<\/td>\n2,0323<\/td>\n0,0688<\/td>\n<\/tr>\n
Al\u00fcminyum \u0130letkenler (Yuvarlak Kesitli)<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
2,5<\/td>\n12,22<\/td>\n14,69<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n14,69<\/td>\n<\/tr>\n
4<\/td>\n7,64<\/td>\n9,18<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n9,18<\/td>\n<\/tr>\n
6<\/td>\n5,09<\/td>\n6,12<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n6,12<\/td>\n<\/tr>\n
10<\/td>\n3,06<\/td>\n3,67<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n3,67<\/td>\n<\/tr>\n
16<\/td>\n1,91<\/td>\n2,29<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n2,29<\/td>\n<\/tr>\n
25<\/td>\n1,20<\/td>\n1,44<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n1,44<\/td>\n<\/tr>\n
35<\/td>\n0,868<\/td>\n1,043<\/td>\n0,0001<\/td>\n0,0000<\/td>\n1,0001<\/td>\n1,039<\/td>\n<\/tr>\n
50<\/td>\n0,641<\/td>\n0,770<\/td>\n0,0001<\/td>\n0,0000<\/td>\n1,0001<\/td>\n0,767<\/td>\n<\/tr>\n
70<\/td>\n0,443<\/td>\n0,532<\/td>\n0,0003<\/td>\n0,0000<\/td>\n1,0003<\/td>\n0,530<\/td>\n<\/tr>\n
95<\/td>\n0,320<\/td>\n0,384<\/td>\n0,0006<\/td>\n0,0001<\/td>\n1,0007<\/td>\n0,383<\/td>\n<\/tr>\n
120<\/td>\n0,253<\/td>\n0,304<\/td>\n0,0009<\/td>\n0,0003<\/td>\n1,0012<\/td>\n0,303<\/td>\n<\/tr>\n
150<\/td>\n0,206<\/td>\n0,248<\/td>\n0,0014<\/td>\n0,0007<\/td>\n1,0020<\/td>\n0,247<\/td>\n<\/tr>\n
185<\/td>\n0,164<\/td>\n0,197<\/td>\n0,0021<\/td>\n0,0017<\/td>\n1,0038<\/td>\n0,197<\/td>\n<\/tr>\n
240<\/td>\n0,125<\/td>\n0,150<\/td>\n0,0037<\/td>\n0,0050<\/td>\n1,0086<\/td>\n0,151<\/td>\n<\/tr>\n
300<\/td>\n0,100<\/td>\n0,120<\/td>\n0,0057<\/td>\n0,0120<\/td>\n1,0177<\/td>\n0,122<\/td>\n<\/tr>\n
400<\/td>\n0,0778<\/td>\n0,0935<\/td>\n0,0094<\/td>\n0,0322<\/td>\n1,0416<\/td>\n0,0970<\/td>\n<\/tr>\n
500<\/td>\n0,0605<\/td>\n0,0727<\/td>\n0,0155<\/td>\n0,0838<\/td>\n1,0993<\/td>\n0,0796<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

90\u00baC i\u015fletme s\u0131cakl\u0131\u011f\u0131na sahip kablolar i\u00e7in AC Diren\u00e7 tablosu a\u015fa\u011f\u0131dad\u0131r. (Tablo-2)<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Kesit<\/td>\nDC Diren\u00e7 20\u00baC<\/td>\nDC Diren\u00e7 90\u00baC<\/td>\nDeri Etkisi Katsay\u0131s\u0131<\/td>\nYakla\u015f\u0131m Etkisi Katsay\u0131s\u0131<\/td>\nAC Diren\u00e7 Katsay\u0131s\u0131<\/td>\nAC Diren\u00e7 90\u00baC<\/td>\n<\/tr>\n
mm2<\/td>\nOhm\/km<\/td>\nOhm\/km<\/td>\n<\/td>\n<\/td>\n<\/td>\nOhm\/km<\/td>\n<\/tr>\n
Bak\u0131r \u0130letkenler (Yuvarlak Kesitli)<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
1,5<\/td>\n12,1<\/td>\n15,4<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n15,4<\/td>\n<\/tr>\n
2,5<\/td>\n7,41<\/td>\n9,45<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n9,45<\/td>\n<\/tr>\n
4<\/td>\n4,61<\/td>\n5,88<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n5,88<\/td>\n<\/tr>\n
6<\/td>\n3,08<\/td>\n3,93<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n3,93<\/td>\n<\/tr>\n
10<\/td>\n1,83<\/td>\n2,33<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n2,33<\/td>\n<\/tr>\n
16<\/td>\n1,15<\/td>\n1,47<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n1,47<\/td>\n<\/tr>\n
25<\/td>\n0,724<\/td>\n0,923<\/td>\n0,0001<\/td>\n0,0000<\/td>\n1,0001<\/td>\n0,923<\/td>\n<\/tr>\n
35<\/td>\n0,524<\/td>\n0,668<\/td>\n0,0002<\/td>\n0,0000<\/td>\n1,0002<\/td>\n0,668<\/td>\n<\/tr>\n
50<\/td>\n0,387<\/td>\n0,493<\/td>\n0,0003<\/td>\n0,0001<\/td>\n1,0004<\/td>\n0,494<\/td>\n<\/tr>\n
70<\/td>\n0,268<\/td>\n0,342<\/td>\n0,0007<\/td>\n0,0004<\/td>\n1,0011<\/td>\n0,342<\/td>\n<\/tr>\n
95<\/td>\n0,193<\/td>\n0,246<\/td>\n0,0014<\/td>\n0,0017<\/td>\n1,0030<\/td>\n0,247<\/td>\n<\/tr>\n
120<\/td>\n0,153<\/td>\n0,195<\/td>\n0,0022<\/td>\n0,0042<\/td>\n1,0063<\/td>\n0,196<\/td>\n<\/tr>\n
150<\/td>\n0,124<\/td>\n0,158<\/td>\n0,0033<\/td>\n0,0096<\/td>\n1,0129<\/td>\n0,160<\/td>\n<\/tr>\n
185<\/td>\n0,0991<\/td>\n0,1264<\/td>\n0,0051<\/td>\n0,0233<\/td>\n1,0285<\/td>\n0,1300<\/td>\n<\/tr>\n
240<\/td>\n0,0754<\/td>\n0,0961<\/td>\n0,0088<\/td>\n0,0669<\/td>\n1,0757<\/td>\n0,1034<\/td>\n<\/tr>\n
300<\/td>\n0,0601<\/td>\n0,0766<\/td>\n0,0138<\/td>\n0,1521<\/td>\n1,1660<\/td>\n0,0894<\/td>\n<\/tr>\n
400<\/td>\n0,0470<\/td>\n0,0599<\/td>\n0,0225<\/td>\n0,3318<\/td>\n1,3543<\/td>\n0,0812<\/td>\n<\/tr>\n
500<\/td>\n0,0366<\/td>\n0,0467<\/td>\n0,0367<\/td>\n0,6059<\/td>\n1,6425<\/td>\n0,0767<\/td>\n<\/tr>\n
600<\/td>\n0,0283<\/td>\n0,0361<\/td>\n0,0601<\/td>\n0,9005<\/td>\n1,9606<\/td>\n0,0707<\/td>\n<\/tr>\n
Al\u00fcminyum \u0130letkenler (Yuvarlak Kesitli)<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
2,5<\/td>\n12,22<\/td>\n15,67<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n15,67<\/td>\n<\/tr>\n
4<\/td>\n7,64<\/td>\n9,80<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n9,80<\/td>\n<\/tr>\n
6<\/td>\n5,09<\/td>\n6,53<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n6,53<\/td>\n<\/tr>\n
10<\/td>\n3,06<\/td>\n3,92<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n3,92<\/td>\n<\/tr>\n
16<\/td>\n1,91<\/td>\n2,45<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n2,44<\/td>\n<\/tr>\n
25<\/td>\n1,20<\/td>\n1,54<\/td>\n0,0000<\/td>\n0,0000<\/td>\n1,0000<\/td>\n1,53<\/td>\n<\/tr>\n
35<\/td>\n0,868<\/td>\n1,113<\/td>\n0,0001<\/td>\n0,0000<\/td>\n1,0001<\/td>\n1,107<\/td>\n<\/tr>\n
50<\/td>\n0,641<\/td>\n0,822<\/td>\n0,0001<\/td>\n0,0000<\/td>\n1,0001<\/td>\n0,817<\/td>\n<\/tr>\n
70<\/td>\n0,443<\/td>\n0,568<\/td>\n0,0003<\/td>\n0,0000<\/td>\n1,0003<\/td>\n0,565<\/td>\n<\/tr>\n
95<\/td>\n0,320<\/td>\n0,410<\/td>\n0,0005<\/td>\n0,0001<\/td>\n1,0006<\/td>\n0,408<\/td>\n<\/tr>\n
120<\/td>\n0,253<\/td>\n0,324<\/td>\n0,0008<\/td>\n0,0002<\/td>\n1,0010<\/td>\n0,323<\/td>\n<\/tr>\n
150<\/td>\n0,206<\/td>\n0,264<\/td>\n0,0012<\/td>\n0,0005<\/td>\n1,0017<\/td>\n0,263<\/td>\n<\/tr>\n
185<\/td>\n0,164<\/td>\n0,210<\/td>\n0,0019<\/td>\n0,0013<\/td>\n1,0032<\/td>\n0,210<\/td>\n<\/tr>\n
240<\/td>\n0,125<\/td>\n0,160<\/td>\n0,0032<\/td>\n0,0038<\/td>\n1,0071<\/td>\n0,161<\/td>\n<\/tr>\n
300<\/td>\n0,100<\/td>\n0,128<\/td>\n0,0050<\/td>\n0,0093<\/td>\n1,0144<\/td>\n0,129<\/td>\n<\/tr>\n
400<\/td>\n0,0778<\/td>\n0,0997<\/td>\n0,0083<\/td>\n0,0251<\/td>\n1,0334<\/td>\n0,1025<\/td>\n<\/tr>\n
500<\/td>\n0,0605<\/td>\n0,0776<\/td>\n0,0137<\/td>\n0,0661<\/td>\n1,0798<\/td>\n0,0833<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

<\/h4>\n

Reaktans (X)<\/em><\/h5>\n

Alternatif ak\u0131m, iletken \u00fczerinde ister istemez bir manyetik alan ve elektrik alan olu\u015fturacakt\u0131r. Bu alanlar\u0131n olu\u015fmas\u0131 neticesinde kablo \u00fczerinde end\u00fcktans ve kapasitans de\u011ferleri meydana gelir. Asl\u0131nda bunlar\u0131n direk bir eneji kayb\u0131na yol a\u00e7malar\u0131 s\u00f6z konusu de\u011fildir ama manyetik ve elektrik alanlar\u0131n olu\u015fmas\u0131 i\u00e7in kaynaktan bir miktar ak\u0131m \u00e7ekilmelidir. Bu ak\u0131m\u00a0I2<\/sup>R\u00a0 kayb\u0131nda k\u00fc\u00e7\u00fck de olsa art\u0131\u015fa sebep olur.<\/p>\n

Kablolar\u0131n end\u00fcktans ve kapasitans de\u011ferlerinden toplam reaktans de\u011feri olu\u015fur. Bu de\u011fer, kablonun tesis edildi\u011fi yer, yak\u0131n\u0131ndaki \u00e7elik konstr\u00fcksiyon, ba\u015fka kablolara kom\u015fuluk gibi bir tak\u0131m b\u00fcy\u00fckl\u00fcklere g\u00f6re de\u011fi\u015fir. Ancak her zaman bu detatylara g\u00f6re hassas hesaplama yap\u0131lamayaca\u011f\u0131 i\u00e7in \u00e7e\u015fitli kaynaklarda verilen kablo reaktans de\u011ferlerinden yola \u00e7\u0131k\u0131larak hesaplamalara devam edilir.<\/p>\n

A\u015fa\u011f\u0131daki tabloda baz\u0131 kablolar\u0131n yerle\u015fim tiplerine g\u00f6re reaktans de\u011ferleri verilmi\u015ftir. (Tablo-3)<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Kesit<\/td>\nReaktans<\/td>\nReaktans<\/td>\nReaktans<\/td>\n<\/tr>\n
Ohm\/km<\/td>\nOhm\/km<\/td>\nOhm\/km<\/td>\n<\/tr>\n
mm2<\/td>\nT\u0130P1<\/td>\nT\u0130P2<\/td>\nT\u0130P3<\/td>\n<\/tr>\n
1,5<\/td>\n0,1150<\/td>\n0,2150<\/td>\n0,1484<\/td>\n<\/tr>\n
2,5<\/td>\n0,1068<\/td>\n0,1997<\/td>\n0,1378<\/td>\n<\/tr>\n
4<\/td>\n0,1065<\/td>\n0,1991<\/td>\n0,1374<\/td>\n<\/tr>\n
6<\/td>\n0,1008<\/td>\n0,1886<\/td>\n0,1301<\/td>\n<\/tr>\n
10<\/td>\n0,0946<\/td>\n0,1768<\/td>\n0,1220<\/td>\n<\/tr>\n
16<\/td>\n0,0895<\/td>\n0,1681<\/td>\n0,1166<\/td>\n<\/tr>\n
25<\/td>\n0,0861<\/td>\n0,1615<\/td>\n0,1100<\/td>\n<\/tr>\n
35<\/td>\n0,0820<\/td>\n0,1561<\/td>\n0,1046<\/td>\n<\/tr>\n
50<\/td>\n0,0826<\/td>\n0,1536<\/td>\n0,1021<\/td>\n<\/tr>\n
70<\/td>\n0,0798<\/td>\n0,1486<\/td>\n0,0971<\/td>\n<\/tr>\n
95<\/td>\n0,0795<\/td>\n0,1464<\/td>\n0,0949<\/td>\n<\/tr>\n
120<\/td>\n0,0785<\/td>\n0,1439<\/td>\n0,0924<\/td>\n<\/tr>\n
150<\/td>\n0,0776<\/td>\n0,1426<\/td>\n0,0911<\/td>\n<\/tr>\n
185<\/td>\n0,0779<\/td>\n0,1417<\/td>\n0,0902<\/td>\n<\/tr>\n
240<\/td>\n0,0770<\/td>\n0,1398<\/td>\n0,0883<\/td>\n<\/tr>\n
300<\/td>\n—<\/td>\n0,1370<\/td>\n0,0865<\/td>\n<\/tr>\n
400<\/td>\n—<\/td>\n0,1343<\/td>\n0,0848<\/td>\n<\/tr>\n
T\u0130P1<\/td>\nAyn\u0131 d\u0131\u015f k\u0131l\u0131f i\u00e7inde Bak\u0131r iletken 3+1 damarl\u0131 kablo<\/td>\n<\/tr>\n
T\u0130P2<\/td>\nTek damarl\u0131 bak\u0131r iletkenli kablolar<\/td>\n<\/td>\n<\/tr>\n
3 adedi yanyana yatay d\u00fczlemde yerle\u015ftirilmi\u015f<\/td>\n<\/tr>\n
T\u0130P3<\/td>\nTek damarl\u0131 bak\u0131r iletkenli kablolar<\/td>\n<\/td>\n<\/tr>\n
3 adedi \u00fc\u00e7gen formda yerle\u015ftirilmi\u015f<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

 <\/p>\n

B\u00f6ylece gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131m\u0131z\u0131 yapmam\u0131za sebep olan hat direnci (R) ve hat reaktans\u0131 (X) de\u011ferlerine nas\u0131l ula\u015faca\u011f\u0131m\u0131z\u0131 g\u00f6rm\u00fc\u015f olduk.\u00a0\u00a0\u015eimdi gerilim d\u00fc\u015f\u00fcm\u00fc form\u00fcl\u00fcm\u00fcz\u00fc bir fazl\u0131 devreler ve \u00fc\u00e7 fazl\u0131 devreler \u00f6zelinde tekrar g\u00f6zden ge\u00e7irelim.<\/p>\n

Bir fazl\u0131 devrelerde gerilim d\u00fc\u015f\u00fcm\u00fc<\/h4>\n

Hesaba konu kablonun metre ba\u015f\u0131na R ve X de\u011ferlerini\u00a0 yukar\u0131da verilen tablolardan ald\u0131ktan sonra iletken uzunlu\u011fu ile \u00e7arparak kablonun toplam R ve X de\u011ferleri bulunur.\u00a0 Bundan sonras\u0131 form\u00fclde ilgili de\u011ferleri yerine konarak bulunur.\u00a0(R i\u00e7in alternatif ak\u0131m direncini kullanmay\u0131 l\u00fctfen atlamayal\u0131m, benzer \u015fekilde X de\u011ferini de tablodan kablo yerle\u015fim tipine g\u00f6re almal\u0131y\u0131z.)<\/p>\n

\"res11\"<\/p>\n

Ancak \u015funu vurgulayal\u0131m, bir fazl\u0131 devrelerin gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131nda n\u00f6tr hatt\u0131n\u0131n empedans\u0131n\u0131n da toplam kablo empedans\u0131na eklenmesi gerekmektedir. Bu sebeple L uzunlu\u011funun yani pano ile t\u00fckete\u00e7 aras\u0131ndaki uzunlu\u011fun 2 kat\u0131n\u0131 almam\u0131z gerekir.<\/p>\n

\"res12\"<\/p>\n

B\u00f6ylece bir fazl\u0131 devrelerdeki gerilim d\u00fc\u015f\u00fcm\u00fc form\u00fcl\u00fcm\u00fcz \u015fu hale gelmi\u015f olur:<\/p>\n

\"res13\"<\/p>\n

 <\/p>\n

\u00dc\u00e7 fazl\u0131 devrelerde gerilim d\u00fc\u015f\u00fcm\u00fc<\/h4>\n

\u00dc\u00e7 fazl\u0131 devrelerde, \u015fayet dengeli y\u00fck varsa n\u00f6tr hatt\u0131ndan bir ak\u0131m ge\u00e7mez. B\u00f6ylece n\u00f6tr hatt\u0131nda bir gerilim d\u00fc\u015f\u00fcm\u00fc olamaz ve haliyle sadece faz iletkeninin empedans\u0131 hesaba kat\u0131l\u0131r. Yani bir fazl\u0131 devrelerde yapt\u0131\u011f\u0131m\u0131z gibi uzunlu\u011fu 2 ile \u00e7arpmay\u0131z. <\/span>Sadece pano ile t\u00fckete\u00e7 aras\u0131ndaki uzunluk kullan\u0131l\u0131r. \u0130lgili tablolardan metre ba\u015f\u0131na R ve X de\u011ferleri al\u0131n\u0131p hat uzunlu\u011fu L ile \u00e7arp\u0131larak hatt\u0131n toplam R ve X de\u011ferleri tespit edilir. (R i\u00e7in alternatif ak\u0131m direncini kullanmay\u0131 l\u00fctfen atlamayal\u0131m, benzer \u015fekilde X de\u011ferini de tablodan kablo yerle\u015fim tipine g\u00f6re almal\u0131y\u0131z.)<\/p>\n

\"res14\"<\/p>\n

\u00dc\u00e7 fazl\u0131 sistemlerd gerilim fazaras\u0131 de\u011feriyle an\u0131ld\u0131\u011f\u0131 i\u00e7in gerilim d\u00fc\u015f\u00fcm\u00fc de\u011ferini de fazaras\u0131 de\u011fere d\u00f6n\u00fc\u015ft\u00fcrmek ad\u0131na \u221a3 katsay\u0131s\u0131n\u0131 da eklersek dengeli \u00fc\u00e7 fazl\u0131 devrelerdeki gerilim d\u00fc\u015f\u00fcm\u00fc form\u00fcl\u00fcm\u00fcz \u015fu hale gelmi\u015f olur:<\/p>\n

\"res15\"<\/p>\n

 <\/p>\n

Gerilim d\u00fc\u015f\u00fcm\u00fc ile ilgili \u00f6rnekler<\/h5>\n

 <\/p>\n

\"res16\"<\/p>\n

Yukar\u0131daki 3 fazl\u0131, 400V dengeli olarak \u00e7al\u0131\u015fan bir tali pano beslemesi g\u00f6sterilmi\u015ftir. Bu panoyu besleyen kablo \u00fczerindeki gerilim d\u00fc\u015f\u00fcm\u00fcn\u00fc bulal\u0131m.<\/p>\n

Kullanaca\u011f\u0131m\u0131z gerilim d\u00fc\u015f\u00fcm\u00fc form\u00fcl\u00fc, 3 fazl\u0131 dengeli sistemler i\u00e7in buldu\u011fumuz a\u015fa\u011f\u0131daki form\u00fcl olacak.<\/p>\n

\"res15\"<\/p>\n

Bu ba\u011flant\u0131da ge\u00e7en kablonun birim uzunluk ba\u015f\u0131na R ve X de\u011ferlerini tespit edelim.\u00a0 Kablomuz N2XH tipinde oldu\u011fundan malzemesi bak\u0131r ve i\u015fletme s\u0131cakl\u0131\u011f\u0131 90\u00baC dir.\u00a0 70mm2 bak\u0131r i\u00e7in Tablo 2’den\u00a0 alternatif ak\u0131m direncini R=0,342 ohm\/km olarak al\u0131r\u0131z. Ayr\u0131ca kablomuz 3+1 formunda oldu\u011fundan yani Tablo-3 e g\u00f6re Tip-1 oldu\u011fundan Tablo-3 den X=0,0798 ohm\/km de\u011ferini al\u0131r\u0131z.<\/p>\n

sin(acos(0,8)) ba\u011f\u0131nt\u0131s\u0131n\u0131 kullanarak sin\u00d8=0,6 buluruz.<\/p>\n

Son olarak P=\u00a0\u221a3 Un. In. cos\u00d8 ba\u011f\u0131nt\u0131s\u0131n\u0131 kullanarak In=125000 \/ 400 \/ 0,8 \/\u221a3 = 225,53A de\u011ferini buluruz.<\/p>\n

Bundan sonra de\u011ferleri gerilim d\u00fc\u015f\u00fcm\u00fc form\u00fcl\u00fcm\u00fcze yerle\u015ftirelim.<\/p>\n

dU=\u00a0\u221a3 . 100 . 225,53 . (0,342 . 0,8 \/1000+ 0,0798 . 0,6\/1000) = 12,55 V<\/p>\n

e= dU\/Un = 12,55\/400 = %3,13<\/p>\n

olarak bulunur.<\/p>\n

 <\/p>\n

Sonu\u00e7:<\/h4>\n

L\u00fctfen, internet ortam\u0131nda dola\u015fan ve a\u015fa\u011f\u0131da resmini verdi\u011fimiz EMOnun eski usul hesap y\u00f6ntemini kullanmaman\u0131z\u0131 \u00f6neririm. E\u011fer bu eski y\u00f6ntemler gerilim d\u00fc\u015f\u00fcm\u00fc hesab\u0131 yapsayd\u0131k, yukar\u0131daki \u00f6rne\u011fimizde sonu\u00e7 e=%2,21 \u00e7\u0131kacakt\u0131. Yani %34 hatal\u0131 olacakt\u0131.<\/p>\n

\"Eski<\/p>\n

Gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131 bir miktar hata pay\u0131 i\u00e7ermektedir. Ama bu hata pay\u0131 pratikte \u00e7ok ufak oldu\u011fundan g\u00f6zard\u0131 edilmektedir. Fakat sadece DC diren\u00e7 de\u011ferlerini kullanarak gerilim d\u00fc\u015f\u00fcm\u00fc hesab\u0131 yapmaktansa yukar\u0131da detay\u0131n\u0131 verdi\u011fimiz \u015fekilde kablonun AC diren\u00e7 de\u011ferleri ve reaktans\u0131n\u0131 kullanarak hesap yapmak daha do\u011fru sonu\u00e7lar verir.<\/p>\n

Tekrarlamak gerekirse tesisatlarda aranacak gerilim d\u00fc\u015f\u00fcm\u00fc s\u0131n\u0131r\u0131: Yap\u0131 ba\u011flant\u0131 kutusu ile t\u00fcketim ara\u00e7lar\u0131 aras\u0131nda, ayd\u0131nlatma ve priz devreleri i\u00e7in % l,5’i, motor devreler i\u00e7in % 3 ‘\u00fc, ge\u00e7memelidir. Kendi trafosu bulunan tesislerde toplam gerilim d\u00fc\u015f\u00fcm\u00fc, ayd\u0131nlatma ve priz devreleri i\u00e7in %6,5, motor devreleri i\u00e7in %8 olabilir.<\/p>\n

 <\/p>\n

Bu linke t\u0131klayarak<\/a>, gerilim d\u00fc\u015f\u00fcm\u00fc hesaplamas\u0131 yapan online elektronik tablomuza ula\u015fabilirsiniz.<\/p>\n

 <\/p>\n

 <\/p>\n

Kaynaklar:<\/strong><\/p>\n

\u0130sa \u0130lisu,\u00a0TMMOB ELEKTR\u0130K M\u00dcHEND\u0130SLER\u0130 ODASI \u0130STANBUL \u015eUBES\u0130 B\u00dcLTEN\u0130 \u015eubat 2016<\/a><\/p>\n

IEC 60287-1-1 Part1-1<\/p>\n

https:\/\/pdhonline.com\/courses\/e426\/e426content.pdf<\/a><\/p>\n

 <\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Gerilimde bir d\u00fc\u015fme olmasa ne g\u00fczel olurdu. \u0130ncecik kablolarla kayna\u011f\u0131n gerilimini istedi\u011fimiz kadar uza\u011fa ta\u015f\u0131r, en ufak bir enerji kayb\u0131 da ya\u015famazd\u0131k.\u00a0 Ama maalesef b\u00f6yle bir d\u00fcnya yok. En az\u0131ndan s\u00fcper iletken teknolojisi ilerleyene kadar bu gerilim d\u00fc\u015f\u00fcm\u00fc hesaplar\u0131n\u0131 yapmaya devam etmek zorunday\u0131z. Bildi\u011finiz gibi bir gerilim kayna\u011f\u0131n\u0131, bir t\u00fcketece iletkenle ba\u011flar\u0131z. \u0130letkenler de neticede… Gerilim D\u00fc\u015f\u00fcm\u00fc<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":3566,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"neve_meta_sidebar":"","neve_meta_container":"","neve_meta_enable_content_width":"","neve_meta_content_width":0,"neve_meta_title_alignment":"","neve_meta_author_avatar":"","neve_post_elements_order":"","neve_meta_disable_header":"","neve_meta_disable_footer":"","neve_meta_disable_title":"","_themeisle_gutenberg_block_has_review":false,"footnotes":""},"categories":[5],"tags":[63,64,44,65,66,67,68,69,70,71],"class_list":["post-1696","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-elektrik-tesisat","tag-alternatif-akim-direnci","tag-deri-etkisi","tag-elektrik","tag-elektrik-hesaplamalari","tag-gercek-gerilim-dusumu","tag-gerilim-dusumu","tag-gerilim-dusumu-hata-payi","tag-kablo-reaktansi","tag-voltage-drop","tag-yakinlik-etkisi"],"_links":{"self":[{"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/posts\/1696","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/comments?post=1696"}],"version-history":[{"count":0,"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/posts\/1696\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/benga.pro\/index.php\/wp-json\/"}],"wp:attachment":[{"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/media?parent=1696"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/categories?post=1696"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/tags?post=1696"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}