{"id":2099,"date":"2018-09-13T08:39:43","date_gmt":"2018-09-13T05:39:43","guid":{"rendered":"http:\/\/mekanikelektriktesisat.com\/?p=2099"},"modified":"2018-09-13T08:39:43","modified_gmt":"2018-09-13T05:39:43","slug":"basinc","status":"publish","type":"post","link":"https:\/\/benga.pro\/index.php\/2018\/09\/13\/basinc\/","title":{"rendered":"Bas\u0131n\u00e7"},"content":{"rendered":"

Bas\u0131n\u00e7, tan\u0131m olarak bir y\u00fczeye etkiyen toplam kuvvettir.  D\u00fcnyadan yeterice uzak bir noktada, uzay\u0131n bir k\u00f6\u015fesinde durdu\u011fumuzu ve elimize i\u00e7i c\u0131va(Hg) dolu, \u00fcst\u00fc a\u00e7\u0131k bir kap ald\u0131\u011f\u0131m\u0131z\u0131 varsayal\u0131m. Baz\u0131 n\u00f6trino gibi par\u00e7ac\u0131klar\u0131n etkilerini ihmal edersek bu c\u0131van\u0131n \u00fczerine hi\u00e7 bir kuvvet gelmeyecektir. Yani bas\u0131n\u00e7 s\u0131f\u0131rd\u0131r. Bu duruma mutlak vakum ad\u0131 da verilir. Daha sonra elimize bir cam t\u00fcp alal\u0131m. Uzay ortam\u0131nda, t\u00fcp\u00fcn i\u00e7inde de mutlak vakum olacakt\u0131r, yani bas\u0131n\u00e7 s\u0131f\u0131rd\u0131r. T\u00fcp\u00fc c\u0131va dolu kab\u0131n i\u00e7ine ters olarak bat\u0131ral\u0131m. Kaptaki c\u0131va seviyesiyle, t\u00fcp\u00fcn i\u00e7indeki c\u0131va seviyesi ayn\u0131 olacakt\u0131r.<\/p>\n

C\u0131va d\u00fczene\u011fimizle, uzay\u0131n derinliklerinden d\u00fcnyaya do\u011fru yakla\u015fmaya ba\u015flayal\u0131m. Uzun bir s\u00fcre kaptaki c\u0131va seviyesiyle, t\u00fcp\u00fcn i\u00e7indeki c\u0131va seviyesi aras\u0131nda fark olu\u015fmayacakt\u0131r. D\u00fcnyan\u0131n atmosferine girdi\u011fimiz an t\u00fcp\u00fcn i\u00e7indeki c\u0131va seviyesi y\u00fckselmeye ba\u015flayacakt\u0131r. T\u00fcp\u00fcn etraf\u0131ndaki c\u0131vaya atmosferi olu\u015fturan havan\u0131n a\u011f\u0131rl\u0131\u011f\u0131 t\u0131pk\u0131 bir piston gibi \u00e7al\u0131\u015facakt\u0131r ve t\u00fcp\u00fcn i\u00e7indeki c\u0131vay\u0131 y\u00fckseltecektir. C\u0131va d\u00fczene\u011fimizle yery\u00fcz\u00fcne yakla\u015ft\u0131k\u00e7a c\u0131va seviyesi daha da artacakt\u0131r \u00e7\u00fcnk\u00fc art\u0131k t\u00fcp\u00fcn etraf\u0131ndaki c\u0131vaya, atmosferde katetti\u011fimiz mesafeyle orant\u0131l\u0131 olarak daha fazla hava  kuvvet uygulayacakt\u0131r. Yery\u00fcz\u00fcne, yani deniz seviyesine, 20.000m mesafe kald\u0131\u011f\u0131nda c\u0131va seviyesi 32mm y\u00fckselmi\u015f olacakt\u0131r.<\/p>\n

\"ucak.jpg\"Yolcu u\u00e7aklar\u0131n\u0131n u\u00e7tu\u011fu seviyeye geldi\u011fimizde yani yery\u00fcz\u00fcnden 10.000m y\u00fckseklikte c\u0131va seviyesi 198mm’e gelecektir.<\/p>\n

 <\/p>\n

 <\/p>\n

\"everest.jpg\"Everest tepesinin yan\u0131ndan ge\u00e7erken yani 8848m’de c\u0131va seviyesi 241mm olacakt\u0131r. Yery\u00fcz\u00fcnden 1500m y\u00fckseklikte   c\u0131va seviyemiz art\u0131k 634mm’dedir. Bu \u015fekilde yery\u00fcz\u00fcne yakla\u015ft\u0131k\u00e7a t\u00fcpteki c\u0131va seviyesi artacak, artacak ve deniz seviyesine ula\u015ft\u0131\u011f\u0131m\u0131zda t\u00fcpteki c\u0131va seviyesi tam olarak 760mm y\u00fcksekli\u011fe eri\u015fecektir. Bu de\u011fer yery\u00fcz\u00fcnde hepimizin maruz kald\u0131\u011f\u0131 hava bas\u0131nc\u0131d\u0131r.<\/p>\n

Fiziksel olarak 760mm olarak \u00f6l\u00e7\u00fclen bu c\u0131va y\u00fcksekli\u011fine 1 atm (atmosfer) ad\u0131 verilir. Yani etraf\u0131m\u0131zda bulunan havan\u0131n bas\u0131nc\u0131n\u0131n 760mmHg veya 1atm oldu\u011funu s\u00f6yleyebiliriz. Ancak d\u00fcnyada ideal olarak her zaman 760mmHg hava bas\u0131nc\u0131 yoktur. D\u00fcnya genelindeki havan\u0131n hareketleri, hava s\u0131cakl\u0131\u011f\u0131, yerle\u015fim yerlerinin rak\u0131m\u0131 vb. sebeplerle bu de\u011fer de\u011fi\u015fiklik g\u00f6sterir.<\/p>\n

Bu arada hava bas\u0131nc\u0131n\u0131 \u00f6l\u00e7mek i\u00e7in yapt\u0131\u011f\u0131m\u0131z deneyi herhangi bir ba\u015fka s\u0131v\u0131yla da yapabilirdik. Mesela su ile yapsayd\u0131k, 760mm yerine 10330mm y\u00fckseklikle kar\u015f\u0131la\u015facakt\u0131k ki  10 k\u00fcsur metre y\u00fcksekli\u011finde bir d\u00fczene\u011fin haz\u0131rlanmas\u0131 hi\u00e7 pratik olmayacakt\u0131. (Tabi uzaydan d\u00fcnyaya s\u00fcz\u00fcl\u00fcp gelmenin pratikli\u011fini hi\u00e7 kar\u0131\u015ft\u0131rmayal\u0131m:) ) Bu a\u00e7\u0131dan c\u0131va, y\u00fcksek yo\u011funlu\u011fu (1m3 c\u0131va 13596kg a\u011f\u0131rl\u0131\u011fa sahiptir) sebebiyle bas\u0131n\u00e7 \u00f6l\u00e7\u00fcm\u00fc anlam\u0131nda en pratik s\u0131v\u0131d\u0131r.<\/p>\n

Evet! 760mm c\u0131va y\u00fcksekli\u011fi bas\u0131nc\u0131n bir g\u00f6stergesidir ama hem \u00e7abuk alg\u0131lanmas\u0131 hem de hesaplarda kullanma kolayl\u0131\u011f\u0131 yaratmas\u0131 a\u00e7\u0131s\u0131ndan biraz teknik a\u00e7\u0131dan irdeleyelim.<\/p>\n

\n

Atmosfer (Atm), Pascal (Pa)<\/h4>\n

\u00d6ncelikle tan\u0131m olarak Atmosfer (Atm)<\/strong> birimini ortaya koyal\u0131m. 760mm C\u0131va y\u00fcksekli\u011fini veren bas\u0131nca 1 Atm<\/strong> denir.  Bu arada, bas\u0131n\u00e7 i\u00e7in, tan\u0131m olarak bir y\u00fczeye, alana etkiyen toplam kuvvettir, demi\u015ftik. O halde deney t\u00fcp\u00fcm\u00fcz\u00fcn i\u00e7indeki 760mm y\u00fcksekli\u011findeki c\u0131van\u0131n taban\u0131na uygulad\u0131\u011f\u0131 kuvveti bulmaya \u00e7al\u0131\u015fal\u0131m.<\/p>\n

Bas\u0131n\u00e7 (P) = Kuvvet (F) \/ Alan (A)<\/p>\n

\"1.JPG\"<\/p>\n

m: K\u00fctle [kg]<\/p>\n

g : Yer \u00e7ekimi ivmesi [m\/s2<\/sup>]<\/p>\n

V: Hacim [m3<\/sup>]<\/p>\n

\u03c1 : Yo\u011funluk  [kg\/m3<\/sup>]<\/p>\n

A :  Taban Alan\u0131 [m2<\/sup>]<\/p>\n

h :  S\u0131v\u0131 y\u00fcksekli\u011fi [m]<\/p>\n

olmak \u00fczere<\/p>\n

\"2.JPG\"<\/p>\n

\"3.JPG\"<\/p>\n

\"4.JPG\"<\/p>\n

bulunur. Birimlerden yola \u00e7\u0131karak<\/p>\n

\"5.JPG\"<\/p>\n

bas\u0131n\u00e7 birimini N \/m2<\/sup> \u015feklinde buluruz. Bu b\u00fcy\u00fckl\u00fc\u011fe Pascal (Pa)<\/strong> ad\u0131 verilir.<\/p>\n

\"6.JPG\"<\/p>\n

\u0130\u015fin form\u00fcl\u00fcn\u00fc kurdu\u011fumuza g\u00f6re 760mm y\u00fckseli\u011findeki c\u0131vaya bir bakal\u0131m:<\/p>\n

P=0,76 . 13595 . 9,80665<\/p>\n

P=101332 Pa<\/p>\n

Yani 76cm c\u0131va s\u00fctunu taban\u0131na 101332Pa bas\u0131n\u00e7 uygulamaktad\u0131r. 76cm C\u0131va y\u00fcksekli\u011fini veren bas\u0131nca 1 Atm dedi\u011fimize g\u00f6re<\/p>\n

1 Atm = 101332 Pa olmaktad\u0131r.<\/p>\n

\n

Metre Su S\u00fctunu (mSS)<\/strong><\/p>\n

Tesisat sekt\u00f6r\u00fcnde bar  kadar s\u0131kl\u0131kla kullan\u0131lan bu birimi incelemek i\u00e7in tersten bir hesap yapal\u0131m.<\/p>\n

76cm y\u00fcksekli\u011finde c\u0131va 1 Atm bas\u0131n\u00e7 demekti. Di\u011fer bir deyi\u015fle 1 Atm hava bas\u0131nc\u0131 c\u0131vay\u0131 76cm y\u00fcksekli\u011fe \u00e7\u0131karabiliyordu. Peki ayn\u0131 1 Atm hava bas\u0131nc\u0131 suyu ne kadar y\u00fckse\u011fe \u00e7\u0131karabilir? (Suyun yo\u011funlu\u011funu +4C de 1000kg\/m2<\/sup> alaca\u011f\u0131z)<\/p>\n

\"5.JPG\"<\/p>\n

ifadesini kullanarak<\/p>\n

1 Atm = 101332 Pa = 101332 N\/m2<\/sup>  = h .  \u03c1 . g = h . 1000 . 9,80665<\/p>\n

denklemini kurabiliriz. Burada h bilinmeyenini \u00e7ekersek<\/p>\n

h=10,3329m olarak buluruz.<\/p>\n

Yani 1 Atm hava bas\u0131nc\u0131 suyu 10,33 y\u00fcksekli\u011fe kadar \u00e7\u0131karacakt\u0131r. Yani<\/p>\n

1 Atm = 10,3329 mSS   olur.<\/p>\n

\n

Teknik Atmosfer (At)<\/strong><\/p>\n

\u015eimdi de biraz eski birimlerle analiz yapal\u0131m. \u00d6nceleri alan\u0131n birimi cm2<\/sup>, kuvvetin birimi de kg-f olarak al\u0131nm\u0131\u015f ve 1 cm2<\/sup> alan \u00fczerine etkiyen 1kg-f kuvvete 1 teknik atmosfer (At)<\/strong> denilmi\u015fti.<\/p>\n

1 kg-f= 9,80665 N<\/p>\n

1 cm2<\/sup>= 0,0001 m2<\/sup><\/p>\n

Bu d\u00f6n\u00fc\u015f\u00fcmleri kullanarak<\/p>\n

101332 Pa = 101332 N\/m2<\/sup> = 1,03329 kg-f \/cm2<\/sup> = 1,03329 At<\/p>\n

de\u011ferini buluruz. \u00d6zetle<\/p>\n

1 Atm = 1,03329 At   dir.<\/p>\n

Ayn\u0131 zamanda 1 Atm = 10,3329 mSS   oldu\u011fundan<\/p>\n

1 At=10mSS  olmaktad\u0131r.<\/p>\n

 <\/p>\n

\n

 <\/p>\n

Bar<\/strong><\/p>\n

Tesisat sekt\u00f6r\u00fcnde \u00f6zellikle s\u0131v\u0131lar\u0131n bas\u0131nc\u0131 i\u00e7in kullan\u0131lan “bar” birimi Yunanca “Baros” yani a\u011f\u0131rl\u0131k ve “Barus” yani a\u011f\u0131r kelimelerinden t\u00fcretilmi\u015ftir. Zaten bas\u0131n\u00e7 \u00f6l\u00e7\u00fcm\u00fcnde kullan\u0131lan Barometre ad\u0131 da bu k\u00f6kten gelmektedir.<\/p>\n

1 bar = 100 000 Pa olarak tan\u0131mlan\u0131r.<\/p>\n

\n

<\/h4>\n

Mutlak Bas\u0131n\u00e7 ve Efektif Bas\u0131n\u00e7<\/h4>\n

D\u00fcnyada normal ko\u015fullarda her yerde hava bas\u0131nc\u0131 vard\u0131r. Tesisatlara tak\u0131lan manometreler bu bas\u0131nc\u0131 \u00f6l\u00e7mez. E\u011fer sistem hen\u00fcz doldurulmam\u0131\u015fsa bas\u0131n\u00e7 olarak s\u0131f\u0131r bar (0 bar) de\u011ferini g\u00f6sterir. Sistem dolduk\u00e7a 1bar, 2bar… diye bas\u0131n\u00e7 artar. \u0130\u015fte bu de\u011fer efektif bas\u0131n\u00e7 de\u011feridir. “bar” biriminin sonuna “gauge” yani g\u00f6sterge ifadesinin ilk harfi “g” eklenerek “barg” olarak g\u00f6sterilir. Yani 0barg, 1 barg, 2barg… Ancak uygulamada e\u011fer “bar” ifadesi varsa bu” barg” olarak anla\u015f\u0131l\u0131r, efektif bas\u0131n\u00e7tan bahsedildi\u011fi bilinir.<\/p>\n

Asl\u0131nda sistemi hen\u00fcz doldurmam\u0131\u015f bile olsak sistemin i\u00e7inde atmosfer bas\u0131nc\u0131 bulunur. Bu bas\u0131n\u00e7 1 Atm civar\u0131ndad\u0131r ve asl\u0131nda manometrenin 1,013 bar g\u00f6stermesi, sistem dolduruldu\u011funda ise manometrenin 1,013’tan ba\u015flayarak artmas\u0131 gerekirdi. Bu durumda mutlak bas\u0131nc\u0131 okumu\u015f olacakt\u0131k. Fakat \u00f6zellikle mutlak bas\u0131n\u00e7 de\u011feriyle \u00e7al\u0131\u015f\u0131lmayacaksa uygulamada atmosfer bas\u0131nc\u0131 s\u0131f\u0131r kabul edilerek sistemin bas\u0131nc\u0131 s\u0131f\u0131rdan ba\u015flayarak artar. E\u011fer \u00f6zellikle mutlak bas\u0131n\u00e7 de\u011feriyle i\u015flem yap\u0131lacaksa “bar” biriminin sonuna “Absolute” yani mutlak s\u00f6zc\u00fc\u011f\u00fcn\u00fcn ilk harfi olan “a” harfi getirilir ve “bara” olarak kullan\u0131l\u0131r.  Yani \u00f6zetle, mutlak bas\u0131n\u00e7, atmosfer bas\u0131nc\u0131 ile efektif bas\u0131nc\u0131n toplam\u0131d\u0131r.<\/p>\n

Bu konuyla ilgili olarak, “At\u00fc” birimine de de\u011finelim. Teknik atmosfer birimi olan “At” yi efektif bas\u0131n\u00e7 anlam\u0131nda kullanacaksak “\u00fcst de\u011fer” kelimesinin ilk harfi “\u00fc” eklenir ve “At\u00fc” ortaya \u00e7\u0131kar ki zaten \u00e7oklukla kar\u015f\u0131m\u0131za \u00e7\u0131kan birim de budur. E\u011fer mutlak bas\u0131n\u00e7 i\u00e7in kullan\u0131lacaksa “Ata” olur.<\/p>\n

\n

Birimlerin d\u00f6n\u00fc\u015f\u00fcm\u00fc<\/h4>\n\n\n\n\n\n\n\n\n
Bas\u0131n\u00e7<\/strong><\/td>\nAtm<\/strong><\/td>\nAt<\/strong><\/td>\nmSS<\/strong><\/td>\nPa<\/strong><\/td>\nbar<\/strong><\/td>\n<\/tr>\n
Atm<\/strong><\/td>\n1<\/td>\n1,033296<\/td>\n10,33296<\/td>\n101332<\/td>\n1,01332<\/td>\n<\/tr>\n
At<\/strong><\/td>\n0,9677769<\/td>\n1<\/td>\n10<\/td>\n98066,77<\/td>\n0,980667689<\/td>\n<\/tr>\n
mSS<\/strong><\/td>\n0,09677769<\/td>\n0,1<\/td>\n1<\/td>\n9806,677<\/td>\n0,098066769<\/td>\n<\/tr>\n
Pa<\/strong><\/td>\n9,86855E-06<\/td>\n1,01971E-05<\/td>\n0,000101971<\/td>\n1<\/td>\n0,00001<\/td>\n<\/tr>\n
bar<\/strong><\/td>\n0,98685509<\/td>\n1,019713417<\/td>\n10,19713417<\/td>\n100000<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

\n

 <\/p>\n

 <\/p>\n

Bu blog i\u00e7in bir video da haz\u0131rlanm\u0131\u015ft\u0131r. Videoyu izlemek i\u00e7in t\u0131klay\u0131n\u0131z<\/a>.<\/p>\n

 <\/p>\n

 <\/p>\n

\n

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

Bas\u0131n\u00e7 Birimleri<\/a><\/p>\n

 <\/p>\n

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

Bas\u0131n\u00e7, tan\u0131m olarak bir y\u00fczeye etkiyen toplam kuvvettir.  D\u00fcnyadan yeterice uzak bir noktada, uzay\u0131n bir k\u00f6\u015fesinde durdu\u011fumuzu ve elimize i\u00e7i c\u0131va(Hg) dolu, \u00fcst\u00fc a\u00e7\u0131k bir kap ald\u0131\u011f\u0131m\u0131z\u0131 varsayal\u0131m. Baz\u0131 n\u00f6trino gibi par\u00e7ac\u0131klar\u0131n etkilerini ihmal edersek bu c\u0131van\u0131n \u00fczerine hi\u00e7 bir kuvvet gelmeyecektir. Yani bas\u0131n\u00e7 s\u0131f\u0131rd\u0131r. Bu duruma mutlak vakum ad\u0131 da verilir. Daha sonra elimize… Bas\u0131n\u00e7<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":3580,"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":[7],"tags":[202,203,204,205,206,207,208],"class_list":["post-2099","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mekanik-tesisat","tag-atu","tag-bar","tag-barg","tag-basinc","tag-civa","tag-manometre","tag-mss"],"_links":{"self":[{"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/posts\/2099","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=2099"}],"version-history":[{"count":0,"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/posts\/2099\/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=2099"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/categories?post=2099"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/benga.pro\/index.php\/wp-json\/wp\/v2\/tags?post=2099"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}