AKIM ŞEMALARI Dr.Hasip Yeniova
1. Giriş 2. Ak õm Şemalar õnõn Gösterilişi. 2.1 Blok diagramlar 2.2 Resim şeklinde gösterilen ak õm şemalar õ 2.3 Ak õş hõzlar õnõn gösterilişi 2.4 Ak õm şemasõ üzerinde yer alan diğer veriler 2.5 Yerleşim planõ (Layout) 2.6 Verilerin kesinliği ve doğruluğu 2.7 Hesaplamalarda kullanõlan temeller 2.8 Kesikli prosesler 2.9 Yardõmcõ üniteler 2.10 Ekipmanlara kod ve ad verilmesi verilmesi Örnek ak õm şemalar õ. Akrilonitril’in polimerizasyonu Nitrik asit üretim prosesi 3. Hesaplamalar 3.1 Temeller 3.2 Ak õm şemasõ üzerinde yer alan çeşitli üniteler için kütle ve enerji ener ji denklikleri. 3.2.1 Şift konvertörü 3.2.2 Dikloretan üretimi 3.2.3 Nitrik asit üretimi 4. Bilgisayar Destekli Ak õm Şemalar õ. 4.1 Tam ve yatõşk õn hal benzeşim (simulasyon) proğramlar õ. 4.2 Bilgi ak õm diagramlar õ. 4.3 Örnek: Nitrobenzen’den Nitrobenzen’den anilin üretimi 5. Basit Kütle Denkliği Proğramlar õ 5.1 Fraksiyon katsayõlar õ kavramõ ve kütle denkliklerinin matriks formda gösterilmesi. 5.2 Örnek: İzopropil alkolden aseton üretimi. 6. Borulandõrma ve Enstrümantasyon (PNI) ANKARA Eylül 2003 1
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AKIM ŞEMALARI 1. GİR İŞ İŞ Ak õm şemalar õ, proses tasar õmõnda anahtar döküman olarak adlandõr õlacak kadar önemli bir konudur. Ak õm şemalar õ üzerinde tasar õmda yer alan ekipmanlar, ak õm hatlar õ, ak õmlar õn hõzlar õ, bileşimleri ekipmanlar õn işletme koşullar õ yer alõr. Hazõrlanan bir ak õm şemasõ proje grubunda yer alan uzman tasar õmcõlar õn kullanõmõna sunulur. Uzman tasar õmcõlar õn yapacaklar õ işler arasõnda borulandõrma ve enstrümantasyon, ekipman tasar õmõ, yerleşim planõnõn hazõrlanmasõ gibi işler yer alõr. İşletme personeli de işletme k õlavuzlar õnõn hazõrlanmasõ ve işletme teknisyenlerinin eğitimi için gerekli dökümanlar õn hazõrlanmas õ amacõyla ak õm şemalar õnõ kullanõrlar. Ak õm şemalar õ, işletmeye alma çalõşmalar õ (start up) sõras õnda prosesin işletme performansõnõn tasar õm değerleriyle k õyaslanmasõ amacõyla da kullanõlõr. Ak õm şemalarõ, prosesde yer alan her bir ünite için ve tüm proses için yapõlan kütle ve enerji denklikleri temel alõnarak oluşturulur. Ak õm şemalar õ üzerinde yer alan niceliklerin hesaplanmasõ, büyük ve kompleks tesisler için oldukça güç ve zaman alõcõdõr. Bu nedenle proses şemalar õ, günümüzde bilgisayar destekli olarak hazõrlanmaktad õr. Bilgisayar destekli olarak çalõşmanõn diğer bir üstünlüğü ise alternatif ak õm şemalar õnõn oluşturulmasõ, en uygun prosesin seçimi ve en uygun proses koşullar õnõn saptanmasõnda getirdiği kolaylõklardõr. Ak õm şemalar õnõn hazõrlanmas õnda yararlanõlan bazõ bilgisayar proğramlar õn adlar õ ve özellikleri ve basit bir ak õm şemasõ hazõrlanmas õnda kullanõlan bir bilgisayar proğramõnõn detaylar õ (Massball) Sinnot, RK.1983, Chemical Engineering Design adlõ kaynakta verilmiştir. Bir prosesin tasar õmõnda, ak õm şemasõnõn ortaya konulmasõndan sonraki adõm borulandõrma ve enstrümantasyon (P&I) şemasõnõn hazõrlanmasõdõr. Bu şema, MÜHENDİSLİK AKIM ŞEMASI veya MEKANİK AKIM ŞEMASI olarak da adland õr õlõr.
2. AKIM ŞEMALARININ GÖSTER İLİŞİ Ak õm şemasõ, prosesi tanõmlayan bir döküman olmasõ olmas õ nedeniyle açõk, anlaşõlõr, doğru ve tam olmalõdõr. Çeşitli tipte ak õm şemalar õ vardõr. Bunlar hakk õnda k õsa bilgiler aşağõda verilmiştir.
2.1 Blok diağramlar Blok diagram en basit gösterim şeklidir. Çizimde yer alan her bir blok bir ekipmanõ veya prosesin belirli bir adõmõnõ simgeler. Bugüne değin okumuş olduğunuz birçok dersin içeriğinde kar şõlaşmõş olduğunuz prosesler size blok diagramlarla gösterildi. Kare, dikdörtgen, daire gibi Şekillerden oluşan ve hatta bazõlar õnda ak õmlar õn ak õş hõzlar õnõn ve bileşimlerininde gösterildiği bu diagramlar õn mühendisler için pek fazla kullanõmõ ve yarar õ yoktur. Bu diagramlar, ön raporlar õn hazõrlanmasõnda ve eğitim amac õyla kullanõlõr. 2.2 Resim Şeklinde gösterilen ak õm Şemalarõ Ayr õntõlõ ak õm şemalar õnda, ekipmanlar genelde belirli bir stile (tarza) uygun olarak çizilirler. Bu
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2.3 Ak õş hõzlarõnõn gösterilişi Her bir bileşenin ak õş hõzõ, toplam ak õş hõzlar õ ve ak õmlar õn yüzde bileşimleri ak õm şemalar õ üzerinde çeşitli Şekillerde gösterilebilir. En basit yöntem, Şekil-1 de akrilonitrilin polimerizasyon prosesi için gösterildiği gibi sadece önemli üniteleri göz önüne alarak ak õm hatlar õ boyunca yerleştirilen bloklar õn içerisine verilerin yazõlmasõdõr. Daha iyi bir ak õm şemasõ Şekil-2 de nitrik asit üretim prosesi için verilmiştir. Bu şemadan görülebileceği gibi her ak õm hattõna bir numara verilerek ak õmlar õn özellikleri şemanõn alt õndaki çizelgeye yazõlmõşt õr. Bu yöntem profesyonel tasar õmcõlar taraf õndan tercih edilir 2.4 Ak õm Şemasõ üzerinde yer alan diğer veriler Ak õm şemalar õ üzerine yazõlan diğer bilgiler müşterinin talebine ve tasar õmcõnõn pratiğine bağlõdõr. Şema üzerine yazõlan bilgileri iki grupta toplayabiliriz: to playabiliriz: Zorunlu bilgiler. 1. Ak õmõn bileşimi; i) her bir bileşenin ak õş hõzõ kg/st olarak veya ii) ak õmõn ağõrl õkça yüzde bileşimi verilir. 2. Ak õmõn toplam ak õş hõzõ kg/st olarak verilir. 3. Ak õmõn sõcaklõğõ oC olarak yazõlõr. 4. Nominal işletme basõncõ (istenilen işletme basõnc õ) yazõlõr. Seçimli bilgiler 1. Molar yüzde bileşimler. 2. Ak õmlar õn ortalama olarak fiziksel özellikleri, örneğin; i) yoğunluk, kg/m3 ii) viskozite, mNs/m2 3. Bir iki kelimeyle tan õmlanabilecek Şekilde ak õmlara birer ad verilmesi. ver ilmesi. 4. Ak õmlar õn entalpisi, kJ/st. Ak õmlar õn fiziksel özelliklerinin, ak õm şemasõnõn hazõrlanmasõndan sorumlu proses mühendisi taraf õndan hesaplanmasõ, ünite tasar õmlar õnõ yapan uzman tasar õmcõlara kolaylõk sağlar. Ayr õca tasar õm ekibinde yer alan tüm gruplar aynõ fiziksel özellikleri kullanmõş olur.
2.5 Yerleşim planõ (LAYOUT) Prosesde yer alan temel ekipmanlar õn ak õm şemasõ üzerinde sõrasõyla gösterilmesidir. Bazen gerekli olduğunda, õsõ değiştirici pompa gibi yard õmcõ ekipmanlarda şema üzerinde sõrasõyla gösterilir. Amaç madde ak õmõnõ gerçekteki şekline uygun olarak gösterebilmektir. Şema üzerinde, temel ekipmanlar gerçek büyüklüklerine k õyasla daha büyük bir orantõda, yardõmcõ ekipmanlar ise daha küçük bir orantõda gösterilirler. Fazla sayõda ünite içeren kompleks proseslerin ak õm şemalar õ tek bir kağõda sõğdõr õlamadõğõ durumda biribirini izleyen birden fazla sayõda kağõt üzerine çizim yap õlõr. Ak õm hatlar õ numaralandõr õlõr ve bir önceki sayfadan gelen veya bir sonraki sayfaya devam eden ak õm hattõnõn numarasõ çift daire içinde gösterilir. Bu ak õm hattõnõn hangi sayfadan geldiği veya hangi sayfaya devam ettiği çift dairenin alt õna sayfa numarasõ yazõlarak belirtilir. Ak õmlar õn özellikleri ve diğer bilgiler ekipman yerleşim planõnõn alt õnda veya üzerinde yer alan
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2.6 Verilerin kesinliği ve doğruluğu Proses ak õm şemalar õ üzerine yazõlan verilerin yüksek bir doğruluk derecesinde olmasõ gerekmez. Verileri virgülden sonra bir basamaklõ olarak yazmak yeterlidir. Eğer ak õm veya bileşenlerden birisi için yazõlacak bir veri uyulmasõ gereken hassasiyet derecesinden çok küçükse bu verinin daha yüksek bir hassasiyet derecesinde yazõlmas õ gerekir. Çok daha küçük değerler ise 'eser' miktarda olarak yazõlõr. Sözü edilen bu çok küçük değer prosesi k õsõtlayan bir öneme sahipse 'ppm' olarak belirtilmelidir. Bazõ durumlarda eser miktardaki maddeler çok önemlidir. Örneğin; katalizör zehirlenmesine neden olan ve malzeme seçimini etkileyen maddeler söz konusu olduğunda bu maddelerin miktarlar õ belirtilmelidir.
2.7 Hesaplamalarda kullanõlan temeller Ak õm şemasõ üzerine yazõlan değerlerin hesaplanmasõnda kullanõlan temeller ak õm şemasõ üzerinde belirtilmelidir. Bu temeller içerisinde; yõllõk işletme süresi, reaksiyon verimi ve fiziksel verimler, enerji denkliğinde kullanõlan ortam sõcaklõğõ, hesaplamalarda kullanõlan varsayõmlar yer alõr.
2.7 Kesikli prosesler Kesikli prosesler için hazõrlanan ak õm şemalar õnda gösterilen değerler tek bir dolum (batch) için gerekli miktarlardõr. E ğer sözü edilen kesikli proses, sürekli bir prosesin bir parçasõ ise ona ait ak õm şemasõ sürekli prosesin ak õm şemasõ içerisinde gösterilir ve sõnõrlar õ belirtilir. Sürekli proseslerde kg/st olarak yazõlan miktarlar keskli proses için kg/dolum olarak yazõlõr. Örneğin sürekli bir polimerizasyon prosesi için gerekli katalizörün hazõrlanmasõ kesikli bir prosesdir.
2.9 Yardõmcõ üniteler Bir kar õşõklõğa neden olmamasõ amacõyla yardõmcõ ünitelerden gelen veya yardõmcõ ünitelere giden ak õm hatlar õ aynõ ak õm şemasõ üzerinde gösterilir ve bu hatlar õn üzerine ne olduklar õ belirtilir.
2.10 Ekipmanlara kod ve ad verilmesi Ak õm şemasõnda yer alan her ekipman bir ad ve kod numarasõ verilerek tanõmlanõr. Ekipmanõn tanõmõ, genellikle bir harf ve bir kaç basamaklõ sayõdan oluşur. Örneğin; R : Reaktör, H: Isõ değiştirici, C: Kolon
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İç k õsõtlamalar: Prosese ve ekipmanlar õn fonksiyonlar õna bağlõ olan k õsõtlamalard õr. i. Proses stokiometrisi, dönüşüm oranõ ve verim ii. iii. iv. v.
Kimyasal denge S õvõ-s õvõ ve gaz-sõvõ ayõrma işlemlerinde faz dengesi Azeotrop kar õşõmlar Enerji denkliğinde kar şõlaşõlan k õsõtlamalar. Örneğin, flash damõtmada olduğu gibi enerji ve kütle denkliklerinin birbirini etkilemesi. etk ilemesi. vi. Ekipman tasar õmõnda kar şõlaşõlan k õsõtlamalard õr. Kompleks proseslerin detay tasar õmõ yapõlmadan önce basit bir ak õm şemasõ oluşturulur. Bu şema tasar õm ekibinde çalõşanlar arasõnda iletişimin kolay sağlanmasõ açõsõndan yararlõdõr.
3.1 TEMELLER 1. Zaman: Hiçbir tesis sürekli olarak işletilemez. Bak õm için gerekli önceden planlanmõş duruşlar olmalõdõr. Ekipmanlar õn temizlenmesi, katalizörlerin yenilenmesi, kolon dolgu maddelerinin değiştirilmesi ve benzeri işler bak õm zamanlar õnda yapõlõr. Duruş sürelerinin sõklõğõ ve bu nedenle ortaya çõkan üretim kayb õ prosesin türüne bağlõdõr. Çoğu kimya ve petrokimya tesisinde yõllõk işletme süresi, bir yõlõn % 90-95 i aras õndadõr. Genellikle 8000 saat kabul edilir. 2. Ölçek faktörü: Ak õm şemasõndaki hesaplamalar õ prosesde yer alan ekipman sõras õna uyarak yapmak kolaylõk sa ğlar. Ham maddeden (girdi) son ürüne doğru bir yol izlenir. İstenilen üretim hõz õ girdi üzerinden değil ürün üzerinden verilir. Bu nedenle hesaplamalar için bir temel seçilmelidir. Örneğin, 100 kmol/st hammadde temel olarak alõnabilir. Bu durumda ak õmlar õn gerçek değeri; her bir ak õmõ, istenilen üretim hõzõ üzerinden hesaplanmõş ölçek faktörüyle faktörüyle çarparak elde edilir. Ölçekfaktörü =
Mol/st ürün 100 kmol hammadde için elde dilen ürün miktar õ (mol))
3.2 AKIM ŞEMASINDA YER ALAN ÇEŞİTLİ ÜNİTELER İÇİN YAPILAN HESAPLAMALAR Reaktörlerde ve denge kademelerinde, sabit ak õmlar õn bileşimlerinin hesaplanmasõnda, enerji ve
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2 1 500oK Temel: 100 kmol/st kuru gaz (girdi) Girdi ak õmõndaki su buhar õ = 3x11 = 33 kmol ; Karbonmonoksidin dönüşüm oranõ, C ise tepkimeye giren karbonmonoksid mol sayõsõ, tepkimenin stokiometrisi de göz önüne alõndõğõnda 11C dir. Aşağõda stokiometrik tablo ve bileşenlerin özgül õsõlar õ verilmiştir. Bileşen CO2 CO H2O H2
Girdi ak õmõ, Mol sayõsõ 8.5 11.0 33.0 76.5
Çõkan ak õm Mol sayõsõ 8.5 + 11C 11(1-C) 33 – 11C 76.5+11C
Cpo A 19.80 30.87 32.24 27.14
b 7.34 E-2 -1.29 E-2 19.24 E-4 9.29 E-3
kJ/kmol K c d -5.6 E-5 17.15 E-9 27.9 E-6 -12.72 E-9 10.56 E-6 -13.60 E-9 -13.81 E-6 7.65 E-9
Konvertörden çõkan gaz ak õmõnõn ideal olduğu ve termodinamik dengeye ulaşt õğõ varsayõlmaktadõr. Ayr õca tepkimenin stokiometrisinden dolayõ denge sabiti üzerine basõncõn etkisi yoktur.
K p =
PCO 2 PH2 PCO PH 2
K p =
(8.5 + 11C) (76.5 + 11C) 11(1 − C) (33 − 11C)
(1)
K p, aynõ zamanda sõcaklõğa bir sabittir. Çeşitli tepkimelerin s õcaklõğa bağlõ olarak denge sabitleri Değişik kaynaklardan yararlanarak bulunabilir. Örneğin tepkimenin serbest enerji değişimi , ∆G biliniyorsa, ∆G = -RTlnK p bağõntõsõndan yararlanarak hesaplanabilir. Veya termodinamik kitaplar õnda 1/T ; logK p şeklinde verilen nomogramlardan yararlanarak bulunabilir *. Bu örnekte, Technical Data on Fuel, Spiers adlõ kaynaktan CO2 + H 2 = CO + H2O tepkimesi için s õcaklõğa bağlõ olarak verilen K p değerleri kullanõlmõşt õr. O nedenle; 1 nolu ba ğõntõ,
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makinesi kullanarak çözüme ulaşmağa çalõşõnõz. Sonuçlar: Dönüşüm Çõkan ak õm Çõkan ak õmõn 2 Oranõ K p.10 Sõcaklõğõ o C K CO CO2 0.88 1.86 550 1.32 18.18 0.79 3.69 600 2.31 17.19 0.68 6.61 650 3.52 15.98
bileşimi,
Mol
H2O 23.32 24.31 25.52
H2 86.18 85.19 83.98
Aktar õlan Isõ, Q -175268 76462 337638
o
Referans sõcakl õk 298 K al õnmõ şt õr .
Yapõlan üç iterasyondan sonra Q; T grafiği çizilerek, Q= 0 için T= 580oK bulunur. C C C
10 20 23 24 25 26 27 28 29 35 40 50 60 70 80 90 100 110 120 130 140
KISA ENERJİ DENKLİĞİ PROGRAMI SİSTEME VER İLEN veya SİSTEMDEN UZAKLAŞTIRILAN ISI MİKTARININ HESAPLANMASI COMMON A(10),B(10),C(10),D(10) A(10),B(10),C(10),D( 10) WRITE(*,*)’ ‘ WRITE(*,*)’ ***** ENERJİ ******* ‘ WRITE(*,*)’ WRIT E(*,*)’ Enerji Denkliği Programõ ‘ WRITE(*,*)’ WRITE(*,*) ’ AKIŞ HIZLARI KMOLl/ST, SICAKLIK’, K, REFERANS SICAKLIK 298 K’ WRITE(*,*) WRITE(*,*) ’******BİLEŞEN SAYISINI GİR İ N NİZ*****’ READ(*,*) N1 WRITE(*,*) ‘ ISI KAPASİTESİ VER İLER, A+BT+CT**2+DT**3’ DO 110 I=1,N1 WRITE(*,*) I,’.BİLEŞEN İÇİ N N A,B,C,D KATSAYILARINI YAZINIZ’ READ(*,*) A(I) , B (I) , C (I) , D (I) WRITE(*,*)A(I) , B (I) , C (I) , D (I) CONTINUE H4=H5=H6=Q=0 WRITE(*,*)’*******SİSTEME GİREN AKIM SAYISINI YAZINIZ****** READ(*,*) S1
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C
C
280 290 300 310 320 330 340 345 350 360 365 366 370 380 390 400 410 420 430 440 450 460 470 480 490 500 510 520 530 535
C C
540 545
CALL DUYISI(I,T1,N2,H4) WRITE(*,*) ‘AKIMIN ‘AKIMIN DUYULAN ISISI=’ ,H4, ‘KJ/ST’ URUN AKIMLARININ TOPLAM DUYULAN ISISI H6=H6+H4 CONTINUE WRITE(*,*) ‘ ******REAKSIYON VE FAZ DEGISIM SAYISINI GIRINIZ’ READ(*,*) N4 IF (N4.EQ.0) GOTO 440 WRITE(*,*) ‘ WRITE(*,*) ‘TEPKIME ISISI/DUYULAN ISI (KJ/MOL) VE DEGISIME’, ‘ UGRAYAN MIKTARI (MOL/ST) GIRINIZ’ GIRINIZ’ WRITE(*,*) ‘SISTEMDE ACIGA CIKAN ISI POZITIF OLARAK ALINACAK’ WRITE(*,*) ‘ABSORBLANAN ISI NEGATIF’ DO 420 I=1,N4 WRITE(*,*) ‘BIR SONRAKI TEPKIME/FAZ DEGISIMI’ READ(*,*) R,F2 H7=F2*R Q1=Q1+H7 CONTIUE ISI DENKLIGI Q=H6-H5-Q1 IF (Q.LT.0) GOTO 480 WRITE(*,*) ‘SISTEME VERILMESI GEREKEN ISI MIKTARI=’ ,Q,’KJ/ST’ GOTO 490 WRITE(*,*) ‘SISTEMIN DISARIYA VERDIGI ISI MIKTARI=’ ,-Q,’KJ/ST’ WRITE(*,*) ‘TEKRAR HESAPLAMA YAPMAK ISTERMISINIZ?’ ‘EVET ISE 1 HAYIR ISE 0 GIRINIZ READ(*,*) P1 IF (P1.EQ.0) GOTO 530 GOTO 120 STOP END ALT PROGRAM AKIMLARIN DUYULAN ISILARININ HESAPLANMASI SUBROUTINE DUYISI (I, T1, N2, H4)
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Örnek 2 Bu örnekte, kütle denkliği hesaplamalar õnda faz dengesi bağõntõlar õnõn (buhar-sõvõ) kullanõlmas õ amaçlanmõşt õr. Etilenin oksihidroklorinasyonu ile dikloroetan (EDC) üretiminde reaktörü terk eden ürün kar õşõmõna seyreltik hidroklorik asit kar õştõr õlarak reaksiyon durdurulur. Bu işleme QUENCH (söndürme), işlemin yap õldõğõ ekipmana ise QUENCH KULESI adõ verilir. Quench kulesinden çõkan gaz ak õmõ bir yoğuşturucuya gönderilerek burada yoğunlaşmayan gaz ak õmõ reaktöre devir ettirilir.
Reaktör
Quench kulesi Yoğuşturucu (Ürün)
4 bar basõnç alt õnda işletilen yo ğuşturucuya giren gaz ak õmõnõn özellikleri aşağõda verilmiştir. Yoğuşturucuyu terk eden ak õmlar õn bileşimlerini hesaplayõnõz. 1 Gaz Girişi EDC 6350 kg/st Etilen 150
1
3 Geri dönen gaz
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Yoğunlaşmayan bileşenler (etilen ve inertler) ba bağlant õ bile şeni olarak alõnõr. Gaz fazõn ideal olduğu ve yoğuşmuş olan EDC ve Suyun birbirleri ile kar õşmadõğõ kabul edilir. (Yoğuşmayanlar õn k õsmi basõncõ) = Toplam basõnç - EDC’nin buhar bas õncõ + Suyun buhar basõncõ = 4 - 0.16 - 0.055 = 3.79 bar Buhar içindeki EDC’nin ak õş hõzõ =
=
EDC'nin buhar bas õncõ (Yoğuşmayanlar õn ak õş hõzõ) Yoğuşmayanlar õn k õsmi basõncõ
0.16 213.4 = 9 kmol/st 3.79
Benzer şekilde buhar buhar fazdaki suyun ak õş hõzõ =
0055 . 213.4 = 3.1 kmol/st 3.79
Bu hesaplamalar õn sonucu olarak gaz ak õmõnõn bileşimi aşağõdaki çizelgede verilmi ştir. EDC H2O Inertler C2H4
Kmol Kmol/s /stt 9 3.1 208 5.4
% mol kg/s kg/stt 4.0 891 1.4 56 92.3 6640 2.3 150
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Kütle denkliği sonuçlar õ. Ak õm No Bileşen EDC H2O Etilen Inertler Toplam Sõcaklõk OC Basõnç Bar
Ak õş hõzlar õ (kg/st) 1 2 Kondenser Kondensat beslemesi 6350 5459 1100 1044 150 Eser 6640 14,240 6503 95 35 4 4
3 Geri dönen gaz 891 56 150 6640 7737 35 4
Örnek .3 Bu örnekte, bile şen kütle dengesi hesaplamalar õnda sõvõ-sõvõ faz dengesinin kullan õlmasõ açõklanmõşt õr. Örnek.2’de tanõmlanan kondenserden ç õkan kondensat ak õmõ, yoğuşmuş su ve dikloretan õ (EDC) ayõrmak için bir dekantöre beslenmektedir. Dekantör ç õk õş ak õmõnõn bileşimini hesaplay õnõz. 2 1
Su fazõ
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0.16 × 5459 = 8.73 kg/st 100
EDC içinde çözünmü ş su miktar õ
=
Dekantörü terkeden su miktar õ
= 1075 – 8.73 = 1066.3 kg/st 1066.3 = × 0.86 = 9.2 kg/st 100 = 5459 - 9.2 = 5449.8 kg/st kg/ st
Sõvõ faz içerisinde çözünmü ş EDC miktar õ Organik faz õn ak õş hõzõ tekrar hesaplan õrsa Organik faz içerisindeki su miktar õ
= (5449.8/100) x 0.16 = 8.72 kg/st
Görüldüğü gibi sonuç ilk yaklaşõmdan çok farklõ değildir. Sonuçlar: Ak õm no Bileşen EDC H2O Toplam
1 Dekantöre Giren ak õm 5459 1075 6534
2 Organik Faz 5449.8 8.7 5458.5
3 Sulu Faz 9.2 1066.3 1075.5
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Örnek. 4 NİTR İK ASİT ÜRETİM PROSESİ Bu örnekte, çe şitli proses üniteleri içeren nitrik asit üretim prosesi için kütle ve enerji denkliklerinin kurulmas õ ve bilgisayar kullan õlmaksõzõn, hesap makinas õyla yapõlan çözümü aç õklanmõşt õr. Amaç; Susuz amonyaktan ç õkarak yõlda 20 000 ton %100 safl õkta nitrik asit üretmek ve daha sonra bu asidi %50-60 sulu çözeltisi haline getirmektir. Proses için ak õm şemasõnõn çizilmesi isteniyor (tabi ki kütle ve enerji denkliklerinin sonuçlar õnõ da içeren bir ak õm şemasõ). Nitrik asit üretimi; esas olarak, amonyağõn oksidasyonuyla olu şan gazlar içerisindeki azot monokisidin suyla absorbe edilerek nitrik asite dönü ştürülmesinden ibarettir. ibarett ir. Kaynaklarda, nitrik asit üretimi için üç de ğişik proses önerilmktedir. Bunlar; 1. Atmosferik basõnçta oksidasyon ve absorbsiyon, 2. Yüksek bas õnçta oksidasyon ve absorbsiyon (yakla şõk 8 atm) 3. Atmosferik basõnçta oksidasyon ve yüksek bas õnçta absorbsiyon. Bu çalõşmada 2 nolu alternatif yani yüksek bas õnçta oksidayon ve absorbsiyon’un yap õldõğõ proses tercih edilmiştir. Başlangõç olarak, sadece prosesde yer alan ekipmanlar õ gösteren ve blok diagramlarla çizilmiş bir ak õm şemas õ oluşturulur. Hava
Hava
Su
Ürün NH3
∼ % 60 HNO3
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Kütle denklikleri için temel olarak reaktöre giren amonyak = 100 kmol olarak al õnõr.
Oksidasyon ünitesi: NH3
Hava
1
3
4
2
1 nolu tepkimenin verimi %96 oldu ğuna göre; oluşan NO = 100 x (96/100) = 96 kmol kmol gerekli oksijen = 96 x (5/4) = 120 kmol oluşan su = 96 x (3/2) = 144 kmol Girdideki amonya ğõn %4 ü ise 2 nolu tepkimeyle azot oluşumu için tüketilmektedir. 1 mol azot
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Yukar õdaki ak õm şemasõ hazõrlanõrken; giren hava ak õmõ içerisindeki su eser miktarda oldu ğu için ihmal edilmiştir. Aynõ şekilde ürün ak õmõ içerisindeki NO 2 miktar õ da çok az oldu ğu için ihmal edilmiştir.
Atõk õsõ kazanõ ve yoğuşturucu için kütle denklikleri: Oksidasyon ünitesinden ç õkan gazlar õn sõcaklõğõ, atõk õsõ kazan õ (WHB) ve yo ğuşturucu ünitelerinden ünitelerinden geçirilerek dü şürülür. At õk õsõ kazanõnda herhangi bir ak õm ayr õlmasõ olmadõğõ için giren ve ç õkan ak õmõn bileşimleri ve toplam kütlesel ak õş hõzlar õ biribirine e şittir. O nedenle WHB için kütle denkliği yapõlmasõna gerek yoktur. Ancak giren ak õm içerisinde bulunan NO ortam s õcaklõğõ azaldõkça NO2 ye oksitlenir. Yo ğuşturucudan ç õk õncaya kadar oksidasyon tamamlanarak NO in tamamõ NO2 ye dönüşmüş olur. Yoğuşturucuya giren gaz ak õmõ içerisindeki su da yo ğunla şarak ağõrlõkça %40-50 HNO3 içeren seyreltik nitrik asit çözeltisi olu şturur. Yo ğuşturucu için kütle denkliği hesaplamalar õ aşağõda gösterilmiştir.
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Oluşan HNO3 = (Girdideki her 100 kmol kmol NH3 başõna oluşan su miktar õ) ( Oluşan 100 kmol HNO 3 i %45 lik çözelti haline getirmek için gerekli su miktar õ) = 100 x (2700/8600) = 31,4 kmol 5 nolu tepkimeyle tüketilen NO 2 miktar õ = 31,4 x (3/2) = 47,1 kmol Tepkimeye giren NO miktar õ = 31,4 x(1+2) = 15,7 kmol Tepkimeye giren su miktar õ = 15,7 kmol Yoğunla şan, fakat NO2 ile tepkimeye girmemi ş olan su miktar õ = 150 – 15,7 = 134,3 kmol Yoğuşturucudan ç õkan gaz ak õmõ içerisindeki NO miktar õ, yoğuşturucuya giren ak õm içerisindeki NO ve NO2 miktarlar õ ile ak õmõn yoğuşturucuda kalma süresine ba ğlõdõr. Çözümü basitle ştirmek amacõyla bu ayr õntõlar õ göz ard õ ederek, yo ğuşturucudan ç õkan gaz ak õmõ içerisindeki NO miktar õnõ, kondensat içerisinde absorplanarak HNO 3 oluşturan NO2 miktar õna eşit alabiliriz.Bu nedenle Yoğuşturucudan ç õkan gaz ak õmõ içerisindeki NO miktar õ = 15,7 kmol
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Yoğuşturucuyu terkeden ak õm içindeki su miktar õ = Suyun mol kesri x Toplam ak õş hõzõ Bu ak õmõn toplam ak õş hõzõ (su buhar õ hariç) = 804,6 kmol -3 Suyun bu ak õm içerisindeki mol kesri = 4,77x10 -3 Su buhar õ miktar õ = 4,77x10 x 804,6 = 3,8 kmol kmol Yoğunla şan su buhar õ miktar õ = 134,3 – 3,8 = 130,5 kmol 6 nolu ak õmdaki su miktar õ = 3,8 kmol = 68,4 kg 7 nolu ak õmdaki su miktar õ = 134,3 – 3,8 130,5 kmol = 2349 kg 6 ve 7 nolu ak õmlar õn ak õş hõzlar õ kmol ve kg olarak a şağõdaki çizelgede özetlenmi ştir. Ak õm no Bileşen
6 (gaz ak õmõ) kmol kg NO
15,7
471
7 (asit ak õmõ) kmol kg Eser E ser
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Bu ak õm içerisindeki NO/NO 2 oranõnõn kesin olarak bilinmemesi gerekli oksijen miktar õnõn hesaplanmas õna etki etmez. Giren ak õm içindeki azotmonoksidi NO 2 ye okside etmek için gerekli oksijen miktar õ (4 nolu tepkimeden yararlan õlarak) = 15,7 / 2 = 7,85 kmol dür. Dolayõsõyla giren ak õm içerisindeki serbest oksijen miktar õ = 20 – 7,85 = 12,15 kmol (4) ve ve (5) nolu nolu tepkimelerin tepkimeler in toplam õ absorbsiyon absorbsi yon ünitesinde NO 2 den HNO3 üretimi amac õyla kullan õlan tepkimeyi verir . Bu da (6) nolu tepkime olarak a şağõda gösterilmiştir.
4 NO 2 + 2 H 2 O + O 2
→ 4 HNO 3
(6 )
Bu tepkimeden yararlanarak, absorbsiyon ünitesinde NO i okistlemek için gerekli oksijen miktar õ hesaplanabilir. NO in oksidasyonu için gerekli oksijen miktar õ = 6 nolu ak õmdaki (NO + NO2 )miktar õ x (1/4) = (48,9 + 15,7) x (1/4) = 16,15 kmol kmol
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Atõk gazlar, absorbsiyon kolonuna gönderilen su ak õmõnõn kolona giriş sõcaklõğõnda su ile doygun hale geçeceklerdir. Su giri ş sõcaklõğõ 25oC ise bu sõcaklõkta suyun k õsmi basõncõ = 0,032 atm atm dir. Absorbsiyon ünites ünitesinin inin bas bas õncõ takriben 8 atm olup -3 Suyun mol kesri = 0,032 0 ,032 / 8 = 4.10 -3 Atõk gaz ak õmõ içerisindeki su buhar õ miktar õ = (831,8 + 25,7) x 4.10 = 3,4 kmol
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13 nolu asit ak õmõnõn bileşimi; 13 Asit akõ mõ kmol kg 94,3 5940,3 HNO3 277,3 4990,8 H2O Toplam 10931,7
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Nitrik asit prosesi için enerji denklikleri Temel: 1 saat Kompresör Kompresör gücü ve enerji gereksinimi hesab õ:
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1, 64 64 −1 1,64 1 , 64 − 1 = 3072,9 kJ / kmol ( ) Birinci bölümişi, giriş 15 C = 1 × 288 × 8, 31 314 × 2 , 8 3 1,64 − 1 o
1, 64 64 −1 1,64 1 ,64 − 1 = 3552 ,6 kJ / kmol ( ) İkinci bölümi şi, giriş 60 C = 1 × 333 × 8,31 314 × 2 , 8 3 1,64 − 1 o
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