BAB I TEORI DASAR
1.1.Pertukaran Kalor Radiasi antara Benda Tak Hitam Radiasi pada benda hitam cukup mudah karena semua radiasi akan diserap langsung. Bila benda tersebut tidak hitam, masalah radiasi akan menjadi lebih sulit dihitung, karena tidak semua energi dari radiasi yang jatuh di dipermukaan tidak semuanya diserap, tetapi sebagian dipantulkan, dapat dipantulkan ke permukaan benda lain aaupun keluar sistem. Terdapat dua definisi baru ada radiasi antar benda arak hitam, yaitu :
G = iradiasi (irradiation) = total radiasi yang menimpa suatu permukaan per satuan waktu per satuan luas
J = radiositas (radiocity) = total radiasi yang meninggalkan suatu permukaan per satuan waktu per satuan luas Pada penyelesaian soal-soal kita sering mengansumsikan bahwa :
Permukaan yang dianalisis bersifat baur
Suhu permukaan yang dianalisis seragam
Sifat-sifat refleksi dan emisinya konstan pada seluruh permukaan
Iradiasi dan radiositas seragam pada setiap permukaan Radiositas dinyatakan dalam rumus :
Dimana ϵ merupakan emisivitas dan E b adalah daya emisi benda hitam. Transmisivitas dapat kita katakana bernilai nol, dan refleksivitas dapat dinyatakan :
Sehingga,
Energi netto yang meninggalkan permukan adalah selisih radiositas dengan iradiasi, yaitu :
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Rumus diatas dimodifikasi menjadi :
1.2. Faktor Bentuk Radiasi Beberapa persamaan yang menghubungkan berbagai faktor bentuk dapat diperoleh dengan memperhatikan gambar dibawah ini.
Gambar 1. Sketsa yang Menunjukkan Beberapa Hubungan Antara Faktor Bentuk
(Sumber: J. P. Holman, 2010) Faktor bentuk radiasi dari bidang A3 ke A1,2 adalah:
(6) (7)
Dengan menggunakan hubungan resprositas:
(8)
atau dengan kata lain
di mana radiasi total yang mencapai permukaan 3 adalah jumlah radiasi dari permukaan 1 dan 2. Misalnya, ingin diketahui nilai faktor bentuk F1-3 untuk permukaan pada gambar 2a
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dengan menggunakan faktor bentuk yang diketahui, untuk siku empat tegak lurus dengan sisi bersama. Persamaan yang diperoleh adalah:
(9)
Selanjutnya, apabila kondisi yang terjadi sesuai dengan gambar 2b, perlu diketahui nilai
F1-4 yang menggunakan faktor bentuk yang sudah diketahui untuk dua siku empat tegak lurus dengan sisi bersama.
Gambar 2 a dan b. Beberapa Hubungan Antar Factor Bentuk
(Sumber: J. P. Holman, 2010) Persamaan yang diperoleh adalah:
di mana:
(10)
dan
sehingga:
(11)
Sebuah persamaan yang tak kalah pentingnya adalah sebagai berikut.
(12)
Hamilton dan Morgan menyajikan persamaan umum untuk siku empat sejajar atau tegak lurus satu sama lain dengan menggunakan faktor bentuk Dua situasi yang menarik perhatian ditunjukkan pada gambar 3 dan 4.
Gambar 3. Segi Empat Tegak Lurus
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(Sumber: J. P. Holman, 2010)
Gambar 4. Segi Empat Sejajar
(Sumber: J. P. Holman, 2010) 2010 )
Untuk siku empat tegak lurus dalam gambar 3 dapat dibuktikan bahwa hubungan resprositas di bawah ini berlaku:
(13)
Dengan menggunakan hubungan resprositas, faktor bentuk radiasi F13 dapat dinyatakan dengan:
(6.25)
di mana suku-suku K didefinisikan sebagai:
(14)
Generalisasi susunan siku empat sejajar terlihat pada gambar 16. Hubungan resprositas
yang berlaku adalah:
(15)
Dengan memanfaatkan hubungan tersebut, diperoleh faktor bentuk F19 sebagai berikut.
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1.3.Radiasi Gas Radiasi yang terjadi antara permukaan suatu benda dengan gas jauh lebih kompleks dibanding pada zat padat. Absorpsi suatu radiasi di lapisan gas dapat di ilustrasikan sebagai berikut.
Gambar 5. Ab Abso sorr si Pad Pada a Suat Suatu u La is isan an Gas Gas
(Sumber : Heat Transfer, 10th Edition, Holman, 2010) 2010) Gambar 5 menunjukkan suatu sinar monokromatis radiasi dengan intensitas
yang ditembakkan
ke lapisan gas dengan ketebalan dx. Penurunan intensitas akibat peristiwa absobrsi sebanding dengan ketebalan lapisan gas dan d an intensitas radiasi pada titik tersebut.
….(16) ….(16) Persamaan 16 dikenal sebagai Hukum Beer, dan transmisivitas monokromatisnya dirumuskan …………..(17) …………..(17) Untuk gas yang tidak memantulkan berlaku persamaan : …………(18) …………(18) Persamaan 16 dan 18 mendeskripsikan variasi intensitas dan absorptivitas untuk lapisan gas dengan ketebalan x ketebalan x.. Emisivitas gas CO2 dan uap air dapat dilihat pada Gambar 12-36 buku Heat buku Heat Transfer 2nd Edition Cengel pada hal. 643. Selain itu, emisivitas gas juga dipengaruhi oleh tekanan. Apabila gas tidak berada pada tekanan 1 atm maka diperlukan faktor koreksi untuk menghitung nilai emisivitasnya. Faktor koreksi ini dapat dilihat melalui Gambar 12-37 pada
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buku Heat buku Heat Transfer 2nd Edition Cengel pada hal. 643. Nilai emisivitas gas juga bergantung pada panjang sinar rata-rata ( Le), Le), yang dirumuskan sebagai :
.......(19) dengan V volume gas dan A luas permukaan total benda yang mengadakan kontak dengan gas. Radiasi Radiasi pada Gas di B enda Abu -Abu Tertu tup
Laju perpindahan panas total secara radiasi pada benda abu-abu lebih kompleks dari benda hitam, namun Hottel telah menemukan korelasinya dengan laju radiasi di benda hitam tertutup yaitu:
.........(20) 1.4.Evaporasi dan Unit Operasi Evaporator Evaporasi merupakan peristiwa penguapan pelarut dari campuran yang terdiri dari zat terlarut dan pelarut yang mudah menguap. Pada contoh proses evaporasi, kebanyakan pelarutnya adalah air. Tujuan dari evaporasi adalah memekatkan konsentrasi larutan yang menjadi zat terlarut sehingga didapatkan larutan degan konsentrasi zat terlarut yang lebih tinggi. Pemisahan evaporasi berbeda dengan distilasi. Pada evaporasi, uap yang dihasilkan biasanya komponen tunggal, dan jika uap yang dihasilkan masih berupa campuran, tidak ada usaha untuk memisahkannya menjadi fraksi-fraksi. Dalam distilasi, uap yang dihasilkan masih memiliki komponen yang lebih dari satu, dan terkadang masih dipisahkan lagi menjadi fraksifraksi. Evaporasi dengan pengeringan juga memiliki perbedaan-perbedaan, yaitu:
Produk akhir yang diinginkan (Evaporasi: zat cair berkonsentrasi tinggi; pengeringan: zat padat)
Kuantitas zat yang diuapkan (Evaporasi : relatif banyak; P engeringan : relatif sedikit)
Media untyuk memanaskan (Evaporasi : Alat pemanas; Pengeringa: Gas)
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Penyebab air/pelarut berpindah/keluar (Evaporasi: Perbedaan titik didih; Pengeringan: Perpindahan massa akibat perbedaan konsentrasi dan/atau tekanan) Unit operasi yang digunakan untuk proses evaporasi disebut evaporator. Evaporator
umumnya memiliki empat komponen dasar, yaitu tabung penguapan, alat penukar panas, kondensor, dan sistem untuk menjaga tekanan vakum. Evaporator banyak digunakan pada industri kimia dan mineral, karena unit operasi tersebut merupakan unit yang vital. Pada industri kimia, contohnya adalah pada industri penghasil garam. Garam diperoleh dari air asin jenuh yang diolah di dalam evaporator. Evaporator mengubah air sebagai pelarut menjadi uap, dan menyisakan residu mineral dalam evaporator. Contoh lainnya adalah pada produksi air mnum. Pada industri ini, sebaliknya, uap air yang menjadi produk yang diinginkan dari evaporator, dimana uap air yang sudah murni karena dipisahkan dari zat kontaminan di evaporator dikondensasi menjadi air. Evaporator memiliki beberapa tipe, yaitu : 1. Evaporator vertikal tabung panjang a. Aliran ke atas (film-panjat) b. Aliran ke bawah (film-jatuh) c. Sirkulasi Paksa 2. Evaporator film-aduk Design Design evaporator sangat bergantung kepada karakteristik dari zat-zat yang dijadikan ), kepekaan bahan feed bahan feed terhadap terhadap suhu, kerak, feed , yaitu konsentrasi, pembentukan busa ( foaming foaming ), dan banyak lagi karakteristik penting lainnya. Selain itu, karakteristik zat-zat yang dijadikan feed dijadikan feed juga mempengaruhi pemiihan bahan konstruksi yang akan digunakan evaporator. 1.5. Neraca Massa dan Panas Evaporator Efek Tunggal
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Gambar 6. Evaorator Efek Tunggal
(Sumber : Anonim) Anonim) Asumsi : Steam yang digunakan adalah steam jenuh Neraca entalpi sekitar evaporator yaitu
Dimana
Sehingga persamaan menjadi :
Perpindahan panas dari steam ke permukaan permukaan pemanas :
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BAB II JAWABAN SOAL PEMICU
2.1.Soal 1 Terdapat 2 buah bidang abu-abu yang keadaannya: o
Bidang I : T1 = 1540 F, ϵ1 = 0,8 o
Bidang II : T2 = 540 F, ϵ2 = 0,5 2
Hitunglah jumlah kalor yang dipindahkan antara bidang I dan bidang II (q = BTU/J.ft ) jika : a. Kedua bidang sangat luas, tetapi letaknya berdekatan satu dengan yang lain. b. Kedua bidang masing-masing berukuran 1 ft x 20 ft dan berjarak 5 ft satu sama lain. c. Bidang I berukuran 3 ft x 9 ft, bidang II berukuran 6 ft x 9 ft, yang letaknya saling tegak lurus satu sama lain, berimpit pada sisi 9 ft. Jawab : o
o
T1 = 1540 F -> 2000 R
Dik :
o
o
Bidang I : T1 = 1540 F / 2000 R
= 0,8
ϵ1
ϵ1 = 0,8. o
o
Bidang II : T2 = 540 F / 1000 R o
ϵ2 = 0,5 Dit. :
o
T2 = 540 F -> 1000 R ϵ2
2
q (BTU/J.ft )
= 0,5
a. Kalor yang dipindahkan jika kedua bidang sangat luas, dan letaknya berdekatan . Digunakan rumus 8-40 pada buku J.P Holman, yaitu :
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b. Kalor yang dipindahkan jika kedua bidang masing-masing berukuran 1 ft x 20 ft dan berjarak 5 ft. Soal b. memiliki cara yang mirip dengan soal sebelumnya, hanya saja terdapat nilai F12 karena luas benda telah diketahui. Nilai F12 dapat dicari dengan grafik 8-12 pada buku J.P Holman, dimana kita harus mengetahui nilai X/D dan Y/D dari benda untuk mencari F12 pada grafik 8-12. Nilai X=panjang benda, Y=lebar benda, dan D=jarak antar benda
Mencari nilai X/D dan Y/D pada benda
Mencari nilai F12 melalui grafik 8-12.
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Gambar 7. Grafik untuk mencari F12 pada bidang datar abu-abu.
(Sumber: Holman, J.P. . 2010. Heat Transfer, 10th Edition. New York: McGraw-Hill Companies, Inc.) Karena tidak ada garis Y/D = 4 pada grafik diatas, maka dilakukan interpolasi dari nilai F12 garis Y/D = 5 dan garis Y/D = 3. -
F12 garis Y/D = 5 = 0.09 F12 garis Y/D = 3 = 0.08
Interpolasi untuk mendapatkan F12 garis Y/D = 4
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c. Kalor yag dipindahkan jika bidang I berukuran 3 ft x 9 ft, bidang II berukuran 6 ft x 9 ft, letaknya saling tegak lurus, dan berimpit pada sisi 9 ft. o
o
T1 = 1540 F -> 2000 R = 0,8
ϵ1
o
o
T2 = 540 F -> 1000 R ϵ2
= 0,5
Asumsi : Tidak ada pengaruh lingkungan pada sistem -
2
A1 = 3ft x 9ft = 27 ft 2 A2 = 6 ft x 9ft = 54 ft
Diketahui jika F11 dan F22 = 0, lalu kita mendapatkan :
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-
Pada benda II
2.2. Soal 2 Hitunglah perpindahan panas radiasi yang terjadi antara gas hasil pembakaran dengan dinding dapur jika diketahui:
Gas hasil pembakaran terdiri atas 10% CO2, 20% H2O, dan sisanya inert (%mol). Tekanan total = 1 atm o
o
Suhu gas = 2040 F, suhu dinding dalam dapur = 540 F
Dinding dapur merupakan benda abu-abu abu-abu dengan ε = 0,9
Dapur berbentuk kubus dengan sisi 5 ft
Faktor dimensi karakteristik untuk radiasi ke seluruh permukaan kubus dianggap = 0,6 L
Jawaban
Pada soal ini, diketahui data-data berikut:
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Jawaban: Pertama-tama marilah kita gambar sistem pada soal nomor 2 ini
Gambar 8. Sistem Pada Nomor 2 (Dinding Dapur Berbentuk Kubus Dengan Sumber Radiasi Berasal Dari Dalam Dinding Dapur)
(Sumber: Dokumentasi Pribadi) Sebelum kita melakukan perhitungan untuk menjawab soal ini, asumsi yang dipakai adalah:
Semua gas pada campuran dalam sistem adalah gas ideal
Hanya CO2 dan H2O yang dianggap berada pada sistem, karena gas lain yang berada pada campuran inert (tidak reaktif)
Emisivitas yang ditentukan adalah emisivitas rata-rata radiasi yang diemisikan ke seluruh permukaan dinding dapur
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Grafik 1. Emisivitas gas CO2 dan H2O di dalam sebuah campuran dengan gas-gas lain pada P total = 1 atm
(Sumber: Cengel, Heat Transfer, 2002) Dari grafik, didapatkan nilai , εc ≈ 0,09 dan εw ≈ 0.15 Kemudian, mencari Δε dengan memplot (Pc.L + Pw.L) vs (Pw/(Pc+Pw)) pada grafik 12-38 dari buku Heat buku Heat Transfer; A Practical Approach karangan Cengel. Pc.L + Pw.L = 0,3 atm.ft + 0,6 atm.ft = 0.9 atm.ft
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Dida pat, Δε = 0,037 dengan menggunakan grafik yang diperuntukkan untuk T = 1200 K and above karena Tg = 1388,56 K. Maka, εgas = εw + εc – Δε – Δε = 0,09 + 0,15 – 0,037 0,037 = 0,203
Menghitung Absorptivitas
Langkah selanjutnya adalah dengan mencari nilai absorptivitas, sebagai berikut.
Plot
vs Ts dan
Practical Approach karangan Cengel.
vs Ts pada grafik 12-36 dari buku Heat Transfer; A
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Kemudian, kita menghitung αc dan αw
Kemudian, Kemudian, mencari Δε dengan memplot (Pc.L + Pw.L) vs (Pw/(Pc+Pw)) pada grafik 12-38 dari buku Heat buku Heat Transfer; A Practical Approach karangan Cengel. Pc.L + Pw.L = 0,3 atm.ft + 0,6 atm.ft = 0.9 atm.ft
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Maka,
α α α α gas =
w +
– Δ c – Δ
= 0,136 + 0,362 – 0,362 – 0,037 0,037 = 0,461
Luas permukaan dari dinding dapur berbentuk kubus, sehingga luas permukaannya adalah
Sehingga, perpindahan kalor secara radiasi yang terjadi pada permukaan abu-abu adalah:
Jadi, perpindahan panas radiasi yang terjadi antara gas hasil pembakaran dengan dinding dapur
adalah sebesar 538.904,48 W atau sebesar 0,39 MW
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Ditanya : Panjang pipa (L) agar perpindahan panas sempurna. Asumsi : -
Perpindahan kalor hanya terjadi secara radiasi
-
Radiasi terhadap gas inert diabaikan
-
Pipa adalah benda hitam 0
Tg2 = 1000 F Inlet
outlet
0
Tw2 = 600 F
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Dari gambar 8-34, J.P. Holman halaman 382, diperoleh nilai
c
pada 1366,3 K yaitu 0,04,
sedangkan dari gambar 8-35, J.P. Holman halaman 383, diperoleh nilai 0,014 dengan persamaan
pada
1366,3 K yakni
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Maka nilai εg adalah
g
Langkah selanjutnya mencari
Keterangan :
yang dirumuskan secara matematis menjadi :
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= (0,103) (1atm) = 0,103 atm
034 034 atm-ft
Kemudian kita analisis untuk tiap bagian pada outlet: Dari gambar 8-34, J.P. Holman halaman 382, diperoleh nilai
c pada
Dan dari gambar 8-35, J.P. Holman halaman 383, diperoleh nilai
1366,3 K yaitu 0,05.
pada
1366,3 K yakni
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Keterangan :
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2.4. Soal 4 Suatu evaporator digunakan untuk mengkonsentrasikan 4536 kg/jam (10000 lb/jam) larutan o
o
NaOH 20% di dalam air, yang masuk pada suhu 60 C (140 F) untuk menghasilkan produk yang mengandung 50% zat padat. Tekanan steam Tekanan steam jenuh jenuh yang digunakan adalah 172,4 KPa (25 psia) dan tekanan di bagian uap evaporator adalah 11,7 KPa (1,7 psia). Koefisien transfer panas 2
2o
menyeluruh adalah 1560 W/m K (275 BTU/J.ft F). Hitung kebutuhan steam steam yang digunakan, 2
, dan luas permukaan pemanasan dalam m
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Kuantitas yang menguap :
Konsumsi uap
-
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-
o
Hf saat fraksi 20% dan suhu 140 F o
-
H saat fraksi 50% dan suhu 190 F
-
Hc saat suhu 190 F, 1,7 lbf/in , dan pada fase superheated fase superheated Lihat tabel superheated tabel superheated
o
2
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