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Dastek 911 views Apple Cooling qfrig Share Like Download ...

Muhammad Luthfan, Divisi KWU at Himalogista Follow Published on May 29, 2015

Materi kuliah tentang Dastek. Cari lebih banyak di; http:// INTRODUCTION • Food process engineering: includes the part of human activity to convert raw material to be ready or proces... ... Published in: Science 0 Comments 0 Likes Statistics Notes

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Theaimsofthefoodindustry • 1. To extend the period during which a food remains wholesome (the shelf life) by preservation ... Share your thoughts… Post Be the first to comment Be the first to like this No Downloads Views Total views 911 On SlideShare 0 From Embeds 0 Number of Embeds 12 Actions Shares 0Food processes are usually schematized by means of flow charts diagrams indicate different manufacturing steps, as well as... Downloads 17 Comments 0 Likes 0 Embeds 0 No embeds No notes for slide

Dastek 1. 1. DasarKeteknikanPengolahan Pangan Sudarminto Setyo Yuwono 2. 2. Apple Cooling qfrig 3. 3. INTRODUCTION • Food process engineering: includes the part of human activity to convert raw material to be ready or processed foods • The main objective: to study the principles and laws governing the physical, chemical, or biochemical stages of different processes, and the apparatus or equipment by which such stages are industrially carried out • The study of process engineering is an attempt to combine all forms of physical processing into a small number of basic operations, which are called unit operations • Food processes may seem bewildering in their diversity, but careful analysis will show that these complicated and differing processes can be broken down into a small number of unit operations • Important unit operations in the food industry are fluid flow, heat transfer, drying, evaporation, contact equilibrium processes (which include distillation, extraction, gas absorption, DIMENSIONSANDUNITS • All engineering deals with definite and measured quantities, and so depends on the making of measurem... crystallization, and membrane processes), mechanical separations (which include filtration, centrifugation, sedimentation and sieving), size reduction and mixing. 4. 4. Theaimsofthefoodindustry • 1. To extend the period during which a food remains wholesome (the shelf life) by preservation techniques which inhibit microbiological or biochemical changes and thus allow time for distribution, sales and home storage. • 2. To increase variety in the diet by providing a range of attractive flavours, colours, aromas and textures in food (collectively known as eating quality, sensory characteristics or organoleptic quality); a related aim is to change the form of the food to allow further processing (for example the milling of grains to flour). • 3. To provide the nutrients required for health (termed nutritional quality of a food). • 4. To generate income for the manufacturing company 5. 5. Food processes are usually schematized by means of flow charts diagrams indicate different manufacturing steps, as well as the flow of materials and energy in the process. There are different types of flow charts; the most common use “blocks” or “rectangles.” connected by arrows to indicate the way in which the materials flow. 6. 6. DIMENSIONSANDUNITS • All engineering deals with definite and measured quantities, and so depends on the making of measurements • To make a measurement is to compare the unknown with the known • record of a measurement consists of three parts: the dimension of the quantity, the unit which represents a known or standard quantity and a number which is the ratio of the measured quantity to the standard quantity 7. 7. 1.2 BESARAN DAN SATUAN1.2 BESARAN DAN SATUAN Besaran : Sesuatu yang dapat diukur dinyatakan dengan angka (kuantitatif) Contoh : panjang, massa, waktu, suhu, dll. Mengukur : Membandingkan sesuatu dengan sesuatu yang lain yang sejenis yang ditetapkan sebagai satuan. contoh : panjang jalan 10 km Besaran Fisika baru terdefenisi jika : ada nilainya (besarnya) ada satuannya nilai satuan 1.4 8. 8. Satuan : Ukuran dari suatu besaran ditetapkan sebagai satuan. Contoh : Sistem satuan : ada 2 macam 1. Sistem Metrik : a. mks (meter, kilogram, sekon) b. cgs (centimeter, gram, sekon) 2. Sistem Non metrik (sistem British) Sistem Internasional (SI) Sistem satuan mks yang telah disempurnakan yang paling banyak dipakai sekarang ini. Dalam SI : Ada 7 besaran pokok berdimensi dan 2 besaran pokok tak berdimensi meter, kilometer satuan panjang detik, menit, jam satuan waktu gram, kilogram satuan massa dll. 1.5 9. 9. NO Besaran Pokok Satuan Singkatan Dimensi 1 Panjang Meter m L 2 Massa Kilogram kg M 3 Waktu Sekon s T 4 Arus Listrik Ampere A I 5 Suhu Kelvin K 6 Intensitas Cahaya Candela cd j 7 Jumlah Zat Mole mol N 7 Besaran Pokok dalam Sistem internasional (SI)7 Besaran Pokok dalam Sistem internasional (SI) NO Besaran Pokok Satuan 1.2 BESARAN DAN SATUAN1.2 BESARAN DAN SATUAN Besaran : Sesuatu yang dapat diukur dinyatakan dengan angka (kuantitatif)... Singkatan Dimensi 1 Sudut Datar Radian rad - 2 Sudut Ruang Steradian sr - Besaran Pokok Tak Berdimensi 1.6 10. 10. Dimensi Cara besaran itu tersusun oleh besaran pokok. Besaran Turunan Besaran yang diturunkan dari besaran pokok. 1. Untuk menurunkan satuan dari suatu besaran 2. Untuk meneliti kebenaran suatu rumus atau persamaan - Metode penjabaran dimensi : 1. Dimensi ruas kanan = dimensi ruas kiri 2. Setiap suku berdimensi sama - Guna Dimensi : 1.7 11. 11. Contoh : a. Tidak menggunakan nama khusus NO Besaran Satuan 1 Kecepatan meter/detik 2 Luas meter 2 b. Mempunyai nama khusus NO Besaran Satuan Lambang 1 Gaya Newton N 2 Energi Joule J 3 Daya Watt W 4 Frekuensi Hertz Hz 1.8 12. 12. Besaran Turunan dan Dimensi NO Besaran Pokok Rumus Dimensi 1 Luas panjang x lebar [L]2 2 Volume panjang x lebar x tinggi [L]3 3 Massa Jenis [m] [L]-3 4 Kecepatan [L] [T]-1 5 Percepatan [L] [T]-2 6 Gaya massa x percepatan [M] [L] [T]-2 7 Usaha dan Energi gaya x perpindahan [M] [L]2 [T]-2 8 Impuls dan Momentum gaya x waktu [M] [L] [T]-1 massa volume perpindahan waktu kecepatan waktu 1.9 13. 13. Faktor Penggali dalam SI NO Faktor Nama Simbol 1 10 -18 atto a 2 10 -15 femto f 3 10 -12 piko p 4 10 -9 nano n 5 10 -6 mikro µ 6 10 -3 mili m 7 10 3 kilo K 8 10 6 mega M 9 10 9 giga G 10 10 12 tera T 1.10 14. 14. 1. Tentukan dimensi dan satuannya dalam SI untuk besaran turunan berikut : a. Gaya b. Berat Jenis c. Tekanan d. Usaha e. Daya Jawab : b. Berat Jenis = = = = MLT -2 (L-3 ) = ML-2 T-2 satuan kgm-2 berat volume Gaya Volume MLT -2 L3 a. Gaya = massa x percepatan = M x LT -2 = MLT -2 satuan kgms-2 c. Tekanan = = = MLT -2 satuan kgm-1 s-1 gaya luas MLT -2 L2 d. Usaha = gaya x jarak = MLT -2 x L = ML 2 T -2 satuan kgm-2 s-2 e. Daya = = = ML 2 T -1 satuan kgm-2 s-1 usaha waktu ML 2 T -2 T Contoh SoalContoh Soal 1.11 15. 15. 2. Buktikan besaran-besaran berikut adalah identik : a. Energi Potensial dan Energi Kinetik b. Usaha/Energi dan Kalor Jawab : a. Energi Potensial : Ep = mgh Energi potensial = massa x gravitasi x tinggi = M x LT-2 x L = ML2 T-2 Energi Kinetik : Ek = ½ mv2 Energi Kinetik = ½ x massa x kecepatan2 = M x (LT-1) 2 = ML2 T-2 Keduanya (Ep dan Ek) mempunyai dimensi yang sama keduanya identik b. Usaha = ML2 T-2 Energi = ML2 T-2 Kalor = 0.24 x energi = ML2 T-2 Ketiganya memiliki dimensi yang sama identik 1.12 16. 16. Dimensionless Ratios • It is often easier to visualize quantities if they are expressed in ratio form and ratios have the great advantage of being dimensionless • For example, specific gravity is a simple way to express the relative masses or weights of equal volumes of various materials. The specific gravity is defined as the ratio of the weight of a volume of the substance to the weight of an equal volume of water • SG = weight of a volume of the substance/ weight of an equal volume of water . Dimensionally, SG=[F]/ [L]-3 divided Satuan : Ukuran dari suatu besaran ditetapkan sebagai satuan. Contoh : Sistem satuan : ada 2 macam 1. Sistem Metrik : ... by[F]/ [L]-3 = 1 • it gives an immediate sense of proportion • This sense of proportion is very important to food technologists as they are constantly making approximate mental calculations for which they must be able to maintain correct proportions • Another advantage of a dimensionless ratio is that it does not depend upon the units of measurement used, provided the units are consistent for each dimension • Dimensionless ratios are employed frequently in the study of fluid flow and heat flow. These dimensionless ratios are then called dimensionless numbers and are often called after a prominent person who was associated with them, for example Reynolds number, Prandtl number, and Nusselt number 17. 17. Suhu dan komposisi • C, F, K • Fraksi mol, konsentrasi • Suatu wadah berisi 50g air dan 50 g NaOH, berapa fraksimol masing-masing • Albumin 2% berat memiliki densitas 1,028g/cm3 . Berat molekul albumin 67000 g/g mol. Berapa fraksi mol masing-masing komponen 18. 18. Neraca Massa • Sangat penting dalam menentukan efisiensi proses dan memprediksi hasil akhir proses • Rumus umum => massa in = massa out + akumulasi • Neraca massa: – Proses-proses yang tidak terjadi reaksi kimia – Proses-proses yang terjadi reaksi kimia 19. 19. Proses yang tidak terjadi reaksi kimia • Proses yang tidak mengalami reaksi kimia: – Pengeringan, – pembekuan, – pemekatan, – kristalisasi, – Pencampuran, dsb • Reaksi kimia mungkin terjadi pada proses tersebut namun tidak terlalu mempengaruhi massa total 20. 20. Tahapan perhitungan • Gambar diagram • Tulis reaksi kimia jika ada • Tulis dasar-dasar perhitungan • Hitung neraca massanya 21. 21. Contoh neraca massa • Larutan soda api (NaOH), sebanyak 1000 kg/jam mengandung 10% NaOH di pekatkan pada evaporator sehingga kadarnya menjadi 60%. Hitung larutan NaOH pekat yang dihasilkan. • Cabe 100 kg berkadar air 80% dikeringkan hingga kadar air 10%. Berapa kilogram cabe kering yang dihasilkan 22. 22. Dikerjakan dan dikumpulkan • Proses produksi selai buah dilakukan dengan cara memekatkan bubur buah dari kadar padatan 10% menjadi 30%. Pemekatan dilakukan dalam 2 tahap evaporator. Pada evaporator yang pertama kadar padatan meningkat menjadi 22%. Hitung selai buah yang dihasilkan untuk tiap 100 kg/jam bubur buah yang dipakai. 23. 23. Tugas dikerjakan dan dikumpulkan • Adonan biskuit diperoleh dengan mencampurkan Terigu sebanyak 60% berat, gula 10%, telur 10%, garam 3%, mentega 12% dan air 5%. Jika diketahui kadar protein terigu dan telur sebesar 10% dan 15% berapa kadar protein adonan. • Proses pembuatan daging burger dilakukan dengan mencampurkan daging sapi dengan lemak sapi. Daging sapi memiliki kadar protein 15%, lemak 20% dan air 63%, sedangkan lemak sapi berkadar protein 3%, lemak 80%, air 15%. Berapa daging sapi dan lemak sapi yang ditambahkan untuk memperoleh adonan daging burger sebanyak 100 kg dengan kadar lemak 25%? 24. 24. Batas proses (boundary) • Batas proses dapat digunakan untuk menyederhanakan suatu proses • Dapat diperluas atau diperkecil NO Besaran Pokok Satuan Singkatan Dimensi 1 Panjang Meter m L 2 Massa Kilogram kg M 3 Waktu Sekon s T 4 Arus Listrik Amper... 25. 25. Dikerjakan • Niratebu1000kg/jamberkadargula20% dipekatkanhinggakadargula60%. Nira pekatselanjutnyadikristalisasi padasuhu 20o C. Konsentrasi kejenuhangulapadasuhu 20o Csebesar40%. Berapakg/jamkristalgula yangdihasilkan?Diasumsikankristalgula tidakmengandungair 26. 26. Proses pencampuran • Draw a diagram and set up equations representing total mass balance and component mass balance for a system involving the mixing of pork (15% protein, 20% fat, and 63% water) and backfat (15% water, 80% fat, and 3% protein) to make 100 kg of a mixture containing 25% fat. 27. 27. • Draw a diagram and set up a total mass and component balance equation for a crystallizer where 100 kg of a concentrated sugar solution containing 85% sucrose and 1% inert, water- soluble impurities (balance, water) enters. Upon cooling, the sugar crystallizes from solution. A centrifuge then separates the crystals from a liquid fraction, called the mother liquor. The crystal slurry fraction has, for 20% of its weight, a liquid having the same composition as the mother liquor. The mother liquor contains 60% sucrose by weight. 28. 28. Neraca massa jika terjadi reaksi kimia • Beberapa proses pengolahan kemungkinan terjadi reaksi kimia – Fermentasi – Pembakaran – Netralisasi • Dasar perhitungan bukan dari massa tetapi dari perubahan mol • Setelah itu baru dikonversikan ke massa 29. 29. contoh • Pembakaran C • Pembuatan sodium sitrat C6H5Na3O7 dari asam sitrat C6H8O7 dengan NaOH • Gas LPG : Propana (C3H8) dan Butana (C4H10), serta sejumlah kecil Etana (C2H6,) dan Pentana (C5H12). 30. 30. Tahapan • Konversikan semua massa menjadi mol • Dari reaksi kimia hitung jumlah mol yang dibutuhkan serta mol produk • Neraca massa diperoleh dengan mengkonversi mol bahan dan mol produk menjadi massa 31. 31. Harap dikerjakan • Larutan NaOH diproduksi dengan cara menambahkan larutan Na2CO3 berkadar 10% ke dalam aliran bubur Ca(OH)2 yang berkadar 25%. Bagaimana komposisi bubur akhir (komponen dan kadarnya) jika reaksi 90% sempurna. Gunakan dasar 100 kg/jam aliran bubur Ca(OH)2 • Ca(OH)2 + Na2CO3 => 2NaOH + CaCO3 • MR Ca(OH)2= 74,1; MR Na2CO3 = 106 32. 32. contoh • Bahan bakar mengandung 5 %mol H2, 30 %mol CO, 5 %mol CO2, 1 %mol O2, dan 59 %mol N2. Dibakar dengan media udara. Untuk 100 kg mol bahan bakar hitung mol gas buang dan komponennya, jika : • A. Pembakaran sempurna, udara pas • B. Pembakaran 90% sempurna, udara pas • C. Udara berlebih 20%, pembakaran sempurna 80% 33. 33. • Bubur susu berkadar air 80%. Pada proses fermentasi bubur susu, Laktosa C12H22O11 dioksidasi • Untuk 100g bubur susu, jika sebanyak 1 g laktosa yang dioksidasi, berapa Dimensi Cara besaran itu tersusun oleh besaran pokok. Besaran Turunan Besaran yang diturunkan dari besaran pokok. 1. U... kadar air bubur susu setelah fermentasi 34. 34. • Selai dibuat dengan formulasi 45 bagian adalah buah dan 55 bagian adalah gula.. Untuk menghasilkan gel yang baik, maka kandungan padatan terlarut selai minimal 65%. Proses pembuatan meliputi pencampuran bubur buah, gula, dan pektin lalu dievaporasi sehingga diperoleh selai. Pektin yang ditambahkan pada pembuatan selai adalah pectin 100 grade (untuk tiap 1 kg pektin memerlukan gula 100kg). Jumlah pektin yang ditambahkan bergantung pada jumlah gula yang digunakan. Jika bubur buah mengandung padatan terlarut 10% hitung kebutuhan bubur buah, gula dan pektin yang ditambahkan untuk menghasilkan 100 kg selai. Pektin tidak mengandung padatan terlarut. 35. 35. • Sodium sitrat (Na2C6H6O7) dibentuk dengan mereaksikan larutan asam sitrat (C6H8O7) 10% (berat) dengan bubur NaOH 50% (berat). Untuk tiap 100 kg larutan asam sitrat, buat neraca massanya (reaksi berlangsung sempurna) • Berat atom O : 16; C : 12; H : 1; Na : 23 36. 36. C12H22O11 + 12O2 =>12CO2 + 11H2O • Suatu larutan asam sitrat (C6H8O7) 12%(berat) direaksikan dengan NaOH sehingga terbentuk Sodium Sitrat (Na2C6H6O7). Sodium sitrat yang terbentuk dipekatkan sehingga diperoleh larutan dengan konsentrasi 35% berat. Larutan lalu didinginkan pada suhu 15o C untuk mengkristalkan sodium sitrat. Jika kelarutan sodium sitrat pada suhu 15o C sebesar 20% berat, hitung kristal sodium sitrat yang diperoleh, untuk setiap 100 kg asam sitrat • Asumsi : - Sodium sitrat dalam bentuk anhydrous • Berat atom O : 16; C : 12; H : 1; Na : 23 • Reaksi berlangsung secara sempurna 37. 37. Recycle • Proses pengulangan ke tahap sebelumnya dengan tujuan memperbaiki sifat produk sesuai kebutuhan • Banyak digunakan pada proses – Evaporasi – Kristalisasi – Fermentasi • Tahap : – perluas batasan proses – Hitung yang direcycle 38. 38. Contoh Recycle dikerjakan • Pada suatu proses produksi sodium sitrat, 1000kg/jam larutan sodium sitrat berkadar 10% dipekatkan di suatu evaporator bersuhu 353K sehingga diperoleh kadar 40%. Larutan lalu dimasukkan ke kristalizer yang bersuhu 303K sehingga diperoleh kristal Na sitrat berkadar air 5%. Larutan jenuh yang mengandung 30% Na sitrat lalu direcycle ke evaporator. Hitung berapa laju aliran recycle dan produk yang dihasilkan. 39. 39. Harap dikerjakan • Pada industri gula, larutan gula 1000 kg/jam berkadar 25% dipekatkan hingga berkadar 55%. Larutan tersebut lalu dimasukkan ke kristalizer sehingga diperoleh kristal gula berkadar air 15%. Larutan jenuh berkadar gula 40% selanjutnya direcycle ke evaporator lagi. Kristal gula yang dihasilkan lalu dikeringkan hingga berkadar air 5%. Hitung jumlah larutan yang direcycle dan gula yang dihasilkan. 40. 40. • Evaporator berkapasitas menguapkan air sebanyak 10 kg/jam sehingga kadar padatan berubah dari 5,5% menjadi 25%. Untuk meningkatkan kualitas produk, sebagian Contoh : a. Tidak menggunakan nama khusus NO Besaran Satuan 1 Kecepatan meter/detik 2 Luas meter 2 b. Mempunyai nama khusu... konsentrat di-recycle dan dicampurkan dengan bahan masuk dengan menggunakan pompa berkapasitas 20 kg campuran/jam. Hitung berapa banyak aliran konsentrat yang dihasilkan serta aliran re-cycle nya. 41. 41. FLUID FLOW THEORY • Many raw materials for foods and many finished foods are in the form of fluids. • Thin liquids - milk, water, fruit juices, Thick liquids - syrups, honey, oil, jam, Gases - air, nitrogen, carbon dioxide, Fluidized solids - grains, flour, peas. • The study of fluids can be divided into: – the study of fluids at rest - fluid statics, and – the study of fluids in motion - fluid dynamics. 42. 42. FLUID STATICS • very important property : the fluid pressure • Pressure is force exerted on an area • force is equal to the mass of the material multiplied by the acceleration due to gravity. • mass of a fluid can be calculated by multiplying its volume by its density • F = mg = Vg • F is force (Newton) or kg m s-2 , m is the mass, g the acceleration due to gravity, V the volume and the density. 43. 43. The force per unit area in a fluid is called the fluid pressure. It is exerted equally in all directions. • F = APs + ZAg • Ps is the pressure above the surface of the fluid (e.g. it might be atmospheric pressure • total pressure P = F/A = Ps + Zg • the atmospheric pressure represents a datum P = Zg 44. 44. EXAMPLE . Total pressure in a tank of peanut oil • Calculate the greatest pressure in a spherical tank, of 2 m diameter, filled with peanut oil of specific gravity 0.92, if the pressure measured at the highest point in the tank is 70 kPa. 45. 45. • Density of water = 1000 kg m-3 Density of oil = 0.92 x 1000 kg m-3 = 920 kg m-3 Z =greatest depth = 2 m and g = 9.81 m s-2 Now P = Zg = 2 x 920 x 9.81 kg m-1 s-2 = 18,050 Pa = 18.1 kPa. • To this must be added the pressure at the surface of 70 kPa. • Total pressure = 70 + 18.1 = 88.1 kPa. • the pressure depends upon the pressure at the top of the tank, the depth of the liquid 46. 46. Expressing the pressure • absolute pressures • gauge pressures • head 47. 47. EXAMPLE. Head of Water • Calculate the head of water equivalent and mercury to standard atmospheric pressure of 100 kPa. • Density of water = 1000 kg m-3, Density of mercury = 13,600 kg m-3 g = 9.81 m s-2 and pressure = 100 kPa = 100 x 103 Pa = 100 x 103 kg m-1s-2. Water Z = P/ g = (100 x 103 )/ (1000 x 9.81) = 10.2 m Mercury Z = (100 x 103 )/ (13,600 x 9.81) = 0.75m Besaran Turunan dan Dimensi NO Besaran Pokok Rumus Dimensi 1 Luas panjang x lebar [L]2 2 Volume panjang x lebar x tinggi [... 48. 48. FLUID DYNAMICS • In most processes fluids have to be moved • Problems on the flow of fluids are solved by applying the principles of conservation of mass and energy • The motion of fluids can be described by writing appropriate mass and energy balances and these are the bases for the design of fluid handling equipment. 49. 49. Mass Balance • 1A1v1 = 2A2v2 • incompressible 1 = 2 so in this case • A1v1 = A2v2 (continuity equation) • area of the pipe at section 1 is A1 , the velocity at this section, v1 and the fluid density 1 , and if the corresponding values at section 2 are A2, v2, 2 50. 50. EXAMPLE. Velocities of flow • Whole milk is flowing into a centrifuge through a full 5 cm diameter pipe at a velocity of 0.22 m s-1 , and in the centrifuge it is separated into cream of specific gravity 1.01 and skim milk of specific gravity 1.04. Calculate the velocities of flow of milk and of the cream if they are discharged through 2 cm diameter pipes. The specific gravity of whole milk of 1.035. 51. 51. Solving • 1A1v1 = 2A2v2 + 3A3v3 • where suffixes 1, 2, 3 denote respectively raw milk, skim milk and cream. • since the total leaving volumes equal the total entering volume • A1v1 = A2v2 + A3v3 • v2 = (A1v1 - A3v3 )/A2 • 1A1v1 = 2A2(A1v1 – A3v3)/A2 + 3A3v3 • 1 A1v1 = 2 A1v1 - 2 A3v3 + 3 A3v3 • A1v1(1 - 2 ) = A3v3(3 - 2 ) 52. 52. • A1 = (/4) x (0.05)2 = 1.96 x 10-3 m2 • A2 = A3 = (/4) x (0.02)2 = 3.14 x 10-4 m2 v1 = 0.22 m s-1 1 = 1.035, 2 = 1.04, 3 = 1.01 • A1v1(1 - 2 ) = A3v3(3 - 2 ) • -1.96 x 10-3 x 0.22 (0.005) = -3.14 x 10-4 x v3 x (0.03) v3 = 0.23 m s-1 • v2 = (A1v1 - A3v3 )/A2 • v2 = [(1.96 x 10-3 x 0.22) - (3.14 x 10-4 x 0.23)] / 3.14 x 10-4 = 1.1m s-1 53. 53. Energy Balance • Referring Fig. before we shall consider the changes in the total energy of unit mass of fluid, one kilogram, between Section 1 and Section 2. • Firstly, there are the changes in the intrinsic energy of the fluid itself which include changes in: (1) Potential energy = Ep = Zg (J) (2) Kinetic energy = Ek = v2 /2 (J) (3) Pressure energy = Er = P/ (J) • Secondly, there may be energy interchange with the surroundings including: (4) Energy lost to the surroundings due to friction = Eƒ (J). (5) Mechanical energy added by pumps = Ec (J). (6) Heat energy in heating or cooling the fluid • In the analysis of the energy balance, it must be remembered that energies are normally measured from a datum or reference level. • Ep1 + Ek1 + Er1 = Ep2 + Ek2 + Er2 + Ef - Ec. • Z1g + v1 2 /2 + P1/1 = Z2g + v2 2 /2 + P2/2 + Ef - Ec. • Zg + v2 /2 + P/ = k Persamaan Bernouilli 54. 54. Water flows at the rate of 0.4 m3 min-1 in a 7.5 cm diameter pipe at a pressure of 70 kPa. If the pipe reduces to 5 cm diameter calculate the new pressure in the pipe. Density of water is 1000 kg m-3 . • Flow rate of water = 0.4 m3 min-1 = 0.4/60 m3 s-1 . • Area of 7.5 cm diameter pipe = (/4)D2 = ( /4)(0.075)2 = 4.42 x 10-3 m2 . So velocity of flow in 7.5 cm diameter pipe, v1 = (0.4/60)/(4.42 x 10-3 ) = 1.51 m s-1 • Area of 5 cm diameter pipe = (/4)(0.05)2 = 1.96 x 10-3 m2 and so velocity of flow in 5 cm diameter pipe, v2 = Faktor Penggali dalam SI NO Faktor Nama Simbol 1 10 -18 atto a 2 10 -15 femto f 3 10 -12 piko p 4 10 -9 nano n 5 10 -6 mik... (0.4/60)/(1.96 x 10-3 ) = 3.4 m s-1 • Now • Z1g + v1 2 /2 + P1 /1 = Z2g + v2 2 /2 + P2 / 2 and so 0 + (1.51)2 /2 + 70 x 103 /1000 = 0 + (3.4)2 /2 + P2/1000 0 + 1.1 + 70 = 0 + 5.8 + P2/1000 P2/1000 = (71.1 - 5.8) = 65.3 P2 = 65.3k Pa. 55. 55. Water is raised from a reservoir up 35 m to a storage tank through a 7.5 cm diameter pipe. If it is required to raise 1.6 cubic metres of water per minute, calculate the horsepower input to a pump assuming that the pump is 100% efficient and that there is no friction loss in the pipe. 1 Horsepower = 0.746 kW. • Volume of flow, V = 1.6 m3 min-1 = 1.6/60 m3 s-1 = 2.7 x 10-2 m3 s-1 • Area of pipe, A = (/4) x (0.075)2 = 4.42 x 10-3 m2, • Velocity in pipe, v = 2.7 x 10-2 /(4.42 x 10-3 ) = 6 m s-1 , • And so applying eqn Z1g + v1 2 /2 + P1/1 = Z2g + v2 2 /2 + P2/2 + Ef - Ec. • Ec = Zg + v2 /2 • Ec = 35 x 9.81 + 62 /2 = 343.4 + 18 = 361.4 J • Therefore total power required • = Ec x mass rate of flow = EcV = 361.4 x 2.7 x 10-2 x 1000 J s-1 = 9758 J s-1 • and, since 1 h.p. = 7.46 x 102 J s-1 , • required power = 13 h.p. 56. 56. VISCOSITY • Viscosity is that property of a fluid that gives rise to forces that resist the relative movement of adjacent layers in the fluid. • Viscous forces are of the same character as shear forces in solids and they arise from forces that exist between the molecules. • If two parallel plane elements in a fluid are moving relative to one another, it is found that a steady force must be applied to maintain a constant relative speed. This force is called the viscous drag because it arises from the action of viscous forces. 57. 57. If the plane elements are at a distance Z apart, and if their relative velocity is v, then the force F required to maintain the motion has been found, experimentally, to be proportional to v and inversely proportional to Z for many fluids. The coefficient of proportionality is called the viscosity of the fluid, and it is denoted by the symbol µ (mu). From the definition of viscosity we can write F/A = µ v/Z 58. 58. Unit of Viscosity • N s m-2 = Pascal second, Pa s, • The older units, the poise and its sub-unit the centipoise, • 1000 centipoises = 1 N s m-2 , or 1 Pa s. • the viscosity of water at room temperature 1 x 10-3 N s m-2 • acetone, 0.3 x 10-3 N s m-2 ; • tomato pulp, 3 x 10-3 ; • olive oil, 100 x 10-3 ; • molasses 7000 N s m-3 . • Viscosity is very dependent on temperature decreasing sharply as the temperature rises. For example, the viscosity of golden syrup is about 100 N s m-3 at 16°C, 40 at 22°C and 20 at 25°C. 59. 59. Newtonian and Non-Newtonian Fluids • F/A = µ v /Z = µ(dv/dz) = " = k(dv/dz)n power-law equation • Newtonian fluids (n = 1, k = µ ) • Non-Newtonian fluids (n ≠ 1) • (1) Those in which n < 1. The viscosity is apparently high under low shear forces decreasing as the shear force increases. Pseudoplastic (tomato puree) • (2) Those in which n > 1. With a low apparent viscosity under low shear stresses, they become more viscous as the shear rate rises. Dilatancy (gritty slurries such as crystallized sugar solutions). • Bingham fluids have to exceed a particular shear stress level (a yield stress) before they start to move. • Food : Non-Newtonian 1. Tentukan dimensi dan satuannya dalam SI untuk besaran turunan berikut : a. Gaya b. Berat Jenis c. Tekanan d. Usaha e. D... 60. 60. STREAMLINE AND TURBULENT FLOW • STREAMLINE, flow is calm, in slow the pattern and smooth • TURBULENT, the flow is more rapid, eddies develop and swirl in all directions and at all angles to the general line of flow. v2 D/µv = Dv/µ =Reynolds number (Re), dimensionless • D is the diameter of the pipe • For (Re) < 2100 streamline flow, For 2100 < (Re) < 4000 transition, For (Re) > 4000 turbulent flow. 61. 61. EXAMPLE . Flow of milk in a pipe Milk is flowing at 0.12 m3 min-1 in a 2.5-cm diameter pipe. If the temperature of the milk is 21°C, is the flow turbulent or streamline? • Viscosity of milk at 21°C = 2.1 cP = 2.10 x 10-3 N s m-2 Density of milk at 21°C = 1029 kg m-3 . Diameter of pipe = 0.025 m. Cross-sectional area of pipe = (/4)D2 = /4 x (0.025)2 = 4.9 x 10-4 m2 Rate of flow = 0.12 m3 min-1 = (0.12/60)m3 s-1 = 2 x 10m3 s-1 • So velocity of flow = (2 x 10-3 )/(4.9 x 10-4 ) = 4.1 m s-1 , and so (Re) = (Dv/µ) = 0.025 x 4.1 x 1029/(2.1 x 10-3) = 50,230 and this is greater than 4000 so that the flow is turbulent. 62. 62. ENERGY LOSSES IN FLOW • Friction in Pipes • Energy Losses in Bends and Fittings • Pressure Drop through Equipment • Equivalent Lengths of Pipe 63. 63. Friction in Pipes • Eƒ : the energy loss due to friction in the pipe. • Eƒ : proportional to the velocity pressure of the fluid and to a factor related to the smoothness of the surface over which the fluid is flowing. • F/A = f v2 /2 • F is the friction force, A is the area over which the friction force acts, is the density of the fluid, v is the velocity of the fluid, and f is a coefficient called the friction factor (depends upon the Reynolds number for the flow, and upon the roughness of the pipe). • P1 - P2 = (4fv2 /2)(L1 - L2)/D ΔPf = (4fv2 /2) x (L/D) (Fanning-D'Arcy equation) • Eƒ = ΔPf/ = (2fv2 )(L/D) • L = L1 - L2 = length of pipe in which the pressure drop, ΔPf = P1 - P2 is the frictional pressure drop, and Eƒ is the frictional loss of energy. 64. 64. Friction factors in pipe (Moody graph) 65. 65. predicted f • f = 16/(Re) streamline flow, Hagen-Poiseuille equation 0 < (Re) < 2100 • ƒ = 0.316 ( Re)-0.25 /4 (Blasius equation for smooth pipes in the range 3000 < (Re) < 100,000) • roughness ratio = Roughness factor ()/pipe diameter (turbulent region) • 66. 66. ROUGHNESS FACTORS FOR PIPES Material Roughness factor () Material Roughness factor () Riveted steel 0.001- 0.01 Galvanized iron 0.0002 Concrete 0.0003 - 0.003 Asphalted cast iron 0.001 Wood staves 0.0002 - 0.003 Commercial steel 0.00005 Cast iron 0.0003 Drawn tubing Smooth 67. 67. EXAMPLE Pressure drop in a pipe Calculate the pressure drop along 170 m of 5 cm diameter horizontal steel pipe through which olive oil at 20°C is flowing at the rate of 0.1 2. Buktikan besaran-besaran berikut adalah identik : a. Energi Potensial dan Energi Kinetik b. Usaha/Energi dan Kalor Jawa... m3 min-1 • Diameter of pipe = 0.05 m, Area of cross-section A = (/4)D2 = /4 x (0.05)2 = 1.96 x 10-3 m2 • From Appendix 4, • Viscosity of olive oil at 20°C = 84 x 10-3 Ns m-2 and density = 910 kg m-3 , and velocity = (0.1 x 1/60)/(1.96 x 10-3 ) = 0.85 m s-1 , • Now (Re) = (Dv/µ) • = [(0.05 x 0.85 x 910)/(84 x 10-3 )] = 460 • so that the flow is streamline, and from Fig. moody, for (Re) = 460 • f = 0.03. • Alternatively for streamline flow from f = 16/(Re) = 16/460 = 0.03 as before. • And so the pressure drop in 170 m, • ΔPf = (4fv2 /2) x (L/D) • = [4 x 0.03 x 910 x (0.85)2 x 1/2] x [170 x 1/0.05] = 1.34 x 105 Pa = 134 kPa. 68. 68. Thermal conductivity Specific heat Density Viscosity Temperature (J m-1 s-1 °C-1 ) (kJ kg-1 °C-1 ) (kg m-3 ) (N s m-2 ) (°C) Water 0.57 4.21 1000 1.87 x 10-3 0 4.21 987 0.56 x 10-3 50 0.68 4.18 958 0.28 x 10-3 100 Sucrose 20% soln. 0.54 3.8 1070 1.92 x 10-3 20 0.59 x 10-3 80 60% soln. 6.2 x 10-3 20 5.4 x 10-3 80 Sodium chloride 22% soln. 0.54 3.4 1240 2.7 x 10-3 2 Olive oil 0.17 2.0 910 84 x 10-3 20 Rape-seed oil 900 118 x 10-3 20 Soya-bean oil 910 40 x 10-3 30 Tallow 900 18 x 10-3 65 Milk (whole) 0.56 3.9 1030 2.12 x 10-3 20 Milk (skim) 1040 1.4 x 10-3 25 Cream 20% fat 1010 6.2 x 10-3 3 30% fat 1000 13,8 x 10-3 3 69. 69. Energy Losses in Bends and Fittings • energy losses due to altering the direction of flow, fittings of varying cross-section • This energy is dissipated in eddies and additional turbulence and finally lost in the form of heat. • Eƒ = kv2 /2 Losses in fittings • Ef = (v1 - v2)2 /2 Losses in sudden enlargements • Ef = kv2 2 /2 Losses in sudden contraction 70. 70. FRICTION LOSS FACTORS IN FITTINGS k Valves, fully open: gate 0.13 globe 6.0 angle 3.0 Elbows: 90° standard 0.74 medium sweep 0.5 long radius 0.25 square 1.5 Tee, used as elbow 1.5 Tee, straight through 0.5 Entrance, large tank to pipe: sharp 0.5 rounded 0.05 LOSS FACTORS IN CONTRACTIONS D2/D1 0.1 0.3 0.5 0.7 0.9 k 0.36 0.31 0.22 0.11 0.02 71. 71. FLUID-FLOW APPLICATIONS • Two practical aspects of fluid flow in food technology : • measurement in fluids: pressures and flow rates, and • production of fluid flow by means of pumps and fans. • Pumps and fans are very similar in principle and usually have a centrifugal or rotating action • a gas : moved by a fan, • a liquid: moved by a pump. 72. 72. MEASUREMENT OF PRESSURE IN A FLUID • Method : – Piezometer ("pressure measuring") tube – U-tube – Pitot tube – Pitot-static tube – Bourdon-tube • P = Z11g 73. 73. EXAMPLE. Pressure in a vacuum evaporator The pressure in a vacuum evaporator was measured by using a U-tube containing mercury. It was found to be less than atmospheric pressure by 25 cm of mercury. Calculate the extent by which the pressure in the evaporator is below atmospheric pressure (i.e. the vacuum in the evaporator) in kPa, and also the absolute pressure in the evaporator. The atmospheric pressure is 75.4 cm of mercury and the specific gravity of mercury is 13.6, and the density of water is 1000 kg m-3. • We have P = Zg = 25 x 10-2 x 13.6 x 1000 x 9.81 = 33.4 kPa • Therefore the pressure in the evaporator is 33.4 kPa below atmospheric pressure and this is the vacuum in the Dimensionless Ratios • It is often easier to visualize quantities if they are expressed in ratio form and ratios have the ... evaporator. • For atmospheric pressure: • P = Zg • P = 75.4 x 10-2 x 13.6 x 1000 x 9.81 = 100.6 kPa Therefore the absolute pressure in the evaporator = 100.6 - 33.4 = 67.2 kPa 74. 74. MEASUREMENT OF VELOCITY IN A FLUID • Pitot tube and manometer : • Z1g + v1 2 /2 + P1/1 = Z2g + v2 2 /2 + P2/1 • Z2 = Z + Z' • Z' be the height of the upper liquid surface in the pipe above the datum, • Z be the additional height of the fluid level in the tube above the upper liquid surface in the pipe; • Z' may be neglected if P1 is measured at the upper surface of the liquid in the pipe, or if Z' is small compared with Z • v2 = 0 as there is no flow in the tube • P2 = 0 if atmospheric pressure is taken as datum and if the top of the tube is open to the atmosphere • Z1 = 0 because the datum level is at the mouth of the tube. • v1 2 /2g + P1/1 = (Z + Z')g » Z. • Pitot-static tube • Z = v2 /2g 75. 75. EXAMPLE . Velocity of air in a duct Air at 0°C is flowing through a duct in a chilling system. A Pitot-static tube is inserted into the flow line and the differential pressure head, measured in a micromanometer, is 0.8 mm of water. Calculate the velocity of the air in the duct. The density of air at 0°C is 1.3 kg m-3. • Z = v1 2 /2g " 1Z1 = 2Z2. • Now 0.8 mm water = 0.8 x 10-3 x 1000 1.3 = 0.62 m of air • v1 2 = 2Zg = 2 x 0.62 x 9.81 = 12.16 m2 s-2 • Therefore v1 = 3.5 m s-1 76. 76. Venturi and orifice meters • v1 2 /2 + P1/1 = v2 2 /2+ P2/2 (Bernouilli's equation) • A1v1 = A2v2 (mass balance, eqn) • 1 = 2 = • v1 2 /2 + P1/ = (v1A1/A2)2 /2 + P2/ v1 2 = [2(P2 -P1)/] x A2 2 /(A2 2 -A1 2 ) • (P2 -P1)/ = gZm / • Z = (P2 -P1)/m g • v1 = C √[2(P2 -P1 )/]x A2 2 /(A2 2 -A1 2 ) • In a properly designed Venturi meter, C lies between 0.95 and 1.0. 77. 77. Pompa dan Fan • mechanical energy from some other source is converted into pressure or velocity energy in a fluid. • The food technologist is not generally much concerned with design details of pumps, but should know what classes of pump are used and something about their characteristics. • The efficiency of a pump is the ratio of the energy supplied by the motor to the increase in velocity and pressure energy given to the fluid. 78. 78. Jenis Pompa • Positive Displacement Pumps • the fluid is drawn into the pump and is then forced through the outlet • Positive displacement pumps can develop high- pressure heads but they cannot tolerate throttling or blockages in the discharge. 79. 79. Jet Pumps • a high-velocity jet is produced in a Venturi nozzle, converting the energy of the fluid into velocity energy. • This produces a low-pressure area causing the surrounding fluid to be drawn into the throat • Jet pumps are used for difficult materials that cannot be satisfactorily handled in a mechanical pump. • They are also used as vacuum pumps. • Jet pumps have relatively low efficiencies but they have no moving parts and therefore have a low initial cost. 80. 80. Air-lift Pumps • air or gas can be used to impart energy to the liquid • The air or gas can be either provided from external sources or produced by boiling within the liquid. Suhu dan komposisi • C, F, K • Fraksi mol, konsentrasi • Suatu wadah berisi 50g air dan 50 g NaOH, berapa fraksimol masing... Examples of the air-lift principle are: • Air introduced into the fluid as shown in Fig. 4.3(e) to pump water from an artesian well. Air introduced above a liquid in a pressure vessel and the pressure used to discharge the liquid. Vapours produced in the column of a climbing film evaporator. In the case of powdered solids, air blown up through a bed of powder to convey it in a "fluidized" form. • A special case of this is in the evaporator, where boiling of the liquid generates the gas (usually steam) and it is used to promote circulation. Air or gas can be used directly to provide pressure to blow a liquid from a container out to a region of lower pressure. • Air-lift pumps and air blowing are inefficient, but they are convenient for materials which will not pass easily through the ports, valves and passages of other types of pumps. 81. 81. Propeller Pumps and Fan • Propellers can be used to impart energy to fluids • They are used extensively to mix the contents of tanks and in pipelines to mix and convey the fluid. • Propeller fans are common and have high efficiencies. • They can only be used for low heads, in the case of fans only a few centimetres or so of water 82. 82. Centrifugal Pumps and Fans • The centrifugal pump converts rotational energy into velocity and pressure energy • The fluid to be pumped is taken in at the centre of a bladed rotor and it then passes out along the spinning rotor, acquiring energy of rotation. This rotational energy is then converted into velocity and pressure energy at the periphery of the rotor. • Centrifugal fans work on the same principles. These machines are very extensively used and centrifugal pumps can develop moderate heads of up to 20 m of water. They can deliver very large quantities of fluids with high efficiency. 83. 83. Gambar jenis-jenis pompa 84. 84. • EXAMPLE Centrifugal pump for raising water Water for a processing plant is required to be stored in a reservoir to supply sufficient working head for washers. It is believed that a constant supply of 1.2 m3 min-1 pumped to the reservoir, which is 22 m above the water intake, would be sufficient. The length of the pipe is about 120 m and there is available galvanized iron piping 15 cm diameter. The line would need to include eight right-angle bends. There is available a range of centrifugal pumps whose characteristics are shown in Fig. 4.4. Would one of these pumps be sufficient for the duty and what size of electric drive motor would be required? 85. 85. Reynold number • Assume properties of water at 20°C are density 998 kg m-3, and viscosity 0.001 N s m-2 • Cross-sectional area of pipe A = (/4)D2 = /4 x (0.15)2 = 0.0177 m-2 Volume of flow V = 1.2 m3 min-1 = 1.2/60 m3 s-1 = 0.02 m3 s-1. • Velocity in the pipe = V/A = (0.02)/(0.0177) = 1.13 ms-1 • Now (Re) = Dv/µ • = (0.15 x 1.13 x 998)/0.001 = 1.7 x 105 so the flow is clearly turbulent. Neraca Massa • Sangat penting dalam menentukan efisiensi proses dan memprediksi hasil akhir proses • Rumus umum => massa i... 86. 86. friction loss of energy From Table 3.1, the roughness factor is 0.0002 for galvanized iron and so roughness ratio /D = 0.0002/0.15 = 0.001 So from Fig. 3.8, ƒ = 0.0053 Therefore the friction loss of energy = (4ƒv2 /2) x (L/D) = [4ƒv2 L/2D] = [4 x 0.0053 x (1.13)2 x 120]/(2 x 0.15) = 10.8 J. 87. 87. TABLE 3.1 RELATIVE ROUGHNESS FACTORS FOR PIPES Material Roughness factor () Material Roughness factor () Riveted steel 0.001- 0.01 Galvanized iron 0.0002 Concrete 0.0003 - 0.003 Asphalted cast iron 0.001 Wood staves 0.0002 - 0.003 Commercial steel 0.00005 Cast iron 0.0003 Drawn tubing Smooth 88. 88. Friction factors in pipe 89. 89. TABLE 3.2 FRICTION LOSS FACTORS IN FITTINGS k Valves, fully open: gate 0.13 globe 6.0 angle 3.0 Elbows: 90° standard 0.74 medium sweep 0.5 long radius 0.25 square 1.5 Tee, used as elbow 1.5 Tee, straight through 0.5 Entrance, large tank to pipe: sharp 0.5 rounded 0.05 90. 90. • For the eight right-angled bends, from Table 3.2 we would expect a loss of 0.74 velocity energies at each, making (8 x 0.74) = 6 in all. velocity energy = v2/2 = (1.13)2/2 = 0.64 J • So total loss from bends and discharge energy = (6 + 1) x 0.64 = 4.5 J There would be one additional velocity energy loss because of the unrecovered flow energy discharged into the reservoir. Energy loss from bends and discharge 91. 91. Energy to move 1 kg water • Energy to move 1 kg water against a head of 22 m of water is E = Zg = 22 x 9.81 = 215.8 J. • Total energy requirement per kg: Etot = 10.8 + 4.5 + 215.8 = 231.1 J 92. 92. energy requirement of pump • and theoretical power requirement = Energy x volume flow x density = (Energy/kg) x kgs-1 = 231.1 x 0.02 x 998 = 4613 J s-1. • Now the head equivalent to the energy requirement = Etot/g = 231.1/9.81 = 23.5 m of water, • and from Fig. 4.4 this would require the 150 mm impeller pump to be safe, and the pump would probably be fitted with a 5.5 kW motor. 93. 93. Energy Balance 94. 94. Gas and Vapour • naturally associated with foods and food-processing systems: – Equilibrium between food and water vapor determines temperatures achieved during processing. – Dissolved gases in foods such as oxygen affect shelf life. – Gases are used to flush packages to eliminate oxygen and prolong shelf life. – Modified atmospheres in packages have been used to prolong shelf life of packaged foods. – Air is used for dehydration. – Gases are used as propellants in aerosol cans and as refrigerants. • The distinction between gases and vapors is very loose because theoretically all vapors are gases. • The term “vapor” is generally used for the gaseous phase of a substance that exists as a liquid or a Proses yang tidak terjadi reaksi kimia • Proses yang tidak mengalami reaksi kimia: – Pengeringan, – pembekuan, – pemekatan... solid at ambient conditions. 95. 95. Kinetika Gas • The postulates of the kinetic theory – Gases are composed of discreet particles called molecules, which are in constant random motion,colliding with each other and with the walls of the surrounding vessel. – The force resulting from the collision between the molecules and the walls of the surrounding vessel is responsible for the pressure of the gas. – The lower the pressure, the farther apart the molecules, thus, attractive forces between moleculeshave reduced influence on the overall properties of the gas. – The average kinetic energy of the molecules is directly proportional to the absolute temperature 96. 96. Absolute Temperature and Pressure • pressure : force of collisions of gas molecules against a surface in contact with the gas. • pressure is proportional to the number of gas molecules and their velocity (absolute pressure). • Pressure is often expressed as gauge pressure when the measured quantity is greater than atmospheric pressure, and as vacuum when below atmospheric. • Unit: psig, psia, kPa absolute, kPa above atmospheric, atmospheres (atm) • standard atmosphere, the mean atmospheric pressure at sea level, equivalent to 760 mm Hg, 29.921 in. Hg, 101.325 kPa, or 14.696 lbf/in.2 • Temperature (T) is a thermodynamic quantity related to the velocity of motion of molecules 97. 97. Absolute and gage pressure 98. 98. Conversion factor 99. 99. Calculate the absolute pressure inside an evaporator operating under 20 in. Hg vacuum. Atmospheric pressure is 30 in. Hg. Express this pressure in SI and in the American Engineering System of units. • Pabsolute =Patmospheric − Pvacuum =(30 − 20) in. Hg=10 in. Hg • From the table of conversion factors, the following conversion factors are obtained: 100. 100. The Ideal Gas Equation • Pressure, the force of collision between gas molecules and a surface, is directly proportional to temperature and the number of molecules per unit volume. • PV = nRT the ideal gas equation. • R is the gas constant and has values of 0.08206 L(atm)/(gmole.K); or 8315 N(m)/ (kgmole.K) or 1545 ft(lbf)/(lbmole.R). • a fixed quantity of a gas that follows the ideal gas equation undergoes a process where the volume, temperature, or pressure is allowed to change, the product of the number of moles n and the gas constant R is a constant 101. 101. Calculate the quantity of oxygen entering a package in 24 hours if the packaging material has a surface area of 3000 cm2 and an oxygen permeability 100 cm3 /(m2 )(24 h) STP (standard temperature and pressure = 0o C and 1 standard atmosphere of 101.325 kPa). • Jawaban: Tahapan perhitungan • Gambar diagram • Tulis reaksi kimia jika ada • Tulis dasar-dasar perhitungan • Hitung neraca massanya 102. 102. Calculate the volume of CO2 in ft3 at 70o F and 1 atm, which would be produced by vaporization of 1 lb of dry ice. 103. 103. Calculate the density of air (M = 29) at 70F and 1 atm in (a) American Engineering and (b) SI units. 104. 104. Suatu proses memerlukan debit udara bertekanan 2 atm sebesar 10 m3 /s pada suhu 20C. Hitung debit kompresor pada STP yang harus diberikan. • Kondisi STP adalah suhu (T) = 0o C (273 K), tekanan (P) = 1 atm atau 101,325 kPa • Debit 1 (V1) = 10 m3 /s; T1 = 293 K; P1 = 2 atm atau 202,65 kPa; • V = (P1V1T)/(T1P) = (2 x 293 x 273)/(293 x 1) = 18,64 m3 /s 105. 105. An empty can was sealed in a room at 80o C and 1 atm pressure. Assuming that only air is inside the sealed can, what will be the vacuum after the can and contents cool to 20o C? • Solution: 106. 106. Gas Mixtures • If components of a gas mixture at constant volume are removed one after the other, the drop in pressure accompanying complete removal of one component is the partial pressure of that component • Pt = Pa + Pb + Pc + . . . Pn (Dalton’s law of partial pressures) • PaV = naRT 107. 107. Hitung kuantitas udara pada headsapce kaleng yang bersuhu 20o C jika tekanan pada headspace sebesar 10 in Hg. Tekanan atmosfer sebesar 30 in Hg. Volume headspace sebesar 15 ml berisi uap jenuh dan udara. • vapor pressure of water at 20o C = 2336.6 Pa. • Pt (tekanan absolut dalam kaleng) = 30-10 in Hg = 20 in Hg • = 20 x 3386,38 = 67727,6 Pa • Pudara = Pt - Puap = 67727,6 – 2336,6 = 65391 Pa • V = 15 x 10-6 m3 • T = 20 + 273 = 293 K • Nudara = (Pudara V)/(RT) • = (65391 x 1,5 x 10-5 )/(8315 x 293) • =4,03 x 10-7 kgmol 108. 108. • Assume there are no dissolved gases in the product at the time of sealing, therefore the only gases in the headspace are air and water vapor. The vapor pressure of water at 20C and 80C are 2.3366 and 47.3601 kPa, respectively. In the gas mixture in the headspace, air is assumed to remain at the same quantity in the gaseous phase, while water condenses on cooling 109. 109. soal • Proses penutupan kaleng dilakukan pada suhu 80o C dan tekanan 1 atm. Di bagian headspace hanya ada udara dan uap. Setelah dilakukan sterilisasi, kaleng lalu didinginkan hingga suhu 20o C. Hitung berapa tekanan di headspace? Diasumsikan jumlah udara di headspace tetap dan uap air mengkondensasi pada saat pendinginan. 110. 110. A gas mixture used for controlled atmosphere storage of vegetables contains 5% CO2, 5% O2, and 90% N2. The mixture is generated by mixing appropriate quantities of air Contoh neraca massa • Larutan soda api (NaOH), sebanyak 1000 kg/jam mengandung 10% NaOH di pekatkan pada evaporator sehing... and N2 and CO2 gases. 100 m3 of this mixture at 20o C and 1 atm is needed per hour. Air contains 21% O2 and 79% N2. Calculate the volume at which the component gases must be metered into the system in m3 /h at 20o C and 1 atm. • All percentages are by volume. No volume changes occur on mixing of ideal gases. Because volume percent in gases is the same as mole percent, material balance equations may be made on the basis of volume and volume percentages. Let X = volume O2, Y = volume CO2, and Z = volume N2, fed into the system per hour. • Oxygen balance: 0.21(X) = 100(0.05); X = 23.8 m3 • CO2 balance: Y = 0.05(100); Y = 5 m3 • Total volumetric balance: X + Y + Z = 100 • Z = 100 − 23.8 − 5 = 71.2 m3 111. 111. Ruangan penyimpanan buah segar diatur sehingga komposisi gas yang masuk ke ruang penyimpanan menjadi 88% N2, 6% O2 dan 6% CO2. Udara yang bersuhu 25o C tekanan 1 atm mengalir ke dalam ruang penyimpanan dengan debit 80 m3/jam. Hitung kebutuhan udaranya jika komposisi udara adalah 79% N2 dan 21% O2. 112. 112. Campuran gas terdiri dari 5% CO2, 5% O2, and 90% N2 digunakan pada CAS untuk buah. Gas campuran dibuat dengan mencampur udara, N2 dan CO2. Komposisi udara adalah 21% O2 and 79% N2. Gas campuran dibutuhkan sebanyak 100m3/jam. Hitung volume komponen gas yang harus diatur kedalam CAS pada suhu 20oC dan tekanan 1 atm. • Semua persentase dalam bentuk volume, tidak ada perubahan volume dalam pencampuran gas. • Persen volume sama dengan persen mol • Neraca massa total U + 113. 113. Udara 1 m3 bertekanan 5 atm dijenuhkan dengan uap air pada suhu 50C. Jika udara tersebut diturunkan tekanannya menjadi 1 atm dan suhu 20C, hitung jumlah uap air yang mengkondensasi. • The vapor pressure of water at 50C and 20C are 12.3354 and 2.3366 kPa, respectively. • Basis:1 m3 air at 5 atm pressure and 50C. The number of moles of air will remain the same on cooling Moles water condensed = 0.004593 - 0.004344 = 0.000249 kg moles. 114. 114. Tekanan parsial uap air di udara pada 25C dan 1 atm adalah 2,520 kPa. Jika udara ditekan hingga 5 atm pada suhu 35C, hitung tekanan parsial uap air di udara. • Increasing the total pressure of a gas mixture will proportionately increase the partial pressure of each component • for the mixture and for the water vapor, let V1 = the volume of the gas mixture at 25C and 1 atm; Pt = total pressure; Pw = partial pressure of water vapor. • The total number of moles of air and water vapor is Assuming no condensation, the ratio, nt/nw will be the same in the low-pressure and high-pressure air, therefore: 115. 115. Temperature Pressure Enthalpy (sat. vap.) Latent heat Specific volume (°C) (kPa) (kJ kg-1 ) (kJ kg-1 ) (m3 kg-1 ) 20 2.34 2538 2454 57.8 22 2.65 2542 2449 51.4 24 2.99 2545 2445 45.9 26 3.36 2549 2440 40.0 28 3.78 2553 2435 36.6 30 4.25 2556 2431 32.9 40 7.38 2574 2407 19.5 50 12.3 2592 2383 12.0 60 19.9 2610 2359 7.67 70 31.2 2627 2334 5.04 80 47.4 2644 2309 3.41 90 70.1 2660 2283 2.36 100 101.35 2676 2257 1.673 105 120.8 2684 2244 1.42 110 143.3 2692 2230 1.21 115 169.1 2699 2217 1.04 120 Dikerjakan dan dikumpulkan • Proses produksi selai buah dilakukan dengan cara memekatkan bubur buah dari kadar padatan 10%... 198.5 2706 2203 0.892 125 232.1 2714 2189 0.771 130 270.1 2721 2174 0.669 135 313.0 2727 2160 0.582 140 361.3 2734 2145 0.509 150 475.8 2747 2114 0.393 160 617.8 2758 2083 0.307 116. 116. PROPERTIES OF SATURATED AND SUPERHEATED STEAM • Steam and water are the two most used heat transfer mediums in food processing. • Saturated Liquid:. Liquid water in equilibrium with its vapor. If the total pressure above a liquid equals the vapor pressure, the liquid is at the boiling point. • Saturated Vapor: saturated steam and is vapor at the boiling temperature of the liquid. Lowering the temperature of saturated steam at constant pressure by a small increment will cause vapor to condense to liquid. The phase change is accompanied by a release of heat. If heat is removed from the system, temperature and pressure will remain constant until all vapor is converted to liquid. Adding heat to the system will change either temperature or pressure or both. • Vapor-Liquid Mixtures: Steam with less than 100% quality. Temperature and pressure correspond to the boiling point; therefore, water could exist either as saturated liquid or saturated vapor. Addition of heat will not change temperature and pressure until all saturated liquid is converted to vapor. Removing heat from the system will also not change temperature and pressure until all vapor is converted to liquid. • Steam Quality: The percentage of a vapor-liquid mixture that is in the form of saturated vapor. • Interpolation: data 117. 117. If 1 lb of water at 100 psig and 252F is allowed to expand to 14.7 psia, calculate (a) the resulting temperature after expansion and (b) the quantity of vapor produced. • The absolute pressure= 100 + 14.7 = 114.7 psia. At 252F, water will not boil until the pressure is reduced to 30.9 psia. The water therefore is at a temperature much below the boiling point at 114.7 psia and it would have the properties of liquid water at 252F. • (a) After expansion to 14.7 psia, the boiling point at 14.7 psia is 212F. Part of the water will flash • to water vapor at 212F and the remaining liquid will also be at 212F. • (b) The enthalpy of water at 252F is (hf at 252F) 220.62 BTU/lb. • Basis: 1 lb H2O. Heat content = 220.62 BTU. When pressure is reduced to 14.7 psia, some vapor will be formed, but the total heat content of both vapor and liquid at 212F and 14.7 psia will still be 220.62 BTU. 118. 118. How much heat would be given off by cooling steam at 252F and 30.883 psia to 248F, at the same pressure? • First, check the state of water at 30.883 psia and 252F and 248F. From steam tables, the boiling point of water at 30.883 psia is 252F. Therefore, steam at 252F and 30.883 psia is saturated vapor. • At 30.883 psia and 248F, water will be in the liquid state, because 248F is below the boiling temperature at 30.883 psia. • Heat given off = q = hg at 252F − hf at 248F • From steam tables, • hg at 252F = 1164.78 BTU/lb • hf at 248F = 216.56 BTU/lb • q = 1164.78 − 216.56 = 948.22 BTU/lb • Saturated steam is a very efficient heat transfer medium. Note that for only a 4F change in temperature, 948 BTU/lb of steam is given off. The heat content of saturated vapors come primarily from the latent heat of vaporization, and it is possible to extract this heat simply Tugas dikerjakan dan dikumpulkan • Adonan biskuit diperoleh dengan mencampurkan Terigu sebanyak 60% berat, gula 10%, telur... by causing a phase change at constant temperature and pressure. 119. 119. Superheated Steam Tables • Superheated Steam: Water vapor at a temperature higher than the boiling point. The number of degrees the temperature exceeds the boiling temperature is the degrees superheat. Addition of heat to superheated steam could increase the superheat at constant pressure or change both the pressure and temperature at constant volume. Removing heat will allow the temperature to drop to the boiling temperature where the temperature will remain constant until all the vapor has condensed. • A superheated steam table: Both temperature and absolute pressure must be specified to accurately define the degree of superheat. • From the temperature and absolute pressure, the specific volume v in ft3/lb and the enthalpy h in BTU/lb can be read from the table 120. 120. How much heat is required to convert 1 lb of water at 70F to steam at 14.696 psia and 250F? • First determine the state of steam at 14.696 psia and 250F. At 14.696 psia, the boiling point is 212F. Steam at 250F and 14.696 psia is superheated steam. From the superheated steam table, h at 250F is 1168.8 BTU/lb. • Heat required = hg at 250F and 14.696 psia − hf at 70F • = 1168.8 BTU/lb − 38.05 BTU/lb • = 1130.75 BTU/lb 121. 121. How much heat would be given off by cooling superheated steam at 14.696 psia and 500F to 250F at the same pressure? • Basis: 1 lb of steam. • Heat given off = q = h at 14.696 psia and 500F − hg at 14.696 psia and 250F • = 1287.4 − 1168.8 = 118.6 BTU/lb • Superheated steam is not a very efficient heating medium. Note that a 250F change in temperature is accompanied by the extraction of only 118.6 BTUs of heat. 122. 122. Soal • Campuran gas terdiri dari 5% CO2, 5% O2, and 90% N2 digunakan pada CAS untuk buah.Gas tersebut dibuat dengan mencampur udara, N2 dan CO2. Komposisi udara adalah 21% O2 and 79% N2. CAS membutuhkan gas campuran sebanyak 100m3 /jam. Hitung volume komponen gas yang harus diatur kedalam CAS pada suhu 20o C dan tekanan 1 atm. • Campuran Udara dan uap air 1 m3 bertekanan 5 atm absolut bersuhu 50o C. Jika udara tersebut diturunkan tekanannya menjadi 1 atm dan suhu 20o C, hitung jumlah uap air yang mengkondensasi 123. 123. Harap dikerjakan • Proses produksi Sodium Sitrat (Na2C6H6O7) dilakukan dengan mereaksikan larutan asam sitrat (C6H8O7) 10%(berat) dengan NaOH. Larutan Sodium sitrat yang terbentuk dipekatkan sehingga diperoleh larutan dengan konsentrasi 30% berat. Larutan lalu didinginkan pada suhu 15o C untuk mengkristalkan sodium sitrat. Jika kelarutan sodium sitrat pada suhu 15o C sebesar 20% berat, hitung kristal sodium sitrat yang diperoleh, untuk setiap 100 kg larutan asam sitrat yang digunakan. – Asumsi : - Sodium Batas proses (boundary) • Batas proses dapat digunakan untuk menyederhanakan suatu proses • Dapat diperluas atau diperkecil sitrat dalam bentuk anhydrous – Berat atom O : 16; C : 12; H : 1; Na : 23 – Reaksi berlangsung secara sempurna • Ekstraksi menggunakan supercritical CO2 beroperasi pada tekanan 30 MPa dan suhu 60o C di wadah ekstraksi (extraction chamber). Debit gas CO2 meninggalkan ekstraktor pada tekanan 101,3 kPa dan suhu 20o C sebesar 10 L/menit. Hitung waktu tinggal (residence time) dari CO2 dalam extraction chamber jika diketahui chamber berbentuk tabung dengan diameter 5 cm dan tinggi 45 cm. Waktu tinggal adalah volume chamber/debit gas dalam chamber. 124. 124. Temperature Pressure Enthalpy (sat. vap.) Latent heat Specific volume (°C) (kPa) (kJ kg-1 ) (kJ kg-1 ) (m3 kg-1 ) 20 2.34 2538 2454 57.8 22 2.65 2542 2449 51.4 24 2.99 2545 2445 45.9 26 3.36 2549 2440 40.0 28 3.78 2553 2435 36.6 30 4.25 2556 2431 32.9 40 7.38 2574 2407 19.5 50 12.3 2592 2383 12.0 60 19.9 2610 2359 7.67 70 31.2 2627 2334 5.04 80 47.4 2644 2309 3.41 90 70.1 2660 2283 2.36 100 101.35 2676 2257 1.673 105 120.8 2684 2244 1.42 110 143.3 2692 2230 1.21 115 169.1 2699 2217 1.04 120 198.5 2706 2203 0.892 125 232.1 2714 2189 0.771 130 270.1 2721 2174 0.669 135 313.0 2727 2160 0.582 140 361.3 2734 2145 0.509 150 475.8 2747 2114 0.393 160 617.8 2758 2083 0.307 125. 125. Heat • Sensible heat is defined as the energy transferred between two bodies at different temperatures, or the energy present in a body by virtue of its temperature. • Latent heat is the energy associated with phase transitions, heat of fusion, from solid to liquid, and heat of vaporization, from liquid to vapor. • Enthalpy, is an intrinsic property, the absolute value of which cannot be measured directly. • However, if a reference state is chosen for all components that enter and leave a system such that at this state the enthalpy is considered to be zero, then the change in enthalpy from the reference state to the current state of a component can be considered as the value of the absolute enthalpy for the system under consideration. • The reference temperature (Tref) for determining the enthalpy of water in the steam tables is 32.018F or 0.01C. 126. 126. Specific Heat • The specific heat (Cp) is the amount of heat that accompanies a unit change in temperature for a unit mass. • The specific heat, which varies with temperature, is more variable for gases compared with liquids or solids. • Most solids and liquids have a constant specific heat over a fairly wide temperature range. 127. 127. specific heat J/(kg K) 128. 128. Estimation of Cp • Cavg = 3349M+ 837.36 in J/(kg K) for fat free plant material • Cavg = 1674.72 F + 837.36 SNF + 4l86.8M in J/(kg K) • the mass fraction fat (F), mass fraction solids non-fat (SNF), and mass fraction moisture (M) • Example: Calculate the heat required to raise the temperature of a 4.535 kg roast beef containing 15% protein, 20% fat, and 65% water from 4.44C to 65.55C • Solution: • Cavg = 0.15(837.36) + 0.2(1674.72) + 0.65(4186.8) = 3182 J/ (kg K) • q = 4.535 kg[3182 J/(kg K)] (65.55 − 4.44)K = Dikerjakan • Niratebu1000kg/jamberkadargula20% dipekatkanhinggakadargula60%. Nira pekatselanjutnyadikristalisasi padasuhu ... 0.882 MJ 129. 129. Specific heat of gas and vapor • whereCpm is mean specific heat from the reference temperature To to T1. Tabulated values for the mean specific heat of gases are based on ambient temperature of 77F or 25C, as the reference temperature. 130. 130. Contoh • Hitung kebutuhan panas untuk menaikkan suhu udara pengering pd tekanan 1 atm dari suhu ruang 25o C ke suhu pengeringan 50o C jika tiap menit dialirkan udara sebanyak 100m3 • q= mCp (50-25) • m=PVM/RT • R = 0.08206 m3 atm/kg mole K 131. 131. PROPERTIES OF SATURATED AND SUPERHEATED STEAM • Steam and water are the two most used heat transfer mediums in food processing. Water is also a major component of food products. The steam tables that list the properties of steam are a very useful reference when determining heat exchange involving a food product and steam or water. At temperatures above the freezing point, water can exist in either of the following forms. • Saturated Liquid:. Liquid water in equilibrium with its vapor. The total pressure above the liquid must be equal to or be higher than the vapor pressure. If the total pressure above the liquid exceeds the vapor pressure, some other gas is present in the atmosphere above the liquid. If the total pressure above a liquid equals the vapor pressure, the liquid is at the boiling point. • Saturated Vapor: This is also known as saturated steam and is vapor at the boiling temperature of the liquid. Lowering the temperature of saturated steam at constant pressure by a small increment will cause vapor to condense to liquid. The phase change is accompanied by a release of heat. If heat is removed from the system, temperature and pressure will remain constant until all vapor is converted to liquid. Adding heat to the system will change either temperature or pressure or both. • Vapor-Liquid Mixtures: Steam with less than 100% quality. Temperature and pressure correspond to the boiling point; therefore, water could exist either as saturated liquid or saturated vapor. Addition of heat will not change temperature and pressure until all saturated liquid is converted to vapor. Removing heat from the system will also not change temperature and pressure until all vapor is converted to liquid. • Steam Quality: The percentage of a vapor-liquid mixture that is in the form of saturated vapor. • Superheated Steam: Water vapor at a temperature higher than the boiling point. The number of degrees the temperature exceeds the boiling temperature is the degrees superheat. Addition of heat to superheated steam could increase the superheat at constant pressure or change both the pressure and temperature at constant volume. Removing heat will allow the temperature to drop to the boiling temperature where the temperature will remain constant until all the vapor has condensed. 132. 132. Steam table • The saturated steam table consists of entries under the headings of temperature, absolute pressure, specific volume, and enthalpy. 133. 133. Temperature Pressure Enthalpy (sat. vap.) Latent heat Specific volume (°C) (kPa) (kJ kg-1 ) (kJ kg-1 ) (m3 kg-1 ) 20 2.34 2538 2454 57.8 22 2.65 2542 2449 51.4 24 2.99 Proses pencampuran • Draw a diagram and set up equations representing total mass balance and component mass balance for a ... 2545 2445 45.9 26 3.36 2549 2440 40.0 28 3.78 2553 2435 36.6 30 4.25 2556 2431 32.9 40 7.38 2574 2407 19.5 50 12.3 2592 2383 12.0 60 19.9 2610 2359 7.67 70 31.2 2627 2334 5.04 80 47.4 2644 2309 3.41 90 70.1 2660 2283 2.36 100 101.35 2676 2257 1.673 105 120.8 2684 2244 1.42 110 143.3 2692 2230 1.21 115 169.1 2699 2217 1.04 120 198.5 2706 2203 0.892 125 232.1 2714 2189 0.771 130 270.1 2721 2174 0.669 135 313.0 2727 2160 0.582 140 361.3 2734 2145 0.509 150 475.8 2747 2114 0.393 160 617.8 2758 2083 0.307 134. 134. contoh • Pada tekanan vakum berapa sehingga air mendidih pada suhu 80o C, nyatakan dalam kPa dan dalam cm Hg – Lihat tabel uap= 47,4 kPa abs – Tekanan vakum = 101 47,4 = 53,6 kPa – Tekanan vakum = (53,6/101) x 76 = 40,3 cmHg • Sterilisasi dilakukan pada suhu 120o C, berapa tekanan yang terbaca pada manometer yang menggunakan satuan psi? – Dari tabel pada suhu 120o C tekanan uap = 198,5 kPa, maka tekanan pada manometer = 198,5 – 101 = 97,5 kPa – 1 atm = 14,7 psi = 101 kPa – Tekanan pada manometer = (97,5/101) x 14,7 = 14,2 psig 135. 135. Freezing Points of Food Products Unmodified from the Natural State • the heat to be removed during freezing of a food product : sensible heat and latent heat. • determining the amount of heat by calculating the enthalpy change. • calculating enthalpy change below the freezing point (good only for moisture contents between 73% and 94%) is the procedure of Chang and Tao (1981). In this correlation, it is assumed that all water is frozen at 227 K (−50 F). 136. 136. Calculate the freezing point and the amount of heat that must be removed in order to freeze 1 kg of grape juice containing 25% solids from the freezing point to −30 C. • Solution: • Y = 0.75. • for juices: Tf = 120.47 + 327.35(0.75) − 176.49(0.75)2 = 266.7 K • Hf = 9792.46 + 405,096(0.75) = 313, 614 J • a = 0.362 + 0.0498(0.02) − 3.465(0.02)2 = 0.3616 • b = 27.2 − 129.04(0.1316) − 481.46(0.1316)2 = 1.879 • Tr = (−30 + 273 − 227.6)/(266.7 − 227.6) = 0.394 • H = 313,614[(0.3616)0.394 + (1 − 0.3616)(0.394)1.879 ]= 79, 457 J/kg • The enthalpy change from Tf to −30 C is • ΔH = 313,614 − 79, 457 = 234, 157 J/kg Recommended • Draw a diagram and set up a total mass and component balance equation for a crystallizer where 100 kg of a concentrated ...

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ITP UNS Semester 3, Satuan Operasi 2: Pemisahan secara mekanik Fransiska Puteri English Español Português • Sodium sitrat (Na2C6H6O7) dibentuk dengan mereaksikan larutan asam sitrat (C6H8O7) 10% (berat) dengan bubur NaOH 50% (be... Français Deutsch About Dev & API Blog Terms Privacy Copyright Support

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Contoh Recycle dikerjakan • Pada suatu proses produksi sodium sitrat, 1000kg/jam larutan sodium sitrat berkadar 10% dipeka...

Harap dikerjakan • Pada industri gula, larutan gula 1000 kg/jam berkadar 25% dipekatkan hingga berkadar 55%. Larutan terse...

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• Evaporator berkapasitas menguapkan air sebanyak 10 kg/jam sehingga kadar padatan berubah dari 5,5% menjadi 25%. Untuk me...

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FLUID FLOW THEORY • Many raw materials for foods and many finished foods are in the form of fluids. • Thin liquids - milk,...

FLUID STATICS • very important property : the fluid pressure • Pressure is force exerted on an area • force is equal to th...

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You just clipped your first slide! Clipping is a handy way to collect important slides you want to go back to later. Now customize the name of a clipboard to store your clips. The force per unit area in a fluid is called the fluid pressure. It is exerted equally in all directions. • F = APs + ZAg... Name* Best of Slides



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EXAMPLE . Total pressure in a tank of peanut oil • Calculate the greatest pressure in a spherical tank, of 2 m diameter, f...

• Density of water = 1000 kg m-3 Density of oil = 0.92 x 1000 kg m-3 = 920 kg m-3 Z =greatest depth = 2 m and g = 9.81 m s...

Expressing the pressure • absolute pressures • gauge pressures • head

EXAMPLE. Head of Water • Calculate the head of water equivalent and mercury to standard atmospheric pressure of 100 kPa. •...

FLUID DYNAMICS • In most processes fluids have to be moved • Problems on the flow of fluids are solved by applying the pri...

Mass Balance • 1A1v1 = 2A2v2 • incompressible 1 = 2 so in this case • A1v1 = A2v2 (continuity equation) • area of the ...

EXAMPLE. Velocities of flow • Whole milk is flowing into a centrifuge through a full 5 cm diameter pipe at a velocity of 0...

Solving • 1A1v1 = 2A2v2 + 3A3v3 • where suffixes 1, 2, 3 denote respectively raw milk, skim milk and cream. • since the...

• A1 = (/4) x (0.05)2 = 1.96 x 10-3 m2 • A2 = A3 = (/4) x (0.02)2 = 3.14 x 10-4 m2 v1 = 0.22 m s-1 1 = 1.035, 2 = 1.04...

Energy Balance • Referring Fig. before we shall consider the changes in the total energy of unit mass of fluid, one kilogr...

Water flows at the rate of 0.4 m3 min-1 in a 7.5 cm diameter pipe at a pressure of 70 kPa. If the pipe reduces to 5 cm dia...

Water is raised from a reservoir up 35 m to a storage tank through a 7.5 cm diameter pipe. If it is required to raise 1.6 ...

VISCOSITY • Viscosity is that property of a fluid that gives rise to forces that resist the relative movement of adjacent ...

If the plane elements are at a distance Z apart, and if their relative velocity is v, then the force F required to maintai...

Unit of Viscosity • N s m-2 = Pascal second, Pa s, • The older units, the poise and its sub-unit the centipoise, • 1000 ce...

Newtonian and Non-Newtonian Fluids • F/A = µ v /Z = µ(dv/dz) = " = k(dv/dz)n power-law equation • Newtonian fluids (n ...

STREAMLINE AND TURBULENT FLOW • STREAMLINE, flow is calm, in slow the pattern and smooth • TURBULENT, the flow is more rap...

EXAMPLE . Flow of milk in a pipe Milk is flowing at 0.12 m3 min-1 in a 2.5-cm diameter pipe. If the temperature of the mil...

ENERGY LOSSES IN FLOW • Friction in Pipes • Energy Losses in Bends and Fittings • Pressure Drop through Equipment • Equiva...

Friction in Pipes • Eƒ : the energy loss due to friction in the pipe. • Eƒ : proportional to the velocity pressure of the ...

Friction factors in pipe (Moody graph)

predicted f • f = 16/(Re) streamline flow, Hagen-Poiseuille equation 0 < (Re) < 2100 • ƒ = 0.316 ( Re)-0.25 /4 (Blasius eq...

ROUGHNESS FACTORS FOR PIPES Material Roughness factor () Material Roughness factor () Riveted steel 0.001- 0.01 Galvaniz...

EXAMPLE Pressure drop in a pipe Calculate the pressure drop along 170 m of 5 cm diameter horizontal steel pipe through whi...

Thermal conductivity Specific heat Density Viscosity Temperature (J m-1 s-1 °C-1 ) (kJ kg-1 °C-1 ) (kg m-3 ) (N s m-2 ) (°...

Energy Losses in Bends and Fittings • energy losses due to altering the direction of flow, fittings of varying cross-secti...

FRICTION LOSS FACTORS IN FITTINGS k Valves, fully open: gate 0.13 globe 6.0 angle 3.0 Elbows: 90° standard 0.74 medium swe...

FLUID-FLOW APPLICATIONS • Two practical aspects of fluid flow in food technology : • measurement in fluids: pressures and ...

MEASUREMENT OF PRESSURE IN A FLUID • Method : – Piezometer ("pressure measuring") tube – U-tube – Pitot tube – Pitot-stati...

EXAMPLE. Pressure in a vacuum evaporator The pressure in a vacuum evaporator was measured by using a U-tube containing mer...

MEASUREMENT OF VELOCITY IN A FLUID • Pitot tube and manometer : • Z1g + v1 2 /2 + P1/1 = Z2g + v2 2 /2 + P2/1 • Z2 = Z +...

EXAMPLE . Velocity of air in a duct Air at 0°C is flowing through a duct in a chilling system. A Pitot-static tube is inse...

Venturi and orifice meters • v1 2 /2 + P1/1 = v2 2 /2+ P2/2 (Bernouilli's equation) • A1v1 = A2v2 (mass balance, eqn) • ...

Pompa dan Fan • mechanical energy from some other source is converted into pressure or velocity energy in a fluid. • The f...

Jenis Pompa • Positive Displacement Pumps • the fluid is drawn into the pump and is then forced through the outlet • Posit...

Jet Pumps • a high-velocity jet is produced in a Venturi nozzle, converting the energy of the fluid into velocity energy. ...

Air-lift Pumps • air or gas can be used to impart energy to the liquid • The air or gas can be either provided from extern...

Propeller Pumps and Fan • Propellers can be used to impart energy to fluids • They are used extensively to mix the content...

Centrifugal Pumps and Fans • The centrifugal pump converts rotational energy into velocity and pressure energy • The fluid...

Gambar jenis-jenis pompa

• EXAMPLE Centrifugal pump for raising water Water for a processing plant is required to be stored in a reservoir to suppl...

Reynold number • Assume properties of water at 20°C are density 998 kg m-3, and viscosity 0.001 N s m-2 • Cross-sectional ...

friction loss of energy From Table 3.1, the roughness factor is 0.0002 for galvanized iron and so roughness ratio /D =...

TABLE 3.1 RELATIVE ROUGHNESS FACTORS FOR PIPES Material Roughness factor () Material Roughness factor () Riveted steel 0...

Friction factors in pipe

TABLE 3.2 FRICTION LOSS FACTORS IN FITTINGS k Valves, fully open: gate 0.13 globe 6.0 angle 3.0 Elbows: 90° standard 0.74 ...

• For the eight right-angled bends, from Table 3.2 we would expect a loss of 0.74 velocity energies at each, making (8 x 0...

Energy to move 1 kg water • Energy to move 1 kg water against a head of 22 m of water is E = Zg = 22 x 9.81 = 215.8 J. • T...

energy requirement of pump • and theoretical power requirement = Energy x volume flow x density = (Energy/kg) x kgs-1 = 23...

Energy Balance

Gas and Vapour • naturally associated with foods and food-processing systems: – Equilibrium between food and water vapor d...

Kinetika Gas • The postulates of the kinetic theory – Gases are composed of discreet particles called molecules, which are...

Absolute Temperature and Pressure • pressure : force of collisions of gas molecules against a surface in contact with the ...

Absolute and gage pressure

Conversion factor

Calculate the absolute pressure inside an evaporator operating under 20 in. Hg vacuum. Atmospheric pressure is 30 in. Hg. ...

The Ideal Gas Equation • Pressure, the force of collision between gas molecules and a surface, is directly proportional to...

Calculate the quantity of oxygen entering a package in 24 hours if the packaging material has a surface area of 3000 cm2 a...

Calculate the volume of CO2 in ft3 at 70o F and 1 atm, which would be produced by vaporization of 1 lb of dry ice.

Calculate the density of air (M = 29) at 70F and 1 atm in (a) American Engineering and (b) SI units.

Suatu proses memerlukan debit udara bertekanan 2 atm sebesar 10 m3 /s pada suhu 20C. Hitung debit kompresor pada STP yang...

An empty can was sealed in a room at 80o C and 1 atm pressure. Assuming that only air is inside the sealed can, what will ...

Gas Mixtures • If components of a gas mixture at constant volume are removed one after the other, the drop in pressure acc...

Hitung kuantitas udara pada headsapce kaleng yang bersuhu 20o C jika tekanan pada headspace sebesar 10 in Hg. Tekanan atmo...

• Assume there are no dissolved gases in the product at the time of sealing, therefore the only gases in the headspace are...

soal • Proses penutupan kaleng dilakukan pada suhu 80o C dan tekanan 1 atm. Di bagian headspace hanya ada udara dan uap. S...

A gas mixture used for controlled atmosphere storage of vegetables contains 5% CO2, 5% O2, and 90% N2. The mixture is gene...

Ruangan penyimpanan buah segar diatur sehingga komposisi gas yang masuk ke ruang penyimpanan menjadi 88% N2, 6% O2 dan 6% ...

Campuran gas terdiri dari 5% CO2, 5% O2, and 90% N2 digunakan pada CAS untuk buah. Gas campuran dibuat dengan mencampur ud...

Udara 1 m3 bertekanan 5 atm dijenuhkan dengan uap air pada suhu 50C. Jika udara tersebut diturunkan tekanannya menjadi 1 ...

Tekanan parsial uap air di udara pada 25C dan 1 atm adalah 2,520 kPa. Jika udara ditekan hingga 5 atm pada suhu 35C, hit...

Temperature Pressure Enthalpy (sat. vap.) Latent heat Specific volume (°C) (kPa) (kJ kg-1 ) (kJ kg-1 ) (m3 kg-1 ) 20 2.34 ...

PROPERTIES OF SATURATED AND SUPERHEATED STEAM • Steam and water are the two most used heat transfer mediums in food proces...

If 1 lb of water at 100 psig and 252F is allowed to expand to 14.7 psia, calculate (a) the resulting temperature after ex...

How much heat would be given off by cooling steam at 252F and 30.883 psia to 248F, at the same pressure? • First, check ...

Superheated Steam Tables • Superheated Steam: Water vapor at a temperature higher than the boiling point. The number of de...

How much heat is required to convert 1 lb of water at 70F to steam at 14.696 psia and 250F? • First determine the state ...

How much heat would be given off by cooling superheated steam at 14.696 psia and 500F to 250F at the same pressure? • Ba...

Soal • Campuran gas terdiri dari 5% CO2, 5% O2, and 90% N2 digunakan pada CAS untuk buah.Gas tersebut dibuat dengan mencam...

Harap dikerjakan • Proses produksi Sodium Sitrat (Na2C6H6O7) dilakukan dengan mereaksikan larutan asam sitrat (C6H8O7) 10%...

Temperature Pressure Enthalpy (sat. vap.) Latent heat Specific volume (°C) (kPa) (kJ kg-1 ) (kJ kg-1 ) (m3 kg-1 ) 20 2.34 ...

Heat • Sensible heat is defined as the energy transferred between two bodies at different temperatures, or the energy pres...

Specific Heat • The specific heat (Cp) is the amount of heat that accompanies a unit change in temperature for a unit mass...

specific heat J/(kg K)

Estimation of Cp • Cavg = 3349M+ 837.36 in J/(kg K) for fat free plant material • Cavg = 1674.72 F + 837.36 SNF + 4l86.8M ...

Specific heat of gas and vapor • whereCpm is mean specific heat from the reference temperature To to T1. Tabulated values ...

Contoh • Hitung kebutuhan panas untuk menaikkan suhu udara pengering pd tekanan 1 atm dari suhu ruang 25o C ke suhu penger...

PROPERTIES OF SATURATED AND SUPERHEATED STEAM • Steam and water are the two most used heat transfer mediums in food proces...

Steam table • The saturated steam table consists of entries under the headings of temperature, absolute pressure, specific...

Temperature Pressure Enthalpy (sat. vap.) Latent heat Specific volume (°C) (kPa) (kJ kg-1 ) (kJ kg-1 ) (m3 kg-1 ) 20 2.34 ...

contoh • Pada tekanan vakum berapa sehingga air mendidih pada suhu 80o C, nyatakan dalam kPa dan dalam cm Hg – Lihat tabel...

Freezing Points of Food Products Unmodified from the Natural State • the heat to be removed during freezing of a food prod...

Calculate the freezing point and the amount of heat that must be removed in order to freeze 1 kg of grape juice containing...

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