Dasar Keteknikan Pengolahan Pangan

Sudarminto Setyo Yuwono

Apple Cooling

qfrig

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, crystallization, and membrane processes), mechanical separations (which include filtration, centrifugation, sedimentation and sieving), size reduction and mixing.

The aims of the food industry• 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

Food processes are usually schematized by means of flow charts

diagrams indicate different manufacturing steps, as wellas 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 thematerials flow.

DIMENSIONS AND UNITS

• 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

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

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

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 θ

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    Singkatan Dimensi  

1 Sudut Datar Radian rad -

2 Sudut Ruang Steradian sr -

Besaran Pokok Tak Berdimensi

1.6

Dimensi Cara besaran itu tersusun oleh besaran pokok.

Besaran TurunanBesaran yang diturunkan dari besaran pokok.

1. Untuk menurunkan satuan dari suatu besaran2. Untuk meneliti kebenaran suatu rumus atau persamaan

- Metode penjabaran dimensi :

1. Dimensi ruas kanan = dimensi ruas kiri2. Setiap suku berdimensi sama

- Guna Dimensi :

1.7

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

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

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

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-2T-2 satuan kgm-2

berat

volume

Gaya

Volume

MLT -2

L3a. Gaya = massa x percepatan

= M x LT -2

= MLT -2 satuan kgms-2

c. Tekanan = = = MLT -2 satuan kgm-1s-1

gaya

luas

MLT -2

L2

d. Usaha = gaya x jarak = MLT -2 x L = ML 2 T -2 satuan kgm-2s-2

e. Daya = = = ML 2 T -1 satuan kgm-2s-1 usaha

waktu

ML 2 T -2

T

Contoh SoalContoh Soal

1.11

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 = ML2T-2

Energi Kinetik : Ek = ½ mv2

Energi Kinetik = ½ x massa x kecepatan2

= M x (LT-1) 2

= ML2T-2

Keduanya (Ep dan Ek) mempunyai dimensi yang sama keduanya identik

b. Usaha = ML2T-2

Energi = ML2T-2

Kalor = 0.24 x energi = ML2T-2

Ketiganya memiliki dimensi yang sama identik

1.12

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 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

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

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

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

Tahapan perhitungan

• Gambar diagram

• Tulis reaksi kimia jika ada

• Tulis dasar-dasar perhitungan

• Hitung neraca massanya

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

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.

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%?

Batas proses (boundary)• Batas proses dapat digunakan untuk menyederhanakan suatu

proses• Dapat diperluas atau diperkecil

Dikerjakan

• Nira tebu 1000 kg/jam berkadar gula 20% dipekatkan hingga kadar gula 60%. Nira pekat selanjutnya dikristalisasi pada suhu 20oC. Konsentrasi kejenuhan gula pada suhu 20oC sebesar 40%. Berapa kg/jam kristal gula yang dihasilkan? Diasumsikan kristal gula tidak mengandung air

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.

• 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.

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

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).

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

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

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%

• 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 kadar air bubur susu setelah fermentasi

• 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.

• 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

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 15oC untuk mengkristalkan sodium sitrat. Jika kelarutan sodium sitrat pada suhu 15oC 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

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

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.

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.

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

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.

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 = Vρg • F is force (Newton) or kg m s-2, m is the mass, g the

acceleration due to gravity, V the volume and ρ the density.

The force per unit area in a fluid is called the fluid pressure. It is exerted

equally in all directions. • F = APs + ZρAg

• Ps is the pressure above the surface of the fluid (e.g. it might be atmospheric pressure

• total pressure P = F/A = Ps + Zρg

• the atmospheric pressure represents a datum P = Zρg

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.

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

Now P = Zρg            = 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

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.

• Density of water = 1000 kg m-3, Density of mercury = 13,600 kg m-3 g = 9.81 m s-2and pressure   = 100 kPa                        = 100 x 103 Pa   =  100 x 103 kg m-1s-2.

Water       Z     = P/ ρ g                        = (100 x 103)/ (1000 x 9.81)                        = 10.2 m

Mercury   Z      = (100 x 103)/ (13,600 x 9.81)                        = 0.75m

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.

Mass Balance

• ρ1A1v1 = ρ2A2v2 • incompressible

ρ1 = ρ2 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 ρ1 , and if the corresponding values at section 2 are A2, v2, ρ2

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.

Solving

• ρ1A1v1 = ρ2A2v2 + ρ3A3v3

• 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   

• ρ1A1v1 = ρ2A2(A1v1 – A3v3)/A2 + ρ3A3v3

• ρ1 A1v1 = ρ2 A1v1 - ρ2 A3v3 + ρ3 A3v3

• A1v1(ρ1 - ρ2 ) = A3v3(ρ3 - ρ2 )             

• A1 = (π/4) x (0.05)2 = 1.96 x 10-3 m2

• A2 = A3 = (π/4) x (0.02)2 = 3.14 x 10-4 m2

 v1 = 0.22 m s-1

ρ1 = 1.035, ρ2 = 1.04, ρ3 = 1.01

• A1v1(ρ1 - ρ2 ) = A3v3(ρ3 - ρ2 )

• -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

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/ρ  (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/ρ1 = Z2g + v22/2 + P2/ρ2 + Ef - Ec.

• Zg + v2/2 + P/ρ  = k Persamaan Bernouilli 

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 = (π/4)D2

                                           = (π /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    = (π/4)(0.05)2

                                           = 1.96 x 10-3 m2

and so velocity of flow in 5 cm diameter pipe,                                       v2 = (0.4/60)/(1.96 x 10-3) = 3.4 m s-1

• Now•            Z1g + v1

2/2 + P1 /ρ1 = Z2g + v22/2 + P2 / ρ2

and so0 + (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.

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 = (π/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 + v12/2 + P1/ρ1 = Z2g + v2

2/2 + P2/ρ2 + 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

    = EcVρ    = 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.

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.

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  

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.

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

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.

v2D/v = Dv/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.

EXAMPLE . Flow of milk in a pipeMilk 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 10 m3 s-1

• So velocity of flow = (2 x 10-3)/(4.9 x 10-4)= 4.1 m s-1,and so (Re) = (Dv/)= 0.025 x 4.1 x 1029/(2.1 x 10-3)

= 50,230and this is greater than 4000 so that the flow is turbulent.

ENERGY LOSSES IN FLOW

• Friction in Pipes

• Energy Losses in Bends and Fittings

• Pressure Drop through Equipment

• Equivalent Lengths of Pipe

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 Pf = (4fv2/2) x (L/D) (Fanning-D'Arcy equation)• Eƒ = Pf/ = (2fv2)(L/D)

• L = L1 - L2 = length of pipe in which the pressure drop, Pf = P1 - P2 is the frictional pressure drop, and Eƒ is the frictional loss of energy.

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 equation for smooth pipes in the range 3000 < (Re) < 100,000)

• roughness ratio = Roughness factor ()/pipe diameter (turbulent region)

•

ROUGHNESS FACTORS FOR PIPES

MaterialRoughness factor ()

MaterialRoughness 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

EXAMPLE Pressure drop in a pipeCalculate 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 m3 min-1

• Diameter of pipe = 0.05 m,Area of cross-section A                          = (π/4)D2                         = π /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) = (Dvρ/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, •                  Pf = (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.

 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

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 = kv22/2 Losses in sudden contraction

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

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.

MEASUREMENT OF PRESSURE IN A FLUID

• Method :– Piezometer ("pressure measuring") tube – U-tube– Pitot tube– Pitot-static tube – Bourdon-tube

• P = Z11g

EXAMPLE. Pressure in a vacuum evaporatorThe 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 evaporator.

• For atmospheric pressure:

• P = Zg

• P = 75.4 x 10-2 x 13.6 x 1000 x 9.81 = 100.6 kPaTherefore the absolute pressure in the evaporator = 100.6 - 33.4 = 67.2 kPa

MEASUREMENT OF VELOCITY IN A FLUID • Pitot tube and manometer :

• Z1g + v12/2 + 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.

• v12/2g + P1/1 = (Z + Z')g Z.

• Pitot-static tube

• Z = v2/2g

EXAMPLE . Velocity of air in a ductAir 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 = v12/2g

1Z1 = 2Z2.

• Now 0.8 mm water = 0.8 x 10-3 x 1000

1.3

= 0.62 m of air

• v12 = 2Zg

= 2 x 0.62 x 9.81

= 12.16 m2s-2

• Therefore v1 = 3.5 m s-1

Venturi and orifice meters

• v12/2 + P1/1 = v2

2/2+ P2/2 (Bernouilli's equation)

• A1v1 = A2v2 (mass balance, eqn)

1 = 2 =

v12/2 + P1/ = (v1A1/A2)2/2 + P2/

v12 = [2(P2 -P1)/] x A2

2/(A22 -

A12)

(P2 -P1)/ = gZm /

Z = (P2 -P1)/m g

v1 = C √[2(P2 -P1 )/]x A22/(A2

2 -A12 )

In a properly designed Venturi meter,

C lies between 0.95 and 1.0.

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.

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.

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.

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. 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.

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

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.

Gambar jenis-jenis pompa

• 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?

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 = (π/4)D2

                                             = π /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) = Dvρ/µ•                                              = (0.15 x 1.13 x 998)/0.001

                                             = 1.7 x 105

so the flow is clearly turbulent.

friction loss of energy

From Table 3.1, the roughness factor ε is 0.0002 for galvanized ironand so              roughness ratio   ε /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)= [4ƒv2L/2D]= [4 x 0.0053 x (1.13)2 x 120]/(2 x 0.15)= 10.8 J.

TABLE 3.1RELATIVE 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

Friction factors in pipe

TABLE 3.2FRICTION 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

• 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 JThere would be one additional velocity energy loss because of the unrecovered flow energy discharged into the reservoir.

Energy loss from bends and discharge

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

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.

Energy Balance

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 solid at ambient conditions.

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

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

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. 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:

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

Calculate the quantity of oxygen entering a package in 24 hours if the packagingmaterial has a surface area of 3000 cm2 and an oxygen permeability 100 cm3/(m2)(24 h) STP (standard temperature and pressure = 0oC and 1 standard atmosphere of 101.325 kPa).

• Jawaban:

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

Calculate the density of air (M = 29) at 70◦F 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 20◦C. Hitung debit kompresor

pada STP yang harus diberikan.

• Kondisi STP adalah suhu (T) = 0oC (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

An empty can was sealed in a room at 80oC and 1 atm pressure. Assuming that onlyair is inside the sealed can, what will be the vacuum after the can and contents cool to 20oC?

• Solution:

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

Hitung kuantitas udara pada headsapce kaleng yang bersuhu 20oC 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 20oC = 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

• 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 20◦C and 80◦C 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

soal

• Proses penutupan kaleng dilakukan pada suhu 80oC dan tekanan 1 atm. Di bagian headspace hanya ada udara dan uap. Setelah dilakukan sterilisasi, kaleng lalu didinginkan hingga suhu 20oC. Hitung berapa tekanan di headspace? Diasumsikan jumlah udara di headspace tetap dan uap air mengkondensasi pada saat pendinginan.

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 and N2 and CO2 gases. 100 m3 of this mixture at 20oC 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 20oC 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

Ruangan penyimpanan buah segar diatur sehingga komposisi gas yang masuk ke ruang penyimpanan menjadi 88% N2, 6% O2 dan 6% CO2. Udara yang bersuhu 25oC 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.

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 +

Udara 1 m3 bertekanan 5 atm dijenuhkan dengan uap air pada suhu 50◦C. Jika udara tersebut diturunkan tekanannya menjadi 1 atm dan suhu 20◦C, hitung jumlah uap air

yang mengkondensasi.

• The vapor pressure of water at 50◦C and 20◦C are 12.3354 and 2.3366 kPa, respectively.

• Basis:1 m3 air at 5 atm pressure and 50◦C. The number of moles of air will remain the same on cooling

Moles water condensed = 0.004593 - 0.004344 = 0.000249 kg moles.

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

Temperature Pressure Enthalpy (sat. vap.)

Latent heatSpecific volume

(°C) (kPa) (kJ kg-1) (kJ kg-1) (m3 kg-1)20 2.34 2538 2454 57.822 2.65 2542 2449 51.424 2.99 2545 2445 45.926 3.36 2549 2440 40.028 3.78 2553 2435 36.630 4.25 2556 2431 32.940 7.38 2574 2407 19.550 12.3 2592 2383 12.060 19.9 2610 2359 7.6770 31.2 2627 2334 5.0480 47.4 2644 2309 3.4190 70.1 2660 2283 2.36

100 101.35 2676 2257 1.673105 120.8 2684 2244 1.42110 143.3 2692 2230 1.21115 169.1 2699 2217 1.04120 198.5 2706 2203 0.892125 232.1 2714 2189 0.771130 270.1 2721 2174 0.669135 313.0 2727 2160 0.582140 361.3 2734 2145 0.509150 475.8 2747 2114 0.393160 617.8 2758 2083 0.307

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

If 1 lb of water at 100 psig and 252◦F 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 252◦F, 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 252◦F.

• (a) After expansion to 14.7 psia, the boiling point at 14.7 psia is 212◦F. Part of the water will flash• to water vapor at 212◦F and the remaining liquid will also be at 212◦F.• (b) The enthalpy of water at 252◦F is (hf at 252◦F) 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 212◦F and 14.7 psia will still be 220.62 BTU.

How much heat would be given off by cooling steam at 252◦F and 30.883 psia to

248◦F, at the same pressure?

• First, check the state of water at 30.883 psia and 252◦F and 248◦F. From steam tables, the boiling point of water at 30.883 psia is 252◦F. Therefore, steam at 252◦F and 30.883 psia is saturated vapor.

• At 30.883 psia and 248◦F, water will be in the liquid state, because 248◦F is below the boiling temperature at 30.883 psia.

• Heat given off = q = hg at 252◦F − hf at 248◦F

• From steam tables,

• hg at 252◦F = 1164.78 BTU/lb

• hf at 248◦F = 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 4◦F 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 by causing a phase change at constant temperature and pressure.

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

How much heat is required to convert 1 lb of water at 70◦F to steam at 14.696 psia

and 250◦F?

• First determine the state of steam at 14.696 psia and 250◦F. At 14.696 psia, the boiling point is 212◦F. Steam at 250◦F and 14.696 psia is superheated steam. From the superheated steam table, h at 250◦F is 1168.8 BTU/lb.

• Heat required = hg at 250◦F and 14.696 psia − hf at 70◦F

• = 1168.8 BTU/lb − 38.05 BTU/lb• = 1130.75 BTU/lb

How much heat would be given off by cooling superheated steam at 14.696 psia

and 500◦F to 250◦F at the same pressure?

• Basis: 1 lb of steam.

• Heat given off = q = h at 14.696 psia and 500◦F − hg at 14.696 psia and 250◦F

• = 1287.4 − 1168.8 = 118.6 BTU/lb

• Superheated steam is not a very efficient heating medium. Note that a 250◦F change in temperature is accompanied by the extraction of only 118.6 BTUs of heat.

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 20oC dan tekanan 1 atm.

• Campuran Udara dan uap air 1 m3 bertekanan 5 atm absolut bersuhu 50oC. Jika udara tersebut diturunkan tekanannya menjadi 1 atm dan suhu 20oC, hitung jumlah uap air yang mengkondensasi

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 15oC untuk mengkristalkan sodium sitrat. Jika kelarutan sodium sitrat pada suhu 15oC sebesar 20% berat, hitung kristal sodium sitrat yang diperoleh, untuk setiap 100 kg larutan asam sitrat yang digunakan.

– Asumsi : - Sodium 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 60oC di wadah ekstraksi (extraction chamber). Debit gas CO2 meninggalkan ekstraktor pada tekanan 101,3 kPa dan suhu 20oC 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.

Temperature Pressure Enthalpy (sat. vap.)

Latent heatSpecific volume

(°C) (kPa) (kJ kg-1) (kJ kg-1) (m3 kg-1)20 2.34 2538 2454 57.822 2.65 2542 2449 51.424 2.99 2545 2445 45.926 3.36 2549 2440 40.028 3.78 2553 2435 36.630 4.25 2556 2431 32.940 7.38 2574 2407 19.550 12.3 2592 2383 12.060 19.9 2610 2359 7.6770 31.2 2627 2334 5.0480 47.4 2644 2309 3.4190 70.1 2660 2283 2.36

100 101.35 2676 2257 1.673105 120.8 2684 2244 1.42110 143.3 2692 2230 1.21115 169.1 2699 2217 1.04120 198.5 2706 2203 0.892125 232.1 2714 2189 0.771130 270.1 2721 2174 0.669135 313.0 2727 2160 0.582140 361.3 2734 2145 0.509150 475.8 2747 2114 0.393160 617.8 2758 2083 0.307

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 32.018◦F or 0.01◦C.

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.

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 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 4.44◦C to 65.55◦C

• 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 = 0.882 MJ

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 77◦F or 25◦C, as the reference temperature.

Contoh

• Hitung kebutuhan panas untuk menaikkan suhu udara pengering pd tekanan 1 atm dari suhu ruang 25oC ke suhu pengeringan 50oC jika tiap menit dialirkan udara sebanyak 100m3

• q= mCp (50-25)

• m=PVM/RT

• R = 0.08206 m3 atm/kg mole K

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.

Steam table• The saturated steam table consists of entries under the headings of

temperature, absolute pressure, specific volume, and enthalpy.

Temperature Pressure Enthalpy (sat. vap.)

Latent heatSpecific volume

(°C) (kPa) (kJ kg-1) (kJ kg-1) (m3 kg-1)20 2.34 2538 2454 57.822 2.65 2542 2449 51.424 2.99 2545 2445 45.926 3.36 2549 2440 40.028 3.78 2553 2435 36.630 4.25 2556 2431 32.940 7.38 2574 2407 19.550 12.3 2592 2383 12.060 19.9 2610 2359 7.6770 31.2 2627 2334 5.0480 47.4 2644 2309 3.4190 70.1 2660 2283 2.36

100 101.35 2676 2257 1.673105 120.8 2684 2244 1.42110 143.3 2692 2230 1.21115 169.1 2699 2217 1.04120 198.5 2706 2203 0.892125 232.1 2714 2189 0.771130 270.1 2721 2174 0.669135 313.0 2727 2160 0.582140 361.3 2734 2145 0.509150 475.8 2747 2114 0.393160 617.8 2758 2083 0.307

contoh

• Pada tekanan vakum berapa sehingga air mendidih pada suhu 80oC, 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 120oC, berapa tekanan yang terbaca pada manometer yang menggunakan satuan psi?– Dari tabel pada suhu 120oC 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

Freezing Points of Food Products Unmodifiedfrom 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).

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