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Dasar Keteknikan Pengolahan Pangan

Dasar Keteknikan Pengolahan Pangan. Sudarminto Setyo Yuwono. q frig. Apple Cooling. INTRODUCTION. 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

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Dasar Keteknikan Pengolahan Pangan

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  1. Dasar Keteknikan Pengolahan Pangan Sudarminto Setyo Yuwono

  2. qfrig Apple Cooling

  3. INTRODUCTION • 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.

  4. 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

  5. Dimensions • These dimensions are length, mass, time and temperature. • For convenience in engineering calculations, force is added as another dimension • Dimensions are represented as symbols by: length [L], mass [M], time [t], temperature [T] and force [F].

  6. example: • Length = [L], area = [L] 2, volume = [L] 3. • Velocity   = length travelled per unit time=[L]/[t] • Pressure      = force per unit area=[F]/[L] 2 • Energy         = force times length=[F] x [L]. • Power          = energy per unit time=[F] x [L]/[t] • h=[F] x [L]/[L]2 [t] [T] =[F] [L] -1 [t] -1 [T] -1

  7. Units • the metre (m) is defined in terms of the wavelength of light; • the standard kilogram (kg) is the mass of a standard lump of platinum-iridium; • the second (s) is the time taken for light of a given wavelength to vibrate a given number of times; • the degree Celsius (°C) is a one-hundredth part of the temperature interval between the freezing point and the boiling point of water at standard pressure; • the unit of force, the newton (N), is that force which will give an acceleration of 1 m sec-2 to a mass of 1kg; • the energy unit, the newton metre is called the joule (J), • the power unit, 1 J s-1, is called the watt (W).

  8. 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

  9. Neraca Massa • Tahapan perhitungan: • Gambar diagram • Tulis reaksi kimia jika ada • Tulis dasar-dasar perhitungan • Hitung neraca massanya

  10. Mass Balance • Proses produksi soda api (NaOH), sebanyak 1000 kg/jam larutan mengandung 7,5% NaOH di pekatkan pada evaporator yang pertama sehingga kadarnya menjadi 20%. Larutan ini lalu dipekatkan pada evaporator yang kedua yang menghasilkan pekatan berkadar 60% NaOH. Hitung NaOH yang dihasilkan.

  11. Neraca massa jika terjadi reaksi kimia • Beberapa proses pengolahan kemungkinan terjadi reaksi kimia • Fermentasi • Pembakaran • Dasar perhitungan bukan dari massa tetapi dari perubahan mol • Setelah itu baru dikonversikan ke massa

  12. 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 kamponennya, jika : • A. Pembakaran sempurna, udara pas • B. Pembakaran 90% sempurna • C. Udara berlebih 20%, pembakaran sempurna 80%

  13. contoh • Larutan NaOH diproduksi dengan cara menambahkan larutan Na2CO3 10% ke dalam aliran bubur Ca(OH) 2 25%. Bagaimana komposisi bubur akhir jika reaksi 90% sempurna dan 100% sempurna. Gunakan dasar 100 kg aliran bubur Ca(OH)2 • Na2CO3 + Ca(OH) 2 => 2NaOH + CaCO3 • Ca(OH) 2= 74,1; Na2CO3 = 106

  14. Recycle • Pada suatu proses produksi sodium sitrat, 1000kg/jam larutan sodium sitrat berkadar 20% dipekatkan di suatu evaporator bersuhu 353K sehingga diperoleh kadar 50%. Larutan lalu dimasukkan ke kristalizer yang bersuhu 303K sehingga diperoleh kristal Na sitrat berkadar 95%. Larutan jenuh yang mengandung 30% Na sitrat lalu direcycle ke evaporator. Hitung berapa laju aliran recycle dan produk yang dihasilkan.

  15. Harap dikerjakan • Pada industri gula, larutan gula 1000 kg/jam berkadar 30% dipekatkan hingga berkadar 60%. Larutan tersebut lalu dimasukkan ke kristalizer sehingga diperoleh kristal gula berkadar air 5%. Larutan jenuh berkadar gula 40% selanjutnya direcycle ke evaporator lagi. Hitung jumlah larutan yang direcycle dan gula yang dihasilkan.

  16. 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.

  17. 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.

  18. 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

  19. 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.

  20. 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-2Now 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

  21. Expressing the pressure • absolute pressures • gaugepressures • head

  22. 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

  23. 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.

  24. 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

  25. 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.

  26. 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 = ρ2A1v1 - ρ2A3v3 + ρ3 A3v3 • A1v1(ρ1 - ρ2) = A3v3(ρ3 - ρ2)

  27. 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-1ρ1 = 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

  28. 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 + v12/2 + P1/ρ1 = Z2g + v22/2 + P2/ρ2 + Ef - Ec. • Zg + v2/2 + P/ρ  = k Persamaan Bernouilli 

  29. 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 m2and 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 + v12/2 + P1 /ρ1 = Z2g + v22/2 + P2 / ρ2and 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) = 65.3P2 = 65.3k Pa.

  30. 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 eqnZ1g + v12/2 + P1/ρ1 = Z2g + v22/2 + 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.

  31. 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.

  32. 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 writeF/A = µv/Z

  33. 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.

  34. Newtonian and Non-Newtonian Fluids • F/A = µv /Z = µ(dv/dz) = t • t = k(dv/dz)npower-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

  35. 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. rv2D/mv = Dvr/m =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.

  36. 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-2Density of milk at 21°C = 1029 kg m-3.Diameter of pipe = 0.025 m.Cross-sectional area of pipe = (p/4)D2= p/4 x (0.025)2= 4.9 x 10-4 m2Rate 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) = (Dvr/m)= 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.

  37. ENERGY LOSSES IN FLOW • Friction in Pipes • Energy Losses in Bends and Fittings • Pressure Drop through Equipment • Equivalent Lengths of Pipe

  38. 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 rv2/2 • F is the friction force, A is the area over which the friction force acts, r 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 = (4frv2/2)(L1 - L2)/D • DPf = (4frv2/2) x (L/D) (Fanning-D'Arcy equation) • Eƒ = DPf/r = (2fv2)(L/D) • L = L1 - L2 = length of pipe in which the pressure drop, DPf= P1 - P2 is the frictional pressure drop, and Eƒ is the frictional loss of energy.

  39. Friction factors in pipe (Moody graph)

  40. 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 (e)/pipe diameter (turbulent region)

  41. ROUGHNESS FACTORS FOR PIPES

  42. 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, • DPf = (4frv2/2) 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.

  43. 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

  44. FRICTION LOSS FACTORS IN FITTINGS LOSS FACTORS IN CONTRACTIONS

  45. 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.

  46. MEASUREMENT OF PRESSURE IN A FLUID • Method : • Piezometer ("pressure measuring") tube • U-tube • Pitot tube • Pitot-static tube • Bourdon-tube • P = Z1r1g

  47. 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 = Zrg= 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 = Zrg • 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

  48. MEASUREMENT OF VELOCITY IN A FLUID • Pitot tube and manometer : • Z1g + v12/2 + P1/r1 = Z2g + v22/2 + P2/r1 • 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/r1 = (Z + Z')g»Z. • Pitot-static tube • Z = v2/2g

  49. 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 • r1Z1 = r2Z2. • 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

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