Thiết kế hệ thống sấy thùng quay để sấy cà phê nhân đầu vào 1200kg The drying technique has been researched for a long time and is now being widely applied in the current drying processes. For coffee products, people have many different drying products such as: horizontal blister drying, drying tunnel, or rotary drum drying, etc., in which the rotary drum drying system is being applied quite a lot because it has many superiority advantages over other methods.
HO CHI MINH UNIVERSITY OF TECHNOLOGY AND EDUCATION FACULTY OF HIGH QUALITY TRAINING FOOD TECHNOLOGY PROJECT REPORT DESIGN OF A ROTARY DRUM DRYING SYSTEM USING HOT AIR FOR DRYING GREEN COFFEE BEANS Course : Project of Food Processing and Machinery Student Name Student ID Lê Đức Huy Hoàng 19116041 Phan Đặng Hồng Nhơn 19116034 Instructor : PhD Phạm Văn Hưng HO CHI MINH CITY - 10/2022 EVALUATION FROM THE INSTRUCTOR Ho Chi Minh city, ./ / Sign of Instructor EVALUATION FROM MEMBERS OF COMMITTEE Ho Chi Minh city, ./ / Signs of Commitee Members TABLE OF CONTENTS FOREWORD CHAPTER 1: OVERVIEW 1.1 Overview of Drying technique 1.1.1 Drying definition 1.1.2 Classification of Drying methods 1.1.3 The purposes of Drying 1.1.4 Drying agents 1.2 Overview of Drying Material (Coffee Beans) 1.2.1 Introduction of Coffee 1.2.2 Coffee structure and compositions 1.2.3 Production process of green coffee beans CHAPTER 2: DESIGN A ROTARY DRUM DRYING SYSTEM 10 2.1 Rotary drum drying system for drying green coffee beans 10 2.2 Process explaination 10 CHAPTER 3: MAIN EQUIPMENT CALCULATION OF THE PROCESS 13 3.1 Establish balance of drying material 13 3.1.1 Parameter symbols 13 3.1.2 Initial parameters 13 3.1.3 General balance material calculation 13 3.2 Main device calculation 14 3.2.1 Drying time 14 3.2.2 Volume of the rotating barrel 14 3.2.3 Determine the length and diameter of rotating barrel 14 3.2.4 The number of revolutions of the drying barrel (n) 15 3.2.5 Equipment capacity 15 3.3 Establish heat balance equation 15 3.3.1 Outer air state: 15 3.3.2 Calculation of the input air of calorife 16 3.3.3 Calculation of output air of the dryer 16 3.3.4 Calculate dewpoint temperature 17 3.4 Heat balance equation 18 3.4.1 Heat input of the dryer 19 3.4.2 Heat output of the dryer 19 3.5 Real drying process construction 25 3.5.1 Determination of parameters during real drying 25 3.5.2 Calorific equilibrium equation: 26 CHAPTER AUXILIARY EQUIPMENTS CALCULATION OF THE PROCESS 27 4.1 Calorife 27 4.1.1 Select metrics and size calculation 27 4.1.2 Determination of heat transfer surfaces 28 4.2 Cyclon 31 4.2.1 Calculation 32 4.2.2 Basic size of cyclone ЦH – 24 32 4.3 Fan 33 4.3.1 Resistance due to friction in each pipeline 34 4.4 Local resistance calculation 37 4.4.1 Local resistance by “đột mở” to calofife 38 4.4.2 Resistance by “đột thu” from calorife 38 4.4.3 Local resistance by “đột mở” to material tank 38 4.4.4 Local resistance by “đột mở” from material tank to pipeline 38 4.4.5 Local resistance by cyclone 39 4.5 The resistance from fan 39 4.5.1 Resistance from push fan 39 4.5.2 Resistance from exhausted fan 39 4.6 The resistance from rotating barrel 39 4.7 The resistance of calorife 39 4.8 Fan calculation 40 4.8.1 Push fan 40 4.8.2 Exhausted fan 40 CHAPTER CONCLUSION 41 CHAPTER REFERENCES 42 FOREWORD In the recent years, coffee has become an important agricultural export commodity of Vietnam, with an export value of nearly millions USD by the end of 2021 Therefore, coffee gradually becomes an important position in the economy of our country Currently, Vietnamese coffee has appeared in many continents from North America, Western Europe, Eastern Europe to Australia and South Asia, etc The quality of Vietnamese coffee beans depends on many different factors, including the processing process The preservation process greatly affects the quality, especially for dry goods The duration of this preservation process is long or short depending on the moisture content of the food Therefore, the process of drying coffee makes an important and very important contribution affect the quality of the product – coffee beans The requirement for drying technology for agricultural products and food is to make the material evaporate moisture down to the necessary level for storage and to reach the strict requirements such as: no strange odors from drying agents , loss of compounds in the product, reaching the optimum dryness for each type And another quite important factor is to attract the senses of consumers The drying technique has been researched for a long time and is now being widely applied in the current drying processes For coffee products, people have many different drying products such as: horizontal blister drying, drying tunnel, or rotary drum drying, etc., in which the rotary drum drying system is being applied quite a lot because it has many superiority advantages over other methods In this project, we have the task of “Designing a rotary drum drying system for drying green coffee beans” with a capacity of 1200 kg/h Because this is the first time we have access to a specialized drying system design project, with limited knowledge and references, we cannot avoid mistakes in the design process We would like to say thank you to our instructor - Mr Pham Van Hung, for helping and giving enthusiastic suggestions so that we could successfully complete the project CHAPTER 1: OVERVIEW 1.1 Overview of Drying technique 1.1.1 Drying definition Drying is the oldest method of preserving food Throughout history, the sun, the wind, and a smoky fire were used to remove water from fruits, meats, grains, and herbs By definition, food dehydration is the process of removing water from food by circulating hot air through it, which prohibits the growth of enzymes and bacteria During the drying process, the water is separated from the material by the principle of evaporation or sublimation Samples of raw materials are usually in solid state, but samples of materials to be dried can also be in liquid or suspension form The products obtained after the drying process is always in solid or powder form The temperature and the surrounding humid air environment have a great and direct influence on the drying speed The essence of the drying process is to transfer the amount of water present in the material from the liquid phase to the vapor phase, which occurs when the partial pressure of water vapor on the surface of the material is greater than the partial pressure in the surrounding air around the material From there, we have to study both sides of the drying process: Statics and Kinetics - Statics: based on material balance and heat balance, we will find the relationship between the first and last parameters of the drying material, the drying agent From there determine the material composition, the amount of drying agent, the amount of heat required for the drying process - Kinetics: the study of the relationship of the variation of material moisture with drying time and process parameters such as: properties, structure, size of the drying materials and the drying hydrodynamics conditions of the drying agent to determine the appropriate drying mode, speed and time 1.1.2 Classification of Drying methods - Natural drying: Evaporation is carried out by natural energy such as sun, wind energy also known as natural drying This method saves heat energy, but does not actively adjust the process speed according to technical requirements, low productivity, etc - Artificial drying: Usually carried out in various types of drying equipment to provide heat for moist materials Artificial drying has many forms, depending on the method of heat transfer, drying techniques can be classified as follows: + Convection drying: Using hot air as the drying agent The raw material sample will be directly exposed to the hot air in the drying chamber, a part of the moisture in the material will be evaporated + Contact drying: the sample of the material to be dried is placed on a heated surface, whereby the temperature of the material will increase and cause some of the moisture in the material to evaporate and escape to the outside environment The material to be dried will be heated according to the principle of heat conduction + Radiation drying: Using a radiant heat source to provide samples of materials to be dried The most commonly used radiation source today is infrared In the radiation drying method, the raw material sample is heated by the radiation phenomenon, while the moisture discharge from the raw material sample to the outside environment will occur according to the convection principle + Cold drying: is a drying method in which the temperature and humidity of the drying agent is much lower than that of the low-temperature environment to ensure the sensory properties of the product, while the humidity is low to create a moisture difference The moisture in the material will escape easily + Sublimation drying: The sample of raw materials first will be frozen so that a part of the moisture in the material turns into a solid state, then a vacuum pressure is created and the temperature is slightly raised so that the water sublimes, that is, the water will change from solid state to vapor state without going through liquid state 1.1.3 The purposes of Drying - Prolong storage time: the drying process reduces the activity value of water in the raw materials, so it inhibits the microflora and some enzymes, helping to prolong the shelf life of the product - Increase the sensory properties of food: the drying process transforms the raw materials and creates many characteristic properties for the product - Cooking and shaping food products: the drying process helps to partially ripen and shape the product according to the manufacturer's requirements - Convenience for transportation: increase shipping volume for each shipment, thereby reducing transportation costs 1.1.4 Drying agents 1.1.4.1 Definition - Drying agents are substances used to transport the moisture removed from the drying device out of the drying device during the drying process - The mechanism of the drying process consists of two stages: heating the dried material to humidify the vapor and bringing moisture from the surface of the object into the environment If moisture escapes from the material without being taken away in time, it will affect the moisture removal process from the drying object and even stop the moisture drainage process 1.1.4.2 Functions of Drying agents Depending on the characteristics of the drying process, the drying agents have the following tasks: - Transporting moisture separated from the drying object out of the drying device during the drying process - Provide heat for moist objects: In the method of convection drying - Protect the damp object from letting its temperature rise above the allowable temperature 1.1.4.3 Classification of Drying agents In convection drying, the role of the drying agent is especially important because it acts as both a heat carrier and a moisture carrier Commonly used drying agents are hot air and flue gas, superheated steam, liquid, etc - Humid air: is the most common type of drying agent that can be used for most products Using moist air does not contaminate the product after drying and change its taste However, using moist air as a drying agent, it is necessary to equip an additional air heater (gas calorifer, gas evaporator or smoke), the drying temperature is not too high, usually less than 500℃ because if the temperature is too high, the heat exchangers must be made of high-cost alloy steel or ceramics - Furnace smoke: The drying agent has a wide temperature range, which can raise the drying temperature to 1000℃ without a heating device, but the drying material is polluted causing smoke smell Therefore, smoke is only used for materials that are not afraid of pollution such as wood, pottery, some shelled nuts - Superheated steam: This drying agent is used for products that are prone to explosions and are generally resistant to high temperatures Drying by superheated steam with a temperature greater than 100℃ (drying at atmospheric pressure) Total length without ribbed Lkg = l - Lg = 1.2 – 0.24 = 0.96 m 4.1.2 Determination of heat transfer surfaces − The amount of air required for returnable drying (according to practical calculations) Where: l = 55.556 (kg/kg of dry air) L = l x W = 55.556 x 290.11 = 16117.35(kg/h) − The air temperature after passing the calorife is t1 = 82oC − Specific volume of air: + V820C = 1.006 (m3/kg) + V370C = 0.878 (m3/kg) + V250C = 0.844 (m3/kg) 𝑉𝑉𝑡𝑡𝑡𝑡 = 𝑣𝑣820 𝐶𝐶 +𝑣𝑣250 𝐶𝐶 = 1.006+0.844 − The amount of dry air entering the calorife: = 0,925 (kg/h) V = L x Vtb = 16117.35 x 0.925 = 14908.54875 (m3/h) = 4.141 (m3/s) − Heat supply coefficient α1: + Average heat difference of air in calorife: ttb = thn - ∆ttb With Δttb = Δtd −Δtc Δt ln Δtd c + Choosing the temperature of the steam when entering: thnđ = 130oC + Choosing the temperature of the steam when it comes out: thnc = 110oC ∆tđ = thnđ – tmt = 130 – 25 = 105 oC ∆tc = thnc – tc = 110 – 82 = 28 oC Δttb = 105−28 105 ln 28 = 58.526 oC ttb = thn - ∆tb = 120 – 70.109 = 49.891 oC + For the value ttb = 58.256oC we have the following table of values: Density Thermal ρ =1.0024 Kg/m conductivity coefficient λ = 2.984x10-2 W/m0C Air velocity λ = 2.984x10-6 m2/s Pran's constant Pr = 0.69 28 + The inner area of the pipe: Ftr = π x dtr x l = 3.14 x 0.025 x 1.2 = 0.094 (m2) + Outer surface area of the pipe: Fng = π x dng x l = 3.14 x 0.03 x 1.2 = 0.113 (m2) + Area of the outer surface part of the pipe: Fbm = Fgân + Fkgân 𝜋𝜋 𝜋𝜋 𝐹𝐹𝑔𝑔â𝑛𝑛 = π𝐷𝐷𝑔𝑔 𝐿𝐿𝑔𝑔 + 𝐷𝐷𝑔𝑔2 − 𝑑𝑑𝑛𝑛𝑛𝑛 4 = 3.14x0.042x0.24 + 3.14 (0.042)2 − 3.14 (0.03)2 = 0.032 (m2) 𝐹𝐹𝑘𝑘𝑘𝑘â𝑛𝑛 = π𝑑𝑑𝑛𝑛𝑛𝑛 𝐿𝐿𝑘𝑘𝑘𝑘 = 0.96 x 3.14 x 0.03 = 0.090 (m2) So, 𝐹𝐹𝑏𝑏𝑏𝑏 = 0.032 + 0.090 = 0.1220 (m2) + Choose the number of pipes to line up: i = 30 + Distance between pipes: 0.007 (m) + Distance of outermost pipe to calorife: 0.01 (m) + The free area of calorific value is Ftd • The length of calorife: Lx = 2s + I x dng + (i – 1)r = = x 0.01 + 30 x 0.042 + (30 - 1) x 0.007 = 1.483 (m) • Horizontal cross-section area of calorife: Fx = Lx x hcao = 1.48 x 1.2 = 1.780 (m) • Resistance area of the ribbed: Fcg = Dg x Lgxi = 0.042 x 0.24 x 30 = 0.302 (m) • Resistance area of the pipe: Fcơ = dn x Lkg x i = 0.03 x 0.96 x 30 = 0.864 (m) So the area of the free part: Ftd = Fx - Fcô - Fcg = 1.780– 0.864 – 0.302 = 0.614 (m) + The velocity of air: ωkk = V Ftd = 3.832 0.614 = 6.241 (m/s) + Heat supply coefficient from saturated steam to the horizontal surface of the pipe: α1 = 2.04xA � Where: r HΔt 0.25 � (W/m2C) H: tube height, choose H = 1.2 r: hidden thermochemical vaporization J/kg (at head temperature 130) r = 2179.103 (J/kg) (Bảng I.250 – sổ tay QT&TB – tập – trang 312) A coefficient whose value depends on tm (membrane temperature) Select: tT = 129.740C inner tube wall temperature of the pipe So, t m = tT +tbh = 129.74+103 = 129.87 oC 29 Look up the table QT&TB Handbook Volume Page 29, we have A = 192,389 ∆𝑡𝑡 = t hnđ − t T = 130 − 129.74 = 0.26 oC 2179x103 Instead of numbers, we can calculate: α1 = 2.04x192.389 � 20176,046 (W/m2C) 1.2x0.26 0.25 � = => So the specific heat is: q1 = α1x∆t =20176.046x0.26 = 5245.772(W/m2) − Calculate the coefficient of heat supply from the pipe surface to the moving air in calorife α2: − The fluid flows through the outside of the ribbed beam tube: Nu = CxRe Where: n dng xPr 0.4 � � bg −0.54 ℎ �b � g −0.54 dng: The outer diameter of the pipe, dng = 0.03m bg: Step of ribble, bg = 0.01m h: ribbed height, hg = 0.006m Pr = 0.69 C, n: the quantities depend on how the pipe is stacked, for the pipe aligns C = 0.116; n = 0.72 - Reynolds number: Re = ωkk xbg Nu = 0.016x3089.604 γ 0.72 = 6.241x0.01 x0.69 20.2x106 0.4 = 3089.604 0.03 −0.54 0.006 −0.54 = 19.326 � � � � 0.01 0.01 => Convective heat supply coefficient: 𝛼𝛼2 = (W/m2oC) 𝑁𝑁𝑁𝑁𝑁𝑁 𝑏𝑏𝑔𝑔 = 19.326x2.984x10−2 0.01 = 57.669 + Actual heat supply coefficient: αT = 40 (Actual heat supply coefficient: αT = 40) 𝐾𝐾 = Where: 1 𝐹𝐹 + x 𝑏𝑏𝑏𝑏 + 𝑟𝑟𝑡𝑡 𝛼𝛼 𝑇𝑇 𝛼𝛼1 𝐹𝐹𝑡𝑡𝑡𝑡 = 1 0.130 0.0025 + x + 40 20176.046 0.094 385 = 39.880 Fbm: the entire outer surface of the pipe including the ribbed surface calculated for one unit of pipe length (m) Ftr: inner surface of pipe calculated for one unit of pipe length m2 α 1: heat supply coefficient in tube (W/m2) 30 𝑟𝑟𝑡𝑡 = Where : 𝛿𝛿 𝜆𝜆 rt: total heat resistance of walls and layers of sediment δ: length of pipe, m λ : thermal conductivity coefficient of the pipe material, W/m.degree − Specific heat: q2 = k.tm = 39,880,129.87 = 5179,216 (W/m2) − Calculation of error: Δq = q1 − q 5245.772 + 5179.217 x100 = x100 = 1.268% < 5% 5245.772 q1 So the sizes selected above are acceptable − The amount of heat provided by calorife: Qs = L x (I2 - I0) = 14912.397x(126.806 – 68.218) = 873687.515 (KJ/h) − Calorife efficiency taken η = 0.9 Qt = Qs η = 873687.515 0.9 = 970763.907 − The actual amount of heat transferred from the steam in the tube to the wall of the tube: Qt = 3.6 x k x F x ∆tb − Heat transfer surface of calorife: F = 115,533 (m2) − Average heat transfer surface: Ftb = Qs 3.6xKx∆txη Fbm −Ftr = = 0.13+0.094 n = − Number of heat transfer tubes in calorife: 𝑛𝑛 = (pipes) − Number of tubes arranged horizontally: 𝑚𝑚 = 873687.515 3.6x39.880x58.526x0.9 𝑖𝑖 F Ftb = 1032 30 = = 0.112 (m2) 115.533 0.112 = 1032 = 35(pipes) − Length of calorife: Lx = 0.01x2 + (30 - 1)x0.007 + 30x0.03= 1.123 (m) − Width of calorife: Bx= 0.01x2 + (35 - 1)x0.007 + 35x0.03 = 1.308 (m) − High of calorife: Hx = H + 2xHch = 1.2 + 2x0.15 = 1.5(m) with Hch=0.15m 4.2 Cyclon 31 − A Cyclon is a equipment used for recovering fly-dried goods Hot air flows quickly throughout the drying process, thus some of the drying material will follow the air out A cyclone is inserted into the hot air output pipe for cleaner separation in order to recover exhaust gas and drying material 4.2.1 Calculation − At a temperature of 37oC the specific volume of air is: V35 = 1.139 − Air = 0.878 (m3/h) flow out of the drying room: 𝜌𝜌37 = V2 = L x v37 = m3 16117.35 x 0.878 = 1411.0333 (m3/h) = 3.9308( s ) The flow of air into the Cyclon is the flow of the drying agent out of the drying tank Vcyclon = V2 = 1411.0333 (m3/h) − To call ∆Pcyclone the impedance of cyclons: 540 < (QT&TB Handbook Volume 1- Page 524) Choose ΔPcyclon ρk = 700 So, ∆P=700 x 1.139 = 797.3 (N/m2) − The conventional speed is Wq: Wq = � (m/s) − The � 2∆ ξpk diameter 13092.980 0.785 x 4.830x 3600 of the cyclone is: = 0.979 (m) 𝛥𝛥𝑃𝑃𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝜌𝜌𝑘𝑘 x 842.86 =� 60 x 1.139 𝐷𝐷 = � < 740 = 4.830 V2 0.785 x Wq x 3600 = Based on the standard diameter we choose D = 1000 mm (in accordance with the diameter of the cyclone ЦH - 24 in the range of 400 - 1000mm) 4.2.2 Basic size of cyclone ЦH – 24 Entrance height (interior a = 1.11 x D = 1.11 x = 1.11 dimensions) (m) Central pipe height with flange h1= 2.11 x D = 2.11 x 1= 2.11 (m) Cone section height h2= 2.11 x D = 2.11 x 1= 2.11 (m) Cylindrical section height h3= 1.75 x D = 1.75 x 1= 1.75 (m) 32 Center tube outer section height h3= 0.4 x D = 0.4 x 1= 0.4 (m) General height H= 4.26 x D = 4.26 x 1= 4.26 (m) The outer diameter of the pipe d1= 0.6 x D = 0.6 x 1= 0.6 (m) The inner diameter of the pipe d2= 0.4 x D = 0.a x 1= 0.4 (m) Entrance width 𝑏𝑏1 The inlet of the inlet pipe Distance from the end to the flange 𝑏𝑏 = 0.26𝐷𝐷 0.2𝐷𝐷 = 1.3 (m) l = 0.6 x D = 0.6 x = 0.6 (m) h5= 0.24 x D = 0.24 x 1= 0.24 (m) Angle of inclination of the inlet α = 240 cover Diameter of cyclone D=1 Cyclon resistance coefficient ξ = 60 4.3 Fan − A fan is a component used to transport air and create pressure for the air flow to enter calorific equipment, dryers, pipes and cyclones The energy generated by the fan provides the kinetically pressurized air flow to move and partially overcomes the resistance on the conveying pipeline − During the drying process, we will use fans: + Push fan: placed at the top of the system, responsible for providing air for calorife to heat and then push into the drying barrel + Exhaust fan: located at the end of the system, responsible for sucking the dried air out of the drying barrel and through the cyclone to collect dust Pressure caused by fan (Δp): ∆P = ∆Pm + ∆Pcb + ∆Pcyl + ∆Pclo + ∆Pqh + ∆Pqd + ∆Ps Where: ΔPm: Resistance due to friction in each pipeline ΔPcb: Local resistance due to “đột mở” and “đột thu” ΔPcyl: Local resistant by cyclon ΔPclo: Local resistant by calorife ΔPqh : Resistance due to dynamic pressure at the outlet of exhaust fan 33 ΔPqd: Resistance due to dynamic pressure at the outlet of the push fan ΔPs: Resistant of rotating barrel (Unit : N/m2) Before entering After existing After Calorife Calorife 25 82 t ( 0C ) existing drying barrel 37 1.185 0.994 1.139 v= (1/ρ) (m3/kg dry 0.844 1.006 0.878 13603.0434 16214.0541 14151.0333 = 3.7786 = 4.5039 = 3.9308 16.7853x10-6 29.4942x10-6 18.0295x10-6 ρ (Kg /m3 ) air) V=L.v (m3/h) (m3/s) v ( m2/ s ) Table Parameters 4.3.1 Resistance due to friction in each pipeline - Choose the diameter of the pipelines : dtd = 0.45 (m) 4.3.1.1 Resistance from air filter to push fan ΔPm0 = λ0 x Where: L0 x ω20 x ρ0 x dtd λ: coefficient of friction dtd: diameter of pipe segments (m) L0: the length from the air filter to the calorife: L0 = 1.2 (m) ρ0: Density of gas (kg/m3) ω0: Velocity of the air in the tube (m/s) ω0 = V0′ F1′ With Vo’= 13603.0434 (m3/h) = 3.7786 (m3/s) Where : F1’ = => ω0 = V0′ F1′ = 𝜋𝜋 𝑥𝑥 𝑑𝑑12 3.7786 0.2826 = 3.14 x 0.062 = 0.2826 (m/s) = 13.3708 (m/s) 34 - Reynolds number: Re = dtd x ω0 V0 = 0.45 x 13.3708 16.7853x10−6 = 3.58x105 > 104 (Tangled flow mode) - The coefficient of friction is calculated by the formula: �λ0 6.81 0.9 = −2 x lg[� => λ0 = 0.016 � Re + ε 3.7 x dtd ]=7.8675 Where ε is the roughness of the pipe material Choose ε = 10-4 (m) - Resistance from air filter to push fan: ΔPm0 = λ0 x = 0.016 x L0 x ω20 x ρ0 x dtd 1.2x 13.07382 x 1.185 x 0.45 = 4.3209 (N/m2) 4.3.1.2 The resistance from the push fan to the elbow angle - Choose the length of pipe is L1 = 1.2 (m) Applying the formular: ΔPm1 = λ1 x While ω0 = L1 x ω21 x ρ1 V0′ x dtd F1′ => ω0 = ω1 = 13.3708 (m/s) - Reynolds number: Re = dtd x ω0 V0 = λ1 = λ0 = 0,016 0.45 x 13.3708 16.7853x10−6 ΔPm1 = 0.016 x = 3.5 x 105 > 104 (Tangled flow mode) 1.2x 13.07382 x 1.185 x 0.45 = 4.3209 (N/m2) 4.3.1.3 The resitance from the elbow angle to calorife - Choose the length of pipeline L2 = 1.3 (m) While ω2 = ω0= 13.3708 (m/s) - Reynolds number: Re = dtd x ω0 V0 = 3.58x105 > 104 (Tangled flow mode) - Coefficient of friction: λ2 = λ0 = 0.016 - The resitance from the elbow anle to calorife: ΔPm2 = λ0 x L2 x ω20 x ρ0 x dtd = 4.6810 (N/m2) 4.3.1.4 The resistance from calorife to material inlet tank 35 - Choose the length of pipeline L3 = 1.2 (m) 𝑉𝑉1 ω3 = F1′ = 4.5039 0.159 = 28.3264(m/s) - Reynolds number: Re = dtd x ω3 v1 = 0.45 x 28.3264 29.4942 x 10−6 - Coefficient of friction: �λ3 6.81 0.9 = −2 x lg[� Re � = 4.32 x 105 > 104 (Tangled flow mode) + ε 3.7 x dtd ] = 7.9356 => λ3 = 0.0159 - The resistance from calorife to material inlet tank: ΔPm3 = λ3 x L3 x ω23 x ρ1 x dtd = 16.9084 (N/m2) 4.3.1.5 Frictional resistance in the drying barrel - Average volumetric flow in the drying tank: Vtb = 𝑉𝑉82+ 𝑉𝑉37 = 15182.5437 (m3/h) = 4.2174 (m3/s) - Air velocity in the barrel: ω4 = Where: Vtb Ftb = 1.6603 Ftd : free cross-section of the drying barrel At the average temperature in the drying drum 59.5℃, ρ = 1,062 (kg/m3), v = 18,918.10-6 (m2/s) Re = Dt x ω4 v Dt x 1.6603 = v - Coefficient of friction: �λ4 = 1.66 x 105 > 104 (Tangled flow mode) 6.81 0.9 = −2 x lg[� => λ4 = 0.0176 Re � + ε 3.7 x dtd ] 4.3.1.6 The resistance from material tank to cyclone - Choose the length of pipeline L5 = 1.2 (m) ΔPm5 = λ5 x L5 x ω25 x ρ2 x dtd With V2 = 3.9308 (m3/s) 36 V2 ω5 = F1′ = 24.72 - Reynolds number: Re = dtd x ω5 V2 = 6.16 x 105 > 104 (Tangled flow mode) - Coefficient of friction: �λ5 6.81 0.9 = −2 x lg[� => λ5 = 0.01765 Re � + ε 3.7 x dtd ]=7.5253 - The resistance from material tank to cyclone: ΔPm5 = λ5 x L5 x ω25 x ρ2 x dtd = 13.6496 (N/m2) 4.3.1.7 The resistance from cyclone tank to elbow angle - Choose the length of pipeline L6 = 0.6 (m) - We have: ω6 = ω7 =24.72 - Reynolds number: = 6.16 x 105 > 104 (Tangled flow mode) => λ6 = λ5 = 0.01765 ΔPm6 = λ5 x L6 x ω25 x ρ2 x dtd = 8.189 (N/m2) 4.3.1.8 The resistance of the elbow angle to the exhausted fan - Choose the length of L7 = 1.2 (m) ω7 = ω5 = 24.72 - Reynolds number: Re = (4.3.1.6 result) => λ7 = λ5 =0.01765 ΔPm7 = λ5 x L7 x ω25 x ρ2 x dtd = 13.6496 (N/m2) => Total resistance by friction: ΔPm = 65.7194(N/m2) 4.4 Local resistance calculation 37 4.4.1 Local resistance by “đột mở” to calofife - We have: dtd = Where: 4𝐹𝐹0 𝜋𝜋0 F0: cross-sectional of the pipe (m2), F0 = F1 = 0.159 (m2) π0: The cross-sectional perimeter of the pipe π0 = x π x r =1.413 => dtd = 0.45 (m) - Inner cross-sectional area of calorife: Fcal = Lx x H = 1.483 x 1.2 = 1.780 (m2) => F0 Fcal Re = = 0.089 dtd x ω0 V0 = 3.58x105 >1000 => ζ=0.98 Then, resistance by “đột mở” to calorife: ΔPcb1 = ζ ω20 x ρ0 = 103.8 4.4.2 Resistance by “đột thu” from calorife Re = dtd x ω3 V1 4.32 x 105 > 3,5.103 => ζ= 0,470 ΔPcb2 = ζ ω23 x ρ1 = 187.42 4.4.3 Local resistance by “đột mở” to material tank Select the length of the material tank: 2.6 (m) Width of supply tank: 1.5 (m) Cross-sectional area of the material tank: Ftl = 2.6 x 1.5 = 3.9 (m2) => F0 𝐹𝐹𝑡𝑡𝑡𝑡 Re = = 0.04 dtd x ω3 V1 ΔPcb3 = ζ = 4.32 x 105 > 1000 => ζ=0.924 ω23 x ρ1 = 321.56 4.4.4 Local resistance by “đột mở” from material tank to pipeline Re = dtd x ω5 V2 ΔPcb4 = ζ = 6.16 x 105 > 3.5 x 103 => ζ= 0.49 ω25 x ρ2 = 170.52 (N/m2) => So, total local resistance : ΔPcb = 783.3 38 4.4.5 Local resistance by cyclone Call ΔPcyc is the resistance of cyclone 540 < ΔP cyc ρ2 < 740 ΔP cyc Choose ρ2 = 590 Then ΔP = 600 x 1.139 = 619.38 (N/m2) 4.5 The resistance from fan 4.5.1 Resistance from push fan ΔPqd = ω20 x ρ0 = 105.9 ω22 x ρ2 = 101.81 4.5.2 Resistance from exhausted fan ΔPqh = 4.6 The resistance from rotating barrel Allow the range of ΔPs = 20-30% ΔPcb Choose ΔPs = 20% = 0.2 => ΔPs = 0.2 x ΔPcb = 156.66 4.7 The resistance of calorife - The average of hot air in calorife: ttb = 82+25 = 53.5 (℃) Search the value with the table I.255/319 Handbook of QTTB volume 1, we have: ρ =1,081 (kg/m3) λ =2,854.10-2 (W/m.°) v =18,30.10-6 (m2/s) - Velocity of air inside calorife: ω= 𝐿𝐿 3600 𝑥𝑥 𝜌𝜌 𝑥𝑥 𝐹𝐹 = 2.11 (m/s) Where: F = Bx x Hx = 1.962 - Reynolds number: Re = dtd x ω𝑎𝑎𝑎𝑎𝑎𝑎 v = 1.53 x 105 > 104 (Motion in vortex mode) Because pipes are arranged in a corridor style, then: 39 𝑠𝑠 ζ = (6+9m) x ( )−0.23 x 𝑅𝑅𝑅𝑅 −0.26 Where : 𝑑𝑑 m: Number of beam sequences in the direction of motion, m = 30 d: Outer pipe diameter, dng = 0.03 (m) s: Distance between pipe axes along the direction of motion S= 0.03 + 0.006 + 0.007 = 0.028 (m) 0.028 => ζ = (6+9 x 30) x � 0.03 � x 1.53 x 105 = 12.572 => So, total resistance are: ΔP = ΔPm + ΔPcb + ΔPcyc + ΔPqđ + ΔPqh + ΔPs + ΔPclo = 1832.7694 (N/m2) 4.8 Fan calculation 4.8.1 Push fan - Push fan productivity: Q1 = V1 =13603.0434 (m3/h) Choose reserve factor: K3 = 1.4 => Real productivity of push fan : Qđ = K3 x Q1 = 19044.26 (m3/h) - Loss of push fan needs to be fixed: ΣΔP = 4.8.2 Exhausted fan ΣΔP = 916.3847 (N/m2) - Push fan productivity: Q2 = V2 = 16214.0541 Choose reserve factor K3 = 1.4 => Real productivity of push fan : Qh = K3 x Q2 = 22699.67 - Loss of push fan needs to be fixed: ΣΔP = ΣΔP = 916.3847 40 CHAPTER CONCLUSION The designed rotary drum dryer can work with the specifications as followed: - Input capacity: 1200 kg/h - Humidity 31% -> 9% - The temperature of the drying agent into the device: 82°C - The temperature of the drying agent out of the device: 37°C For, the design and calculation relies heavily on experimental formulas and theoretical calculations given in many different documents The use of such formulas and figures cannot avoid errors in the design process To be able to design correctly, we need to set up a test operation system to check and choose the optimal working mode At the same time, the system design is based on many theoretical documents and practical experience Therefore, in the process of implementing the topic, there are still many unreasonable places, we are looking forward to receiving the guidance and suggestions of teachers to improve the system 41 CHAPTER REFERENCES [1] GS TSKH Nguyễn Bin, Sổ tay QT & thiết bị cơng nghệ hóa chất – Tập 1, NXB khoa học kĩ thuật, 2005 [2] Nguyễn Bin, Sổ tay QT & thiết bị cơng nghệ hóa học – Tập 2, NXB Khoa học Kĩ thuật, 2003 [3] PGS-TSKH Trần Văn Phú - Tính tốn thiết kế hệ thống sấy, NXB Giáo Dục, 2001 [4] PGS-TSKH Trần Văn Phú - Kỹ thuật sấy – NXB Giáo dục, 2008 [5] Nguyễn Bin – Các q trình, thiết bị cơng nghệ hóa chất thực phẩm – Tập – NXB KHVKT, 2006 [6] Trịnh Xuân Ngọ - Cà phê kỹ thuật chế biến, 2009 [7] Lê Văn Việt Mẫn, Công nghệ chế biến thực phẩm – NXB Đại học Quốc gia TPHCM, 2011 42 ... DRYING SYSTEM 2.1 Rotary drum drying system for drying green coffee beans Air Green coffee beans Push pan Bucket load Air Material input Calorifer Drying tank Condensed Water Conveyor Exhausted Fan... has many superiority advantages over other methods In this project, we have the task of “Designing a rotary drum drying system for drying green coffee beans? ?? with a capacity of 1200 kg/h Because... beverages in the world Currently, the two most planted coffee species in the world are Arabica coffee (Coffea Arabica) and Robusta coffee (Coffea Canopera) In which, Arabica coffee accounts for about