Food Processing FST 4304 Fall 2024 Sterilization PDF

Food Processing FST 4304 Fall, 2024 Sterilization Most Important Concepts Definition of ‘sterilization’ and ‘commercial sterilization’ Commercial sterilization systems  Batch retort systems  Continuous retort systems  Aseptic processing systems...

Food Processing FST 4304 Fall, 2024 Sterilization Most Important Concepts Definition of ‘sterilization’ and ‘commercial sterilization’ Commercial sterilization systems  Batch retort systems  Continuous retort systems  Aseptic processing systems 2 Sterilized foods 3 Sterilization Sterilization is a unit operation in which foods are heated at a sufficiently high temperature and for a sufficiently long time to destroy vegetative microbial cells, spores, and enzymes. 4  Sterilization v.s. Pasteurization Pasteurization Sterilization Mild heat treatment (< 100 ºC) Rigorous heat treatment (> 100 ºC) Microbial and enzyme Microorganisms and enzymes activities are reduced are destroyed to a greater extent Shelf-life – several days to Shelf-life – more than six few weeks months Insignificant loss of nutrient Substantial loss of nutrients and sensory quality and sensory quality Still need refrigeration Does not need refrigeration, shelf-stable foods 5 Sterilization Complete destruction of microorganisms 121 ºC for 15 min or more Considered to be very harsh treatment Aim to completely kill all microorganisms and spores 6 Commercial sterilization In Practice, If food that contains more heat-resistant spoilage microorganisms are given a 12D process, it would result in over-processing and excessive loss of quality “Commercial sterilization” Heat treated to eliminate all pathogenic organisms and to reduce spoilage organisms to achieve a shelf-stable foods Spoilage organisms (spores) do not produce a health hazard or reduce the quality of a food product. 7 Commercially “sterile” foods All pathogens and toxin-forming organisms are destroyed as well as other organisms that could cause spoilage under normal conditions. May contain a small number of heat-resistant bacterial spores, but these will normally not multiply in the food. Long shelf-life 8 Commercial Sterilization Systems In-container Processing  Batch retort systems  Continuous retort systems Ultra-high-temperature (UHT)/Aseptic processing (out-of-container Processing) 9 In-container sterilization process  Food is filled and sealed in cans before sterilization.  Food and cans are sterilized together  Traditional way of sterilization Sterilization 10 In-container processing The heating time required to sterilize a food is influenced by Heat resistance of microorganisms or enzymes in the foods Heating conditions pH of foods Size and shape of the container Physical state of food 11 In-container sterilization - Batch Retort systems The typical batch system for commercial sterilization is called still retort (pressure cooker). Cooling Water line A crate of cans Steam line12 Vertical and Horizontal Retorts Vertical Retorts Horizontal Retorts Vertical: Traditional design, require less floor space Horizontal: Easier to load and unload and have facilities for agitating, but require more floor space 13 In-container sterilization - Continuous Retort systems Hydrostatic Retorts Constant process temperature Continuous container-conveyor Steam or cascading water with overpressure Produce more uniform products Allow close control over processing conditions High on-process stock lost if a breakdown occurred. Hot water Cold Water 14 Commercial Sterilization Systems In-container Processing  Batch retort systems  Continuous retort systems Ultra-high-temperature (UHT)/Aseptic processing (out-of-container Processing) 15 In-container sterilization process  Food is filled and sealed in cans before sterilization.  Food and cans are sterilized together  Traditional way of sterilization Sterilization 16 Ultra-high-temperature (UHT)/Aseptic processing system Higher processing temperature at shorter time are possible (130-135 ºC for a a few seconds) Food and packaging materials are sterilized separately Empty Container Food Product Sterilize Sterilize Container Food Fill Sterile Food Into Container Hermetically Seal Container 17 Comparison of conventional retorting and aseptic processing Criteria Retorting (in-container) Aseptic processing Stability Shelf stable Shelf stable Nutrient loss High Relatively low Sensory quality Not suited to heat-sensitive Suitable for heat-sensitive foods foods Energy efficiency Low > 30% saving Labor/handing costs High Low Downtime Minimal Re-sterilization needed Aseptic processing has now almost completely replaced in-container sterilization of liquid foods. Aseptic processing gaining popularity for foods that contain small particles. For foods with big particles or solid foods, still need in-container processing 18 Limitations of UHT/Aseptic processing High processing cost and complexity of the plant Necessary to sterilize packaging materials and associated pipe/heat exchanger Need maintenance of sterile air and surface in working zone The higher skill levels of workers required to operation 19  How to sterilize foods in aseptic processing? Indirect heating with heat exchangers Plate heat Shell-and-tube Scraped surface exchanger heat exchanger heat exchanger High heat transfer rate Viscous foods Most viscous foods Low viscosity foods Relative low capital cost High capital cost Operating pressure low Operating pressure high 20 Aseptic processing of protein drink by indirect heat exchangers 3D view Cross-section view 21  Indirect heat exchangers Large regeneration capacity Low operating cost Control less complex Stable operation Food grade steam not required Product may foul surfaces 22 How to sterilize foods in aseptic processing? Direct heating with steam mixing Steam injection Steam infusion Extremely rapid heating, resulting in high sensory and nutritional properties Suitable for heat sensitive foods Require food-grade steam (expensive steam) Vacuum cooling often required Less regeneration of energy ( h of natural convection  Natural convection (air): 5 -25 w/m2 oC  Forced convection (air): 10 -200 w/m2 oC  Natural convection (water): 20 -100 w/m2 oC  Forced convection (water): 50 -10,000 w/m2 oC Freezing food with still air: long freezing time Freezing food with blast air: short freezing time 29 Example The surface of a meat slab is 20 oC, a chilling air at 3 oC is blowing on the meat fast enough to make the heat transfer coefficient 100 W/m2 oC. What is the heat transfer rate between the meat and the air if the area of the meat is 0.1 m2. Step 1: q = hA(T1 - T2 ) Step 2: Step 3: q = hA(T1 - T2 ) h =100W / m2 oC W A = 0.1m2 =100 2 ´ 0.1m 2 ´ 17C mC T1 - T2 = 20 oC - 3o C =17o C =170W Heat radiation  Radiation is heat transfer by electromagnetic waves  Heat radiation requires no physical medium for transferring heat  All objects at a temperature above 0 K emit thermal radiation. 31 Heat Radiation 32 Summary 1. How heat flows in physical systems in terms of conduction, convection, and radiation. 2. How to apply the concepts of heat transfer in food processing systems. 3. How to calculate the heat transfer rate in simple systems. 33  Food Processing FST 4304 Fall, 2024 Heat Exchangers heat exchangers Shell and tube heat exchangers Plate heat exchangers 2 Heat exchangers everywhere in the food industry Heat Heat Exchangers Exchangers Using steam Using cold water/glycol 3  Milk Processing line Cream Whole Milk Milk pasteurization 5 Heat exchangers (HX) Shell and tube heat exchangers Plate heat exchangers 6 Shell and tube heat exchangers Also called ‘tube in tube’ or ‘concentric tube’ heat exchangers A low-cost heat exchanger Particularly suitable for more viscous non-Newtonian foods e.g., tomato ketchup, mayonnaise Water/steam out Food in Food out Water/steam in 7 Co-current and counter-current flow Co-current flow Counter-current flow Cold liquid Hot liquid Cold food Hot food Cold food Hot food Hot liquid Cold liquid Temperature Heating liquid Temperature food 8 Position Position  Hot liquid T3 Temperature Cold food T1 Hot food T2 Cold liquid T4 Heat gain in food: 𝑄𝑓 = 𝑚𝑓 × 𝐶𝑝𝑓 × (𝑇2 − 𝑇1 ) 𝑘𝑔 𝑘𝐽 𝑚𝑓 : 𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑓𝑜𝑜𝑑 𝐶𝑝𝑓 : 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑜𝑓 𝑓𝑜𝑜𝑑 𝑠 𝑘𝑔 °𝐶 Heat loss in liquid: 𝑄𝑙 = 𝑚𝑙 × 𝐶𝑝𝑙 × (𝑇3 − 𝑇4 ) 𝑘𝑔 𝑘𝐽 𝑚𝑙 : 𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝐶𝑝𝑙 : 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑠 𝑘𝑔 °𝐶 𝑄𝑓 = 𝑄𝑙 𝑚𝑓 × 𝐶𝑝𝑓 × (𝑇2 − 𝑇1 ) = 𝑚𝑙 × 𝐶𝑝𝑙 × (𝑇3 − 𝑇4 ) 9  A heat exchanger is to be used to heat orange juice from 18 ºC to 80 ºC at a flowrate of 0.5 kg/s. A counter-current heat exchanger is used, and hot water is available at 95 ºC to pass through the annular pipe at a flow rate of 1.5 kg/s. Calculate the existing temperature of the hot water. The specific heat of the 𝑘𝐽 𝑘𝐽 juice is 3.89 , the specific heat of water is 4.18 𝐾𝑔 𝐶 𝐾𝑔 𝐶 Hot water T3 = 95 C Cold juice T1 = 18 C Hot juice T2 = 80 C Cold water T4 = ? 𝑘𝑔 𝑘𝐽 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑗𝑢𝑖𝑐𝑒 𝑄𝑓 = 𝑚𝑓 × 𝐶𝑝𝑓 × 𝑇2 − 𝑇1 = 0.5 × 3.89 × (80 𝐶 − 18 𝐶) 𝑠 𝐾𝑔 𝐶 = 120.59 𝑘𝐽 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑗𝑢𝑖𝑐𝑒 𝑄𝑓 = Heat loss by hot water 𝑄𝑙 𝑄𝑙 = 120.59 𝑘𝐽 𝑘𝑔 𝑘𝐽 𝑄𝑙 = 𝑚𝑙 × 𝐶𝑝𝑙 × 𝑇3 − 𝑇4 = 1.5 × 4.18 × 95 − 𝑇4 𝑠 𝐾𝑔 𝐶 𝑇4 = 76.0 𝐶 Exercise A heat exchanger is to be used to heat milk at 25 ºC at a flowrate of 0.5 kg/s. A counter-current heat exchanger is used, and hot water is available at 95 ºC to pass through the annular pipe at a flow rate of 1.0 kg/s, the exiting temperature of water is measured at 70 ºC. Calculate the existing temperature of the heated milk. The 𝑘𝐽 𝑘𝐽 specific heat of the milk is 3.93 , the specific heat of water is 4.18 𝐾𝑔 𝐶 𝐾𝑔 𝐶 Shell-and-Tube heat exchanger Multi-tubes inside of one shell to increase capacity 12 Shell-and-Tube heat exchanger Multistage shell-and-tube heat exchangers 13 Plate heat exchanger Consists of a series of thin vertical stainless-steel plates The plate form parallel channels for liquid food and heating medium Each plate is fitted with a rubber gasket to produce a watertight seal that prevents leakage Very widely used in food industry for pasteurization 14 Plate heat exchanger 15  Corrugated plates The corrugations on plates induce turbulence in the liquids Reduce fouling Increase heat transfer coefficients  8000 – 11500 W/m2 C (plate heat exchanger)  Less than 2500 W/m2 C (shell-and-tube heat exchanger) 16 Summary of Chapters 10 A & B The basic concept of three ways of heat transfer o Heat Conduction o Heat Convection o Heat Radiation Apply equations to estimate the heat transfer rate of heat conduction and convection Types of heat exchanges and fundamental of energy balance during heat exchange 17  Food Processing FST 4304 Fall, 2024 Effect of heat on microorganisms Heat Treatment to Preserve Food Heat Treatment Blanching Pasteurization Sterilization 2 Heat Treatment to Preserve Food Heat Treatment Effect on foods Reduce (kill) Inactivate May destruct microorganisms enzymes nutrients and flavors 3 Effect of heat on microorganisms Microorganism survivor curve 4 Effect of heat on microorganisms Microorganism survivor curve at semi-log scale 5 Microorganism death rate curve Decimal Reduction Time (D-value): time required to destroy 90% (i.e., a one-log reduction) of a specific microbial population at a specific temperature log 1000 = log 103 = 3 D-value = 5 min Microbial population log 100 = log 102 = 2 log 10 = log 101 = 1 log 1 = log 100 = 0 Heating time 6 Decimal Reduction Time – D-value D-value = 5 min D-values differ for different microbial species. D-values are temp. dependent The higher the D-value, the greater the heat resistance. 7 Calculation of D-values From a data set of # surviving organisms vs. heating time: D = (T2 – T1)/(log(N1) –log(N2))  Tx = heating time X in minutes  Nx = no. of surviving organisms at time x  Units on D are minutes 8 Example Data Set Calculate the D-value for the microorganism when given the following thermal resistance data for a spore suspension: Time (minutes) Number of Organisms 0 105 10 103 20 101 9  Death rate curve Number of surviving cells 105 103 D = 5 min 101 0 5 10 15 20 25 Heating time (min) D = (T2 – T1)/(log(N1) –log(N2)) D = (20-0)/(log(105)–log(101)) D = 20/(5-1) = 5 min 10 Decimal Reduction Time – D-value Microorganism D-value Temperature Typical food (min) (ºC) Vegetative cells E. coli O111: B4 5.5 - 6.6 57.2 Milk Listeria monocytogenes 0.22 – 0.58 63.3 Milk Salmonella senftenberg 276 – 480 71 Chocolate Spores Bacillus subtilis 30.2 88 0.1% NaCl Clostridium botulinum 6.8 - 13 74 Seafood 11 Practice The following data were obtained from a thermal resistance experiment counted on a spore suspension at 112 ºC. Determine the D-value of the bacterial spores? D = (T2 – T1)/(log(N1) –log(N2)) D = (12-4)/(log(5 x 105)–log(5 x 103)) D = 8/(5.7-3.7) = 4 min 12 Decimal Reduction Time – D-value Microorganism D-value Temperature Typical food (min) (ºC) Vegetative cells E. coli O111: B4 5.5 - 6.6 57.2 Milk Listeria monocytogenes 0.22 – 0.58 63.3 Milk Salmonella senftenberg 276 – 480 71 Chocolate Spores Bacillus subtilis 30.2 88 0.1% NaCl Clostridium botulinum 6.8 - 13 74 Seafood 13 Example It requires 6 min to reduce 1-log of Bacillus Botulinus at 110 º C in vegetable products If we want to achieve a 5 log reduction of B. botulinus the required time is 5D = 5 × 6 min = 30 min If we want to achieve a 12 log reduction of B. botulinus the required time is 12D = 12 × 6 min = 72 min 14 Survivor Curves at Different Temp. Microorganism cells die more rapidly at higher temperature. 15 Thermal resistance constant: Z-values Thermal resistance Decimal reduction time (D-value) (min) constant (Z-value) is the temperature change required to achieve 90% (one-log, or ten times) reduction in the D-value 11 ºC Temperature (ºc) 16 Example –how to determine z value The decimal reduction times D for a spore suspension were measured at several temperatures, as follows: Determine the thermal resistance constant z for the spores. 17  𝑇2 −𝑇1 116−104 𝑧= = = 11 ºC log 𝐷1 −log 𝐷2 log 27.5 −log 2.2 18 Z-value can be used to estimate heat treatment time It takes 5 min to reduce 1 log of E. coli O157:H7 in ground beef at 57 ºC. The z-value of E. coli O157:H7 is 5 ºC. It means that if we increase temperature by 5 ºC, the required heat treatment time to achieve 1-log reduction of E. coli is reduced by 10-fold. When the temperature is 62 ºC, the required time is 0.5 min When the temperature is 67 ºC, the required time is 0.05 min What would happen if the temperature is decreased by 5 to 52 ºC? 19 Effect of heat on nutrition of foods Destruction of many vitamins, aroma compounds and pigments by heat follow similar first-order deactivation to microorganisms However, Z-values of nutritional compounds are higher than those of microorganisms. 20 Effect of heat on nutrition of foods Components Z-value (degree C) Bacterial cells 4-8 Bacterial spores 7-12 Vitamins 25-30 Proteins 15-37 Pigments 24-50 Nutrients are less sensitive to temperature increase than bacteria. As a results, nutrients are better retained by using higher temperatures and shorter times during heat processing. This concept forms the basis of high-temperature-short-time (HTST) pasteurization and ultra-high-temperature (UHT) sterilization. 21 Pasteurization of milk for 12-log bacteria reduction Standard pasteurization (63 ºC, 30 min) causes larger changes to flavor and higher loss of vitamins because of longer time. High-Temperature-Short-Time (HTST) process (71.8 ºC, 15 s) preserve more of nutrients and vitamins Ultra- pasteurization: 137.8 ºC for 1-2 s (shelf life: 2-3 months) 22  Process Lethality, F-value, TDT (Thermal Death Time) Goal: to characterize a thermal process so that we can assess its effectiveness in terms of sterilization At different situations, different temperature/time treatments may be received by the food Process in Virginia Process in California 90 oC, 10 min 120 oC, 2 min 23  Process Lethality, F-value, TDT (Thermal Death Time) Process in Virginia Process in California 90 oC, 10 min 120 oC, 2 min Need a procedure to evaluate and compare effect of different heating treatments to reduce microbial population For this purpose, we determine process lethality, also called F-value or TDT (thermal death time) 24  Thermal Death Time (TDT): F-value F-value is number of minutes required to accomplish a given reduction in the microbial population at a specific temperature. Represents the combined effect of temperature and time duration on the microbial inactivation achieved during a thermal processing Used to design and evaluate thermal processes to ensure sufficient destruction of harmful microorganisms Process in Virginia Process in California 90 oC, 10 min 120 oC, 2 min 25 Thermal Death Time (TDT): F-value Thermal resistance constant z Process temperature T Thus, 𝐹𝑇𝑍 is the thermal death time for a temperature T and a thermal resistance constant z 8 𝐹98 means thermal death time at a temperature of 98 ºC with a z-value of 8 ºC. 26 Reference thermal death time (F0) Thermal resistant constant z Process temperature T If T is selected as a reference temperature, we could write Tref = 121 oC (250 F); Z= 10 oC (18 F) For an easy reference purpose, a commonly used reference thermal 10 death time is 𝐹121 (temperature to kill Clostridium botulinum spores) Simply written as F0 Represents the number of minutes required to destroy a specified number of Clostridium botulinum spores at 121 oC when z = 10 oC. 27 Lethal rate Lethal rate is a relative term that compares the microbial kill effect at a measured temperature to one minute at the reference temperature (e.g.,121 C) Lethal rate =  FR is the known thermal death time at a reference temperature  TR is the known reference temperature (e.g.,121 C)  T is the new temperature  F is the new thermal death time  Z is the thermal resistant constant (e.g., 10 C) 28 Lethal rate 10 The most important reference: 𝐹0 = 𝐹121 𝐹𝑅 Lethal rate = = 10(𝑇−𝑇𝑅 )/𝑧 = 10(𝑇−121)/10 𝐹 For example, if a product is processed at 131 ºC, the lethal rate would be 𝐹𝑅 131−121 Lethal rate = = 10 10 = 10 𝐹 Heating at 131 ºC for 1 min is equivalent to heating at 121 C for 10 min for achieving the same lethal (sterilization) effect For example, if a product is processed at 111 ºC, what is the lethal rate? 111−121 𝐹𝑅 Lethal rate = = 10 10 = 0.1 𝐹 Heating at 111 ºC for 1 min is equivalent to heating at 121 C for 0.1 min for achieving the same lethal (sterilization) effect 29  Thermal Death Time (TDT): F-value Process in Virginia Process in California 90 oC, 10 min 120 oC, 2 min Lethal rate = 10(𝑇−𝑇𝑅 )/𝑧 Lethal rate = 10(𝑇−𝑇𝑅 )/𝑧 = 10(90−121)/10 = 10(120−121)/10 = 0.0008 = 0.79 Heating at 90 ºC for 1 min is equivalent to Heating at 120 ºC for 1 min is equivalent to heating at 121 C for 0.0008 min heating at 121 C for 0.79 min Heating at 90 ºC for 10 min is equivalent to Heating at 120 ºC for 2 min is equivalent to heating at 121 C for 0.008 (0.0008*10) min heating at 121 C for 1.58 (0.79 *2) min The process in California has more lethal effect than that in Virginia to kill microorganisms. 30 The General method for process calculation In a real life situation, a product experiences a range of different temperatures as it heats up and cooling down during the process. The general method allows determine the lethality of the heating process experiencing different temperatures/times 115 C for 10 min 140 Temperature (C) 120 100 80 10 min 60 40 20 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Time (min) 31 The General method for process calculation Steam off Steam on 32 The General method for process calculation The practical question: How could we evaluate the lethality (bacterial kill effect) of the process with temperature changes at different time? 33 Steps for the general method Step 1. Determine temperature curve using a temperature sensor located at the slowest heating location (coldest location) of food (in a container). Thermocouple Thermocouple 1/3 of height Conduction Convection (solids food) (food with liquid) 34 The General method for process calculation Steam off Steam on 35 Steps for the general method Step 2. Convert each point of measured temperature into lethal rate using the lethal rate equation: 𝐹𝑅 (𝑇−121)/10 Lethal rate = = 10(𝑇−𝑇𝑅 )/𝑧 = 10 𝐹 111−121 𝐹𝑅 Example: at temperature of 111 C Lethal rate = = 10 10 = 0.1 𝐹 Temperature curve Lethal rate curve 36 Steps for the general method Step 3. Determine the total lethality of the process by summing lethal rate (effect) at each time interval (1 minute): the area under the lethal rate curve 37  Example: A product experiences a range of different temperatures as it heats up and cooling down during the process. 140 Time (min) Temperature (degree C) Temperature (C) 120 0 25 100 1 min 98 80 2 min 103 60 3 min 109 40 4 min 116 20 5 min 120 0 6 min 120.5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 7 min 121.5 Time (min) 8 min 122 9 min 120 10 min 119 11 min 105 12 min 100 13 min 95 Thermometer 14 min 25 1/3 from bottom 38  The General method for process calculation Question: How do we calculate the lethal rate and lethality of the thermal process? 98−121 Time (min) Temperature Time (min) Temperature Lethal RateRate Lethal Equivalent to Lethal rate = 10 10 = 0.005 (ºC) (ºC) heating at 121 ºC 0 25 25 0.0000.000 0.000 min 1 min 98 98 0.0050.005 0.005 min Equivalent accumulated heating 2 min 103 103 0.0160.016 0.016 min time at 121 ºC is 5.93 min. 3 min 109 109 0.0630.063 0.063 min 10 = 5.93 min) (𝐹121 4 min 116 116 0.3160.316 0.316 min 5 min 120 120 0.7940.794 0.794 min 6 min 120.5120.5 0.8910.891 0.891 min The accumulated impact of time and 7 min 121.5121.5 1.1221.122 1.122 min temperature on a food product is 8 min 122 122 1.2591.259 1.259 min defined as lethality, expressed as 9 min 120 120 0.7940.794 0.794 min time at the reference temperature. 10 min 119 119 0.6310.631 0.631 min 11 min 105 105 0.0250.025 0025 min 12 min 100 100 0.0080.008 0.008 min The total lethality of the whole 13 min 95 95 0.0030.003 0.003 min process is equivalent to 5.93 min at 14 min 25 25 0.0000.000 0.000 min 121 ºC when z = 10 C. Total 5.93 min 10 = 5.93 min; or F = 5.93 min 𝐹121 0 39  Summary D-value (Decimal reduction time) : the time required for one-log reduction in surviving microorganisms D = (T2 – T1)/(log(N1) –log(N2)) z-value (thermal resistant constant) : the temperature change required for one-log reduction in D z = (T2 – T1)/(log(D1) –log(D2)) F-value (thermal death time): is the time required to achieve a stated reduction in the microbial population Lethal rate 𝐹𝑅 Lethal rate = = 10(𝑇−𝑇𝑅 )/𝑧 𝐹 The general method for process calculation 40 Chapter 4. Mixing Content 1 3.1 Theory of solids mixing 2 2 3.2 Theory of liquids mixing 3 3.3 Equipment Definition: Mixing (or blending) is a unit operation in which a uniform mixture is obtained from two or more components, by dispersing one within the other(s). Cocoa Powder Chocolate Protein Shakes The larger component is sometimes called the “continuous phase” and the smaller component the “dispersed phase” by analogy with emulsions, but these terms do not imply emulsification when used in this context. Mixing 3.1 Theory of solids mixing In contrast with liquids and viscous pastes it is not possible to achieve a completely uniform mixture of dry powders or particulate solids. The degree of mixing that is achieved depends on: 1) The relative particle size, shape and density; 2) The moisture content, surface, and flow characteristics; 3) The tendency of the particles to aggregate; 4) The efficiency of a particular mixer for mixing those components. 3.1 Theory of solids mixing In some mixtures, uniformity is achieved after a given period and then unmixing begins, and it is therefore important in such cases to time the mixing operation accurately The uniformity of the final product depends on the equilibrium achieved between the mechanisms of mixing and unmixing, which in turn is related to the type of mixer, the V-Cone Blender operating conditions and the component foods. 3.1 Theory of solids mixing As mixing proceeds, the composition of each sample becomes more uniform and approaches the average composition of the mixture. One method of determining the changes in composition is to calculate the standard deviation of each fraction in successive samples: σ∞ Where σm = standard deviation, n= number of samples, c= concentration of the component in each sample and cത= the mean concentration of sample. Lower standard deviations are found as the uniformity of the mixture increases.  Different mixing indices are available to monitor the extent of mixing and to compare alternative types of equipment: σ∞= the standard deviation of a “perfectly mixed” sample σ0 = the standard deviation of a sample at the start of mixing V= the average fractional volume or σm= the standard deviation of a sample taken during mixing mass of a component in the mixture 3.1 Theory of solids mixing The mixing time is related to the mixing index using: where K = mixing rate constant, which varies with the type of mixer and the nature of the components, and tm (s) = mixing time.  3.1 Theory of solids mixing During the preparation of a dough, 700 g of sugar is mixed with 100 kg of flour. Then 100 g samples are taken after 1, 5, and 10 min and analyzed for the percentage sugar. The results are as follows: Percentage after 1 min 0.21 0.32 0.46 0.17 0.89 1.00 0.98 0.23 0.10 0.14 Sugar Dough Percentage after 5 min 0.85 0.80 0.62 0.78 0.75 0.39 0.84 0.96 0.58 0.47 Percentage after 10 min 0.72 0.69 0.71 0.70 0.68 0.71 0.70 0.72 0.70 0.70 Calculate the mixing index for each mixing time. Assume σ∞ = 0.0001. 3.1 Theory of solids mixing Calculate the mixing index for each mixing time. Assume σ∞ = 0.0001. σ0 σm σ∞ (After 1 min, calculate σm) σ∞ = 0.0001 Percentage after 1 min 0.21 0.32 0.46 0.17 0.89 1.00 0.98 0.23 0.10 0.14 After 1 min mean cത of the samples = (0.21+0.32+0.46+0.17+0.89+1.00+0.98+0.23+0.10+0.14)/10 = 0.45 1 1 = = 0.11 𝑛−1 10−1 σm = 0.11 ∗ 1.197= 0.13167 = 0.003629 σ 𝑐 − 𝑐ҧ 2 = (0.21-0.45)2 + (0.32-0.45)2 + (0.46- 0.45)2 + (0.17-0.45)2 + (0.89-0.45)2 + (1.00-0.45)2 + (0.98-0.45)2 + (0.23-0.45)2 + (0.10-0.45)2 + (0.14-0.45)2 = 1.197 (After 1 min, calculate σ0) Average fractional mass V1 of sugar in the mix σ0 = 0.08337 σm = 0.003629 700 V1 = = 7*10-3 σ∞ = 0.0001 (100∗103 +700） log 0.003629 −log (0.0001) M2= σ0= [7∗10−3(1− 7∗10−3] = 0.08337 log 0.08337 −log(0.0001) = 0.533 (After 5 min) σ∞ = 0.0001 σ0 = 0.08337 Percentage after 5 min 0.85 0.80 0.62 0.78 0.75 0.39 0.84 0.96 0.58 0.47 After 1 min mean cത of the samples = (0.85+0.80+0.62+0.78+0.75+0.39+0.84+0.96+0.58+0.47)/10 = 0.704 1 1 σm = 0.11 ∗ 0.29824= 0.032806 = 0.001811 𝑛−1 = 10−1 = 0.11 log 0.001811 −log(0.0001) σ 𝑐 − 𝑐ҧ 2 = (0.85-0.704)2 + (0.80-0.704)2 + M2= (0.62-0.704)2 + (0.78-0.704)2 + (0.75-0.704)2 + log 0.08337 −log (0.0001) (0.39-0.704)2 + (0.84-0.704)2 + (0.96-0.704)2 + = 0.429 (0.58-0.704)2 + (0.47-0.704)2 = 0.29824 (After 10 min) σ∞ = 0.0001 σ0 = 0.08337 Percentage after 10 min 0.72 0.69 0.71 0.70 0.68 0.71 0.70 0.72 0.70 0.70 After 10 min mean cത of the samples = (0.72+0.69+0.71+0.70+0.68+0.71+0.70+0.72+0.70+0.70)/10 = 0.703 σm = 0.11 ∗ 0.00141= 0.000155 = 0.00125 σ 𝑐 − 𝑐ҧ 2 = (0.72-0.703)2 + (0.69- 0.703)2 + (0.71-0.703)2 + (0.70-0.703)2 log 0.00125 −log (0.0001) + (0.68-0.703)2 + (0.71-0.703)2 + M2= log 0.08337 −log (0.0001) (0.70-0.703)2 + (0.72-0.703)2 + (0.70- 0.703)2 + (0.70-0.703)2 = 0.00141 = 0.031 3.2 Theory of liquids mixing The component velocities induced by a mixer in low-viscosity liquids are as follows A) Longitudinal velocity (parallel to the mixer shaft) B) Rotational velocity (tangential to the mixer shaft) C) Radial velocity that acts in a direction perpendicular to the mixer shaft Component velocities in fluid mixing: A, longitudinal; B, rotational; C, radial. 3.2 Theory of liquids mixing Most liquid foods are non-Newtonian, and the viscosity changes with rate of shear. The most common types fall into one of the following categories: Dilatant Fluids— Viscosity increases as shear rate increases (shear thickening). Cornstarch & Water mixtures 3.2 Theory of liquids mixing Pseudoplastic Fluids—Viscosity decreases as shear rate increases (shear thinning). Ketchup Syrups Molasses 3.2 Theory of liquids mixing Viscosity (cp) 3.2 Theory of liquids mixing The rate of mixing is characterized by a mixing index. The mixing rate constant depends on the characteristics of both the mixer and the liquids. The effect of the mixer characteristics on K is given by: Proportional D (m) = the diameter of the agitator, N (rev s-1) = the agitator speed Dt(m) = the vessel diameter and z (m) = the height of liquid Rev s−1: (Revolutions per Second) Rev s−1 = 60 RPM Rev min−1: (RPM, Revolutions per Minute) The power requirements of a mixer vary according to the nature, amount and viscosity of the foods in the mixer and the position, type, speed and size of the impeller. Liquid flow is defined by a series of dimensionless numbers: the Reynolds number, Re, P (W): the power transmitted via the agitator ρm (kg m-3): the density of the mixture μm (N s m-2):the viscosity of the mixture the Froude number, Fr D: (m) the diameter of the agitator N: (rev s-1) the agitator speed the Power number, Po: If we know the Power number Po, we can 𝑃 = 𝑃𝑂 𝜌𝑚 𝑁 3𝐷 5 calculate the actual power of the agitator. 3.2 Theory of liquids mixing ρm = 𝑉1ρ1 + 𝑉2ρ2 μm = μ1 μ2 where V= the volume fraction of the ingredient. μ = viscosity of the liquid The subscripts 1 and 2 are the continuous phase and dispersed phase, respectively (or ingredient 1 and ingredient 2)  1: Six-blade turbine with disk 2: Six-blade open turbine Po 3: Pitched six-blade turbine with pitch angle 45 ̊ Power number 4: Pitched three-blade turbine with pitch angle 45 ̊ 0.38 5: Propeller 2000 Reynolds number 6: High shear stress impeller Practice Olive oil and rapeseed oil are blended in a ratio of 1 to 4 (by volume) by a propeller agitator 20 cm in diameter operating at 750 rpm in a cylindrical tank 1 m in diameter at 20 ℃. Calculate the power (kw) of the motor required. Viscosity Density (Ns m-2) (kg m-3) Olive Oil 0.084 910 Rapeseed Oil Rapeseed Oil 0.118 900 Olive Oil ρm = 𝑉1ρ1 + 𝑉2ρ2 μm = μ1𝑉1 μ2𝑉2 Volume Fraction of ingredient 1 (Olive oil) 1 : V1 = = 0.2 1 + 4 Volume Fraction of ingredient 1 (Rapeseed 4 oil) V2 = = 0.8 1+4 μm = μ1𝑉1 ×μ2𝑉2 μm = 0.0840.2 ×0.1180.8 =0.110 N s m-2 ρm = 𝑉1ρ1 + 𝑉2ρ2  1: Six-blade turbine with disk 2: Six-blade open turbine Power number 3: Pitched six-blade turbine with pitch angle 45 ̊ 4: Pitched three-blade turbine 0.38 with pitch angle 45 ̊ 5: Propeller 4100 Reynolds number Re=4100, Po = 0.38 6: High shear stress impeller 𝑘𝑔 750 𝑟𝑝𝑚 3 𝑃 = 𝑃𝑂 𝜌𝑚𝑁 3𝐷5 = 0.38 × 902 3 ×( ) × (0.2 𝑚)5 = 214 𝑚 60 W In-class Quiz Honey and water are blended in a ratio of 1 to 3 (by volume) by a Six-blade open turbine agitator 30 cm in diameter operating at 60 rpm in a cylindrical tank 1.5 m in diameter at 20 ℃. Calculate the power (kw) of the motor required. Viscosity Density (Ns m-2) (kg m-3) Water 0.001 1000 Water Honey 5 1400 Honey ρm = 𝑉1ρ1 + 𝑉2ρ2 μm = μ1𝑉1 μ2𝑉2 Volume Fraction of ingredient 1 (honey) : V1 1 = = 0.25 1 + 3 Volume Fraction of ingredient 1 (water) V2 3 = = 0.75 1+3 μm = μ1𝑉1 ×μ2𝑉2 μm = 50.25 × 0.001 0.75 =0.0084 N s m-2 ρm = 𝑉1ρ1 + 𝑉2ρ2  1: Six-blade turbine with disk 4.0 2: Six-blade open turbine Power number 3: Pitched six-blade turbine with pitch angle 45 ̊ 4: Pitched three-blade turbine with pitch angle 45 ̊ 5: Propeller 10,164 Reynolds number Re=4100, Po = 4.0 6: High shear stress impeller 𝑘𝑔 60 𝑟𝑝𝑚 3 𝑃 = 𝑃𝑂 𝜌𝑚𝑁 3𝐷5 = 4.0 × 1100 3 ×( ) × (0.3 𝑚)5 = 10.7 𝑚 60 W 3.3 Equipment There is a very wide range of mixers available, due to the large number of mixing applications. The selection of the correct type of mixer for a particular application depends on the type and amount of food being mixed per hour or per batch (the throughput) and the time needed to achieve the required degree of mixing. Table 1 Factors in mixer selection 1) The viscosity of the food; 2) The mixer capacity; 3) The shear rate required; 4) The energy consumption; Mixers can be grouped into types that are suitable for Dry powders or particulate solids Low- or medium-viscosity liquids High-viscosity liquids and pastes Water Olive Oil Honey Low Viscosity Medium Viscosity High Viscosity Dispersion of powders in liquids 3.3.1 Dry powders or particulate solids These mixers have two basic designs: 1) Tumbling action inside rotating vessels (Tumbling Mixers) 2) Positive movement of materials in screw-type mixers (Screw-Type Mixers) Tumbling Mixers V-Cone Blender Y-Cone Blender Double Cone Blender Drum Blender In the Y-cone mixer, the powders are divided into two portions each time the arms of the ‘Y’ are lowered and remixed when they are raised during the next rotation Screw-Type Mixers Vertical-screw Mixer Ribbon Mixer Flow pattern of double reversing agitator 3.3.2 Low- or medium-viscosity liquids To adequately mix low-viscosity liquids, turbulence must be induced throughout the bulk of the liquid to entrain slower-moving parts within faster-moving parts. Many designs of agitator are used to mix liquids in baffled or unbaffled vessels. Six-blade turbine Six-blade open Pitched Six-blade Pitched Three-blade Propeller with disk turbine turbine turbine 3.3.2 Low- or medium-viscosity liquids The simplest paddle agitators are wide, flat blades, which measure 50-75% of the vessel diameter and rotate at 20-150 rpm. Paddle Agitators Turbine agitators are impeller agitators that have four or more blades mounted together. Their size is 30-50% of the diameter of the vessel and they operate at 30-500 rpm. Turbine Agitators Impellers that have short blades (less than a quarter of the diameter of the vessel) are known as propeller agitators and these operate at 400-1500 rpm. Propeller Agitators 3.3.2 Low- or medium-viscosity liquids Table 2 Advantages and limitations of selected liquid mixers 3.3.3 High-viscosity liquids and pastes Efficient mixing is achieved by creating and recombining fresh surfaces in the food, but because the material does not easily flow, it is necessary to either move the mixer blades throughout the food or to move the food to the mixer blades. The basic design in this group is the ‘anchor and gate’ agitator. Some complex designs have arms on the gate that intermesh with stationary arms on the anchor to increase the shearing action. The basic design has three separate agitators including: 1) Anchor; 2) High-speed disperser; 3) Rotor-stator homogenizer. Mixer with multi-shaft helical anchor design  High-speed rotor blades create a Centrifugal force pushes materials low-pressure zone, pulling liquid outward for shearing into the stator head Rotor-Stator Mixers Materials exit stator slots at high velocity, Expelled materials shoot outward undergoing shearing for size reduction as fresh material flows in Encapsulated VA Powder Formation Gum Arabic Pump In VA Powder Pump In 60 ℃ VA Powder Pump Out 3.3.4 Dispersion of powders in liquids Z-blade Mixers Planetary Mixer Summary 3.1 Theory of solids mixing 3.2 Theory of liquids mixing Dry powders or particulate solids Low- or medium-viscosity liquids 3.3 Equipment High-viscosity liquids and pastes Dispersion of powders in liquids V-cone Blender Ribbon Mixer Rotor-Stator Mixers