Postharvest Technology-units Operations Notes

July 13, 2017 | Author: ntjanv | Category: Pump, Shear Stress, Heat, Viscosity, Stress (Mechanics)
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class notes on unit operation in food industry. describes different aspects of food processing from material and energy ...



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Ntwali Janvier Department of Biotechnologies INES, Email: [email protected]


Fundamental concepts

Material balance

Energy balances

Fluid flow theory and viscosity

Heat transfer and its application in food industry



Mechanical separation


Size reduction


Useful references 1. George D. Saravacos Z B M, Maroulis Z B. Food Process Engineering Operations. CRC Press INC, 2011. 2. Toledo R T. Fundamentals of Food Process Engineering. Springer, 2010. 3. White F M. Fluid Mechanics . McGraw-Hill, 2008. 4. Heldman D R, Lund D B. Handbook of Food Engineering, Second Edition. Taylor & Francis Group, 2007. 5. Ibarz A, Barbosa-Cánovas G V. Unit Operations in Food Engineering. Taylor & Francis Group, 2003. 6. L. R. Earle M D E. Unit Operations in Food Processing . 1983


Fundamental concept of unit operations in food processing

Fundamental concept of unit operations in food processing Basic definitions Food process engineering can also be defined as “the science of conceiving, calculating, designing, building, and running the facilities where the transformation processes of agricultural products, at the industrial level and as economically as possible, are carried out.” Thus, an engineer in the food industry should know the basic principles of process engineering and be able to develop new production techniques for agricultural products. He should also be capable of designing the equipment to be used in a given process. The main objective of food process engineering is 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 processing of most food products involves raw materials; cleaning; separating; disintegrating; forming, raw; pumping; mixing; application methods (formulations, additives, heat, cold, evaporation, drying, fermenting, etc.); combined operations; and forming, finished product. We discuss some of these as units of operations. Certain items - heating, cooling, sanitation, quality control, packaging, and similar procedures - are discussed as separate topics rather than as units of operations. Unit operation: unit operation is defined as a basic step in a food processing operation. In food industry unit operations consists of bringing physical change like separation, evaporation, crystallization, etc. Unit operation classification: unit operations are classified depending on the nature of the transformation performed. There are physical, chemical and biochemical processes. •

In physical processes we have: grinding, sieving, mixture, fluidization, sedimentation, flotation, filtration, rectification, absorption, extraction, adsorption, heat exchange, drying evaporation, etc.

Chemical processes include: refining, chemical peeling, chemical browning…

Biochemical processes are fermentation, sterilization, enzymatic peeling, blanching, pasteurization…

Food processes are usually schematized by means of flow charts. These are diagrams of all processes that indicate different manufacturing steps, as well as the flow of materials and energy in the process. In these charts each stage of the process is represented by a block or rectangle connected by arrows to indicate the way in which the materials flow. The stage represented is written within the rectangle. An example of a process flow chart is shown in the fig 1.1


Fundamental concept of unit operations in food processing

FIGURE 1.1 Process flow showing the production of fruit concentrated juices.


Material balance

Chapter I Material balance Material balance are calculations employed in tracing the inflow and outflow of material in a process and thus establish quantities of components or the whole process stream. The procedures are useful in formulating products to specified compositions from available raw materials, evaluating final compositions after blending, evaluating processing yields, and evaluating separation efficiencies in mechanical separation systems. Material balances are undertaken before a food processing unit is installed to determine the quantity of raw material and product to be handled. Compositions of raw material, product, and by-product streams can be evaluated by using material balances Material balances are statements on the law of conservation of mass. If there is no accumulation, what goes into a process equals what comes out. Inflow = outflow + accumulation

1.1 Conservation of mass Even in the case of a chemical reaction, the composition of mass of a reactant and the product before and after the reaction may be different, but the mass of the total system remains unaltered. In a unit operation, whatever its nature, the mass of material going in, must balance with the mass of material going out. This is true for batch operation as well as continuous operation in a certain interval of time The fig. 1.2 shows flow of a liquid in a tank. If the level of the liquid in the tank remains constant, then the flow rate at the inlet must be equal to the flow rate at exit. Thus, the rate at the inlet is 1kg/s and the rate at the exit is also 1kg/s.

Figure 1.2: Liquid flow in and out of a tank


Material balance

Although only one inlet and one exit stream is shown in the figure above, there may be more than one stream entering and exiting a control volume. Therefore, for a general case, the rate of mass flow entering the system is



Where subscript i denotes the inlet, and n is the number of inlets to a system. The mass accumulation in the system as function of time is



Where subscript e denotes the exit and p is the number of exits from a system. The rate of mass accumulation within the system boundary, expressed as a function of time, is



The conservation of mass principle can then be expressed in terms of an equation



1.2 Useful steps in solving a material balance problem In order to conduct material balance in an organized manner, one should: •

Collect all known data on mass and composition at all inlet and exit streams

Draw a block diagram indicating the process, with inlet and exit streams identified, show the system boundary

Write all available data on the block diagram

Select a suitable basis (e.g.: mass, time) for calculations

Using Equation (1.4), write material balances in terms of the selected basis for calculating unknowns

Solve material balances to determine the unknowns


Material balance

1.3 Composition and concentration For a stream that contains more than one component, the mass fraction of a component is simply the mass of that component expressed as a fraction of the total mass of the mixture containing that component. For a mixture of components of mass mA and mB of components A and B, the mass fraction of the component A is:



Similarly, a mole fraction is defined as a number of moles a component expressed as a fraction of the total number of moles in the mixture as illustrated in the equation 1.6



1.4 Moisture content Moisture content expresses the amount of water present in a moist sample. Two bases are widely used to express moisture content; namely, moisture content wet basis and moisture content dry basis. Moisture content wet basis (MCwb) is the amount of water per unit mass of moist (or wet) sample. MCwb=mass of water/mass of moist sample Moisture content dry basis (MCdb) is the amount of water per unit mass of dry solids (bone dry) present in the sample. MCdb=mass of water/mass of dry solids Thus, moisture content wet basis is !"



$$ & !

$$ & ! '


$$ & ') $



1.5 Component Material Balances For a system with four streams as shown in the fig. 1.3, suppose that each of the four streams contains water, at a steady state, the combined mass flow rate of water in all streams entering the system, must equal the combined mass flow rate of water leaving the system.


Material balance

Figure 1.3: Material balance: a process with four streams, m1+m2=m3+m4

If x is the mass fraction of water in a stream of total mass flow rate m, the mass flow rate of water is xm. And according to the low of conservation of mass we get the equation 1.8. Equation 1.8 is called component material balance, and any material balance problem, other than the most trivial, will require the simultaneous solution of an overall balance and at least one component balance equation.


+ *,


= *-


+ *.



Examples 1. Consider a problem in crystallization. The problem may be stated as: Determine the amount of sugar (water-free basis) that can be produced from 100 kg of sugar solution that contains 20% by weight of sugar and 1% of a water-soluble un-crystallizable impurity. The solution is concentrated to 75% sugar, cooled to 20◦C, centrifuged, and the crystals dried. From what we know, the process of sugar crystallization can be represented in the fig. 1.4.

Figure 1.4: process flow diagram for crystallization

But that figure alone is not sufficient; we need to add data from the problem statement. That’s why a more detailed figure is drawn, showing different stages with their inflows and outflows based on what was given in problem statement. Figure 1.5 is a flow diagram of the same process, but after taking into consideration how components partition in various steps in the process, additional streams leaving the system are drawn in the diagram. To concentrate a 20% solution to 75% requires the removal of water. Thus, water leaves the system at the evaporator. The process of cooling does not alter the mass;


Material balance

therefore, the same process stream enters and leaves the crystallizer. Centrifugation separates most of the liquid phase from the solid phase, and the crystals, the solid phase containing a small amount of retained solution, enter the drier. A liquid phase leaves the system at the centrifuge. Water leaves the system at the drier.

Figure 1.5: Process flow diagram for a crystallization problem showing inflows and outflows and boundaries enclosing subsystems for analyzing sections of the process.

2. A wet food product contains 70% water. After drying, it is found that 80% of original water has been removed. Determine (a) mass of water removed per kilogram of wet food and (b) composition of dried food. Given Initial water content = 70% Water removed = 80% of original water content Solution 1. Select basis =1 kg wet food product 2. Mass of water in inlet stream = 0.7 kg 3. Water removed in drying = 0.8(0.7) = 0.56 kg/kg of wet food material 4. Write material balance on water, Water in dried food = 0.7(1) - 0.56 = 0.14 kg 5. Write balance on solids, 0.3(1) = solids in exit stream Solids = 0.3 kg 6. Thus, the dried food contains 0.14 kg water and 0.3 kg solids.


Material balance

3. A membrane separation system is used to concentrate total solids (TS) in a liquid food from 10% to 30%. The concentration is accomplished in two stages with the first stage resulting in release of a low-total-solids liquid stream. The second stage separates the final concentration product from a low-total-solids stream, which is returned to the first stage. Determine the magnitude of the recycle stream when the recycle contains 2% TS, the waste stream contains 0.5% TS, and the stream between stages 1 and 2 contains 25% TS. The process should produce 100 kg/min of 30% TS. Given: Concentration of inlet stream= 10% Concentration of outlet stream = 30% Concentration of the recycle stream = 2%

Figure 1.6: A schematic arrangement of equipment described in Example 3

Concentration of the waste water stream = 0.5% Concentration of the stream between the 2 stages= 25% Mass flow rate of the exit stream= 100kg/min Solution: 1. Select 1 min as a basis 2. For the total system F = P +W FxF=Pxp+WxW F (0.1) =100(0.3) +W (0.005) Were x is the solids fraction 3. For the first stage F+R=W+B FxF+ RXR=WxW+ BxB F90.1) +R90.02) =W (0.005) +B (0.25) 4. From step (2) (100+W)(0.1)=30+0.005W


Material balance

0.1W-0.005W=30-10 0.095W=20 W=210.5Kg/min F=310.5Kg/min 5. From step (3) 310.5+R=210.5+B B=100+R 310.5(0.1) +0.02R=210.5(0.005) +0.25B 31.05+0.02=1.0525+0.25B 4.9975=0.23R R=21.73kg/min 6. The results show that the recycle will be flowing at a rate of 21.73Kg/min 4. In a furnace, 95% of carbon is converted to carbon dioxide and the remainder to carbon monoxide. By material balance, predict the quantities of gases appearing in the flue gases leaving the furnace. Given Carbon converted to CO2 = 95% Carbon converted to CO=5% Solution 1. Basis is 1 kg of carbon 2. The combustion equations are C + O2 → CO2 C + ½O2 →CO 3. From these equations, 44kg of carbon dioxide is being formed by combustion of 12kg of carbon, and 28kg of carbon monoxide is being produced by combustion of 12kg carbon 4. The amount of CO2 produced is #../0 123 (#4.67/0 1 " ' ,/0 1 " '


= 3.43 :;
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