Postharvest Technology-units Operations Notes

T : +250 788 90 30 30, +250

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

POSTHARVEST

TECHNOLOGY

INSTITUT D’ENSEIGNEMENT SUPÉRIEUR DE RUHENGERIB.P. 155, Ruhengeri, Rwanda

+250 788 90 30 32, W : www.ines.ac.rw, E : [email protected]


POSTHARVEST

TECHNOLOGY

STITUT D’ENSEIGNEMENT SUPÉRIEUR DE RUHENGERI

[email protected]


POSTHARVEST

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COURSE OUTLINE

• Fundamental concepts

• Material balance

• Energy balances

• Fluid flow theory and viscosity

• Heat transfer and its application in food industry

• Evaporation

• Dehydration

• Mechanical separation

• Mixing

• Size reduction

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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

http://www.nzifst.org.nz/unitoperations/index.htm(English).

Fundamental concept of unit operations in food processing

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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


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

Material balance

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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

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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

= ∑ (1.1)

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

= ∑ (1.2)

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

= (1.3)

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

− = (1.4)

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

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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:

=

(1.5)

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

=

(1.6)

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

!" = #$$ & !'($$ & !'$$ & ') $$ (1.7)

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

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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.

* + *,, = *-- + *.. (1.8)

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 20C, 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

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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

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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

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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 :; <,

5. Similarly the amount of CO produced #28:; <(#0.05:; ABCDEF(

12:; ABCDEF = 0.12 :; <

6. Thus, the flue gasses contain 3.4 kg of CO2 and 0.12 kg CO for every kg of carbon burned

5. Potato flakes (moisture content=75% wet basis) are being dried in a concurrent flow

drier. The moisture content of the air entering the drier is 0.08 kg of water per 1 kg dry air. The moisture content of air leaving the drier is 0.18 kg water per 1 kg of dry air. The air flow rate in the drier is 100 kg dry air per hour. As shown in Figure below,

Material balance

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50 kg of wet potato flakes enter the drier per hour. At steady state, calculate the following

1. What is the mass flow rate of “dried potatoes”? 2. What is the moisture content, of “dried potatoes” exiting the drier?

Given

Weight of potato flakes entering the drier F = 50 kg Time = 1 h

Solution

1. Basis = 1 h 2. M ass of air entering the drier + mass of dry air + mass of water

I=100+100x0.8 I=108

3. Mass of the air leaving the drier = mass of air + mass of water E=100+100x0.18 E=118

4. Total balance of the drier I+F=E+P P=I+F E P=108+100 118 P=40kg

5. Solid balance on the drier We know that =

⇒ XB = 1 XA

In our case, XA is the mass fraction of water in the moist product; XB is the mass fraction of solid in the product. Thus, mass of solids is

mA + mB #1 XA( mA + mB = F, XA = 75%=0.75

Mass of solid content in the feed = F (1-0.75) If y is the solid fraction in the product stream P, then solid balance on the drier gives: 0.25 × F = y × P

Material balance

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X = 4.,7×74.4

y = 0.3125 Thus moisture content in the product feed is 1 y=0.3125=0.6875 or 68.75% In 40kg of the product there is 40×0.6875= 27.5 kg of water.

6. An experimental engineered food is being manufactured using five stages, as shown in Figures below. The feed is 1000 kg/h; various streams have been labeled along with the known composition values on the diagram. Note that the composition of each stream is in terms of solids and water only (TS=total solids). Stream C is divided equally into streams E and G. Product P, with 80% solids, is the desired final product. Stream K produces a by-product at the rate of 450 kg/h with 20% solids. Calculate the following:

1. The mass flow rate of P 2. The mass flow rate of the recycle stream A 3. The mass flow rate of the recycle stream R

F D

W

P K

System total

Material balance

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System boundary

Feed F

1000 kg/h

15% TS

85% water

D

150 kg/h

0% TS

A 5% TS

II

V

III IV

Product

p

80% TS

W

water

K

450 kg/h

20% TS

G

C

B

R

L

E 10% TS

I

Flow chart of the experimental food manufacturing

Energy balances

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Chapter II Energy balances

2.1 General principles

Increasing energy costs have forced people to recognize and appreciate the value of energy more than ever before. Energy conservation is being stressed not only in industrial operations but also in almost all aspects of an individual’s daily activities. Energy balance calculations can be used to account for the various forms of energy involved in a system. Energy audits are essential in identifying effectiveness of energy conservation measures and in identifying areas where energy conservation can be done, or in designing systems involving heating and cooling

2.2 Laws of thermodynamics

The first law of thermodynamics is a statement of the conservation of energy. The law states:

The energy of an isolated system remains constant.

Stated in other words,

Energy can be neither created nor destroyed but can be transformed from one form to another.

Energy can be either stored within an object or transferred to another one, such as in the form of thermal or mechanical energy. If we increase the elevation of an object, its potential energy will increase. The increased potential energy will remain stored in the object until we move it again. Similarly, we can increase the thermal energy of an object by transferring heat into it and observing an increase in temperature.

Energy can also be transformed from one form to another. For example, in a hydroelectric plant, as water falls from a high elevation onto the blades of a turbine, the potential energy of water is converted into mechanical energy in the turbine, and the generator then converts the mechanical energy into electrical energy. The electrical energy is transmitted to homes or factories where it is further converted to other useful forms such as thermal energy in electric heaters.

The second law of thermodynamics is useful in examining the direction of energy transfer or conversion. The following two statements of the second law are by Rudolf Clausius and Lord Kelvin, respectively.

No process is possible whose sole result is the removal of heat from a reservoir (system) at one temperature and the absorption of an equal quantity of heat by a reservoir at a higher temperature.

Energy balances

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No process is possible whose sole result is the abstraction of heat from a single reservoir and the performance of an equivalent amount of work.

The second law of thermodynamics helps explain why heat always flows from a hot object to a cold object; why two gases placed in a chamber will mix throughout the chamber, but will not spontaneously separate once mixed; and why it is impossible to construct a machine that will operate continuously while receiving heat from a single reservoir and producing an equivalent amount of work.

The second law of thermodynamics assigns both quantity and quality to energy. The

importance of this law is evident in any process—the path of a process is always toward

that of decreasing quality. For example, a hot bowl of soup left to itself on a table cools down. In this case, the quality of energy degrades. Energy of a higher quality (at a higher temperature) transfers from the soup to the surroundings and converts into less useful forms of energy.

2.3 Energy

Energy is a scalar quantity. It was first hypothesized by Newton to express kinetic and potential energies. We cannot observe energy directly, but we can measure it using indirect methods and analyze its value. Energy may be in different forms, such as potential, kinetic, chemical, magnetic, or electrical.

Potential energy of a system is by virtue of its location with respect to the gravitational field. If an object has a mass m, located at elevation h, and acceleration due to gravity is g, then the potential energy is

EPE=mgh (2.1)

Kinetic energy of an object is due to its velocity. If an object is moving with a velocity u, and it has mass m, then its kinetic energy is:

^_` = , B, (2.2)

Both kinetic and potential energies are macroscopic; that is, they represent energy of a system due to its entire being. This is in contrast to internal energy, which is due to the microscopic nature of a system. At the molecular scale, the atoms of a substance are continuously in motion. They move in random direction, collide with each other, vibrate, and rotate. Energies related to all these movements, including energy of attraction between the atoms is combined into one lump sum and is called the internal energy.

Energy balances

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Internal energy is an extensive property, and it is independent of the path of a process. Although we cannot measure an absolute value of internal energy, we can relate changes in internal energy to other properties such as temperature and pressure.

In many engineering systems, one or two forms of energy may dominate whereas others can be neglected. For example, when a sugar beet is dropped from a conveyor into a bin, the potential and kinetic energy of the sugar beet changes, but other energy forms such as chemical, magnetic, and electrical do not change and may be neglected in the analysis. Similarly, when tomato juice is heated in a hot-break heater, the potential or kinetic energy of the juice does not change, but the internal energy will change as temperature increases.

The total energy of a system can be written in the form of an equation as

^ = ^_` + ^a` + ^' + ^b0 + ^1c … + (2.3)

Where E i is the internal energy, kJ.

If the magnitudes of all other energy forms are small in comparison with the kinetic, potential, and internal energies, then

= ^_` + ^a` + (2.4)

2.4 Enthalpy

Enthalpy (H) is an extensive property, expressed as the sum of internal energy (Ei) and the product of pressure and volume (PV).

H = Ei + PV (2.5)

In some special cases, enthalpy can be an energy quantity. These are cases where there is fluid entering and leaving an open system. In this case the product of volume and pressure represents the flow energy. In this case, enthalpy of the system is the sum of internal energy and flow energy

2.5 Equation of state and perfect gas law

The thermodynamic properties of a simple system are established by any two independent properties. A functional relationship between the properties of a system is called an equation of state. Values of two properties of a system help establish the value of the third property.

Energy balances

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For a perfect gas, an equation of state is a relationship between pressure, volume, and temperature. The equation may be written as

PV’ = RTA or P = ρRTA (2.6)

Where P is absolute pressure (Pa), V’ is specific volume (m3/kg), R is the gas constant (m3Pa/[kg oK]), TA is absolute temperature (oK) and ρ is density.

2.6 Heat

Heat transfer between a system and its surroundings is probably the most prevalent form of energy that we observe in many food engineering systems. Heat plays a major role in cooking, preservation, and creating new food products with unique properties. Heat is an energy form that is easy to sense because of its association with temperature. Heat transferred to a product can be defined into two types:

Sensible heat is defined as the energy transferred between two bodies at different temperatures, or the energy present in a body by virtue of its temperature.

Latent heat is the energy associated with phase transitions, heat of fusion, from solid to liquid, and heat of vaporization, from liquid to vapor.

Unit of heat in SI is Joule (J). If we consider heat transfer per unit of time, then we express it in terms of rate of heat transfer q, with the units J/s or Watt (W)

Thermal energy for a specific compound can be determined if the heat capacity c is known. Thus:

j = k lFm3n (2.7)

t1 is the initial temperature, while t2 is the final temperature

If the path of energy transfer is under constant pressure, the heat capacity constant is the one under constant pressure cp:

j = k l Fm3n (2.8)

For constant volume energy transfer, the heat capacity is the one for constant volume cv:

j = k loFm3n (2.9)

The numeric values of c p and c v are similar for solids and liquids; however, they may be considerably different for gases.

Energy balances

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2.7 Work

Work includes all interactions between a system and its surroundings that are not a result of temperature difference. There are many such interactions. Examples include the movement of a piston in an engine, an electric wire conducting electric current across a system boundary, and a shaft transmitting energy from a motor to another piece of equipment that is enclosed in the system boundary.

Examples of such interactions include a piston moving in a cylinder, a shaft transmitting energy from a motor to equipment in the system

The sign for work is W and the SI unit is joules (J). Work can be expressed in mathematical terms as follows: consider a system where an object moves a short distance ds due to the force F. the work done by the system is then

dW = -Fds (2.10)

For an object moving from position 1 to position 2 the work is:

pq, = − k rFs, = r#s − s,( (2.11)

The sign convention used for work (W) is that any time work is done by a system, W is positive; when work is done on a system, W is negative. This is opposite to that of heat transfer where Q is positive when heat is applied to a system from the surroundings and is negative when heat flows from the system to the surroundings. This is why in the formula of work above the sign was negative because the force is applied to the system. The work interaction between a system and its surroundings can be attributed to several mechanisms, such as a moving boundary, gravitational forces, acceleration, and shaft rotation.

2.7.1 Work Due to a Moving Boundary

Consider the case of a piston and a cylinder as shown in the figure 2.1.

Figure 2.1: Work due to moving boundary

Energy balances

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As the gas heats it expands and causes the piston to move from position 1 to position 2 as the system boundary is flexible, it expands and the piston moves outward. In this case, work is being done by the expending gas. When the piston moves by a small distance, ds, then the differential amount of work done by the system is force, F, times the distance, ds:

W=Fds (2.12)

But we know that force/area=pressure, if the cross-section of the cylinder is A, then:

W=PAds=PdV (2.13)

The work done to move the piston from position 1 to position 2 is:

pq, = k tFu, (2.14)

2.7.2 Work Due to Gravitational Forces

Force of gravity F=mg. If the mass m is lifted a small distance dh then the corresponding small work is

dW=Fdh ⇒ dW=mgdh To lift an object from location 1 to 2, we have

k Fp, = k ;Fℎ,

⇒ W = mg#h, − h( #2.15( 2.7.3 Work Due to Change in Velocity

For an object moving at velocity u1 to move at velocity u2, there is need to apply force. Force is

F=ma

Where F is force, m is mass and a is acceleration.

y = and B = $

Work is W=Fds

Substituting in this equation of work

p = BFz (2.16)

After simplifying and integrating, we obtain:

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p = k BFB, thus =

, #B,, − B,( (2.17)

Thus, the work done in changing velocity is equal to the change in kinetic energy of the system

2.7.4 Work Due to Shaft Rotation

A typical example is in an electric motor a rotating shaft provides a source of mechanical energy to the connected equipment

Figure 2.2: work due to rotating shaft

As shown in Figure 2.2, the torque Ω (rotating force) applied to a shaft is determined if the force F and radius r are known,

Ω= Fr

If the movement along the circumference is distance s, and the number of revolutions is n, then the distance traveled along the circumference during n revolutions is

s=(2πr)n

Since the product of force and distance is work, we get

= |' #2C(D = ~#2(D (2.18)

2.7.5 Work Due to Frictional Forces

If there is friction in the system, work must be done to overcome it. If we denote friction

energy by Ef then work due to friction force is: W=-Ef (2.19)

Energy balances

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2.8 Energy Balance

The first law of thermodynamics states that energy can be neither created nor destroyed. Therefore, when a process is undergoing in a system, energy entering the system minus energy leaving the system must equal the energy variation in the system:

− ^ = ∆ $)$ (2.20)

This energy balance per unit of time can be written as a rate expression:

− = ∆ $)$ (2.21)

The dot added on top of the E is to show that the units are per unit of time

2.8.1 Energy balance in a closed system

In a closed system there is no mass transfer between the system and the surrounding. Energy can be transferred between the system and the surrounding only through heat and different forms of work. According to the first law of thermodynamics, the total change in energy of a closed system is equal to the heat added to the system minus the work done by the system

∆^ = j − p (2.22)

The total change of energy in the system ∆E is composed of internal energy Ei, Kinetic energy EKE and potential energy EPE, assuming that other forms of energy are too small to be considered which is the case in many systems we get:

∆ + ∆^_` + ∆^a` = j − p (2.23)

Based on the explanations on expansion of gas in a cylinder, the complete equation for work may be written as:

p = k tFu − ^_` − ^a` − & (2.24)

Substituting W in the equation (2.22), and eliminating ∆EKE and ∆EPE, we obtain:

= j + & − k tFu (2.24)

From an elementary theorem of calculus, we know that

d#PV(=dPV+PdV ⇒ ΔPV=kPdV+kVdP Thus

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kPdV= ΔPV- kVdP Substituting in the energy balance equation:

+ tu = j + & − k uFt ⇒ = j + & − k uFt (2.25)

∆H being variation in enthalpy; for heating processes at constant pressure, friction is absent and we remain with

∆H=Q

Constant pressure processes are encountered most commonly in food processing applications. The change in enthalpy shown in the equation above is simply called heat content.

2.8.2 Energy balance for an open system

For an open system, in addition to heat transfer and work, mass is transferred across the boundaries of the system. Mass entering or leaving the system carries a certain amount of energy therefore; change in energy due to the mass flow has to be considered. The work associated to mass flow is referred to as flow work

Figure 2.3: movement of a liquid volume

We can calculate flow work by determining the work involved in pushing a certain mass through the system boundary. Consider a differential element of fluid of uniform properties entering an open system (Fig. 2.3). If the cross-sectional area of the element is A and the pressure of the fluid is P, the force required to push this element through the system boundaries is

F = PA

Energy balances

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If the element is pushed to a distance L then the work done on the fluid is

p$$ &! = r = t = tu (2.26)

As we have seen before, the total energy of the fluid element in the fig. 2.3 includes the kinetic energy, potential energy and internal energy. In addition we have to account for the energy associated with the flowing of the fluid or flow work.

= ^_` + ^a` + + tu (2.26)

By substituting each term by its value we get

= , B, + ;ℎ + + tu (2.27)

Energy Balance for Steady Flow Systems: When a system is in a steady state, its properties do not change with time, there is no change in the energy of the system with time. We have:

= = ∆ $)$ = 0 ⇒ = (2.28)

2.8.3 Total energy balance

Substituting all terms presented in previous sections into the equation (2.28) of conservation of energy

= We get:

j + p + ∑ ^, + 3

, + ;ℎ + tu = j +p + ∑

^′, + 3, + ;ℎ + tu′ (2.29)

In the above equation Ei’ is internal energy per unit mass and V’ is specific volume. p and q are numbers of inlet and exit streams respectively. For a system where there is only one inlet (position 1) and one exit (position 2) streams,

j = 33, + ;ℎ, + a3

3 − n3

, + ;ℎ + ann

+ ^,, − ^, + p (2.30)

Energy balances

25

Where Qm and Wm are the transfer of heat and work per unit mass, respectively. The specific volume V’ is replaced with 1/ρ where ρ (rho) is the density

Examples

1. A tubular water blancher is being used to process beans (Fig. 2.4 ). The product mass flow rate is 860 kg/h. It is found that the theoretical energy consumed for the blanching process amounts to 1.19 GJ/h. The energy lost due to lack of insulation around the blancher is estimated to be 0.24 GJ/h. If the total energy input to the blancher is 2.71 GJ/h,

Figure 2.4 schematic of the water blancher

a. Calculate the energy required to reheat water. b. Determine the percent energy associated with each stream.

Given

Product mass flow rate 860 kg/h

Theoretical energy required by product 1.19 GJ/h

Energy lost due to lack of insulation 0.24 GJ/h

Energy input to the blancher 2.71 GJ/h

Approach

We will first write an energy balance and then solve for the unknowns

Solution

1. Select 1 hour as basis

2. Energy balance can be written as follows:

Energy input to the blancher = energy output with the product + energy loss due to lack of insulation + energy out with water

Energy balances

26

3. Substituting appropriate values in the energy balance,

2.71 = 1.19 + 0.24+Ew

We get:

EW = 1.28Gj/h

Thus, the blancher requires 2.71 -1.28 = 1.43 GJ/h to both reheat water and maintain it at conditions necessary to accomplish the blanching process.

4. These values can be converted to percentage of total thermal input as follows:

Energy out with product = #.6(,. #100( = 43.91%

Energy loss due to lack of insulation = #4.,.(#44(,. = 8.86%

Energy out with water = #.,(#44(,. = 47.23%

5. The results indicate that this water blancher operates at about 44% thermal energy efficiency

2. Five kilograms of ice at 10oC is heated to melt it into water at 0oC; then additional heat is added to vaporize the water in the stream. The saturated vapors exit at 100oC. Calculate the different enthalpy values involved in the process. Specific heat of ice is Ci=2.05kj/kg oK, specific heat of water is Cw=4.182kj/kg ok, latent heat of fusion is 333.2 kj/kg, latent heat of vaporization at 100oC is 2257.06kj/kg.

Given:

The fig 2.5 shows a plot of temperature versus enthalpy. Note that temperature remains constant in regions that involve latent heat.

Figure 2.5: A plot of temperature versus enthalpy for melting of ice and vaporization of water

Energy balances

27

Solution:

The enthalpy calculations are made separately for each zone in the fig. 2.5

1. Zone A-B

= 5(kg)×2.05(kj/kg oC)×(0+10)oC

= 102.5kj/kg

2. Zone B-C

∆HBC=mHlatent

= 5kg×333.2kj/kg

=1666kj

3. Zone C-D

=5kg×4.1822.05(kj/kg oC)×(100-0)oC

=2091kj

3. Steam is used for peeling potatoes in a semicontinuous operation. Steam is supplied at the rate of 4 kg per 100 kg of unpeeled potatoes. The unpeeled potatoes enter the system with a temperature of 17oC, and the peeled potatoes leave at 35oC. A waste stream from the system leaves at 60oC. The specific heats of unpeeled potatoes, waste stream, and peeled potatoes are 3.7, 4.2, and 3.5 kJ/(kg oK), respectively. If the heat content (assuming 0oC reference temperature) of the steam is 2750 kJ/kg, determine the quantities of the waste stream and the peeled potatoes from the process.

Figure 2.6: A block diagram showing various streams described in Example above

Fluid flow theory and Viscosity

28

Chapter III Fluid flow theory and viscosity

Fluids are substances that flow without disintegration when pressure is applied. This definition of a fluid includes gases, liquids, and certain solids. A number of foods are fluids. In addition, gases such as compressed air and steam are also used in food processing and they exhibit resistance to flow just like liquids. The resistance of fluids to flow and its implications in the design of a fluid handling system will be discussed here.

3.1 Liquid transport system

Fluids are substances that flow without disintegration when pressure is applied. This definition of a fluid includes gases, liquids, and certain solids. In any commercial food processing plant, the movement of liquid foods from one location to another is an essential operation. Various types of systems are used for moving raw liquid foods as well as processed liquid products before packaging. In food industry, transport systems are of particular design because of the important need of sanitation in order to maintain the product clean.

From our daily experience with handling different kinds of fluids, we know that if pressure at one location within a fluid system is higher than another, the fluid moves toward the region of lower pressure. Gravity causes the flow of fluid from higher to lower elevations. A fluid moving to a lower elevation undergoes a decrease in its potential energy, while its kinetic energy increases. With the presence of thermal gradients, heated fluids experience a decrease in density, causing lighter fluid to rise while denser fluid takes its place.

Figure 3.1: A typical liquid food processing system, illustrating pipelines and pipeline components


29

A typical transport system consists of four basic components: tanks, pipelines, pumps and fittings. Tanks serve in keeping the liquid products prior to or after processing. Pipelines are used to transport the product from one processing station to another. Fittings consist of elbows and tee that serve in changing direction of the product; and valves that serve in control of the flow rate. Apart from regions where liquids are transported by gravitational force, pumps are the ones that produce mechanical energy for transport

3.2 Pipelines for processing plants

Fluids (liquids and gases) in food processing plants are transported mostly in closed

conduits—commonly called pipes if they are round, or ducts if they are not round.

Although sometimes used in processing plants, open drains generally are avoided, for sanitation reasons. The pipelines used for liquid foods and their components have numerous unique features. Probably the most evident feature is the use of stainless steel for construction. This metal provides smoothness, cleanability, and corrosion resistance.

Typical pipeline system for liquid food transport contains several essential components. In addition to the straight lengths of the pipe, which may vary in diameter from 2 to 10 cm, elbows and tees are essential for changing the direction of product movement. These components are welded into the pipeline system and may be used in several different configurations. Another component is the valve used to control the flow rate; an air-actuated valve is illustrated in Figure 3.2. This valve may be operated remotely, often based on some type of preset signal.

Figure 3.2: An air-actuated valve for liquid foods.


30

3.3 Types of pumps

Except for situations where gravity can be used to move liquid products, some type of mechanical energy must be introduced to overcome the forces opposing transport of the liquid. This mechanical energy is supplied by pumps. There are numerous types of pumps used in the industry. As shown in Figure 3.3, pumps may be classified as centrifugal or positive displacement. There are variations within each of these types, as shown in the figure.

Figure 3.3 Classification of pumps

3.3.1 Centrifugal pumps

The basic principle of these pumps is the use centrifugal force to increase pressure of the fluid. The pump consists of a motor-driven impeller enclosed in a case. The product enters the pump at the center of impeller rotation and, due to centrifugal force, moves to the impeller periphery. At this point, the liquid experiences maximum pressure and moves through the exit to the pipeline. These pumps are most efficient in high viscosity foods like milk and juices.


31

Figure 3.4: Exterior view of a centrifugal pump

Figure 3.4: A centrifugal pump with components

Figure 3.5: Principle of operation of a centrifugal pump


32

3.3.2 Positive displacement pumps

By application of direct force to a confined liquid, a positive displacement pump produces the pressure required to move the liquid product. Product movement is related directly to the speed of the moving parts within the pump. Thus, flow rates are accurately controlled by the drive speed to the pump. The mechanism of operation also allows a positive displacement pump to transport liquids with high viscosities.

Figure 3.6: examples of positive displacement pumps. Above left is reciprocating piston pump, above left is Peristaltic (flexible membrane rotary) pump and below is Lobe pump

3.4 Properties of liquids

The transport of a liquid food by one of the systems described above is directly related to liquid properties, primarily viscosity and density. These properties will influence the power requirements for liquid transport as well as the flow characteristics within the pipeline. An understanding of the physical meaning associated with these properties is necessary in order to design an optimal transport system.


33

Fluid flow takes place when force is applied on a fluid. Force per unit area is defined as stress. When force acting on a surface is perpendicular to it, the stress is called normal stress. More commonly, normal stress is referred to as pressure. When the force acts parallel to the surface, the stress is called shear stress, σ. When shear stress is applied to a fluid, the fluid cannot support the shear stress; instead the fluid deforms, or simply stated, it flows.

The influence of shear stress on solids and liquids leads to a broad classification of such materials as plastic, elastic, and fluid.

In the case of an elastic solid, when shear stress is applied, there is a proportional finite deformation, and there is no flow of the material. On removal of the applied stress, the solid returns to its original shape.

A plastic material, on the other hand, deforms continuously on application of shear stress; the rate of deformation is proportional to the shear stress. When the shear stress is removed, the object shows some recovery. Examples include some types of cheese, gelatin deserts

A fluid deforms continuously on application of shear stress. The rate of deformation is proportional to the applied shear stress. There is no recovery; that is, the fluid does not retain or attempt to retain its original shape when the stress is withdrawn.

When normal stress or pressure is applied on a liquid, there is no observed appreciable effect. Thus, liquids are called incompressible fluids, whereas gases are compressible fluids, since increased pressure results in considerable reduction in volume occupied by a gas.

3.4.1 Density

The density of a liquid is defined as its mass per unit volume and is expressed as kg/m3 in the SI unit system. In a physical sense, the magnitude of the density is the mass of a quantity of a given liquid occupying a defined unit volume. The most evident factor is that the magnitude of density is influenced by temperature. For example, the density of water is at maximum at 4oC and decreases consistently with increasing temperature. Densities of liquids are most often measured by a hand hydrometer. This instrument measures specific gravity, which is the ratio of the density of the given liquid to the density of water at the same temperature. The measuring instrument is a weighted fl oat attached to a small-diameter stem containing a scale of specific-gravity values. The fl oat will sink into the unknown liquid by an amount proportional to the specific gravity, and the resulting liquid level is read on the stem scale.


34

Figure 3.7: Density of water as a function of temperature

3.4.2 Viscosity

A fluid may be visualized as composed of different layers. The fluid begins to move as soon as a force acts on it. The relative movement of one layer of fluid over another is due to the force, commonly called shearing force, which is applied in a direction parallel to

the surface over which it acts. From Newton’s second law of motion, a resistance force

is offered by the fluid to movement, in the opposite direction to the shearing force, and it must also act in a direction parallel to the surface between the layers. This resistance

force is a measure of an important property of fluids called viscosity.

With different types of fluids, we commonly observe a wide range of resistance to movement. For example, honey is much more difficult to pour out of a jar or to stir than water or milk. Honey is considerably more viscous than milk. With this conceptual framework, we will consider a hypothetical experiment.

Consider a fluid element sheared in one plane by a single shear stress σ (tau) as shown in fig. 3.8. The shear strain angle δθ will continuously grow with time as long as the stress σ is maintained, the upper surface moving at speed of δu larger than the lower. When the shearing stress increases, the rate of shear will increase in direct proportion:

∝ ) (3.1)

and

tan ¡ = ¢¢¢) (3.2)


35

Figure 3.8: illustration of a fluid element straining at a rate of δθ/δt

For a small angular deflection

tanδθ≈ δθ (3.3)

Therefore

¡ = ¢¢¢) ⇒ ¢£

¢ = ¢¢) (3.4)

Replacing δθ/δt by its value in the equation (3.1) and introducing a constant µ in order to remove the proportionality we get:

= ¤ ¢¢) = ¤

) (3.5)

µ is the coefficient of viscosity, or simply viscosity, of the fluid. It is also called “absolute” or “dynamic” viscosity and is expressed in Pa s of poise (1 poise=0.1 Pa s). Liquids that follow Equation (3.5), exhibiting a direct proportionality between shear rate and shear stress, are called Newtonian liquids. When shear stress is plotted against shear rate, a straight line is obtained passing through the origin.

Figure 3.9: Shear stress vs. shear rate behavior for a Newtonian fluid


36

Figure 3.10: Rheological behavior of various fluids

The slope of the line gives the value for viscosity, µ. Water is a Newtonian liquid; other foods that exhibit Newtonian characteristics include honey, fluid milk, and fruit juices. Table 3.1 gives some examples of coefficients of viscosity. Viscosity is a physical property of the fluid and it describes the resistance of the material to shear-induced flow. Furthermore, it depends on the physicochemical nature of the material and the temperature.

Fluid Approximate viscosity in Pa s

Air 10-5

Water 10-3

Olive oil 10-1

Glycerol 1

Liquid honey 10

Golden syrup 102

Glass 104

Table 3.1: The Viscosity of Some Common Materials at Room Temperature


37

In some books the symbol τ is used to symbolize shear stress but the Society of Rheology recommends the use of symbol σ.

3.4.2.1 Methods of measuring viscosity

Instruments used for measuring flow properties of fluids are called viscometers. The most common viscometers are the capillary tube and the rotational viscometer shown in the figure below.

Figure 3.11: Diagram of a glass capillary viscometer and a rotary viscometer.

3.5 Liquid transport in food industry

In food industry, transport of liquid food from one processing equipment to another is usually achieved by using pumps. Depending on velocity, internal viscous and inertial forces, different types of flow characteristics are observed.

Flow patterns of liquids can be viewed in different ways, will consider flow of fluid along a streamline which is a line everywhere tangent to the velocity vector at a given instant. When bunched together into a stream tube, stream lines provide a good indication of the instantaneous flow of a fluid as shown in fig. 3.12


38

Figure 3.12: a stream tube formed by a closed collection of streamlines

3.5.1 The Continuity Equation

Consider a fluid flowing in a pipeline as shown in the fig. 3.13. Suppose that in time space δt the fluid occupying the space XX’ has moved to the space YY’.

Figure 3.13: Fluid flow through a pipe of varying cross-section. Cross-section areas of the two and are different in order to prove that calculations to follow are applicable to these variations

Note that the amount of fluid in space YX’ is the same in the initial and final stage of the experiment thus mass of fluid in space XY must equal mass fluid in space X’Y’. Therefore:

ρ1A1δx1 = ρ2A2δx2 (3.6)

Dividing by time step δt:

¢n¢ = ,,

¢3¢ ⇒ B¥ = ,,B¥, (3.7)

Where ū is the average velocity


39

The equation 3.7 is called equation of continuity. It can also be expressed in terms of mass flow since:

σAū = ṁ (3.8)

For incompressible fluids the density remains constant thus σ1 = σ2 and we remain with

B¥ = ,B¥, yDF B¥ = u (3.9)

Where ⩒ is the volumetric rate

The volumetric rate is a product of the cross-section area A of the pipe and the fluid velocity ū.

Example

The volumetric flow rate of beer flowing in a pipe is 1.8 l/s. the inside diameter of the pipe is 3 cm. the density of beer is 1100 kg/m3. Calculate the average velocity of beer and its mass flow rate in kg/s. what is the mass flow rate if another pipe with 1.5 cm is used, and what is the velocity for the same volumetric flow rate?

Given: Pipe diameter: 3 cm = 0.03 m Volumetric flow rate: 1.8 l/s = 0.0018 m3/s Density: 1100 kg/l

Solution: • Velocity ū is

ū = 0.0018¨-/s© × 0.03,

4 ¨,©= 2.55/s

• Mass flow rate ṁ:

= 1100¨:;/-© × × 0.03,

4 ¨,© × 2.55¨/s© ṁ = 1.98 kg/s

• New velocity if the diameter is halved and the volumetric flow rate is kept constant:

ū = 0.0018¨-/s© × 0.015,

4 ¨,©= 10.19 /s

• Note that by halving the diameter of the pipe, the average velocity is increased four times


40

3.5.2 Reynolds Number

In order to visualize the flow characteristics of a liquid in a pipe suppose that a dye is injected in a liquid flowing in a pipe as shown in fig. 3.14

Figure 3.14: Laminar, transitional, and turbulent flow in a pipe

At low flow rates, the dye moves in a straight-line manner in the axial direction. As the flow rate increases to some intermediate level, the dye begins to blur at some distance away from the injection point. The blurring is caused by movement of some of the dye in radial direction. At high flow rate, the dye becomes blurred immediately after injection.

The straight-line flow observed at low flow rates is called laminar flow; at intermediate

flow rates, the flow is called transitional flow; and the erratic flow obtained at higher

flow rates is called turbulent flow.

The flow characteristics for laminar flow are influenced by liquid properties, flow rate and the dimensions of liquid solid interfaces. As the mass flow rate increases, the force of momentum or inertia increase; but these forces are resisted by friction of viscous forces within the flowing liquid. The opposing forces reach a certain balance, changes in the flow characteristics occur. It has been proved that the inertial forces are a function of liquid density ρ, pipe diameter D and average fluid velocity ū. The viscous forces on the other hand are a function of liquid viscosity.

A dimension less number called Reynolds number (NRE) is defined as the ratio of the inertial to the viscous forces:

ª«` = ' &'$o$$ &'$ ⇒ ª«` = ū¬

(3.10)

If instead of average velocity ū, mass flow rate ṁ is given, then:

ª«` = .ṁ¯¬ (3.11)

A Reynolds number is most useful in quantitatively describing the flow characteristics of a fluid

For a NRE ≤ 2100: the flow is laminar For a 2100 ≤NRE ≤ 4000: the flow is transitional


41

For a NRE ≥ 4000: the flow is turbulent.

Example

A 3 cm inside diameter pipe is being used to pump liquid food into a buffer tank. The tank is 1.5 m diameter and 3 m high. The density of the liquid is 1040 kg/m3 and viscosity is 1600 × 10-6 Pa s.

a. What is the minimum time to fill the tank with this liquid food if it is fl owing under laminar conditions in the pipe?

b. What will be the maximum time to fill the tank if the flow in the pipe is turbulent

Given Pipe diameter: 3 cm = 0.03 m Tank height: 3 m Tank diameter: 1.5 m Density of liquid: 1040 kg/m3 Viscosity of liquid: 1600 ×10-6 Pa s =1600 ×10-6 kg/m s

Solution Part a:

1. The maximum velocity under laminar flow is:

ū = 2100¤°± = 2100 × 1600 × 10q²¨:;/ s©

1040 ¨:;/-© = 0.108/s

Volumetric flow rate is:

u = × 0.03,¨,©4 × 0.108¨/s© = 7.63 × 10q7¨-/s©

2. The volume of the tank is:

u = #F³yEzEC(,#ℎE³;ℎz(4 = × 1.5,¨,© × 3¨©

4 = 5.3-

3. The minimum time to fill the tank:

Volume/volumetric flow rate = 5.3 [m3]/7.63×10-5 [m3/s]=6.95×104 s= 19.29h

Part b: 4. Minimum velocity under turbulent flow conditions is:

ū = 4000¤°± = 4000 × 1600 × 10q²¨:;/ s©

1040 ¨:;/-© = 0.205/s

The volumetric flow rate:


42

u = × 0.03,¨,©4 × 0.205¨/s© = 1.449 × 10q.¨-/s©

5. The maximum time to fill the tank: volume of tank/volumetric flow rate: = 5.3 [m3]/1.449×10-4 [m3/s]=3.66×104 s= 10.16h

3.5.3 Entrance region and fully developed flow

When a liquid enters a pipe, there is a certain initial length of the pipe, called the entrance region, where the flow characteristics of the liquid are quite different from those in the following length of pipe. As shown in Figure 3.15, immediately at the entrance to the pipe, the liquid has a uniform velocity profile, identified by the same length of arrows in the diagram. As it begins to move into the pipe, the liquid next to the inside wall of the pipe is held back by friction between the liquid and the wall surface. The velocity of the liquid is zero at the wall and increases toward the central axis of the pipe. Therefore, the boundary (or the wall surface) begins to influence the velocity profile.

Figure 3.15: V elocity profi le in a fl uid fl owing in a pipe.

As shown in Figure 3.15, in the entrance region, the boundary layer develops from location X to Y. At location Y, the effect of the boundary layer on the velocity profile extends all the way to the central axis. The cross-sectional velocity profile at Y is parabolic in shape (as we will mathematically derive in the following section). From X to Y, the region is called the entrance region, and the liquid flow in the region beyond Y is commonly referred to as fully developed flow. Using dimensional analysis, it has been shown that the dimensionless entrance length, Le/D, is a function of the Reynolds number. Thus for laminar flow:

Le/D=0.06NRE (3.12)


43

And for turbulent flow:

Le/D=4.4(NRE)1/6 (3.13)

3.5.4 Velocity profile in a liquid flowing under fully developed flow conditions

Consider fluid flow taking place under steady and fully developed conditions in a constant-diameter pipe. Forces due to pressure and gravity cause the fluid flow in the fully developed region. In the case of a horizontal pipe, gravitational effects are negligible. Therefore, for the purpose of this analysis, we will consider only forces due to pressure. When a viscous liquid (a liquid with viscosity greater than zero) flows in a pipe, the viscous forces within the liquid oppose the pressure forces. Application of pressure is therefore necessary for flow to occur, as it overcomes the viscous forces opposing the flow.

Furthermore, the flow takes place without accelerating; the velocity profile within the fully developed flow region does not change with location along the x-axis. For the flow to be steady, a balance must exist between the pressure and viscous forces in the liquid. We will conduct a force balance to analyze this case.

Figure 3.16: Force balance for a liquid flowing in a pipe

Using Newton’s second law of motion, we can describe forces acting on the small cylindrical element of liquid of radius r and length L as shown in fig. 3.16. The pipe’s diameter is D. From the location A-B at time t, the element has moved to A’-B’ after a small lapse of time ∆t. The ends of A’B’ are shown distorted, indicating that under fully developed flow conditions, velocity at the central axis is maximum and deceases with increasing r.


44

Since the pipe is horizontal, we neglect the gravitational forces. The pressure varies from one axial location to another, but remains constant along any vertical cross-section of the pipe. Let’s assume that pressure forces at the faces A and B are respectively PA and PB and the decrease in pressure is ∆P

∆P= PA- PB

The pressure forces acting on the element are

at location A: =PA πr2

at location B: #PA-ΔP( πr2

On the circumferential area, 2πrL viscous forces opposing pressure forces =σ π 2 rL

Where σ is the shear stress

According to Newton’s second law, Fx = max, but under fully developed flow conditions, ax=0, thus Fx=0. Thus, all forces acting on the element must balance.

PA πr2- #PA-ΔP( πr2- σπ2rL=0 Or simplifying,

·a¸ = ,¹

' (3.14)

For Newtonian liquids, the shear stress is related to viscosity. For pipe flow, we rewrite this equation in cylindrical coordinates as

= −¤ ' (3.15)

Note that du/dr is negative in case of a pipe flow, the velocity decreasing with increasing radial distance, r, as seen in Figure 3.16. Therefore, we have introduced a negative sign in Equation (3.15) so that we obtain a positive value for shear stress, σ. Substituting Equation (3.14) in Equation (3.15),

' = − ·a

,¸ C (3.16)

integrating,

k FB = − ·a,¸ k CFC (3.17)


45

Or

B#C( = − ·a.¸ C, + (3.18)

Where C1 is a constant

For a viscous fluid flowing in a pipe, u=0 at r=R; therefore

= ·a.¸ », (3.19)

Therefore, the velocity profile for a laminar, fully developed flow, in a horizontal pipe, is:

B#C( = ·a.¸ #», − C,( (3.20)

This is equal to

B#C( = ·a«3.¸ ¼1 − '

«,½ (3.21)

The equation (3.21) is an equation of a parabola. Therefore, for a fully developed flow, we obtain a parabolic velocity profile. Furthermore, substituting r=0, the maximum velocity umax is obtained at the pipe center. It is equal to

B = ·a«3.¸ (3.22)

Next, let us determine the volumetric flow rate by integrating the velocity profile across the cross-section of the pipe. First, we will examine a small ring of thickness dr with an area dA, where dA=2πrdr, as shown in Figure 3.17. The velocity, u, in this thin annular ring is assumed to be constant.


46

Figure 3.17: small ring of thickness dr

Then the volumetric flow rate through the annular ring, ⩒ring is:

u'0 = B#C(F = B#C(2CFC (3.23)

The volumetric flow rate for the entire pipe cross-section is obtained by integration as follows:

u = k B#C(F = k B#C(2CFC'«'4 (3.24)

or, substituting Equation (3.21) in (3.4),

u = ,¯·a«3.¸ k ¼1 − '

«,½«4 CFC (3.25)

This is equal to:

u = ¯«¾·a¸ (3.26)

The equation above is called Poiseuille’s Law. The mean velocity, ū, is defined as the volumetric flow rate divided by the cross-sectional area of the pipe, πR2, or

ū = ¿¯«3 = ·a«3

¸ (3.27)

If we divide the equation of maximum velocity by the equation of average velocity we obtain:


47

ūÀÁ

= 0.5 (3.28)

From Equation (3.28), we observe that the average velocity is half the maximum velocity for fully developed laminar flow conditions. Furthermore, the radius (or diameter) of the pipe has a dramatic influence on the flow rate, as seen in Equation (3.26). Doubling the diameter increases the volumetric flow rate 16-fold.

For turbulent flow in the fully developed region, the mathematical analysis necessary to obtain a velocity profile is complex. Therefore, the following empirical expression is generally used.

ū#'(ÀÁ

= 1 − '« ⁄

(3.29)

Where j is a function of the Reynolds number; for most applications j=7 is recommended. Under turbulent conditions, the velocity profile may be obtained from

ū#C( = B 1 − '« ⁄

(3.30)

Equation (3.30) is also called the Blassius 1/7th power law.

A volumetric flow rate under turbulent conditions may be obtained in a similar manner as for laminar conditions. Substituting Equation (3.29) in Equation (3.24)

u = k B 1 − '« ⁄ 2CFC'«

'4 (3.31)

Integrating the equation (3.31)

u = 2B'33

#(#,( (3.32)

A relationship between average and maximum velocity may be obtained by substituting Equation (3.27) in Equation (3.32) and we get,

ūÀÁ

= ,3#(#,( (3.33)

Substituting the value 7 for j in Equation (3.33),

ūÀÁ

= 0.82 (3.34)


48

Thus, in the case of turbulent flow, the average velocity is 82% of the maximum velocity. The maximum velocity occurs at the central axis of the pipe.

3.5.5 Forces Due to Friction

The forces that must be overcome in order to pump a liquid through a pipe derive from several sources. As seen previously, viscous forces are important in liquid flow; these forces occur due to the movement of one layer over another. The other important forces are due to friction between the liquid and the surface of the wall of a pipe. When a fluid flows through a pipe, some of its mechanical energy is dissipated due to friction. It is common to refer to this dissipated energy as a frictional energy loss.

The friction forces vary with conditions such as flow rates, as described with a Reynolds number, and surface roughness. The influence of the friction forces is expressed in the form of a friction factor f. The following mathematical development is for laminar flow conditions.

The friction factor is the ratio between the shear stress at the wall, σw, to the kinetic energy of the fluid per unit volume.

Ã = ¹Äū3/, (3.35)

As seen earlier, the shear stress at the wall is

! = ¬·a.¸ (3.36)

Substituting the equation (3.36) into equation (3.35)

Ã = ¬·a,¸ū3 (3.37)

By rearranging terms in Equation (3.27), the pressure drop in fully developed laminar flow conditions is determined as

ū = ·a«3¸ ⟹ t = -,ū¸

¬3 (3.38)

Substituting ∆P value in the friction factor equation, we get:

Ã = -,ū¸¬,¸ū3¬3 = ²

ū¬ (3.39)

Where f is called the Fanning friction factor; Note that many civil and mechanical engineering textbooks refer to a different friction factor, called Darcy 3 friction factor,


49

with the same symbol, f. The Darcy friction factor is four times the Fanning friction factor. In chemical engineering literature, the Fanning friction factor is used more commonly, that’s why I preferred it over the Darcy friction factor.

The previous calculations leading to Fanning friction factor are for laminar flow conditions only. In cases of transitional and turbulent flow conditions, the mathematical derivations become highly complex. For situations involving flow conditions other than laminar flow, we will use a graphical chart that presents friction factor as a function of the Reynolds number. This chart, called a Moody chart, is shown in Figure 3.18. The Moody chart presents the friction factor as a function of the Reynolds number for various magnitudes of relative roughness of the pipes. At low NRE (<2100) the curve is described by the equation (3.39) and is not influenced by surface roughness ε. In the transition from laminar to turbulent flow or critical region, either set of curve can be used. Most often the friction factor is selected for turbulent flow since it ensures that the pressure loss due to friction will not be underestimated. From the Moody chart there is always a certain friction loss when a fluid flows in a pipe. The following equation is recommended for calculating the Fanning friction factor for a turbulent region:

Æ& ≈ −3.6 log ¼ ².6

ÉÊË+ Ì/¬

-. .½ (3.40)

Figure 3.18: The Moody diagram for the Fanning friction factor. Equivalent roughness for new pipes (ε in meters) Cast iron: ε= 259x10-6, Drawn tubing: ε=1.5235x10-6 , Galvanized iron: ε=152x10-6, Steel or wrought iron: ε=45.7x10-6


50

Examples:

1. A fluid is flowing under laminar conditions in a cylindrical pipe of 2 cm diameter. The pressure drop ∆P is 330 Pa, the viscosity of the fluid is 5 Pa s, and the pipe is 300 cm long. Calculate the mean velocity and velocity of fluid at different radial locations in the pipe.

Given:

Diameter of the pipe D = 2 cm Length of the pipe L = 300 cm Pressure drop ∆P = 330 Pa Viscosity µ = 5 Pa s

Solution:

B#C( = ·a.¸ #», − C,(

The velocity is calculated at r = 0, 0.25, 0.5, 0.75, and 1 cm r=0 cm u= 0.055 cm/s r=0.25 cm u= 0.0516 cm/s r=0.5 cm u= 0.413 cm/s r=0.75 cm u= 0.0241 cm/s r=1 cm u= 0 cm The mean velocity ū is calculated as 0.0275 cm/s; this value is half of the maximum velocity.

2. Water at 30oC is being pumped through a 30 m section of 2.5 cm diameter steel pipe at a mass flow rate of 2 kg/s. Compute the pressure loss due to friction in the pipe section.

Given:

Density ρ= 995.7 kg/m3 Viscosity µ= 792.377x10-6 Pa s Length of the pipe L= 30 m Diameter of the pipe D= 2.5 cm = 0.025 m Mass flow rate ṁ= 2kg/s

Solution: 1. Compute the Mean velocity:

ū = ,¨/0/$©#667. /0/Í(¨¯#4.4,7 (3/.© = 4.092 /s

2. Compute the Reynolds number:


51

ª«` = #667. /0/Í(#4.4,7 (#..46, /$(#6,.-×4ÎÏ a $( = 28,550

3. Using the given information and the fig. 3.13, relative roughness is:

Ð ±⁄ = .7.×4ÎÏ 4.4,7 = 1.82810q-

4. The friction factor f is obtained from the fig. 3.18:

f= 0.006

5. Ã = ¬·a,¸ū3 ⇒ ·a

= 2#Ã( ¸ū3¬ = 2#0.006( #..46, /$(3#-4 (

4.4,7 = 241.12 ,/s,

Note that 1 J = 1 kg m2/s2 Thus:

·a = 241.12 ,/s, = 241.12 Ñ/:;

∆P/ρ represents the energy consumed due to friction on a per-unit-mass basis 6. The pressure loss is calculated as

ΔP=#241.12 J/kg(#995.7 kg/m3(= 240.08 kg/#ms2( ΔP= 240.08 kPa

Heat transfer in food processing

52

Chapter IV Heat Transfer in Food Processing

4.1 Introduction

The most common processes found in a food processing plant involve heating and cooling of foods. In the modern industrialized food industry, we commonly find unit operations such as refrigeration, freezing, thermal sterilization, drying, and evaporation. The study of heat transfer is important because it provides a basis for understanding how these food processes operate.

4.2 Systems for heating and cooling food products

In a food processing plant, heating and cooling of foods is conducted in equipment called heat exchangers. The fig. 4.1 shows the classification of different heat exchangers. There are contact heat exchanger and non contact heat exchangers. As the name implies, in noncontact-type heat exchangers, the product and heating or cooling medium are kept physically separated, usually by a thin wall. On the other hand, in contact-type heat exchangers, there is direct physical contact between the product and the heating or cooling streams. We will analyze them, see the mechanisms of heat transfer and then see the application of heat transfer in food industry

Figure 4.1: Classification of commonly used heat exchangers

4.2.1 Plate Heat Exchanger

The plate heat exchanger invented more than 70 years ago has found wide application in the dairy and food beverage industry. A schematic of a plate heat exchanger is shown in Figure 4.2. This heat exchanger consists of a series of parallel, closely spaced stainless-steel plates pressed in a frame. Gaskets, made of natural or synthetic rubber, seal the plate edges and ports to prevent intermixing of liquids. These gaskets help to direct the heating or cooling and the product streams into the respective alternate gaps. The direction of the product stream versus the heating/cooling stream can be either parallel flow (same direction) or counter flow (opposite direction) to each other. The plates used in the plate


53

heat exchanger are constructed from stainless steel: Special patterns are pressed on the plates to cause increased turbulence in the product stream, thus achieving better heat transfer. Plate heat exchangers are suitable for low-viscosity (< 5 Pa s) liquid foods. If suspended solids are present, the equivalent diameter of the particulates should be less than 0.3 cm.

Figure 4.2: Plate heat exchanger: the first picture shows the outer view; the second one shows the fluid flow between the plates and the third one shows individual plates

Plate heat exchangers offer the following advantages:

• The maintenance of these heat exchangers is simple, and they can be easily and quickly dismantled for product surface inspection.

• The plate heat exchangers have a sanitary design for food applications.

• Their capacity can easily be increased by adding more plates to the frame.

• With plate heat exchangers, we can heat or cool product to within 1oC of the adjacent media temperature, with less capital investment than other noncontact-type heat exchangers.

• Plate heat exchangers offer opportunities for energy conservation by regeneration.

4.2.2 Tubular Heat Exchanger

The simplest noncontact-type heat exchanger is a double-pipe heat exchanger, consisting of a pipe located concentrically inside another pipe. The two fluid streams flow in the annular space and in the inner pipe, respectively. The streams may flow in the same direction (parallel flow) or in the opposite direction (counter flow). Figure 4.3 is a schematic diagram of a counter flow double-pipe heat exchanger. A slight variation of a double-pipe heat exchanger is a triple-tube heat exchanger. In this type of heat exchanger, product flows in the inner annular space, whereas the heating/cooling medium flows in the inner tube and outer annular space. The innermost tube may contain specially designed obstructions to create turbulence and better heat transfer.


54

Figure 4.3: Schematic illustration of a tubular heat exchanger showing a double and a triple tube heat exchanges

Another common type of heat exchanger used in the food industry is a shell-and-tube heat exchanger for such applications as heating liquid foods in evaporation systems. As shown in Figure 4.4, one of the fluid streams flows inside the tube while the other fluid stream is pumped over the tubes through the shell. By maintaining the fluid stream in the shell side to flow over the tubes, rather than parallel to the tubes, we can achieve higher rates of heat transfer.

Figure 4.4: A shell-and-tube heat exchanger


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4.2.3 Scraped-Surface Heat Exchanger

In conventional types of tubular heat exchangers, heat transfer to a fluid stream is affected by hydraulic drag and heat resistance due to film buildup or fouling on the tube wall. This heat resistance can be minimized if the inside surface of the tube wall is scraped continuously by some mechanical means. The scraping action allows rapid heat transfer to a relatively small product volume. A scraped-surface heat exchanger, used in food processing, is shown schematically in Figure 4.5. The inside rotor contains blades that are covered with plastic laminate or molded plastic. The cylinder containing the product and the rotor is enclosed in an outside jacket. The heating/cooling medium is supplied to this outside jacket.

Figure 4.5: A scraped-surface heat exchanger with a cutaway section illustrating various components.

4.2.4 Steam-Infusion Heat Exchanger

A steam-infusion heat exchanger provides a direct contact between steam and the product. Product in liquid state is pumped to the top of the heat exchanger and then allowed to flow in thin sheets in the heating chamber. The viscosity of the liquid determines the size of the spreaders. High rates of heat transfer are achieved when steam contacts tiny droplets of the food. The temperature of the product rises very rapidly due to steam condensation. The heated products with condensed steam are released from the chamber at the bottom. A specific amount of liquid is retained in the bottom of the chamber to achieve desired cooking.

The water added to the product due to steam condensation is sometimes desirable, particularly if the overall process requires addition of water. Otherwise, the added water


56

of condensation can be“ flashed off” by pumping the heated liquid into a vacuum

cooling system. The amount of water added due to condensation can be computed by measuring the temperature of the product fed to the heat exchanger and the temperature of the product discharged from the vacuum cooler.

This type of heat exchanger has applications in cooking and/or sterilizing a wide variety of products, such as concentrated soups, chocolate, processed cheese, ice cream mixes, puddings, fruit pie fillings, and milk.


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Figure 4.6: A steam-infusion heat exchanger

4.3 Thermal properties of foods

4.3.1 Specific heat

It is the quantity of heat that is gained or lost by a unit mass of product to accomplish a unit change in temperature without change of state.

= Ô#·Õ( (4.1)

Where Q is heat gained or lost (kJ), m is mass (kg), ∆T is temperature change in the material (°C or °K), and Cp is specific heat (kJ/[kg °C]).

Specific heat is an essential part of the thermal analysis of food processing or of the equipment used in heating or cooling of foods. For processes that involve change of state like freezing or thawing, apparent specific heat is used which is the specific heat of the product plus the heat necessary for change of state.

4.3.2 Thermal Conductivity

In quantitative terms, this property gives the amount of heat that will be conducted per unit time through a thickness of the material if a unit temperature gradient exists across that thickness.

In SI units, thermal conductivity k is W/ (m oC)

In commonly found materials thermal conductivity varies a lot. For example:

Metals: 50-400 W/ (m oC) Alloys: 10-120 W/ (m oC) Water: 0.597 W/ (m oC) Air: 0.0521 W/ (m oC) Insulating materials: 0.035-0.173 W/ (m oC)

Most high-moisture foods have thermal conductivity values closer to that of water. On the other hand, the thermal conductivity of dried, porous foods is influenced by the presence of air with its low value.

4.3.3 Thermal diffusivity

Thermal diffusivity, a ratio involving thermal conductivity, density, and specific heat, is given as,


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Ö = /1 (4.2)

The units of thermal diffusivity are m2/s

Thermal diffusivity may be calculated by substituting values of thermal conductivity, density, and specific heat in Equation (4.2)

The above properties play an important role in heat transfer.

4.4 Modes of heat transfer

Heat energy is simply the sensible and latent forms of internal energy. Heat content of an object is determined by its mass, specific heat and temperature.

j = ×m (4.3)

Although determining heat content is an important calculation, the knowledge of how heat may transfer from one object to another or within an object is of even greater practical value.

The common heat transfer modes are conduction, convection and radiation

4.4.1 Conductive Heat Transfer

Conduction is the mode of heat transfer in which the transfer of energy takes place at a molecular level. There are two commonly accepted theories:

• According to one theory, as molecules of a solid material attain additional thermal energy, they become more energetic and vibrate with increased amplitude of vibration while confined in their lattice. These vibrations are transmitted from one molecule to another without actual translatory motion of the molecules.

• The second theory states that conduction occurs at molecular level due to the drift of free electrons. These free electrons are prevalent in metals, and they carry thermal and electric energy.

From everyday experience, we know that heat transfer from outside to inside through wall of a chamber depends on:

• The surface area of the wall • The thermal properties of the construction materials • Wall thickness • Temperature difference


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Figure 4.7: Conductive heat flow in a wall

The rate of heat transfer through the wall may be expressed as

Ø ∝ #! $'&(# '' &&'(! c/$$ (4.4)

Or

Ø ∝ (4.5)

By introducing a constant of proportionality:

Ø = −: (4.6)

Where qx is the rate of heat flow in the direction of heat transfer by conduction (W); k is thermal conductivity

(W/[m°C]); A is area through which heat flows (m2); T is temperature (°C); and x is length (m). The equation above is called Fourier’s law. As heat flows from high temperature to low temperature, the gradient dT/dx is negative, because temperature decreases with increasing values of x.

4.4.2 Convective Heat Transfer

When a fluid comes into contact with a solid body such as the surface of a wall, heat exchange will occur between the solid and the fluid whenever there is a temperature difference between the two. During heating or cooling of gases and liquids, the fluid


60

stream exchanges heat with solid surfaces by convection. The magnitude of the fluid motion plays an important role in convection heat transfer.

Figure 4.7: Convective heat flow from the surface of a fl at plate

Depending on whether the flow of the fluid is artificially induced or natural, there are two types of convective heat transfer: forced convection and free (also called natural) convection

Consider heat transfer from a heated flat plate, PQRS, exposed to a flowing fluid, as shown in Figure 4.8. The temperature of the surface is Ts and temperature of the fluid is t∞. Because of the viscous properties of the fluid, a velocity profile is set up within the fl owing fluid, with the fluid velocity decreasing to zero at the solid surface. The rate of heat transfer from the solid to the fluid is proportional to the surface area A and the difference in temperature between the ts and t∞ or

Ø ∝ #ms − m∞( ⇒ Ø = ℎ#ms − m∞( (4.7) Where h is the convective heat transfer coefficient sometimes called surface heat-transfer coefficient.

4.4.3 Radiation Heat Transfer

Radiation heat transfer occurs between two surfaces by the emission and later absorption of electromagnetic waves (or photons). In contrast to conduction and convection, radiation requires no physical medium for its propagation—it can even occur in a perfect vacuum, moving at the speed of light, as we experience everyday solar radiation. Liquids are strong absorbers of radiation. Gases are transparent to radiation, except that some gases absorb radiation of a particular wavelength (for example, ozone absorbs ultraviolet radiation). Solids are opaque to thermal radiation. The exception to this is for microwaves and radio frequency radiation


61

4.5 Applications of heat transfer in food industry

The principles of heat transfer are widely used in food processing in many items of equipment. As we saw before heat transfer in food industry is done using heat exchangers. Deferent processes need heat transfer:

4.5.1 Sterilization

The object of sterilization is to destroy all microorganisms, that is, bacteria, yeasts and moulds, in the food material to prevent decomposition of the food, which makes it unattractive or inedible. Also, sterilization prevents any pathogenic (disease-producing) organisms from surviving and being eaten with the food. Some microorganisms are heat resistant than others, thus the heat required to kill them varies. Generally foodstuff are treated at 121oC for efficiently sterilize them.

Studies of the microorganisms that occur in foods, have led to the selection of certain types of bacteria as indicator organisms. A frequently used indicator organism is Clostridium botulinum. This particular organism is a very important food poisoning organism as it produces a deadly toxin and also its spores are amongst the most heat resistant. Thus, a heat process that can’t kill this microorganism is considered as inefficient. A combination of heat and time is used to inactivate the microorganisms efficiently

4.5.2 Pasteurization

Pasteurization is a heat treatment applied to foods, which is less drastic than sterilization, but which is sufficient to inactivate particular disease-producing organisms of importance in a specific foodstuff. The most common application is pasteurization of liquid milk.

4.5.3 Refrigeration: chilling and freezing

Rates of decay and of deterioration in foodstuffs depend on temperature. At suitable low temperatures, changes in the food can be reduced to economically acceptable levels. The growth and metabolism of micro-organisms is slowed down and if the temperature is low enough, growth of all microorganisms virtually ceases. To reach temperatures low enough for deterioration virtually to cease, most of the water in the food must be frozen. The effect of chilling is only to slow down deterioration changes.

• Chilling of foods is a process by which their temperature is reduced to the desired holding temperature just above the freezing point of food, usually in the region of -2 to 2°C. Many commercial chillers operate at higher temperatures, up to 10-12°C. The effect of chilling is only to slow down deterioration changes and the reactions are temperature dependent. So the time and temperature of holding the chilled food determine the storage life of the food. Rates of chilling are governed


62

by the laws of heat transfer which have been described in previous sections. It is an example of unsteady-state heat transfer by convection to the surface of the food and by conduction within the food itself. The medium of heat exchange is generally air, which extracts heat from the food and then gives it up to refrigerant. The rate of heat transfer gets bigger with increase of air velocity.

• Any process below that is freezing. Water makes up a substantial proportion of almost all foodstuffs and so freezing has a marked physical effect on the food. In the freezing process, added to chilling is the removal of the latent heat of freezing. This latent heat has to be removed from any water that is present. Since the latent heat of freezing of water is 335 kJ/kg, this represents the most substantial thermal quantity entering into the process. There may be other latent heats, for example the heats of solidification of fats which may be present, and heats of solution of salts, but these are of smaller magnitude than the latent heat of freezing of water.

Evaporation

63

Chapter V Evaporation

5. 1 Introduction

Historically, the most important unit operation in concentrating liquid foods has been evaporation. Solvent (usually water) is removed from the liquid food by boiling to concentrate the food solids to desired concentration. The requirements for an optimal evaporation include:

• Adequate heat transfer • Efficiency vapor liquid separation • Efficient energy use • Proper product treatment • Pressure at which the evaporation take place

Some examples of use of evaporation include:

• concentration of fruit juices, • evaporation of fluid milk products, • refining sugar from beets or canes, etc.

5.2 Evaporation in food processes

5.2.1 Boiling point elevation:

The presence of solute alters the vapor pressure of the solution and thus affects the equilibrium boiling point. Salts and, to a lesser extent, sugars are mainly responsible for increasing the boiling point of foods. Boiling point elevation is determined in several ways one of which is the use of Dühring rule. This rule states that the boiling point of a solution is a linear function of the boiling point of pure water at the same pressure. The linear relationship does not hold over a wide range of temperatures, but over moderate temperature ranges, it is quite acceptable. Fig. 5.1 shows Dühring lines for a sodium chloride–water system

Evaporation

64

Figure 5.1: Dühring lines illustrating the influence of solute concentrations on boiling point elevation of NaCl

5.2.2 Loss of volatiles

Most of foods contain flavor and aroma compounds that will be lost from the product as the water is evaporated. This is extremely important for the quality of final product. The volatiles must be put back in the product.

5.2.3 Thermal sensitivity of foods:

Some food components undergo thermally induced modifications during processing. Some negative effects include: denaturation of proteins, production of off colors and off flavors, vitamin degradation, etc. During evaporation, this modifications have to be minimized and design techniques have been developed for minimization of these effects during evaporation:

• Use of low pressure, low temperature evaporation • Minimization of the time under high temperature

5.2.4 Physical and chemical properties of foods

Several properties of foods must be accounted for designing an efficient evaporation system

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65

• Viscosity: sugars, starches, proteins at high concentration • Changes in density, thermal conductivity and heat capacity limit the heat transfer

5.2.4 Fouling during heat transfer:

Heat induced reactions in the food can cause formation of compounds that deposit on the heat transfer surface. These fouling layers limit the rate of heat transfer and reduce the capacity of the evaporator. Examples of foods that can cause fouling include milk.

5.2.5 Cleaning and sanitation:

Periodically, the evaporator must be shutdown and cleaned. The cleaning cycle depends on the rate of fouling.

5.3 Evaporator components

5.3.1 Heat exchange device (calandria):

Consists of a tube bundle inside which the product to be concentrated is evenly distributed as a thin film. Steam is condensed outside the tubes causing vaporization of the film of products inside the tubes.

5.3.2 Distribution device:

The wetting rate of an evaporator is the proportion of product feed to the heat transfer surface area. The distribution device must deliver the product to the tube bundle in an appropriate manner. The feed must be distributed evenly in different tubes.

5.3.3 Vapor liquid separator:

There are different types like gravity separator, baffle separator, centrifugal separator.

5.3.4 Condenser:

Residual vapors that will not be further utilized in the evaporation must be removed and condensed. A cooling medium is required and usually water is used but sometimes air or another refrigerant may be used. Condensation may be done in heat exchangers or by direct contact between the vapor and the cooling medium.

5.3.5 Vacuum production:

Vacuum is needed to lower the boiling point. This is partially accomplished by removal of condenser fluid, but usually a mechanical pump is used.

5.3.6 Pre heaters:

Evaporation

66

To ensure maximal efficiency in the evaporator, the feed needs to be preheated so that it enters the evaporator at or near the boiling point.

5.4 Single effect evaporator

The typical evaporator is made up of three functional sections: the heat exchanger, the evaporating section, where the liquid boils and evaporates, and the separator in which the vapor leaves the liquid and passes off to the condenser or to other equipment. In many evaporators, all three sections are contained in a single vertical cylinder. In the centre of the cylinder there is a steam heating section, with pipes passing through it in which the evaporating liquors rise. At the top of the cylinder, there are baffles, which allow the vapors to escape but check liquid droplets that may accompany the vapors from the liquid surface. Fig 5.2 shows a typical diagram of single effect evaporator

Figure 5.2: Schematic diagram of a single-effect evaporator.

5.5 Multiple effect evaporation

This design is used in order to reduce steam consumption. Vapors are used as primary steam in a subsequent evaporation.

Evaporation

67

For these vapors to provide a temperature difference, the operation pressure of subsequent evaporator is reduced resulting in a continually decreasing pressure and subsequent boiling temperature through this type of design. The great the number of effects, the greater steam economy but also higher the expenses of installation.

Figure 5.3: A two-effect evaporator with feed preheating, below is a tree effects evaporator

5.6 Thermal vapor recompression evaporator

In this process, high pressure steam passing through a nozzle draws vapors from the evaporator chamber. The resulting steam has higher energy content than the evaporator vapors and is then fed back into the steam chest to supply the energy for evaporation

Evaporation

68

Dehydration

69

Chapter VI Dehydration

6.1 Introduction

Drying is one of the oldest methods of preserving food. Dried foods can be stored for long periods without deterioration occurring. The principal reasons for this are that the microorganisms which cause food spoilage and decay are unable to grow and multiply in the absence of sufficient water and many of the enzymes which promote undesired changes in the chemical composition of the food cannot function without water. Preservation is the main purpose of drying, but in some cases drying is done in conjunction with other processes.

For example in the baking of bread, application of heat expands gases, changes the structure of the protein and starch and dries the loaf.

6.2 Heat and Mass Transfer

The removal of moisture from a food product involves simultaneous heat and mass transfer. Heat transfer occurs. Within the product structure and is related to the temperature gradient between the product surface and the water surface at some location within the product. As temperature rises, vapor is transported from the water inside the product to the surface of the product.

The heat and the mass transfer within the product structure occurs at the molecular level, with heat transfer being limited by thermal conductivity of the product structure, whereas mass transfer is proportional to the molecular diffusion of water vapor in air. At the product surface, simultaneous heat and mass transfer occurs but is controlled by convective processes.

6.3 Psychrometry

The psycrometrics are the study of the properties of gas-vapor mixtures. The mixture most commonly encountered in food processing is air-water vapor.

Absolute humidity or specific humidity H, is the weight of water present in the water air mixture per unit weight of dry air.

= bÚaÛb#aqaÛ( (6.1)

Where: Ma is molecular weight of air Mw is molar weight of water Pv is partial pressure of water vapor in the system P is total pressure of the system

Dehydration

70

Saturation humidity, Hs, is the absolute humidity of saturated air. In saturated air, the water vapor pressure equals the vapor pressure of water at air temperature.

$ = bÚaÚb#aqaÚ( (6.2)

Where Pw is vapor pressure of water at air temperature

Percentage absolute humidity Hp is the ratio of the absolute humidity to the saturation humidity at a given temperature.

a = 100 × aÛ#aqaÛ(aÚ#aqaÚ( (6.3)

It is expressed in terms of percentage.

Percentage relative humidity HR is the ratio of the partial pressure of water vapor to vapor pressure of water at a given temperature.

« = 100 × aÛaÚ

(6.4)

Dew point or saturation temperature is the temperature at which a give air water vapor mixture becomes saturated, if cooled at constant humidity.

The humid heat CH the amount of heat needed to increase by 1°C the temperature of 1 kg of dry air plus the quantity of water it contains.

Ü = + ¿ (6.5) Where CA= specific heat of dry air

Cv= specific heat of water vapor

Temperature volume VH is the volume of a mass of mixture containing a unit of mass of dry air at one atmosphere pressure and at a given temperature.

uÜ = ,,..,- #¡ + 273(

b+ Ü

bÚ (6.6)

Where θ is the temperature of the mixture

When H=0, VH= specific volume of dry air, H=Hs VH= specific volume of saturated air.

Adiabatic saturation temperature: if a stream of moist air passes concurrently through water sprays at temperature θs under adiabatic conditions, so that air leave the system in equilibrium with the water and at temperature θs, the temperature θs is said to be the

Dehydration

71

adiabatic saturation temperature. The line relating the temperature and humidity of the air during the adiabatic cooling is known as adiabatic cooling (or saturation) line. The equation of this line is

$⎯ = # Ü/$(#¡⎯¡$( (6.7) Wet bulb temperature, θw, is the steady state (dynamic equilibrium) temperature attained by a small quantity of water evaporating under adiabatic condition in an air stream. Wet bulb temperature is measured by means of temperature sensing element covered by a wick which is maintained saturated with water.

6.4 Psycrometric chart

The properties of air water vapor mixtures may be displayed on a chart at one atmosphere pressure. This chart shows the humidity as a function of temperature at varying degrees of saturation. This chart is very useful in determining the change in temperature and humidity. From the psychrometric charts different properties that we have seen earlier can be determined.

Figure 6.1: Diagram showing the use of a psychometric chart for determining properties of air-water mixtures

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72

The fig. 6.1 shows the way psychrometric chart is used. The figure A shows humidity, the figure B is for wet bulb temperature, C is for percentage relative humidity and the figure D shows how humidity and dew point are determined from wet and dry bulb temperature.

The psycrometric chart is useful when heating or cooling air, mixing of air, drying.

6.5 Stages of drying

During drying of wet solid in heated air, the air supplies the necessary sensible heat which is converted to latent heat for vaporization of the moisture and also acts as carrier gas for removal of the vapor from the vicinity of the evaporating surface. A number of stages are to be considered during drying process:

1. The first stage is called the settling down period during which the solid conditions come into equilibrium with the drying air. It is often a negligible fraction

2. Constant rate period of drying is a period during which the solid surface remains saturated with water. This is due to the fact that the movement of water within the solid to the surface takes place at the same rate as the evaporation from the surface.

3. Then comes the stage when the surface starts to dry out as the drying process proceeds. This stage is called the falling rate period. During this stage the temperature at the surface increases

6.6 Types of driers

Cabinet dryers:

These types of drying systems use trays or similar product holders to expose the product to heated air in an enclosed space. Air movement over the product surface is at relatively high velocities to ensure that heat and mass transfer will proceed in an efficient manner.

Tunnel Dryers:

In this type of design, products move on trays that circulate in a tunnel in which air is blown at high velocity either in the same direction as the product or in opposite direction.

Fluidized-Bed Drying:

In this system, the product pieces are suspended in the heated air throughout the time required for drying.

Spray drying:

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73

Spray drying is a process where a liquid droplet is rapidly dried as it comes in contact with a stream of hot air. The small size of the liquid droplets allows very rapid drying and the residence time of the material inside the spray drier is in the order of seconds. The dried material is separated from air in a cyclone separator. The dried material is continuously withdrawn and cooled.

Figure 6.2: Schematic illustration of a spray-drying system and a pressure nozzle

Freeze-Drying:

Freeze-drying is accomplished by reducing the product temperature so that most of the product moisture is in a solid state, and by decreasing the pressure around the product, sublimation of ice can be achieved. When product quality is an important factor for consumer acceptance, freeze-drying provides an alternative approach for moisture removal.

Sublimation is the direct transition from solid state to gaseous state without melting. Sublimation occurs at a definite range of temperatures and pressures, depending on the substance in question. The phase diagram of pure water (Figure 6.3) indicates that sublimation of water ice can occur only if the vapor pressure and temperature are below those of the triple point of water, i.e. below 611.73 Pa and 0.01°C respectively.

Theoretically, freeze drying should be possible at atmospheric pressure, provided that the partial pressure of water vapor is very low, i.e. the air is very dry. Atmospheric freeze-drying occurs in nature, for example when snow disappears without melting in cold, dry weather. In practice, however, freeze drying is carried out at very low total pressure (typically 10 to 50 Pa).

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74

The advantages of the freeze-drying process are superior product quality resulting from low temperature during sublimation and the maintenance of product structure. These advantages are balanced against the energy-intensive aspects of the product freezing and vacuum requirements.

Figure 6.3: Phase diagram of water showing conditions for sublimation

Mechanical separation

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Chapter VII Mechanical separation

7.1 Filtration

Filtration is defined as the unit operation in which the insoluble solid component of a solid liquid suspension is separated from the liquid component by passing the latter through a porous membrane or septum which retains the solid particles in the upstream surface. The solid liquid suspension is called feed slurry, the liquid component that passes through the membrane is called the filtrate and the membrane is called the filtering medium. The separated solids are known as the filter cake.

Filter media

Filter media may be rigid or flexible. Rigid filter media may be loose such as sand or gravel, diatomaceous earths or charcoal. There are also rigid fixed media like porous carbon, porcelain, fused alumina, perforated metal prate,…

Flexible media may consist of woven fabrics of cotton, silk, wool and jute. Many synthetic materials are finding application, including nylon, polypropylene, polythene, polyvinylchloride…

Different filtration equipments are known. These include:

Pressure filters where a pressure high than atmospheric pressure is maintained upstream of the medium to induce the flow of the filtrate. There may be vertical plate pressure filter and horizontal plate filters.

Vacuum filters where low pressure is maintained downstream of the medium. There are: continuous rotary drum vacuum filter, rotary vacuum disc filter.

7.2 Ultrafiltration

Ultrafiltration is widely employed in the increasingly important biotechnological industry for separation of fermentation products, particularly enzymes. Its largest commercial use is in the dairy industry for recovery of proteins from cheese whey and for pre-concentration of milk for cheese making. This process used membranes with a very small pore size but a bit larger than for reverse osmosis.

7.3 Reverse osmosis

It is well known that when a plant or an animal membrane separates two solutions of different concentrations, pure water passes through the membrane. The movement of water occurs from a solution with high concentration of water to a solution with low concentration of water. This process is called osmosis and it is driven by osmotic pressure. If a pressure higher than the osmotic pressure is applied to the side where the


76

concentration of water is higher, the process of osmosis is reversed. This process is called reverse osmosis.

7.4 Centrifugation

Centrifugation is a unit operation involving the separation of materials by the application of centrifugal force.

7.4.1 Separation of immiscible liquids:

The centrifugal force Fc acting on an object of mass m, rotating in a circular path of radius R at an angular velocity of ω is:

r = »Þ, If two liquids with different densities are placed in a cylinder rotating about a central axis, the more dense liquid will tend to move towards the wall thus separating from the other liquid.

Figure 7.1: Liquid separation in a centrifuge

7.4.2 Separation of insoluble solids from liquids:

This process is usually called centrifugal clarification where small quantities of insoluble solids are removed from a liquid by centrifugation.

7.4.3 Centrifugal filtration:

This tem describes another method of separating insoluble solids from liquids involving centrifugal force. However, in this case the mechanism of separation is filtration and the flow of the filtrate is induced by centrifugal force. In this case, the centrifugal bowl is leaned with filtration medium.


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7.5 Sieving:

This is a mode of mechanical separation where particles are separated base on their size. The separating medium is called a sieve. The material is shaken or agitated above a mesh or cloth screen; particles of smaller size than the mesh openings can pass through under the force of gravity.

The volume of material handled by the sieve depends on: • nature and the shape of the particles, • frequency and the amplitude of the shaking, • methods used to prevent sticking or bridging of particles in the apertures of the

sieve and • tension and physical nature of the sieve material.

Industrial sieves include rotary screens which are horizontal cylinders either perforated or covered with a screen, into which the material is fed. Other sieves are vibrating screen generally vibrated by eccentric weights.

7.6 Sedimentation

• Sedimentation uses gravitational forces to separate particulate material from fluid streams.

• The particles are usually solid, but they can be small liquid droplets, and the fluid can be either a liquid or a gas.

• Sedimentation is very often used in the food industry for separating dirt and debris from incoming raw material, crystals from their mother liquor and dust or product particles from air streams. In sedimentation, particles are falling from rest under the force of gravity.

Mixing

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Chapter VIII Mixing

Mixing is the dispersing of components, one throughout the other. It occurs in innumerable instances in the food industry and is probably the most commonly encountered of all process operations.

Mixing can be, particles mixing where small particles are mixed to make an homogeneous mixture, it can also be particle mixing in liquids it can also be about mixing of different liquids.

8.1 Mixing equipments:

Liquid Mixers:

The propeller mixer is probably the most common and the most used in liquid mixing.

Powder and Particle Mixers

The essential feature in these mixers is to displace parts of the mixture with respect to other parts.

Figure 8.1: Mixers (a) ribbon blender, (b) double-cone mixer

Dough and Paste Mixers

Dough and pastes are mixed in machines that have, of necessity, to be heavy and powerful. Because of the large power requirements, it is particularly desirable that these machines mix with reasonable efficiency, as the power is dissipated in the form of heat, which may cause substantial heating of the product.

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Figure 8.2: Example of a Kneader

Size reduction

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Chapter IX Size reduction

Raw materials often occur in sizes that are too large to be used and, therefore, they must be reduced in size. This size-reduction operation can be divided into two major categories depending on whether the material is a solid or a liquid. If it is solid, the operations are called grinding and cutting, if it is liquid, emulsification or atomization.

9.1 Grinding and cutting

These processes reduce the size of solid materials by mechanical action, dividing them into smaller particles. Perhaps the most extensive application of grinding in the food industry is in the milling of grains to make flour, but it is used in many other processes, such as in the grinding of corn for manufacture of corn starch, the grinding of sugar and the milling of dried foods, such as vegetables.

9.1.1 Grinding Equipments

Crushers:

In crusher, the material is fed in between two heavy jaws, one fixed and the other reciprocating, so as to work the material down into a narrower and narrower space, crushing it as it goes.

Figure 9.1: Crushers

Hammer mills:

In a hammer mill, swinging hammerheads are attached to a rotor that rotates at high speed inside a hardened casing

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Figure 9.2: Hammer mills

Roller mills:

Roller mills have smooth or finely fluted rolls, and rotate at differential speeds. They are used very widely to grind flour.

Figure 9.3: roller mills

Cutters

Cutting machinery is generally simple, consisting of rotating knives in various arrangements. A major problem often is to keep the knives sharp so that they cut rather than tear.

9.2 Emulsification

Emulsions are stable suspensions of one liquid in another, the liquids being immiscible. Stability of the emulsion is obtained by dispersion of very fine droplets of one liquid, called the disperse phase, through the other liquid, which is called the continuous phase.

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The stability of an emulsion is controlled by • interfacial surface forces, • size of the disperse phase droplets, • viscous properties of the continuous phase and • density difference between the two phases.

The terms emulsification and homogenizers are often used interchangeably for equipment used to produce emulsions. Emulsification is the formation of a stable emulsion by the intimate mixing of 2 or more immiscible liquids. Homogenization is the reduction in size and increase in number of solid or liquid particles of the dispersed phase, by the application of intense shearing forces, to increase the intimacy and stability of the tow substances. Homogenization is therefore a more severe operation than emulsification. Both operations are used to change the functional properties or eating quality of liquid foods. They have little or no effect on nutritional value or shelf life.

Equipments used in homogenization include high pressure homogenizers, high speed mixers, colloid mills, ultrasonic homogenizers …

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