FDE201 MASS AND ENERGY BALANCESProf.Dr. Mehmet ÖzkanProf.Dr.Ferruh ErdoğduDepartment of Food EngineeringAnkara UniversityPhone: 203 3300/3621e-mail: mozkan@ankara.edu.tr, ozkanm64@gmail.comOffice Hours: 09:00–11:00 (Thursday)

Teaching Assistant Fatmagül HamzaoğluDepartment of Food EngineeringAnkara UniversityPhone: 203 3300e-mail: fatma-gul1990@hotmail.com

CONTENTS

I. Units and dimensions

Definition of dimensions

Systems of measurements

SI system

Conversion of units

Dimensional consistency of equations

II. Some basic physical properties and their units in various systems

Concentration

Density

Temperature

Pressure

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III. Mass balances

Basic principles

Process flow diagrams

Total mass balance

Component mass balance

Mass balance problems involved in fruit juice/nectar preparations

Mass balance problems involved in jam/marmalade preparations

Mass balance problems involved in crystallization, dilution, concentration, dehydration and filtration

IV. Energy balances

Basic principles

Energy terms (Enthalpy, determination of specific heat of solids, liquids and gaseous)

Properties of saturated and superheated steams

Energy balance problems involved in various food applications

Suggested references

1) Toledo RT. 1994. Fundamentals of Food Process Engineering. 2nd ed., Chapman & Hall, New York, NY.

Chapter 2: Units and dimensions, p.51–65.Chapter 3: Material balances, p.66–108.Chapter 5: Energy balances, p.132–159.

2) Özkan M, Cemeroğlu B and Türkyılmaz M. 2011. Gıda Mühendisliğinde Kütle ve Enerji Denklikleri, 264 s., Gıda Teknolojisi Derneği Yayınları No: 43, Bizim Grup Basımevi, Ankara.

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1) UNITS AND DIMENSIONS

Dimension: A physical quantity, which can be observed or measured, is defined qualitatively

by a dimension. For example; time, length, area, volume, mass, force, temperature and

energy are all considered dimensions. Some dimensions are considered as primary and some

are considered as secondary. Primary dimensions vary from one unit system to another.

Time, length, mass and temperature are considered as primary dimensions. Primary

dimensions are also called as base unit. Secondary dimensions are a combination of various

dimensions. For example, volume is length cubed, velocity is distance divided by time and

force includes the dimensions of mass, length and time.

Unit: The quantitative magnitude of a dimension is expressed by a unit. For example, a unit

of length may be measured as a meter, centimeter or a millimeter and a unit of mass may be

measured as a kilogram, gram or a microgram.

Engineering units

Physical quantities are measured using various systems. The most common unit systems are:

The Imperial (English) system (English Engineering System, ees)

Centimeter, gram, second (cgs) system

Meter, kilogram, second (mks) system

English system is used primarily by American and British chemical and food industries.

Outside the United States and Britain, industry has adopted the mks system, while the

sciences have adopted cgs system. The confusion that resulted from the use of various units

led to the development of a common system of units. As a result of international agreement

in 1960 by the General Conference on Weights and Measures, the “ Systeme International

d’Unities” or the SI units, have emerged. SI is designed to meet the needs of both science

and industry. Although the use of SI units is gaining momentum, it is still necessary to

convert data from one system to another. The various systems in use are shown in Table 1.

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Table 1–Systems of measurements

System Use Length Mass Time Temperature Force Energy

English Industrial Foot Pound mass

Second °F Pound force

BTU

Metric

Cgs Scientific Centimeter Gram Second °C Dyne Calorie

Mks Industrial Meter Kilogram Second °C Kilogram force

Kilo calorie

SI Universal Meter Kilogram Second K Newton Joule

The SI system is based on 7 units. These units, along with their symbols are summarized in

Table 2.

Table 2–Examples of SI-derived units expressed in terms of base units

Quantity Name Symbol

Length meter m

Mass kilogram kg

Time second s

Electric current ampere A

Temperature kelvin K

Amount of substance mole mol

Luminous intensity candela cd

Derived units are algebraic combinations of base units expressed by means of multiplication

and division. Definitions of some commonly used derived units are as follows:

Newton (N) : The force that gives to a mass of 1 kg an acceleration of 1 m/s2.

(1 kg’lık kütleye m/s2 ivme kazandıran kuvvete 1 newton denir.)

1 (N) = 1 (kg) x 1 (m s–2)

4

Joule (J) : The work done when a force of 1 N is displaced by a distance of 1 m in the

direction of force.

(1 N’luk kuvvetin kendi doğrultusunda 1 m yol almasıyla yapılan işe, 1 joule denir.)

1 (J) = 1 (N) x 1 (m)

kg m = –––––– x m s2

= kg m2 s–2 (Joule)

Heat, energy and work are all in the same dimension.

Watt (W) : The power that gives rise to the production of energy at the rate of 1 J/s.

(Birim zamanda (s) yapılan işe (J) güç ya da birimi ile watt denir.)

J N m kg m m kg m2

1 (W) = –––– = –––––– = –––––– x –––– = ––––––– s s s2 s s3

Electrical power unit 1 (W) = 1 (A) x 1 (V) equals to the heat and work units 1 (W) = 1 (J/s)

Pascal (Pa) : The force exerted per unit area is called pressure.

(Birim alana etki eden kuvvete basınç denir)

N kg m s–2 kg1 (Pa) = ––––– = –––––––– = –––––– (Pascal) m2 m2 m s2

Since the value of Pascal is very small, kpa (1000 times larger) or bar (105 times larger) units

are used.

1 bar = 105 Pa

Example of SI-derived units expressed in terms of base units, SI-derived units with special

names, and SI-derived units expressed by means of special names are given in Table 3, 4 and

5, respectively.

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Table 3–Examples of SI-derived units expressed in terms of base units

SI unit

Quantity Name Symbol

Area Square meter m2

Volume Cubic meter M3

Speed (velocity) Meter per second m/s

Acceleration Meter per square second m/s2

Density Kilogram per cubic meter kg/m3

Concentration Mole per cubic meter mol/m3

Specific volume Cubic meter per kilogram m3/kg

Table 4–Examples of SI-derived units with special names

Quantity Name Symbol Expression in terms of other units

Expression in terms of SI base

unitsForce newton N m kg s–2

Pressure Pascal Pa N m–2 kg m–1 s–2

Energy, work,

heat

joule J N m kg m2 s–2

Power watt W J s–1 kg m2 s–3

Table 5–Examples of SI-derived units expressed by means of special names

Quantity Name Symbol Expression in terms of SI base units

Viscosity Pascal second Pa s kg m–1 s–1

Heat capacity Joule per kelvin J K–1 kg m2 s–2 K–1

Specific heat capacity Joule per kilogram

kelvin

J kg–1 K–1 m2 s–2 K–1

Thermal conductivity Watt per meter kelvin W m–1 K–1 m kg s–3 K–1

Prefixes Recommended for use in SI system

Prefixes for general use are shown in Table 6.

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Table 6–Prefixes recommended for use in SI

Prefix Multiple Symbol

tera 1012 T

giga 109 G

mega 106 M

kilo 1000 k

hekto 102 h

deka 101 da

deci 10–1 d

centi 10–2 c

mili 10–3 m

micro 10–6 μ

nano 10–9 η

pico 10–12 p

femto 10–15 f

Symbols for the prefixes are written in capital letters when the multiplying factor is 10 6 and larger. Prefixes designating multiplying factor less than 106 are written in lower case letters.

1 g = 103 mg = 106 μg = 109 ng = 1012 pg

A dimension expressed as a numerical quantity and a unit must be such that the numerical

quantity is between 0.1 and 1000.

Examples:

1) 10,000 cm should be 100 m.

2) 0.0000001 m should be 1 μm.

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3) 10,000 Pa should be 10 kPa.

Dimensional equation

The magnitude of a numerical quantity should be written both the number and its unit. An

equation that contains both numerals and their units is called a dimensional equation. The

units in a dimensional equation are treated just like algebraic terms. All mathematical

operations applied on the numerals must also be applied on their corresponding units.

Examples:

(4 m)2 = 16 m2

J J . kg . K 5 [––––––] (10 kg) (5 K) = 5 (10) (5) [–––––––––] = 250 J

kg . K kg . K

Addition and subtraction of numerals and their units also follow the rules of algebra, i.e., only

same terms can be added or subtracted. Thus, 5 m – 3 m = (5 – 3) m = 2 m, but 5 m– 3 cm

cannot be simplified unless their units are expressed in the same units.

Equations must be dimensionally consistent. Thus, if the dimension of the left-hand side of

an equation is “length,” the dimension of the right-hand side must also be “length;”

otherwise, the equation is incorrect.

Conversion of units

During unit conversion, first prepare an equation containing, the unit being converted

(çevrilecek birimi)”, the unit of final answer (ulaşılmak istenen birimi) and conversion factors

(çevirme katsayıları). For “conversion factor,“ we need a unit conversion table given in

Appendix Table A.1.

Steps in conversion of units

a) Place the units of the final answer on the left-hand side of the equation and the number

being converted and its unit are the first entry on the right hand side of equation.

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Example 1: Thermal conductivity of stainless steel in EES is 10 BTU/h ft °F. Convert

this value in EES to SI unit system (J/m s K).

First we need to prepare an equation, which is given below:

(CF) [J/(m s K)] = BTU/(ft h °F) (CU)

where;

CF : Conversion factor (çevirme katsayısı) which will be calculated,

CU : Conversion units from conversion table.

b) Then, put an equal sign between two expressions.

c) Conversion factors are found from conversion tables for the units to be converted in

the right hand side of equation. In this example, conversion units were found from

table and given below:

1 BTU = 1055 J,

1 ft = 0.3048 m,

1°C = 1.8°F (where, temperature difference occurs),

1°C = 1 K,

1 h = 3600 s.

d) Set up the conversion factors as a ratio, using Appendix Table 1. In this example, the

rations were given below:

1055 J/BTU, ft/0.3048 m, 1.8°F/°C and h/3600 s

e) Sequentially multiply the conversion factors such that the original units are

systematically eliminated by cancellation replacement with the desired unit.

J BTU 1055 J ft 1.8°F °C h (CF) – = – – – ––––– m s K h ft °F BTU 0.3048 m °C K 3600 s

J 1055 BTU J ft °F h J

9

(CF) – = –– – = 1.73 m s K 0.3048 x 5/9 x 360 h ft °F BTU m K s m s K

f) After cancellation of units in the right hand side of equation, appropriate conversion

factor is calculated. After cancellation, the numerical value for conversion factor (CF)

is 1.73. The conversion factor calculated is put in the equation:

1 BTU/(h ft °F) = (1.73) J/(m s K)

g) The numerical value in front of the unit converted is now taken into consideration.

This numerical value is put on both side of equation.

10 BTU/(h ft °F) = 10 (1.73) J/(m s K)

Result : 10 BTU/(h ft °F) = 17.3 J/(m s K)

Example 2: Specific heat of orange juice concentrate with 45% water soluble solid content is

cp= 0.64 BTU/ lbm °F. Express this value in SI unit system.

Example 3: Heat transfer coefficient of salami at 21°C is hs = 210 BTU/ft2 h °F. Express this

value in SI unit system.

10

Example 4: The density of cow milk is 64.5 lbm/ft. Express this value in SI unit system.

Example 5: The enthalpy of veal containing 70% water at –20°C is 5257 x 10–2 BTU/lbm.

Express this value in SI unit system

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Example 6: For a fluid passing through a pipe, type of flow (laminer, kararsız, turbulant)

depends on the pressure (ρ V2) ve viscosity forces (µV/D) of fluid. And, the ratio of these

values gives Reynolds number. Reynolds number is used to determine the type of flow. For

example; if a Reynolds number is above 4000, then the flow will be turbulant. The equation

defining Reynolds number is given below:

ρ V2 ρ V D Re = –––––– = –––––– µ V/D µ

where;D : diameter, m,

V : velocity, m/s, ρ : density, kg/m3, μ : viscosity, kg/m s.

In a milk processing plant, the type of flow of milk is found to be turbulant (Re=50000). The

velocity (V) of milk flowing in a diameter (D) of 1 in. of a pipe is 13.5 ft/s and the density (ρ)

at 294 K is 64.3 lbm/ft3. Determine the viscosity (µ) of milk at 294 K in:

a) SI unit system,

b) cgs unit system.

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Example 7 : A tube is filled with a fruit juice with the height of h cm and density of ρ. The

pressure exerted to the base of the tube by this fruit juice is P atmosphere. Calculate the

density of fluid in the SI system.

13

Example 8: In a milk processing plant, once the milk is brought to the plant, first, the milk is

placed in a pre-storage tank and then centrifuged to remove somatic cells, leucosits, blood

coagulates and some microorganisms (clarification process). During this clarification process,

the milk is pumped to the centrifuge which is 5 m higher than the storage tank at a velocity of

120 kg/min. During transportation, the friction loss in pipes is 45 J/kg. Calculate the

necessary power of pump is “SI” unit system.

Example 9: The heat loss through the walls of an electrical oven is 6500 BTU/h. If the oven

is operated for 2 h, how many kilowatt-hours of electricity will be used just to maintain the

oven temperature (heat input = heat loss)?

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Example 10: The height of a pomegranate juice with a density of 1.068 g/cm3 in mercury

column is 8.325 in. Calculate the pressure applied by pomegranate juice to mercury column

in the SI system.

Length units in English unit system

1 in = 2.54 cm

1 foot = 12 in

3 feet = 1 yard

3.28 feet = 1 m

Volume units in English unit system

1 gal = 3.79 L

1 gal = 4 quarts

1 quart = 2 pints

15

1 pint = 16 fluid ounce (fl oz)

1 quart = 32 fluid ounce

Example 11: Calculate your height in “feet” and “in.” (173 cm)

Example 12: Calculate the volume of a car tank with the capacity of 40 L in “gal” and pint.

Example 13: Convert “BTU / (lb °F)” to “J / (g K).”

Example 14: Convert the following units to SI units.

16

a) A density value of 60 lbm/ft3.

b) A pressure value of 14.69 psig.

Example 15: Express the viscosity values given below in the unit asked.

Viscosity is the resistance of fluids to movement (Sıvıların akışa direncini gösteren fiziksel

özelliklerinden birisidir.). It is expressed as:

g cgs → ––––– = poise

cm s

kg SI → –––––

17

m s

lbm

İngiliz → ––––– ft s

a) Express 20 cp in “Pa s.”

b) Express 15 cp in “EES.”

18

c) Express “30 lbm/(ft h)” in “SI” unit system.

Example 16: Pressure is expressed in the SI system in Pascal’s. Calculate the English

equivalent of 8 Pa.

19

Example 17: Calculate the power required for peach nectar which flows down the raceway

of a reservoir at a rate of 525 lbm/min from a height of 12.3 ft.

Relationship between lbm and lbf

In the English Engineering System of measurement, the units of force and the units of mass

are both expressed in pounds. To eliminate confusion between these units, the pound force is

written as lbf and the pound mass is written as lbm.

Force = mass x acceleration

Substituting lbf for force, lbm for mass and ft/s2 for acceleration results in a dimensional

equation for force that is not dimensionally consistent.

ftlbf = lbm x ––––

20

s2

To make this equation dimensionally consistent, we need to add a dimensional constant (gc).

ftlbf = lbm x –––– x gc

s2

To make lbf numerically equal to lbm when acceleration due to gravity is 32.174 ft/s2, it is

obvious that the numerical value of gc will be 32.174 and gc will be a denominator in the force

equation.

ft 1

lbf = lbm x –––– x –––– s2 gc

ft lbm

gc = ––––––– lbf s2

ft lbf = lbm x 32.174 ––––

s2

Example 18: Pressure is expressed in the SI system in Pascal’s. Calculate the English

equivalent of 8 Pa by not using “lbf = 4.44823 N” conversion.

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Reliability of Analyses

The reliability of an analytical method depends on its 1) accuracy, 2) precision, and 3)

sensitivity

Accuracy

Accuracy of an analytical method is defined as the deviation from the ideal value (known

value). To test the accuracy of measurement, the mean of a number of determination is

compared with a known value. Accuracy depends on proper calibration of an instrument.

∑ (x – “known value”)2

√ –––––––––––––––––––– (n – 1)

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It is given as “known value ± precision value.” (7.02 ± 0.03)

Precision (synonymous with reproducibility)

Precision of an analytical method is defined as the deviation of the measurements from mean.

Precision is equal to standard deviation

∑ (x – “arithmetic mean”)2

√ –––––––––––––––––––––– (n – 1)

Both accuracy and precision are often expressed as a plus or minus term (“arithmetic mean ±

precision value”). (7.02 ± 0.03)

Example 12: Two laboratory technicians were pipetted distilled water in a beaker with 5-mL

pipettes and weighed out distilled water on an analytical balance (±0.1 mg).

a) Which technician’s measurements are more accurate?

b) Which technician’s measurements are more precise?

Technician A Technician B

4.8000 g 5.0010 g

4.8000 g 4.9920 g

4.8000 g 5.0010 g

HOMEWORK

1. Convert 25 lbm/gal into lbm/ft3. (Answer : 187)

2. Convert 7 g/cm3 into lbm/gal. (Answer : 58.5)

3. Convert 16 cal/s into W. (Answer : 67)

4. The amount of heat required to change the temperature of a material from T1 to T2 is

given:

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q = m Cp (T2 – T1)

where q = BTU, m = mass of material in lbm, Cp = specific heat of material in BTU / (lbm °F)

and T1 and T2 = initial and final temperatures in °F.

a) How many BTUs heat are required to cook a roast weighing 10 lbm from 40°F to

130°F? Cp for meat = 0.8 cal / (g °C). (Answer : 720)

b) Convert the number of BTUs of heat into watt-hours. (Answer : 211)

c) If this roast is heated in a microwave oven having an output of 200 watts, how long

will it take to cook the roast? (Answer : 1.05)

5. How many kilowatt-hours of electricity will be required to heat 100 gal of water from

60°F to 100°F? (Answer : 9.8)

gdwater = 1 –––––

cm3

kcalCpwater = 1 ––––––

kg °C

6. Calculate the power requirements for an electric heater necessary to heat 10 gal of water

from 70°F to 212°F in 10 min. Express this in J/min and in kilowatts.

(Answer : 1.25 x 106 and 20.9)

2- SOME IMPORTANT PHYSICAL PROPERTIES Some of the important physical properties of foods are density, specific heat, thermal

conductivity and viscosity. The physical properties of foods change in certain limits

depending on the chemical and physical properties of foods. As the chemical composition of

foods change as result of many factors, physical properties of foods also change depending on

the chemical composition. Therefore, there would be different values in the literature for a

given physical property.

The most accurate way to determine the physical properties of foods is to determine the

physical property experimentally. However, the physical properties of foods can also be

determined by using equations developed for this purpose. Some of these equations were

24

given in below. In literature, the physical property values of foods are frequently given by

both calculating from equations and determining experimentally.

2.1 Concentration

Concentration is a measure of the amount of dissolved solute per unit of volume or weight.

Terms used to express the concentration are:

Molarity

Normality

Molality

ppm

ppb

Mass fraction and mass ratio

Mole fraction and mole ratio

ppm, ppb, mass and mole fractions are not only used to express the concentration of solutions,

but also used in solid materials.

I) Concentrations based on volume

a) Molarity or molar concentration (M) : The number of moles of solute per liter of

solution.

The unit for molarity is “mole/L.” And, this unit is called as “Molar (M).” To calculate the

molarity, we need to know the weight of dissolved solute and its molecular weight. The

following equations (2.1 and 2.2) were used to calculate the molarity. The equation of 2.1 is

used to calculate the mole number (n) of solutes in solution and the equation of 2.2 is used for

molarity (M) of the solution.

WMole number (n) = ––––––––– (2.1)

Mw (Aw)

nMolarity (M) = ––––– (2.2)

25

V

where;

W: the weight of solutes, g

Mw (Aw) : Molecular or atomic weight, g/g-mole

V: Volume of solution, L

n: mole number.

Dilute solutions are often expressed in terms of millimolarity, micromolarity, and so on, where:

1 mmole = 10–3 moles

1 μmole = 10–6 moles

1 nmole = 10–9 moles

1 pmole = 10–12 moles

Therefore :

1 mM = 10–3 M = 1 mmole/L

1 μM = 10–6 M = 1 μmole/L

1 nM = 10–9 M = 1 nmole/L

1 pM = 10–12 M = 1 pmole/L

b) Normality (N) : The number of equivalents of solute per liter of solution. To calculate the

normality, we need to know the weight of dissolved solute and its equivalent weight.

The following equations (2.3, 2.4 and 2.5) are used to calculate the normality o f a solution.

W Equivalents = –––––––––––––––– (2.3) Equivalent weight

Mw Equivalent weight = –––– (2.4) f

Equivalent weightN = –––––––––––––––– (2.5)

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V

where;

W : the weight of solutes, g

V : Volume, L

f: the equivalence factor; the number of replaceable H+ or OH– per molecule (for acids and bases) or the number of electrons lost or gained per molecule (for oxidizing and reducing agents)

f: NaOH = 1

f: H2SO4 = 2

f: H3PO4 = 3

The relation between normality and molarity are given by the equation of 2.6.

W/(Mw/f) n (f)N = ––––––––– = –––––– = M (f)

V V

N = M (T.D.) (2.6)

For example, a 0.01 M solution of H2SO4 is 0.02 N.

c) Molality (m) : The molality is defined as the number of moles of solute per 1000 g of

solvent.

It is calculated from the equation of 2.7. The unit of molality is “molal” and shown as “m.”

For dilute aqueous solutions, m and M will be quite close.

n (mole number)m = ––––––––––––––– (2.7)

1000 g

Example 1: Calculate the molality of concentrated stock HCL solution (%28).

Solution:

27

d) parts per million (ppm) : “ppm” means “parts per million.” For example, 1 ppm means 1

mg of solute in 1 million mg of solution or food. Since 1 million mg is equal to 1 kg, most

commonly ppm means the weight in “mg’s of a solute per 1 kg of solution or 1 kg of material

(food). This concentration unit is used dilute aqueous solutions. Ppm can also be expressed

with the following terms:

1 g = 103 mg = 106 μg = 109 ng = 1012 pg

mg ng pg μgppm = –––– = –––– = –––– = ––––

kg mg μg g

Example 2 : Calculate the concentration of Ca++ ions in water in ppm.

Solution:

a) 1 mg Ca++ ions in 1 L water.

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b) 1 mg Ca++ ions in 100 mL water.

c) 1 mg Ca++ ions in 80 mL water.

e) parts per billion (ppb) : “ppb” is commonly expressed as the weight in μg of a solute per 1

kg of solution. This concentration unit is also used for dilute aqueous solutions (typical

example is aflatoxins in. In other word, “ppb” means “parts per billion.” For example, 1 ppb

means 1 μg of solution per 1 billion μg. Since 1 billion μg is equal to 1 kg, most commonly

ppb means the weight in “μg”s of a solute per 1 kg of solution or 1 kg of material. This

concentration unit is also used for dilute aqueous solutions (typical example is aflatoxins

formed by molds in dried foods). ppb can also be expressed with the following terms:

μg pg ng ppb = –––– = –––– = ––––

kg mg g

Example 3 : Calculate the concentration of Ca++ ions in water in ppm.

a) 1 mg Ca++ ions in 1 L water.

b) 1 mg Ca++ ions in 100 mL water.

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c) 1 mg Ca++ ions in 80 mL water.

f) Weight / Volume Percent (%w/v) : The weight in “g” of a solute per 100 mL of solution.

For example, 10% (w/v) sugar solution means 10 g of sugar per 100 mL of sugar solution.

II) Concentrations based on weight

a) Weight / Weight Percent (%w/w) : The weight in “g” of a solute per 100 g of solution.

The concentrations of many commercial acids (HCl, H2SO4) are given in terms of % w/w. In

order to calculate the volume of the stock solution required for a given preparation, we need

to know the density of stock solution.

Example 4 : Describe the preparation of 2 L of 0.4 M HCl starting with a concentrated HCl

solution (28% w/w, ρ=1.15 g/cm3).

Solution:

30

P.S. Add acid on water. Never add water on acid.

b) Degrees brix (°Bx) : Degrees Brix is the water soluble solutes per 100 g of an aqueous

solution (food). One degree Brix is 1 gram of water soluble solids in 100 grams of solution

and represents the strength of the solution as percentage by weight (% w/w) (strictly speaking,

by mass). The most important water soluble solutes in foods are sugars, acids and salts which

are not volatile. The °Bx gives the dissolved solid content. The °Bx is traditionally used in

the wine, sugar, fruit juice, and honey industries. The °Bx is calculated from the equation

given in 2.8.

Mass of water soluble solutes (g)Brix = –––––––––––––––––––––––––––– (2.8)

100 g of solution (food)

Example 5: In 74 g of pomegranate juice, there are 11.65 g of sugar, 1.05 g of organic acid

and 0.17 g of salts of organic acids. Find out the brix of pomegranate juice.

Solution:

31

P.S.: The brix of fluid or fluid foods is measured by refractometer. This equipment is

calibrated to sucrose solutions. For example, the refractometer shows 10% (w/w) of sucrose

solution exactly as 10%. In example 5, the brix of pomegranate juice containing the water

soluble sugars, organic acids and the salts of organic acids at a concentration of 17.39 g/100 g

will be close to this value (17.39).

III) Dimensionless concentration forms

a) Mass fraction and mass ratio:

Mass fraction (XA) of solution or material (A) consisting of two components is calculated

from the equation 2.8.

weight of solute in solutionMass fraction (XA) = ––––––––––––––––––––––––

weight of solution

WA XA = –––––––––– (2.9)

WA + WB

If the third component (C) was dissolved in the same solution, the mass ratio of first

component to the third component is calculated from the equation 2.9.

weight of solute in solutionMass ratio = –––––––––––––––––––––––––––––

weight of another solute in solution

WA

Mass ratio = ––––– (2.9) WC

where,

WA : The mass of soluble solute A, g

WB : The mass of solvent B, g

WC : The mass of soluble solute C, g

32

b) Mole fraction and mole ratio:

Mole fraction (YA) of A component in the solution or material consisting of A and B can be

calculated from the equation 2.10.

the number of moles of solute in solutionMole fraction (YA) = ––––––––––––––––––––––––––––––––––––

total number of moles in solution

nA

Mole fraction (YA) = ––––––– (2.10) nA + nB

If the third component (C) was dissolved in the same solution, the mass fraction of first

component to the third component is calculated from the equation 2.11.

the number of moles of solute in solutionMole ratio (YA) = ––––––––––––––––––––––––––––––––––––––––

the number of moles of another solute in solution

nA

YA = ––– (2.11) nC

where;

nA : The mole number of soluble solute A, mole

nB : The mole number of solvent B, mole

nC : The mole number of soluble solute C, mole

Example 6 : Calculate the mole fraction of HCl in concentrated stock HCL solution (%28).

Solution :

33

Example 7: Calculate the mass and mole fractions of the air with the following composition.

Solution:

Mean molecular weight (of gas mixtures): Mean molecular weight of gas mixtures is

calculated from the equations 2.12 and 2.13. If the mass fractions are used, then the equation

2.12 is used. If the mole fractions are used, then the equation 2.132 is used.

–––MW = YA MWA + YB MWB + ……………. + Yi MWi = ∑ Yi MWi (2.12)

1 XA XB Xi Xi

–––– = –––––– + –––––– + ………………… + –––––– = ∑ ––––– (2.13) ––– MW MWA MWB MWi MWi

Example 8 : Calculate the mean molecular weight of air by using:

a) Mass fraction

b) Mole fraction

Solution:

34

Example 9: The density of sugar solution prepared by dissolving 43 kg sucrose in 100 kg

water is 1127 kg/m3. Calculate the following terms;

a) Mass fraction of sugar,

b) Mass/volume fraction of sugar in sugar syrup,

c) Mole fraction of sucrose,

d) Molal concentration.

Solution:

35

HOMEWORK

1) Calculate the mean molecular weight of the gas mixture with the following composition.

15% oxygen (O2)

65% nitrogen (N2)

10% sulfur dioxide (SO2)

65% carbon monoxide (CO)

2) Develop a spreadsheet on a computer to calculate concentration units for a sugar solution.

The sugar solution is prepared by dissolving 10 kg of sucrose in 90 kg of water. The density

of this sugar solution is 1040 kg/m3. Determine:

a) Concentration, weight per unit weight (w/w)

b) Concentration, weight per unit volume (w/v, kg/L)

36

c) °Brix

d) molarity

e) mole fraction

f) molality

g) Using the spreadsheet, recalculate the concentration values from the spreadsheet you

formed if a) sucrose solution contains 20 kg of sucrose in 80 kg of water, and the density

of the solution is 1,083 kg/m3, b) sucrose solution contains 30 kg of sucrose in 70 kg of

water, and the density of the solution is 1,129 kg/m3.

2.2. Density

Density is defined as mass per unit of volume, with dimensions (mass)/(length)3.

mρ = –––– (2.14) V

where;

ρ : Density, kg/m3,

m : The mass of the substance, kg

V : The volume of the substance, m3.

The unit of density in various systems is given in Table 2.1.

37

Table 2.1. Examples of SI-derived units expressed in terms of base units

System Unit

SI kg/m3

English lbm/ft3

mks kg/L

cgs g/cm3 (g/mL)

The volume of almost all of the substances changes depending on the temperature. This

causes to change in the density depending on the temperature. The volume of all substances

increases with temperature. Only exception for this rule is water. Increasing the temperature

of water from 0°C to 3.98oC will increase the density of water. However, after 3.98oC, the

increasing temperature of water will result in the decrease in density. While the density of a

substance is given, the temperature should be specified due to the effect of temperature on the

volume. This is especially important in gases whose volumes significantly change with

temperature. The densities of gases are given on the basis of 0°C and 760 mm Hg which are

defined as standard conditions.

In gases;

As the temperature increases, the pressure decreases.

As the pressure decreases, the volume increases and thus the density decreases.

For gases: If height↑, then pressure ↓; If pressure ↓, then volume ↑; and then density ↓

P.S. : 1. The density of pure substances is always the same at constant temperature and

pressure.

2. Substances which had different density at the same temperature and pressure are

absolutely different substances.

Example 10 : Calculate the density at 12°C of pure water by using Table 2.2, according to :

a) SI unit system,

b) EES.

Table 2.2 Relation between temperature and density of pure water

Temperature (°C) ρ (kg/m3)

38

0 0.99987

3.98 1.00000

5 0.99999

10 0.99973

15 0.99913

40 0.99224

80 0.97183

100 0.95838

P.S.: As given in Table 2.2, the density at 3.98°C of pure water which contains no air is ρmaks= 1 g/cm3.

Solution :

For this purpose, Table 2.2 which indicates the changes in the density of pure water

depending on temperature is used. However, there was no data at 12°C in temperature column

of Table 2.2. In this case, interpolation process is needed.

Interpolation is a method for the calculation of a value which is not found in a table by taking

into consideration of changing trend of values.

Interpolation is carried out with a three-step procedure:

1) In the temperature column, while the temperature increases from 10 to 15, i.e., 5 unit

increase in temperature, the density decreases 6 x 10–4 (0.99973 – 0.99913).

2) Meanwhile, while the temperature increases from 10 to 12, i.e., 2 unit increase in

temperature, find out the decrease in density.

3)

5 unit increase in temperature 6 x 10–4 unit decrease in density

2 unit increase in temperature x––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

x = 2.4 x 10–4 unit decrease

ρ12°C = 0.99973 – 2.4 x 10–4

ρ12°C = 0.99949 g/cm3

a) In SI unit system, the unit of density is kg/m3.

39

b) In EES, the unit of density is lbm/ft3.

2.2.1 Types of densities

The only way to determine the density of foodstuffs is experiment. Developed equations for

this purpose have not given true result. The densities of foodstuffs are depending on their

compositions, especially water and water-soluble solid contents. For example, there are strong

relation between the density of milk with content and type of the dissolved colloidal

substances in it. Thus, the density increases or decreases depending on these factors. While

increase in fat content of milk leads to decrease in the density of milk, increases in contents of

protein, sugar and salt lead to increase the density of milk. However, other factors also play a

role on the density of foodstuffs. For example, intercellular and other spaces in fruits and

vegetables make a significant impact on their densities. For this reason, the densities of fruits

and vegetables are less than that of water, although it is expected that the densities of fruits

and vegetables are higher than that of water. The density of foods is expressed with three

different density concepts. These are:

Solid density

Particle density

Bulk density

40

The values of these densities depend on how the pore spaces in a food material are

considered.

2.2.1.1. Solid density : Solid density is calculated from the equation 2.15. Disregards all the

pore spaces (both the spaces in the particles and between particles).

Mass of a particleSolid density = –––––––––––––––––––––––––––– (2.15)

Volume (disregarding all space)

The solid density of most food particles is 1.4–1.6 g/cm3 (or 1400–1600 kg/m3), except for

water, fat and salt. Solid densities of major food ingredients are given in Table 2.2. If the

type of the density is not specified, then the given density is taken as solid density.

Table 2.2. Solid densities of major food ingredients

Ingredient g/cm3 Ingredient g/cm3

Glucose 1.56 Citric acid 1.54

Sucrose 1.59 Fat 0.90

Starch 1.50 Salt 2.16

Cellulose 1.27–1.61 Water 1.00

Protein 1.4 Ethyl alcohol 0.79

2.2.1.2. Particle density : Particle density is calculated from the equation 2.16. Accounts for

the presence of internal pores in the food particles, but disregards the pores between particles.

Mass of a particleParticle density = –––––––––––––––––––––––––––––––––––––––––– (2.16)

Volume (including the internal pores in particles, disregarding the pores between particles)

Particle density is defined as a ratio of the actual mass of a particle to its actual volume.

2.2.1.3. Bulk density : Bulk density is calculated from the equation 2.17. This measurement

accounts for the all void spaces. Accounts for the presence of internal pores in the food

particles and the void spaces between the particles.

41

Mass of a particleBulk density = –––––––––––––––––––––––––––––––––––––––––– (2.17)

Volume (including the internal pores in particles and the pores between particles)

The void space in food materials can be described by determining the porosity, which is

expressed as the volume not occupied by the solid material. Porosity is calculated from the

equations 2.18 and 2.19.

Bulk densityPorosity = 1 – ––––––––––––– (2.18)

Solid density

Bulk densityInterparticle porosity = 1 – –––––––––––––– (2.19)

Particle density

P.S. The term “porosity” implies all void spaces, while the term “interparticle porosity”

implies only the void spaces between particles)

Example 11 : Calculate the density of apple in SI unit with the following composition.

85% water

14.4% sugar

0.4% fat

0.2% protein

Solution: Since no information is given about the type of the density, the density in this

question must be the solid density. For this purpose, the equation 2.14. is used.

Total mass (∑W1 + …………….. + Wn)ρfood = ––––––––––––––––––––––––––––––––––––––––

W1 Wn

Total volume (∑–––– + …………….. + ––––) d1 dn

We will use 100 g apple as a basis. Then;

42

Although the calculated density is higher than the density of water (1000 kg/m3 or 1 g/cm3),

apples stay on the water. This is because apples contain high amount of air in both

intercellular and intracellular. Accordingly, particle density of the apple is quite below 1000

kg/m3. During storage of fruit and vegetable as well as many other foods, the bulk density is

taken into account for calculating of the storage volume.

2.2.2. Specific gravity (sp-gr): The ratio of the density of a given substance at 4°C to the

density of water at the same temperature. The density of water is the highest at 4°C (1000

kg/m3 or 1 g/cm3). Specific gravity is calculated from the equation 2.20. Specific gravity is

dimensionless since it is the ratio of densities.

Density of a material at a given temperatureSpecific gravity = ––––––––––––––––––––––––––––––––––––– (2.20)

Density of water at the same temperature

In the calculation of specific gravity of gases, air is taken as reference, not water. To state this

in a different manner, the specify gravity of a gas indicates that how many times the density

of air is. The specific gravity of gases is determined under standard conditions (0°C and 760

mm-Hg), as in their densities, because the volume of gases varies widely depending on

temperature and pressure. The density of dry air is 1.293 kg/m3 and its specific gravity is 1.0.

Example 12 : The density of ammonia is 0.769 kg/m3. Calculate the specific gravity of this

gas.

Solution :

43

2.2.3. Specific volume : The reciprocal of the density of a given substance. The specific

volume is a very important concept in particular related issues on gases. In the IS unit system,

specific volume is defined as m3 of gas per kg. Specific volume is calculated from the

equation 2.21.

1Specific volume (v) = ––– (2.21) ρ

Example 13 : Calculate the specific volume of dry air under standard conditions (0°C and

760 mm-Hg).

Solution : Under standard conditions, the specific volume of dry air is as follows:

2.3. Temperature

Atoms and molecules that make up an item are always moving, not static. Although atoms

and molecules in solid matter compared with gases and liquids are not in a position to move

freely, atoms and molecules in them without leaving from their location can move by

vibration back and forth. This is valid even at low temperatures. According to Newton's

second law of motion, kinetic energy of a moving object depends on the mass and speed of

the object. If atoms and molecules of a matter are moving, they must have the kinetic energy

at a certain level depending on the movement. The kinetic energy is internal energy of the

matter. Kinetic energy which is resulted from movement of all molecules in a matter makes

up the temperature of the matter. In simple terms, the higher the temperature, the greater is the

movement of molecules. In other words, as the average kinetic energy of the atoms and

44

molecules increase, material temperature also increases. To raise the temperature of an object,

an additional energy such as heat must be given to the object. Temperature is a measure of the

amount of heat energy possessed by an object.

Thermal equilibrium : We generally perceive temperature as a measure of our physiological

response to “hotness” or “coldness.” However, physiological response is subjective, and, it

does not provide us with an objective measure. For example, holding a block of steel at 40°C

gives us a much colder sensation than holding a block of wood also at 40°C. If two objects

defined as “cold” and “hot” are contacted with each other, after a period of time, the

temperatures of the objects will be the same. The object with the higher temperature cools,

while the cooler object becomes warmer. When the two objects reach to the same

temperature, they will also reach to "thermal equilibrium". During the thermal equilibrium,

heat transfer from heat object to cold object occurs. In that case, temperature determines the

direction of heat flow. Accordingly, the heat always flows from the object at high temperature

to the object at lower temperature.

Thermal equilibrium should be measured scientifically, not with a sense of touch. For this

purpose, a third object (thermometer) should be used. The basis of this measurement is based

on zeroth law of thermodynamics. According to the zeroth law, if each of (A) and (B) objects

are separately in thermal equilibrium with (C) object (i.e. thermometer), (A) and (B) objects

are also in thermal equilibrium with each other (Figure 2.1). For this, a more comprehensive

description can be expressed as follows. Temperature is such a feature of a system that when

the system is contacted with other system, both systems ultimately reach the same

temperature.

45

AHot object

BCold object

CThermometer

Heat flow

Figure 2.1 Thermal equilibrium

Temperature measurement : An accurate measure of temperature is possible because of the

way the properties of many materials change due to heat or cold. These changes can be the

volume of a liquid, the length of a metal rod or the pressure of a gas at constant volume. The

materials whose physical properties change depending on the temperature is called

"thermometric object".

If the thermometric object is a liquid in a capillary tube, increase in liquid volume depending

on the temperature leads to change in height of liquid. Thus, scale of measurement equipment

is arranged by taking into account the length of the liquid column. However, it must be

emphasized that this change depending on the temperature must be linear. For example, for a

mercury thermometer column, regardless of the initial temperature, change by a unit in the

temperature degree must always result in the same change in length in mercury column.

Typically, in a glass thermometer, alcohol as well as mercury is also used. In the alcohol

column as well as mercury column, the changes in the temperature always lead to the changes

in the same length.

To create the thermometer scale, firstly, a fixed point must be selected. No matter what type

of thermometer, all thermometers show the same value at this point. To arrange the

thermometer scale, the triple point of water is selected as the fixed point . This point

corresponds to the sign of 0.01°C in Celsius scale and 273.16 K in Kelvin scale (Figure 2.2.).

46

Temperature (t), °C

gas

solid Cold liquid supe

rhea

ted

stea

m

V = constant

Pres

sure

(p)/a

tm

Figure 2.2 A graphic showing the triple point of water

2.3.1. Temperature Scales

There are mainly two thermometer scales such as Fahrenheit and Celsius scales. Of these, a

special thermometer scale which was manufactured in 1724 by Gabriel Daniel Fahrenheit

who was the German manufacturer of measuring instruments is called "Fahrenheit" scale. In

this scale, mercury in a capillary tube as the thermometric object is used, the level of mercury

at the freezing point of water is signed as 32, that at the boiling point of water is signed as 212

and the unit is shown with °F symbol. Thus, a scale which consists of 180 (212 ‒ 32 = 180)

units between freezing and boiling points of water is formed. There was no physical meaning

of 0 degree in Fahrenheit scale. This temperature scale is still widely used in the U.S. and UK.

Although it is not known exactly who made the Celsius scale, it is believed that it was

arranged in 1742 by the Swedish scientist Andreas Celsius. In this scale, the freezing point of

water is signed as 0, the boiling point of water is signed as 100. This scale is also called

"centigrade" (Celsius, cents: 1/100; Grade: grade) and the unit is shown with °C symbol. In

Celsius scale, temperature of an object which is colder than the freezing point of water is

measured in the negative numbers. Despite the Fahrenheit scale has been found earlier, the

Celsius scale takes its place in the scientific research. The main reason for this is that the

metric system (SI, MKS and cgs) is arranged on the basis of 10 (decimal). As well as

scientific studies, nowadays, the Celsius scale is widely used in many countries.

There are two more scales for measuring temperature. One of these is Kelvin and the

other is Rankine scales. Kelvin scale was devised in 1854 by Lord William Kelvin (1824-

1907) who was a Scottish physicist. The symbol of Kelvin scale is K and degree symbol (°)

is not used. The Kelvin scale uses the same graduation as the Celsius scale. Thus, 1°C is

equal to 1 K. However, 0 K accepted as fixed point in Kelvin scale corresponds to ‒273°C

in Celsius scale.

Rankine scale was devised in 1859 by William John Macquorn Rankine who was a

Scottish engineer and physicist. The symbol of Rankine scale is R and degree symbol (°) is

not used. Similar relationship between Kelvin and Celsius scales is valid between Fahrenheit

and Rankine scale. In this case, 0 R in Rankine scale corresponds to ‒460°F in Fahrenheit

47

scale. Thus, 1 R is equal to 1 °F. There are no negative values in Rankine and Kelvin scales

(Figure 2.3.).

Figure 2.3 Scales of thermometer

0 points in Kelvin and Rankine scales are the absolute (actual) zero points and to

measure lower temperatures than the absolute zero point is not theoretically possible. Due to

these features, these scales are widely used in scientific research to measure very low

temperatures. While Rankine scale is preferred by American and English engineering fields,

Kelvin scale is preferred by all over the world.

As seen, zero value of Kelvin and Rankine scales is arranged to show the absolute

zero. Zero value of these two scales is the lowest temperature degree which is measurable in

nature. Therefore, the scales are called as “absolute temperature scales.” Especially, Kelvin

scale is the main scale used in the science and is also called as “absolute temperature scale of

thermodynamic.”

Absolute zero : Absolute zero is the theoretical temperature and is explained as “accessible

minimum temperature.” "Absolute zero” is the temperature of atoms, electrons and molecules

when they have the lowest energy level". Thus, at this temperature, all molecular motion

48

Freezing point of water

Boiling point of water

100 equal parts

180 equal parts

0°C 100 equal parts

273 K

–273°C

0°F

0 K

–460°F

0 R

100°C

373 K

212°F

672 R492 R

32°F

Absolute zero point

180 equal parts

stops and no discernible energy can be detected. At absolute zero point, the pressure of an

ideal gas reaches to zero and the product of the pressure and volume (P V) of a gas is directly

proportional to the temperature, thus if the pressure is zero, then the temperature will also be

zero.

These concepts and definitions are based on a scientific experiment result. When

measured pressure values of a gas at a constant volume value during cooling are plotted

against temperature values in a graph, all points are located on a straight line. Once the

temperature falls below a certain level, gas liquefies and then an experimental data cannot be

obtained. Despite of this, when the straight line passing from the experimental points is

extended to the direction in which the experimental data cannot be taken, it is shown that this

straight line cuts the temperature axis at exactly 273.16°C. Independently from the type of gas

used in the experiment, it always reaches to –273.16°C which is absolute zero point. The scale

which was arranged by marking –273.16°C point in Celsius scale as zero is called as Kelvin

scale. Triple point of water is equal to 273.16 K in the Kelvin scale and 0.01°C in Celsius

scale.

The above description is shown in Figure 2.4. If the experiment described above is performed

as measurement of volume values at constant pressure and temperature, it is shown that

obtained curve again cuts the temperature axis at −273.16 °C.

Figure 2.4. Determination of absolute zero temperature by “pressure-temperature” curve of gases (Allen, 1964)

49

Decrease by a degree at a temperature leads to decrease by 1/273 of pressure at 0°C in gas pressure.

Regardless of the initial pressure, a straight line which reachs to zero pressure at ‒273°C is always obtained.

If the line of “pressure-temperature” is extended to the pressure of zero, it cuts the temperature axis at −273 °C.

Pres

sure

Temperature, °C

2.3.2. Conversions between different scales

Conversion of different temperature units to each other is actually a simple process. However,

the biggest mistake is made when the conversion of a measured temperature value from one

scale to another scale is done using the same way with the conversion of the temperature

differences (ΔT) from one scale to another scale. For example, if temperature of an object is

measured as 90°F, the calculating its equivalent in the Celsius scale can be done by using the

simple equation. However, when the temperature of an object is increased by 90°F, how

many centigrades of the temperature of the object increased are different situations. And, the

example of this type of conversion will be given later.

In fact, the conversion between different scales is not different from interpolation process.

Taking into account the data in Table 2.4., the temperature units can easily be converted to

each other.

Table 2.4 Temperature scales and equivalents

Celsius (oC) Fahrenheit (oF) Kelvin (K) Rankine (R)

Boiling point (of water) 100 212 373.15 671.67

Freezing point (of water) 0 32 273.15 491.67

Absolute zero –273.15 –459.67 0 0

Reference : Geankoplis, 2008.

Example 14. Taking into account the data in Table 2.4., convert 5°C into °F.

Solution : The data used for the solution of the problem are given in Table 2.5.

Table 2.5 Temperature data used in the solution of example

Celsius Fahrenheit

Freezing point 0°C 32°F

Boiling point 100°C 212°F

50

5°C ? °F

1) In the temperature column, while the temperature increases from 10 to 15, i.e., 5 unit

increase in temperature, the density decreases 6 x 10–4 (0.99973 – 0.99913).

2) Meanwhile, while the temperature increases from 10 to 12, i.e., 2 unit increase in

temperature, find out the decrease in density.

3)

5 unit increase in temperature 6 x 10–4 unit decrease in density

2 unit increase in temperature x––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

x = 2.4 x 10–4 unit decrease

ρ12°C = 0.99973 – 2.4 x 10–4

ρ12°C = 0.99949 g/cm3

1) In Celsius scale, while the temperature increases from 0 to 100°C, i.e., 100 unit

increase in temperature, the temperature in Fahrenheit scale increases 180 (212 – 32 =

180) unit.

2) In Celsius scale, while the temperature increases from 0 to 5°C, i.e., 5 unit increase in

temperature, proportion given in third step is used for calculating increase in the

temperature in Fahrenheit scale.

3)

100 unit increase in Celsius scale 180 unit increase in Fahrenheit scale

5 unit increase in Celsius scale x––––––––––––––––––––––––––––––––––––––––––––––––––––––––––‒‒‒‒‒‒‒–

x = 9 unit increase in Celsius scale

5°C = 32°F + 9°F

5°C = 41°F

In conversion of Fahrenheit and Celsius scales into each other, one of the most important

points is that while the initial of Celsius scale is “0”, that of Fahrenheit scale is “32”. The

Fahrenheit and Celsius scales are also related by the following formula.

To convert Fahrenheit to Celsius degree:

51

°F = (9/5 x °C) + 32 or; °F = 1.8 x °C + 32 (2.22)

To convert Celsius to Fahrenheit degree:

°C = 5/9 (°F – 32) (2.23)

Source of the coefficients of 9/5 or 1.8 and 5/9 in the formulas:

(212 – 32)°F 180°F 9 ––––––––––– = ––––––– = ––– = 1.8

(100 – 0)°C 100°C 5

(100 – 0)°C 100°C 5 ––––––––––– = –––––– = –––

(212 – 32)°F 180°F 9

During conversion of Kelvin and Celsius scales into each other, it should not be out of

sight that zero points of the scales are different. The Kelvin and Celsius scales are also

related by the following formula.

K = °C + 273.15 (2.24)

In the calculations, the value of 273 is usually rounded to the value of 273.15.

During conversion of Rankine and Fahrenheit scales into each other, it should not be out

of sight that the beginning points of the scales are different. The Rankine and Fahrenheit

scales are also related by the following formula.

R = °F + 459.67 (2.25)

In the calculations, the value of 459.67 is usually rounded to the value of 460.

As mentioned above, the actual scale division in the Kelvin and Celsius scales with that in the

Rankine and Fahrenheit scales are the same. Therefore, the temperature differences in Celsius

with Kelvin and Rankine with Fahrenheit scales will be the same.

In this case:

52

ΔT (K) = ΔT (°C)

ΔT (R) = ΔT (°F)

To reveal the above explanation, the following example has been given. If specific

heat of a food is 3.5 kJ/(kg °C), this means that to increase the temperature of the food by

1°C, 3.5 kJ of the heat is required. Therefore, whenever we have temperature in the

denominator, we are actually considering a unit difference in temperature, since 1° change in

the Celsius scale is the same as a unit change in the Kelvin scale. Therefore, specific heat of

the given liquid food may also be reported as 3.5 kJ/(kg K). However, the actual scale

division in the Celsius and Fahrenheit scales is different. Therefore, when "temperature

difference" is taken into consideration, the following conversion factor should be used.

(212 – 32)°F 180°F (Δ)1.8°F––––––––––– = –––––– = –––––––– (100 – 0)°C 100°C (Δ)°C

Temperature difference of 1 degree in Celsius scale leads to temperature difference of

1.8 degree in Fahrenheit scale.

Example 15 : When water at 40°C has been heated to 95°C:

a) Find out the initial and final temperatures of water in Kelvin scale.

b) Find out the initial and final temperatures of water in Fahrenheit scale.

c) Find out the temperature differences in Celsius (°C), Fahrenheit (°F) and

Kelvin (K) scales.

Solutions :

53

Example 16 : Temperature of a frozen food (A) has been measured as –20°C, while that of

another (B) has been measured –20°F. Find out the temperature of A food as “°F”, and that of

B food as “°C”.

Solution :

P.S. : In a physical law, absolutely the unit of absolute temperature scale is used.

2.4 Pressure

Pressure is expressed as force per unit area. The dimensions of pressure are (mass)/(time)2

(length)1. In SI system, the units are N/m2. This unit is also called a Pascal (named after

Blaise Pascal, a French philosopher and mathematician, 1623–1662). The pressure units in SI

and English systems are given in Table 2.6.

Table 2.6 Pressure units

Units

Force Area Pressure

SI N m2 N/m2

English lbf in2 lbf/in2

In English system, the pressure is expresses as psi, i.e., pound per square inch (lbf/in2).

Pressure is sometimes expressed as kg/cm², kgf/cm2, kp/cm2, atm, at, ata and atu.

Technical atmosphere: Pressure exerted on an area of 1 cm2 by a standard mass of 1 kg. It

is shown as “at”, “kg/cm2”, “kgf/cm2” or “kp/cm2”. “kg/cm2”, “kgf/cm2” and “kp/cm2” are the

same units. Kilopound (kp) is the weight of standard mass under the effect of standard gravity

acceleration. When the weight is used as a force unit, 1 kp is equal to kgf. (f) sub-suffix is used

for showing that it is a force unit.

Pressure is often expressed as a height of a particular fluid (e.g., inches of water). For

example, the air at 0°C and sea level raises the height of mercury in a column as 760 mm. If

water is used instead of mercury, the air at the same conditions raises the height of water in

54

the column as 10.33 m. Therefore, standard atmospheric pressure can be either shown as 760

mm Hg or 10.33 m water. However, expression the pressure by the height of the liquid

column is generally used for hydrostatic pressures. Hydrostatic pressure is the pressure which

results from the weight of the liquid column.

Standard (normal) atmospheric pressure: Pressure produced by a column of mercury to

760 mm high. It is approximately equal to typical air pressure at sea level (1 atm). 1 atm is

equivalent to the hydrostatic pressure occurred by standard gravity acceleration on mercury

column of 760 mm Hg whose density is 13.5951 g/mL under standard conditions. The

standard atmospheric pressure can be expressed, using different systems of units, as

1 atm = 14.696 lbf/in2 = 1.01325 bar =101 325 Pa

Meanwhile;

1 atm = 1.0332 kgf/cm2 (at). This value is calculated as follows.

ρHg = 13.5991 g/cm3 = 0.0135951 kg/cm3

During the conversions of units, when the numerical values of dimensions are written to the

equation, result is obtained without calculation of CF :

As known, blood pressure is measured in millimeters of mercury in the most of the world, and

lung pressures in centimeters of water are still common.

Example 17 : Atmospheric pressure in Ankara is 690 mm Hg. Find out the equivalent of

this pressure in kPa and lbf/in2.

Solution :

Conversion to kPa :

55

Conversion to lbf/in2 :

Example 18 : Convert the pressure of 690 mm Hg into lb f/in2 without the use of “14.696

lbf/in2.”

Solution :

Example 19 : Convert the technical atmosphere of 1.0332 kg/cm2 into standard atmosphere.

Solution :

As mentioned earlier, the unit of pressure in the SI unit system is the Pascal, and is expressed

as given below.

56

N kg m/s2

Pa = ––––– = ––––––– = kg/m s2

m2 m2

Since Pascal is a very little unit, kPa (which is 103 times higher than Pa), bar (which is 105

times higher than Pa) or MPa (which is 106 times higher than Pa) are mostly used instead of

Pa. In fact, since 1 bar is approximately equal to 1 atm (exactly 1 atm = 1.01325 bar), bar unit

is commonly used.

Pressure measurement is based on two different principles. One of them is that standard

atmosphere pressure is accepted as “zero” and pressures which are higher than this pressure is

measured. The pressure measured according to this basis is called as “effective pressure”. The

other is based on “absolute zero pressure” concept. Accordingly, normal pressure is higher

than absolute zero pressure and is expressed with 1atm (abs).

The (abs) of expression is also always used with the other pressure units. For example, … bar

(abs) or …kPa (abs) etc.

If a system pressure is lower than 1 atm (abs), there is a vacuum in there.

Accordingly ;

Vacuum = Atmosphere pressure – Absolute pressure

Example 20: The pressure in a hermetically closed container is measured as 0.605 atm

(abs.). Find out the vacuum in this container.

Solution :

Instruments that are used to measure pressure are called "gauge (manometer)". In gauge,

normal atmospheric pressure is marked as "zero". Thus, the gauge measures the pressure

57

which is greater than normal atmospheric pressure by ignoring the atmospheric pressure. This

pressure is known as “effective pressure” or widely “gauge pressure”. Accordingly, sum of

"gauge pressure" and "atmospheric pressure" gives the "absolute pressure".

In the light of all of the explanation, while a pressure value is given, the pressure type such as

absolute (abs) or gauge (effective/manometer) pressure must be explained. For example, an

explanation such as 2 atm (abs) or 2 atm (eff) is required. If there is no an explanation, it is

thought that the given pressure is gauge (effective) pressure.

In English engineering system, pressure is expressed as force of 1 lbf per 1 in2 and is shown as

“psi” (pound per square inch). “psia” is used for “absolute pressure”, whereas “psig” is used

for gauge (effective, manometer) pressure. The equivalent of physical atmosphere (atm) and

technical atmosphere (at) with the other units are shown below:

1 atm = 760 mm Hg 1 at = 735.56 mm Hg

1 atm = 101.3 kPa 1 at = 98 kPa

1 atm = 1.033 at 1 at = 0.968 at

1 atm = 14.69 psi 1 at = 14.22 psi

1 atm = 1.013 bar 1 at = 0.98 bar

Absolute (perfect) vacuum: Zero pressure characterizes absolute vacuum. However, it is

impossible to achieve.

Absolute pressure: When pressure is measured relative to absolute vacuum, it is called

absolute pressure.

Gauge pressure: As mentioned before, when we use a pressure measuring devise such as a

manometer, it is calibrated to read zero at one atmospheric pressure. Therefore, these devices

are actually reading the difference between the absolute pressure and the local atmospheric

pressure. A pressure measured by a gauge (manometer) is called as gauge pressure, and it is

related to atmospheric pressure based on the following expression:

Pabsolute = Patmosphere + Pgauge (In English system, psia = Patm + psig)

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(for pressures greater than Patmosphere)

Pvacuum = Patmosphere – Pabsolute

(for pressures below Patmosphere)

Example 21: The relationship between absolute and gauge pressure is explained by “psia =

psig + 14.7” Convert the pressure value of 14.7 psig to kPa.

Solution:

Example 22 : An evaporator works on a vacuum of 20 in Hg. Find out absolute pressure of

the evaporator in :

a) SI and

b) English systems.

Solution :

59

Pressure is the term used to express this property for liquids and gases. For solids, we use the

term normal stress instead of pressure. In situation involving fluid flow, pressure is often

expressed in terms of height or “head” of a fluid and is expressed as the following expression:

P = ρ g h (2.26)

where :

P : absolute pressure (Pa),

ρ : fluid density (kg/m3),

g : the acceleration due to gravity (9.81 m/s2)

h : the height of fluid (m).

This pressure is also called as hydrostatic pressure, i.e., the pressure due to the weight of a

fluid.

Example 23: Find out the height (mm) of milk in a tank filled with milk with 0.2 atm

pressure. The density of milk is 1.028 g/cm3.

Solution : Firstly, the density of milk is converted to SI unit system (kg/m3) :

60

Consider a tank filled with cold water to a height of 7 m. The pressure exerted by the water

on the bottom of tank is independent of the diameter of the tank, but depends on the height of

the water in the tank. The height of water in the tank is called static head.

Example 24: A spherical tank whose diameter is 2 m has been filled with olive oil whose

density is 0.915 g/cm3. The pressure at the top of the tank is 75 kPa. Thus, calculate the

highest pressure in the tank according to SI units.

Solution :

Example 25 : A pressure of 0.69 bar (gauge) is applied the bottom of a tank filled with water

by the water. Calculate the static head of water in SI units

Solution : Firstly, the density of water is converted to SI unit system (kg/m3) :

61

Example 26 : The tank in Example 2.25 is filled with ethyl alcohol. A pressure of 0.69 bar

(gauge) is applied the bottom of the tank by the ethyl alcohol. If the specific gravity of ethyl

alcohol is 0.79, calculate the static head of the ethyl alcohol in SI units.

Solution :

62

Example 27 : An unclosed tank whose diameter is 1 m will be filled with tomato juice (6%

soluble solid content). However, the pressure in the tank must not greater than 120 kPa. Thus,

calculate the highest content of tomato juice which can be filled to the tank. The density of

tomato juice is 1025 kg/m3.

Solution :

63

3. MATERIAL BALANCE

Material balance calculations are employed in tracing the inflow and outflow of material in a

process. Therefore, these calculations help to calculate the quantities of various materials in

each process stream. The procedure is useful in making formulations, determination of final

compositions after blending, calculation of yields and evaluating separation efficiencies in

mechanical separation systems.

Basic principles

Material balances are based on the principles of the law of conservation of mass. Thus, in any

process, a material balance can be made as follows:

Inflow = outflow + accumulation

The principles of law of conservation of mass states that

Mass can be neither created nor destroyed. However, its composition can be altered

from one form to another.

Even in 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.

If accumulation is 0, inflow = outflow and the process is at steady state. If the accumulation

term is not 0, then the concentration of components in the system could change with time and

the process is at unsteady state. Steady state systems can be applied even to the liquid flows.

For example, if the level of milk in a tank remains constant, and the milk flow rate at the inlet

pipe is 1 kg/s, then the flow rate of milk at the exit must also be 1 kg/s.

Before preparing the material balance, first the process is put together in a flow diagram and

the boundaries of the system (process) is determined. A system is any region prescribed in

space by a boundary, either real or imaginary. The boundary of a system can be real, such as

the walls of a tank, or it can be an imaginary surface that encloses the system. Furthermore,

the boundary may be stationary or movable. For example, the system boundary can include a

64

tank, piping and a valve. If our analysis had concerned only the valve, we could have drawn

the system boundary just around the valve. Once we choose the boundaries of a system, then

everything outside the boundary becomes the surroundings. The analysis of a given problem

is often simplified by how we select a system and its boundaries; therefore, proper care must

be exercised in selecting the system and its boundaries. A system boundary may even enclose

an entire food processing system.

A system can be open or closed. In a closed system, the boundary of the system is impervious

to flow of mass. In other words, a closed system does not change mass with its surroundings.

A closed system may exchange heat and work with its surroundings, which may result in a

change in energy, volume, or other properties of the system, but its mass remains constant.

For example, a tank which is impervious to the flow of matter is a closed system. If the

system consists of tank and piping, then we are dealing with an open system, where the liquid

flows in and out of the system, thus changing the mass.

When a system does not change mass, heat or work with its surroundings, it is called as an

isolated system. For example, if we carry out a chemical reaction in an insulated vessel (no

heat exchange with its surroundings and no change in mass), then we have an isolated system.

If no exchange of heat takes place in a system with its surroundings, the system is called as

adiabatic system. When a process occurs at a constant temperature, often with an exchange

of heat with its surroundings, then we have an isothermal system.

Most of the time, system contains subsystems. For subsystems, the same principals are valid

for systems. For material balance calculations, one should first describe the system and

define the boundaries of the system. After defining the system, the basis should be chosen for

the calculations. A basis is useful in problems where no initial quantities are given. For

example, if the fruit juice is processed into concentrate in an evaporator, the basis can be

selected by 4 different ways. These are given below:

1) The amount of fruit juice entering the evaporator (for instance; 100 kg or 1000 kg),

2) The amount of fruit juice concentrate leaving the evaporator (for instance; 100 kg or

1000 kg),

65

3) The ratio of the amount of concentrate leaving the evaporator to the amount of fruit

juice entering the evaporator (for instance; 1:4 or 1:6),

4) The ratio of the amount of fruit juice entering the evaporator to the amount of

concentrate leaving the evaporator (for instance; 4:1 or 6:1),

The same principles are applied for material balance calculations of batch and continuous

systems. In batch system, the substance enters the system at one time. On the other hand,

material balance in a continuous flow system is obtained by assuming as a basis a fixed time

of operation (for instance, 1 h), and then the material balances are conducted exactly the same

way as the batch system.

The following steps should be useful in conducting material balance in an organized manner:

a) Collect all known data on mass and composition of an inlet and exit streams from the

statement of the problem.

b) Draw a block diagram (process flow diagram), indicating the process, with inlet and

exit streams properly identified. Draw the system boundary.

c) Write all available data on the block (flow) diagram.

d) Select the suitable basis* (such as mass or time) for calculations. The selection of

basis depends on the convenience of computations.

e) Write total and component mass balances. For each unknown, an independent

material balance is required.

f) Solve material balance to determine the unknowns.

Process flow diagrams

Before writing material balance equation, visualize the process and determine the boundaries

of the system for which the material balance is to be made. The process flow diagram for a

crystallization problem is given in Figure 3.1.

66

Figure 3.1 Flow diagram for a crystallization problem showing input and exit streams and boundaries of whole system and sub-systems

Figure 3.1 shows how the boundaries of the system can be moved to facilitate solution of the

problem. If the boundary completely encloses the whole process, two streams will be

entering and four streams leaving the system. The boundary can also be set just around the

evaporator. In this case, there will be one stream entering and two streams leaving. The

boundary can also be set around the centrifuge or around the drier. A material balance can be

carried out around any of these sub-systems or around the whole system. The material

balance equation may be a total mass balance or a component mass balance.

The following example explains all these points.

Example 3.1: Plot the flow diagram for the determination of the yield in tomato paste

production and determine the basis in material balance calculations.

Solution:

In tomato production, hot-break or cold-break techniques are applied.

Hot break: heated to about 85°–95°C; pectin is preserved; therefore, thick paste is

obtained.

Cold break: heated to about 65°–75°C; color and flavor is preserved.

However, hot-break is the most commonly applied method. In this method, the processing

steps as follows:

1) Washing of tomatoes,

2) Sorting of tomatoes,

3) Chopping of tomatoes,

4) Heating of chopped tomatoes,

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5) Obtaining refined juice or pulp from the heated tomato pulp (fiber, juice, skin and

seeds) by pulper

6) Concentration of refined juice through an evaporator into paste,

7) Heating of paste in an tubular heating at usually 93°C, depending on hot- or cold-

break paste production,

8) Hot filling at 93°C of paste in metal containers,

9) Cooling of metal containers,

10) Packaging of metal containers in carton packages,

11) Storage.

For tomato paste production, there are 11 production steps. If one puts each processing step

side by side or from top to bottom, then the whole process flow diagram is obtained. If the

purpose is how much of the paste is obtained from a certain amount of tomatoes and during

this process, how much of water should be evaporated, then only the steps which affect the

material balance calculations should be taken into consideration. Such diagram is prepared in

Figure 3.2.

Figure 3.2 Process flow diagram in tomato paste production

Compared to 11 steps in the production of tomato paste, only 3 steps are found in Figure 3.2.

Washing steps do not cause any significant mass changes; therefore, it was excluded from the

diagram. Chopping and heating processes also do not cause any mass changes. Similarly,

heating of paste in tubular heating, filling of containers, cooling of containers, packaging of

68

carton packages and storage will all be excluded form material balance calculations. This is

because the yield can be calculated from the steps up to the obtaining the paste. If the

question would how many containers could be obtained from a certain amount of tomatoes,

then the diagram should include the filling step.

As seen in Figure 3.2, there are only one inflow stream and 4 outflow streams in the system.

For the calculations, the basis of 100 kg or 1000 kg of tomatoes entering the system can be

taken. The other basis can be how much of paste is obtained from certain amount of tomatoes.

Total Mass Balance: Total weight of each stream entering or leaving a system represents a

total mass balance. The following examples illustrate how total mass balance equations are

formulated for systems and sub-systems.

Example 3.2: In an evaporator, dilute material enters and concentrated material leaves the

system. Water is evaporated during the process. If I is the weight of the dilute material

entering the system, W is the weight of water vaporized, and C is the weight of the

concentrate, write an equation that represents the total mass balance for the system. Assume

that a steady-state exists.

Solution: The diagram for the process is shown in Figure 3.3.

Fig. 3.3 Input and exit streams in an evaporation process

The total mass balance is:

Inflow = outflow + accumulation

I = W + C (Accumulation is 0 in a steady-state system)

69

Example 3.3: Drier is fed with W kg/min wet material and A kg/min hot air and dried

material is left the system with D kg/min. Plot the flow diagram fort his process and write

down the material balance equations for this process.

Solution: In this process, the system is the drier. The wet material and hot air enter the drier

and dried material exists from the drier. In addition, the air containing the moisture also exits

the drier. Therefore, the moisture from the material is carried through the air. The diagram

for the process is shown in Figure 3.4.

Figure 3.4 Flow diagram for the hot air drier

As seen in Figure 3.4, there are two subsystems in this system, one of which ,s air and the

other is the material dried.

Total material balance for whole system: W + A = Wet air + D (1)

Material balance around air subsystem: A + Su = Wet material (2)

Material balance around drying material: W = S + D (3)

Example 3.4: Orange juice concentrate is made by concentrating single-strength juice to

68% solids (C) in an evaporator followed by dilution of the concentration to 45% solids using

single-strength juice. The concentrate of 68oBrix is poor in aroma; therefore, the freshly

squeezed orange juice is added to this concentrate to enrich the aroma. This process is called

70

as cut-back. Draw a diagram for the system, and set up mass balances for the whole system

and for as many sub-systems as possible.

Solution: The diagram for the process is shown in Figure 3.5. A part of freshly squeezed

orange juice (F) is concentrated in an evaporator, and the rest of orange juice (A) is directly

mixed with this concentrate. Therefore, the whole orange juice (S) is assumed as the dividing

into two streams in an subsystem. Therefore, the whole system contains 3 subsystems.

Figure 3.5 Diagram of an orange juice process involving evaporation and blending of concentrate with freshly squeezed juice

Total material balances:

Overall:

Proportionator:

Evaporator:

Blender:

Component Mass Balance : The same principles apply as in the total mass balance, except

that the components are considered individually. If there are n components, n independent

equations can be formulated; one equation for total mass balance and n – 1 component mass

balance equations. For the calculation of component material balance, the amounts and mass

percentages of the inflow and outflow streams are written in the flow diagrams. It is often

necessary to establish several equations and solve these equations simultaneously to find out

the unknowns. In a material balance, use mass units and concentration in mass fraction or

mass percentage. Mass percentages are used for the most of the times

71

mass of component AMass percentage = ––––––––––––––––––––––––––––– (100)

total mass of mixture containing A

If the quantities are expresses in volume units, convert them to mass units using density. This

is because, the volumes change during heating.

In the preparation of component material balance equations, the other way is to use mass

fraction equation. This equation is especially used in the solving of problems involving with

concentration and dilution.

mass of component AMass fraction = ––––––––––––––––––––––––––––––

total mass of mixture containing A

If 80 kg of fruit juice contains 12% sugar, sugar (component) balance is set up as:

Example 3.5: 7 kg of food material with 8.1% soluble solid content is dried in an air dryer

and the soluble solid content is raised to 90%. Find out the weight of dried material.

Solution:

72

Example 3.6: The brix of fruit juice containing 12% soluble solid and 0.8% acid was

reconstituted to 15% by adding sugar syrup with 66% sugar content. Find out the weight of

syrup and the mass percentage of acid content of reconstituted juice.

Solution:

Example 3.7: 500 kg of water is removed in an evaporator at each hour. Fruit juice with

12% soluble solid content enters the evaporator and leaves the system as concentrate with

45% solid. Find out the concentrate production rate.

Solution:

73

Example 3.8: Determine the quantity of sucrose crystals that will crystallize out of 100 kg of

a sucrose solution after cooling to 15°C. The mother liquor contains 75% sucrose. And also,

calculate the mass of syrup after cooling.

Solution:

Example 3.9: In a crystal sugar producing plant, the sugar crystals are obtained from 100 kg

of a concentrated sugar solution containing 85% sucrose and 1% inert, i.e., water-soluble

impurities. Upon cooling, the sugar crystallizes from solution. A centrifuge then separates

the crystals from a liquid fraction called the mother liquor. The mother liquor leaving the

centrifuge contains 60% sucrose by weight. The crystal slurry fraction has, for 20% of its

weight, a liquid with the same composition as the mother liquor. Find out the mass of the

crystals and concentrated sugar solution.

Solution:

74

Example 3.10: Determine the amounts beef and back fat (iç yağı) that must be used to make

a 100 kg of frankfurter (sausage) formulation. The beef contains 15% protein, 20% fat, and

63% water, and back fat contains 3% protein, 80% fat and 15% water. The frankfurter

contains 25% fat.

Solution:

Material balance calculations in the preparation of fruit juice and nectars

In fruit and vegetable processing, material balance calculations are the most frequently used

in the preparation of fruit juice and nectars. Therefore, there is some information given below

about the fruit juice and nectar preparations. In drink industry including fruit juice industry,

sugar syrup is commonly used and therefore, before material balance calculations in fruit

juice and nectars, the preparation of sugar syrup will be discussed.

Fruit juice or juice concentrate is reconstituted by blending the juice or concentrate with

water, sugar and acid to standardize the juice or to bring the juice in a drinkable state. The

fruit drinks prepared by the addition of water, sugar and acid are called as nectar. Sour

cherry, apricot and peach are the example of nectars. On the contrary, apple, orange and

grape are commonly prepared without the addition of water, sugar and acid; therefore, they

are called as fruit juice.

The dilution of concentrates with water to their original brix is called as reconstitution.

However, the addition of water, sugar and acid to fruit juices to bring the fruit juice to the

drinkable state is also called as reconstitution. For reconstitution, demineralized water or at

least the drinkable water should be used. Sugar can be directly added to the mix or added as

sugar syrup. The sugar syrup is the nest way for adding sugar. As an acid, the most of the

time citric acid is used. Acid is also used as a solution (50%).

For the reconstitution of fruit juices, the volumes are usually used in the production plants.

However, all mass balance calculations must be based on the weight. Therefore, the

75

density of juice and concentrate is needed. The density of juices is given in Table 2. For the

density of juice concentrates, the Table 1 for sugar syrups will be used.

Preparation of sugar syrup

The sugar syrup used in juice industry is prepared at 65–70% sugar content by dissolving

crystal sugar in warm water. Before use, the sugar syrup is filtered. The amount of sugar and

water for the preparation of sugar syrup at a given concentration is either directly found form

the table prepared from the experimental data or the series of calculations are carried out by

taking into consideration of the density of sugar. The experimentally obtained table for the

preparation of sugar syrup is given in Table 1.

Example 3.11: 25 kg of sugar syrup at 66% sugar content is diluted with water to bring the

soluble solids of syrup to 11%. Calculate the weight of sugar syrup obtained.

Solution:

Example 3.12: For the preparation of 500 kg of sugar syrup with 66% sucrose, calculate the

mass of sugar and water, and the mass of sugar solution obtained.

Solution:

76

Example 3.13: Solve the example 13 by using the density of sucrose 1.61 g/mL and not using

sugar syrup table. Find out the density of sugar syrup obtained as "kg/L."

Solution:

Example 3.14: Sugar syrup with 66% sugar content will be prepared by adding sugar into

750 L of water. Calculate the mass of sugar needed and the mass and volume of syrup

obtained.

Solution :

77

Fruit juice reconstitution: Fruit juice or juice concentrate is reconstituted by blending the

juice or concentrate with water, sugar and acid to standardize the juice or to bring the juice in

a drinkable state. For the reconstitution of fruit juices, the volumes are usually used in the

production plants. However, all calculations must be based on the weight. Therefore, the

density of juice and concentrate is needed. The density of juices is given in Table 2. For the

density of juice concentrates, the Table 1 for sugar syrups will be used. There are some

samples given below for the reconstitution of fruit juices and the preparation of fruit nectars.

Example 3.15: The fruit juice is reconstituted by blending sugar syrup at 66% sugar content

with fruit juice at 12% soluble solid and 0.8% acid content. The final soluble solid content of

reconstituted juice will be 15%. Calculate the weight of sugar syrup added and the final acid

concentration of reconstituted juice.

78

Example 3.16: 10 000 L of apple juice at 12% solids will be produced from the apple juice

concentrate at 72% solids. Calculate the amount of water and concentrate required in weight

(kg) and volume (L).

Solution:

79

Example 3.16: The apples are processed to juice at 10.5% solids in fruit juice plant. This

juice is reconstituted to 12% solids by blending with the apple juice concentrate at 72% solid

content. Calculate the amount of concentrate required in weight (kg) and volume (L) to be

added to 5000 L of apple juice at 10.5% solids.

Solution:

80

Example 3.17: In a fruit juice plant, the sour cherries are processed to juice at 10.5% solids

and 2% acid. This juice is reconstituted to 14% solids and 0.8% acid by adding water and

sugar. For the 4000 L of sour cherry juice:

a) Calculate the weight of sugar syrup, water and reconstituted fruit juice.

b) In case crystal sugar used, find out the weight of sugar, water and reconstituted fruit

juice.

Note: The solid content of sugar syrup is 68%.

Solution:

81

Example 3.18: 100 000 boxes (200 mL/box) of sour cherry nectar will be produced form the

sour cherry concentrate containing 68% solids and 7.1% acids. The sugar syrup will be used

in this reconstitution will be at 65% solid content. The reconstituted juice will contain 14%

solids and 0.8% acid. Calculate the weight of concentrate, sugar syrup and water.

Solution:

82

Example 3.19: Find out how much aroma concentrate should be added to the sour cherry

nectar. Mechanical concentration of sour cheery aroma is 1:250.

Mechanical concentration indicates that how much aroma concentrate is obtained from how

many liters of fruit juice.

Solution:

Example 3.20: The peach nectar at 14°Brix and 0.6% acid content will be prepared from

4500 kg peach pulp at 12% solids and 0.5% acid content. The sugar syrup used in this

reconstitution will be at 66% sucrose content. Calculate the weight of sugar syrup, citric acid

and water content as well as the peach nectar obtained in this process.

Solution:

83

Material balance calculations for jams and marmalade preparations

One of the most commonly used are of material balance calculations is the preparation of

jams and marmalades. Before the examples on this topic given, the definition of jams,

marmalades and similar products will be discussed.

Jam: Jam contains pieces of the fruit's flesh. Properly, jam refers to a product made with

whole fruit or fruit pieces. The fruit is heated with water and sugar to activate the pectin

(gelling agent) in the fruit. If the fruit do not contain enough pectin and acid, then pectin and

acid (such as citric acid) are added. In jam making, whole berries (such as sour cherries) are

most frequently used without crushing. Larger fruits such as apricots, peaches, or plums are

also used after cutting into small pieces. Good jam has a soft even consistency, a bright color,

a good fruit flavor and a semi-jellied texture that is easy to spread but has no free liquid.

Preserve: The term preserves is usually interchangeable with jam. Preserve is usually

defined as cooked and gelled whole fruit and it includes a significant portion of the fruit.

Marmalade : Refers to a sweet preserve, traditionally with a bitter tang (flavor: taste and

smell), made from citrus fruits rind (kabuk, most popularly oranges), sugar, water, and

pectin (in some commercial brands).

Fruit spread: Refers to a jam or preserve with no added sugar.

Jelly : Refers to a type of clear fruit spread made of heating the fruit juice with pectin. In

jelly making, the fruit pulp is filtered out after the initial heating. Good jelly is clear and has a

fresh flavor of the fruit from which it is made. It is tender (not hard) enough to vibrate when

moved.

Example 3.21: The standard identity for jams and preserves specifies that the ratio of fruit to

added sugar in the formulation is 45 parts fruit to 55 parts sugar. A jam must have a soluble

solid content of at least 65% to produce a satisfactory gel. The standard identity requires

soluble solids of at least 65% for fruit preserves from apricot, peach, pear, nectarine, plum,

figs, and quince. The process of making fruit preserves involves mixing the fruit and sugar in

the required ratio, adding pectin, and the concentrating the mixture by boiling in a vacuum

84

and atmospheric pressure in a steam jacketed kettle until the soluble solid content is at least

65%. The amount of pectin added is determined by the amount of sugar used in the

formulation and the by the grade of the pectin. If the fruit contains 10% soluble solids and

100 grade pectin is used, calculate the weight of the fruit, sugar, and pectin necessary to

produce 100 kg of fruit preserve. For quality control purposes, soluble solids are those which

change the refractive index and can be measured on a refractometer. Thus only the soluble

solids and sugar are considered soluble solids in this context; pectin excluded.

Pectin grade: A 100 grade pectin is one that will form a satisfactory gel at a ratio of 1 kg

pectin to 100 kg sugar. A 150 grade pectin is one that will form a satisfactory gel at a ratio of

1 kg pectin to 150 kg sugar.

Solution:

85

Example 3.22: The Turkish standard identity for apricot marmalade specifies that the final

product should contain 40% apricot pulp (in essence “pure fruit”). The marmalade of 100 kg

with 68% soluble solids will be produced from the fruit at 10% soluble solids. To produce 100

kg of apricot marmalade, calculate:

a) the weight of the apricot pulp, sugar, acid and pectin,

b) the weight of the sugar and water to prepare pectin solution (3%),

c) the amount of water to be evaporated in this process.

Data:

For the production of 100 kg of marmalade, 400 g of citric acid will be used in

addition the acid naturally present in the fruit. Citric acid will be added at 50%

aqueous solution.

150 grade pectin will be used. The pectin used in this process will be mixed with 5

times its weight of sugar. The pectin will added at 3% of its weight.

The glucose syrup will be added at a concentration that the final product will contain

glucose syrup at 5% of its weight. The soluble solid content of glucose syrup will be

80%.

Solution:

86

Material balance calculations for drying of foods

Material balance calculations are frequently used during the calculations air mixture and

drying yield of fruits and vegetables. Some examples are given below.

Example 3.23: After drying of food initially containing 70% moisture, 80% of its moisture

content was removed. Find out:

a) the moisture removed on the basis of 1 kg of wet material.

b) The composition of dried food.

Solution:

87

Example 3.24: How much weight reduction would result when apricots are dried from 80%

moisture to 18% moisture?

Solution:

Example 3.25: 2080 kg/min of the hot air left from a dryer with the absolute humidity

(mutlak nem) of H1 = 0.04 is mixed with 9135 kg/min of the air taken form atmosphere with

the absolute humidity of H2 = 0.015. By this process, after heating this mixture, the relatively

hot and dry air is sent back to the dryer. Find out the absolute humidity of air mixture (H3).

Solution:

88

Example 3.26: Fresh carrots with 88% moisture are dried to the moisture content of 4%. Hot

air enters the dryer with the absolute humidity of “0.01 kg water/kg dry air,” and then leaves

the dryer with the absolute humidity of “0.0225 kg water/kg dry air.” Find out the masses of

dried carrots leaving the dryer, and the hot air entering the dryer as dry and wet air.

Solution:

89

Material balance calculations involved with dilution and mixing

Example 3.27: How many kilograms of a solution containing 10% NaCl can be obtained by

diluting 15 kg of a 20% solution with water?

Solution:

Volume changes on mixing

When two liquids are mixed, the volumes are not always additive. This is true with most

solutions and miscible liquids. NaCl solution, sugar solutions, and ethanol solution all exhibit

volume changes on mixing. Because of volume changes, material balances must be based on

the mass rather than the volume of the components. Concentrations on a volume basis must

be converted to a mass basis before the material balance equations are formulated.

Example 3.28: Alcohol contents of beverages are reported as percentage by volume. The

density of absolute ethanol is 0.7893 g/cm3. The density of a solution containing 60% by

weight of ethanol is 0.8911 g/cm3. Calculate the volume of absolute ethanol which must be

diluted with water to produce 1 L of 60% by weight ethanol solution. Also calculate alcohol

content of 60% ethanol solution by volume.

Solution:

90

Continuous vs Batch: Material balance calculations are the same regardless of whether a

batch or a continuous process is being evaluated. In a batch system, the total mass considered

includes what entered or left the system at one time. In a continuous system, a basis of a unit

time of operation may be used, and the material balance will be made on what entered or left

the system during that period of time. If the process is continuous, the quantities will all be

mass/time, e.g., kg/h. If the basis used is 1 h of operation, the problem is reduced to the same

form as a batch process.

Example 3.29: An evaporator has a rated evaporation capacity of 500 kg/h of water.

Calculate the rate of production of juice concentrate containing 45% total solids from raw

juice containing 12% solids.

Solution:

91

Recycle

Example 3.30: A pilot plant model of a falling film evaporator has an evaporation capacity

of 10 kg/h of water. The system consists of a heater through which the fluid flows down in a

thin film, and the heated fluid discharges into a collecting vessel maintained under vacuum in

which flash evaporation reduces the temperature of the heated fluid to the boiling point. In a

continuous operation, a recirculating pump draws part of the concentrate from the reservoir,

mixes this concentrate with feed, and pumps the mixture through the heater. The recirculating

pump moves 20 kg/h of fluid. The fluid in the collecting vessel should be at the desired

concentration for withdrawal from the evaporator at any time. If feed enters at 5.5% solids

and a 25% concentrate is desired, calculate;

a) the feed rate

b) the concentrate production rate

c) the amount of concentrate recycled

d) the concentration of the mixture of feed and recycled concentrate

Solution:

92

Material Balance problems involved with blending food ingredients

These problems involve setting up total mass and component balances and solving several

equations simultaneously.

Example 3.31: Determine the amount of a juice concentrate containing 65% solids and

single-strength juice containing 15% solids that must be mixed to produce 100 kg of a

concentrate containing 45% solids.

Solution:

93

Determinants (matrices): Coefficients of linear equations may be set up in matrix and the

matrices resolved to determine the values of the variables.

In a system of equations:

a1 x + b1 y + c1 z = d1

a2 x + b2 y + c2 z = d2

a3 x + b3 y + c3 z = d3

The values of x, y and z are determined as follows:

a1 b1 c1 d1

a2 b2 c2 d2

a3 b3 c3 d3

d1 b1 c1

d2 b2 c2

d3 b3 c3

x = ––––––––––––––––– a1 b1 c1

a2 b2 c2

a3 b3 c3

a1 d1 c1

a2 d2 c2

a3 d3 c3

y = ––––––––––––––––– a1 b1 c1

a2 b2 c2

a3 b3 c3

a1 b1 d1

94

a2 b2 d2

a3 b3 d3

z = ––––––––––––––––– a1 b1 c1

a2 b2 c2

a3 b3 c3

A 2 x 2 matrix is resolved by cross-multiplying the elements in the array (a display) and

subtracting one from the other. The position of an element in the matrix is designated by the

subscript ij, with i representing the row and j representing the column. The element in the

first column whose subscript adds up to an odd number is assigned a negative value. A 2 x 2

matrix and its value are shown below :

a11 a12

= a11 a22 – a21 a12a21 a22

A 3 x 3 matrix is evaluated by multiplying each of the elements in the first column with the 2

x 2 matrix left-utilizing elements in the second and third columns other than those in the same

row as the multiplier. As with the 2 x 2 matrix above, the multiplier whose subscript adds up

to an odd number is assigned a negative value. The multiplier and the 2 x 2 matrices are

determined as follows :

a11 a12 a13 a11 a12 a13 a11 a12 a13

a21 a22 a23 a21 a22 a23 a21 a22 a23

a31 a32 a33 a31 a32 a33 a31 a32 a33

Thus, the 3 x 3 matrix resolves into :

a11

a22 a23

– a21

a12 a13

+ a31

a12 a13

a32 a33 a32 a33 a22 a23

95

Example 3.32: Determine the values of x, y and z in the following equations :

x + y + z = 100

0.8 x + 0.62 y + z = 65

0.89 x + 0.14 y = 20

Solution:

96

Example 3.33: Determine the amounts of lean beef, back fat, and water that must be used to

make 100 kg of a frankfurter formulation by using matrices. The composition of the raw

materials and the frankfurter are:

Lean beef – 14% fat, 67% water, 19% protein

Back fat – 89% fat, 8% water, 3% protein

Frankfurter – 20% fat, 65% water, 15% protein

Solution:

97

Example 3.34: A food fix is to be made which would balance the amount of methionine

(MET), a limiting amino acid in terms of food protein nutritional value, by blending several

types of plant proteins. Corn which contains 15% protein has 1.2 g MET/100 g protein; soy

flour with 55% protein has 1.7 g MET/100 g protein, and nonfat dry milk with 36% milk

protein has 3.2 g MET/100 g protein. How much of each of these ingredients must be used to

produce 100 kg formula which contains 30% protein and 2.2 g MET/100 g protein?

Solution:

98

Concentration with membrane system

Example 3.35: A liquid food is concentrated from 10% solids to 30% solids. The process of

concentration takes place in two steps. The liquid with low percentage of solids obtained

from the second step is recycled to the first step of concentration. The recycled liquid

contains 2% solids, waste contains 0.5% solids, and the semi-concentrate obtained from first

concentration step contains 25% solids. To obtain 100 kg of the concentrated product with

30% solids, determine the rate of recycled feed per h.

Data:

Solid content of feed: 10%,

Solid content of final concentrate: 30%,

Solid content of recycled liquid : 2%,

Solid content of waste: 0.5%,

Solid content of concentrate obtained in the first step of concentration: 25%,

Rate of feed: 100 kg/min.

Solution:

99

Concentration with ultrafiltration

Example 3.36: Whey (peynir altı suyu) proteins are concentrated by ultrafiltration system

from 1.7% solids to 14% solids. Ultrafiltration system has a membrane area of 0.75 m2 and

water permeability of 180 kg water /m2 h. It works under the pressure of 1.033 Mpa. The

system is fed by a pump which delivers 230 kg/h, and the appropriate concentration of solids

in the product is obtained by recycling some of the product through the membrane. The

concentrate contained 11% lactose and the unprocessed whey contained 5.3% lactose. There

is no protein in the permeate.

a) The production of rape of concentrate (25% solids) through the system,

b) The amount of product recycled per h,

c) The amount of lactose removed in the permeate per h,

d) The concentration of lactose in the mixture of fresh and recycled whey entering the

membrane unit,

e) The average rejection factor by the membrane for lactose based on the average lactose

concentrations entering and leaving the unit. The rejection factor (Fr) of solute

through a membrane is defined by:

Xf – X p

Fr = ––––––––Xf

Xf : Solute concentration on the feed side of the membrane, which may be considered

as the mean of the solute concentrations in the fluid entering and leaving the

membrane unit,

Xp : Solute concentrations in the permeate.

Solution:

Multistage processes

100

Problems of this type require drawing a process flow diagram and moving the system

boundaries for the material balance formulations around parts of the process to find out the

unknown quantities.

Example 3.37: In solvent extractions, the material to be extracted is thoroughly mixed with a

solvent. An ideal system is one which the component to be extracted dissolves in the solvent

and the ratio of solute to solvent in the liquid phase equals the ratio of solute to solvent in the

liquid absorbed in the solid phase. This condition occurs with thorough mixing until

equilibrium is reached and if sufficient solvent is present such that the solubility of solute in

the solvent is not exceeded.

Meat (15% protein, 20% fat, 64% water, 1% inert insoluble solids) is extracted with 5 times

its weight of a fat solvent that is miscible in all proportions with water. At equilibrium, the

solvent mixes with water and fat dissolves in the mixture. Assume that there is sufficient

solvent to allow all the fat dissolve.

After through mixing, the solid is separated from the liquid phase by filtration and is dried

until all the volatile material is removed. The weight of the dry cake is only 50% of the

weight of the cake leaving the filter. Assume that none of the fat, protein and inert is removed

from the filter cake by drying and that nonfat solids are in the liquid phase leaving the filter.

Calculate the fat content in the dried solids.

4. ENERGY BALANCE

101

Energy

Energy was first hypothesized by Newton to express kinetic and potential energies. We

cannot observe energy directly, but we can measure energy using indirect methods and

analyze its value. Energy may be in different forms, such as potential, kinetic, chemical

mechanical or electrical.

Potential energy of an object is due to its height. Potential energy is calculated from the

equation 4.1.

P.E. = m g h (4.1)

where;

P.E.: Potential energy, J

m : Mass, kg

g : Acceleration due to gravity, m/s2

h: Height, m.

Kinetic energy of an object is due to its velocity. Kinetic energy is calculated from the

equation 4.2.

1K.E. = ––– m v2 (4.2)

2

where;

K.E.: Kinetic energy, J

m : Mass, kg

v : Velocity, m/s.

The atoms and molecules forming 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, are called as the internal

energy. 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.

102

In many engineering systems, one or two forms of energy may dominate while 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 and electrical do not change and may be neglected in the analysis. Similarly, when

tomato juice is heated in a heater, the potential or kinetic energy of the juice does not change,

but the internal energy will change as temperature increases.

The total energy (kJ) of a system can be written in the form of an equation (4.3).

ETotal = EKE + EPE + EElectrical + EMagnetic + EChemical + ………. + EInternal (4.3).

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

and internal energies, then the total energy can be calculated from the equation 4.4:

ETotal = EKE + EPE + EInternal (4.4)

Energy is not static; it is always in flux. Even under steady-state conditions, an object absorbs

energy from its surroundings and at the same time emits energy to its surroundings at the

same rate. When there is imbalance between the energy absorbed and emitted, the steady-

state is altered, molecular energy in the system may increase, new compounds may be formed,

or work may be performed.

Energy balance calculations can be used to account for the various forms of energy involved

in a system. Mechanical (work), electrical and thermal energies can all be reduced to the

same units. Mechanical energy is used to overcome friction in the system, electrical or

electromagnetic energy, such as microwave energy, is used to increase the heat content of the

system.

An energy balance around a system is based on the first law of thermodynamics, that is, the

law of conservation of energy. This energy law states that the overall energy in an isolated

system does not change, indicating that the energy do not form or disappear, only the energy

in one form can be converted to another form. The boundaries of the system are determined

as in the case of mass balances. The basic energy balance equation is given in the equation

4.5:

103

Energy in = energy out + accumulation (4.5)

In the above equation, all energy terms in the system must be accounted for. The heat content

is expressed as enthalpy. If the system is steady-state, the accumulation term is zero. In this

case, the energy equation will be:

Energy entering the system = Energy leaving the system (4.6)

If the system involves only an exchange of energy between two components, the energy

balance will be:

Energy gain by component 1 = Energy loss by component 2 (4.7)

Energy balances are essential in identifying the effectiveness of energy conservation measures

and identifying areas where energy conservation can be done. These equations are also very

useful in the design of processing systems involving heating or cooling to ensure that fluids

used for heat exchange are adequately provided and the equipment is sized adequately to

achieve the processing objectives. When energy exchange involves a change in mass due to

evaporation and condensation, energy balance can be used during formulation so that, after

processing, the product will have the desired composition.

Enthalpy

Enthalpy is the sum of internal energy and the multiplication of pressure and specific volume

and it can be calculated from the equation 4.8.

H = E + P V’ (4.8)

where;

H: Enthalpy, kJ/kg

E: Internal energy, kJ/kg

P: Pressure, kPa

V’: Specific volume, m3/kg

104

Enthalpy is used to describe the internal energy of the system in some certain situation. For

example, when the enthalpy of an air in a room is described, then this enthalpy includes the

internal energy. On the other hand, if a liquid enters and leaves the open system, then the

enthalpy will be equal to the multiplication of pressure and specific volume ( flow energy). In

this case, the enthalpy of the liquid will be equal to the sum of internal energy and flow

energy.

Enthalpy is an instrinc property which cannot be measured directly. Enthalpy values should

always be given on the reference conditions (in particular temperature). In that case, the

reference temperature (Tref) for determining the enthalpy of water in the steam tables is

0.01oC. The enthalpy of the system which would be calculated from the following equation:

H = cp (T – Tref) (4.9)

Heat

Heat is an energy form that is easy to sense because of its association with temperature. We

know that heat transfers from a hot object to a cold one because of the temperature difference.

Heat exchange between a system and its surroundings is temperature driven. Heat transfer

between a system and its surroundings is probably the most prevalent (widespread) form of

energy that we observe in many food engineering systems. Heat plays a major role in cooking

and preservation (pasteurization and sterilization).

The common units of heat are calorie, kilocalories, BTUs and joule. The definitions of these

units are given:

Calorie: Heat required raising the temperature 1 g of water from 14.5°C to 15.5°C.

Kilo-calorie: Heat required raising the temperature 1 kg of water from 14.5°C to 15.5°C.

BTU: Heat required raising the temperature 1 lbm of water from 63°F to 64°F.

Joule: The equivalent of mechanical energy in heat energy is joule. Joule is defined as the

work done by a force of 1 N in its direction for 1 m.

The equivalents of “joule” in other heat energy units are given below:

105

1 kJ = 0.23889 cal

1 kJ = 0.948 BTU

1 kcal = 4.186 kJ

1 kcal = 3.968 BTU

1 BTU = 0.252 kcal

1 BTU = 1055 J

We will denote heat with a symbol Q, with the unit of joule (J). A sign convention is used in

thermodynamics regarding transfer of heat across a system boundary. If heat transfer is from

a system to its surroundings, then Q is negative. On the other hand, if heat is transferring into

a system from its surroundings (such as in heating of a potato), then heat transfer Q is

positive.

Specific heat: The specific heat is the amount of heat gained or lost for a unit mass of a

substance, accompanying a unit change in temperature but not involving with phase change.

The specific heat which varies with temperature is more variable for gases than for liquids and

solids. Most solids and liquids have a constant specific heat over a fairly wide temperature

range.

The numerical values of cp and cv are similar for solids and liquids; however, they are

considerably different for gases. Since most food processes are carried out under constant

pressure, the specific heat values under constant pressure are used for calculations.

Q = m cp ΔT (4.10)

where;

Q: Heat gained or lost (kJ or kcal),

M: Mass (kg),

ΔT: Change in temperature (K or °C)

The enthalpy change of a material with mass m is calculated from 4.11.

106

T2

Q = m ∫ c dT (4.11) T1

where;

Q: Heat gained or lost, kJ

M: Mass, kg

c: Specific heat, kJ/(kg K)

Δ T: Change in temperature of the substance, K

If the path for energy transfer is under constant pressure, then the heat energy is calculated

from the equation 4.12.

T2

Q = m ∫ cp dT (4.12) T1

where;

cp: Specific heat at constant pressure, J/(kg K).

If the path for energy transfer is under constant volume, then the heat energy is calculated

from the equation 4.13.

T2

Q = m ∫ cv dT (4.13) T1

where;

cp: Specific heat at constant volume, J/(kg K).

For constant pressure processes, the change in enthalpy (ΔH) is equal to heat content (Q).

ΔH = Q

cal kcal BTUFor water → cp = 1 ––––– = 1 –––––– = 1 ––––––

107

g °C kg °C lbm °F

cal kcal BTUFor ice → cp = 0.5 ––––– = 0.5 –––––– = 0.5 ––––––

g °C kg °C lbm °F

Forms of Heat

There are two forms of heat energy: sensible heat (hissedilir ya da duyarlı ısı) and latent heat

(gizli ısı).

Sensible heat is defined as the heat energy transferred between two bodies (systems or

objects, such as foods) at different temperatures. Latent heat is the energy associated with

phase transitions (no temperature changes) such as heat of fusion (erime) from solid to liquid,

and heat of vaporization from liquid to vapor. The combination of sensible heat and latent

heat of the system is called as enthalpy.

If the water in a container at 20°C is heated (or heat energy is given), the temperature of water

increases if we continuously give heat energy. The heat gained by water is the sensible heat.

The amount of sensible heat can be calculated from the following equation (4.10) by using the

temperature difference and specific heat of water:

Q = m Cp ΔT (4.10)

If 1 kg water at 20°C is heated up to 80°C, then

Q = (1 kg) (1 kcal/kg °C) (80 – 20)°C

Q = 60 kcal

1 kg water gained 60 kcal of sensible heat.

If we continue to heat the water, the temperature of water will also increase and finally water

boils. If we still heat the boiling water, the water molecules will start to transfer to vapor

phase. During this process, the temperature of vapor will still be 100°C. Therefore, the heat

energy given to water is used to phase transition, that is, the conversion water molecules from

108

liquid phase into vapor phase. In order to evaporate the water molecules, we need to give the

latent heat of vaporization. The latent heat of condensation is equal to latent heat of

evaporation.

The condensation is the phase change from vapor phase to liquid phase. If the heat is

removed from vapor, then the distance between the vapor molecules will decrease and the

vapor molecules will condensate in droplets. The heat removed during this process is called

as the latent heat of condensation.

If the water in liquid state changes to solid phase, the heat released will be the latent heat of

freezing. During the melting of a frozen material, the same amount of heat absorbed and this

heat is called as latent heat of fusion.

The boiling point of water depends on the pressure. Under normal atmospheric pressure (1

atm), the water boils at 100°C. If the pressure is increased, then the boiling point will also

increase. The latent heat of vaporization of water also depends on the boiling point of water.

The relationship between the pressure, boiling water and latent heat of vaporization is given

in Table 4.1.

Table 4.1 The relationship between pressure, boiling water and latent heat of vaporization

Pressure (atm) Boling point of water (°C) Latent heat of vaporization (kcal/kg)

1 100 540

1.5 110 532.5

0.2 60 563

After all water molecules transfer to vapor phase, the heat absorbed by vapor molecules will

cause the increase in the temperature of vapor molecules. As a result, the sensible heat of

vapor molecules increases.

If the heat is removed from water molecules, then the water begins to cool down. The cooling

of water molecules will continue until the temperature of water molecules drops to 0°C. After

this point, the water molecule begins to freeze. At this point, the temperature of water or ice

molecules stays stable until all water molecules freeze. The heat removed is used for the

109

transformation of water molecules to ice molecules. This heat is called as latent heat of

freezing. The temperature of ice formed is equal to the freezing point of water and this

temperature is 0°C. Contrary to boiling, the pressure has no effect on the freezing point of

water. The water will always freeze at 0°C regardless of the pressure. After all water

molecules freeze, the heat removed will cause the decrease in the temperature of ice

molecules. Therefore, the heat removed during the cooling of ice molecules is the sensible

heat.

Various forms of latent heat

liquid → vapor : latent heat of evaporation is added (about 600 cal per gram)

vapor → liquid : latent heat of condensation is released

liquid → ice : latent heat of freezing is released (about 80 cal per gram)

ice → liquid : latent heat of fusion (melting) is added

During calculations of heat requirements, the change in enthalpy is taken into consideration,

not the absolute value of enthalpy.

Example 4.1: Calculate the enthalpy change during heating of 1 kg water from 20°C to 90°C.

Example 4.2: 1 kg of water at 20°C was boiled under 1.5 atm of pressure and 110°C, and

then transformed to vapor. Calculate the enthalpy change for the vapor 110°C.

110

Example 4.3: The temperature of 1 kg water at 20°C increases to 110°C after heating under

1.5 atmosphere constant pressure. After continuous heating, the temperature of the resulting

vapor at 110°C increases to 125°C. Calculate the enthalpy change during transforming 1 kg

water to vapor at 125°C. The specific heat of vapor is 0.445 kcal/kg °C.

The final enthalpy (629.2) is not the enthalpy of vapor at 125°C. It is the total heat or

enthalpy needed to transfer 1 kg water to 1 kg vapor at 125°C. If we want to calculate the

enthalpy of vapor at 125°C, we need to take into consideration of the enthalpy of water

between 0° and 20°C. The reference point in enthalpy calculation is 0°C.

Then the enthalpy of vapor at 125°C will be the sum of ……………………………

Example 4.4: The ice at –10°C was melted by heating and then under normal atmospheric

pressure, the resulting water molecules were transferred to vapor at 100°C. Calculate the

enthalpy change during transforming 1 kg ice to vapor. The specific heat of ice is 0.5 kcal/kg

°C and latent heat of fusion is 80 kcal/kg.

111

Example 4.5: The fruit at 25°C is frozen to –20°C. The specific heat of fruit above freezing

point is 0.95 kcal/kg °C and the freezing point of fruit is -3°C, the specific heat of fruit below

freezing point is 0.460 kcal/kg °C and latent heat of freezing is 68 kcal/kg °C. Calculate the

enthalpy change during freezing of 1 kg fruit from 25°C to –20°C.

Example 4.6: 6 kg of water at 25°C was boiled to 104°C under 1 atm of pressure. And then,

the steam was heated to 130°C.

a) Find out the enthalpy change in “kcals” during the heating of water at 25°C to vapor at

130°C.

b) Find out the enthalpy of vapor at 130°C in SI unit system by taking into consideration

of enthalpy calculations for water.

112

Calculation of specific heat of solids and liquids

Most solids and liquids have a constant specific heat over a fairly wide temperature range.

For gases, the specific heat varies with temperature. Specific heat of foods very much

depends on the composition and more specifically on the moisture content. If the

composition of a given food is known, then the specific heat of the food can be approximately

determined by using the equations developed for this purpose. However, since the

composition of foods depends on the many factors, the best way is to determine the specific

heat of the foods is experiment. However, in practice, the equations are mostly used for the

calculation of specific heat of the foods.

Since the specific heat of foods mostly depend on the moisture content and the specific heat

of water in liquid and solid state is quite different, the specific heat of foods are calculated

above and below freezing points.

For fat-free fruits and vegetables, purees, and concentrates of plant origin, Siebel (1918)

observed that the specific heat varies with moisture content and that the specific heat can be

determined from the specific heat of water and the specific heat of the solids. For a fat-free

plant material with a mass fraction of water M, the specific heat of water above freezing is 1

BTU/(lb °F) or 4186 J/(kg K), and the specific heat of nonfat solids above freezing is 0.2

BTU/(lb °F) or 837.36 J/(kg K). Since the mass fraction of nonfat solids is (1 – M), the

specific heat of material (food or food product) above freezing is:

Specific heat above freezing:

Cfood = 1 (M) + 0.2 (1 – M) (4.14)

where;

ms : Mass fraction of water in food

1 : Specific heat of water, kcal/(kg C) or BTU/(lbm F)

1 – ms : Mass fraction of non-fat solids in food

Cfood : Specific heat of food, kcal/(kg C) or BTU/(lbm F)

113

The equation 4.14 is reorganized to obtain the equation 4.15.

Cfood = 1 ms + 0.2 – 0.2 ms

Cfood = ms – 0.2 ms + 0.2

Cfood = (1 – 0.2) ms + 0.2

Cfood = 0.8 ms + 0.2 (4.15)

The equation 4.15 in metric or EES unit system is converted to SI in equation 4.16.

Cfood = 3.349 ms + 0.83736 (4.16)

where:

Cfood : Specific heat of food above freezing point, kJ/(kg K)

ms : Mass fraction of water

The conversion of numerical values is given below:

0.8 (4.1868) = 3.349

0.2 (4.1868) = 0.83736

Example 4.7: Calculate the specific heat of orange juice concentrate having a solid content

of 45% in SI unit system.

114

Example 4.8: Calculate the specific heat of strawberries having a brix of 9 in SI unit system.

For fruits and vegetables and the products obtained from fruits and vegetables, only water and

water soluble solids are taken into consideration for the calculation of specific heat. However,

for the animal based products which contain considerable amount of fat, the specific heat of

food should be estimated by taking into consideration of the mass fraction of fat (F). When

the fat is present, the specific heat should be estimated from the mass fraction of fat (F), mass

fraction of non-fat solids (NFS), and mass fraction of moisture (M) as follows:

Cfood = 0.4 F + 0.2 NFS + 1 M ; in BTU/(lb °F) or cal/(kg °C) (4.17)

Cfood = 1.67472 F + 0.83736 NFS + 4.1868 M ;in kJ/(kg K) (4.18)

where;

Cfood : Specific heat of food containing considerable amount of fat above freezing point, kJ/kg K

F : Mass fraction of fat in food

NFS : Mass fraction of non-fat solid in food

M : Mass fraction of water in food

115

Example 4.9: Calculate the specific heat of beef roast containing 15% protein, 20% fat and

65% water in SI unit system.

Since non-fat solids contain carbohydrates, protein and ash, the specific heat of food can be

calculated by taking into consideration of these 3 solids (4.19).

Cfood = 1.424 C + 1.549 P + 1.675 F + 0.837A + 4.187 W (4.19)

where:

Cfood : Specific heat, kJ/(kg K),

m : Mass fraction of the substances in the composition of food,

Notes:

(1) C, P, F, A and W are carbohydrate, protein, fat, ash and water, respectively.

(2) The numerical values in front of the mass fractions are the specific heat of the

substances in their pure state.

Example 4.10: Calculate the specific heat of food containing 30% of carbohydrate, 10% of

protein, no fat, 0.4% of ash, 56% of water in SI unit system.

116

Specific heat below freezing:

For the calculations of specific heat of foods below freezing point, only the change in specific

heat of water into ice is taken into consideration. This is because the specific heat of non-

water substances does not change during phase change. As in the case of specific heat of

food above freezing point, the specific heat of food below freezing point is calculated by 3

different equations depending on the composition.

Cfood = 0.5 ms + 0.2 (1 – ms)

Cfood = 0.5 ms + 0.2 – 0.2 ms

Cfood = 0.3 ms + 0.2 (4.20)

The equation 4.20 in metric or EES unit system is converted to SI in equation 4.21.

Cfood = 1.256 ms + 0.83736 (4.21)

When the fat is present, the specific heat should be estimated from the mass fraction of fat (F),

mass fraction of non-fat solids (NFS), and mass fraction of moisture (M) as follows:

Cfood = 0.4 F + 0.2 NFS + 0.5 M ; in BTU/(lb °F) or cal/(kg °C) (4.22)

Cfood = 1.67472 F + 0.83736 NFS + 2.0934 M ;in kJ/(kg K) (4.23)

Cfood = 1.424 C + 1.549 P + 1.675 F + 0.837 A + 2.0934 W (4.24)

Example 4.11: Calculate the specific heat of strawberries having a brix of 9 below freezing

point in SI unit system.

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Example 4.12 (Homework): Calculate the specific heat of apple juice having a soluble solid

content of 12% in SI unit system above and below freezing points.

Answer:

J1,94 –––––

kg K

The specific heat of 1.94 kJ/kg K below freezing point is the specific heat of apple juice after

all water has been frozen. Freezing of apple juice starts below 0oC and during freezing, the

specific heat of apple juice decrease and finally reaches to 1.94 kJ/kg K. Although the

specific heat of apple juice above freezing is 3.78 kJ/kg K, this value decreases during the

freezing of apple juice and after the freezing of “almost” all water, the specific heat of apple

juice reaches to 1.94 kJ/kg K. Specific heat of various fruit products is given in Table 4.2. As

seen in Table 4.2, the specific heat of foods changes with changing the water content of foods.

As the water content of food increases, then the specific heat of food increases.

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Example 4.13: Calculate the heat required to raise the temperature of a 4.535 kg roast

containing 15% protein, 20% fat and 65% water from 4.44°C to 65.55°C. Express this energy

in a) BTUs, b) kilojoules and c) watt-hour.

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Table 4.2 Specific heat of various food products

Product % water cp

Dairy products Butter 14 2050 Cream, sour 65 2930 Milk, skim 91 4000Fresh meat, fish, poultry and eggs Codfish 80 3520 Chicken 74 3310 Egg white 87 3850 Egg yolk 48 2810 Pork 60 2850Fresh fruits, vegetables and juices Apples 75 3370 Apple juice 88 3850 Cabbage, white 91 3890 Carrots 88 3890 Cucumber 97 4103 Orange juice, fresh 87 3890 Plums 76.5 3500 Spinach 87 3800 Strawberries 91 3805Other products Bread, white 44 2720 Bread, whole wheat 48.5 2850 Flour 13 1800

The specific heat calculated by Siebel’s equation is used by the American Society for Heating,

Refrigerating, and Air Conditioning Engineers in one of the most comprehensive tabulated

values for specific heat of foodstuffs. Although Siebel equation is commonly used for the

calculation of foods, there are two weak points of Siebel equation:

Siebel’s equation assumes that all types of nonfat solids have the same specific heat. This

is not correct. Siebel equation assumes that all non-fat solids have the specific heat of 0.2

BTU/lbm °F.

Siebel’s equation for specific heat below the freezing point assumes that all the water is

frozen, and this is most inaccurate.

Specific heats of solids and liquids may also be estimated using correlations (equations)

obtained from Choi and Okos (1987). The procedure is quite unwieldy (difficult to handle

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due to its size). To overcome this difficulty, the computer software is commonly used. The

specific heats, in J/(kg K), as a function of temperature (°C), for various components of

foods are as follows:

Protein: Cpp = 2008.2 + 1208.9 x 10–3 T – 1312.9 x 10–6 T2 (4.26)

Fat: Cpf = 1984.2 + 1473.3 x 10–3 T – 4800.8 x 10–6 T2

Carbohydrate: Cpc = 1548.8 + 1962.5 x 10–3 T – 5939.9 x 10–6 T2

Fiber: Cpfi = 1845.9 + 1930.6 x 10–3 T – 4650.9 x 10–6 T2

Ash: Cpa = 1092.6 + 1889.6 x 10–3 T – 3681.7 x 10–6 T2

Water above freezing:

Water: Cpwat = 4176.2 + 9.0862 x 10–5 T – 5473.1 x 10–6 T2

The specific heat of the mixture above freezing is:

Cavg = P (Cpp) + F (Cpf) + C (Cpc) + Fi (Cpfi) + A (Cpa) + M (Cpwat) (4.27)

Where P, F, C, Fi, A and M represents the mass fraction of protein, fat, fiber, ash and

moisture, respectively.

Example 4.14: Calculate the specific heat of a formulated food product which contains 15%

protein, 20%starch, 1% fiber, 0.5% ash, 20% fat and 43.5% water at 25°C. (Answer: Cpavg =

2865 J/(kg K))

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Values for Cp calculated using Choi and Okos’ (1987) correlations are generally higher than

those calculated using Siebel’s equations at high moisture contents. Siebel’s equations have

been found to agree closely with experimental values when M > 0.7 and when no fat is

present. Choi and Okos’ (1987) correlation is more accurate at low moisture contents and for

a wider range of product composition. However, the simplicity of Siebel’s equations appeals

to most users.

For enthalpy change calculations, Choi and Okos’ (1987) equations for specific heat must be

expressed as an average over the range of temperatures under consideration. The mean

specific heat, C*, over a temperature range T1 to T2, where (T2 – T1) = δ, (T22 – T1

2) = δ2, and

(T23 – T1

3) = δ3 is:

T2

C*pp = 1/δ ∫ cp dT (4.29)

T1

where;

δ: Temperature difference (ΔT = T2 – T1)

Thus the equations for the mean specific heats of the various components over the

temperature range δ become:

Protein: C*pp = (1/δ) [2008.2 (δ) + 0.6045 (δ2) – 437.6 x 10–6 (δ3)] (4.30)

Fat: Cpf = (1/δ) [1984.2 (δ) + 0.7367 (δ2) – 1600 x 10–6 (δ3)]

Carbohydrate: Cpc = (1/δ) [1548.8 (δ) + 0.9812 (δ2) – 1980 x 10–6 (δ3)]

Fiber: Cpfi = (1/δ) [1845.9 (δ) + 0.9653 (δ2) – 1500 x 10–6 (δ3)]

Ash: Cpa = (1/δ) [1092.6 (δ) + 0.9448 (δ2) – 1227 x 10–6 (δ3)]

Water: Cpwat = (1/δ) [4176.2 (δ) + 4.543 x 10–5 (δ2) – 1824 x 10–6 (δ3)]

The specific heat of the mixture above freezing is:

C*avg = P (C*

pp) + F (C*pf) + C (C*

pc) + Fi (C*pfi) + A (C*

pa) + M (C*pwat) (4.31)

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Example 4.15: Calculate the mean specific heat of the formulated food product in Example

10 in the temperature range of 25°C and 100°C. (Answer: Cpavg = 2904 J/(kg K))

Enthalpy change with a change in phase

When considering the heat to be removed during freezing of a food product, a change phase is

involved and the latent heat of freezing must be considered. Not all water in a food changes

into ice at the freezing point. Some unfrozen water exists below the freezing point; therefore,

Siebel’s equations for specific heat below the freezing point are very inaccurate. The best

method for determining the amount of heat which must be removed during freezing, or the

heat input for thawing, is to calculate the enthalpy change. One method for calculating the

enthalpy change below the freezing point (good only for moisture contents between 73 and

94%) is the procedure of Chang and Tao (1981). In this calculations, it is assumed that all

water is frozen at 227.6K (–45.4°C).

A reduced temperature (Tr) is defined as:

T – 227.6Tr = ––––––––– (4.32)

Tf – 227.6

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where;

Tf : the freezing point temperature

T : the temperature at which the enthalpy is being determined

Two parameters, a and b, have been calculated for different products as a function of the mass

fraction of moisture in the product, M. The correlation equations are:

Meats:

a = 0.316 – 0.247 (M – 0.73) – 0.688 (M – 0.73)2 (4.33-a)

b = 22.95 – 54.68 (a – 0.28) – 5589.03 (a – 0.28)2 (4.33-b)

Vegetable, fruits, juices:

a = 0.362 – 0.0498 (M – 0.73) – 3.465 (M – 0.73)2 (4.34-a)

b = 27.2 – 129.04 (a – 0.23) – 481.46 (a – 0.23)2 (4.34-b)

The freezing point (Tf ), in K, is:

Meats: Tf = 271.18 x 1.47 M (4.35-a)

Fruits and vegetables: Tf = 287.56 – 49.19 M + 37.07 M2 (4.35-b)

Juices: Tf = 120.47 + 327.35 M – 176.49 M2 (4.35-c)

The enthalpy at the freezing point, Hf, in J/kg, relative to 227.6 K, is:

Hf = 9792.46 + 405,096 M (4.36)

The enthalpy at temperature T relative to 227.6 K is determined by:

H = Hf [a Tr + (1 – a) Trb] (4.37)

Example 4.16: Calculate the amount of heat which must be removed in order to freeze 1 kg

of grape juice containing 25% solids from the freezing point to –30°C.

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Specific heats of gases and vapors

The specific heat of gases depends upon whether the process is carried out at constant

pressure or at constant volume. If a gas is heated by blowing air across heating element, the

process is a constant-pressure process. The specific heat is designated by Cp, the specific heat

at constant pressure. The heat required to raise the temperature of a gas with mass m at

constant pressure equals the change in enthalpy, ΔH. The enthalpy change associated with a

change in temperature from a reference temperature To to T2 is:

T2

ΔH = m ∫ Cp dT (4.38) To

ΔH = m Cpm (T2 – To)

Cpm is the mean specific heat over the temperature range To to T2, and Cp is the expression for

specific heat as a function of T.

If a gas is heated from any temperature T1 to a final temperature T2, the change in enthalpy

accompanying the process must be calculated as follows:

ΔH = m Cpm (T2 – To) – m C’pm (T1 – To) (4.39)

where;

C’pm : mean specific heat from the reference temperature To to T1.

Tabulated values for the mean specific heats of gases are based on an ambient temperature of

77°F or 25°C as the reference temperature. Table 1 lists the mean specific heats of common

gases in the American Engineering System of units, and Table 2 lists the specific heats of the

same gases in SI unit. The values for Cpm in Table 1 and 2 change very little at the

temperatures ordinarily used in food processes. Therefore, Cpm based on the temperature

range from To to T2 can be used for ΔH between T1 and T2, with very little error in comparison

with the use of previous equation.

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Example 4.17: Calculate the heating requirement for drying the apples in an air drier that

uses 2000 ft3/min air at 1 atm and 170°F if ambient air at 70°F is heated to 170°F for use in

the process.

Example 4.18: How much heat would be required to raise the temperature of 10 m3/s air at

50°C to 120°C at 1 atm?

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Properties of saturated and superheated steam

Steam and water are the most frequently utilized heat transfer media in food processing.

Water is also a major component of food products. Depending on the heat and water content,

there are three types of steam:

a) Saturated steam or vapor

b) Vapor-liquid mixtures

c) Superheated vapor

These vapors differ from each other, depending on their heat and water contents.

Saturated vapor: This is also known as saturated steam and is the vapor at the boiling

temperature of water. Saturated vapor do not contain water droplets (damlacık). Lowering

the temperature of saturated steam at a constant pressure by a small increment (amount) will

cause the vapor to condense to liquid. The phase change is accompanied by a release of heat.

If heat is removed from the system, temperature and pressure will remain constant until all

vapors are converted to liquid. Adding heat to the system will change either the temperature

or the pressure or both.

At atmospheric pressure, 1 kg of saturated steam occupies 1675 L of volume and contains 539

kcal of latent heat of vaporization. If cooled, 539 kcal of latent heat of condensation is

released from 1 kg of steam.

Vapor-Liquid mixtures: This is the steam containing some water. In other words, steam

with less than 100% quality. In this mixture, steam and water are at the same temperature.

Steam with 95% quality implies that the mixture contains 95 kg of dry steam and 5 kg of

water. Vapor-liquid mixtures are obtained from the cooling of saturated steam; therefore, the

heat content of mixtures is less than that of saturated steam. For example, steam with 95%

quality contains the heat of 539 x 0.95 = 512 kcal/kg.

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Since the mixture contains both vapor and water, addition of heat will not change the

temperature and pressure until all water is converted to vapor. Removing heat from the

system will also not change temperature and pressure until all vapors are converted to liquid.

Superheated steam: Steam at a temperature above the boiling point water. This is obtained

by increasing the temperature of saturated steam at a constant pressure. If the superheated

steam is cooled at a constant pressure, first saturated steam and then vapor-liquid mixture are

obtained. Superheated steam is used in food industry when high temperatures of steam are

needed. For example, it is used to remove the peels of the onions and peppers as well as to

heat the frying oil.

Saturated steam is the most common form of steam used in food industry.

Vapor-water mixtures are not used during steam infusion and injection (in these

applications, food is in direct contact with steam). This is because the vapor-water

mixture will cause to increase the water content of food.

The purity of steam is as important as its water content. Steam should not contain

other gases. If the steam contains corrosive matters, steam can corrode the pipes and

equipment in which it passes through.

Steam tables: The steam tables, which list the properties of steam, are a very useful

reference when determining heat exchange involving a food product and steam or water. The

steam tables are tabulated values for the properties of saturated and superheated steam.

The saturated steam table: The saturated steam table consists of entries under the headings

of temperature, absolute pressure, specific volume and enthalpy. A saturate steam table is in

the Appendix (Table A.3). The temperature and absolute pressure correspond to the boiling

point, or the temperature and pressure under which steam can be saturated.

The entries under in this table are given for saturated liquid, evaporation and saturated steam.

Saturated liquid gives the enthalpy and specific volume of water at the indicated

temperature.

128

Evaporation gives the enthalpy and specific volume during the phase transformation

and numerically, it is calculated from the difference between the properties of

saturated vapor and saturated liquid.

Saturated vapor gives the enthalpy and specific volume of steam at the boiling point.

Specific volume is the reciprocal of the density. It is the volume in cubic feet occupied by 1

lbm of water or steam under the conditions given.

Enthalpy is the heat content of a unit mass of steam or water at the indicated temperature and

pressure. Enthalpy values in the steam tables are calculated from a base temperature of 0°C.

The change in the heat content of steam is equal to the difference between the initial and final

enthalpies of steam.

Example 4.19: At what vacuum would water boil at 80°F? a) Express this in inches of

mercury vacuum. b) absolute pressure in kilopascals.

129

Example 4.20: How much heat would be given off by cooling steam at 252°F and 30.883

psia to 248°F at the same pressure?

Result: Saturated steam is a very efficient heat transfer medium. Note that only a 4°F

change in temperature, 948 BTU/ lbm of steam is given off. The heat content of saturated

vapors comes primarily from the latent heat of vaporization, and it is possible to extract this

heat simply by causing a phase change at constant temperature and pressure.

130

The superheated steam table: The superheated steam table is presented in Appendix

(Tables A.2). Both temperature and absolute pressure must be specified to define the degree

of superheat accurately. From the temperature and absolute pressure, the specific volume and

enthalpy can be read from the table.

Example 4.21: How much heat is required to convert 1 lbm of water at 70°F to steam at

14.696 psia and 250°F?

Example 4.22: How much heat would be given off by cooling superheated steam at 14.696

psia and 500°F to 250°F at the same pressure?

131

Double interpolation from superheated steam table

Since the entries in the steam table do not cover all conditions, it may be necessary to

interpolate between entries to obtain the properties under a given set of conditions. In the

case of superheated steam where both temperature and pressure are necessary to define the

state of the system, double interpolation is sometimes necessary.

Example 4.23: Calculate the enthalpy of superheated steam at 320°F and 17 psia.

132

Heat balances

Heat balance calculations are treated in the same manner as material balances. The amount of

heat entering a system must be equal the amount of heat leaving the system, or: Heat in =

Heat out + accumulation. At a steady state, the accumulation term is zero and heat entering

the system must equal that leaving the system.

Example 4.24: Calculate the amount of water that must be supplied to a heat exchanger that

cools 100 kg/h of tomato paste from 90° to 20°C. The tomato paste contains 40% solids. The

increase in water temperature should not exceed 10°C while passing through the heat

exchanger. There is no mixing of water and tomato paste in the heat exchanger.

Solution:

133

Example 4.25: Calculate the amount of steam at 121.1°C that must be added to 100 kg of a

food product with a specific heat of 3559 J/(kg K) to heat the product from 4.44°C to 82.2°C

by direct steam injection.

Solution:

134

Example 4.26: Steam is used for peeling of potatoes in a semi-continuous 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 17°C, and the peeled potatoes leave at 35°C. A waste

stream from the system leaves at 60°C. The specific heats of unpeeled potatoes, waste

stream, and peeled potatoes are 3.7, 4.2 and 3.5 kJ/(kg K), respectively. If the heat content

(assuming 0°C reference temperature) of the steam is 2750 kJ/kg, determine the quantities of

the waste stream and the peeled potatoes from the process.

Solution:

135

Example 4.27:  The milk is heated in cross-flow heat exchanger (ters akımlı ısı değiştirici) at

a mass flow rate of 1000 kg/h from 42°C to 70°C. The water used for heating the milk enters

the heat exchanger at 95°C and leaves the system at 80°C. If the heat is emitted from the

system to its surroundings is 1 kW, then find out the mass flow rate of water used to heat the

milk. The specific heat of milk is 3.9 kJ/kg °C.

136