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Transcript
```FDE201 MASS AND ENERGY BALANCES
Prof.Dr. Mehmet Özkan
Prof.Dr.Ferruh Erdoğdu
Department of Food Engineering
Ankara University
Phone: 203 3300/3621
e-mail: [email protected], [email protected]
Office Hours: 09:00–11:00 (Thursday)
Teaching Assistant Fatmagül Hamzaoğlu
Department of Food Engineering
Ankara University
Phone: 203 3300
e-mail: [email protected]
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
1
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
concentration, dehydration and filtration
IV.
crystallization,
dilution,
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
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.
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.
3
Table 1–Systems of measurements
System
English
Use
Length
Industrial Foot
Mass
Pound
mass
Time
Temperature
Force
Energy
Second
°F
Pound
force
BTU
Calorie
Metric
Cgs
Scientific Centimeter Gram
Second
°C
Dyne
Mks
Industrial Meter
Kilogram
Second
°C
Kilogram Kilo
force
calorie
Universal Meter
Kilogram
Second
K
Newton
SI
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
candela
cd
Luminous intensity
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.)
4
1 (N) = 1 (kg) x 1 (m s–2)
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
kg
1 (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
5
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.
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
units
m kg s–2
Force
newton
N
Pressure
Pascal
Pa
N m–2
kg m–1 s–2
Energy, work,
joule
J
Nm
kg m2 s–2
watt
W
J s–1
kg m2 s–3
heat
Power
Table 5–Examples of SI-derived units expressed by means of special names
Quantity
Name
Symbol
Viscosity
Pascal second
Pa s
Expression in terms of SI
base units
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
J kg–1 K–1
m2 s–2 K–1
W m–1 K–1
m kg s–3 K–1
kelvin
Thermal conductivity
Watt per meter kelvin
6
Prefixes Recommended for use in SI system
Prefixes for general use are shown in Table 6.
Table 6–Prefixes recommended for use in SI
Prefix
Multiple
Symbol
tera
1012
T
giga
109
G
mega
106
M
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
kilo
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.
7
2) 0.0000001 m should be 1 μm.
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.
8
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.
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.
9
J
BTU
1055 J
ft
1.8°F
°C
h
(CF) – = – – –  ––––– 
ms K
h ft °F
BTU
0.3048 m
°C
K
3600 s
J
1055
BTU J ft °F h
J
(CF) – = –– – = 1.73 
msK
0.3048 x 5/9 x 360 h ft °F BTU m K s
msK
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.
10
Example 3: Heat transfer coefficient of salami at 21°C is hs = 210 BTU/ft2 h °F. Express this
value in SI unit system.
Example 4: The density of cow milk is 64.5 lbm/ft. Express this value in SI unit system.
11
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
Example 6: For a fluid passing through a pipe, type of flow (laminer, kararsız, turbulant)
2
depends on the pressure (ρ V ) 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:
2
ρV
ρVD
Re = –––––– = ––––––
µ V/D
µ
where;
D : diameter, m,
V : velocity, m/s,
3
ρ : density, kg/m ,
μ : viscosity, kg/m s.
12
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.
13
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.
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.
14
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)?
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.
15
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
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.
16
Example 13: Convert “BTU / (lb °F)” to “J / (g K).”
Example 14: Convert the following units to SI units.
a) A density value of 60 lbm/ft3.
b) A pressure value of 14.69 psig.
17
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 → –––––
ms
lbm
 İngiliz → –––––
ft s
a) Express 20 cp in “Pa s.”
18
b) Express 15 cp in “EES.”
c) Express “30 lbm/(ft h)” in “SI” unit system.
19
Example 16: Pressure is expressed in the SI system in Pascal’s. Calculate the English
equivalent of 8 Pa.
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.
20
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.
ft
lbf = lbm x ––––
s2
To make this equation dimensionally consistent, we need to add a dimensional constant (gc).
ft
lbf = 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
21
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.
22
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)
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
23
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:
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)
g
dwater = 1 –––––
cm3
kcal
Cpwater = 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)
24
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
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.
25
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.
W
Mole number (n) = –––––––––
Mw (Aw)
(2.1)
n
Molarity (M) = –––––
V
(2.2)
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.
26
The following equations (2.3, 2.4 and 2.5) are used to calculate the normality o f a solution.
W
Equivalents = ––––––––––––––––
Equivalent weight
(2.3)
Mw
Equivalent weight = ––––
f
(2.4)
Equivalent weight
N = ––––––––––––––––
V
(2.5)
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.
27
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 = –––––––––––––––
1000 g
(2.7)
Example 1: Calculate the molality of concentrated stock HCL solution (%28).
Solution:
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:
28
1 g = 103 mg = 106 μg = 109 ng = 1012 pg
mg
ng
pg
μg
ppm = –––– = –––– = –––– = ––––
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.
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
29
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.
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
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
31
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 = ––––––––––––––––––––––––––––
100 g of solution (food)
(2.8)
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:
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 solution
Mass fraction (XA) = ––––––––––––––––––––––––
weight of solution
WA
XA = ––––––––––
WA + W B
(2.9)
32
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 solution
Mass ratio = –––––––––––––––––––––––––––––
weight of another solute in solution
WA
Mass ratio = –––––
WC
(2.9)
where,
WA : The mass of soluble solute A, g
WB : The mass of solvent B, g
WC : The mass of soluble solute C, g
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 solution
Mole fraction (YA) = ––––––––––––––––––––––––––––––––––––
total number of moles in solution
nA
Mole fraction (YA) = –––––––
nA + nB
(2.10)
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 solution
Mole ratio (YA) = ––––––––––––––––––––––––––––––––––––––––
the number of moles of another solute in solution
nA
YA = –––
nC
(2.11)
where;
33
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 :
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
34
(2.12)
1
XA
XB
Xi
Xi
–––– = –––––– + –––––– + ………………… + –––––– = ∑ –––––
–––
MW
MWA
MWB
MWi
MWi
Example 8 : Calculate the mean molecular weight of air by using:
a) Mass fraction
b) Mole fraction
Solution:
35
(2.13)
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:
36
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)
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.
37
2.2. Density
Density is defined as mass per unit of volume, with dimensions (mass)/(length)3.
m
ρ = ––––
V
(2.14)
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.
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.
38
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
ρ (kg/m3)
Temperature (°C)
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 ρ maks= 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).
39
2) Meanwhile, while the temperature increases from 10 to 12, i.e., 2 unit increase in
temperature, find out the decrease in density.
3)
6 x 10–4 unit decrease in density
5 unit increase in temperature
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.
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
40
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
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 particle
Solid density = ––––––––––––––––––––––––––––
Volume (disregarding all space)
(2.15)
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
Water
1.00
Ethyl alcohol
0.79
Cellulose
Protein
1.27–1.61
1.4
41
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 particle
Particle density = ––––––––––––––––––––––––––––––––––––––––––
Volume (including the internal pores in particles,
disregarding the pores between particles)
(2.16)
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.
Mass of a particle
Bulk density = ––––––––––––––––––––––––––––––––––––––––––
Volume (including the internal pores in particles
and the pores between particles)
(2.17)
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 density
Porosity = 1 – –––––––––––––
Solid density
(2.18)
Bulk density
Interparticle porosity = 1 – ––––––––––––––
Particle density
(2.19)
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
42
 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;
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 temperature
Specific gravity = –––––––––––––––––––––––––––––––––––––
Density of water at the same temperature
43
(2.20)
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 :
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.
1
Specific volume (v) = –––
ρ
(2.21)
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:
44
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
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
45
the system is contacted with other system, both systems ultimately reach the same
temperature.
C
Thermometer
A
Hot object
Heat flow
B
Cold object
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
46
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.).
solid
Cold liquid
superheated steam
Pressure (p)/atm
V = constant
gas
Temperature (t), °C
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
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
47
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 (18241907) 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
scale. Thus, 1 R is equal to 1 °F. There are no negative values in Rankine and Kelvin scales
(Figure 2.3.).
48
Boiling
point of
water
Freezing
point of
water
Absolute
zero
point
100 equal parts
–273°C
0°C
0K
273 K
100 equal parts
100°C
373 K
180 equal parts
–460°F
0°F
32°F
212°F
180 equal parts
0R
492 R
672 R
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
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
49
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.
Decrease by a degree at a
decrease by 1/273 of
pressure at 0°C in gas
pressure.
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.
Temperature, °C
Figure 2.4. Determination of absolute zero temperature by
“pressure-temperature” curve of gases (Allen, 1964)
50
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)
Boiling point (of water)
Freezing point (of water)
Absolute zero
Fahrenheit (oF)
Kelvin (K)
Rankine (R)
100
212
373.15
671.67
0
32
273.15
491.67
–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
Freezing point
0°C
Boiling point
100°C
Fahrenheit
5°C
32°F
212°F
? °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).
51
2) Meanwhile, while the temperature increases from 10 to 12, i.e., 2 unit increase in
temperature, find out the decrease in density.
3)
6 x 10–4 unit decrease in density
5 unit increase in temperature
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:
°F = (9/5 x °C) + 32
or;
°F = 1.8 x °C + 32
52
(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:
ΔT (K) = ΔT (°C)
ΔT (R) = ΔT (°F)
53
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 :
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.
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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
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
55
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 :
 Conversion to lbf/in2 :
56
Example 18 : Convert the pressure of 690 mm Hg into lbf/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.
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.
57
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
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
58
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)
(for pressures greater than Patmosphere)
Pvacuum = Patmosphere – Pabsolute
(for pressures below Patmosphere)
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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 :
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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=ρgh
(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) :
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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) :
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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 :
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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 :
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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
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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),
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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.
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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 coldbreak 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
69
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
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)
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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
71
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
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mass of component A
Mass 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 A
Mass 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:
73
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:
74
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:
75
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
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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:
77
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 :
78
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.
79
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:
80
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:
81
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:
82
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:
83
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:
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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
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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:
86
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
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:
87
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:
88
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:
89
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:
90
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:
91
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:
92
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:
93
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:
94
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
95
a1
b1
d1
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
a21
a22
=
a11 a22 – a21 a12
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
a32
a33
– a21
a12
a13
a32
a33
96
+ a31
a12
a13
a22
a23
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:
97
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:
98
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:
99
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:
100
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 m 2 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:
101
Multistage processes
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.
102
4. ENERGY BALANCE
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.
1
K.E. = ––– m v2
2
(4.2)
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.
103
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 steadystate 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
104
as in the case of mass balances. The basic energy balance equation is given in the equation
4.5:
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
105
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.
106
The equivalents of “joule” in other heat energy units are given below:
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)
107
The enthalpy change of a material with mass m is calculated from 4.11.
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
108
cal
kcal
BTU
For water → cp = 1 ––––– = 1 –––––– = 1 ––––––
g °C
kg °C
lbm °F
cal
kcal
BTU
For 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
109
energy given to water is used to phase transition, that is, the conversion water molecules from
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
110
molecules stays stable until all water molecules freeze. The heat removed is used for the
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.
111
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.
112
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.
113
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)
114
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.
115
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)
Cfood = 1.67472 F + 0.83736 NFS + 4.1868 M
;in kJ/(kg K)
(4.17)
(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
116
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.
117
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 nonwater 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)
Cfood = 1.67472 F + 0.83736 NFS + 2.0934 M
;in kJ/(kg K)
Cfood = 1.424 C + 1.549 P + 1.675 F + 0.837 A + 2.0934 W
(4.22)
(4.23)
(4.24)
Example 4.11: Calculate the specific heat of strawberries having a brix of 9 below freezing
point in SI unit system.
118
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.
J
1,94 –––––
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.
119
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.
120
Table 4.2 Specific heat of various food products
Product
Dairy products
Butter
Cream, sour
Milk, skim
Fresh meat, fish, poultry and eggs
Codfish
Chicken
Egg white
Egg yolk
Pork
Fresh fruits, vegetables and juices
Apples
Apple juice
Cabbage, white
Carrots
Cucumber
Orange juice, fresh
Plums
Spinach
Strawberries
Other products
Flour
% water
cp
14
65
91
2050
2930
4000
80
74
87
48
60
3520
3310
3850
2810
2850
75
88
91
88
97
87
76.5
87
91
3370
3850
3890
3890
4103
3890
3500
3800
3805
44
48.5
13
2720
2850
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
121
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))
122
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 – T12) = δ2, and
(T23 – T13) = 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)
123
(4.31)
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 – 227.6
Tr = –––––––––
Tf – 227.6
(4.32)
124
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
b = 22.95 – 54.68 (a – 0.28) – 5589.03 (a – 0.28)2
(4.33-a)
(4.33-b)
Vegetable, fruits, juices:
a = 0.362 – 0.0498 (M – 0.73) – 3.465 (M – 0.73)2
b = 27.2 – 129.04 (a – 0.23) – 481.46 (a – 0.23)2
(4.34-a)
(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
Juices: Tf = 120.47 + 327.35 M – 176.49 M2
(4.35-b)
(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.
125
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.
126
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 m 3/s air at
50°C to 120°C at 1 atm?
127
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.
128
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.
129

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.
130
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.
131
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?
132
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.
133
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:
134
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:
135
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:
136
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.
137
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