Fde 201 lecture notes

NEAR EAST UNIVERSITY
ENGINEERING FACULTY
FOOD ENGINEERING DEPARTMENT
FDE 201
MATERIAL & ENERGY
BALANCES
LECTURE NOTES
PREPARED BY : Filiz ALSHANABLEH
NICOSIA – 2012

Table of Content
Chapter 1
Dimensions, Units, and Unit Conversion 1
Chapter 2
Introduction to Process Variables and Basic Food Engineering Calculations 21
Chapter 3
Introduction to Material Balances 41
Chapter 5
Material Balances for Multiple Units 87
Chapter 6
Energy Balances 121
Chapter 7
Material & Energy Balances of Some Selected Unit Operations 157
APPENDIX 1 : SYMBOLS, UNITS AND DIMENSIONS 182
APPENDIX 2 : UNITS AND CONVERSION FACTORS 186

FDE 201-LECTURE NOTES-1
Chapter 1:
Dimensions, Units, and Unit Conversion

Chapter 1
Learning Objectives
Upon completing this Chapter, you should be able to:
• understand the definitions and physical meanings of dimensions and units
• perform operations (i.e. addition, subtraction, multiplication, and division) of numbers
accompanied by corresponding units
• identify units commonly used in engineering and scientific calculations, including those in
cgs, SI, and American Engineering (AE) systems
• convert one set of units (or one unit) associated with numbers (or a number) or in an
equation into another equivalent set of units (or another unit), using a given conversion
factor
• explain and utilize the concept of dimensional homogeneity (consistency) of equations to
identify units of specific numbers in those equations
• appreciate the importance and rationale of dimensionless groups (quantities)
James Clark Maxwell, a Scottish Mathematician and Theoretical Physicist (1831– 1879) expressed the
definition of unit that “Every physical quantity can be expressed as a product of a pure number and a
unit, where the unit is a selected reference quantity in terms of which all quantities of the same kind
can be expressed”.
Physical Quantities
• Fundamental quantities
• Derived quantities
Fundamental Quantities
• Length
• Mass
• Time
• Temperature
• Amount of substance
• Electric current
• Luminous intensity

Examples of Derived Quantities
• Area [length × length or (length)2
]
• Volume [(length)3
]
• Density [mass/volume or mass/(length)3
]
• Velocity [length/time]
• Acceleration [velocity/time or length/(time)2
]
• Force [mass × acceleration or (mass × length)/(time)2
]
Dimension
A property that can be measured directly (e.g., length, mass, temperature) or calculated, by
multiplying or dividing with other dimensions (e.g., volume, velocity, force)
Unit
A specific numerical value of dimensions.
Systems of Units (commonly used in engineering and scientific calculations)
• cgs (centimetre, gram, second)
• SI (Le Système International d’ Unitès)
• American Engineering (AE) (or fps: foot, pound, second)
Units & Dimensions of Fundamental Quantities
UNIT SYSTEM
cgs
QUANTITY
SI AE
DIMENSION
• Length
• Mass
• Time
• Temperature
• Amount of Substance
• Electric current
• Luminous intensity
cm
g
s
o
C
_
A
_
m
kg
s
K
mol
A
can*
ft
lbm
s
o
F
_
A
_
L
M
t or θ
T
n
I
* stands for “Candela”

Examples of the Dimensions of Derived Quantities
• Area (A) [ A = L×L = L2
]
• Volume (V ) [ V = L×L×L = L3
]
• Density (ρ ) [ρ = M / V = M / L3
]
• Velocity ( ) [ = L / t ]
• Acceleration (a) [ a= V / t = ( L / t) /( t ) = L / t2
]
Example 1.1 Sir Isaac Newton (English physicist, mathematician, astronomer, philosopher, and
alchemist: 1643–1727) established a second law of motion equation that the force (F ) is the product
of mass (m) and its acceleration (m), which can be described in the equation form as follows:
F=ma
What is the dimension of F (force)?
From a previous page, the dimensions of
• mass (m) ≡ M
• acceleration (a) ≡ L / t2
The dimension of force (F ) can, thus, be expressed by those of fundamental quantities as follows
F = ma
≡ ( M ) ( L / t2
)
F ≡ ( M )( L ) / ( t2
)
Example From a Physics or Chemistry courses, pressure (P ) is defined as “the amount of force (F )
exerted onto the area (A) perpendicular to the force”
What is the dimension of P ?
• Dimension of force F ≡ ( M )( L ) / ( t2
)
• Dimension of area (A) ≡ L2
Hence, the dimension of pressure is
Dimensions and Units
The “dimension” is the property that can be measured experimentally or calculated, and in order to
express the physical quantity of a dimension, we use a pure number and its corresponding unit

For example, a ruler has a dimension of “length” (L ), its physical quantity can be expressed as
1 foot (ft)
or 12 inches (in)
or 30.45 centimetres (cm)
Another example, Americans express their normal freezing point of water as 32 o
F, while Europeans
say that the normal freezing point of water at 1atm is at 0 o
C
We can see that a physical property [e.g., length (L ) or temperature (T )] with the same dimension
may be expressed in different numerical value if it is accompanied with different unit.
Units of Derived Quantities and Alternative Units
The units of fundamental quantities of different unit systems are summarized on Page 3.
What are the units for derived quantities?
We can assign the unit (in any unit system) to each individual derived quantity using its dimension
For instances,
• the unit of area (A), in SI system, is m2
, since its dimension is L2
• the unit of volume (V ), in AE system, is ft3
Example 1.2. Determine the units of density, in cgs, SI, and AE systems?
Since the dimension of density (ρ ) is
its corresponding units in
• cgs unit system is g / cm3
• SI unit system is kg/ m3
• AE unit system is lbm / ft3
What are “alternative” units?
From our previous example we have learned that the dimension of force (F ) is
F ≡ ( M )( L ) / ( t2
)

Hence, its corresponding units in
• SI system is (kg)(m) / (s2
)
• AE system is (lbm)(ft) / (s2
)
Have you ever heard that the unit of force (F ) is (kg)(m) / (s2
)?
To honour Sir Isaac Newton (1643-1727), who established the 2nd law of motion, a community of
scientists gave the name of the unit of force as “Newton (N)”, which is defined as
Newton (N) is an example of an alternative unit.
Accordingly, instead of expressing the unit of pressure (P), in SI system, as
since the dimension of pressure is
we can, alternatively, write the unit of pressure, in SI system, as
This comes from the fact that and that the units of force (F) and area (A), in SI system, is
N and m2
, respectively.
However, the unit of pressure in SI system is expressed as “Pascal (Pa)”, which is defined as
1 Pa ≡ 1
(try proving it yourself that 1 Pa ≡ 1 )
In AE system, the unit of pressure is expressed as or psi
(note that lbf is the unit of force in AE system, not a unit of mass, and that “psi” stands for pound
force per squared inches”
1 N ≡ 0.224809 lbf
Example 1.3. Work (W ) is defined as “force (F ) acting upon an object to cause a displacement (L )”.
What are the dimension and the corresponding unit of work, in SI system?
• Dimension of force F ≡ ( M )( L ) / ( t2
)

• Dimension of a displacement = L
Hence, the dimension of work is W = FL
≡ ( M )( L 2
) / ( t2
)
Accordingly, the unit of work, in SI system, is
Alternatively, the unit of work, in SI system, can be expressed as W = FL
≡ (N)(m)
Commonly, the unit of work, in SI system is expressed as “Joule (J)”, in which
1 J ≡ 1 (N)(m)
Units of Work, Energy, and Heat
We have just learned that the unit of work, in SI system, is J or (N)(m) or
and from Physics courses, we learned that work, energy, and heat are in the same unit
Is it true?
From the definition of work: “force acting upon an object to cause a displacement” the unit of work
can be expressed as J or (N)(m) or ,as mentioned above.
How about the unit of “Energy”?
Energy:
• Potential Energy; Ep =mgL where g is an acceleration (a) caused by gravitational force
Thus, in SI system, potential energy has the unit of
Ep =mgL

• Kinetic Energy; Ek = ½(m)(V)2
Hence, the unit of kinetic energy, in SI system, is Ek = ½(m)(V)2
It is clear that EP and EK are in the same unit, i.e.
, and we have already got the fact that
1 J ≡ 1 (N)(m) ≡
Accordingly, we can conclude that work and energy are in the same unit Since, from Physics or
Chemistry courses, both work and heat are the form of energy transferring between a system and
surroundings, the unit of heat is as same as that of work.
Units of Temperature
A unit of temperature used in any calculations must be an absolute temperature unit
Absolute temperature unit
• SI K (Kelvin)
• AE R (Rankine)
T(K)=T( o
C)+273.15* (1.1)
T(R)=T( o
F)+459.67** (1.2)
* For convenience, the value of 273 is used
** For convenience, the value of 460 is used
The Conversion of the Temperature Units between o
C and o
F
Principle
where
Tnb = normal boiling point of water
Tnf = normal melting/freezing point of water
Hence,

(1.3)
Example 1.4. The specific gravity of liquid is normally reported at 60 o
F in AE system. What is the
equivalent temperature in SI system?
Employing Eq. 1.3 yields
T(o
F)=1.8T(o
C)+32
15.6 o
C
Temperature Difference (ΔT )
Consider the following example:
ΔT ( o
C) = 15 o
C – 10 o
C
= 5 o
C
ΔT (K) = (273+15) K – (273+10) K
= 5 K
Thus, it can be concluded that
ΔT ( o
C) = ΔT (K) (1.4)
When considering in the same manner for o
F and R, we shall obtain the fact that
ΔT ( o
F) = ΔT (R) (1.5)
T(o
F)=1.8T(o
C)+32

How about the relationship between ΔT ( o
C) and ΔT ( o
F) or between ΔT (K) and ΔT (R)?
The temperatures of 10 and 15 o
C are equivalent to the temperatures in AE system of (using Eq. 1.3)
T ( o
F) =1.8(o
C) + 32
T ( o
F) =1.8(10) + 32 and T ( o
F) =1.8(15) + 32 respectively
Hence, the temperature difference between 15 and 10 o
C, in AE system, is
[1.8(15) + 32] − [1.8(10) + 32] = [1.8(5)] o
F
Accordingly, the relationship between ΔT ( o
F) and ΔT ( o
C) can be written in a general form as follows
(1.6)
When doing the same for ΔT (R) and ΔT (K), we obtain the following relationship
(1.7)
Units of Pressure
In addition to Pa and psi, pressure can be presented in the unit of, e.g.,
• atm
• bar
• mm Hg
• in H2O
As same as temperature, the unit of pressure used in any calculations must be an absolute pressure
unit.
Operations with Units
What is the correct answer for 9+5?
• Is 14 the only correct answer?
• Could 9 + 5 be 2?
It is certain that 9 apples + 5 apples = 14 apples
9 oranges + 5 oranges = 14 oranges
but 9 oranges + 5 apples = ?
A fruit salad!!
ΔT ( o
F) =1.8ΔT ( o
C)
ΔT (R) =1.8ΔT (K)

Can 1 + 1 be 13 or 1 – 1 = 99?
These are some examples of confusion caused by writing numbers without units.
Some other worse examples
Consider the following news
“A Chinese air-traffic controller at Shanghai international airport directed the pilots of a Korean
Airlines plane to take the plane to the altitude of 1,500 meters (the plane was at the altitude of
~1,000 meters at the time), but the pilots thought that it was 1,500 feet, which is equivalent to 455
meters. So, instead of climbing up, the pilots lowered the altitude of the plane. This
misunderstanding in “unit” caused the plane to crash, which killed all crewmembers and another five
people on the ground”
(Modified from Wall Street Journal, June 6th, 2001, Page A22)
“In 1999, the Mars Climate Orbiter was crashed to the Martian surface, because engineers forgot to
convert units in SI system to AE one. This damage cost ~US$ 125 million!”
(Modified from Basic Principles and Calculations in Chemical Engineering (7th ed.), Page 15)
By attaching units to all numbers when performing any calculations, you can get the following
benefits:
• reduce/diminish the possibility of errors in your calculations
• a logical approach to solve the problem rather than remembering a formula
• easy interpretation of the physical meaning of the number you are dealing with
Example 1.5. You want to calculate the mass (m) of substance A, when you are given a volume (V )
and a density (ρ ) of 0.2 m3
and 1,250 kg/m3
, respectively, but you totally forgot the relationship
between m, V, and . How would you do?
Addition & Subtraction
You can add or subtract numerical quantities only when they are in the same dimension. On top of
that, to obtain the correct answer, the units of those numerical values must be the same.
For example: You cannot carry out the following addition/subtraction;
• 5 kg – 3 N (mass vs force)
• 45 m3
/kg + 250 m3
(specific volume vs volume)
as the numbers in operations are in different dimensions.

You, however, can do the following addition/subtraction
• 5 km + 5 mi
• 50 m3
/kg – 12 ft3
/lbm
BUT you have to do the UNIT CONVERSION before carrying out an addition/a subtraction (unit
conversion will be discussed in the next section)
Multiplication & Division
You can multiply or divide any units, but you cannot cancel the units unless they are identical
For example:
You can do the followings
• ( 15 kg) (9.81 ) = 147.2 or 147.2 N
• = 600 or 600 Pa
• = 30
but you cannot do the following
• = 30
• = 4
• = 2
since the units are different.
Unit Conversion
In engineering calculations, there are TWO commonly used unit systems: SI and American
Engineering (AE)
As a prospective engineer, you must be careful of handling all sorts of unit systems and be able to
convert a given unit to another competently.

The advantages of SI system over AE one
Consider the unit increment in AE system
Length:
12 inches = 1 foot
3 feet = 1 yard
1,760 yards = 1 mile
Mass:
16 ounces (oz) = 1 pound (lbm)
14 pounds (lbm) = 1 stone
It is evident that the increment in unit is NOT systematic, which usually leads to confusion and errors
On the other hand, the unit increment in SI system is systematic
For example
Length:
100 cm = 1 m
1,000 m = 1 km
Mass:
1,000 mg = 1 g
1,000 g = 1 kg
You can see that the unit increment, in SI system, is in the power of 10, and the incremental
patterns are the same for (almost) all kinds of quantities.
Can you think of any quantity in SI system whose unit increment is NOT in the power of 10?
The power of 10 can be expressed by prefixes, and some commonly used prefixes are
• centi- (c) = 10-2
• milli- (m) = 10-3
• micro- (micro- or μ ) = 10-6
• nano- (n) = 10-9
• deci- (d) = 10-1
• kilo- (k) = 103
• mega- (M) = 106
• giga- (G) = 109

Examples
1 GBytes = 109
Bytes
15 MW = 15 × 106
Watts
30 kN = 30 × 103
N
35 cm = 35 × 10-2
m
Example 1.6. Find the length in ft that is equivalent to 47.25 cm
1 m ≡ 100 cm
and 1 m ≡ 3.28084 ft
Thus,
Example 1.7. An example of nano-sized semiconductor is ZnS (in a semiconductor plant, chemical
engineers produce this kind of semiconductor). If its size is 1.8 nanometres (nm), what is the size in
inches (in.)
Example 1.8 At 4 o
C, water has a density of 1 g/cm3
. Liquid A has a density at the same temperature
of 60 lbm/ft3
. When water is mixed with liquid A, which one is on the upper layer?
Can you answer that which substance is on the upper layer?
Example 1.9. Convert the mass flux of 0.04 g / (min. m2
) to that in the unit of lbm /( h.ft2
)

Dimensional Consistency (Homogeneity)
As stated previously, the answers of adding and/or subtracting numerical quantities can be obtained
only when the unit of each quantity is identical This is a basic principle of “dimension homogeneity
(consistency)”
The basic principle states that, in order to add, subtract, or equate any terms, each term must be in
the same dimension and unit.
By employing this principle, it leads to a conclusion that the numerical values in any non-linear forms
(e.g., log, exp) must be dimensionless (i.e. have no unit)
Example 1.10 What is the unit of R in an ideal-gas equation of state (EoS)?
An ideal-gas equation of state (EoS) can be written as follows
PV = nRT (1.8)
where
P has a unit of Pa (or N/m2
)
V has a unit of m3
n has a unit of mol
T has a unit of K
From the principle of dimensional homogeneity, it is required that each side of equation (i.e. EoS in
this case) must be in the same unit
Hence, the unit of R can be calculated by substituting the unit of each quantity into Eq. 1.8 and, then,
rearranging the equation, as follows
PV = nRT
(Pa)(m3
) = (mol) R (K)
Performing further unit conversions yields

Example 1.11. One of the real-gas EoS is that proposed by van der Waals (called van der Waals
equation of state), which can be written as follows:
Use the principle of dimensional homogeneity to determine the unit of constants a and b (note that
the units of each quantity is as described in the previous Example). The principle of dimensional
consistency states that, to add, subtract, and equate the terms, each term must have the same unit.
Hence, the term
• must be in the same unit as P
• b must have the same unit as V
Accordingly, b is in the unit of m3
. Since the term must have the unit of Pa (i.e. the unit of P ),
the unit of a can be calculated as follows;
Dimensionless Groups or Quantities
Quantities or properties that have NO unit. Mostly, used in “process design” and/or “scaling
up/down”
An example of dimensionless groups
Reynolds number (Re) (used to describe the flow of fluid), which is defined as
where
ρ = density (unit in SI ≡ )[ dimension ≡ ]
D = diameter (unit in SI ≡ m)[dimension ≡ L]
V = velocity (unit in SI ≡ ) [dimension ≡ ]
μ = viscosity (unit in SI ≡ )[dimension ≡ ]

Dimension of Re = ?
Unit of Re = ?
or (-1)
Example 1.12. Use the principle of dimensional consistency to determine the units of the numerical
values of 70.5 and 8.27 × 10-7
in the following empirical formula:
ρ = 70.5exp(8.27 ×10−7
P) (1.10)
where
ρ = density (lbm/ft3
)
P = pressure (lbf/in2
)
The left hand side (LHS) of Eq. 1.10 has the unit of lbm/ft3
Thus, the RHS must have the same unit; i.e. lbm/ft3
Since the number in an exponential (exp) form must be dimensionless, the term exp(8.27 ×10−7
P )
has no unit
Accordingly, the numerical value of 70.5 must have the unit of lbm/ft3
, in order to enable each side of
equation to have the same unit

The term 8.27 ×10−7
P must be dimensionless. Hence, the unit of 8.27 ×10−7
can be determined as
follows
8.27 ×10−7
P
8.27 ×10−7
8.27 ×10−7
In summary, the numerical values of
• 70.5 is in the unit of
• 8.27 ×10−7 is in the unit of

Problem Set 1
PROB 1.
a) In the Olympics, on average, those who enter the final round of the 100 m (men) track event can
run 100 m within 10 seconds. Some said that these men run faster than the car travelling at the
speed of 40 km/h. Is this comment correct?
b) You are in the purchasing division, which is considering buying a new car. The prices of the cars A
and B are almost identical. Other options are also almost the same. However, the fuel consumption
rates are presented in different unit systems. The car A consumes the fuel at the rate of 28 miles/gal,
while the fuel consumption rate for the car B is 9 km/L. Which car is you going to recommend the
company to purchase, and why?
c) The current prices of gasoline (ULG-95) in some countries are as follows: 3.361 U.S. dollars/U.S.
gallon in California, U.S.A.; 4.93 Danish Krones/Litre in Denmark; 384.0 Japanese Yens/U.S. gallon;
and 32.49 Thai Bahts/L. Compare the price of each country in the unit of “Euro/L”. Note that the
current foreign exchange rates per U.S. dollar for Danish Krone, Japanese Yen, Thai Baht, and Euro
are 5.09, 111.72, 33.81, and 0.706, respectively.
d) To lower the fuel cost used by an airplane, it is instructed that the velocity if to be cut down from
525 mi/h to 475 mi/h, which leads to a cut in fuel consumption from 2,200 to 2,000 U.S. gallons/h.
How many litre of fuel that can be saved for the distance of 1,500 km.
PROB 2. Perform the unit conversions for the following questions
a) 554 m4
/(day . kg) = ?? cm4
/(min g)
b) 38.1 ft/s = ?? km/h
c) 921 kg/m3
= ?? lbm/ft3
d) 42 ft2
/h = ?? cm2
/s
PROB 3. A waste-water-treatment basin has the dimension: width × length × depth of
15 m × 50 m × 2 m, and the density of the waste water is 75.6 lbm/ft3
. Calculate the total weight, in
kg, of the waste water, if the basin is completely filled with it.

PROB 4. When a fluid flows through an area that causes any friction, pressure drop arises, and the
pressure drop is proportional to the flow rate or the velocity of the fluid. One the of relationship
between the pressure drop and the velocity of the fluid is proposed by Calvert, as follows:
ΔP = 5×10−5
2L ,
where
ΔP = pressure drop, in psi or lbf/in2
= gas velocity, in ft/s
L = liquid flow rate, in US gallon/min
a) What is a unit of the constant 5 × 10-5
b) What is the dimension of the constant 5 × 10-5
PROB 5. In a manual from a U.S. company, it is said that the unit of a material’s thermal conductivity
(k) is
, but in SI system, the unit of k is reported as . Is this possible?

Chapter 2:
Introduction to Process Variables and
Basic Food Engineering Calculations

Chapter 2
Introduction to Process Variables and Basic Food Engineering Calculations
Learning Objectives
Upon completing this Chapter, you should be able to:
• define what mole is
• define gram-mole, kilogram-mole, and pound-mole
• convert from moles to mass and vice versa
• define density and specific gravity
• calculate a density of a substance given its specific gravity and vice versa
• convert a composition of a mixture from one concentration unit to another
• define mass, molar, and volumetric flow rates; and velocity
Before we can proceed to the main content of this course (i.e. learn to carry out material and energy
balance calculations), some technical terms from basic Chemistry and Physics courses necessary for
material and energy balance calculations should be reviewed, in order to enable you to have a
decent understanding of the materials in the remaining chapters.
Mole
Wilhelm Ostwald (Latvian/German chemist and a 1909 Nobel prize laureate: 1853–1932) introduced
the word “mole” in 1896 as follows
one mole as molecular weight of a substance in grams
In 1969, International Committee on Weights and Measures defined the “mole” as “the amount of a
substance that contains as many elementary entities as there are atoms in 0.012 kg of carbon 12”
(note that the word “entities” can be atoms, molecules, ions, or other particles)
Element vs Compound
Element = a substance that cannot be broken down into smaller substances (e.g., O, H, P, Na, S)
Compound = a substance that comprises atoms of more than one elements (e.g., H2O, H3PO4, SO2)
Entities for • element = atoms
• compound = molecules
From a basic Chemistry course:
1 mole ≡ 6.02214 × 1023
atoms/molecules ≡ 22.4 dm3
or L at STP

Atomic Weight or Molecular Weight vs Molar Mass
To be exact,
• atomic weight (AW) (in some textbooks, atomic mass) is “the compositional average mass of
an element, averaged over the distribution of its isotopes in nature” (for example, atomic
weight of CARBON = 12.011, which is averaged from the atomic weights of C-12 (AW = 12.0),
C-13, and C-14)
• molecular weight (MW) is “a summation of the weights of the atoms in a compound”
Note that both AW and MW are dimensionless or have no unit
• Molar mass (in some textbooks, molar mass and MW is used interchangeably) is the mass in
grams equal to atomic/ molecular weight of one mole of a substance (either element or
compound)
Molar mass (or MW) has a unit of g/mole (g/mol)
Example 2.1.
• Atomic weight of O = 16.00 • Atomic weight of V = 50.94 (V = Vanadium)
Hence, molar mass of • oxygen (O) = 16.00 g/mol , • vanadium (V) = 50.94 g/mol
Example 2.2. A superconductivity material is the material that does not have an electrical
resistance; an example of the superconductivity material is Ba2Cu16O24Y
Calculate the molar mass (or MW) of this superconductor
Given AW of Ba = 137.34, Cu = 63.55, O = 16.00, and Y (Yttrium) = 88.91
MW is the summation of AW, and molar mass of each substance is equal to MW, but has a unit of,
e.g., g/mol
Hence, molar mass (or MW) of Ba2Cu16O24Y can be calculated as follows:
[2×137.34] + [16× 63.55] + [24×16.00] + [1× 88.91] = 1,764.39 g/mol
Example 2.3. How many moles of MeOH if it weighs 32.04 kg? (do you know what MeOH is?)
MeOH is, in fact, “methanol”, CH3OH (CH4O)
AW of C = 12.01, H = 1.008, and O = 16.00
Hence, MW of CH3OH = CH4O = [1×12.01] + [4×1.008] + [1×16.00] = 32.04
Thus, molar mass of MeOH = 32.04 g/mol

It is given that the mass of MeOH is 32.04 kg, in which 32.04 kg = 32.04 × 103
g
32.04 × 103
g = ? mole
= 1 × 103
mol
= 1 kmol
= 1 kg-mol
Example 2.4. How many moles of MeOH if it weighs 32.04 lbm?
From a conversion-factors table:
32.04 lbm = 32.04 × (4.536 × 102
) g
Thus, for MeOH, 32.04 g ≡ 1 mol
= 1× (4.536×102
) mol
= 1× (4.536×102
) g-mol
= 1 lbm-mol
So, in conclusion
• g mol (or g-mol)
• kg kmol (or kg-mol)
• lbm lbm-mol
Hence, the units of molar mass could be
• g/mol or g/g-mol
• kg/kmol or kg/kg-mol
• lbm/lbm-mol
Proposed (New) Method for Calculations Concerning “Mole” and “Mass”
Example 2.5. How many moles of C2H5OH in 100 kg of C2H5OH ? (MW of C2H5OH = 46.07)
From the given data, molar mass of

C2H5OH (or EtOH) is 46.07 kg/kg-mol; in other words,
1 kg-mol of EtOH ≡ 46.07 kg
Thus,
= 2.17 kg-mol EtOH
= 2.17 × 103
mol
= 2,170 mol
Example 2.6. How many lbm-mol of C6H12O6 in 1,000 g of C6H12O6 (MW of C6H12O6 = 180.16)
From the given information, molar mass of C6H12O6 (monosaccharide; e.g., glucose) is
180.16 lbm/lbm −mol
Hence,
= 0.0122 lbm-mol C6H12O6
Density
Density a “ratio of mass per volume”; in other words, it is mass of a unit volume of a specified
substance. Hence, a density can be written in an equation form as follow
(2.1)
Note that, a specific volume (v ) is defined as volume of a unit mass
Accordingly,
(2.2)
From Eqs. 2.1 & 2.2, dimensions of
•
•
Thus, the units of density and specific volume are, e.g.,

• for density:
• for specific volume:
Specific Gravity (SG) of LIQUIDS
Specific gravity of liquid is defined as follows
(2.3)
Normally, T 2= Tref = 15.6 o
C or 60 o
F or 288.7 K (in some textbooks, Tref = 4 o
C)
Note that SG has no unit; in other words, SG is dimensionless
In addition to SG, there are a number of scales that represent specific gravity and/or density, e.g.,
• Twaddell (Tw ) = 200 (SG288.7/288.7 −1)
• o
Baume ( o
Be)
o For liquid heavier than water
o For liquid lighter than water
• (used mostly for petroleum products)
• (commonly used in sugar industry)

Density of GASES
Let’s start with an equation of state of ideal gases:
(2.4)
(Note that, in some textbooks, Rv is used instead of )
We have just learned recently that molar mass (or AW/MW) = m / n (2.5)
where m = mass and n = mole of a selected substance
Hence,
(2.6)
Combining Eq. 2.6 with Eq. 2.4 gives
(2.7)
(note that ≡ gas constant)
Rearranging Eq. 2.7 results in
(2.8)
Combining Eq. 2.2:
with Eq. 2.8 yields
(2.9)

Rearranging Eq. 2.9 gives
(2.10)
This indicates that densities of gases depends directly on T & P
So, it is necessary to specify T & P when dealing a density of any gas.
Specific Gravity (SG) of GASES
“Specific gravity (SG) of gas is the ratio between density of any gas and that of air at the same T & P”
Combining Eq. 2.10:
with the definition that
where = universal gas constant (in some textbooks, may be written as Rν ) yields
(2.12)
Following the same procedure, the density of air at given T & P can be obtained as follows
(2.13)
(2.12)/(2.13) results in SG of any gas, as follows
(2.14)

Example2.7. Determine molar volume and density of ethane (MW = 30.07) at 101.3 kPa (1 atm) and
25o
C (NTP – normal T & P)
Given Rν or = 8.314 m3
⋅kPa/(kmol)(K)
Rearranging an ideal-gas EoS or Eq. 2.4:
yields
(2.15)
Note that ≡ molar volume ( V / n)
Molar volume (v ) of ethane can thus be computed, using Eq. 2.15, as follows
The density of ethane can be calculated by substituting corresponding numerical values into Eq. 2.10
(or Eq. 2.12), as follows
Example 2.8. Calculate the density of air at NTP. Given MW of air = 28.97
The density of air can be calculated using Eq. 2.12, as follows

From the recent Examples, it is worth noting that
1) molar volumes at the same T & P are identical for all gases, as, according to Eq. 2.15 a molar
volume (v ) depends only on T and P (this statement is valid only when all gases are assumed to be
ideal gases)
2) the density of air (@ NTP – 25 o
C & 1 atm) is ~1/1,000 times that of water
(note that ρwater = 1,000 kg/m3
)
Mixtures & Solutions
Normally, chemical (food) engineers usually deal with not only pure substance, but we also deal with
systems or processes that comprise more than one substance, which are called “mixtures” or
“solutions”
Hence, it is necessary to learn how to express the concentration of each substance (or species) in a
mixture or solution, and how to calculate one form of concentration from the given form of
concentration.
From a basic Chemistry course, we have learned that a concentration or each species in a mixture
and solution (soln) can be expressed as, e.g.,
• wt%
• vol%
• %wt/vol
• molarity (molar or M) mol of solute/L of soln: or mol/L)
• molality (mol of solute/1,000 g of solvent)
In this course, in addition to the above concentration expressions, we are going to learn some more
concentration expressions commonly used in chemical processes.
Such forms of concentration include:
• mole fraction and percentage
• mass fraction and percentage
• ppm & ppb
To begin with, we will review how to convert one form of concentration to another.

Example 2.9. Convert a concentration of a 10 wt% H2SO4 solution to the unit of molarity
(mol H2SO4/L soln: mol/L)
From the definition of wt%, we can write the following relationship:
In this Example, we want to convert
To convert
• g H2SO4 to mol (in “g-mol”) H2SO4
• g H2SO4 solution to volume (in L or mL) of soln
what kind of “data” do we need?
• g H2SO4 g-mol H2SO4 H2SO4
data needed = MW of H2SO4
• g solution L solution
data needed = density of solution
• MW of H2SO4 = 98.08 g/g-mol
• ρ of 10 wt% H2SO4 soln = 1.064 g/mL (@ 25 o
C)
Accordingly,
Thus,
In conclusion, a sulphuric acid solution with the concentration of 10 wt% is equivalent to that of
1.085 g-mol/L .

Example 2.10. Convert the concentration of a 10 wt% H2SO4 solution to the units of vol%
From a basic Chemistry course, we can formulate the following equation regarding vol% as follows:
From our recent calculations ,
100 g H2SO4 soln ≡ 93.98 mL H2SO4 soln
In order to convert g H2SO4 to mL H2SO4,
what kind of data do we needed?
DENSITY of H2SO4
From “Perry’s Chemical Engineers’ Handbook”
Density of H2SO4 = 1.834 g/mL
Hence,
Thus, vol% of a 10 wt% H2SO4 solution is
= 5.80 vol%
Mass & Mole Fractions/Percentages
Mass fraction/percentage and mole fraction/percentage are defined as follows
(2.16)
(2.17)

(2.18)
(2.19)
Example 2.11. An industrial drain cleaner contains 6.0 kg of water (H2O: MW = 18.02) and 4.0 kg of
NaOH (MW = 40.00).
Compute: (a) mass fraction and (b) mole percentage of each species in the mixture
(a) The total mass of all species in the mixture is
6.0 kg H2O + 4.0 kg NaOH = 10.0 kg mixture
Hence,
(b) In order to calculate mol% of the mixture, mass of each species must be converted to “mole” first
To convert mass to mole, what kind of data do we need?
MOLAR MASSES (with the unit of, e.g., kg kg -mol)
MW of •H2O = 18.02 kg/kg-mol
•NaOH = 40.00 kg/kg-mol
Hence,
Thus, total moles of all species in the mixture is
0.33 kg-mol H2O + 0.10 kg-mol NaOH = 0.43 kg-mol mixture

Hence, mole percentages of
Proposed Method for Calculations Concerning Mass & Mole Fractions/ Percentages
From a recent Example, a new way of calculating mass fraction/percentage and mole
fraction/percentage, in a table format, is proposed as follows
Species Mass Mass
fraction
Molar
mass
Mole Mole
percentage
H2O 6.0 0.60 18.02 0.33 76.7
NaOH 4.0 0.40 40.00 0.10 23.3
TOTAL 10.0 1.0 0.43 100.0
Example 2.12 A mixture of gases from coal gasification has the following composition
• CO2 5.3 mol%
• CO 27.3%
• H2 16.6%
• CH4 3.4%
• N2 47.4%
What is the composition in mass fractionof this mixture?
Assuming that the total number of moles of all species in this mixture is 100 g-mol (in other words,
we set the basis of calculation as 100 g-mol of the mixture)
Hence, the mixture comprises
CO2 = 5.3 g-mol, CO = 27.3 g-mol, H2 = 16.6 g-mol, CH4 = 3.4 g-mol, N2 = 47.4 g-mol
Species Moles
(g-mol)
Molar
mass
Mass
(g)
Mass
fraction
CO2 5.3 44.01 233.25 0.097
CO 27.3 28.01 764.67 0.317
H2 16.6 2.016 33.47 0.014
CH4 3.4 16.04 54.54 0.022
N2 47.4 28.02 1,328.15 0.550
TOTAL 100.0 2,414.08 1.000

Average MW ( )
In order to compute molecular weight (or molar mass) of a mixture (called “average molecular
weight”: ), the following equation is employed
(2.20)
Example 2.13. Calculate average MW of air, which comprises 79 and 21 mol% of N2 and O2,
respectively (note that this is an approximate composition of air). Given MW: of N2 = 28.02 and O2 =
32.00
To compute an average MW of air ( ) air M , we employ Eq. 2.20 in the table format, as follows
Species Mol % Mole fraction( ) Molar mass ( )
N2 79 0.79 28.02 22.136
O2 21 0.21 32.00 6.72
TOTAL 10.0 1.0 28.856
Thus, an average MW of air = 28.86 ( the exact and widely-used value of is 28.97)
Example2.14. A mixture of liquefied mixture has the following composition:
• n-C4H10 50 wt%
• n-C5H12 30%
• n-C6H14 20%
Calculate an average MW of the mixture?
Basis: 100 g of the mixture
Hence, the mixture contains
• n-C4H10 50 g
• n-C5H12 30 g
• n-C6H14 20 g
Since this is a “mass” basis, or the composition of the mixture is given as “mass percentage”, in order
to compute an average MW of this mixture, we have to convert mass fraction to “mole fraction”,
before being able to use Eq. 2.20, as follows
Species Mass (g) Molar mass ( ) Mole (g-mol) Mole fraction( )
C4H10 50 58.12 0.860 0.570 33.13
C5H12 30 72.15 0.416 0.276 19.91
C6H14 20 86.17 0.232 0.154 13.27
TOTAL 100 1.508 1.0 66.31

Part per million (ppm) and Part per billion (ppb)
“The units of ppm and ppb are used to express the concentrations of trace species”
For instances, the concentrations of gaseous pollutants (e.g., SO2, NO, CO) inthe air, or the
concentrations of heavy metals (e.g., Pb, Hg, Cd) in the waters (e.g., river, cannel, ocean)
The concentrations in the forms of “ppm” and “ppb” can be expressed in equation forms as follows
(2.21)
(2.22)
Accordingly,
(2.23) & (2.24)
(2.25) & (2.26)
In case of “gaseous” solutions, ppm (or ppb) may mean
(2.27)
(2.28)
Thus, for clarification, they are denoted as “ppmv” & “ppbv”, respectively
Flow rate
By saying “rate”, it means “how fast” or it means “per unit time” Thus,
• Mass flow rate has a dimension of and the unit of mass flow rate can be,
e.g., kg/s (SI) or lbm/s (AE)
• Volumetric flow rate is in the dimension of or , and has the unit of,
e.g., cm3
/s (cgs), or m3
/s (SI)

Example A volumetric flow rate of diesel oil in a pipe is found to be 80 US gallons per minute. Find
the mass flow rate in kg/s ( ρdiesel = 0.84 kg/L)
Volumetric flow rate & Velocity
Since volumetric flow rate and area have the dimensions of t and , respectively, we obtain the
fact that
(2.29)
Example 2.15. Velocity of natural gas (NG) in a 300-mm diameter pipe is found to be 5 m/s (using an
anemometer). Find
(a) volumetric flow rate (in ft3
/min) &
(b) mass flow rate (in kg/s).
Given
ρNG ( @25 o
C & 5 atm = 3.28 kg/m3
)
Assume that the pipe has a constant diameter (of 300 mm)
The cross-sectional area of the pipe can thus be calculated as follows
Thus, by rearranging Eq. 2.29 and substituting numerical values into the resulting equation, we
obtain
Volumetric flow rate = Velocity x Area
Performing a unit conversion yields

Mass flow rate of NG can be computed, using the data of volumetric flow rate and density of NG, as
follows
Example 2.16. The “molar” flow rate of octane (C8H18) in a 3-in. diameter pipe is 35 lbm-mol/min. Find
(a) the mass flow rate, in kg/s and (b) velocity, in m/s.
Given: ρC8H18 = 703 kg/m3
and MWC8H18 = 114.22
(a)
Given MW = 114.22 means that a molar mass of C8H18 = 114.22 lbm/lbm-mol
Hence,
Performing a unit conversion gives
Hence, the volumetric flow rate of octane is
(b) Performing a unit conversion for a pipe diameter yields
Hence, the cross-sectional area of the pipe can be computed as follows
Thus, the velocity of octane can be calculated as follows

Problem Set 2
Introduction to Process Variables and Basic Engineering Calculations
PROB 1. Determine that NH4Cl with the mass of 267.5 kg has
a) how many g-mol?
b) how many lbm-mol?
c) how many molecules?
PROB 2. How many kilograms (kg) of silver nitrate (AgNO3) are there in
a) 13.0 lbm-mol of AgNO3?
b) 55.0 g-mol of AgNO3?
PROB 3. Magnesite is the mixture of MgCO3 81.0 %, SiO2 14.0%, and H2O 5.0% by mol. Find the
composition of the magnesite in the form of mass fraction
PROB 4. Glass comprises Na2O 7.65%, MgO 10.57%, ZnO 7.25%, Al2O3 1.19%, B2O3 7.43%, and SiO2
65.91% by mass. Calculate the composition of this glass in the form of mole percent
PROB 5. The gaseous mixture from a polymer-producing plant contains C2H4 30.6%, C6H6 24.5%, O2
1.3%, CH4 15.5%, C2H6 25.0%, and N2 3.1% by volume. Determine
a) average MW (M) of the gaseous mixture
b) density, in the unit of kg/m3
, of this mixture
(Given that 1 mol of the gaseous mixture has the volume of 0.0224 m3
@ STP)
PROB 6. Liquid A flows in the tube with a diameter of 2.5 cm at the flow rate of 30 cm/s. Calculate
the flow rate of liquid A in the units of
a) L/h
b) kg/min
(Given the density of liquid A of 1,553 kg/m3
)

PROB 7. The mixture of methanol and methyl acetate comprises 15 wt.% methanol. If the flow rate
of methyl acetate is found to be 100 kg-mol/h, calculate the flow rate, in g-mol/s, of methanol.
(Given MW of methanol and methyl acetate of 32.04 and 74.08, respectively)
PROB 8. Flue gas from the combustion of coal is at the temperature and pressure of 190 o
C and 100
kPa, respectively, and contains CO2 10.0%, O2 7.96%, N2 82.0% and SO2 0.04% by volume. Find
a) an average MW (M) of the flue gas
b) a concentration of SO2, in the units of ppmv and ppm
PROB 9. A gas mixture comprises CH4 30 mol%, H2 10%, and N2, 60%. How many kilograms (kg) of this
mixture when it is in the amount of 3 lbm-mol?

Chapter 3:
Introduction to Material Balances

Chapter 3
Introduction to Material Balances
Learning Objectives
Upon completing this Chapter, you should be able to
• understand what process flow sheet (PFS) or process flow diagram (PFD) is
• know a standard symbol for each of some important process equipments
• draw a simple process flow diagram (or a “block flow diagram” or a flow chart) for the given
problem
• make a necessary assumption/necessary assumptions pertaining the given problem
• set up an appropriate “basis of calculation” (basis) of the given problem
• understand what “steady-state” process is and how it affects the establishment/set-up of
material-balance equations
• understand what “overall” and “species” balances are
• establish overall- and species-balance equations for the given problem
• solve simple material-balance problems
One of the main responsibilities of food engineers is to create/construct/ analyse chemical
processes (or, at least, to understand the existing processes) The layout of a chemical process is
called “process flow sheet (PFS)” or “process flow diagram (PFD)”. PFS or PFD can be for just a single
process unit of for the whole process, either simple or complicated process.
PFD for a water-softening by ion-exchange process (from Stoichiometry, 4th ed; Bhatt et.al, 2004)

Normally, a PFS or a PFD comprises
• all major process equipments/units
• lines entering or leaving the process/unit and/or lines connecting two or more process
equipments/units (these lines are called “streams”)
• flow rate of each stream
• composition of each stream
• operating conditions of each stream and/or unit/equipment (e.g., T, P)
• energy/heat needed to be added to and/or removed from any particular part of the process
or the entire process
Some important symbols of process equipment are illustrated as follows
(from “Introduction to Chemical Processes” by Murphy, 2007)
In order to be able to create or to understand PFS or PFD, the knowledge concerning MATERIAL &
ENERGY BALANCES is required.
As a food engineer, you need to be able to perform “material and energy balances” for any
particular process or for the entire process efficiently/competently. We start our learning by doing
MATERIAL BALANCES, using an underlying knowledge of “law of conservation of mass”

Principles of MATERIAL BALANCE
In the case that there is NO Chemical Rxn.
However, the “material balance” problem will be more complicated (but not too difficult – believe
me!) when there is/are a Rxn./Rxns. in the process/unit, as follows:
From the principles above, the following equations can be written:
In the case that there is NO Rxn.
where
= initial mass of a system
= final mass of a system
In the case that there is/are a Rxn./Rxns.
Normally, chemical processes are continuous, the change in mass should, therefore, be written in the
“rate” form (i.e. it changes with time)
Eqs. 3.1 & 3.2 can, then, be re-written, as follows
Total mass entering a process/unit – Total mass leaving a process/unit
= Mass accumulation in a process/unit
Total mass entering a process/unit – Total mass leaving a process/unit +
+ Mass generating from a Rxn/Rxns
= Mass accumulation in a process/unit

Example 3.1. One thousand (1,000) kilograms of a mixture of benzene (B) and toluene (T), containing
40% by mass of B is to be separated into two streams in a distillation column. The top output stream
of the column contains 375 kg of B and the bottom output stream contains 515 kg of T.
(a) Perform the mass balance for B & T
(b) Determine the composition of the top and bottom streams
Standard Procedure:
1) Make any necessary assumption(s)
For instance, in this case, we make an assumption that the process is “steady state”
2) Draw a flow chart of the process/unit
3) Set a “basis of calculations”
In this case, we should set a basis as “1,000 kg of mixture”
4) Determine the numbers of unknowns
In this case, there are 3 unknowns:
x = ?
y = ?
% of T of an input stream = ?
5) Establish “material balance” equations
In order to be able to solve for unknowns, it is necessary that the # of Eqs. must be equal to the # of
unknowns.
In this Example, since the input stream consists of only 2 components, we obtain the following
equation:
wt% of B + wt% of T = 100

Accordingly,
% T = 100 - % B
100 = 40
% T = 60%
Note that one unknown is eliminated (the # of unknowns are now only 2)
From the basis of calculation we have set (in Step 3) and from the percentages of benzene and
toluene (the percentage of toluene (%T) has been solved in Step 5), we obtain the information that
the input stream comprises:
Benzene (B) = (1,000 kg) = 400 kg
Toluene (T) = (1,000 kg) = 600 kg
In a general form, let
mF = mass of the input stream (feed)
mB = mass of benzene in the input stream
mT = mass of toluene in the input stream
yB = mass fraction of benzene in the input stream
yT = mass fraction of toluene in the input stream
We can, then, write the following equations
For the output streams, let
mtop = mass of the top stream
mbottom = mass of the bottom stream
Since the process is steady-state (as we made an assumption in Step 1) and has no Rxn., the mass
balance equation can be written as follows

From the process flow chart (on Page 4), we obtain the facts that
and that
Thus,
Eq. 3.5 is an example of an “overall mass balance” equation
Substituting corresponding numerical values into Eq. 3.5 yields
In order to solve for 2 unknowns in Eq.3.6 (i.e. top m & bottom m ), only 1 equation is NOT enough
We need to have 2 equations; since we have already got one, we, therefore, need another equation
To obtain another equation, we need to do a species balance
In this Example, we shall perform a “benzene” balance, as follows
Let
(mB)top = mass of benzene in the top output stream
(mB)bottom = mass of benzene in the bottom output stream
we can, then, write the following equation:
(3.7)
or (3.8)

Eqs. 3.7 & 3.8 are the examples of species balance equations (in this case, it is called “benzene
balance” equations)
Substituting numerical values into Eq. 3.8 gives
It is given that
Hence,
We can also perform a toluene balance, as follows
(3.9)
or (3.10)
It is given that
Thus, we obtain
We can summarise our calculations as illustrated in the following Table
Species Input Output (kg)
(kg) Top Bottom
Benzene 400 375 25
Toluene 600 85 515
TOTAL 1,000 460 540
Mole Balance
Performing “mole balance” is similar to that of the mass balance, but “overall” mole balances are
applicable only for the processes that have no Rxns
Thus, in the case that there is NO Rxn.:

Overall balance
Species balance
where
= initial total # of moles of all species in the system
= final total # of moles of all species in the system
= # of moles of species j in the system at the initial state
= # of moles of species j in the system at the final state
Eqs. 3.11 & 3.12 can also be written in the “rate” form (i.e. with respect to “time”), as follows
Overall balance
Species balance
In the case that there is a Rxn/are Rxns, mole balance is still applicable, but only for species
balances (NOT for an overall balance), as follows
or, in the rate form

Example 3.2. An experiment on the growth rate of organisms requires an environment of humid air
enriched in oxygen. Three input streams are fed into an evaporation chamber:
• liquid water, fed at the volumetric flow rate of 20 cm3
/min
• air (21.0 vol% O2 & 79 vol% N2)
• pure O2, with the molar flow rate of one-fifth (1/5) of the stream of air to produce an output
stream with the desired composition
The output gas is analysed and is found to contain 1.5 mol% of water
Calculate all unknowns
(Data: Density of water is 1.0 g/cm3
; and MW of water = 18.02, of O2 = 32.00, & of N2 = 28.02)
Assumption: Steady-state process
Flow chart:
Since the process is steady-state, (as we made an assumption),
From Eq. 3.13
is then reduced to
From the flow chart, we can write an overall mole balance equation, as follows

Let
Thus, Eq. 3.17 can be re-written as follows
It is given that
It is also given that the volumetric flow rate of stream 3 = 20 cm3
/min
Thus, the mass flow rate of stream 3 can be calculated as follows
which can be converted to the molar flow rate as follows
Thus,
Performing species balances:
H2O balance:
where, = mole fraction of H2O in stream 4
Substituting corresponding numerical values into Eq. 3.21:

Substituting & into Eq. 3.20:
Thus, the flow rate of stream 2 can be computed, using Eq. 3.19, as follows
N2 balance:
= 64.8 mol%
(note that, for gases, vol% = mol% )
In other words, mol% of N2 in the output stream (stream 4) is 64.8%
Thus, mol% of O2 in the output stream (stream 4) or the value of y can be computed as follows
mol% O2 = 100 - mol% H2O - mol% N2
y = 100 – 1.5 – 64.8
= 33.7 %mol
Finally, we can summarise our calculations in the following Table:
* Flow rate of each stream is in the unit of g-mol/min

Chapter 4:
Material Balances for a Single Unit

Chapter 4
Material Balances for a Single Unit
Learning Objectives
• analyse a problem statement and organise the solution strategy, by applying the principles
and procedures learned from the previous Chapter (Introduction to Material Balances)
• solve material-balance problems without chemical reactions
• write and balance chemical reactions
• determine the stoichiometric quantities of reactants and products on molar and mass bases
of given chemical reactions
• define excess and limiting reactants, conversion, degree of completion, yield, selectivity, and
extent of reaction
• calculate excess percentage of excess reactants, conversion fraction or percentage, degree of
completion, yield, selectivity, and reaction extent for given problems
• solve material-balance problems involving chemical reactions
Material Balances for Processes without Chemical Reactions
In the previous Chapter (Introduction to Material Balances), we have learned how to solve simple
material-balance problems, by following a standard material-balance problem solving procedure.
In this Chapter, we extend the principles obtained from the previous Chapter to solve more
complicated material-balance problems, involving or without chemical reactions.
Mixing Process
A mixing process involves two or more streams and the resulting combined stream, for which its
concentration is to be solved
Let’s consider the following Examples

Example 4.1 Two methanol + water mixtures are contained in the separate tanks. The first mixture
contains 40 wt% methanols, and the second contains 30 wt% water. Mix 200 g of the first mixture
with 150 g of the second one. What are the total mass and composition of the product?
Assumptions:
• Steady-state process (i.e. NO accumulation and NO leaking in the mixing tank)
• NO chemical reaction between water and methanol
Flow chart:
Unknowns:
• wt% of water in Stream 1
• wt% of MeOH in Stream 2
• Mass of the product (x )
• wt% of MEOH and water in the product stream (y and z )
Overall mass balance:
m1 + m2 = mproduct
mproduct = 200 g + 150 g
Thus, the mass of product (x) is 350 g
Since Streams 1 & 2 comprise only MeOH & water,
• wt% of water in Stream 1 = 100 – 40.0 = 60.0 wt%
• wt% of MeOH in Stream 2 = 100 – 30.0 = 70.0 wt%

Species balance:
We can perform a species balance for either MeOH or water.
In this Example, we shall do the MeOH balance, as follows
Substituting numerical values into Eq. 4.1 gives
Once again, since the product stream contains only MeOH and water, the mass fraction of water in
the product stream can, then, be calculated as follows
)
(Try performing“water” balance to cross-check the answer)
Hence, in conclusion,
1) total mass of product is 350 g
2) the composition of the product (in mass fraction) is
• = 0.529 (52.9%)
• = 0.471 (47.1%)
Example 4.2. You are asked to make 1,000 kg of the mixed acid containing 60 wt% H2SO4, 32 wt%
HNO3, and 8 wt% water, by mixing
i) mixed acid containing 11.3% HNO3, 44.4% H2SO4, and 44.3% water,
ii) 90% HNO3 aqueous solution, and
iii) 98% H2SO4 aqueous solution
Calculate mass of Streams i, ii, and iii
Assumptions:
• Steady-state process (i.e. NO accumulation and NO leaking in the mixing tank)
• NO chemical reactions between water and acids and between H2SO4 and HNO3

Basis: 1,000 kg of the product (i.e. the resulting mixed acid solution)
Flow chart:
Unknowns:
• Mass of Stream i (x )
• Mass of Stream ii (y)
• Mass of Stream iii (z )
Hence, we need 3 (three) equations
Overall balance:
We still need another 2 (two) equations.
In order to get another 2 Eqs., we have to perform species balances
Species balance:
H2SO4 balance:
HNO3 balance:
Substituting given numerical values into Eqs. 4.3 & 4.4 results in
(0.444) x + (0)y + (0.98)z = (0.60)(1,000) (4.5)
(0.113) x + (0.90)y + (0)z = (0.32)(1,000) (4.6)
So, we have 3 equations:
x + y + z =1,000 (4.2)

(0.444) x + (0)y + (0.98)z = 600 (4.5)
(0.113) x + (0.90)y + (0)z = 320 (4.6)
Solving Eqs. 4.2, 4.5, & 4.6 simultaneously yields
x = 76.4 kg
y = 346.0 kg
z = 577.6 kg
Comments:
In this case, there are 3 unknowns (i.e. x, y, & z) We, however, need to solve for only two, out of
three, unknowns. The last one (i.e. the 3rd unknown) is to be obtained from the difference between
1,000 (the total amount) and the summation of the other two solved unknowns.
For example, when we obtained x = 76.4 kg and y = 346.0 kg, we can, therefore, get the value of z
from
1,000 – (76.4 + 346.0) = 577.6 kg
or when we got the values of y = 346.0 kg and z = 577.6 kg, we shall get the value of x from
1,000 – (346.0 + 577.6) = 76.4 kg
In such case, we can say that the “degree of freedom (dof)” of this question is 2 (two). In other
words, we need to solve for only 2 unknowns, and the 3rd one will “automatically” be obtained from
the difference of the total value and the summation of the other TWO solved unknowns.
Mixing Point
“The principle of a mixing point is the same as that of the mixing tank”, except that there is/are no
REAL (or ACTUAL) mixing tank/tanks
Example 4.3 Stream 1, containing only substance A, is mixed with Stream 2, which contains
substances A & B, at the mixing point. The flow rates of Streams 1 & 2 are 60.0 & 30.0 kg/h,
respectively. The mass fraction of A in the Stream 2 is 0.35. Determine the composition of the output
stream.
Assumptions:
• Steady-state process
• NO chemical reaction between species A and B

Flow chart:
Basis: 1 h of operation
(Note: it is common for a continuous process to set up a “basis” as the time of the operation)
Thus,
m1 = 60.0 kg (containing A only)
m2 = 30.0 kg (containing A & B)
Unknowns:
• Wt% of species B in Stream 2
• Mass of the output stream (mo)
• Wt% of species A and B in the output stream (y, and (100 − y) wt%)
Since Stream 2 contains only A & B, when mass fraction of A = 0.35 (or wt% of A = 35%), mass
percentage of B = 65% (or yB = 1.00 – 0.35 = 0.65)
Overall balance
m1 +m2 =mo
60.0 + 30.0 = mo
mo = 90.0 kg
Species B balance

= 78.3 %
Hence, we can conclude that
• the flow rate of the output stream (mo) is 90.0 kg/h
• the composition of the output stream is A = 78.3 wt%
B = 21.7 wt% (100 – 78.3)
Drying & Concentrating Processes
Drying and concentrating processes involve two species, e.g., solutes and solvent.
Normally, a solute is to be concentrated by removing (e.g., drying) a solvent.
Example 4.4 Evaporator is used to concentrate a dilute 4 wt% caustic soda (CS) solution to 25 wt%.
Calculate the amount of water to be evaporated per 100 kg of feed.
Assumption: Steady-state process (i.e. NO accumulation and NO leaking in the evaporator)
Basis: 100 kg of feed
(a 4 wt% caustic soda solution)
Flow chart:
Unknowns:
• Mass of the product (i.e. 25 wt% CS solution) (x)
• Mass of water evaporated (y )
Overall balance:

Caustic soda balance:
= 16 kg
Then, = 100 − 16
= 84 kg
Hence, we have to evaporate water in the amount of 84 kg.
Note that the product has a total mass of 16 kg and contains 25% of CS (i.e. the product contains CS
in the amount of
(0.25) 16 = 4 kg,
which means that the amount of CS in the feed and in the product are the same!!
Thus, we can alternatively solve this problem as follows:
Since we know that caustic soda is a SOLID dissolved in water, evaporation will NOT cause the
change in mass of caustic soda (in other words, only water is evaporated)
Accordingly, the mass of caustic soda of the feed stream is identical to (or exactly the same as) that
of the product stream.
The mass of caustic soda (CS) in the feed stream can be calculated as follows:
Since the basis of calculation is “100 kg of feed” and the concentration of CS in the feed is 4% (i.e.
, the mass of caustic soda in the feed stream is
(0.04)(100 kg) = 4 kg
Thus, mass of water in the feed stream is
100 – 4 = 96 kg
In the product stream, the concentration of CS is 25 wt%, meaning that
Since the mass of CS in the product stream is equal to that of the feed stream,
= 4 kg
Substituting mass of CS gives

12 kg
Hence, the amount of water needed to be evaporated is
96 – 12 = 84 kg
which is identical to the answer solved previously.
This kind of problem is called “material-balance problem with a TIE component”
What is a TIE component?
It is a component (i.e substance or species) whose mass (or mole(s)) is(are) constant for both inlet
and outlet streams.
Example 4.5. A centrifuge is used to separate blood pellets from blood fluid. In a continuous
separation of blood fluid, 1,000 g/h of blood fluid is fed into a centrifuge, and the feed contains 2,500
ppm of blood pellets.
It is required that the product must contain 80 wt% of blood pellets. Calculate the amount of blood-
pellet- free fluid has to be discharged per hour.
“blood-pellet-free fluid” = ?
Assumption: Steady-state process (i.e. NO accumulation and NO leaking in the centrifuge)
Hence, the amount of blood fluid fed into the centrifuge is 1,000 g.
Since the fluid contains 2,500 ppm of blood pellets, the amount of blood pellets in the feed is
= 2.5 g
Blood pellets are solid. Thus, their mass is constant; i.e. 2.5 g.
In other words, blood pellets are the tie component. Hence, the feed contains the bloodpellet-
free fluid

The product is required to contain blood pellets in the amount of 80 wt%, implying that
Substituting = 2.5 g gives
Then
= 0.625 g
Hence, we need to remove the “bloodpellet-free” fluid in the amount of
= 997.5 – 0.625 = 996.875 g
Since the basis is 1 h of operation, the rate at which we need to remove the bloodpellet- free fluid
from the blood fluid is 996.875 g/h
Separation Process
Separation processes are the processes in which one stream is divided into two or more streams.
Let’s consider the following Example:
Example 4.6 Membrane separation is a new technology for separating gases. A nano-porous
membrane is used to concentrate the amount of oxygen in the product stream as illustrated on the
next Page.
If the total mole of waste stream is 75% of that of the input stream, what is the molar composition
(percentage) of the waste stream?
Flow chart:

Assumptions:
• Steady-state process (i.e. NO accumulation in the membrane)
• NO chemical reactions between N2 and O2 in the membrane
Basis: 100 g-mol of the input stream
(nInput =100 g-mol)
Thus, the input stream comprises
• O2 21 g-mol
• N2 79 g-mol
Given:
• Composition of the input stream (21% O2)
• Composition of the output stream (30% O2)
• Total mole of the Waste ( = 0.75 n Input) ; hence, n waste =( 0.75)( 100 g-mol) = 75 g-mol
Unknowns:
• wt% of O2 in the Waste
• wt% of N2 in the Waste
Overall mole balance:
ninput = noutput + nwaste
( 100 g-mol ) = noutput + (75 g-mol)
noutput = 25 g-mol
O2 mole balance:
Substituting numerical values
= 0.18 (18%)
Thus, the molar composition (in percentage) of the waste stream is
• O2 18%
• N2 82% (100 – 18)

Material Balances for Processes involving Chemical Reactions
For the processes involving chemical reactions, there is an additional term in the material-balance
equation, as already stated in the “Introduction to Material Balances” chapter.
The details of performing material balances for processes with chemical reactions are described as
follows;
Examples of chemical Rxns.
CH4 + 2O2 CO2 + 2H2O
N2 + 3H2 2NH3
CO + 2H2 CH3OH
From the examples of chemical, or “stoichiometric”, reactions above, we obtain the following
information:
• What are the reactants and what are the products of each reaction
• A mole ratio/mole ratios of reactants reacting with each other/or one another)
• A mole ratio/mole ratios of a resulting product/resulting products per that/those of a
reactant/reactants
A “VALID” stoichiometric equation is a chemical reaction in which a number of atoms of each of
every element in the reactant side and that in the product side are equal to each other.
CO + O2 CO2 NOT valid
2CO + O2 2CO2 VALID
Note that “valid” stoichiometric equation/reaction is, in fact, “balanced” chemical reaction. The
numbers in front of each species (either in the reactant or the product side) in a “valid”
stoichiometric equation are called “stoichiometric coefficients”.
The ratio of stoichiometric coefficients of any two species in a balanced chemical reaction is called
stoichiometric ratio
Example
C7H16 + 11O2 7CO2 + 8H2O
Stoichiometric ratios:

In general, we can write a valid stoichiometric chemical reaction in a general form as follows
Normally, since the reactants are consumed, while the products are generated, as the reaction
proceeds, the coefficients of
• reactants are of negative (−) sign
• products are of positive (+) sign
For instance, for the “valid” chemical reaction:
C7H16 + 11O2 7CO2 + 8H2O
we obtain the fact that
•
•
Limiting & Excess Reactants
Let’s consider the following reaction:
2SO2 + O2 2SO3
This is a “valid” or “balanced” chemical reaction, as # of moles of each element (i.e. S or O on the
reactant and product sides are equal to each other).
From the above chemical reaction, we obtain the information that, for the reactant side,
“2 mol of SO2 is stoichiometrically reacted with 1 mol of O2”
which can also be written as
2 mol SO2 ≡ 1 mol O2
If we feed SO2 in the amount lower than 2 mol (say 1.8 mol), whereas the feeding amount of O2 is
still at 1 mol, the following situation take place
• SO2 is completely consumed

• O2 is left over
It is defined that:
• a species that is completely consumed = limiting reactant
• a species that is left over = excess reactant
The further question is that “how many moles of SO3 is to be produced”
The amount of product(s) generated will be according to the amount of a limiting reactant
consumed.
Fractional Excess
Let
nfed = # of moles of a reactant actually fed into a system
nstoich = # of moles of a reactant consumed according to a “valid” stoichiometric reaction
We can then defined the fractional excess as
(4.7)
Excess percentage = 100(fractional excess) (4.8)
Example 4.7 If 10 g of N2 reacts with 10 g of H2 and NH3 is formed in the reactor,
(a) which species is the limiting reactant?
(b) which species is the excess reactant?
(c) what is the maximum number of grams of NH3 that can be produced?
(d) calculate the excess percentage of the excess reactant
The first thing we MUST do is to write a “valid” stoichiometric chemical reaction, as follows
N2 + 3H2 2NH3
Thus, the stoichiometric ratio for the reactants is

Alternatively, we can write
The # of moles of N2 and H2 actually fed to the reactor can be computed as follows
Thus,
When comparing
it is clear that, with the same # of moles of N2 (i.e. 1 g-mol), the # of moles of H2 actually fed is
higher than that needed stoichiometrically (or theoretically). Hence, some amount of H2 fed must
be left over.
Accordingly, we can conclude that, in this Example,
• N2 is a limiting reactant
• H2 is an excess reactant
Since we have just known that; H2 is an excess reactant
we can then calculate the fractional excess or excess percentage of H2 as follows:
From the stoichiometric ratio of the reactants:
since # of N2 (limiting reactant) consumed is 0.357 g-mol, the # of mol of H2 required
stoichiometrically can, thus, be calculated as follows

but ( H2 )actually fed n = 4.96 g-mol
Thus, the excess percentage of H2 can be calculated as follows:
% Excess 363.6% = 363.6 %
The amount of NH3 generated (or produced) must be computed using the amount of limiting
reactant consumed.
In this Example, N2 is a limiting reactant.
Thus, the amount of NH3 generated has to be calculated based on the amount of N2 as follows:
The stoichiometric ratio between NH3 and N2 is
Since the # of mol of N2 (a limiting reactant) consumed is 0.357 g-mol, the amount of NH3 generated
can be calculated as follows:
and the “mass” of NH3 generated can then be computed as follows:
Fractional conversion, f
It is defined that

that
Fractional unreacted = 1− f (4.10)
that
Percentage conversion = 100 × f (4.11)
and that
Percentage unreacted = 100 – percentage conversion (4.12)
Example 4.8. The combustion of n-heptane is
C7H16 + 11O2 7CO2 + 8H2O
Ten (10) kg of n-heptane is reacted with an excess amount of O2, and 14.4 kg of CO2 is formed.
Calculate the conversion percentage of n-heptane.
Since it is stated that O2 is in excess, n-heptane is, therefore, a limiting reactant.
The # of moles of 10 kg of C7H16 fed and 14.4 kg of CO2 generated can be computed as follows
From the given chemical reaction, the stoichiometric ratio between CO2 and C7H16 can be written as
follows:
(Note that the positive and negative signs represent “generation” and “consumption”, respectively;
but from this point forward, only numerical values will be presented in the calculations)
Since generated is known, the amount of consumed can be calculated as follows:
Accordingly, the # of moles of C7H16 actually consumed is 0.0467 kg-mol
Thus, the conversion percentage can be calculated as follows:

Conversion percentage = 100 x
= 100 x 0.468
Conversion percentage = 46.8%
Fractional Completion
Let’s consider the following simple chemical reaction
A B
Suppose that it takes a very long time for the above reaction to reach the completion of the reaction
(i.e. A is completely consumed)
– it may be 3 hrs, or 3 days, or even 3 yrs
Practically, we are NOT going to wait that long, we just wait until a reactant (a limiting reactant)
partially convert to a certain extent, which means that some amounts of unconverted reactant are
still remained (existed) in the system.
The fraction of a limiting reactant that is “partially” reacted is called “fractional completion”
(note that, sometimes, “fractional completion” can be used interchangeably with “fractional
conversion”)
For more understanding of this term, let’s consider the following Example:
Example 4.9. Rxn.: A B has a fractional completion (or a fractional conversion) of species A at
any instant of time as shown in the following Figure:

An engineer considers that it is too long to wait until this reaction reaches the completion.
He/She then decides to take a reactant (i.e. species A) off a reactor at the time of 200 min.
If 50 kg-mol of species A is fed into the reactor,
• how many kg-mol of B is to be produced at t = 200 min?
• how many kg-mol of unreacted A remain in the reactor?
From the given graph, at t = 200 min, fractional completion = 0.7
This means that the amount of A actually reacted at t = 200 min is
50 kg-mol × 0.7 = 35 kg-mol
Thus, the amount of unreacted A can be calculated as follows
50 kgmol × (1 – 0.7) = 15 kg-mol
From the given reaction, we can write the following stoichiometric ratio:
Thus, the amount of B generated is ;
Extent of Reaction (ε )
The extent of reaction is the number indicating how much reaction occurs.
It can be calculated by dividing the change in number of moles of a species, either reactant or
product, takes place in a specific reaction, by its corresponding stoichiometric coefficient, as follows
where
= moles of species i at the final state
= moles of species i at the initial state
= stoichiometric coefficient of species i

Example 4.9. In the production of ammonia (NH3) according to the following chemical reaction:
N2 + 3H2 2NH3
the feed containing 100 g of N2, 50 g of H2 and 5 g of NH3 is fed into the reactor. The amount of mass
of NH3 at the end of the operation is 90 g. Determine the extent of reaction.
The extent of reaction can be calculated from any species. In this Example, since the data of NH3 are
available, we shall compute the extent of reaction from the information of NH3.
The # of moles of NH3 at the initial and final states can be computed as follows:
The stoichiometric coefficient of NH3 for the “balanced” chemical reaction is +2. Thus, the extent of
reaction can be calculated using Eq. 4.13 as follows
Yield & Selectivity
There are other important technical terms for the processes involving chemical reactions. Such
technical terms include “yield” and “selectivity”.
Let’s consider the following Examples to get to know what these two terms are:
Example 4.10. The dehydrogenation of C2H6 comprises the following reactions:
C2H6 C2H4 + H2
C2H6 + H2 2CH4
C2H4 + C2H6 C3H6 + CH4
Of these reactions’ products, C2H4 is a desired product, whereas CH4 & C3H6 are by-products and
undesired.

For the following reactions:
A + B C
A + B D
C is a desired product, and D is a by-product (and unwanted)
For the above Examples, “yield” and “selectivity” are defined as follows:
Example 4.11. Given the Rxn.:
C2H2 + 2H2 C2H6
At the beginning of the process, 20 kg-mol of acetylene, 40 kg-mol of hydrogen, and 20 kg-mol of
C2H6 are charged into the reactor. Twenty (20) minutes later, 30 kg-mol of hydrogen reacts.
Determine the compositions of the input stream and of a mixture in the reactor at the 20th
minute.

Input:
• C2H2 20 kg-mol
• H2 40 kg-mol
• C2H6 20 kg-mol
From the given data, mole percentage of the Input can be calculated as illustrated in the following
Table:
Species kg-mol Mole fraction(xi) % mol(100xi)
C2H2 20 0.25 25
H2 40 0.5 50
C2H6 20 0.25 25
Total 80 1.00 100
To determine which reactant is a limiting reactant, we perform the following calculations.
A stoichiometric ratio for the reactants of the given reaction:
C2H2 + 2H2 C2H6
The mole ratio of C2H2 : H2 from the given data is
By comparing these two ratios, it is clear that the amount of C2H2 fed into the reactor is equal to that
required stoichiometrically. Thus, there is NO limiting reactant, or, in other words, both reactants
are limiting reactants.
Let are the # of moles of C2H2 , H2, and C2H6 at the initial state,
respectively
Also let and are the # of moles of C2H2 , H2, and C2H6 at t = 20 min, respectively.
It is given that, at t = 20 min, 30 kg-mol of H2 is reacted/consumed.
By using the following stoichiometric ratios:

we obtain the following:
= 15 kg-mol
Thus, at t = 20 min,
= 40 – 30
= 10 kg-mol
= 20 – 15
= 5 kg-mol
= 20 + 15
= 35 kg-mol
The composition of a mixture in the reactor at t = 20 min can then be summarised as follows
Species kg-mol Mole fraction
(xi)
% mol
(100xi)
C2H2 5 0.10 10
H2 10 0.20 20
C2H6 35 0.70 70
Total 50 1.00 100

The fractional conversion ( f ) of the limiting reactant can be computed as follows:
Since a limiting reactant (i.e. H2 & C2H2) is NOT completely consumed, a fractional conversion can be
calculated as follows:
(note that this calculation is for H2; if you perform the calculation of C2H2, the answer would be the
same – try it yourself)
Example 4.12. The following Rxns.:
C2H6 C2H4 + H2 (4.16)
C2H6 + H2 2CH4 (4.17)
in which Rxn. 4.17 is unwanted, take place in a steady-state continuous reactor. The feed contains
85 mol% C2H6 and the balance inerts. The fractional conversion ( f ) of C2H6 is 0.50, and the yield of
C2H4 is 0.47. Calculate the molar composition of the product and the selectivity of ethylene to
Basis: 100 g-mol of feed
Hence, the feed contains
• 85 g-mol C2H6
• 15 g-mol inerts
Let
x = g-mol of C2H6 reacted in Rxn. 4.16
y = g-mol of C2H6 reacted in Rxn. 4.17
Thus, the # of moles of C2H6 totally reacted is x + y g-mol
It is given that
• fractional conversion ( f ) of C2H6 = 0.50
• yield of C2H4 = 0.47
Hence,

(4.18)
The feed contains 85 g-mol of C2H6, and if there was NO side reaction (i.e. no Rxn. 4.17) and a
limiting reactant (i.e. C2H6) is completely consumed (meaning all of 85 g-mol of C2H6 fed to the
reactor is consumed), the amount of C2H4 produced could be calculated using the following
stoichiometric ratio:
However, the amount of moles of C2H6 actually consumed in Rxn. 4.16 is x g-mol. Thus, the amount
of C2H4 actually generated in Rxn. 4.16 is, therefore, x g-mol.
= 39.95 g-mol
Substituting the resulting value of into Eq. 4.18 yields
The amount of C2H4 actually generated (in Rxn. 4.16) can be calculated as follows:

The amount of CH4 actually generated in Rxn. 4.17 can be calculated as follows:
The amount of H2 actually generated (in Rxn. 4.16) can be calculated as follows:
The amount of H2 actually consumed (in Rxn. 4.17) can be calculated as follows:
Thus, at t = 20 min,
• – = 85 – 39.95 – 2.55
= 42.5 g-mol
• = 0 + 39.95
= 39.95 g-mol

• = 0 + 2(2.55)
= 5.10 g-mol
• = 0 – 39.95 – 2.55
= 37.4 g-mol
•
= 15 g-mol
Hence, the molar composition of the product can be tabulated as follows:
Species g-mol Mole fraction
(xi)
% mol
(100xi)
C2H6 42.5 0.304 30.4
C2H4 39.95 0.286 28.6
H2 37.4 0.267 26.7
CH4 5.1 0.036 3.60
Inerts 15 0.107 10.7
Total 139.95 1.00 100
Selectivity can be computed from Eq. 4.15
as follows
• C2H4 is a desired product, and
= 39.95 g-mol
• CH4 is an undesired product, and
= 5.10 g-mol
Thus, by using Eq. 4.15,
= 7.83

Atomic Balances
Since atom can NOT be created or destroyed, for any process, a number of atoms entering the
process must be equal to that leaving the process. We can employ the above “atomic balances”
principle to solve material-balance problems, as illustrated in the following Example.
Example 4.13.
Determine the values of n1 and n2
Note that we have done this Example before. The only difference is that, in this Example, the
chemical Rxn. is NOT given (and there is no CH4 generated)
In the case that we do NOT know the chemical Rxn. in the reactor, what would we do?
Performing “atomic balances” is an alternate way of solving problems involving chemical reactions,
especially when specific chemical Rxns. are NOT known or given.
In this Example, the given data (as shown as a diagram) are as follows:
• C2H6 is fed into the reactor in the amount of 100 g-mol
• the amount of H2 leaving the reactor is 40 g-mol
• C2H4 is another product
• C2H6 is NOT completely consumed
By utilising the principle of atomic balances, we obtain the following:
# of atoms of C (or H) entering the process = # of atoms of C (or H) leaving the process
INPUT
1 molecular mol of C2H6 comprises
• C = 2 atomic mol
• H = 6 atomic mol
Since the amount of C2H6 fed into the reactor is 100 molecular mol, the amounts of
• C = 2 × 100 = 200 atomic mol
• H = 6 × 100 = 600 atomic mol
OUTPUTS
1) 1 molecular mol of H2 comprises

Since the amount of H2 leaving into the reactor is 40 molecular mol, the amount of
• H = 2 × 40 = 80 atomic mol
2) 1 molecular mol of C2H4 comprises
Since the amount of C2H4 leaving into the reactor is n1 molecular mol, the amounts of
• C = 2 × n1 = 2n1 atomic mol
• H = 4 × n1 = 4n1 atomic mol
3) 1 molecular mol of C2H6 comprises
Since the amount of C2H6 leaving into the reactor is n2 molecular mol, the amounts of
• C = 2 × n2 = 2n2 atomic mol
• H = 6 × n2 = 6n2 atomic mol
Performing C and H atomic balances yields
C balance:
Σ(C)input = Σ(C)output , Substituting corresponding numerical values
200 = 2n1 + 2n2
100 = n1 + n2 (4.19)
H balance:
Σ(H)input = Σ(H)output , Substituting corresponding numerical values
600 = 80 + 4n1 + 6n2
260 = 2n1 + 3n2 (4.20)
Solving Eqs. 4.19 & 4.20 simultaneously yields;
• n1 = 40 g-mol
• n2 = 60 g-mol ( )
Thus, the total # of moles of the output stream is
= 140 g-mol

Problem Set 3
Introduction to Material Balances & Material Balances for a Single Unit
PROB 1 . A thickener in a waste disposal unit of a plant is used to remove water from wet sewage
sludge, as shown in Figure 1. If the process is steady state, how many kilograms of water leave the
thickener per 100 kg of wet sludge entering the thickener?
Figure 1: for Problem 1
PROB 2. A new engineer has performed a material balance for a mixing tank as summarised in
Figure 2. Check if this material balance correct, for both overall and species balances.
Figure 2: for Problem 2
PROB 3. A 20 wt% sodium hydroxide (NaOH) solution is to be diluted to the concentration of 8 wt%.
Determine the amount of water needed.
PROB 4. A mixed acid is prepared from mixing 99 wt% sulphuric acid solution (A), 95% nitric acid
solution (B), and pure water (C) together. If 1,000 kg of the mixed acid, containing 50 wt% H2SO4, 40
wt% HNO3, and 10 wt%, water is needed, calculate the amounts of A, B, and C (in kg).

PROB 5. A 20 wt% H2SO4 solution (density = 1.139 g/cm3
) is mixed with a 60 wt% H2SO4 solution
(density = 1.498 g/cm3
), in order to obtain a H2SO4 solution with the concentration of 4.0 mol/L.
Determine
a) the ratio between the 20 wt% H2SO4 solution and 60 wt% H2SO4 solution, in order to get the acid
solution with the desired concentration
b) the amount of the 60 wt% H2SO4 solution needed if 1,250 kg of 4.0 mol/L H2SO4 solution is
required. (MW of H2SO4 and of water are 98.08 and 18.02, respectively)
PROB 6. When humid air containing water vapour and air (21 vol.% O2 and 79% N2) is passed through
a condenser, water vapour in the air is condensed. A condenser can condense 95 vol% of the water
vapour entering the condenser. If the flow rate of the outgoing condensed water is 225 L/h, calculate
a) the ratio between the humid air entering the condenser and that leaving the condenser
b) the composition (in mole fraction) of the air leaving the condenser.
PROB 7. Humid air containing 4 mol% water vapour is passed through a column filled with a
desiccant. It is found that 97 mol% of water vapour is adsorbed onto the desiccant. The weights of
the desiccant before and after the operation are 3.40 and 3.54 kg, respectively. If the operation lasts
5 hour, determine
a) the molar flow rate (in g-mol/h) of the humid air entering the column
b) the mole fraction of water vapour adsorbed onto the desiccant, compared with the total amount
of water entering the column.
PROB 8. Ammonia (NH3) is burned with air (comprising O2 21% and N2 79% by vol.) in a reactor, and
the products are nitric oxide (NO) and water (H2O), as illustrated in the following reaction:
4NH3 + 5O2 4NO + 6H2O
a) If ammonia is fed into the reactor at the feed rate of 100 kmol/h, calculate the flow rate of air in
order to have the percentage excess of O2 of 40%
b) If NH3 and air are fed into the reactor at the flow rates of 50 and 330 kg/h, respectively. Determine
which reactant is a limiting reactant, and calculate the fractional excess of the excess reactant
c) If the percentage conversion of this reaction is 55%, and the NO formation rate is limited not to
exceed 2 kg/h, determine the flow rates of NH3 and air when the percentage excess of O2 is 30%

PROB 9. Highly-concentrated sugar (widely used in food industry) contains sugar 4 out of 5 parts (by
wt.). The remainder is water. The highly-concentrated sugar is to be concentrated (despite the fact
that it is already concentrated) in an evaporator, where water is heated and then evaporated. The
evaporator is able to evaporate 85 wt% of water in the highlyconcentrated sugar. If it is required that
1,000 tonnes/day of highly-concentrated sugar be produced, determine the water evaporation rate
(in kg/h).
PROB 10. Butane (C4H10) is combusted with air. Assume that no CO is produced. Calculate the
composition of the product, when
a) C4H10 is completely consumed and the excess-air percentage is 20%
b) conversion percentage of C4H10 is 90% and the excess-air percentage is still at 20%
PROB 11. Monochloroacetic acid (MCA: CH2ClCOOH) is produced from the reaction between acetic
acid (CH3COOH) and chlorine gas (Cl2) by having PCl3 as a catalyst. The resulting monochloroacetic
acid, however, may, in turn, react with chlorine gas (Cl2) and thus form dichloroacetic acid (DCA:
CHCl2COOH), as shown in the following reactions:
CH3COOH + Cl2 CH2ClCOOH + HCl
CH2ClCOOH + Cl2 CHCl2COOH + HCl
If MCA is to be produced in the amount of 5,000 kg/day, it is required that Cl2 in the amount of 4,536
kg/day is needed and that DCA in the amount of 263 kg/day is to be separated from other products,
using a crystalliser. If the excess reactant is acetic acid, calculate
a) % conversion
b) % yield
c) selectivity of MCA when compared with DCA
PROB 12. The chlorination reaction of methane (CH4) is as follows:
CH4 + Cl2 CH3Cl + HCl
If the feed comprises CH4 40 mol%, Cl2 50%, and N2 10% and the percentage conversion of the
limiting reactant is 67%, determine the composition of the product.

PROB 13. To produce ethanol (C2H5OH) using from glucose (C6H12O6) by an anaerobic fermentation
using Saccharomyces cerevisiae yeast, the following reactions take place
C6H12O6 2C2H5OH + 2CO2
C6H12O6 2C2H3COOH + 2H2O
The feed, a glucose solution with the concentration of 12 wt%, is fed into a bioreactor with the total
mass of 4,000 kg. After the fermentation is ended, 120 kg of CO2 is generated, while 90 kg of glucose
remains unreacted. Determine the mass percentages of C2H5OH and C2H3COOH in the product.

Chapter 5:
Material Balances for Multiple Units

Chapter 5
Learning Objectives
• analyze a problem statement and organize the solution strategy for the problems pertaining
to material balances involving multiple units, with and/or without chemical reactions
• determine the possibility of solving for the unknown variables in each unit or group of units
using a degree-of-freedom-analysis technique
• solve material-balance problems involving more than one units
• solve material-balance problems involving several serially connected units
• solve material-balance problems involving recycling, bypassing, and purging
• solve material-balance problems involving combustion reactions
• understand the meaning of stack, exhaust, and flue gas; Orsat analysis; dry basis; wet basis;
theoretical or stoichiometric air (oxygen); and excess air (oxygen)
• calculate the amount of theoretical or stoichiometric air (oxygen) and excess air (oxygen)
when given the information of the amounts of fuel and air (oxygen) and vice versa, and the
amount of air (oxygen) actually fed when given an appropriate set of data
In the previous Chapters, you have learned how to solve material-balance problems for a single unit
However, it is uncommon for any chemical processes to have only one single unit; in a real chemical
plant or chemical process, there are a large number of processing units interconnected together.
The basic knowledge obtained from the previous chapters is essential for solving the problems
involving multiple units.
However, some necessary additional techniques/knowledge must also be studied, in order to
enable/prepare students to solve these kinds of materialbalance problems (and you have to be able
to do it competently), which are likely to be encountered in the real engineering life.
In this Chapter, the learning technique would be the problem-based style, or students are going learn
or understand how to solve material-balance problems through a number of examples.

Example 5.1. Acetone vapour is considered toxic to the environment (as well as human health). As
an engineer in the chemical plant, you are asked to design an acetonerecovery system having the
flow sheet illustrated on the next page. In this Example, to make the calculation simple, all
concentrations of both gases and liquids are in presented in weight (mass) percent (please note,
however, that, normally, for gas phase, the concentration is presented in “volume” and “mole” basis)
Calculate the values of A, F, W, B, and D in kg per hour, if given G = 1,400 kg/h
Hence, from the basis, G = 1,400 kg
and the “Feed” stream contains
• Acetone: (0.030)( 1,400) = 42 kg
• Air: (0.950) (1,400) = 1,330 kg
• Water: (0.020) (1,400) = 28 kg
The main principle of performing material balances for a system comprising multiple units is that;
We need to carefully choose the unit/group of units that we are going to do the balances such that
the # of equations obtained from the available data/information equals that of the unknowns.
1) Try performing the balances around the absorber unit

Overall balance
G+W = A+F
1,400 +W = A+F
There are 3 unknowns, thus, we need 3 equations. (i.e. we need another TWO equations., in
addition to the overall-balance equation)
The question is
“Is it possible to obtain another 2 equations, by performing the balances around the absorber
column?”
Since the absorber column involves 3 species (i.e. acetone, air, and water), we can have up to 2
“independent” species balances
Hence,
the total # of equations = 3 (1 overall-balance equation and 2 species-balance equation)
while the # of unknowns = 3
Accordingly, the degree of freedom (DoF), which is defined as follows:
DoF = (# of unknowns) – (# of equation available)
for doing the material balances around the absorber column is DoF = (3) – (3) = 0
When DoF = 0, we can solve for all unknowns
Note that:
• when DoF = 0 (or when # of unknowns = # of Eq. available), all of the unknowns can be
solved
• when DoF > 0 (or when # of unknowns > # of Eq. available), we cannot solve for Unknowns
• when DoF < 0 (or when # of unknowns < # of Eq. available), all of the unknowns can be
solved, but the technique is not called “solving”, it is rather called “optimizing” or
“performing regression analysis”
The technique of comparing the number of unknowns with that of the available equations is called
“degree of freedom analysis”
Hence, to solve for all unknowns around the absorber unit (W ,A, & F ), we perform the following
species balances:
Acetone balance
Performing a species balance for acetone yields

(0.03)(1,400) = (0. 19)
221.1 kg
Air balance
Doing the same for air results in
(0.95) (1,400) = (0.995) ( )
= 1,336.7 kg
Substituting F = 221.1 kg & A = 1,336.7 kg into 1,400 +W = A+F
and solving for W gives
1,400 + W = 1,336.7 + 221.1
W = 157.8 kg
Thus, by performing the balances around the absorber unit, we obtain the values of the following
unknowns:
W = 157.8 kg
A = 1,336.7 kg
F = 221.1 kg
We still have another 2 unknowns (i.e. D and B) to solve for.
To solve for B, we try doing the following
2) Doing the balances around the distillation column

It is evident that it is NOT a good idea to choose to do the balances around the distillation column,
since we do NOT have any information of the stream leaving the top of the distillation column.
The distillation column involves 2 species: acetone and water
Hence, when we do not know the data of the stream leaving the top of the column, we have 3
unknowns in total, i.e.
• the value of B (or the mass flow rate of the Bottom stream)
• the mass flow rate of the top stream leaving the distillation column
• the concentration, in wt%, of either acetone or water of the top stream leaving the
distillation column
but we can set up at the maximum of 2 equations, i.e.
• 1 overall balance
• 1 species balance
The DoF is, therefore, +1, meaning that we cannot solve for the unknowns Thus, we need to choose
the NEW location to do the balances.
3) Performing the balances around the distillation column & condenser together
Overall balance
F = B+D
Substituting the value of F, obtained previously, gives
221.1 = B+D
There are 2 unknowns, thus, we need another equation (in order have 2 equations, in order to solve
or the unknowns – B & D
Acetone balance

(0.19 (221.1) = (0. 99) + (0.040)
42.01 = 0.99D + 0.04B
We have also 221.1 = B+D
Solving two equations simultaneously yields
B = 186.2 kg
D = 34.9 kg
To check the validity of all unknowns we have just solved for, we perform the overall balance for the
whole process as follows
G+W = A+B+D
Substituting corresponding numerical values, results in
1,400 + 157.8 = 1,336.7 + 186.2 + 34.9
1,557.8 = 1,557.8
This verifies that the balances performed previously is correct.
Alternatively, we can solve for the values of B & D (after we have obtained the values of W, A, and
F) by performing the balances around the whole process as follows
Overall balance
G+W = A+B+D
1,400 + 157.8 = 1,336.7 + B + D
B+D = 221.1
There are 2 unknowns; we, therefore, need another ONE equation
Acetone balance
(0.030) (1,400) = (0. 99) ( ) + 4.0 ( )
42 = 0.99D + 0.04B
We have also B+D = 221.1
Solving these equations simultaneously yields
B = 186.2 kg
D = 34.9 kg

which is exactly the same as done previously.
The composition, in mass fraction and in mass flow rate, of each stream, can be summarised in the
following Table
Material Balances at a Separating Point
The separating point is the point where there are ONE input stream, but there are two or more
output streams, as illustrated below (1 input, but 2 outputs):
Since there is NO Rxn. at the separating point, the compositions of Streams 1, 2, & 3 are identical
However, the flow rates of Streams 1, 2, & 3 may be different

Material Balances of Processes with Recycling, Bypassing, and Purging
Example 5.2 The flow chart of a steady-state process to recover crystalline potassium chromate
(K2CrO4) from the aqueous solution is shown below
Forty-five hundred kilograms per hour of 33.3% K2CrO4 solution is joined by a recycle stream of 36.4%
K2CrO4 solution, and the combined stream is fed into an evaporator. The concentrated stream leaving
the evaporator contains 49.4% K2CrO4. This stream is then fed into a crystallizer where it is cooled,
causing crystals of K2CrO4 to come out of the solution.
The resulting solution and K2CrO4 crystals are separated by a filter, dividing the product into 2
streams: the filter cake, which consists of K2CrO4 crystals and 36.4% K2CrO4 solution; and the filtrate,
which is the 36.4% K2CrO4 solution. The K2CrO4 crystals account for 95% of the total mass of the filter
cake. The filtrate is recycled to mix with the feed.
Calculate (a) the rate of evaporation, or the rate of water evaporated at the evaporator, and (b) the
recycle ratio [i.e. (mass flow rate of recycle)/(mass flow rate of fresh feed)]
For our convenience, let
K2CrO4 = K , Water = W
Basis 1 h of operation
Specifying unknowns to appropriate points in the given flow chart yields

To answer Question (a), i.e. how many kg of water we need to evaporate, we start performing the
mass balances around the evaporator, as illustrated below:
However, if we do so, there would be 4 unknowns (m2, m3, m4, & x) to be solved for, but we can
have, at best, only 2 equations (1 overall balance equation & 1 species balance equation)
Thus, we need to choose the new point/ location to do the balances.
We, then, try performing the balances for the whole process as follows:
Overall balance
m1=m3+m5+m6
The problem states that
m5 = 0.95 (m5+m6 )
Hence, by re-arranging the above equation,
m6 = 0.0526m5
m1 = 4,500 kg
Overall balance; m1=m3+m5+m6
4,500 = m3 + m5 + 0.0526m5
4,500 = m3 + 1.0526m5

Note that, now, we have 2 unknowns to be solved for
K2CrO4 balance
(0.333) ( 4500) = ( 1) m5 + (0. 364)m6
(0.333) ( 4500) = ( 1) m5 + (0. 364)(0.0526m5)
m5 = 1,470.6 kg
Thus, m6 = 0.0526m5
= 0.0526 (1,470.6)
m6 = 77.4 kg
Then, using equation for overall balance, we can solve for m3 as follows
4,500 = m3 + 1.0526m5
4,500 = m3 + 1.0526 (1,470.6)
m3 = 2,952.0 kg
Hence, the rate of water evaporation is 2,952.0 kg/h
Recycle ratio is defined as follows
Hence, in this Example,
To obtain the value of or m7 , we perform the balances around the crystalliser & filter

Overall balance
m4 = m5 + m6 + m7
Substituting the values of m5 & m6
m4 = (1,470.6) + (77.4) + m7
m4 = (1,548.0) + m7
K2CrO4 balance
(0.494) m4 = (1)m5 + (0.364) m6 + (0.364)m7
(0.494) m4 = (1)(1,470.6) + (0.364) (77.4) + (0.364)m7
(0.494) m4 = (1,498.8) + (0.364)m7
From overall eqn. we have m4 = (1,548.0) + m7
Solving two equations simultaneously m4 & m7 will be
m4 = 7,194.8 kg
m7 = 5,646.8 kg
Thus,
Example5.3 In the preparation of the feedstock for a plant manufacturing gasoline, iso- and normal-
pentane are produced from process shown below. Calculate the fraction of feed that passes through
the iso-pentane tower.

In order to obtain the “fraction of feed that passes through the iso-pentane tower”, we need to solve
for the value of m1 and then compare it with that of mF (= 100 kg)
Performing the mass balances around the whole process (as indicated on previous page) yields
Overall balance
mF = mS + mP
100 = mS + mP
iso-pentane balance
(0.20) mF = (1) mS + (0.10) mP
(0.20)( 100) = (1) mS + (0.10) mP
Also we have 100 = mS + mP
Solving these equations simultaneously gives
mP = 88.9 kg
mS = 11.1 kg
The principle of separating point states that the compositions of all streams entering and leaving
the separating point are IDENTICAL. Hence, the composition of Stream 2 (& Stream 1, too) is exactly
the same as that of the Feed; i.e. the Stream 2 (& Stream 1) contains;
• n-C5H12 80%
• i-C5H12 20%
Performing the mass balances around the mixing point yields
Overall balance
m2 +mB =mP
m2 +mB = 88.9

n-pentane balance
(0.80) m2 + (1) mB = (0.90) mP
(0.80) m2 + (1) mB = (0.90) 88.9
0.8 m2 + mB = 80.01
Also we have m2 +mB = 88.9
Solving equations simultaneously results in
m2 = 44.45 kg
mB = 44.45 kg
Performing the overall balance around the separating point gives
mF = m1 + m2
Substituting corresponding numerical values into the equation yields
100 = m1 + 44.45
m1 = 55.55 kg
Thus, the “fraction of feed that passes through the iso-pentane tower” is
Example 5.4. Redo the Example 5.2. once again, but consider the case that there is no recycle of
the filtrate to be mixed with the fresh feed. Determine
(a) the amount of K2CrO4 produced per hour and
(b) the rate that we need to evaporate water.
Without the recycle of the filtrate, we can re-draw the flow chart as shown on the next page.
We then try performing the balances around the whole process, as done previously.

Overall balance
m1 = m2 + m4 + m5 + m6
4,500 = m2 + m4 + m5 + m6
but, as stated in the problem statement,
m4 = 0.95 ( m4 + m5 )
Thus,
4,500 = m2 +m4 + 0.0526m4 +m6
4,500 =m2 + 1.0526m4 + m6
There are 3 unknowns ( m2 , m4 ,& m6 ), but we can have, at most, another 1 equation (i.e. one
species balance equation – either for K or W ) to add up to only 2 equations.
Hence, we need to choose the new point/ location to perform the mass balances.
Performing the balances around the evaporator gives;
Overall balance
m1 = m2 + m3
4,500 = m2 + m3
Water balance
(0.667)(4,500) = m2 + 0.506 m3
Solving these equations simultaneously yields
m3 = 3,033.4 kg
m2 = 1,466.6 kg
Thus, the rate at which water has to be evaporated is 1,466.6 kg/h (note that it is 2,952.0 kg/h when
the filtrate is recycled to be mixed with the fresh feed)
Doing the balances around the crystalliser & filter gives;

Overall balance
m3 = m4 +m5 +m6 , but
m4 = 0.95 (m4 + m5)
m5 = 0.0526 m4
and, from the mass balances done recently,
m3 = 3,033.4 kg
Hence, we obtain the following equation:
3,033.4 =m4 + 0.0526m4 +m6
3,033.4 = 1.0526m4 +m6
K2CrO4 balance
(0.494)m3 = (1)m4 + (0.364)m5 + (0.364)m6
(0.494)m3 = (1)m4 + (0.364)( 0.0526m4 ) + (0.364)m6
(0.494)(3,033.4) = 1,498.5 = 1.0194m4 + (0.364)m6
Also we have
3,033.4 = 1.0526m4 +m6
Solving equations simultaneously yields
m4 = 619.7 kg
m5 = 0.0526 m4 = 0.0526 (619.7)
m5 = 32.6 kg
m6 = 2.381.1 kg
The rate at which K2CrO4 is produced is 619.7 kg/h (note that it is 1,470.6 kg/h when the filtrate is
recycled)

The comparison between the “recycle” and “not recycle” cases are as follows
Material Balances with Reactions for Multiple-unit Processes
Example 5.5 Propane is dehydrogenated to form propylene in a catalytic reactor:
C3H8 C3H6 + H2
The process is to be designed for a 95% overall conversion of propane. The reaction products are
separated into two streams: the first, which contains H2, C3H6, and 0.555% of C3H8 that leaves the
reactor, is taken off as product; the second, which contains the balance of the unreacted propane
and 5% of the propylene in the first stream, is recycled to the reactor.
Calculate (a) the composition of the product, (b) the ratio: (moles recycled)/(moles fresh feed), and
(c) the “single-pass” conversion.
Basis 100 g-mol of feed (C3H8)
Given data:
1) 95% “overall” conversion of propane (C3H8)
What is an “overall conversion”?
We have learned that
or

(note that “mole(s) of remaining A” is, in fact, “mole(s) of A unreacted”)
Overall conversion A = Fractional conversion of A for the entire process.
Thus, for this Example,
mole(s) of remaining C3 H8 = 0.05 (mole(s) of C3H8 fed)
= 0.05 (100 g-mol)
2) n6 = 0.555% of n3
n6 = 0.00555n3
n3 = 901 g-mol
3) n10 = 5% of n7
n10 = 0.05 n7
Perform the “atomic” balances around the entire process
(as illustrated as dashed lines on previous page)
Overall C (carbon) balance
(C)fresh feed = (C)product
(C in C3H8)fresh feed = (C in C3H8)product + (C in C3H6)product
(3 × 100) = (3 × n6) + (3 × n7)

300 = 3n6 + 3n7
100 = n6 + n7
Since n6 = 5 mol
n7 = 100 – 5 = 95 mol
Overall H (hydrogen) balance
(H)fresh feed = (H)product
(H in C3H8)fresh feed = (H in C3H8)product + (H in C3H6)product
(8 × 100) = (8 × n6) + (6 × n7) + (2 × n8)
Since
n6 = 5 g-mol
n7 = 95 g-mol
then n8 = 95 g-mol
Hence, the product comprises
C3H8 5 g-mol = n6
C3H6 95 g-mol = n7
H2 95 g-mol = n8
which we can summarize in the following Table
The value of n10 can be obtained when we have obtained the value of n7 as follows
n10 = 0.05n7
= 0.05( 95 g-mol)
n10 = 4.75 g-mol

Doing the balances around a separator
Since NO Rxn. is taken place in a separator, the number of moles of each component is conserved,
meaning that the number of moles of each component entering the separator is equal to that
leaving the separator
C3H8 balance
n3 = n6 + n9
901 = 5 + n9
n9 = 896 g-mol
Thus, the total number of moles of the recycle stream is
n9 + n10 = 896 + 4.75 = 900.75 g-mol
Accordingly,
the recycle ratio = (moles recycled)/(moles fresh feed)
= (900.75 g-mol)/(100 g-mol)
= 9.01 ≈9
SINGLE-PASS CONVERSION (?)
“Single-pass conversion” is a fractional conversion of a substance obtained when it passes through a
reactor only, without having passed through any other units of the entire process
Thus, for this Example:
Hence, we need the value of n1 to calculate the value of the single-pass conversion of C3H8
Doing the balances around the mixing point

Since NO Rxn takes place at the mixing point, the number of moles of each component is conserved.
C3H8 balance : 100 + 896 = n1
n1 = 996 g-mol
Thus,
Note that the overall conversion = 95%!
This is why we need to recycle the product of the reactor (or fraction of it) back to the reactor (in
order to obtain higher fractional conversion)
Material Balances for Combustion Processes
Example 5.6 A liquid fuel containing 88 wt% C and 12% H is burned with air, comprising 21 vol% O2
and 79% N2, in a combustor. The combustion product is called “flue gas”, and the flue gas leaving the
combustor is then passed through a desiccant column, to completely removed water from the flue
gas. The resulting dry flue gas is then passed through a gas analyser, to analyse for the composition
of the dry flue gas, and the analytical results are as follows: CO2 13.4% by vol., O2 3.6%, and N2 83.0%.
If the amount of fuel fed into the combustor is 100 g, calculate the amount of the dry flue gas, in g,
and determine the percentage excess of O2 (note that N2 does not react with any other species)
Basis 100 g of feed (liquid fuel)
Thus, the feed comprises
• C 88 g
• H 12 g

Flow chart
Rxns: C + O2 CO2
H + ¼ O2 ½ H2O
Overall C (carbon) balance
C output can be calculated as follows:
From the given data, of n2 g-mol of dry flue gas, there is CO2 in the amount of 13.4% by mol. (note
that, for gas phase, %mol ≡ %vol)
The amount, in g-mol, of CO2 in the dry flue gas is then
(0.134)n2
We know that 1 molecular mol of CO2 contains 1 atomic mol of C and 2 atomic mol of O,
Thus, the amount of C in the dry flue gas is
1x(0.134)n2 = (0.134)n2 atomic mol
Hence,
C output = (0.134)n2
From the atomic balance principle
C input = C output
7.33 g-mol = (0.134)n2
Then n2 = 54.7 g-mol

Thus, the dry flue gas contains
• CO2 = (0.134) (54.7) = 7.33 g-mol
• O2 = (0.036) (54.7) = 1.97 g-mol
• N2 = (0.830) (54.7) = 45.4 g-mol
Since, N2 does not react,
(N2) input = (N2) output
(N2) input = (0.79)n1
(N2) output = 45.4 g-mol
Hence, (0.79)n1 = 45.4
n1 = 57.5 g-mol
Accordingly, the air entering the combustor comprises
• O2 = (0.21) (57.5) = 12.1 g-mol
• N2 = (0.79) (57.5) = 45.4 g-mol
Note: In this Example, N2 can be considered as a tie component (Do you know why?)
From the chemical Rxns.,
C + O2 CO2
H + ¼ O2 ½ H2O
1 g-mol of O2 is needed to react stoichiometrically with 1 g-mol of C, and
¼ g-mol of O2 is required to react stoichiometrically with 1 mol of H.
The amount of C fed is
Thus, O2 in the amount of 1× 7.33 = 7.33 g-mol is needed
The amount of H entering the combustor is
The amount of O2 needed is then (1/4)x 11.90 = 2.98 g-mol
Hence, the total amount of O2 required theoretically is
7.33 + 2.98 = 10.31 g-mol

but the amount of O2 fed into the combustor is 12.1 g-mol
Accordingly, the amount of UNREACTED (or excess) O2 is
12.1 – 10.31 = 1.79 g-mol
From our recent calculations, the amount of O2 in the dry flue gas is 1.97 g-mol
The discrepancy (1.97 vs 1.79 g-mol) may result from an error of the gas analyser; however, these
two numbers are close to each other (so, the error is acceptable).
Summary
Mass of the Dry Flue Gas
Since the dry flue gas comprises
• CO2 = (0.134) (54.7) = 7.33 g-mol
• O2 = (0.036) (54.7) = 1.97 g-mol
• N2 = (0.830) (54.7) = 45.4 g-mol
we can calculate the mass of each species and the total mass of the dry flue gas as illustrated in the
following Table
Excess percentage of O2
From the calculations above, for O2,
nfed = 12.1 g-mol
nstoich = 10.31 g-mol
The percentage excess of O2 can then be calculated as follows
= 17.4%

In the field of combustion, it is customary to call excess percentage of O2 (or %excess O2)
• % excess air (%XA)
• % excess oxygen
Note that %XA is identical to % excess oxygen
Normally, the amount of theoretical O2 is obtained from a valid chemical Rxn. (or valid chemical
Rxns.), and since air contains.
• 21 mol% O2
• 79 mol% N2
the amount of air and N2 can be calculated from the amount of O2 as follows
Note also that, sometimes, flue gas is called “stack gas”.
The flue or stack gas with water (H2O) is called “wet” flue or stack gas; after the water is removed
from the wet flue or stack gas, it becomes “dry” flue or stack gas
Example 5.7 Staged distillation columns are devices used to separate volatile materials by boiling off
more volatile components (as a distillate). In order for a clean separation to take place, these devices
require that at least part of the vapour product from the top of the column be condensed and
returned back to the column. The composition of the vapour product from the top of the column
(called “overhead” is identical to those of the distillate and the stream that returns to the column
(called “reflux”)
Typical Distillation Column

Suppose that the distillation column is employed to separate a three-component (or ternary) mixture
consisting 7 mol% acetone, 61.9% acetic acid, and 31.1% acetic anhydride
The column is designed to yield the bottoms stream containing no acetone and the distillate
containing 10% acetone and 88% acetic acid If the column is operated so that 60% of the overhead is
returned as a reflux. If 700 mol/h of the distillate is to be produced, calculate the flow rates and
compositions of all streams
In this Example, we are dealing with 3 main unit operations:
• a distillation column
• a condenser
• a splitter
To obtain the answers (e.g., the flow rates and compositions of all streams in this Example), we have
to draw a material-balance system boundary for any single unit operation or for a group of unit
operations or for the whole system. The material-balance system boundary must be chosen carefully
and wisely.
The only principle for choosing the materialbalance system boundary is that the system boundary
must be chosen such that the number of unknowns for that boundary system is equal to the number
of equation that the boundary system can have.
If we draw the system boundary for the whole system, we have to determine the unknowns for
stream 1, 3, and 5.
Since this is a 3-component (ternary) system/mixture, specifying concentrations (in mol% or in mole
fraction) of only 2 components (or species or substances) is enough (as the mol% of all components
must be added up to 100 or mole fractions of all species must be added up to 1)
Thus, in this Example, the “mol%” of acetic anhydride in stream 3 is
100 – 10 – 88 = 2 mol%
Hence, the unknowns are

1. the flow rate of the feed
2. the flow rate of the bottoms stream
3. the concentrations of either acetic acid or acetic anhydride
So, we need 3 equations.
Can we have 3 equations, when we draw the material-balance system boundary for the whole
system?
Normally, we can have
• 1 overall-balance equation
• n −1 species-balance equations
(n = # of species)
Hence, in this Example, we can have 2 (i.e. 3 – 1) species-balance equations.
Thus, if for the whole system, we can have, in total, 1 + 2 = 3 equations. Accordingly, we can solve
for all unknowns.
This kind of “degree of freedom” analysis is necessary before we perform the numerical calculations
for any material balance problems.
Thus, the distillate (S3 ) is 700 g-mol
Overall balance
S1 = S3 +S5
S1 = 700 +S5
Species balance
Acetone balance
Thus, from overall balance
S1 = 700 +S5
1,000 = 700 +S5 S5 = 300 g-mol
Acetic acid balance

Hence, the concentration (in mol% of acetic anhydride in stream 5 )
100 – 1 = 99%
In summary, for streams 1, 3, and 5, the flow rate and composition of each stream are as follows;
• Stream 1: 1,000 mol/h
• Acetone 7 mol%
• Acetic acid 61.9 mol%
• Acetic anhydride 31.1 mol%
• Stream 3: 700 mol/h
• Acetone 10 mol%
• Acetic acid 88 mol%
• Acetic anhydride 2 mol%
• Stream 5: 300 mol/h
• Acetic acid 1 mol%
• Acetic anhydride 99 mol%
How can we get the flow rates and compositions of streams 2 and 4?
It is stated that the composition of the overhead (i.e. the vapor stream coming out of the distillation
column) is equal to those of the distillate and the reflux.
Since we have obtained the composition of the distillate (stream 3) from the above calculations, the
compositions of the overhead (stream 2) and the reflux (stream 4) are known (i.e. identical to that of
stream 3)
To obtain the flow rates of streams 2 and 4, we perform the material balance around the condenser
+ the splitter (or we draw the system boundary around the condenser + the condenser)

Since there is NO reactions at the condenser and the splitter, we can perform the mole balance
round this point.
Overall balance
S2 = S3 +S4
S2 = 700 + S4
We have 2 unknowns, but we can have only 1 overall-balance equation, as a species balance does
NOT work for this boundary system (since the compositions of all streams are identical). We need
another equation. It is given, in the problem statement given that
60% of the overhead is returned as a reflux
which can be translated into equation form as follows
S4 = ( 0.60) S2
We have, S2 = 700 + S4
S2 = 700 + ( 0.60) S2
S2 = 1,750 g-mol
Then S4 = ( 0.60) S2
S4 = 1,050 g-mol
Thus, in summary, the flow rates of
• stream 2 = 1,750 mol/h
• stream 4 = 1,050 mol/h
and the compositions of streams 2 and 4
• acetone 10 mol%
• acetic acid 88 mol%
• acetic anhydride 2 mol%
which are identical to that of stream 3
Example 5.9. Normally, fresh orange juice consists of 10-15 wt% dissolved solids in water. In order
to reduce the cost of shipping, the juice is concentrated prior to shipping. Concentration must be
carried out in a specially-designed, short-time evaporator operated at low pressure in order to
reduce the boiling temperature, which, in turn, reduce the loss of flavour and aroma components.
Unfortunately, however, the loss of flavour and aroma is unavoidable, some amount of fresh orange
juice (called “cutback”), by-passed from the evaporator, is mixed with the product from the
evaporator to improve flavour and aroma of the product.

Ten thousand (10,000 kg/h) of the feed containing 12 wt% of dissolved solids is fed to the system,
and 10 wt% of the feed is to be used as a cutback. If the product from the evaporator contains 80
wt% dissolved solids, calculate the evaporation rate and the composition of the final product.
Thus, the feed (F1 ) is 10,000 kg
Performing the overall balance around the splitter yields
F1 = F2 + F3
10,000 = F2 + F3
It is given that F3 = (0.10)F1
Substituting F1 = 10,000 kg results in
F3 = (0.10) 10,000 = 1,000 kg
Thus, from Eqn. 10,000 = F2 + F3
10,000 = F2 + 1,000
F2 = 9,000 kg
The compositions of these 3 streams (i.e. streams 1–3) are identical and are follows
• Dissolve solids 12 wt%
• Water 88 wt%
Performing the balance around the evaporator gives
Overall balance
F2 = F4 + F5
9,000 = F4 + F5

Solids balance
( ysolids )2 F2 = ( ysolids )4 F4 + ( ysolids )5 F5
( 0.12 ) F2 = ( 0.00 ) F4 + ( 0.80 ) F5
0.12 F2 = 0.80 F5
0.12( 9,000) = 0.80 F5
F5 = 1,350 kg
Thus, from Eqn., 9,000 = F4 + F5
9,000 = F4 + 1,350
F4 = 7,650 kg
Thus, the evaporation rate is 7,650 kg/h
Performing the balance around mixer results in
Overall balance
F3 + F5 = F6
Substituting the numerical values of F3 & F5
1,000 + 1,350 = F6
F6 = 2,350 kg
Solids balance
( xsolids )3 F3 + ( xsolids )5 F5 = ( xsolids )6 F6
(0.12 )(1,000) + (0.80 )(1,350) = ( xsolids )6 (2,350)
( xsolids )6 = 0.511 = 51.1 %

Problem Set 4
PROB 1. Determine the following flow chart and calculate numerical values of all unknowns
(indicated by the bold characters).
PROB 2. In order to remove salts from the sea water, the sea water has to pass through a process
called desalinisation. The desalinisation process is;
If the fresh sea water has the flow rate of 500 kg/h, determine
a) the flow rate of brine (in kg/h)
b) the daily production of desalinised water
c) the recycle ratio (i.e. the flow rate of the recycle brine/the flow rate of the fresh feed)

PROB 3. The fruit-drying process is illustrated in the following flow chart. The fresh feed containing
70 wt% dry fruit and 30 wt% water (moisture) is mixed with the recycle stream, which contains
97 wt% dry fruit and 3 wt% moisture in a mixer. The product leaving the mixer contains moisture in
the amount of 15 wt%. It is then fed into a dryer, where moisture is removed from the fruit until the
moisture in the fruit is lower than 3 wt%. The dried fruit from the dryer is divided into two parts: the
first one is shipped to customers, and the second one is recycled to mix with the fresh feed, as
described earlier. Calculate the recycle ratio (i.e. mass of the recycled stream/mass of the fresh feed)
PROB 4. P is a desired product, which can be produced from the substance A, as shown below
2A P
A, in parallel, may be converted to B, which is an undesired product,
A B
The feed consisting of A 90 mol% and inerts (I) 10 mol% is fed into the process, shown in the
following flow chart. The overall conversion of A is 80%, and %yield of P at the reactor is 50%.
Additional data regarding the process are as follows:
• The recycle stream (Stream 2) is adjusted until the Stream 3 has the composition of 84 mol%
A and 16 mol% I
• The Stream 4 contains only B (the undesired product)
• The Stream 6 consists of P (the desired product) only
• The Streams 2, 7, and 8 contain only A and I, and the mole ratios of A and I in these streams
are identical

If the feed has the molar flow rate of 100 g-mol/s, determine the flow rate and composition of each
stream (i.e. Streams 2-8). Also calculate the recycle ratio and the selectivity.
Hints:
• Since I is inerts, the total amount of I entering the process is equal to that leaving the process
• The overall conversion of A can be calculated from the amount of A in the feed (Stream 1)
and that in Stream 8
• %yield of P is obtained at the reactor

Chapter 6:
Energy Balances

Chapter 6
Energy Balances
Learning Objectives
After completing this Chapter, you should be able to
• calculate a specific heat when given an appropriate equation, and initial and final
temperature
• calculate a specific heat when given an appropriate chart, and reference and final
temperature
• calculate sensible and latent heats, when given appropriate sets of data
• select a system boundary suitable for solving a problem, either closed or open (control
volume) system
• write a correct energy-balance equation for a selected system
• simplify a general energy-balance equation to eliminate some unimportant terms, when
given process conditions
• calculate the values of temperature, pressure, heat, work, internal energy, and enthalpy for a
given process
• explain the meaning of standard heat of formation, heat of reaction, and heat of combustion
• compute heats of formation for a given substance in a selected reaction
• calculate the standard heat of reaction from standard heats of formation and combustion
• solve material and energy balance problems involving chemical reactions
• compute the moles or mass, temperature, and/or compositions of the output stream(s),
when given the information on moles or mass, temperature, and or compositions of the
input stream(s), and vice versa
• calculate how much material must be fed/removed into/from a system, in order to provide a
desired quantity of heat transfer for the system
• solve material and energy balance problems involving a combustion process
• determine the adiabatic reaction temperature
As a FOOD ENGINEER, the following are the examples of questions needed to be answered during
working on process design or during operating the existing process(es):
• To send milk from a storage tank to an evaporator, what is the size of a pump required? [or
how many “hp” (energy or power) of a pump do we need?]
• What is the total amount of heat needed to raise the temperature of 2,000 g of water from a
room temperature to 100 o
C?
• If Rxn. A+B C is an exothermic reaction and we want to keep the temperature of the
reactor constant, what should the rate of heat removal (in J/s) be, given the fact that A and B
are fed to the reactor with the mass flow rate of 30 kg/s?
• How can we bring the heat from an exothermic process to an endothermic one?
In order to answer these questions (and many more), we need to perform ENERGY BALANCES.

Consider a closed system below
where
Qin = heat transferred from surroundings to the system (+ sign)
Qout = heat transferred from the system to surroundings (– sign)
Wsurr = work done by surroundings (onto the system) (– sign)
Wsys = work done by the system (+ sign)
and
ΔE = Ef – Ei
Ef = energy of the system at the final state
Ei = energy of the system at the initial state
Note that energy (E ) consists, mainly, of
• internal energy (U )
• kinetic energy ( KE = ½ mV2
)
• potential energy (PE =mgh )
Normally, for most of chemical processes, KE & PE are small, when compared with U.
Thus, they (i.e. KE & PE ) are negligible and, therefore,
From a Thermodynamics (or Physics) course, we can write a relationship between
Q, W, & E as follows
(6.1)
where
i Q f = the amount of heat transferred during the initial state (i ) to the final state ( f )
i Wf = the amount of work occurred during the initial state (i ) to the final state ( f )
When dividing Eq. 6.1 with mass (m) of the system, we obtain

(6.2)
Since, as mentioned previously, KE & PE can be neglected, Eqs. 6.1 & 6.2 can be re-written, as follows
(6.3)
(6.4)
For an open system (or a control volume),as illustrated below
we can apply the law of mass conservation to this system as follows:
Material balance
or
Once again, from a Thermodynamics (or Physics) course, we can apply the 1st law of
thermodynamics to write an energy balance equation as follows:
Energy balance
Note that = flow work (do you know what the flow work is?)

Again, KE & PE are commonly negligible (for most chemical processes), which results in the fact that
and Eq. 6.6 can be re-written as follows
(6.8)
where h = enthalpy per mass
Combining Eqs. 6.7 & 6.8 together gives
Normally, the system we are considering, particularly in this level of study, is in steady state, which
means that
Thus, Eqs. 6.5 & 6.9 can be re-written as follows
and
Energy Balances for Non-Reactive Processes
Eqs. 6.3 (or 6.4) and 6.11 are an energybalance equations for closed systems and open systems
(control volumes), respectively
When considering the term Q (heat), there are 2 types of HEAT:
• Sensible heat (i.e. heat arising from the change in temperature, while the state of a
substance does NOT change)
• Latent heat (i.e. heat occurred when the state of a substance changes, while the temperature
is kept constant)

Calculations of a sensible heat
When there is no work (i.e. W = 0), the sensible heat can be computed using the following equation:
where
cp = specific heat (heat capacity)(kJ/kg.K in SI)
In order to perform the integration (and thus obtain the answer for Eq. 6.12), we need to know the
relationship between cp and T, and the common relationship is in the following form (or equation):
Note that, in fact, there are a number of other forms (or equations) of relationship, which can be
found in the literature.
Combining Eqs. 6.12 & 6.13 together gives
However, for our convenience, we may assume that cp is constant throughout the range of
temperature (from T1 to T2), by using cpmean , as follows

The is an average cp from the reference temperature or T1 (normally 25 o
C or 77 o
F) to the
final temperature (or T2)
The of several gases are as shown below (note that the notation is used as the unit of
the is in “mole” basis)
The of various gases (Ref. T = 77 o
F)

Example 6.1 Calculate of water H2O.
from 77 o
F to 500 o
F = 8.2 Btu/lbm-mol-o
F
from 77 o
F to 1,600 o
F = 9.1 Btu/ lbm-mol-o
F
Applying the rule of integration gives
and
Thus,
Procedure for Energy Balance Calculations for Processes without Chemical Reactions
The following are common procedure for performing energy-balance calculations for non-reactive
processes:
1) Perform material balances (correctly)
2) Write an appropriate energy balance equation (for either closed or open system)
3) Make appropriate assumptions; thus some terms can be dropped
4) Choose an appropriate REFERENCE STATE (e.g., state of a substance (s, l, or g), T, & P)
5) Perform energy balances and/or material balances (by choosing an appropriate system boundary)

REFERENCE STATE
The values of specific internal energy (u) and enthalpy (h ) are NOT absolute values.
The values of these properties are, in fact, relative to those at a reference state.
For instance, for H2O, the triple point (T = 0.01 o
C) is used as a reference state, and the values of
internal energy & enthalpy at this point is set as 0.
The values of internal energy & enthalpy at other points (states) are, therefore,
u1 – uref = u1 – 0 = u1
h2 – href = h2 – 0 = h2
or
Note that, normally, the bar over u and h indicates that u and h are on mole basis (i.e. its unit is, e.g.,
kJ/g-mol or MJ/kg-mol)
Example 6.2 A gas stream, containing 8 mol% CO and 92% CO2 at 500 o
C, is fed to a waste heat
boiler at the rate of 1 kg-mol/h, where the hot gas flows over the outside of the tubes.
Liquid water at 25 o
C is fed to the boiler in a ratio of 0.2 g-mol of liquid water per 1 g-mol of hot gas,
and flows inside the tubes.
Heat is transferred from the hot gas through the tube walls to the liquid water, causing the hot gas to
cool, and the liquid water to heat to its boiling point and evaporate to form steam at 5 bar. The
steam may be used for heating power generation in the plant or as the feed to another process unit
The gas leaving the boiler is burned (flared) and then discharged to the atmosphere. The boiler
operates adiabatically (i.e. all the heat transferred from the hot gas goes into the liquid water)
The flow chart of this process is as follows:
Flow chart

What is the temperature of the off-gas? Given:
Enthalpy of liquid water @ 25o
C = 1,890 kJ/kg-mol
Enthalpy of steam @ 5 bar = 49,455 kJ/kg-mol
No work is taken place in this process
The heat exchanger is working adiabatically, meaning that the heat from the hot gas is thoroughly
transferred into water, or, in other words, there is no heat loss from the system.
Since this is a “control volume” problem, Eq. 6.11 is then used, but enthalpy used in the equation is
on molar basis (not mass basis), Eq. 6.11 must be adapted slightly for the control volume and for the
molar basis as follows
Note that all terms in Eq. 6.16 are on molar basis
From a problem statement,
• Adiabatic process = 0
• No work = 0
Thus, Eq. 6.16 is reduced to
where
Choosing ref. T = 25 o
C (77 o
F) gives

The values of of CO2 and CO can be obtained from the chart in Page 6, but the unit of
of those gases has to be converted from Btu/lbm-mol-o
F to kJ/kg-mol-o
C, before they can be
substituted into the above equations.
Hence, for an input stream, where T = 500 o
C (or 932 o
F),
of CO2 = 10.9 Btu/ lbm-mol-o
F
= 45.6 kJ/kg-mol-o
C
of CO = 7.2 Btu/ lbm-mol-o
F
= 30.1 kJ/kg-mol-o
C
Perform Energy Balances
INPUT
In case of water, since our ref. T = 25 o
C, enthalpy (h ) of water at 25 o
C is, therefore, 0 [i.e. (1,890 –
1,890) kJ/kg-mol = 0]
Thus,
OUTPUT
In order to obtain the values of of CO2 & CO, we need to know Tout, but we have not known
Tout yet. Hence, we need to use the technique called
TRIAL & ERROR
First, we make an initial guess for Tout and then use of the chosen Tout to calculate the value of
If , the chosen Tout is then correct

However, if the resulting , we need to make another guess a new Tout,
and redo the calculations for , until we obtain the value of Tout that makes
Note that, for water, since we chose Tref= 25 o
C, the enthalpy of the outlet water (@ 5 bar) which is
49,455 kJ/kg-mol must be compared with the enthalpy of water at 25 o
C, as follows:
= 49,455 - 1,890
47,565 kJ/kg-mol
To obtain the answer, we make an initial (1st) guess of Tout of 200 o
C (392 o
F)
The corresponding values of at 200 o
C or 392 o
F of CO2 and CO are 9.7 and 7 Btu/lbm-mol-o
F,
respectively, which can be converted to 40.5 and 29.3 kJ/kg-mol-o
C, respectively.
Hence,
which is NOT equal to (i.e.21,071 kJ)
Try to make a new guess of Tout at 300 o
C or 572 o
F, which results in at 300 o
C or 572 o
F of CO2
and CO are 10.1 and 7.1 Btu/lbm-mol-o
F, respectively, which can be converted to 42.2 and 29.7
kJ/kg-mol-o
C, respectively.
Thus,

which is still NOT equal to (i.e.21,071 kJ) – but it is getting closer.
Keep performing the trial & error repeatedly until we get
The answer of Tout for this Example is approximately 305 o
C
Energy Balances on Reactive Processes
Consider the following chemical Rxn.:
CaC2 + 2H2O Ca(OH)2 + C2H2
On the reactant side, the chemical bonds of reactants are broken:
C – Ca – C C Ca C
H – O – H OH H
H – O – H H OH
Breaking these bonds is an endothermic process (heat must be provided)
On the product side, the new chemical bonds are formed as follows
OH – Ca – OH
H – C ≡ C – H
Forming the new bonds is an exothermic process (heat is released)
If the total energy needed to break the bonds > the total energy obtained when the new bonds are
formed endothermic reaction
On the contrary, if the total energy obtained when the new bonds are formed > the total energy
required to break the bonds exothermic reaction
The net amount of heat of a particular reaction is called HEAT of REACTION.
The heat of reaction is, in fact, the change in enthalpy when reactants at given T & P is completely
converted to products at the same T & P.
The “standard” heat of reaction = the heat of reaction at T = 25 o
C and P = 1 atm
For example,
2A + B 3C;
Δ (25 o
C, 1 atm) = −50 kJ/g-mol

is equivalent to
2A + B 3C;
Δ = −50 kJ/g-mol
Note that
+ sign indicates endothermic Rxn.
– sign indicates exothermic Rxn.
Note also that the common unit of “heat of Rxn.” is kJ/mol or kJ/g-mol
From the example on the previous Page,
Δ = −50 kJ/g-mol
the heat of Rxn. is per mol of what substance/species, as there are 3 substances/species in the given
Rxn.?
This answer is as simple as the following:
For the Rxn.:
2A + B 3C Δ = −50 kJ/g-mol
the heat of Rxn. is
Thus, if 30 g-mol of C is generated, the total heat generated from this Rxn. can be calculated as
follows;
Hence, in general, the total amount of heat generated from a particular Rxn. can be calculated using
the following equation:
where
= stoichiometric coefficient of substance i
ni = # of moles of substance i (actually consumed or generated)

Some useful facts of “Heat of Reaction”
1) At low to moderate P, it is safe to assume that heat of Rxn. is only a function of temperature (T)
(i.e. the change in P negligibly affects the change in heat of Rxn.)
2) If, for the following Rxn.,
CH4(g) + 2O2(g) CO2(g) + 2H2O(l);
Δ (25 o
C) = − 890.3 kJ/g-mol
we obtain the following fact:
2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(l)
Δ (25 o
C) = 2x(− 890.3 kJ/g-mol)
= –1,780.6 kJ/g-mol
3) Heat of Rxn. depends on the state of reactant(s) and product(s): for example,
CH4(g) + 2O2(g) CO2(g) + 2H2O(l);
Δ (25 o
C) = − 890.3 kJ/g-mol
CH4(g) + 2O2(g) CO2(g) + 2H2O(g);
Δ (25 o
C) = − 802.5 kJ/g-mol
Example 6.3 Given
C4H10(g) + 13/2 O2(g) 4CO2(g) + 5H2O(l) (a) Δ = − 2,878 kJ/g-mol
Determine
a) Std. heat of rxn. of the Rxn.:
½ C4H10(g) + 13/4 O2(g) 2CO2(g) + 5/2H2O(l) (b)
b) If 80 g-mol of CO2 is generated, calculate of this Rxn.
a) Since Rxn. (b) = ½ Rxn. (a), the std. heat of reaction Δ of Rxn. (b) is
(Δ )Rxn. (b) = 1/2(Δ )Rxn. (a)
Thus,
(Δ )Rxn. (b) = ½ x ( - 2,878 kJ/ g-mol)
= – 1,439 kJ/g-mol

b) Since 80 g-mol of CO2 is generated,
if Rxn. (a) is used,
but if Rxn. (b) is used,
Note that we can use either Rxn. (a) or (b) (or any other equivalent Rxn.) to compute the value of the
total heat of the reaction and obtain the same (or identical) answer
Hess’ Law
Another useful fact of heat of Rxns. İs the Hess’ law ; has same value regardless of steps o rxns
Consider the following Rxn.:
C + O2 CO2 Δ = − 393.1 kJ/g-mol
The heat of Rxn. of the above Rxn. can be measured experimentally, using a calorimeter.
However, the heat of Rxn. of the following Rxn.:
C + ½ O2 CO
(incomplete combustion) cannot be measured directly.
Nevertheless, the heat of Rxn. of the Rxn.:
CO + ½ O2 CO2
can be measured experimentally, in which Δ = − 282.99 kJ/g-mol
By using the Hess’ law, the heat of Rxn. of the Rxn.:
C + ½ O2 CO
can be calculated as illustrated below:
Consider the following Rxns:
C + O2 CO2 Δ = − 393.1 kJ/g-mol (6.19)
CO + ½ O2 CO2 Δ = − 282.99 kJ/g-mol (6.20)
Rxn. (6.19) – Rxn. (6.20) yields

C – CO + ½ O2 0
Rearranging gives
C + ½ O2 CO (6.21)
Thus,
Rxn. (6.21) = Rxn. (6.19) – Rxn. (6.20)
By using Hess’ law, the heat of Rxn. of Rxn. (6.21) can be calculated as follows
= (- 393.51 kJ/g-mol) - (- 282.99 kJ/g-mol)
= - 110.52 kJ/g-mol
TEST YOURSELF
Given
(a) C2H6 + 7/2 O2 2CO2 + 3H2O = – 1,559.8 kJ/g-mol
(b) C + O2 CO2 = – 393.5 kJ/g-mol
(c) H2 + ½ O2 H2O = – 285.8 kJ/g-mol
Determine of the following Rxn.:
(d) 2C + 3H2 C2H6 = ?
Determination of Heat of Reaction from Heat of Formation
The formation reaction of a compound is the reaction in which the compound is formed from its
elemental constituents as they normally occur in nature.
Heat generated (or consumed) from this “formation reaction” is called “HEAT OF FORMATION”
Consider the following formation reaction, in which benzene (C6H6) is formed from carbon (C) and
hydrogen (H2). Note that that, in nature (or its common form), hydrogen appears as a di-atomic
hydrogen gas, whereas carbon appears as a mono-atomic solid carbon.
6C + 3H2 C6H6 = +48.66 kJ/g-mol
Thus, the standard heat of formation ( ) of f C6H6 = +48.66 kJ/g-mol (note that the positive (+) sign
indicates that this Rxn. is endothermic)
Other examples are as follows

N2 + 2H2 + 3/2 O2 NH4NO3 = – 365.14 kJ/g-mol
Hence, the standard heat of formation ( ) of NH4NO3 = –365.14 kJ/g-mol (exothermic Rxn.)
C + 2S CS2(g) = +115.3 kJ/g-mol
Therefore, the standard heat of formation ( ) of CS2(g) = +115.3 kJ/g-mol (endothermic Rxn.)
C + O2 CO2(g) = –393.5 kJ/g-mol
So, the standard heat of formation ( ) of CO2 = –393.5 kJ/g-mol (exothermic Rxn.)
C + ½O2 CO(g) = –110.5 kJ/g-mol
Accordingly, the standard heat of formation ( ) of CO = –110.5 kJ/g-mol (exothermic Rxn.)
Note that the standard heat of formation ( ) of is the heat of formation at 25 o
C and 1 atm.
Normally, each individual substance (either compound or element) has its own (standard) heat
formation ( ) .
It is important to note that the standard heat of formation ( ) of any elements in their natural
form are zero (0).
From the data of heats of formation of substances, the heat of reaction of any reactions can be
computed using the following equation:
Example 6.4 Determine the standard heat of reaction ( ) of the combustion of n-pentane,
assuming H2O(l) is a combustion product
C5H12(l) + 8O2(g) 5CO2(g) + 6H2O(l)
Given of C5H12(l), CO2(g), and H2O(l) = –173.0, –393.5, and –285.8 kJ/g-mol, respectively
Substituting the given into Eq. 6.22 yields
[5(- 393.5)+ 6(- 285.8)]- [1(- 173.0) + 8(0)] = - 3,509 kJ/g-mol

Heat of Combustion
The standard heat of combustion ( ) of a substance is the amount of heat generated from the
reaction of that substance with oxygen (O2) to yield specified products.
For example, for the following reaction:
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
the amount of heat generated from this Rxn. (or the heat of reaction) is the heat of combustion of
C2H5OH(l)
Other examples are as follows:
C8H18 + 25/2 O2 8CO2 + 9H2O
where the heat of reaction of this Rxn. is the heat of combustion of C8H18
H2 + ½ O2 H2O
in which the heat of reaction of this Rxn. is the heat of combustion of H2.
CO + ½ O2 CO2
where the heat of reaction of this Rxn. is the heat of combustion of CO.
Note that not all substances have the data of heat of combustion (as not all substances can react
with O2)
Determination of Heat of Reaction from Heat of Combustion
The heat of Rxn. can also be calculated from the data of heat of combustion of substances using the
following equation:
Example 6.5. Calculate the standard heat of reaction of the following “dehydrogenation” reaction:
C2H6 C2H4 + H2
Given the standard heat of combustion of C2H6, C2H4, and H2 are –1,559.9, –1,411.0, and –285.8
kJ/g-mol, respectively.
Substituting the given into Eq. 6.23 gives

[1(- 1,559.9)]- [1(- 1,411.0)+ 1(- 285.8)] = + 136.9 kJ/g-mol
Material & Energy Balances for Combustion Processes
There are 2 main methods to perform material and energy balances for the combustion processes:
• Heat of reaction method
• Heat of formation method
Heat of Reaction Method
Let’s consider the following reaction (i.e. the combustion of ammonia):
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) –904.7 kJ/g-mol
By the definition of the standard heat of Rxn., the reaction between reactants must take place at
25 o
C and 1 atm, and the products must occur at the same T & P (i.e. 25 o
C & 1 atm)
Practically, however, in the real situation, it is NOT necessary that reactants react with each other (or
one another) at NTP (i.e.25 o
C & 1 atm)
Likewise, it is also NOT necessary (or even impossible) for the products to be generated at NTP.
For example, for a real combustion process, the reaction between reactants may take place at 200-
300 o
C, and the product(s) is (are) formed the at a much higher temperature (e.g., 700-800 o
C).
To calculate heat of Rxn. at other conditions (i.e. at ant T & P that are not 25 o
C and 1 atm), we follow
the following chart:
The actual process follows the path with a solid line, but for the calculations, we shall follow the
paths with dashed lines.
We can do this because we are calculating the change in enthalpy (ΔH ), which is a state function:
i.e. the change in enthalpy depends only on the starting point and the final point, not depends on
the path of the process.

By using the principle that the change in enthalpy depends only on the starting and final points, we
can write the following equation for the chart on the previous Page:
(6.24)
The details of each term in Eq. 6.24 are as follows
where
ni = # of moles of substance i
= specific heat of substance i
And
but
Hence,
and
Since a reference T = 25 o
C (298 K),
Thus, Eqs. 6.27 and 6.28 become
and

Combining Eq. 6.29 with 6.25 and Eq. 6.30 with Eq. 6.26 results in
and
For the term in Eq. 6.24 (see Page 20), we have learned that it can be calculated using Eq. 6.18
(see Page 13)
Combining Eqs. 6.18, 6.31 & 6.32 with Eq. 6.24 gives
Note that “Products” in Eq. 6.33 include the un-reacted reactants and “Reactants” include any
substances in all inputs
Thus, to avoid any confusion, Eq. 6.33 is re-written as follows
Eq. 6.34 (or 6.33) is the energy-balance equation for the combustion process It is noteworthy that,
even though Eq. 6.34 is developed for the combustion process, it can also be used for any other
reactive processes.
Example 6.6 The standard heat of reaction for the oxidation of methane is given below:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) –803 kJ/g-mol

One hundred (100) g-mol CH4 and 300 g-mol O2/s at 25 o
C were fed into a reactor in which CH4 is
completely consumed, and the product is at 500 o
C. Calculate the rate at which heat must be
transferred to or from the reactor, assuming operation at ~1 atm.
Flow chart
First, we have to perform material balances for this process
Rxn.:
CH4 + 2O2 CO2 + 2H2O
From the valid (or balanced) chemical reaction,
1 g-mol CH4 ≡ 2 g-mol O2
From the given data:
• CH4 fed = 100 g-mol/s
• O2 fed = 300 g-mol/s
Thus, from the chemical Rxn., we obtain the fact that
100 g-mol/s CH4 ≡ 200 g-mol/s O2
Since it is given, in the problem statement, that CH4 is completely, it leads to the fact that O2 is an
excess reactant, and the amount of remaining O2 is
300 – 200 = 100 g-mol/s
Also from the balanced reaction, for every 1 g-mol of CH4 consumed,
• 1 g-mol of CO2 is generated
• 2 g-mol of H2O is generated
Thus, for 100 g-mol/s of CH4 consumed,
• 100 g-mol of CO2 is generated
• 200 g-mol of H2O is generated
Accordingly, the amount of moles of each substance in the input and output streams can be
summarised, as follows:

INPUT OUTPUT
CH4 100 g-mol/s O2 100 g-mol/s
O2 300 g-mol/s CO2 100 g-mol/s
H2O 200 g-mol/s
To check whether or not our material-balance calculations are correct, we need to do it on mass
basis, and the calculations for the total mass of the input and output streams are as follows;
The total mass of the “input” stream can be computed as follows:
Performing the same calculations for the “output” stream yields:
This indicates that our material-balance calculations are correct.
We can then proceed to the energy balance calculations.
We start our energy-balance calculations by setting reference states as follows
• T = 25 o
C
• H2O(g) as a combustion product
For our convenience, the enthalpy of each species (substance) for both input and output streams is
tabulated as follows:

Since Tref = 25 o
C, and = 0 (do you know why?)
is an enthalpy of O2 at 500 o
C (773 K), with reference to that at 25 o
C (298 K), or
kJ/g-mol
is an enthalpy of CO2 at 500o
C (773 K), compared with that at 25 o
C, or
kJ/g-mol
is an enthalpy of H2O at 500o
C (773 K), compared with that at 25 o
C, or
kJ/g-mol
Substituting corresponding numerical values into each term on the right hand side (RHS) of Eq. 6.34
gives

Note that, for this Example, i = CH4, thus resulting in the fact that
• ni = 100 g-mol/s
• νi = 1
and, eventually,
( -80,300)+ (7,030)- (0)= -73,270 kJ/s = -73,270 kW
73.3 MW
Since this is an exothermic reaction, heat is transferred out of the system (i.e. system surroundings)
with the rate of 73.3 MJ/s
In the recent Example, there is only one combustion reaction (i.e. the combustion reaction of CH4). In
the case that there are more than one combustion reaction, the term
has to be adapted slightly to be
where is the heat of reaction of the combustion reaction of species i.
Heat of Formation Method
Another method to perform energy balance calculations is the “heat of formation” method. The
details of this method are as illustrated in the following Example

Example 6.7. Methane is oxidised with O2 in the air to produce formaldehyde in a continuous
reactor. Another competing reaction is the combustion of methane to form CO2
CH4(g) + O2 HCHO(g) + H2O(g) (1)
CH4(g) + O2 CO2(g) + H2O(g) (2)
A flow chart of this process is shown below
Determine the rate of heat generated or consumed in this process.
Basis: 100 g-mol CH4 fed to the reactor
Since the mass-balance data are already given, it is, therefore, NOT necessary to perform the
material-balance calculations. However, it should be a good idea to verify that the given material-
balance data are correct. In order to do so, we calculate the total mass of the input and output
streams, and the summary of the calculations as follows:

This confirms that the given material-balance data are correct.
Before proceeding to the energy-balance calculations, we should set reference states. In this
Example, we should set reference states as
• T = 25 o
C
• H2O(g) as a combustion product
Once again, for our convenience, the enthalpy of each species for both input and output streams can
be tabulated as follows
The value of each enthalpy can be computed as follows
kJ/g-mol
kJ/g-mol
kJ/g-mol

kJ/g-mol
kJ/g-mol
kJ/g-mol
kJ/g-mol
kJ/g-mol
kJ/g-mol
Substituting corresponding numerical values into the Table gives

For the case of “heat of formation” method, the amount of heat transfer from/to the reactor can be
calculated using the following equation:
Substituting the corresponding numerical values into Eq. 6.35 give
Note that the main difference between the “heat of reaction” method and the “heat of formation”
method is that, for the “heat of formation” method, in addition to the term (for
mole basis) or (for mass basis) (i.e. the “sensible heat” term of species i ), there
is the term , which is the “heat of formation” term of species i.
Adiabatic Reactor
The principle of “adiabatic” reactor is that there is no heat going into or out of the system, or
= 0
This principle can be used to calculate the maximum temperature of the combustion products for
given conditions.

Example 6.8. The standard heat of reaction for the oxidation of methane is given below:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) –803 kJ/g-mol
One hundred (100) g-mol CH4 and 300 g-mol O2/s at 25 o
C were fed into a reactor in which CH4 is
completely consumed.
If the reactor operates adiabatically, what is the temperature of the combustion products?
Since this Example is the same as that done just recently (Example 6.6), the material-balance
calculations are as same as the previous Example 6.6. The difference is on the ENERGY-BALANCE
calculations.
From the energy-balance equation for this system (Eq. 6.34)
if this is an adiabatic reactor, = 0
Thus, Eq. 6.34 is reduced to
The Rxn. is the same Rxn. as previous Example 6.6. Thus,
Additionally, the conditions of the input streams are still the same; thus, the term does
not change, or
The only change is at the output stream, in which we need to determine the temperature of the
combustion products or T out
The T out we need is the T out that makes the left hand side (LHS) of Eq. 6.36 equal to zero (0), or the
T out that makes
Substituting numerical values into the above equation and re-arranging yields

which means that we need T out that makes
To determine the value of T out, we can follow the following procedure:
1) Make an initial guess of Tout
2) Determine/calculate the value of of each species at the guessed Tout
3) Calculate the value of ( )
4) Determine if ( ),
4.1) if so, the guessed Tout is correct
4.2) if it is not, we need to guess another new Tout, and redo the above procedure all over
again, until we get the value of Tout that makes ( )
Let’s make our initial guess for Tout as 4,000 K
At T = 4,000 K,
• (with ref. to 298 K) = 138.7 kJ/g-mol
Hence,
Thus, we need to make a new guess for Tout and redo the whole procedure once again, until we
obtain the Tout that makes ( )
Eventually, we find that the Tout that makes ( ) is ~4,400 K (4,127 o
C)
The temperature of the combustion products for the case of “adiabatic reactor” is also called “the
adiabatic flame temperature”, and this is the highest possible temperature for given combustion
conditions.
In the recent Example, CH4 is burned with pure O2.

Let’s consider the adiabatic flame temperature for the case when CH4 is burned with air (21 mol% O2
& 79% N2) by repeating the previous Example, but for the case that 100 g-mol/s of CH4 is burned with
300 g-mol/s of O2 that comes into the reactor as air.
As we have learned previously, the amount of N2 that comes along with O2 can be computed as
follows
= 1,128.6 g-mol/s of N2
Since N2 does not react with any other species, all of N2 flows into the reactor must flow out of the
reactor in the same amount.
Thus, the summary of the input and output streams are as follows
INPUT OUTPUT
CH4 100 g-mol/s O2 100 g-mol/s
O2 300 g-mol/s CO2 100 g-mol/s
N2 1,129 g-mol/s H2O 200 g-mol/s
N2 1,129 g-mol/s
Note that the material-balance calculations for other species are still the same as the recent Example
The energy-balance calculations are similar to those for the recent Example, but the enthalpy of N2
must be added to the calculations. For the input stream, since N2 comes with O2, it is fed to the
reactor at the temperature of 25 o
C; thus, its enthalpy is zero (0). For the output stream, N2 must
have the same temperature as other combustion products. To determine the “adiabatic flame
temperature” for this Example, we use the same procedure as the previous Example.
We make a 1st guess of Tout as 2,000 K
At T = 2,000 K,
Hence,

Thus, we have to make a new guess of Tout , and the new guess of Tout is 1,500 K
At T = 1,500 K,
Hence,
Keep iterating making the guess of Tout , and, eventually, we shall find out that the Tout that makes
( ) is,
approximately, 1,800 K or 1,527 o
C, which is much lower than the case when CH4 is burned with pure
O2.

Problem Set 5
Energy Balances
PROB 1. A heat capacity or a specific heat (in mole basis) of CO2 can be written as function of
temperature, as follows:
where is in the unit of J/(kg-mol)(K), and T used in the above equation must be in the unit of K.
Calculate the value of when the temperature is raised from 500 to 1,000 o
C, using the equation:
PROB 2. Redo the Problem 1, but using read from the chart, and compare the
result with that from Problem 1.
PROB 3. Standard heats of reaction of Reactions 1 and 2 are as follows:
2A + B 2C –1,000 kJ/g-mol (1)
A + D C + 3E –2,000 kJ/g-mol (2)
Calculate the heat of reaction of the following reaction:
B + 6E 2D
PROB 4. A standard heat of reaction of the following reaction:
2CO 2C + O2
is +221.0 kJ/g-mol. Determine the standard heat of formation ( of CO.

PROB 5. Natural gas comprising CH4 87.5%, C2H6 7.0%, and C3H8 5.5% by volume is burned with air
(21 vol% O2 and 79% N2) with an excess-air percentage of 10% in an adiabatic combustor. Determine
the outlet temperature of the gaseous products. Chemical Rxns:
CH4 + 2O2 CO2 + 2H2O - 802.4 kJ/mol (1)
C2H6 + 7/2O2 2CO2 + 3H2O -1, 427.9 kJ/mol (2)
C3H8 + 5O2 3CO2 + 4H2O -2, 044.0 kJ/mol (3)
PROB 6. As a safety engineer, you are asked to perform a safety analysis for a power plant boiler.
The maximum design temperature for the boiler is 2,000 o
C, and the worst case scenario is that the
fuel combusts completely and no heat is transferred out of the system; in other words, the system is
operating adiabatically.
The fuel used for this boiler comprises 80 mol% methane (CH4) and 20% ethane (C2H6), and it is
preheated to the temperature of 100 o
C before entering the system. Air (21 mol% O2 + 79% N2),
which is to be burned with the fuel, is also preheated, but to the temperature of 300 o
C, before being
fed into the system. The percentage of excess O2 (or percentage of excess air) is 10%.
Determine whether or not the boiler is going to be exploded. Note that the boiler may be exploded
when the final temperature (for the worst case scenario) of the system exceeds 2,000 o
C.
Given that the heat capacities of these substances (species) are assumed to be constant throughout
the process:
CH4 = 35.4 J/(g-mol-K)
C2H6 = 52.6 J/(g-mol-K)
CO2 = 50.0 J/(g-mol-K)
H2O(g) = 38.5 J/(g-mol-K)
O2 = 33.1 J/(g-mol-K)
N2 = 31.3 J/(g-mol-K)
Additional data (which may or may not be used to solve this Question) Standard heat of formation
( for
CH4 = –74,850 J/g-mol;
C2H6 = –84,670 J/g-mol;
CO2 = –393,500 J/g-mol;
H2O = –241,830 J/g-mol

Chapter 7:
Material &Energy Balances of
Some Selected Unit Operations

Chapter 7
Material & Energy Balances of Some Selected Unit Operations
Learning Objectives
• understand what vapour pressure is
• know how to find the value of vapour pressure at a given temperature from literature
• define saturation and partial saturation
• define absolute saturation (or absolute humidity) and relative saturation (or relative humidity)
• calculate the corresponding value of one form of partial saturation when given the value of one
of the other forms
• define and understand humidity, dry-bulb temperature, wet-bulb temperature, humidity (or
moisture content), dew-point temperature, humid volume (or moist volume), and adiabatic
cooling line
• use humidity chart (or psychrometric chart) to determine the properties of humid or moist air
• calculate enthalpy changes, and solve heating and cooling problems involving humid or moist
air
• use a three-phase diagram to determine the composition of a ternary system
• understand the principle of liquid-liquid (L-L) extraction
• define raffinate and extract phases, tie line, and plait point
• know how to draw a solubility curve for a given set of raffinate- and extract-phase data
• for a given solubility curve, define whether a given ternary system appears in one phase or in
two phases when given a composition of the system
• for a given solubility curve, define the composition of raffinate phase when given the
composition of extract phase, and vice versa
• solve material-balance problems involving ternary systems

Humidity & Saturation
Partial Saturation
Consider the following liquid-vapour system
Liquid water starts vaporising at a certain T, and keeps vaporising until it reaches equilibrium; i.e.
when P above liquid water equals vapour pressure of liquid water at the corresponding T.
At the equilibrium, the “water vapour” above the liquid is now called “saturated vapour”. At any
moment of time when P above liquid is lower than vapour pressure of that liquid at any given T, the
vapour above the liquid is called “unsaturated vapour” partial saturation
The concentration of water vapour in air of “saturated vapour” is normally expressed as “moles (or
mass) of water vapour per mole (or mass) of dry air”.
The concentration of “unsaturated vapour”, however, is NOT expressed as moles (or mass) of water
vapour per mole (or mass) of dry air directly.
Its concentration is normally expressed with reference to saturated vapour as
• relative humidity (RH)
• absolute humidity (AH)
in which
and
At any given temperature, AH and RH can be expressed mathematically as follows

where
= P of water vapour at any instant of time
= vapour pressure at any given T
Pt = total pressure above the liquid
Another way for determining the concentration of water vapour in air (normally, in %RH) is to
measure the temperature of air (called “dry-bulb” temperature: tdb) and that when the air is
saturated with water vapour (called “wet-bulb” temperature: twb).
How to measure a “wet-bulb” T (twb)
In order to measure the temperature of the air “when it is saturated with water vapour” (or “wet-
bulb” temperature), we do the following:
1) Wrap the bulb of the thermometer with cloth or cotton soaked with water
2) Leave the thermometer wrapped with soaked cloth or cotton in the air, until the temperature
reading is constant
3) Record the temperature (and this is called “wet-bulb” temperature: twb)
The dry-bulb and wet-bulb temperatures is then used to read for the corresponding “%RH” or
“humidity” from a humidity (or psychrometric chart)

Humidity Chart (Psychrometric Chart)
Humidity (or Moisture Content)
Humidity is defined as
Thus, the unit of “humidity” is SI unit system is
kg H2O/kg dry air
or
kg H2O/kg “moisture-free” air
Practically, however, it is rather difficult to measure mass of water vapour and mass of dry air
directly.
Normally, the pressures of water vapour and air are measured, and humidity or moisture content can
be calculated as follows:

and
where 18.02 and 28.97 are MW of water vapour and of air, respectively
Relative Humidity ( %RH )
Relative humidity can be expressed in “percentage” form as follows
Dew Point or Dew-point Temperature, tdp
Dew-point temperature is defined as the temperature at which humid air becomes saturated (i.e.
when a liquid drop appears) if it is cooled at constant pressure
Humid Volume, VH [m3
/kg dry air]
Humid volume (VH ) is the total volume of 1 kg of dry air plus the volume of water vapour (or
moisture) in the air at a specified condition
Example 7.1. Determine the following values of air with tdb = 30 o
C and twb = 25 o
C:
1) Humidity
2) % relative humidity (%RH)
3) Dew-point temperature (tdp)
1) To determine the humidity of air, we follow the following procedure:
1. Locate the intersection between tdb and twb
2. Draw a line parallel to the X-axis from the intersection point obtained from 1 to the axis on the
RHS of the chart
3. Read the corresponding humidity (or moisture content), which, in this Example, is found to be
0.018 kg H2O/kg dry air

2) The intersection between tdb and twb (in obtained Question 1) is between the lines of 60% and 70%
relative humidity (RH). Doing the interpolation between 60-70% RH lines gives %RH of air in this
Example of 67%
3) To determine the dew-point temperature of air, we follow the following procedure:
1. Locate the intersection between tdb and twb
2. Draw a line parallel to the X-axis from the intersection point obtained in 1 to the LHS of the
chart until it reaches the 100 %RH line. (note that the line parallel to the X-axis is a “constant-
pressure” or “constant-humidity” line; will be explained in detail later)
3. From that point, draw a downward line parallel to the Y-axis, and read the corresponding dry-
bulb temperature, and this temperature is a dew-point temperature (tdp), which, in this
Example, appears to be 23 o
C
Example 7.2 For the air with the temperature of 41 o
C and with %RH of 10, determine
1) humidity
2) wet-bulb temperature
3) humid volume
4) dew point
5) the amount of water in 150 m3
of air
1) In this Example, to determine the humidity of air, we follow the following procedure:
1. At tdb = 41 o
C, draw a upward line parallel to the Y-axis until it reaches the 10 %RH line – note
that this is the inlet condition of the air in this Example
2. Draw a line parallel to the X-axis from the point obtained in 1 to the axis on the RHS of the
chart and read the corresponding humidity, which, in this Example, is found to be 0.0046 kg
H2O/kg dry air
2) Also start from the inlet-condition point in Question 1 (tdb = 41 o
C and %RH = 10), the wet-bulb
temperature is found to be 19 o
C
4) • Once again start from the inlet-condition point in Question 1
• Draw a line parallel to the X-axis from that point to the LHS of the chart until it hits the 100 %RH
line, and the corresponding tdp is 3 o
C
3) • Start from the intersection of tdb = 41 o
C and %RH = 10 (the inlet-condition point)
• By interpolating between the humid volume (VH) lines of 0.85 and 0.90 m3
/kg dry air, it results in
the humid volume (VH) for this Example of 0.89 m3
/kg dry air
5) From the definition of “humid volume”, the humid volume of 0.89 m3
is volume of 1 kg of dry air +
volume of water vapour (or moisture) in the air. In this example, the amount of moisture (or
moisture content or humidity) in the air is 0.0046 kg/kg dry air.

Thus, the humid volume (VH) of 0.89 m3
is the volume of 1 kg of dry air + the volume of 0.0046 kg of
moisture (or water vapour)
Hence,
VH of 0.89 m3
is from the dry air of 1 kg
VH of 150 m3
is from the dry air of
In this Example, 1 kg of dry air contains 0.0046 kg of moisture (or water vapour). Hence, 168.54 kg of
dry air contains moisture (or water vapour) in the amount of
Example 7.3 Air with the temperature and pressure of 25 o
C and 101.325 kPa, respectively, contains
water vapour (i.e.moisture) with a partial pressure of 2.76 kPa. Determine
1) humidity (H)
2) saturation humidity (Hs)
3) relative humidity percentage (%RH)
Given vapour pressure (P* ) of water at 25 o
C of 3.169 kPa
1) Humidity or moisture content at a given temperature, on mass basis, can be calculated using Eq.
7.4 as follows
Substituting corresponding numerical values into Eq. 7.4 gives
Humidity (H) = 0.0174 kg moisture/kg dry air

2) Saturation humidity is the “maximum” amount of humidity (or moisture content) in the air at a
given temperature. In other words, saturation humidity is the amount of moisture in the air when
pressure of water vapour in the air is equal to the vapour pressure of water at a corresponding T
At 25 o
C, = 3.169 kPa
Thus, by using Eq. 7.4, Saturation humidity can be computed as follows
Saturation Humidity HS = 0.0201 kg H2O/kg dry air
3) %RH can be calculated from Eq. 7.5, as follows
Also from the given data, the following properties can also be obtained from the humidity chart (try
doing them yourself – they may appear in the FINAL EXAM):
• Wet-bulb temperature (twb) = 23 o
C
• Dew-point temperature (tdp) = 22 o
C
• Humid volume (VH) = ~0.86-0.87 m3
/kg dry air
Material & Energy Balance Problems involving HUMIDITY & SATURATION
Normally, the problems can be divided into 2 main categories (or processes)
• Constant PRESSURE process
• ADIABATIC process
Constant Pressure Process
We have learned that, for gases (whose properties can be considered as “ideal” gases), the following
relationship can be established when the process is taken place in a constant-volume container

The constant-P process means that the TOTAL PRESSURE (Pt) of the system is CONSTANT. Hence, as
long as there is no condensation of water vapour, the moisture content (or humidity) is CONSTANT.
Example 7.4. A stream of air at 25 o
C and 101.3 kPa (1 atm) contains 2.5 vol% water vapour
1) Determine the “dew-point” temperature of the air.
2) Calculate the percentage of water vapour that condenses if the air is cooled to 10 o
C at
constant pressure.
Given: Vapour pressures of water vapour at 25 and 10 o
C are 3.17 and 1.23 kPa, respectively
From Eq. 7.5,
and from the definition that
mole fraction of water vapour in the air
= # of moles of water vapour
= # of moles of all species (or # of moles of water vapour + # of moles of dry air)
In this Example,
Pt = 101.3 kPa
= 0.025 (2.5 vol% or 2.5 mol%)
Hence, the partial pressure of water vapour in the air can be calculated, using Eq. 7.8, as follows
= (0.025)( 101.3 kPa)
2.53 kPa
At 25 o
C, = 3.169 kPa

Hence, %RH of air can be calculated, using Eq. 7.5, as follows
From the psychrometric chart, where Tdb = 25 o
C and %RH = 80, the humidity (or moisture content)
of air can be read as 0.016 kg H2O/kg dry air
Alternatively, the humidity of air can be calculated as follows;
The Example states that the concentration of water vapour in the air is 2.5 vol.%, which means that
For a constant pressure process,
Thus,
Reciprocating Eq. 7.9 and re-arranging the resulting equation yields
Since

the humidity of can be computed as follows
= 0.0159 kg moisture/kg dry air
kg moisture/kg dry air
In order to determine the dew-point T (Tdp) of the air, draw a line parallel to the X-axis, where the
moisture content is 0.016 kg H2O/kg dry air, to the LHS, until it reaches the 100 %RH line. From that
point, draw a downward line parallel to Y-axis to read for Tdb, which, in this Example, is ~21 o
C. This T
is the dew-point temperature of the air with the moisture content (or humidity) of 0.016 kg H2O/kg
dry air.
When the temperature is decreased while the pressure (or the total pressure: Pt ) of the system is
kept constant,
• If there is no condensation of water (or the case when %RH 100), the humidity or moisture
content in the air will also be constant (in other words, no change in humidity or moisture
content) – note that the path for this phenomenon (or process) is the line parallel to the X-axis
at a constant humidity (or moisture content).
• If, however, the air is still cooled after it reaches the point where %RH is 100, a certain amount
of moisture (or water) will be condensed, and the path of this kind of process will be along the
100 %RH line.
In this Example, the temperature is kept decreasing from its dew-point temperature (i.e. 21 o
C) to the
final temperature (i.e.10 o
C), while %RH is still at 100%.
The humidity at the point where %RH = 100 and Tdb = 10 o
C is 0.008 kg H2O/kg dry air
This means that the amount of water that is condensed is 0.016 – 0.008 = 0.008 kg H2O/kg dry air,
which is accounted for of the original amount of water in the air
Adiabatic Process
An adiabatic process is the process in which Q = 0, meaning that it is a constant enthalpy (H) process
(i.e. Hin = Hout or hin =hout or = )
Since, on the psychrometric chart, the wet-bulb temperature (or Twb) line is the same line as the
enthalpy line, the path for adiabatic process is, therefore, along the constant-Twb line

Example 7.5. Air with the dry and wet-bulb temperatures of 40 and 30 o
C, respectively is fed into an
adiabatic humidifier, in which water is sprayed onto the incoming air.
If the outlet temperature of the air from the humidifier is 32 o
C, determine the amount of water
added into the air
Flow chart
Data of the inlet AIR stream:
• Tdb = 40 o
C
• Twb = 30 o
C
From the psychrometric chart and the given data above (i.e. Tdb and Twb), we obtain the following
values:
• humidity = 0.023 kg/kg dry air
• %RH ~49
Since the humidifier is working adiabatically, the path of this process is along the constant-Twb line
where Twb = 30 o
C.
Data of the outlet AIR stream
• Tdb = 32 o
C
• Twb = 30 o
C (constant-twb process)
From the intersection between the Twb = 30 o
C and Tdb = 32 o
C, we obtain
• humidity = 0.0265 kg/kg dry air
• %RH ~87%
Thus, the amount of water added into the air is 0.0265 – 0.023 = 0.0035 kg/kg dry air

Example 7.6. In order to remove moisture (water) from wet air (called “dehumidification process”),
air with Tdb and Twb of 40 o
C and 30 o
C, respectively, is passed through an adiabatic humidifier (the
same humidifier used in the previous Example)
However, in this process, cool water, in stead of water at a room temperature, is sprayed onto the
wet air.
The dry-bulb temperature (tdb) of the air leaving the humidifier is 25 o
C.
Surprisingly, it is found that the humidity (or moisture content) of the air leaving the humidifier is
lower than that entering the humidifier.
In other words, air is losing some amount of water while passing through the humidifier, despite the
fact that the air is poured with water!
Prove that this process can actually happen.
The condition of the air entering the humidifier is the same as that in the previous Example.
Thus, the properties of the inlet air stream are as follows;
• humidity = 0.023 kg H2O /kg dry air
• %RH ~49
Additionally, this is the same humidifier as in the previous Example, which is working “adiabatically”.
Thus, the path of the process is on the constant-Twb line where Twb = 30 o
C.
Draw the line on the Twb = 30 o
C line to the left of the chart until it reaches the point where %RH= 100
At this point, the corresponding dry-bulb temperature (tdb) is read as 30 o
C (note that when tdb = twb,
%RH = 100)
In order to reach the outlet temperature of 25 o
C, we must follow the path along the %RH = 100 line
from T = 30 o
C to T = 25 o
C
At the point where %RH = 100 and Tdb = 25 o
C, the humidity can be read as 0.020 kg/kg dry air, and
this is the humidity of the air leaving the humidifier.
This means that water in the amount of 0.023 – 0.020 = 0.003 kg/kg dry air is condensed from the
wet air.
Hence, this proves that it is possible to have such a process that when wet air is passed through a
humidifier, the humidity of air may be decreased, instead of increased, despite the fact that air is
poured with water.
Do you know WHY??
To know WHY this process can take place, try answering the following questions:
1) In this Example, what is the dewpoint temperature of the air entering the humidifier?

2) If the temperature of the air leaving the humidifier is higher than the dewpoint temperature
of the inlet air (obtained from 1), would the humidity of the outlet air be lower or higher
than that of the inlet air?
3) If the temperature of the air leaving the humidifier is lower than the dewpoint temperature
of the inlet air (obtained from 1), would the humidity of the outlet air be lower or higher
than that of the inlet air?
4) What should the dry-bulb temperature (Tdb) of the outlet air be, in order that the humidity of
air leaving the humidifier equals that entering the humidifier?
Ternary Systems
When a system consists of three components (or species), mole (or mass) fraction of each
component (or species) can be presented as a triangular-phase (or three-phase) diagram, which can
be in theform of
• equilateral triangle, or
• right triangle
The following Example is to illustrate how to express the composition of the ternary system.
Example 7.7. A liquid mixture of A, B, and C has the mass fraction of A and B (at equilibrium) at
25 o
C as follows:
• = 0.40
• = 0.30
Locate the exact point of this mixture in a triangular-phase diagram (Point A)
Since we have known that, for the ternary system (3-component system)
must be equal to 1, we can obtain the mass fraction of C as follows
The point A or the composition of the 3-component mixture can be located on the 3-phase diagram
as follows

More Examples
1. yA = 0.4, yC = 0.6 Point P
2. yA = 0.2, yB = 0.5 Point Q
3. yB = 0.25, yC = 0.35 Point R
Trying doing the following problems yourself on the given 3-phase diagram
4) yB = 0.1, yC = 0.9
5) yA = 0.1, yB = 0.2
6) yA = 0.2, yC = 0.35

Liquid-Liquid Extraction
Consider the following solution, where A is a solvent and B is a solute
(can liquid (B) be a solute?)
We need to remove liquid B from liquid A (or we want to separate B from A)
How can we do? or
Which method/technique can be used?
One of the widely-used methods to separate liquid mixtures is
“liquid-liquid extraction (LLE)”
The standard procedure for LLE is as follows:
1) Liquid C is added to the mixture of a solution A+B

2) B (and some small amounts of A) is transferred (from the A+B mixture) into C layer (assuming
that C is lighter than A)
3) At equilibrium, essentially, there are 2 phases of liquids:
Phase I: The mixture of A+B (with some small amounts of C), in which A is the main component.
Phase II: The mixture of C+B (with some small amounts of A), in which C is the main component)
• Phase I (or the A+B (+ small amount of C) mixture) is called a Raffinate Phase
• Phase II (or the C+B (+ small amount of A) mixture) is called an Extract Phase (or a Solvent
Phase)
At equilibrium of a given ternary (threephase) mixture, there are three points on the three-phase
diagram that represent the mixture:
• The composition of the mixture in the raffinate phase
• The composition of the mixture in the extract phase
• The overall composition of the mixture
When we vary the initial concentration of B (in A), then mix C with the A+B mixture, and let the
mixture reaches its equilibrium, as shown below:

The “Solubility Curve” on the three-phase diagram for this system can be established as illustrated
below:
From the three-phase diagram on the previous Page (consider a solid straight line):
• The composition of the raffinate phase is xA = 0.80; xB = 0.10; xC = 0.10
• The composition of the extract phase is xA = 0.15; xB = 0.20; xC = 0.65
• The overall composition of the mixture (Point M) is xA = 0.42; xB = 0.18; xC = 0.40

Note that Point P on the solubility curve is a “Plait Point”
It is the point where xB in the raffinate phase is equal to that of the extract phase (note also that this
point is NOT necessary to be the highest point of the solubility curve)
Tie line is the connecting line between the composition in the raffinate phase and that in the extract
phase
At any single point within the solubility curve, liquids are divided into 2 phases, i.e. raffinate &
extract phases
Any other points beyond the curve (i.e.outside the proximity of the curve), all three components
are homogeneously mixed into a single phase.
Examples of the L-L Extraction Equipment
• Mixer & Settler (Decanter)
• L-L Extractor

Material Balance Problems Related to Ternary Systems
Example 7.8. Referring to the three-phase diagram in Page 17, if (30,000) kg/h of ternary mixture of
40 wt% of A, 20% of B, and 40% of C was fed into a decanter operating at 25 o
C, what would the flow
rates and compositions of two liquid streams leaving the decanter be?
Flow chart
Three-phase diagram
This tie line is used for Example 7.9

From the above three-phase diagram, the given overall composition of the mixture falls within the
solubility curve; thus, at equilibrium, the mixture is separated into 2 phases:
• the raffinate phase
• the extract phase
and the composition of each phase can be read from the diagram, as follows:
Raffinate Phase Extract Phase
A = 78% A = 17%
B = 13% B = 24%
C = 9% C = 59%
Overall Balance:
Input = Output
30,000 kg = R + E (7.10)
Component Balance (“B” Balance):
(30,000)(0.20) = (R)(0.13) + (E)(0.24)
6,000 = 0.13R + 0.24E (7.11)
Solving Eqs. 7.10 & 7.11 simultaneously gives
E = 19,091 kg
R = 10,909 kg
We can verify the above answer by performing the “C” Balance, as follows:
C input = (30,000 kg)(0.4)
= 12,000 kg
C output = (19,091 kg)(0.59) + (10,909 kg)(0.09)
= 12,246 kg
O.K.! Close Enough

Example 7.9 One thousand (1,000) kilograms of 30% of B in A and a stream of pure C are fed into a
decanter, which operates at 25 o
C.
How much (pure) C must be fed into the decanter to reduce B in A from 30% to 5%.
Assume that the settling time is long enough until “equilibrium” is reached
Since it is assumed, in this Example, that the system reaches the equilibrium, the compositions of the
raffinate and extract phases can thus be read from the three-phase diagram.
In this Example, it is required that the mass fraction of B in the raffinate phase must be reduced to
5% or 0.05 in mass fraction.
We then need to locate the point on the solubility curve at which the mass fraction of B in the
raffinate phase is 0.05 (5%), and the composition of that point is as follows: (see the tie line in the
three-phase diagram on Page 19)
A = 85% = 0.85
B = 5% = 0.05
C = 10% = 0.10
We have known also that, at equilibrium, the compositions of the raffinate and extract phases are
connected together by a tie line. Hence, from the composition of the raffinate phase, we can use the
tie line to locate the composition of the extract phase at the other side of the diagram as follows:
A = 10% = 0.10
B = 10% = 0.10
C = 80% = 0.80

Basis: 1,000 kg of feed (A+B)
Overall Balance:
Input = Output
1,000 kg + S = R + E (7.12)
Component Balances
“B” Balance
(1,000 kg)(0.3) + S(0) = (R)(0.05) + (E)(0.10) (7.13)
“A” Balance
(1,000 kg)(0.7) + S(0) = (R)(0.85) + (E)(0.10) (7.14)
Solving Eqs. 7.12-7.14 simultaneously results in
R = 500 kg
E = 2,750 kg
S = 2,250 kg
Thus, we need the solvent C in the amount of 2,250 kg to reduce the percentage of B in A from 30%
(0.30 in mass fraction) to 5% (0.05 in mass fractions).

Problem Set 6
Combined Material & Energy Balances
PROB 1. Moist air with the temperature and total pressure of 38 o
C and 101.3 kPa, respectively, has
the partial pressure of water vapour of 3.59 kPa. Determine
a. moisture content (in the unit of kg water vapour/kg dry air) of this air
b. % absolute humidity (%AH)
c. % relative humidity (%RH)
(Given the vapour pressure of water vapour at 38 o
C as 6.67 kPa)
PROB 2. Reduce the temperature of 100 g of wet air with the relative humidity percentage (%RH) of
30% at 1 atm from 50 o
C to 10 o
C
a. Determine the dew-point temperature of this air
b. Calculate the amount water vapour that is condensed (in g) due to a decrease in air
temperature
PROB 3. Air has a dry bulb temperature and a dew-point temperature of 55 and 19 o
C, respectively.
Determine
a. moisture content (or humidity) of this air
b. % relative humidity
c. humid volume (in ft3/lbm) of dry air
PROB 4. Air with the temperature of 35 o
C and the moisture content of 0.014 kg H2O/kg dry air is fed
into an adiabatic humidifier
a. Initially, what is the %RH of this air?
b. What is the dew-point temperature of this air?
c. If the air leaving the humidifier has %RH of 80%, calculate the amount of water added into
the air (in the unit of kg/kg dry air)

APPENDIX 1
SYMBOLS, UNITS AND DIMENSIONS
a acceleration m s
-2
; [L] [t]
-2
thickness m; [L]
aw water activity p/ps or Y/Ys ; dimensionless
A area m
2
; [L]
2
b height of liquid in a centrifuge m; [L]
(Bi) Biot number hsL/k;hsD/k;hsr/k;hsa/k ; dimensionless
c specific heat kJ kg
-1
°C
-1
; [F] [L] [M]
-1
[T]
-1
,
cp specific heat at constant pressure, cs humid heat
C heat conductance J m
-2
s
-1
°C
-1
; [F] [L]
-1
[t]
-1
[T]
-1
coefficients - discharge, drag, geometric; constant; dimensionless
COP coefficient of performance in refrigeration
d diameter m; [L]
D diameter m; [L]
diffusivity m
2
s
-1
; [L]
2
[t]
-1
sieve aperture m ; [L]
e small temperature difference °C; [T]
;
E energy J; [F] [L]
Ec mechanical pump energy, Ef friction energy, Eh heat energy, Ei Bond's work index in
grinding (energy to reduce unit mass from infinitely large particle size to 100µm), Ek kinetic
energy, Ep potential energy, Er pressure energy
f friction factor; dimensionless
ratio of actual drying rate to maximum drying rate, dimensionless
fc crushing strength of material kg m
-1
s
-2
; [M] [L]
-1
[t]
-2
F force N, kg m s
-2
; [F], [M] [L] [t]
-2
Fc centrifugal force, Fd drag force, Fe external force, Ff friction force, Fg gravitational force; Fs
accelerating force in sedimentation,
Fl mass ratio of liquid to solid in thickener feed; dimensionless
time to sterilize at 121°C min; [t]
(Fo) Fourier number (kt/cρL
2
); dimensionless
(Fr) Froude number (DN
2
/g); dimensionless
F(D) Cumulative particle size distribution, F'(D) particle size distribution; dimensionless
g acceleration due to gravity m s
-2
; [L] [t]
-2
G mass rate of flow kg m
-2
s
-1
; [M] [L]
-2
[t]
-1
(Gr) Grashof number (D
3
ρ2
βg∆t/µ2
); dimensionless
h heat transfer coefficient J m
-2
s
-1
°C
-1
; [F] [L]
-1
[t]
-1
[T]
-1
hc convection, hh condensing vapours on horizontal surfaces, hr radiation, hs surface, hv
condensing vapours on vertical surface
H enthalpy, kJ kg
-1
; [F] [L] [M]
-1
, Hs, enthalpy saturated vapour, Ha, Hb, Hc,enthalpy in
refrigeration system
Henry's Law constant, atm mole fraction
-1,
kPa mole fraction
-1
; [F] [L]
-2
k Constant
constant of proportionality
friction loss factor; dimensionless

thermal conductivity J m
-1
s
-1
°C
-1
; [F] [L]
-1
[t]
-1
[T]
-1
k'g mass-transfer coefficient
kg gas mass-transfer coefficient, k'g mass-transfer coefficient based on humidity difference,
kl liquid mass transfer coefficient (units and dimensions from context)
K constant, K
'
, K
''
, etc.
K' mass-transfer coefficient through membrane, kg m
-2
h
-1
; [M] [L]
-2
[t]
-1
; for ultrafiltration m s
-1
,
for reverse osmosis kg m-2 h-1 kPa
-1
KK Kick's constant m
3
kg
-1
; [L]
3
[M]
-1
KR Rittinger's constant m
4
kg
-1
; [L]
4
[M]
-1
Ks rate constant for crystal surface reactions m s
-1
; [L] [t]
-1
Kd mass transfer coefficient to the interface, m s
-1
; [L] [t]
-1
Kg overall gas mass transfer coefficient
Kl overall liquid mass transfer coefficient
L flow rate of heavy phase kg h
-1
; [M] [t]
-1
half thickness of slab for Fourier and Biot numbers m; [L]
length m; [L]
ratio of liquid to solid in thickener underflow;
Lc thickness of filter cake, equivalent thickness of filter cloth and precoat m; [L]
(Le) Lewis number (hc/k'gcp) or (hc/kgcs) dimensionless
m mass kg; [M]
number, general
(M) mixing index, dimensionless
M molecular weight; dimensionless
molal concentration (kg) moles m
-3
; [M] [L]
-3
n number, general
N number of particles in sample;
rotational frequency, revolutions/minute or s ; [t]
-1
(Nu) Nusselt number (hcD/k); dimensionless
p partial pressure Pa; [F] [L]
-2
pa partial pressure of vapour in air, ps saturation partial pressure
factor in mixing and in grinding, dimensionless; factor in particle geometry in grinding,
fractional content in mixing; dimensionless
P constant in freezing formula; dimensionless;
power N m s
-l, J s-1
; [F] [L] [t]
-1
pressure Pa; [F] [L]
-2
Ps pressure on surface Pa; [F] [L]
-2
(Po) Power number (P/D
5
N
3
ρ); dimensionless
(Pr) Prandtl number (cpµ/k); dimensionless

q heat flow rate J s
-1
; [F] [L] [t]
-1
fluid flow rate m
3
s
-1
; [L]
3
[t]
-1
reduction ratio factor in particle geometry in grinding and mixing; dimensionless
Q quantity of heat J; [F] [L]
r radius m; [L]
rn neutral radius in centrifuge
specific resistance of filter cake kg m
-1
; r' specific resistance of filter cake under 1 atm
pressure [M] [L]
-1
R constant in freezing formulae; dimensionless
resistance to flow through filter; dimensionless
R Universal gas constant 8.314 kJ mole
-1
K
-1
; m
3
kPa mole
-1
K
-1
, [L]
2
[t]
-2
[T]
-1
; 0.08206 m
3
atm
mole
-1
K
-1
(Re)
RH
Reynolds number (Dvρ/µ) and (D
2
Nρ/µ); dimensionless
relative humidity p/p , % ; dimensionless
s compressibility of filter cake; dimensionless
distance m ; [L]
standard deviation of sample compositions from the mean in mixing; dimensionless
so , sr initial and random values of standard deviation in mixing; dimensionless
(Sc) Schmidt number (µ/ρD); dimensionless
(Sh) Sherwood number (K'd/D); dimensionless
SG specific gravity; dimensionless
t time s, h, min (from context) ; [t]
tf , freezing time h
T temperature °C or T K; [T]
Tav mean temperature, Ta air, Ts surface, Tc centre
Tm mean temperature in radiation
U overall heat-transfer coefficient J m
-2
s
-1
°C
-1
; [F] [L]
-1
[t]
-1
[T]
-1
v velocity m s
-1
; [L] [t]
-1
V flow rate of light phase kg h
-1
; [M] [t]
-1
volume m
3
; [L]
3
volumetric flow rate m
3
s
-1
; [L]
3
[t]
-1
w solid content per unit volume kg m
-3
; [M] [L]
-3
mass of dry material kg [M]
weight kg; [F]
W work N m ; [F] [L]
mass of material dried kg; [M]
x concentration in heavy phase kg m
-3
; [M] [L]
-3
distance, thickness m; [L]

fraction, mole or weight, dimensionless
Mean
X moisture content on dry basis ; dimensionless
Xc critical moisture content, Xf final moisture content, Xo initial moisture content;
thickness of slab m ; [L]
y concentration in light phase kg m
-3
; [M] [L]
-3
fraction, mole or weight, dimensionless
Y
Ys, Ya
humidity, absolute kg kg
-1
; humidity difference; dimensionless
humidity of saturated air, humidity of air
z height m; [L]
temperature difference for 10-fold change in thermal death time °C, [T]
Z depth, height of fluid m; [L]
α absorbtivity; dimensionless
β coefficient of thermal expansion m m
-1
°C
-1
; [T]
-1
β1, β2 length ratios in freezing formula; dimensionless
δ thickness of layer for diffusion m; [L]
∆ Difference
∆tm logarithmic mean temperature difference °C; [T]
ε emissivity; dimensionless
roughness factor; dimensionless
η efficiency of coupling of freezing medium to frozen foodstuff
air-drying efficiency, % , dimensionless
λ latent heat kJ kg
-1
; [F] [L] [M]
-1
shape factor for particles, dimensionless
µ viscosity kg s
-1
m
-1
; Pa s, N s m
-2
; [M] [t]
-1
[L]
-1
; [F] [t] [L]
-2
π ratio of circumference to diameter of circle , 3.1416
Π total pressure Pa; [M] [L]
-1
[t]
-2
, [F] [L]
-2
osmotic pressure kPa; [F] [L]
-2
ρ density kg m
-3
; [M] [L]
-3
σ Stefan-Boltzman constant, 5,73 x10
-8
kg m
-2
s
-3
°C
-4 , J m-2 s-1 K-4
; [M] [t]
-3
[T]
-4
or [F] [L]
-1
[t]
-1
[T]
-4
τ shear stress in a fluid Pa; [F] [L]
-2
φ fin efficiency; dimensionless
ω angular velocity radians s
-1
, [t]
-1

APPENDIX 2
UNITS AND CONVERSION FACTORS
Length 1 inch = 0.0254 m
1 ft = 0.3048 m
Area 1 ft
2
= 0.0929m
2
Volume 1 ft
3
= 0.0283 m
3
1 gal Imp = 0.004546 m
3
1 gal US = 0.003785 m
3
= 3.785 litres
1 litre = 0.001 m
3
Mass 1 lb = 0.4536 kg
1 mole molecular weight in kg
Density 1 lb/ft
3
= 16.03 kg m
-3
Velocity 1 ft/sec = 0.3048 m s
-1
Pressure 1 lb/m
2
= 6894 Pa
1 torr = 133.3 Pa
1 atm
= 1.013 x 10
5
Pa
= 760 mm Hg
1 Pa
= 1 N m
-2
= 1 kg m
-1
s
-2
Force
1 Newton
1 lb ft s
-2
= 1 kg m s
-2
= 1.49 kg m s-2
Viscosity 1 cP = 0.001 N s m
-2
= 0.001 Pa s
1 lb/ft sec
= 1.49 N s m
-2
= 1.49 kg m
-1
s
-2
Energy 1 Btu = 1055 J
1 cal = 4.186 J
Power
1 kW
1 W
= 1 kJ s-1
= 1 J s-1
1 horsepower
= 745.7 W = 745.7 J s
-1
= 0.746 kW
1 ton refrigeration = 3.519 kW
Temperature units (°F) = 5/9 (°C) = 5/9 (K)
Heat-transfer coefficient 1 Btu ft
-2
h
-1
°F
-1
= 5.678 J m
-2
s
-1
°C
Thermal conductivity 1 Btu ft
-1
h
-1
°F
-1
= 1.731 J m
-1
s
-1
°C
-1
Constants π 3.1416
σ 5.73 x 10
-8
J m
-2
s
-1
K
-4
e (base of natural
logs)
2.7183
R 8.314 kJ mole
-1
K
-1
or 0.08206 m
3
atm mole
-1
K
-1
(M) Mega = 10
6
,
(k) kilo = 10
3
,
(H) Hecto = 10
2
(m) milli = 10
-3
(µ) micro = 10
-6

Fde 201 lecture notes

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Fde 201 lecture notes