1304 211 Chemical Engineering Principles
and Calculations
1304 211 Chemical
Engineering Principles
and Calculations
By
Assist. Prof. Dr. Wipada Sanongraj
Chapter I : Introduction to
Chemical Engineering
Calculations
By Assist. Prof. Dr. Wipada Sanongraj
1
1304 211 Chemical Engineering Principles
and Calculations
What do chemical engineers
do?
Petroleum
Biotechnology
Consulting
Drugs and
pharmaceuticals
Fats and oils
Fertilizer and agricultural
chemical
Foods and beverages
Government
Lime and cement
Man-made fibers
Metallurgical and metal
products
Paints, varnishes and
pigments
Pesticides and herbicides
Plastic materials and
synthetic resins
1.1 Units and dimensions
Objectives of this section
Add, subtract, multiply, and divide units
associated with numbers
Specify the basic and derived units in the SI and
American Engineering system
Convert one set of units in equation into another
equivalent set for mass, length, area, etc..
Define and know how to use the gravitational
conversion factor, gc
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
1.1 Units and dimensions
Dimensions are basic concepts of
measurement
Such as length, time, mass, temperature
Units are the means of expressing the
dimensions
Such as cm, hour, kg
Rules of operation
Only same units can be added or subtracted
For example, 10 lbs + 5 grams
5 kgs + 3 Joules
1 ft + 3 sec
1 horsepower + 30 watts
Different units can be multiplied or divided
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Example 1.1: Dimensions and
Units
Add the following
A) 1 ft + 3 s
B) 1 horsepower + 300 watts
A) has no meaning since the dimensions
of the two terms are not the same
B) 1 hp = 746 watts,
746 watts + 300 watts = 1046 watts
Example of SI units
Physical Quantity
Name of Unit
Symbol
Definition of Unit
Basic SI Units
Length
meter
m
Mass
kilogram
kg
Temperature
kelvin
K
Time
second
s
Amount of substance
mole
mol
Derived SI Units
Energy
Joule
J
Kg.m2.s-2
Force
Newton
N
Kg.m.s-2
J.m-1
Power
watt
W
Kg.m2.s-3
J.s-1
Density
Kilogram per cubic meter
Kg.m-3
Acceleration
Meter per second squared
m.s-2
Velocity
Meter per second
m.s-1
Pressure
Newton per square
meter, pascal
N.m-2, Pa
Heat Capacity
Joule per
(kilogram.kelvin)
J.kg-1.K-1
Time
minute, hour, day, year
min, h, d, y
Temperature
Degree Celsius
o
Volume
liter
L
Mass
ton, gram
t, g
Alternative Units
By Assist. Prof. Dr. Wipada Sanongraj
C
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1304 211 Chemical Engineering Principles
and Calculations
Example of American
Engineering System Units
Physical Quantity
Name of Unit
Symbol
Basic Units
Length
feet
ft
Mass
pound (mass)
lbm
Force
pound (force)
lbf
Time
second, hour
s, hr
Temperature
degree Rankine
o
R
Derived Units
Energy
British thermal unit, foot pound
(force)
Btu, ft.lbf
Power
horsepower
hp
Density
pound(mass) per cubic foot
lbm/ft3
Velocity
feet per second
ft/s
Acceleration
feet per second squared
ft/s2
Pressure
Pound(force) per square inch
lbf/in2
SI Prefixes
Factor Prefix
109
giga
Symbo Factor Prefix
l
G
10-1
deci
106
mega
M
10-2
centi
c
103
kilo
k
10-3
milli
m
102
hecto
h
10-6
micro
m
101
deka
da
10-9
nano
n
By Assist. Prof. Dr. Wipada Sanongraj
Symbo
l
d
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1304 211 Chemical Engineering Principles
and Calculations
Example 1.2: Conversion of
Units
If a plane travels at twice the speed of
sound (assume that the speed of sound
is 1100 ft/s), how fast is it going in
miles per hour
Solution
2 1100 ft 1 mi 60 s 60 min
s
5280 ft 1 min 1 hr
= 1500 mi/hr (mph)
Example 1.3: Use of Units
Change 400 in3/day to cm3/min
Solution
400 in.3
day
2.543 cm3 1 day 1 hr
1 in3.
24 hr 60 min
= 4.56 cm3/min
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1304 211 Chemical Engineering Principles
and Calculations
At the sea level at 45 °latitude, the numerical value
of the average acceleration of gravity (g) is 32.174
ft/s2.
The acceleration of gravity varies by a few tenths
of 1% from place to place on the surface of the
earth
The inverse of the conversion factor with the
numerical value 32.174 is given the special symbol,
gc
gc = 32.174 (ft)(lbm)/(s2)(lbf)
From gc/g = 1, a one pound mass is to weight one
pound
Weight can be defined as the opposite of the force
required to support a mass
The pound mass and pound force are not the same
units in American Engineering system.
Example 1.4: Use of gc
One hundred pounds of water is flowing through a
pipe at the rate of 10.0 ft/s. What is kinetic energy
of this water in (ft)(lbf)?
Solution
K = 1 100 lbm
2
(10 ft)2
s2
1
32.174 (ft)(lbm)/(s2)(lbf)
= 155 (ft)(lbf)
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1304 211 Chemical Engineering Principles
and Calculations
Example 1.5: Use of gc
What is the potential energy in (ft)(lbf) of a 100 lb drum
hanging 10 ft above the surface of the earth with
reference to the surface of the earth?
Solution
Assume that the 100 lb means 100 lb mass
g = acceleration of gravity = 32.2 ft/s2
P = 100 lbm 32.2 ft
s2
1
10 ft
32.174(ft)(lbm)/(s2)(lbf)
= 1001 (ft)(lbf)
Dimensional Consistency
The equations must be dimensionally consistent in
order to be able to operate.
Each term must have the same net dimensions and
units in the equation.
For example, van der Waals equation
atm.cm6
cm3
(P +
atm
By Assist. Prof. Dr. Wipada Sanongraj
a
)(V - b ) = RT
V2
cm3
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1304 211 Chemical Engineering Principles
and Calculations
1.2 The mole unit
In the SI a mole is composed of 6.02×1023
molecules
In the American engineering system, a
pound mole is composed of
6.02×1023×453.6 molecules
To convert the number of moles to mass,
we use molecular weight
What is the unit of MW?
1.2 The mole unit
(continue)
the g mol = (mass in g)/(molecular weight)
the lb mol = (mass in lb)/(molecular weight)
Or
mass in g = (mol. Wt)(g mol)
mass in lb = (mol. Wt)(lb mol)
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1304 211 Chemical Engineering Principles
and Calculations
Example 1.6: Use of MW
If a bucket holds 2.00 lb of NaOH (mol. Wt. = 40),
how many
a) Pound moles of NaOH does it contain?
b) Gram moles of NaOH does it contain?
Solution
(a) 2.00 lb NaOH
1 lb mole NaOH
40.0 lb NaOH
= 0.05 lb mol
(b) 2.00 lb NaOH 1 lb mol NaOH 454 g mol
40 lb NaOH
1 lb mol
= 22.7 g mol
Example 1.7: Use of MW
How many pounds of NaOH are in 7.5 g mol
of NaOH?
Solution
Basis: 7.50 g mol of NaOH
7.50 g mol NaOH 1 lb mol
40 lb NaOH
454 g mol 1 lb mol NaOH
= 0.661 lb NaOH
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1304 211 Chemical Engineering Principles
and Calculations
1.3 Conventions in Methods of
Analysis and Measurement
1.3.1 Density
Density is the ratio of mass per unit volume
Unit is kg/m3 or lb/ft3
Density of water is 1 g/cm3 or 62.4 lb/ft3
Density of liquid and solid do not change
significantly with changing of pressure but
change with temperature
Density
Liquid density
By Assist. Prof. Dr. Wipada Sanongraj
Density of a mixture of ethyl alcohol
And water as a function of composition
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1304 211 Chemical Engineering Principles
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1.3.2 Specific Gravity
It is the ratio of two densities, density of the
interested substance and density of a
reference substance
Thought of as dimensionless unit
For liquids and solids, a reference substance
is usually water
For gas, a reference substance is air
To be precise, state the temperature at
which each density is chosen
Example 1.8: Density and
SPGR
If dibromopentane (DBP) has a specific
gravity of 1.57, what is the density in (a)
g/cm3 (b) lbm/ft3 and (c) kg/m3
Solution
(a) 1.57 g DBP
1.0 g H2O
cm3
= 1.57 g DBP
cm3
cm3
1.0 g H2O
cm3
(b) 1.57 g DBP 106 cm3
cm3
By Assist. Prof. Dr. Wipada Sanongraj
1
m3
1 m3
35.31
2.20*10-3 lb
ft3
1g
= 97.97 lbm DBP
ft3
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1304 211 Chemical Engineering Principles
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Or 1.57 g/cm3 DBP 62.4 lb/ft3 62.4 lb/ft3
1 g/cm3
1 g/cm3 H2O
62.4 lb/ft3
1 g/cm3
(c) 1.57 g DBP
cm3
(100 cm)3
1 m3
1 kg
1000 g
= 97.97 lbm DBP
ft3
= 1.57 × 103 kg DBP
m3
Or
1.57*103 kg DBP
m3
3
1.0*10 kg H2O
m3
1.0×103 kg H2O
m3
Be Cautious!!!!!
For the mixture system, an average specific
gravity can not be determined by
multiplying the individual component
specific gravities or densities by respective
mass fractions of the components and
summing the products.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Example 1.9: Application of
SPGR
In the production of a drug having a
molecular weight of 192, the exit stream
from the reactor flows at the rate of 10.3
L/min. The drug concentration is 41.2% (in
water), and the specific gravity of the
solution is 1.025. Calculate the
concentration of the drug (in kg/L) in the
exit stream, and the flow rate of the drug in
kg mol/min.
Solution
First, we need to transform the mass
fraction of 0.412 into mass per liter of the
drug. Take 1.00 kg of the exit solution as a
basis
Basis: 1.00 kg solution
Reactor
0.412 kg Drug
0.588 kg Water
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1304 211 Chemical Engineering Principles
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Density of solution = 1.025 g soln
cm3
1.0 g H2O
cm3
1.0 g H2O
cm3
= 1.025 g soln/cm3
Next
0.412 kg drug 1.025 g soln 1 kg soln
1.00 kg soln
1 cm3
103 g soln
103 cm3
1L
= 0.422 kg drug/L soln
To get the flow rate, we take a different basis, 1
minute
Basis: 1 min = 10.3 L solution
10.3 L soln
1 min
0.422 kg drug
1 L soln
1 kg mol drug
192 kg drug
= 0.0226 kg mol/min
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1304 211 Chemical Engineering Principles
and Calculations
1.3.3 Mole Fraction and Mass (Weight) Fraction
Mole fraction is the ratio of moles of particular
substance and the total moles
For example, if the mixture has 2 components, A and B.
The composition of A is 2 moles and B is 4 moles.
xA = moles of A
=
2
= 0.334
total moles
2+4
xB = moles of B
=
4
= 0.667
total moles
2+ 4
XA + XB = 0.334 + 0.667 = 1.00
Mole fraction of A = moles of A
total moles
Mass fraction of A = mass of A
total mass
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1304 211 Chemical Engineering Principles
and Calculations
Example 1.10: Mole fraction
and mass fraction
An industrial-strength drain cleaner contains
5.0 kg of water and 5 kg of NaOH. What are
the mass fraction and mole fraction of each
component in the drain cleaner container?
Solution
Basis: 10 kg of total solution
Component kg
Weight
fraction
Mol. Wt.
kg mol
Mole
fraction
H2O
5.0/10
18.0
0.278
0.278/0.
403
5.0
=0.69
NaOH
5.0
5.0/10
40.0
0.125
0.125/0.
403
=0.31
Total
10.0
By Assist. Prof. Dr. Wipada Sanongraj
1.0
0.403
1.0
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1304 211 Chemical Engineering Principles
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1.3.4 Concentrations
Concentration is the quantity of some solute
per specified amount of solvent, or solution,
in a mixture of two or more components
(1) mass per unit volume (lbm of solute/ft3, g of
solute/L, lbm of solute/bbl, kg of solute/m3)
(2) mole per unit volume (lb mol of solute/ft3, g
mol of solute/L, g mol of solute/cm3)
(3) Parts per million (ppm), parts per billion
(ppb)
(4) Other methods of expressing concentration
e.g., molarity (g mol/L) and normality
(equivalents/L)
Example 1.11: Use of ppm
The current OSHA 8 hour limit for HCN in air is 10.0
ppm. A lethal dose of HCN in air (from the Merck
index) is 300 mg/kg of air at room temperature.
How many mg HCN/kg air is the 10.0 ppm? What
fraction of the lethal dose is 10.0 ppm?
Solution
Basis: 1 kg mol of the air/HCN mixture
(a)
10 ppm = 10 g mol HCN
= 10 g mol HCN
6
10 air + HCN gmol 106 g mol air
10 g mol HCN 27.03 g HCN 1 g mol air 103 mg HCN 103 g air
106 g mol air 1 g mol HCN 29 g air
1 g HCN
1 kg air
= 9.32 mg HCN/ kg air
(b) 9.32/300 = 0.031
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1304 211 Chemical Engineering Principles
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1.4 Basis
To chose the basis, ask yourself these
questions
What do I have to start with?
What answer is called for?
What is the most convenient basis to use?
Example 1.12: Choosing
Basis
The dehydrogenation of the lower alkanes has
been carried out using a ceric oxide catalyst.
What is the mass fraction and mole fraction of
Ce and O in the catalyst?
Solution
No answers for the 1st and 2nd questions, so a
convenient basis would be to take 1 kg mol
because we know the mole ratio of Ce to O in
the compound.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Basis: 1 kg mol CeO
Component
kg mol Mole
fraction
Mol. kg
Wt
Mass
Fraction
Ce
1
0.5
140
140
0.9
O
1
0.5
16
16
0.1
Total
2
1.0
156
1.0
Example 1.13: Changing
Basis
A medium-grade bituminous coal analyzes as follows
Component
%
S
2
N
1
O
6
Ash
11
Water
3
The residuum is C and H in the mole ratio H/C = 9.
Calculate the weight fraction of the coal with the ash
and moisture omitted.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Solution
Basis: 100 kg Coal
The sum of S+N+O+ash+water is 2+1+6+11+3=23
kg
Hence the C and H must be 100-23=77 kg
To determine the kilograms of C and H, we have to
select a new basis. Because C/H is molar ratio
Basis: 100 kg mol of C and H
Component
Mole
Fraction
H
9/10
C
1/10
Total
kg mol
Mol. Wt
kg
0.9
90
1.008
90.7
0.1
10
12
120
1.0
100
210.7
Finally, return to the original basis, we have
H: 77 kg 90.7 kg H
210.7 kg total
120 kg C
210.7 kg total
Summarized table
= 33.15 kg H
C: 77 kg
= 43.85 kg C
Component
kg
Wt. Fraction
C
43.85
0.51
H
33.15
0.39
S
2
0.02
N
1
0.01
O
6
0.07
Total
86.0
1.00
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1304 211 Chemical Engineering Principles
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1.5 Temperature
The temperature is a measure of thermal
state considered in reference surrounding
Units of temperature: Fahrenheit (F),
Celsius (C), Rankine (R), Kelvin (K)
Absolute temperature scales have their
zero point at the lowest possible
temperature that can exist.
Temperature measuring
instruments span the range
from near absolute zero to
beyond 3000 K.
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1304 211 Chemical Engineering Principles
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Temperature scales
Conversion between scales
TR = TF + 460
TK = TC + 273
DF = DR
DC = DK
(DC/DF) = 1.8
(DK/DR) = 1.8
C/5 = (F-32)/9
F-32 = C*1.8
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1304 211 Chemical Engineering Principles
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Example 1.14: Temperature
Conversion
Convert 100 °C to (a) K, (b) ° F, and (c) ° R
Solution
(a) 100 °C + 273 = 373 K
(b) 100 °C /5*9 + 32 = 212 ° F
(c) 212 ° F + 460 = 672 ° R
Or
(373 K)(1.8 oR/1 K) = 672 oR
Example 1.15: Temperature
Conversion
The thermal conductivity of aluminum at 37
°F is 117 Btu/(hr)(ft2)(°F/ft). Find the
equivalent value at 0oC in terms of
Btu/(hr)(ft2)(K/ft)
Solution
117 (Btu)(ft) 1.8 D°F 1 D°C
(hr)(ft2)(°F)
1 D°C
1 DK
= 211 Btu/(hr)(ft2)(K/ft)
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1304 211 Chemical Engineering Principles
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Conversion
The heat capacity of sulfuric acid in a handbook has
the units J/(g mol)(°C) and is given by the relation
heat capacity = 139.1 + 1.56 ×10-1T
Where T is expressed in °C. Modify the formula so
that the resulting expression has the associated units
of Btu/(lb mol)(°R) and T is in ° R
Heat capacity = {139.1+1.56×10-1[(T°R-460-32)/1.8]}*
1 J
1 Btu 454 g mol 1 °C
(g mol)(°C) 1055J
1 lb mol 1.8oR
Heat capacity = 23.06+ 2.07 ×10-2T °R
1.6 Pressure
Pressure is normal force
per unit area
By Assist. Prof. Dr. Wipada Sanongraj
Pressure at the bottom of the
static column of water is
p = F/A = rgh + p0
p = pressure at the bottom of the
column of the fluid
F = force
A = area
r = density of fluid
g = acceleration of gravity
h = height of the fluid column
p0 = pressure at the top of the
column of fluid
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1304 211 Chemical Engineering Principles
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Example 1.17: Pressure
Suppose the cylinder of fluid is a column of mercury that has an
area of 1 cm2 and is 50 cm high. Density of mercury is 13.55
g/cm3. Calculate the force exerted by the mercury alone on the
1 cm2 section of the bottom plate.
Solution
F = 13.55 g 980 cm 50 cm 1 cm2 1kg
1m
1N
3
2
cm
s
1000 g 100cm
(kg)(m)/s2
= 6.64 N
The pressure on the section of the plate covered by the mercury is
the force per area of the mercury plus the pressure of
atmosphere
P = 6.64 N (100 cm)2 (1 m2)(1 Pa) 1 kPa
1 cm2
1 m2
1N
1000 Pa
= 66.4 kPa + p0
Pressure can be expressed by either absolute or
relative scales.
An open-end manometer would measure a
relative pressure (gauge pressure).
Closing off the end of the manometer would
measure an absolute pressure.
Atmospheric pressure measured by barometer is a
barometric pressure
Gauge pressure + barometric pressure = absolute
pressure
Units of pressure: mm Hg, ft H2O, atm, bar, psi,
kgf/cm2, Pascal
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Open-end manometer
Barometer
Absolute pressure manometer
“C” Bourdon
Spiral Bourdon
Visual Bourdon gauge reads zero
pressure when open to the atmosphere.
Pressure sensing device in the Bourdon
gauge is a thin metal tube
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1304 211 Chemical Engineering Principles
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Standard atmosphere is the pressure
equivalent to 1 atm or 760 mm Hg at 0
oC
Atmospheric pressure is a variable and
is measured by a barometer.
Standard atmosphere:
760 mm Hg, 29.92 in Hg
33.91 ft H2O
1 atm
1.013 bars
14.7 psia
1.013*105 Pa or N/m2 or 101.3 kPa
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Example 1.18: Pressure
Conversion
The pressure gauge on a tank of CO2 used
to fill soda-water bottles reads 51.0 psi.
At the same time the barometer reads
28.0 in Hg. What is the absolute
pressure in the tank in psia?
Solution
The pressure gauge is reading psig
Absolute pressure = gauge pressure +
atmospheric pressure
Basis: barometric pressure = 28 in Hg
Atmospheric pressure = 28 in Hg 14.7 psia
29.92 in Hg
= 13.76 psia
Absolute pressure in the tank
51.0 + 13.76 = 64.8 psia
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1304 211 Chemical Engineering Principles
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Example 1.19: Pressure
Conversion
Air is flowing through a duct under a
draft of 4.0 cm H2O. The barometer
indicates that the atmospheric pressure
is 730 mm Hg. What is the absolute
pressure of the gas in inch Hg?
Solution
We can ignore the gas density above the
manometer fluid.
Basis: 730 mm Hg
Atmospheric pressure = 730 mm Hg 29.92 in Hg
760 mm Hg
= 28.7 in Hg
Basis: 4 cm H2O draft (under atmospheric)
4 cm H2O 1 in
1 ft 29.92 in Hg = 0.12 inHg
2.54 cm 12 in 33.91 ft H2O
The reading is 4 cm H2O draft, the absolute reading
in uniform units is
28.7-0.12 = 28.6 in Hg
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1304 211 Chemical Engineering Principles
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Example 1.20:Vacuum
Pressure
Small animals can live at reduced air
pressure down to 20 kPa. In a test a
mercury manometer attached to a tank
reads 64.5 cm Hg and the barometer
reads 100 kPa. Will the mice survive?
Solution
Basis: 64.5 cm Hg below atmospheric
the absolute pressure in the tank is
100 kPa – 64.5 cm Hg 101.3 kPa
76 cm Hg
= 100 – 86 = 14 kPa absolute
The mice will not survive
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1304 211 Chemical Engineering Principles
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Example 1.21: pressure
differences
In measuring the flow of fluids in a
pipeline, a differential manometer can
be used to determine the pressure
difference across the orifice plate. The
flow rate can be calibrated with the
observed pressure drop. Calculate the
pressure drop p1-p2 in pascal for the
manometer in Figure.
Solution
p1-p2 = (rf-r)gd
= (1.1-1)103 kg 9.807 m
m3
s2
22*10-3 m 1 Ns2 1 Pa*m2
kg*m 1 N
= 21.6 Pa
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1304 211 Chemical Engineering Principles
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1.7 The Chemical Equation
and Stoichiometry
C7H16 + 11O2
7CO2 + 8 H2O
Stoichiometric ratios (coefficients)
1 mole of heptane will react with 11 moles
of oxygen to give 7 moles of carbon
dioxide and 8 moles of water
These maybe lb mole, g mole, kg mole.
Example 1.22: Chemical
Equation
if 10 kg of C7H16 react completely with the
stoichiometric quantity of O2, how much kg
of CO2 will be found as products? On the
basis of 10 kg.
1 kg mol of C7H16 will react with O2 to form 7 kg
mol of CO2
10 kg C7H16 1 kg mol C7H16 7 kg mol CO2 44 kg CO2
100 kg C7H16
1 kg molC7H6 1 kg mol CO2
= 30.8 kg CO2
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Example1.23: Chemical
Equation
In the combustion of heptane, CO2 is
produced. Assume that you want to
produce 500 kg of dry ice per hour and that
50% of the CO2 can be converted into dry
ice, as shown in Figure. How many kg of
heptane must be burned per hour
Other products
CO2 gas 50%
CO2 solid dry ice 50%
C7H16 gas
Reactor
500 kg/hr
Solution
Basis: 500 kg of dry ice (equivalent to 1 hr)
C7H16 + 11 O2
7 CO2 + 8 H2O
500 kg dry ice 1 kg CO2
1 kg mol CO2
0.5 kg dry ice 44 kg CO2
1 kg mol C7H16 100 kg C7H16
7 kg mol CO2
1 kg mol C7H16
= 325 kg C7H16
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Example 1.24:
Stoichiometry
A limestone analysis
CaCO3
MgCO3
Insoluble
92.89%
5.41%
1.70%
(a) How many pounds of calcium oxide can
be made from 5 tons of this limestone?
(b) How many pounds of CO2 can be
recovered per pound of limestone?
(c) How many pounds of limestone are
needed to make 1 ton of lime?
Solution
Basis: 100 lb of limestone
CO2
Limestone
Heat
CaCO3
CaO + CO2
MgCO3
MgO + CO2
CaO
MgO
Insoluble
Lime
Substance
CaCO3
MgCO3
CaO
MgO
CO2
Mol. Wt.
100.1
84.32
56.08
40.32
44.0
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1304 211 Chemical Engineering Principles
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Limestone
Component
lb = %
Products
lb mol
Solid
Component
lb
CO2 (lb)
CaCO3
92.89
0.9280 CaO
52.04
40.83
MgCO3
5.41
0.0642 MgO
2.59
2.82
Insoluble
1.70
Total
100.00
92.89 lb CaCO3
5.41 lb MgCO3
1.70
Insoluble
0.9920 Total
56.33
1 lb mol CaCO3
1 lb mol CaO
56.08 lb CaO
100.1 lb CaCO3
1 lb mol CaCO3
1 lb mol CaO
1 lb mol MgCO3 1 lb mol MgO
43.65
= 52.04 lb
40.32 lb MgO
= 2.59 lb
84.32 lb MgCO3 1 lb mol MgCO3 1 lb mol MgO
(a) CaO produced = 52.04 lb CaO 2000 lb 5 ton
= 5200 lb CaO
100 lb stone 1 ton
(b) CO2 recovered = 43.65 lb CO2
= 0.437 lb
100 lb stone
(c) Limestone required = 100 lb stone 2000 lb
56.33 lb lime
By Assist. Prof. Dr. Wipada Sanongraj
= 3550 lb stone
1 ton
36
1304 211 Chemical Engineering Principles
and Calculations
Things to realize !!
In industrial reactors you will rarely find exact
stoichiometric amounts of materials used.
Limiting reactant is the reactant that is present in
the smallest stoichiometric amount.
For example
A + 3B + 2C = P
1.1 mol of A, 3.2 mol of B, and 2.4 mol of C are fed
as reactants in the reactor, we choose A as the
reference substance and calculate
Ratio in feed
Ratio in chemical equation
B/A
3.2/1.1 = 2.91
3/1 = 3
C/A 2.4/1.1 = 2.18
2/1 = 2
B is the limiting reactant relative to A,
and A is the limiting reactant relative to
C, hence B is the limiting reactant
B<A<C
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Excess reactant is a reactant present in
excess of the limiting reactant. The
percent excess of a reactant is based on
the amount of any excess reactant above
the amount required to react with the
limiting reactant
% excess = moles in excess/moles
required to react with the limiting
reactant *100
For example: C7H16 + 11O2 = 7CO2 + 8H2O
If we have 12 mol of O2 and 1 mol of C7H16
Ratio in feed
O2/C7H16 12/1 = 12
Ratio in chemical equation
11/1 = 11
% excess O2 = (12-11)/11*100 = 9.1%
In this example, C7H16 is the limiting reactant
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1304 211 Chemical Engineering Principles
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Conversion is the fraction of the feed or
some key material in the feed that is
converted into products
%conversion = 100* (moles or mass of
feed that react)/(moles or mass of feed
introduced)
For example; if there is 10 kg of C7H16 and 14.4 kg of CO2 are
formed in the reaction of C7H16,
% conversion of C7H16 = mol of C7H16 reacts/mol of C7H16 in
feed
Mol of C7H16 reacts =
14.4 kg CO2 1 kg mol CO2 1 kg mol C7H16
44 kg CO2
7 kg mol CO2 = 0.0468 kg mol
Mol of C7H16 in feed =
10 kg C7H16 1 kg mol
100 kg C7H16 = 0.1 kg mol C7H16
%conversion = 0.0468/0.1*100 = 46.8%
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Selectivity is the ratio of the moles of a
particular product produced to the moles
of another product produced
For example; methanol can be converted
into ethylene or propylene by the
reactions
2CH3OH = C2H4 + 2 H2O
3CH3OH = C3H6 + 3 H2O
Selectivity of C2H4 relative to C3H6 at 80%
conversion of CH3OH = 0.19/0.08 = 2.4
mol C2H4/molC3H6
Products from the conversion of methanol
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Yield is a weight or moles of final
product divided by the weight or moles
of initial or key reactant either fed or
consumed
Example 1.25: Incomplete
Reaction
From this reaction, suppose that 0.6 kg of
stibnite and 0.25 kg of iron are heated
together to give 0.2 kg of SB metal determine
A) the limiting reactant
B) the percentage of excess reactant
C) the degree of completion
D) the percent conversion
E) the yield
Sb2S3 + 3Fe = 2Sb + 3FeS
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Solution
Component
Sb2S3
Fe
Sb
FeS
kg
0.6
0.25
0.2
MW
339.7
55.85
121.8
87.91
g mol
1.77
4.48
1.64
1.77 g mol Sb2S3
4.48 g mol Fe
Reactor
FeS
1.64 g mol Sb
A) to find the limiting reactant, examine ratio of Sb2S3 to
Fe, 1/3 = 0.33. In the actual reaction the ratio is 1.77/4.48
= 0.4, hence Sb2S3 is the excess reactant and Fe is the
limiting reactant. Sb2S3 required to react with the Fe is
4.48/3 = 1.49 g mol
B) The percentage of excess reactant is
% excess = (1.77-1.49)/1.49*100 = 18.8 excess Sb2S3
C) Calculate how much Fe react from 1.64 Sb formed
1.64 g mol Sb 3 g mol Fe
2 g mol Sb = 2.46 g mol Fe
Degree of completion = 2.46/4.48 = 0.55
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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D) percent conversion of Sb2S3
1.64 g mol Sb 1 g mol Sb2S3
2 g mol Sb
= 0.82 g mol Sb2S3
% conversion = 0.82/1.77*100 = 46.3%
E) The yield of Sb formed per kg of Sb2S3 fed to
reaction
Yield = 0.2 kg Sb/0.6 kg Sb2S3
= 0.33 kg Sb/1kg Sb2S3
Chapter 2
Material Balance
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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The objectives in studying this chapter are
* Define a system and draw the system
boundaries for which the material balance is to
be made.
* Explain the different between an open and
closed system.
* Write the general material balance in words
including all terms. Be able to apply the balance
to simple problems.
System : any arbitrary portion or whole of a
process set out specifically for analysis.
System boundary : circumscription of the system
An open (flow) system : the system in which
material is transferred across the system
boundary.
Closed (batch) system : the system in which there
is no material transfer during the time interval of
interest.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Closed system
System boundary
Open system
Mass in
Mass out
System boundary
General Material Balance Equation
é Accumulation
ù é Input through ù é Output through ù
ê within the system ú = êsystem boundary ú - êsystem boundary ú
ë
û ë
û ë
û
éGaneration ù é Consumption ù
ú - ê within
ú
+ êê within
ú ê
ú
êë the system úû êë the system úû
Material balance can refer to a balance on a system
for the
1. Total mass
2. Total moles
Units of the above equation must be mass or moles per time
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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The accumulation term refers to a change in mass
or moles (plus or minus) within the system with
respect to time
The transfer through the system boundaries refers
to inputs to and outputs of the system.
The generation and consumption term refer to the
gain or loss of the mass or moles of the interesting
compound in the system.
Unsteady state : The values of the variables within
the system change with time.
Steady state : The values of the variables within the
system do not change with time.
Therefore, general material balance will be
simplified as:
é Accumulation
ù é Input thorugh ù éOutput through ù
ê within the systemú = ê system boundary ú - ê system boundary ú
ë
û ë
û ë
û
éGaneration ù éConsumption ù
ú - ê within
ú
+ êê within
ú ê
ú
êëthe system úû êëthe system úû
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Example 2.1 : Total Mass Balance
A thickener in a waste disposal unit of a plant
removes water from wet sludge. How many
kilograms of water leave the thickener per 100 kg of
wet sludge that enter the thickener? The process is
in the steady state.
100 kg
70 kg
Thickener
Wet sludge
Dehydrated sludge
Water = ?
First, we have to write the general mass balance.
é Accumulation
ù é Input thorugh ù éOutput through ù
ê within the systemú = ê system boundary ú - ê system boundary ú
ë
û ë
û ë
û
éGaneration ù éConsumption ù
ú - ê within
ú
+ êê within
ú ê
ú
êëthe system úû êëthe system úû
Since there are no generation and consumption, and the
system is under steady state therefore the third and the last
terms on the right hand side and the term on the left hand
side equal to zero.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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é Input thorugh ù éOutput through ù
0=ê
ú - ê system boundary ú
system
boundary
ë
û ë
û
100 kg = 70 kg + kg of water
kg of water = 30 kg
Example 2.2 Mass Balance
From the picture below, what is the accumulation
in the system? If there are no generation and
consumption.
Tank
Qin, Cin
By Assist. Prof. Dr. Wipada Sanongraj
Qout, Cout
48
1304 211 Chemical Engineering Principles
and Calculations
é Accumulation
ù é Input thorugh ù éOutput through ù
ê within the systemú = ê system boundary ú - ê system boundary ú
ë
û ë
û ë
û
éGaneration ù éConsumption ù
ú - ê within
ú
+ êê within
ú ê
ú
êëthe system úû êëthe system úû
Input = QinCin
Output = QoutCout
Accumulation = Vtank(dC/dt)
é Accumulation
ù
dC
ê within the system ú = Vtan k dt = [QinCin ] - [Qout Cout ] + [ 0] - [ 0]
ë
û
Problems
1. Draw a sketch of the following processes and
place a dashed line around the system:
a) Tea kettle
b) Fireplace
c) Swimming pool
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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2. Label the materials entering and leaving the
systems in problem 1. Classify each system as
open or closed.
3. Write down the general material balance in
words. Simplify it for each process in problem 1,
stating the assumptions made in each
simplification.
Strategy for Analyzing Material Balance
Problems
v Read the problem and clarify what is to be
accomplished
vDraw a sketch of the process; define system by a
boundary
vLabel with symbols the flow of each stream and
compositions
vPut all the known values of compositions and
stream flows on the figure
vSelect a basis
vWrite down an appropriate set of balances to solve
vcount the number of independent balances
vSolve the equation
vCheck answers
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Solving Material Balance Problems That Do
Not Involve Chemical Reactions
Example 2.3 : Membrane Separation
Membranes represent a relatively new technology
for the separation of gases. The following figure
illustrates a nanoporous membrane that is made
by coating a very thin layer of the polymer on a
porous graphite-supporting layer. What is the
composition of the waste stream if the stream
amount to 80% of the input?
High-pressure
side
Membrane
21% O2
Input
Low-pressure
side
25% O2
Output
Flow
75% N2
79% N2
O2
waste
By Assist. Prof. Dr. Wipada Sanongraj
N2
stream
51
1304 211 Chemical Engineering Principles
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solution
This is a steady state process without chemical
reaction so that the accumulation term and the
generation and consumption terms are zero. The
system is the membrane. Let xO2 be the mole
fraction of oxygen and xN2 be the mole fraction of
nitrogen, and let nO2 and nN2 be the respective
moles.
F (g mol)
O2
N2
Membrane
Mol fr.
0.21
0.79
1.00
W (g mol)
Mol fr.
O2 xO2
N2 xN2
1.00
P (g mol)
O2
N2
Mol
nO2
nN2
W
Mol fr.
0.25
0.75
1.00
Basis : 100 g mol = F
We know W = 0.80*100 = 80 g mol (because waste
amounts to 80% of input)
Three unknowns exist : P, xO2, and xN2 or P, nO2,
and nN2
Two independent balances are the oxygen and
nitrogen balances either as elements or as
compounds. The third independent balance is xO2
+ xN2 =1.00 or nO2 + nN2 = 80
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1304 211 Chemical Engineering Principles
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The component balances are
In
Out
O2: 0.21(100)
= 0.25(P) + xO2 (80)
N2: 0.79(100)
= 0.75(P) + xN2 (80)
1.00 = xO2 + xN2
Or
In
Out
O2: 0.21(100)
= 0.25(P) + nO2
N2: 0.79(100)
= 0.75(P) + nN2
80 = nO2 + nN2
The solution of these equations is xO2 = 0.20, xN2 =
0.80, and P = 20 gmol
A simpler calculation involves the use of the
total balance and one component balance
because
F= P+W or 100 = P + 80
Gives P = 20 straight off
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Example 2.4 Continuous Distillation
A manufacturer of alcohol is having a bit of
difficulty with a distillation column. Technicians
think too much alcohol is lost in the bottoms
(waste). Calculate the composition of the bottoms
and the mass of the alcohol lost in the bottoms.
Vapor
1000 kg Feed (F)
10% EtOH
90% H2O
Distillation
Heat
Exchanger
Reflux
column
Heat
Bottom (B) kg
EtOH = ?
H2O = ?
Cooling water
Distillate (Product) P = kg
60% EtOH
40% H2O
Wt = 1/10 feed
General Mass Balance Equation:
é Accumulation
ù é Input thorugh ù éOutput through ù
ê within the systemú = ê system boundary ú - ê system boundary ú
ë
û ë
û ë
û
éGaneration ù éConsumption ù
ú - ê within
ú
+ êê within
ú ê
ú
êëthe system úû êëthe system úû
Assuming that the process is in the steady state, no
reaction occurs
Mass in = Mass out
Basis : F = 1000 kg
Total mass balance; F = P + B
By Assist. Prof. Dr. Wipada Sanongraj
(1)
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1304 211 Chemical Engineering Principles
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We are given that P is 1/10 of F, so that P =
0.1*1000 = 100 kg.
Substituting the total mass balance F = P+ B
calculate B by direct subtraction
B = 1000 – 100 = 900 kg
The solution for the composition of the bottom can
be computed directly by subtraction
EtOH
kg feed in
kg distillate out
0.1(1,000) - 0.6 (100)
=
H2O
0.9(1,000) - 0.4 (100)
kg bottom out %
40
4.4
=
860
900
95.6
100.0
Example 2.5 Mixing
Dilute sulfuric acid has to be added to dry charged batteries
at service station to activate a battery. You are asked to
prepare a batch of new 18.63% acid as follows. A tank of old
weak battery acid solution contains 12.43% H2SO4 (the
remainder is pure water). If 200 kg of 77.7% H2SO4 is added
to the tank, and the final solution is to be 18.63% H2SO4,
how many kilograms of battery acid have been made?
Added solution 200 kg = A
H2SO4 77.7%
H2O
22.3%
System
H2SO4 12.43%
H2SO4 18.63%
H2O
H2O
87.57%
Original solution F kg
By Assist. Prof. Dr. Wipada Sanongraj
81.37%
Final solution P kg
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1304 211 Chemical Engineering Principles
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Should the process be treated as an unsteady-state
process or a steady state process?
If the tank is the system, and the tank initially contains
sulfuric acid solution, then a change occurs inside the
system so that accumulation occurs in the system (the
mass increases).
The general mass balance can be reduced to
Accumulation = In - Out
Basis : 200 kg of A
The two unknown quantities are F and P. We can write
two independent mass balances.
Type of Balance
Accumulation in Tank
A in
Out
Initial
Final
H2SO4
P(0.1863) -
F(0.1243)
= 200(0.777)
-
0
H2O
P(0.8137) -
F(0.8757)
= 200(0.223)
-
0
=
-
0
Total
P
-
F
200
Because the equations are linear and only two occur, you
can take the total mass balance, solve it for F, and
substituting for F in the H2SO4 balance to calculate P.
(P-200)(0.1243) + 200(0.777) = P(0.1863)
P = 2110 kg acid
F = 1910 kg acid
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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The problem could also be solved by considering the
mixing to be a steady process with the initial solution F
added to A in a vessel, and the resulting mixture removed
from the vessel.
A in
F in
P out
H2SO4
200(0.777) +
F(0.1243) = P(0.1863)
H2O
200(0.223) +
F(0.8757) = P(0.8137)
Total
A
+
F
= P
You can see by inspection that these equation are no
different than the first set of mass balances except for the
arrangement.
Example 2.6 Drying
In the processing of the fish, after the oil is
extracted, the fish cake is dried in rotary drum
dryers, finely ground, and packed. The resulting
product contains 65% protein. In a given batch of
fish cake that contains 80% water (the remainder is
dry cake), 100 kg of water is removed, and it is
found that the fish cake is then 40% water.
Calculate the weight of the fish cake originally put
into the dryer.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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W = 100 kg H2O
0 kg H2O
A kg
Rotary dryer
Wet fish Cake = ?
B kg
Dry fish Cake = ?
0.8 H2O
0.2 BDC
0.4 H2O
0.6 BDC
Tie component
Basis : 100 kg of water evaporated = W
The unknown stream flows are two: A and B. All the
compositions are known.
Two independent balances can be written so that a
unique solution exists.
The water balance and the total mass balance plus the BDC
balance
The water balance:
0.8A = 0.4B + 100
We will use total mass balance plus the BDC balance rather
than the water balance. Because the BDC balance is
slightly easier to use. The water balance can be used as a
check on the calculations.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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In
Out
Total balance
A
=
B + W = B +100
BDC balance
0.2A
=
0.6B
The BDC balance gives the ratio of A to B : B = 1/3A.
Introduce this relation into the total mass balance to get
A = 150 kg initial cake
Check via water balance:
0.8(150) = 0.4(150)(1/3) + 100
Example 2.7 Crystallization
A tank hold 10,000 kg of a saturated solution
of Na2CO3 at 30 oC. You want to
crystallize from this solution 3000 kg of
Na2CO3.10 H2O without any
accompanying water. To what
temperature must the solution be cooled?
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Saturated solution at 30 oC
Saturated solution at ? oC
Na2CO3
Na2CO3
H2O
H2O
System boundary
Na2CO3.10H2O
3000 kg
Solution
We need solubility data for Na2CO3 as a function of
temperature
Temp (oC)
0
Solubility (g Na2CO3/100 g H2O)
7
10
12.5
20
21.5
30
38.8
At initial, solution is saturated at 30 oC, composition
of initial solution:
38.8 g Na2CO3/(38.8 g Na2CO3 + 100 g H2O)
= 0.280 mass fraction of Na2CO3 and 0.720 mass
fraction H2O
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1304 211 Chemical Engineering Principles
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Next, calculate the composition of crystals.
Basis: 1 g mol Na2CO3.10 H2O
Comp.
Mol
MW
mass
Na2CO3
1
106
106
H2O
10
18
180
286
Basis: 10,000 kg of saturated solution at 30 oC
10,000 kg
Na2CO3 = 0.280
H2O = 0.720
Na2CO3
H2O
System
boundary
Na2CO3
H2O
mass fr
0.371
0.629
1.00
P = ? Kg
mNa2CO3
mH2O
3,000 kg
Na2CO3.10H2O Na CO = 0.371
2
3
H2O = 0.629
System is unsteady state, mass balance equation reduces to
Accumulation = - Out
Mass balance for final state: mNa2CO3 + mH2O = P
Component balance:
Accumulation in tank
Out
Final
Initial
Na2CO3
mNa2CO3
- 10,000(0.28)
= -3000(0.371)
H2O
mH2O
- 10,000(0.72)
= -3000(0.629)
Total
P
- 10,000
= -3000
Solve equations get the compositions of final solution:
mNa2CO3 = 1687 kg, mH2O = 5313 kg, P = 7000 kg
To find temperature of final solution, calculate composition in terms of
g Na2CO3/ 100 g H2O
1687g Na2CO3/ 5313 g H2O = 31.8 g Na2CO3/ 100 g H2O
Do the interpolation from solubility data
30 oC – (38.8-31.8)/(38.8-21.5)*(30-20) oC = 26 oC
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1304 211 Chemical Engineering Principles
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Solving Material Balance Problems that
Involve Chemical Reactions
Objectives of this section
• Define flue gas, stack gas, Orsat analysis, dry
basis, wet basis, theoretical air (oxygen), required
air (oxygen), and excess air (oxygen).
• Given two of the three factors: entering gas
(oxygen), excess air (oxygen), and required air
(oxygen), compute the third factor.
• Understand how to apply the material balance
equation when chemical reactions occur.
• Apply the 10-step strategy to solve problems
involving reactions.
Flue or stack gas : All the gases resulting from a
combustion process including the water vapor,
sometimes known as wet basis.
Orsat analysis or dry basis : all the gases resulting
from the combustion process not including the
water vapor. Another way to state that water vapor
is not included in the gas analysis is to give the
analysis on a dry basis or give the Orsat analysis.
Theoretical air : the amount of air (or oxygen)
required to be brought into the process for
complete combustion. Sometimes this quantity is
called the required air.
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1304 211 Chemical Engineering Principles
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Excess air : the amount of air (or oxygen) in
excess of that required for complete combustion
as computed for theoretical air.
Stack Gas
CO2
CO
O2
N2
SO2
H2O
Flue gas
on SO2
free basis
Flue gas
or Orsat analysis
or dry basis
Comparison of gas analyses on different bases
The calculated amount of excess air does not
depend on how much material is actually burned
but what can be burned. Even if only partial
combustion takes place, as for example, C
burning to both CO and CO2, the excess air is
computed as if the process of combustion
produced only CO2.
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1304 211 Chemical Engineering Principles
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1 C
1 CO2
Combustion
1 O2
C + O2
CO2
The material balance for O2 in moles
Acc.
O2
0
In
= 1 -
Out
0 +
Gen.
0
-
Cons.
1
The mole balances on C and CO2
C
O
The mass balances on C and CO2 (elements)
Acc.
In
Out
Gen.
Cons.
C
O
C
O2
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1304 211 Chemical Engineering Principles
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% excess air = 100
excess O2 / 0.21
excess air
= 100
required air
required O2 / 0.21
% excess air = 100
O2 entering process - O2 required
O2 required
% excess air = 100
O2
excess O2
entering - excess O2
Example 2.8 Excess Air
Fuel for motor vehicles other than gasoline
are being used because they generate
lower levels of pollutants than does
gasoline. Compressed propane has been
suggested as a source of economic power
for vehicles. Suppose that in a test 20 kg
of C3H8 is burned with 400 kg of air to
produce 44 kg of CO2 and 12 kg of CO.
What was the percent excess air?
By Assist. Prof. Dr. Wipada Sanongraj
65
1304 211 Chemical Engineering Principles
and Calculations
Solution
This is a problem involving the following reaction
C3H8 + 5O2
3CO2 + 4H2O
Since the percentage of excess air is based on the
complete combustion of C3H8 to CO2 and H2O
The required O2 is
20 kg C3H8 1 kg mol C3H8 5 kg mol O2
44 kg C3H8
1 kg mol C3H8
= 2.27 kg mol O2
The entering O2 is
400 kg air 1 kg mole air 21 kg mole O2 = 2.9 kgmole O
2
29 kg air
100 kg mole air
The percent excess air is
100 * excess O2 = 100* entering O2 – required O2
required O2
% excess air = 2.9 kg mole O2 – 2.27 kg mole O2 * 100
2.27 kg mole O2
= 28%
By Assist. Prof. Dr. Wipada Sanongraj
66
1304 211 Chemical Engineering Principles
and Calculations
Example 2.9
Suppose a gas containing 80% C2H6 and 20%
O2 is burned in an engine with 200%
excess air. Eighty percent of the ethane
goes to CO2, 10% goes to CO, and 10%
remained unburned. What is the amount
of excess air per 100 moles of the gas?
C2H6 + 7/2O2
2CO2 + 3H2O
Basis: 100 moles of gas
80 moles of C2H6 require 3.5(80) = 280 moles of O2,
However the gas contains 20 moles of O2, so that only
280-20 = 260 are needed in the entering air for complete
combustion. Thus the excess 200% excess O2 (air) is
based on 260 moles of O2
Entering with air
Moles O2
required O2:
260
Excess O2
2(260)= 520
Total O2
3(260) = 780
By Assist. Prof. Dr. Wipada Sanongraj
67
1304 211 Chemical Engineering Principles
and Calculations
Example 2.10: Preventing corrosion
Corrosion of pipes in boilers by oxygen can be
alleviated through the use of sodiumsulfite.
Na2SO3 removes O2 from boiler feed water by
the following reaction:
2Na2SO3 + O2
2Na2SO4
How many pounds of Sodium sulfite are
theoretically required (for complete reaction) to
remove the oxygen from 8,330,000 lb of water
(106 gal) containing 10.0 ppm of dissolved
oxygen and at the same time maintain a 35%
excess of sodium sulfite?
H2O 8,330,000 lb
10 ppm O2
H2O 8,330,000 lb
System
No oxygen
Basis: 8,330,000 lb H2O
The amount of O2 entering is
8,330,000 lb H2O 10 lb O2
(1,000,000-10 lb O2) lb H2O
The O2 balance in lb is simple
In - Out + Generation – Consumption = Accumulation
83.3 – 0 + 0 – mO2 = 0
mO2 = 83.3 lb, then calculate the amount of Na2SO3
83.3 lb O2 1 lb mol O2 2 lb mol Na2SO3 126 lb Na2SO3
32 lb O2
1 lb mol O2
1 lb mol Na2SO3
= 83.3 lb O2
1.35
= 886 lb Na2SO3
By Assist. Prof. Dr. Wipada Sanongraj
68
1304 211 Chemical Engineering Principles
and Calculations
Example 2.11 Combustion
Generation of methane-rich biogas is a way to
avoid high waste-disposal costs, and burning it
can meet up to 60% of the operating costs for
such waste-to-energy plants. Consider the
combustion of methane as shown in figure.
CH4 100%
F = 16 kg
Burner
A = 300 kg
O2 = 21%
N2 = 79%
300 kg A 1 kg mol A
29 kg A
= 10.35 kg mol A in
16 kg CH4 1 kg mol CH4
16 kg CH4
10.35 kg mol A 0.21 kg mol O2
1 kg mol A
10.35 kg mol A 0.79 kg mol N2
1 kg mol A
By Assist. Prof. Dr. Wipada Sanongraj
P = ? (kg mol)
CO2 = ?
N2 = ?
O2 = ?
H2O = ?
= 1 kg mol CH4 in
= 2.17 kg mol O2 in
= 8.18 kg mol N2 in
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1304 211 Chemical Engineering Principles
and Calculations
Basis: 16 kg CH4 = 1 kg mol CH4
The unknowns are P and the four compositions in P:
nPCO2, nPO2, ,nPN2, and nPH2O.
Let use the balance on the elements.
Balance
CH4 in Air in
Out
C:
1
= nPCO2
H2:
2
= nPH2O
O2:
2.17 = 0.5nPH2O+nPO2 + nPCO2
N2:
8.18 = nPN2
nPCO2 + nPH2O + nPN2 + nPO2 = P
Now we can easily solve the set of equations;
nPCO2
nPH2O
nPN2
nPO2
P
=
=
=
=
=
1
2
8.18
2.17 - 0.5(2) – 1 = 0.17
1 + 2 + 8.18 + 0.17 = 11.35
From xiP = (niP)/P
xPCO2 = 0.09
xPH2O = 0.18
By Assist. Prof. Dr. Wipada Sanongraj
xPN2 = 0.72
xPO2 = 0.01
70
1304 211 Chemical Engineering Principles
and Calculations
Example 2.12: Combustion with Nonprecise Data
In a test run, a liquid that is proposed for use as a
fuel in a flare and has the composition of 88% C and
12% H2 is vaporized and burned with dry air to a flue
gas of the following composition on a dry basis.
CO2
O2
N2
13.4%
3.6%
83.0%
100.0%
To help design the equipment of the continuous
steady-state combustion device, determine how
many kg mol of dry flue gas are produced per 100
kg of liquid feed. What was the percentage of
excess air used?
The process is in the steady-state accompanied
by chemical reaction. Let the system be the
flare and associated equipment.
HO
2
1.00
W
H2O
kg mol
mass fr.
C 0.88
H2 0.12
1.00
F
Test liquid
kg
mol fr.
N2 0.79
O2 0.21
1.00
By Assist. Prof. Dr. Wipada Sanongraj
Catalytic
oxidation unit
A
Air (dry)
kg mol
G
kg mol
mol fr.
CO2 0.134
N2 0.830
O2 0.036
1.00
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1304 211 Chemical Engineering Principles
and Calculations
Basis : 100 kg mol dry flue gas = G
The atomic species balances (in moles)
Balance
F in
C:
0.88F
12
A in
W out
+ 0
0
-
- 0.134(100)
W
0
= 0
0.12F
2.016
O2:
0
+ 0.21A - (1/2)W - (0.134 +0.036)100 = 0
N2:
0
+ 0.79A -
0
-
Accum.
H2:
From C :
+ 0
-
G out
- (0.830)100
F = (1304*12/0.88)
From N2 : A = 83.0/0.79
= 0
=
=
=0
182.73 kg
105.06 kg mol
From H2 : W = 182.73*0.12/2.016 = 10.88 kg mol
Use the oxygen balance as a check
105.06(0.21) = 10.88*(1/2) + 17.00
22.06 =
22.44
An exact balance does not occur, but the answers agree
reasonably well here. In many combustion problems,
slight errors in the data will cause large differences in the
calculated flows and percentage of excess air.
By Assist. Prof. Dr. Wipada Sanongraj
72
1304 211 Chemical Engineering Principles
and Calculations
We can now answer the requested questions
G 100 kgmol 54.73 kgmolG
=
=
F 182.73 kg
100 kgF
The percent excess air can be calculated via two routes:
% excess air = 100
or
O2
excess O2
entering - excess O2
% excess air = 100
excess O2
required O2
The required O2 is
C + O2
CO2
H2 + 1/2O2
H2O
182.73*(0.88) = 13.40 kgmol
12
182.73*(0.12) = 5.44 kgmol
2(2.106)
The overall required O2 is 18.84 kgmol
The excess oxygen is
or
22.4 - 18.84 = 3.56
22.06 - 18.84 = 3.22
By Assist. Prof. Dr. Wipada Sanongraj
73
1304 211 Chemical Engineering Principles
and Calculations
Basis: O2 in = 22.44 mol
calculated from the
entering air
Basis: O2 in = 22.06 mol
calculated from the
flue gas
% excess air: 100*3.60 = 19.1%
22.44-3.60
100*3.22 = 16.7%
22.06-3.22
% excess air: 100*3.60 = 19.1%
18.84
100*3.22
18.84
= 16.7%
Example 2.13 : Combustion of Coal
A local utility burns coal having the following
compositions on a dry basis.
Component
C
H
O
N
S
Ash
Total
By Assist. Prof. Dr. Wipada Sanongraj
Percent
83.05
4.45
3.36
1.08
0.70
7.36
100.0
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1304 211 Chemical Engineering Principles
and Calculations
The average Orsat analysis of the gas from the
stack during a 24-hr test was
Component
CO2 + SO2
CO
O2
N2
Total
Percent
15.4
0.0
4.0
80.6
100.0
Moisture in the fuel was 3.9%, and the air on the average
contained 0.0048 lb H2O/lb dry air. The refuse showed
14.0% unburned coal, with the remainder being ash. You are
asked to check the consistency of the data before they are
stored in a data base. Is the consistency satisfactory? What
was the average percent excess air used?
This is a steady-state problem with the reaction.
The system is the furnace.
lb
C 83.05
H 4.45
O 3.36
N 1.08
S 0.70
Ash 7.36
100.00
%
CO2+SO2 15.4
W (lbmol) H2O(g) 100%
CO
0.0
O2
4.0
80.6
Stack gas N2
Coal
furnace
100.0
P (lb mol)
F (lb)
Added H2O 3.9 lb
H: 0.433 lbmol
O: 0.217 lbmol
By Assist. Prof. Dr. Wipada Sanongraj
Air A
(lb mol)
Refuse R (lb)
%
C + H + O +N + S 14.0
Ash
86.0
100.0
Mol fr.
N2 0.79
O2 0.21
1.00
Added H2O 0.0048 lb/lbAir
H: 0.0154 lbmol/lb molA
O: 0.0077 lbmol/lb mol A
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1304 211 Chemical Engineering Principles
and Calculations
Basis : F = 100 lb
We might neglect the C, H, O, N, and S in the
refuse but will include the amount to show what
calculations are necessary if the amounts of the
elements were significant. The ash balance is:
7.36 = R(0.86)
R = 8.56 lbs
The unburned coal in the refuse is:
8.56(0.14) = 1.20 lb
If we assume that the combustibles in the refuse
occur in the same proportions as they do in the coal
(which may not be true), the quantities of the
combustibles in R on an ash-free basis are:
Component
C
H
O
N
S
Total
mass %
89.65
4.80
3.63
1.17
0.76
100.00
lb
1.076
0.058
0.0436
0.014
0.009
1.20
lb mol
0.0897
0.0537
0.0027
0.0010
0.0003
0.1474
The variables whose values are still unknown are
A, W, and P
By Assist. Prof. Dr. Wipada Sanongraj
76
1304 211 Chemical Engineering Principles
and Calculations
Mass balances of the elements in moles
In
Out
F
C+S
H
0
A
W
P
R
83.05 + 0.70 + 0 = 0 + P(0.154) + 0.0897 + 0.0003
12.0
32.0
4.45 + 0.433 + 0.0154A = 2W + 0 +
0.0537
1.008
3.36 + 0.217 + 0.21A(2) + 0.0077A = W + 2P(0.154+0.04) + 0.0027
16.0
N
1.08 +
2(0.79A) = 0 +
2P(0.806) + (0.001)
14.0
Solve the C+S balance to get P = 50. Then solve the N
balance to get A = 45.35. Solve the H balance to get W =
2.746.
Use the O balance to serve as a check: 19.8 =20.0
To calculate the excess air, because of the
oxygen in the coal and the existence of unburned
combustibles, we will calculate the total oxygen in
and the required oxygen:
% excess air = 100
By Assist. Prof. Dr. Wipada Sanongraj
O2 entering process - O2 required
O2 required
77
1304 211 Chemical Engineering Principles
and Calculations
Component Reaction
lb
lb mol
Required
O2 (lb mol)
C
C + O2
CO2
83.05
6.921
6.921
H
H2 + 1/2O2
H2O
4.45
4.415
1.104
0.210
(0.105)
O
-
3.36
N
-
-
S
S + O2
SO2
0.70
0.022
0.022
7.942
The oxygen in the air is 45.35*0.21 = 9.524 lbmol.
% excess air = 100
9.524 - 7.942
= 19.9%
7.942
If you (incorrectly) calculated the % excess air
from the wet stack gas alone, you would get
% excess air = 100
By Assist. Prof. Dr. Wipada Sanongraj
4.00
= 23.8%
15.4 + 2.746 / 2
78
1304 211 Chemical Engineering Principles
and Calculations
Recycle, Bypass, and Purge Calculations
A recycle stream is a term denoting a process
stream that returns material from downstream of a
process unit back to the process unit.
A bypass stream is the one that skips one or more
stages of the process and goes directly to another
down stream stage.
A purge stream is a stream bled off to remove an
accumulation of inert or unwanted material that
might otherwise build up in the recycle stream.
Recycle, R
Fresh
Feed, F
Mixer
Process
Feed
Process
2
3
1
1. About the entire process
Gross
Product
Separator
Net
Product, P
4
2. About the junction point at which the fresh feed
is combined with the recycle stream
3. About the process only
4. About the function point at which the gross
product is separated into recycle and net product
By Assist. Prof. Dr. Wipada Sanongraj
79
1304 211 Chemical Engineering Principles
and Calculations
Example 2.14 : Recycle without Chemical Reaction
A distillation column separates 10,000 kg/hr of a
50% benzene-50% toluene mixture. The product D
recovered from the condenser at the top of the
column contains 95% benzene, and the bottom W
from the column contains 96% toluene. The vapor
stream V entering the condenser from the top of the
column is 8000 kg/hr. A portion of the product from
the condenser is returned to the column as reflux,
and the rest is withdraw for use elsewhere. Assume
that the compositions of the streams at the top of
the column (V), the product withdrawn (D), and the
reflux (R) are identical because the V stream is
condensed completely. Find the ratio of the amount
refluxed R to the product withdrawn (D).
Basis : 1 hr (equal F = 10,000 kg)
Overall Material Balances:
Total material
F=D+W
10,000 = D + W
Component (benzene)
FwF = DwD + WwW
10,000(0.50) = D(0.95) + W(0.04)
Solving for W and D
W = 4950 kg/hr
D = 5050 kg/hr
By Assist. Prof. Dr. Wipada Sanongraj
80
1304 211 Chemical Engineering Principles
and Calculations
Balance around the condenser
Total material:
V=R+D
8,000 = R + 5,050
R = 2,950 kg/hr
R/D = 2950/5050 = 0.58
Example 2.15 : Recycle without chemical reaction
The manufacture of such products as penicillin,
tetracycline, vitamins, and other fine organic
compounds, usually requires separating the
suspended solids from their mother liquor by
centrifuging, and then drying the wet cake. What is
the lb/hr of the recycle stream R?
This is a steady-state problem without reaction and
with recycle.
By Assist. Prof. Dr. Wipada Sanongraj
81
1304 211 Chemical Engineering Principles
and Calculations
R=?
0.286V
0.714 H2O
F
98 lb/hr
20% V
80% H2O
Continuous
Filter
Centrifuge
C=?
60% V
40% H2O
P=?
4% H2O
96 % V
W=?
H2O 100%
Basis : 1 hr (F = 98 lb)
Overall mass balances
In
Out
V:
0.20(98)
H2O : 0.80(98)
Total : 98
= 0
+ 0.96P
= (1.0)W + 0.04P
= W
+ P
Solving for P and W
P = 20.4 lb
By Assist. Prof. Dr. Wipada Sanongraj
W = 98 – 20.4 = 77.6 lb
82
1304 211 Chemical Engineering Principles
and Calculations
Total balance on filter
C=R+P
C = R + 20.4
Component V balance on filter
CwC = RwR + PwP
0.6C = 0.286R + 0.96(20.4)
Solving for R
R = 23.4 lb/hr
Recycle in Processes with
Chemical Reaction
Overall fraction conversion
[Mass (moles) of reactant in fresh feed – mass
(moles) of reactant in output of the overall process]
/ mass (moles) of reactant in fresh feed
Single-pass (once through) fraction conversion
[Mass (moles) of reactant fed into the reactor
(process feed) – mass (moles) of reactant existing
the reactor (gross product)]/ mass (moles) of
reactant fed into the reactor
By Assist. Prof. Dr. Wipada Sanongraj
83
1304 211 Chemical Engineering Principles
and Calculations
Process feed & Gross Product
Recycle
Process feed
Fresh Feed
Gross product
Net product
Example 2.16 : Recycle with a Reaction Occurring
Immobilized glucose isomerase is used as a
catalyst in producing fructose from glucose in a
fixed-bed reactor (water is the solvent). For the
system shown below, what percent conversion of
glucose results on one pass through the reactor
when the ratio of the exit stream to the recycle
stream in mass units is equal to 8.33?
R (kg)
wR,G = ?
wR,F = ?
wR,W = ?
1.00
S = 100 kg
0.4ws,G
0.6ws,,w
1.00
1
By Assist. Prof. Dr. Wipada Sanongraj
T (kg)
Fixed-bed
wT,G = ?
Reactor
wT,F = 0.04
wT,W = ?
2
P = ? (kg)
wR,G = ?
wR,F = ?
wR,W = ?
1.00
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1304 211 Chemical Engineering Principles
and Calculations
Basis : S = 100 kg
Let f be the fraction conversion for one pass through
the reactor. The unknowns are R, F, P, T, wR,G, wR,T,
wR,W, wT,W, wT,G, and f for a total of 9.
The balances are SwR,i = 1, SwT,i = 1, R = P/8.33
Overall balances:
Total :
S = P = 100 kg
R = 100/8.33 = 12 kg
No water is generated or consumed
Water :
100(0.60) = P(wR,W) = 100wR,W
wR,W = 0.6
Mixing point 1, no reaction occurs
Total :
100 + 12 = T = 112
Glucose:
100(0.40) + 12(wR,G) = 112(wT,G)
Fructose:
0 + 12(wR,F) = 112(0.04)
wR,F = 0.373
Because wR,F + wR,G + wR,W = 1
wR,G = 1 – 0.373 – 0.600 = 0.027
And from the glucose balance,
wT,G = 0.360
By Assist. Prof. Dr. Wipada Sanongraj
85
1304 211 Chemical Engineering Principles
and Calculations
Reactor plus Separator 2
Total :
Glucose :
T = 12 + 100 = 112
In
TwT,G
Out
-
Consumed
(R+P)wR,G
- fTwT,G
Accumu.
= 0
112(0.360) – 112(0.027) - f(112)(0.360) = 0
f = 0.93
Example 2.17: Recycle with a reaction occurring
Refined sugar (sucrose) can be converted to glucose and
fructose by the inversion process
C12H22O11 + H2O = C6H12O6 + C6H12O6
The combined quantity glucose plus fructose is called
inversion sugar. If 90% conversion of sucrose occurs on
one pass through the reactor, what would be the recycle
stream flow per 100 lb fresh feed of sucrose solution
entering the process shown in Figure? What is the
concentration of inversion sugar (I) in the recycle stream
and in the product stream? The concentrations of
components in the recycle stream and product stream
are the same
By Assist. Prof. Dr. Wipada Sanongraj
86
1304 211 Chemical Engineering Principles
and Calculations
R
xs,R
xI,R
Xw,R
F = 100 lb
Sucrose 30%
H2O 70%
C
xs,C
xI,C = 0.05
xw,C
Reactor
Solution
Basis: 100 lb = F
Overall balance: F = P = 100 lb
Balance at mixing point:
Total 100 + R = C
Sucrose 100(0.3) + R(xs,R) = C(xs,C)
Inversion 0 + R(xI,R) = C(0.05)
Separator
P
xs,R
xI,R
Xw,R
(a)
(b)
(c)
Make balance on the reactor plus the separator
In
Out
Consumed
Sucrose
C(xs,C) - (R+100)(xs,R) – C(xs,c)(0.9) = 0
(d)
Next do the water balance, but we need to calculate the
pounds of water consumed in the reaction per pound of
sucrose consumed
1 lbmol H2O 1 lbmol S 18 lb H2O
= 0.0526 lb H2O
1 lbmol S
342.35 lb S 1 lbmol H2O
lb S
In
Out
Consumed
Water Cxw,C - (R+100)xw,R - Cxs,C(0.9)(0.0526)
(e)
Xs,R + xI,R + xw,R = 1
(f)
Xs,C + xI,C + xw,C = 1
(h)
Solve equations (a-h), we get R = 20.9 lb and xI,R = 0.313
By Assist. Prof. Dr. Wipada Sanongraj
87
1304 211 Chemical Engineering Principles
and Calculations
Bypass and Purge
A bypass stream- one that skips one or
more stages of the process and goes
directly to another downstream stage
A purge- a stream bled off to remove an
accumulation of inerts or unwanted
materials
Example 2.18: Bypass
Calculations
In the feedstock preparation section of a plant
manufacturing natural gasoline, isopentane is removed
from butane-free gasoline. Assume for purposes of
simplification that the process and components are as
shown in figure. What fraction of the butane-free
gasoline is passed through the isopentane tower? The
process is in the steady state and no reaction occurs.
By Assist. Prof. Dr. Wipada Sanongraj
88
1304 211 Chemical Engineering Principles
and Calculations
Isopentane side stream, S kg
i-C5H12 100 %
Debutanizer
F = 100 kg
n-C5H12 80%
i-C5H12 20%
X kg
Isopentane
tower
Y kg
n-C5H12 100%
To natural gasoline plant, P kg
n-C5H12 = 90%
i-C5H12 = 10%
From diagram part of the butane-free gasoline bypasses
the isopentane tower and proceeds to the next stage in
the natural gasoline plant.
Basis: 100 kg feed
Overall balance
Total material balance: In = Out
100 = S + P
(a)
Component balance (n-C5H12)
100(0.8) = S(0) + P(0.9)
(b)
We get, P = 88.9 kg and S = 11.1 kg
Balance around isopentane tower
Total material balance: x = 11.1 + y
(c)
Component balance (n-C5H12)
x(0.8) = y(1)
(d)
We get, x = 55.5 kg, y = 44.4 kg, therefore the fraction of
butane free gas passed through isopentane tower is
55.5/100 = 0.55
By Assist. Prof. Dr. Wipada Sanongraj
89
1304 211 Chemical Engineering Principles
and Calculations
Another approach is to make a balance at mixing
points 1 and 2
Balance around mixing point 2
Total material balance: (100-x) + y = 88.9
(e)
Component balance (i-C5H12): (100-x)0.2 = 88.9*0.1
(f)
We get, x = 55.5 kg
Example 2.19: Purge
Considerable interest exists in the conversion of coal into
more convenient liquid products for subsequent
production of chemicals. Two of the main gases that can
be generated are H2 and CO. After cleanup, these two
gases can be combined to yield methanol according to
the following equation
CO + 2H2
=
CH3OH
Figure illustrates a steady state process. All compositions
are in the mole fractions or percent. The stream flows
are in moles. A purge stream is used to maintain the CH4
concentration in the exit to the separator at no more than
3.2 mol %. The once-through conversion of the CO in the
reactor is 18%. Compute the moles of recycle, CH3OH,
and purge per mole of feed and purge composition.
By Assist. Prof. Dr. Wipada Sanongraj
90
1304 211 Chemical Engineering Principles
and Calculations
F mol
67.3% H2
32.5% CO
0.2% CH4
Reactor
Separator
Recycle, R
X H2
Y CO
Z CH4
E mol
100% CH3OH
Purge, P mol
Because the problem is presented in terms of moles,
making an overall mass balance is not convenient.
We will use Element balance instead.
Basis: F = 100 mol
x+y+z =1
From limit of CH4 in the reactor, z = 0.032
(a)
(b)
Overall element balance (in moles)
H2: 67.3 + 0.2(2)
= E(2)(1) + P(x+2z)
C: 32.5 + 0.2 = E(1) + P(z + y)
O: 32.5
= E(1) + (y)P
(c)
(d)
(e)
Reactor plus separator balance (in moles)
In
Out
Consumed
CO: 32.5 + R(y) - y(R+P)
= (32.5+Ry)(0.18) (f)
Solve these equations we get
E = 31.25 moles, P = 6.25 moles, R = 705 moles, x = 0.768
y = 0.2, z = 0.032
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Chapter 3
Energy Balance
Introduction
System: 304 12 1 Chemi cal
Closed system: lucose:
I00 = S + P
Open system: otal
E y)
m ta er ila ba l ance: (100
Property: Šr o perty:/a t er
pen system:
Engineering Principles
100(0.40) + 12 (
:
/
nerts
100(0.60) =
Extensive Property: olve
these equa t ions we get
utane fr e e ga s pas sed through
Intensive Property: olve these equt
a ions we get
Šroperty:
State: Entroduction
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Six types of Energy
Work (W): form of
energy that represents
a transfer between the
system and
surroundings
W is positive when
done on the system
s2
W = ò Fds
s1
F is external force in the direction of
s acting on the system
Example 3.1: Work
Suppose that an ideal gas at 300 K and 200 kPa is
enclosed in a cylinder by a frictionless piston, and the
gas slowly forces the piston so that the volume of gas
expands from 0.1 to 0.2 m3. Calculate the work done by
the gas on the piston if two different paths are used to go
from the initial state to the final state:
Path A: The expansion occurs at constant pressure (p =
200 kPa)
Path B: The expansion occurs at constant temperature
(T= 300 K)
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Solution
s2
v2
F
W = - ò Ads = - ò pdV
A
s1
v1
Path A:
W = -p(V2-V1) = - (200*103 Pa)(0.1 m3) = -20 kJ
Path B:
v2
v2
nRT
V
dV = -nRT ln( 2 )
V
V1
v1
W = - ò pdv = - ò
v1
n = (200 kPa)(0.1 m3)/(300 K)(8.314 kPa m3) = 0.00802 kg mol
W = (0.00802 kg mol)(8.314 kJ/kg mol K)(300 K) ln2 = -13.86 kJ
Heat
Part of total energy flow across a system boundary that
is caused by a temperature difference between the
system and the surrounding
Heat is positive when transferred to the system
Heat maybe transfer by conduction, convection, or
radiation
Q = UADT ; A is area for heat transfer, U is an empirical
.
coefficient, DT is the effective temperature difference
between the system and its surroundings, and is the
rate of heat transfer.
Q&
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Kinetic Energy (K)
Energy caused by system’s velocity
relative to the surrounding at rest.
K = 1/2mv2
Specific kinetic energy (k) is the kinetic
energy per unit mass
k = K/m
Example 3.2 : Kinetic
Energy
Water is pumped from a storage tank into
a tube of 3.00 cm inner diameter at the rate
of 0.001 m3/s. See Figure. What is the
specific kinetic energy of the water?
3.00 cm ID
0.001 m3/s
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Solution
Basis: 0.001 m3/s of water
Assume that r = 1000 kg/m3
r = ½(3.00) = 1.50 cm
v = 0.001 m3
s
p(1.50)2 cm2
= 1.415 m/s
(100 cm)2
(1 m)2
k = 1 (1.415)2 1 N
1J
2 (s/m)2 (1 kg m/s2) 1 Nm
= 1.0 J/kg
Potential Energy (P)
Energy caused by body force exerted on
its mass by a gravitational or
electromagnetic field with respect to a
reference surface
P = mgh
Specific potential energy (p) is normalized
by mass
p = P/m = gh
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1304 211 Chemical Engineering Principles
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Example 3.3: Potential
Energy
Water is pumped from one reservoir to
another 300 ft away. The water level in the
second reservoir is 40 ft above the water
level of the first reservoir. What is the
increase in specific potential energy of the
water in Btu/lbm?
Solution
300 ft
40 ft
Let the water level in the first reservoir be the reference plane, then h = 40 ft
p = 32.2 ft/s2 40 ft
32.2 lbm ft
lbf s2
By Assist. Prof. Dr. Wipada Sanongraj
1 Btu
= 0.0514 Btu/lbm
778.2 ft lbf
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1304 211 Chemical Engineering Principles
and Calculations
Internal Energy (U)
Energy of molecule, atom, and subatom.
Internal energy per unit mass (u) can be
calculated from measurable variables such
as pressure, volume, temperature, and
composition.
u = u (T , V )
æ ¶u ö
æ ¶u ö ˆ
du = ç
÷ dT + ç
÷ dV
ˆ
¶
T
è
øVˆ
è ¶V ø T
By definition (¶u / ¶T )Vˆ is the heat capacity at constant
volume and the second term on the right-hand side of
the above equation is so small so that it can be
neglected.
Changes in the internal energy can be computed by
the following equation:
T2
u 2 - u1 = ò C v dT
T1
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1304 211 Chemical Engineering Principles
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Enthalpy (H)
H = U + pV
where p is the pressure and V is
the volume.
æ ¶Hˆ ö
æ ¶Hˆ ö
÷ dT + ç
÷
dHˆ = çç
÷
ç ¶P ÷ dP
è ¶T ø p
è
øT
T2
Hˆ 2 - Hˆ 1 = ò C p dT
T1
Ĥ = enthalpy per unit mass
(neglect second term of above
eq. and (¶Hˆ / ¶T ) is the heat
capacity at constant pressure.)
p
However, in processes operating at high pressures,
the second term on the right hand side of the above
eq. cannot necessarily be neglected.
Calculation of Enthalpy
Changes
Phase transition
enthalpy change
latent heat
Enthalpy changes in a single phase: sensible heat
Enthalpy changes for the phase transition: heat of fusion
and heat of vaporization
Heat of condensation is the negative of the heat of
vaporization.
Solid changes directly to vapor: heat of sublimation
By Assist. Prof. Dr. Wipada Sanongraj
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and Calculations
Enthalpy Computing
Heat capacity equation
Tables
Enthalpy charts
Computer databases
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Heat Capacity Equations
T2
DH = ò C p dT
T1
Units: kJ/(kg mol)(K) or Btu/(lb mol)(F)
For ideal gas mixtures, Cpavg = SxiCpi
Cp = a + bT + cT2
Example 3.4: Heat Capacity
Equation
Heat capacity equation for CO2 gas is Cp =
2.675*104 + 42.27T – 1.425*10-2T2
Cp is in J/kg mol K and T is in K. Convert
this equation into a form so heat capacity
will be in Btu/lb mol F with T in F
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1304 211 Chemical Engineering Principles
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J
1 Btu 1DK
kg mol DK 1055 J 1.8DR
1DR
1DF
0.4536 kg
1 lb
Next, substitute the relation between the
temperature in K and the temperature in F
TK = TR/1.8 = (TF + 460)/1.8
Carry out the mathematical operations, get
Cp = 8.702 + 4.66*10-3TF – 1.053*10-6TF2
Example 3.5: Calculation of DH for a
gas mixture using heat capacity
equation
The conversion of solid wastes to gas can be
accomplished in incinerators. Solid waste can be burned
to a gas of the following composition (on dry basis)
CO2 9.2%, CO 1.5%, O2 7.3%, and N2 82%. What is the
enthalpy difference for this gas per lb mol between the
bottom and the top of the stack if the temperature at the
bottom of the stack is 550 F and temperature at the top is
200 F?
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Solution
The heat capacity equations are ( in Btu/lb mol F)
N2: Cp = 6.895 + 0.7624*10-3T – 0.7009*10-7T2
O2: Cp = 7.104 + 0.7851*10-3T – 0.5528*10-7T2
CO2: Cp = 8.448 + 5.757*10-3T – 21.59*10-7T2 + 3.059*10-10T3
CO: Cp = 6.865 + 0.8024*10-3T – 0.7367*10-7T2
Basis: 1 lb mol of gas
Multiplying these equations by the respective mole fraction of
each component and adding them together can save time in
integration.
N2:Cp = 0.82(6.895 + 0.7624*10-3T – 0.7009*10-7T2)
O2: Cp = 0.073(7.104 + 0.7851*10-3T – 0.5528*10-7T2)
CO2: Cp = 0.092(8.448 + 5.757*10-3T – 21.59*10-7T2 + 3.059*1010T3)
CO: Cp = 0.015(6.865 + 0.8024*10-3T – 0.7367*10-7T2)
Cpnet = (7.053 + 1.2242*10-3T – 2.6124*10-7T2 + 0.2814*10-10 T3)dT
200
DH = ò Cpnet dT
550
DH = -2468.6-160.7+13.8-0.633 = -2616 Btu/lb mol gas
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Example 3.6
Calculate the enthalpy change for 1 kg
mol of N2 gas that is heated at constant
pressure of 100 kPa from 18 to 1100 oC
Solution
Because 100 kPa is 1 atm, the enthalpy of combustion gas
(Appendix D.6) can be used.
At 1100 oC (1373 K): DH = 34,715 kJ/kg mol (by
interpolation)
At 18 oC (291 K): DH = 524 kJ/kg mol
For 1 kg mol of N2
DH = 34,715 – 524 = 34,191 kJ/kg mol
By Assist. Prof. Dr. Wipada Sanongraj
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Example 3.7
Steam is cooled from 640 oF and 92 psia to
480 oF and 52 psia. What is DH in Btu/lb?
Solution
Use steam table and do the double interpolation first
between pressures at fixed temperature, and then
between temperatures at fixed pressures.
Steps:
1st: Get DH at P = 50, 55, 90, and 95 psia at each
temperatures of 600 and 700 oF.
2nd: Interpolate between P = 50 and 55 psia and temp 600
oF to get DH at P = 52 psia and T = 600 oF
3rd: Interpolate between P = 90 and 95 psia and temp 600
oF to get DH at P = 92 psia and T = 600 oF
4th: Interpolate between T = 600 and 700 oF and P = 52
psia to get DH at T = 640 oF and P = 52 psia
5th: Interpolate between T = 600 and 700 oF and P = 92
psia to get DH at T = 640 oF and P = 92 psia
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1304 211 Chemical Engineering Principles
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An example of the interpolation needed at 600 oF is
2*(1328.7-1328.4)/5 = 0.12
So, at p = 92 psia and T = 600 oF , DH = 1328.7-0.12 = 1328.6
The enthalpy change is
DH = 1272.8 – 1348.4 = -75.6 Btu/lb
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Example 3.8
Four kilograms of water at 27 oC and 200
kPa are heated at constant pressure until
the volume of the water becomes 1000
times the original value. What is the final
temperature of the water?
Solution
The initial specific volume is the saturated liquid water
volume at 300 K = 0.001004 m3/kg.
The final volume is = 0.001004*1000 = 1.004 m3/kg
At 200 kPa, using interpolation between 400 and 450 K
which covers the range of the specific volume of 1.004
m3/kg, we find T by solving;
(T2-T1)(DV)/(V2-V1) = DT
0.9024 + (1.025-0.9024)/(450-400)*(T-400) = 1.004 m3/kg
T = 400 + 41 = 441 K
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
General Energy Balance
{accumulation of energy within the
system} = {transfer of energy into system
through system boundary} – {transfer of
energy out of system through system
boundary} + {energy generation within
system} – {energy consumption within
system}
Energy Balance without
reaction for closed system
W=-
W=+ P2
K2
K1
U2
h2
U1
h1
P1
Q=+ Q=DE = Et2 - Et1= DU+DP+DK = Q + W
General process showing the system boundary and
energy transport across the boundary.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Energy Balances for open
system (without chemical
reaction)
Accumulation: DE = mt2(u+k+p)t2-mt1(u+k+p)t1
Energy transfer in with mass flow: (u1+k1+p1)m1
Energy transfer out with mass flow:
(u2+k2+p2)m2
Net transfer by heat flow in: Q
Net transfer by mechanical or electrical work in:
W
Net transfer by work to introduce and remove
mass: p1v1m1-p2v2m2
Application of the general
energy balance without
reaction
1) No mass transfer (closed or batch
system) (m1= m2=0): DE = Q+W
2) No accumulation (DE = 0), no mass
transfer (m1= m2= 0): Q = -W
3) No accumulation (DE = 0), but with mass
flow: Q+W = D(h+k+p)m
4) No accumulation, Q = 0, W = 0, k = 0, p =
0: DH = 0
By Assist. Prof. Dr. Wipada Sanongraj
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Special process names
associated with energy
balance
Isothermal (dT=0): constant-temperature process
Isobaric (dp = 0): constant pressure process
Isometric or isochoric (dV = 0): constant-volume
process
Adiabatic (Q = 0): no heat interchange (i.e., an
insulated system)
System is insulated
Q is very small in relation to the other terms
The process takes place so fast that there is no time
for heat to be transferred
Example 3.9 : Closed
system
10 lbs of CO2 at room temperature (80 oF)
are stored in a fire extinguisher having a
volume of 4 ft3. How much heat must be
removed from the extinguisher so that
40% of the CO2 becomes liquid?
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Solution
Closed system without reaction
Use CO2 chart in Appendix J to get properties
Specific volume of CO2 is 4/10 = 0.4 ft3/lb, hence CO2 is a
gas at the start, pressure is 300 psia and Dh = 160 Btu/lb
Basis: 10 lb CO2
In the energy balance: DE = Q + W
W = 0 (volume of the system is fixed, DK =DP = 0), Q = DU
= Dh – D(pv)
Find Dhfinal from the CO2 chart by following the constantvolume line of 0.4 ft3/lb to the spot where the quality is
0.6.
Dhfinal = 81 Btu/lb, pfinal = 140 psia
Q = (81-160)-{[(140)(144)(0.4)/778.2] –
[(300)(144)(0.4)/778.2]} = -67.2 Btu/lb (heat is removed)
Example 3.10
Argon gas in an insulated plasma deposition
chamber with a volume of 2 L is to be heated by
an electric resistance heater. Initially the gas,
which can be treated as an ideal gas, is at 1.5 Pa
and 300 K. The 1000-ohm heater draws current
at 40 V for 5 minutes (480 J of work is done by
the surrounding). What is the final gas
temperature and pressure at equilibrium? The
mass of the heater is 12 g and its heat capacity is
0.35 J/gK. Assume that the heat transfer to the
chamber from the gas at this low pressure and in
the short time period is negligible.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Solution
Closed system: DK=DP = 0
DE=Q + W = DU
Q = 0 so DE = W = DU
Basis: 5 minutes
Calculate mass of the gas: n = pV/RT
n = 1.5 Pa 2L 10-3 m3 1 gmol K
1L
8.314 Pa m3
300 K
= 1.203*10-6 gmol
Cv = Cp – R and since Cp = 5/2R, so Cv = 3/2R
DU = nCv(T-300)
480 J = (12g)(0.35 J/g K)(T-300) + (1.203*10-6) (3/2)(8.214)(T300)
T = 414.3 K
Final pressure is p2 = p1(T2/T1) = 1.5(414.3/300) = 2.07 Pa
Example 3.11
Air is being compressed from 100 kPa and 255 K
(where it has an enthalpy of 489 kJ/kg) to 1000
kPa and 278 K (where it has an enthalpy of 509
kJ/kg). The exit velocity of the air from the
compressor is 60 m/s. What is the power
required (in kW) for the compressor if the load is
100 kg/hr of air?
225 K
Dh = 489 kJ/kg
v1 = 0
100 kPa
By Assist. Prof. Dr. Wipada Sanongraj
278 K
Dh = 509 kJ/kg
v1 = 60
1000 kPa
112
1304 211 Chemical Engineering Principles
and Calculations
Solution
Basis: 100 kg of air = 1 hr
Simplify energy balance: DE = Q + W – D[(h+k+p)m]
The process is in steady state: DE = 0
m1=m2 = m
D(pm) = 0
Q = 0 by assumption
W = DH + DK
DH = (509-489)kJ/kg*100 kg = 2000 kJ
DK = 1/2m(v22-v12) = ½(100 kg)(602 m2/s2) = 180 kJ
W = 2180 kJ
Convert to power
kW = 2180 kJ/1 hr (=3600 sec) = 0.61 kW
Example 3.12
Water is being pumped from the bottom of a well 15 feet
deep at the rate of 200 gal/hr into a vented storage tank
to maintain a level of water in a tank 165 ft above the
ground. To prevent freezing in the winter a small heater
puts 30,000 Btu/hr into the water during its transfer
from the well to the storage tank. Heat is lost from the
whole system at the constant rate of 25,000 Btu/hr. What
is the temperature of the water as it enters the storage
tank, assuming that the well water is at 35 F? A 2-hp
pump is being used to pump the water. About 55% of
the rated horsepower goes into the work of pumping
and the rest is dissipated as heat to the atmosphere.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Solution
Open system with flow in and out
Material balance is 200 gal enter and 200 gal leave in an
hour
Energy balance is : DE = Q + W – D(h+k+p)m
Process is in the steady state, DE = 0
m2 = m1 = m
DK = 0 because will assume that v1 = v2 = 0, then 0 = Q + W
– D(h + p)m
At the top of the tank, DH = mDh = mCp(T2-35)
Total mass of water pumped is (200 gal/hr)(8.33 lb/1gal) =
1666 lb/hr
Potential energy change = (1666 lb/hr)(32.2
ft/s2)(180ft)/(32.2 ft lbm/s2 lbf)/(778 ft lbf) = 385.5 Btu
Heat lost by system is 25,000 Btu while the heater puts
30,000 Btu into the system, hence the net heat exchange =
Q = 30000 – 25000 = 5000 Btu
Rate of work being done on the water by the pump = w =
(2 hp)(0.55)(33,000 ft lbf/min hp)(60 min/hr)(1 Btu/778
ft lbf) = 2800 Btu/hr
DH can be calculated from: Q + W = DH + DP = 7414 Btu
Heat capacity of liquid water assumed to be constant and
equal to 1.0 Btu/lb F
7414 = DH = 1666(1.0)DT
DT = 4.5 F, T = 39.5 F
By Assist. Prof. Dr. Wipada Sanongraj
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Energy Balance with
Reaction
Standard heat of formation
For single species and mixture, specific
enthalpy change from the reference state
T
DHA = DH ofA +
òC
dT
pA
Tref
s
s
T
DHmixture = åni DH of i + å ò nC
i pi dT
i =1
i =1 Tref
Heat of formation
Sensible heat
For example, suppose we have species 1 and 2
enter system, react, and species 3 and 4 leave.
Then
DHout -DHin = (n3DHof 3 + n4DHof 4 ) - (n1DHof 1 + n2DHof 2 )
Tout
Tin
Tref
Tref
+ ò (n3Cp3 + n4Cp4 )dT -
ò (nC
1 p1
- n2Cp2 )
±Enthalpies associated with phase change
DH rxn = (
å
products
By Assist. Prof. Dr. Wipada Sanongraj
ni DH f i -
å n DH
i
fi
)
reac tan ts
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1304 211 Chemical Engineering Principles
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Example 3.13
Calculate DHrxno for the following reaction of 4 g mol of NH3
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Basis: 4 g mol of NH3
Data
NH3(g)
DHfo per mole-46.191
At 25 C 1atm
(kJ/g mol)
O2(g)
0
NO(g)
+90.374
H2O(g)
-241.826
DHrxno = [4(90.374)+6(-241.826)]-[5(0) + 4(-46.191)]
= -904.696 kJ/4 g mol NH3
Example 3.14
If the standard heat of formation for H2O(l)
is -285.838 kJ/g mol and the heat of
evaporation is 44.012 kJ/g mol at 25 OC
and 1 atm, what is the standard heat of
formation of H2O(g)?
By Assist. Prof. Dr. Wipada Sanongraj
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Solution
DHrxn=SDHf,product-SDHf,reactants
A: H2(g) + 1/2O2(g)
H2O(l)
DHrxno=-285.838 kJ/g mol
B: H2O(l)
H2O(g)
DHvap = 44.012 kJ/g mol
A+B: H2(g) + 1/2O2(g)
H2O(l)
o
o
DHrxn + DHvap = DHrxn H2O(g) = DHfoH2O(g)
= - 241.826 kJ/ gmol
Example 3.15
An iron pyrite containing 85% FeS2 and 15%
gangue (inert dirt, rock) is roasted with an
amount of air equal to 200% excess air according
to the reaction
4FeS2 + 11O2
2Fe2O3 + 8SO2
In order to produce SO2. All the gangue plus the
Fe2O3 end up in the solid waste product (cinder),
which analyzes 4% FeS2. Determine the heat
transfer per kg of ore to keep the product stream
at 25 oC if the entering steams are at 25 oC
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Solution
Basis: 100 kg of pyrite ore
MW of Fe(55.85), Fe2O3(159.7), FeS2(120)
Product:P
Mol%
SO2
x4
O2
x5
N2
x6
200% xs air
Mol%
N2 79
O2 21
Total 100
Ore:F
Wt%
FeS2
85
Gangue 15
Cinder:C
Wt%
Gangue x1
Fe2O3 x2
FeS2
x3
Total x1+x2+x3
The excess air is: Mol FeS2 = (85/120) = 0.7083 kg mol
Required O2 = 0.7083(11/4) = 1.9479 kg mol
Excess O2 = 1.9479(2.0) = 3.8958
Total O2 in = 5.8437 kg mol
Total N2 in = 5.8437(79/21) = 21.983 kg mol
Element mass balance are
In
Out
Gangue
15
x1
N2(kg mol)
21.983 x6
S (kg mol)
2(85/120) x4+ (x3/120)(2)
Fe (kg mol)
1(85/120)
(x2/159)2 + (x3/120)(1)
O2 (kg mol)
5.8437 x4 + x5 + (x2/159.7)(1.5)
We know x3/(x1+x2+x3) = 0.04
By Assist. Prof. Dr. Wipada Sanongraj
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The solution of these equation is
In P
In C
SO2 = 1.368 kg mol
Gangue = 15 kg
O2 = 3.938
Fe2O3 = 54.63 kg = 0.342 kg mol
N2 = 21.983
FeS2 = 2.9 kg = 0.0242 kg mol
General energy balance reduce to DE = 0, DP = 0, DK = 0, W = 0,
so Q = DH. Choose reference state to be 25 oC and 1 atm with
the result that all the sensible heat terms become zero
Q = SniDHio - SniDHio
Products
Comp
10-3 g mol
FeS2
Fe2O3
N2
O2
SO2
0.0242
0.342
21.9983
3.938
1.368
reactants
DHfo (kJ/g
mol)
-177.9
-822.156
0
0
-296.9
niDH fi o(kJ)
-4.305
-281.177
0
0
-406.159
-691.641
10-3 g
mol
0.7083
0
21.983
5.8437
0
DHfo (kJ/g
mol)
-177.9
-822.156
0
0
-296.9
niDH fio
(kJ)
-126.007
0
0
0
0
-126.007
Q = [-691.641-(-126.007)](103) = -565.634*103 kJ/100 kg ore
Q = -5.656*103 kJ/kg ore
By Assist. Prof. Dr. Wipada Sanongraj
119
1304 211 Chemical Engineering Principles
and Calculations
Chapter 4 : Gases, Vapors,
Liquids, and Solids
Gases, Vapors, Liquids, and Solids
The Ideal Gas Law
pV = nRT
Where p = absolute pressure of the gas
V = Total volume occupied by the gas
n = number of moles of the gas
R = ideal gas constant in appropriate units
T = absolute temperature of the gas
Sometimes the ideal gas law is written as:
pVˆ = RT
Vˆ = specific volume of the gas (vol. per mole)
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Phase-rule variables are variables of the kind with
which the phase rule is concerned and they are
intensive properties of the system.
Intensive properties are the ones that do not
depend on the quantity of the material present. The
specific (per unit mass) values are intensive
properties.
Extensive properties are the ones that depend on
how much material we have. The total quantities
are extensive properties.
An example of the use of the phase rule is the ideal
gas law, PV = nRT. In order to be able to determine
the remaining one unknown, you might conclude
that F = 3. However, if we apply the phase rule for
a single phase P = 1 and for a pure gas C = 1, so
that
F=C–P+2=1–1+2=2
How can we explain this apparent paradox with our
previous statement? Since the phase rule is
concerned with intensive properties only, the
following are phase-rule variables in the ideal gas
law:
P, T, and Vˆ
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Thus the ideal gas law would be written as
pVˆ = RT
We can see that by specifying two intensive variables
(F = 2), the third can be calculated.
Example 4.1 Application of the phase rule
Calculate the number of degrees of freedom from
the phase rule for the following materials at
equilibrium:
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
In many processes going from an initial state to a
final state, you can use the ratio of the ideal gas
laws in the respective states and eliminate R as
follows:
p 1V 1
n RT
= 1 1
p 2V 2
n 2 RT 2
or
æ p1
ç
çp
è 2
öæ V1 ö æ n1 öæ T1 ö
÷ç ÷ = ç ÷ç ÷
÷çè V ÷ø çè n ÷øçè T ÷ø
2
2
ø 2
Example 4.2: Use of Standard Conditions
Calculate the volume, in cubic meters, occupied by
40 kg of CO2 at standard conditions.
Solution :
Basis: 40 kg of CO2
40 kg CO2 1 kgmol CO2 22.4 m3 CO2 = 20.4 m3 CO2 at S.C.
44 kg CO2
By Assist. Prof. Dr. Wipada Sanongraj
1 kgmol CO2
123
1304 211 Chemical Engineering Principles
and Calculations
Example 4.3 : Calculation of Gas Density
What is the density N2 at 27 oC and 100 kPa in SI units?
Solution :
Basis: 1 m3 of N2 at 27 oC and 100 kPa
1 m3 273 K 100 kPa
1kgmol
28 kg = 1.123 kg
300 K 101.3 kPa 22.4 m3 1kgmol
density = 1.123 kg/m3 of N2 at 27 oC and 100 kPa
The specific gravity of a gas is usually defined as
the ratio of the density of the gas at a desired
temperature and pressure to that of air at a certain
temperature and pressure.
Example 4.4 : Specific gravity of a gas
What is the specific gravity of N2 at 80 oF and 745
mm Hg compared to air at 80 oF and 745 mm Hg?
Basis : 1 ft3 of air at 80 oF and 745 mm Hg
1 ft3 492 oR 745 mmHg 1lbmol 29 lb = 0.0721 lb/ft3
540 oR 760 mm Hg 359 ft3 1 lbmol
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Basis : 1 ft3 of N2 at 80 oF and 745 mm Hg
1 ft3 492 oR 745 mmHg 1lbmol 28 lb = 0.0697 lb/ft3
540 oR 760 mm Hg 359 ft3 1 lbmol
lbN 2 / ft 3 at 80 o F , 745 mmHg
0.0697
(sp.gr.)N 2 =
= 0.967
0.0721
lbair / ft 3 air at 80 o F ,745 mmHg
Ideal Gas Mixtures and Partial Pressure
The partial pressure of gas i defined by Dalton, pi,
namely the pressure that would be exerted by a
single component in a gaseous mixture if it existed
by itself in the same volume as occupied by the
mixture and at the same temperature of the mixture
is
piVtotal = niRTtotal
piVtotal
n RT
= i total
ptotalVtotal ntotal RTtotal
or
p i = p total
ni
= p total y i
n total
yi is the mole fraction of component i
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Example 4.5 : Calculation of Partial Pressures
from a Gas Analysis
A flue gas analyzes 14% of CO2, 6% of O2, and
80% of N2. It is a t 400 oF and 765 mmHg pressure.
Calculate the partial pressure of each component.
Use pi = ptyi
Basis : 1.00 kg (or lb) mol of flue gas
Component
CO2
O2
N2
Total
kg (or lb) mol
0.140
0.060
0.800
1.000
p (mm Hg)
107.1
45.9
612.0
765.0
Material Balances Involving gases
Example 4.6 : Material Balance with Combustion
A gas produced by gasifying wood chips analyzes
6.4 % CO2, 0.1% O2, 39%CO, 51.8% He, 0.6%CH4,
and 2.1% N2. It enters the combustion chamber at
90 oF and a pressure of 35.0 in Hg. And is burned
with 40% excess air (dry) which is at 70 oF and the
atmospheric pressure of 29.4 in Hg, 10% of the CO
remains unburned. How many cubic feet of air are
supplied per cubic foot of entering gas? How many
cubic feet of product are produced per cubic foot of
entering gas if the exit gas is at 29.4 in Hg and 400
oF?
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
90 oF and 35.0 in Hg
Gas 100 lb mol
400 oF and 29.4 in Hg
Product
Combustion
Comp.
CO2
O2
CO
H2
CH4
N2
% = mol
O2 req.
6.4
0.1
39.0
51.8
0.6
2.1
100.0
(0.1)
19.5
25.9
1.2
46.5
Air A = ?
lb mol
O2 0.21
N2 0.79
1.00
40% xs
70 oF and 29.4 in Hg
P lbmol
CO2 ?
H2O ?
CO ?
O2 ?
N2 ?
Basis : 100 lbmol synthesis gas, % = lbmol
The entering air can be calculated from the
specified 40% excess air, the reactions for
complete combustion are
CO +
1
O 2 ® CO 2
2
1
H 2 + O2 ® H 2 O
2
CH4 + 2O2 ®CO2 + 2H2O
The excess is
Excess O2 : 0.4(46.5) = 18.6 lb mol
Total O2 : 46.5 + 18.6 = 65.10 lb mol
By Assist. Prof. Dr. Wipada Sanongraj
127
1304 211 Chemical Engineering Principles
and Calculations
N2 in is 65.10(79/21) = 244.9 lb mol
Total moles of air in are 244.9 + 65.10 = 310 lb mol.
Element balances in moles
In
Out
N2: 2.1 + 244.9
= nN2
H2: 51.8 + 0.6(2)
= nH2O
C: 6.4+ 39.0+0.6
= nCO2 + 0.10(39.0)
O2: 6.4+0.1+0.5(39)+65.1 = nO2 + nCO2 + 0.5(nH2O
+nCO)
From 10% of CO remains unburned so nCO = 3.9%
The solutions of these equations is
nN2 = 247.0
nCO = 3.9
nCO2 = 42.1
nH2O = 53.0
nO2 = 20.55
The total moles exiting sum up to be 366.55 lbmol.
Then we will convert the lbmol of air and product
into the volumes requested:
Tgas = 90 + 460 = 550 oR
Tair = 70 + 460 = 530 oR
Tproduct = 400 + 460 = 860 oR
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
ft 3 of gas
100 lbmol entering gas (359 ft 3 at S .C.)(550 o R)(29.92 in.Hg )
=
1 lbmol (492 o R)(35.0 in. Hg )
= 343 *102
ft 3 of air
310 lbmol air (359 ft 3 at S .C.)(530 o R)(29.92 in.Hg )
=
1 lbmol (492 o R)(29.4 in. Hg )
= 1220 *102
ft 3 of product
366.55 lbmol P (359 ft 3 at S .C.)(860 o R)(29.92 in.Hg )
=
1 lbmol (492 o R)(29.4 in. Hg )
= 2340 *102
The answers to the questions are
1220 * 10 2
ft 3 air at 530 o R and 29.4 in. Hg
= 3.56 3
343 * 10 2
ft gas at 550 o R and 35.0 in. Hg
2340 *102
ft 3 product at 860 o R and 29.4 in. Hg
= 6.82
343 *102
ft 3 gas at 550 o R and 35.0 in. Hg
By Assist. Prof. Dr. Wipada Sanongraj
129
1304 211 Chemical Engineering Principles
and Calculations
Critical State, Reduced Parameters, and Compressibility
The law of corresponding states expresses the idea
that in the critical state all substances should
behave alike.
The critical state for the gas-liquid transition is the
set of physical conditions at which the density and
other properties of the liquid and vapor become
identical.
This point, for a pure component (only), is the highest
temperature at which liquid and vapor can exist in
equilibrium.
A supercritical fluid, a compound in a state above the
critical point, combines some of the properties of both
gases and liquids. Supercritical fluids are used to
replace the void left by solvents such as
trichloroethylene and methylene chloride.
Figure 1 The region of existence of solid, liquid, gaseous, and supercritical
water. At the triple point solid, liquid, and gas are all in equilibrium
(Himmelblau, 1996)
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Reduced parameters are corrected or normalized
conditions of temperature, pressure, and volume,
normalized by their respective critical conditions.
Tr = T/Tc
Pr = P/Pc
V
r
=
Vˆ
Vˆ c
The common way to modify the ideal gas law to a
real gas law is to insert an adjustable coefficient, z,
the compressibility factor in it. The compressibility
factor is a factor that compensates for the
nonideality of the gas. Thus the ideal gas law
becomes a real gas law, a generalized equation of
state.
If the compressibility factor is plotted for a given
temperature against the pressure for different
gases, we obtain plots as shown in Figure 2.
However, if the compressibility is plotted against the
reduced pressure as a function of the reduced
temperature, then for most gases the
compressibility values at the same reduced temp.
and pressure fall at about the same point.
By Assist. Prof. Dr. Wipada Sanongraj
131
1304 211 Chemical Engineering Principles
and Calculations
Figure 2 (a) Compressibility factor as a function of temperature and pressure
(b) compressibility as a function of reduced temperature and reduced
pressure (Himmelblau, 1996).
Figure 3 (a) Generalized compressibility chart showing the
respective portions of the subsequent expanded charts
(Himmelblau, 1996)
By Assist. Prof. Dr. Wipada Sanongraj
132
1304 211 Chemical Engineering Principles
and Calculations
Figure 3 (b) Generalized compressibility chart, very low
reduced pressure (Himmelblau, 1996).
Figure 3 (c) Generalized compressibility chart, low pressure
(Himmelblau, 1996).
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Figure 3 (d) Generalized compressibility chart, medium
pressure (Himmelblau, 1996).
Figure 3 (e) Generalized compressibility chart, high
pressure (Himmelblau, 1996).
By Assist. Prof. Dr. Wipada Sanongraj
134
1304 211 Chemical Engineering Principles
and Calculations
Figure 3 (f)
Figure 3 shows the generalized compressibility charts
or z-factor chart prepared by Nelson and Obert. These
charts are based on 30 gases. Figure 3 (b) and (c)
represent z for 26 gases (excluding H2, He, NH3, H2O)
with a maximum deviation of 1%, and H2 and H2O within
a deviation of 1.5%. Figure 3 (d) is for 26 gases and is
accurate to 2.5%, while Figure 3 (e) is for 9 gases and
errors can be as high as 5%. For H2 and He only,
corrections to the actual critical constants are used to
give pseudocritical constants which enable us to use
Figure 3 (a-e) for these two gases as well with minimum
error.
Tc' = Tc + 8 K
Pc' = Pc + 8 atm
By Assist. Prof. Dr. Wipada Sanongraj
135
1304 211 Chemical Engineering Principles
and Calculations
Figure 3 (f) is a unique chart that, by having several
parameters plotted simultaneously on it, helps us
avoid trial-and-error solutions or graphical solutions
of real gas problems. One of these helpful
parameters is the ideal reduced volume defined as:
V ri
Vˆ
=
Vˆci
Where Vci is the ideal critical volume (not the
experimental value of the critical volume), or
V ci =
RT
Pc
c
All we need to know to use these charts are the
critical temperature and the critical pressure for a
pure substance. The value of z = 1 represents
ideality and the value z = 0.27 is the compressibility
factor at the critical point.
By Assist. Prof. Dr. Wipada Sanongraj
136
1304 211 Chemical Engineering Principles
and Calculations
Example 4.7 : Use of the compressibility factor
In spreading liquid ammonia fertilizer, the charges
for the amount of NH3 are based on the time
involved plus the pounds of NH3 injected into the
soil. After the liquid has been spread, there is still
some ammonia left in the source tank (volume = 120
ft3 ), but in the form of a gas. Suppose that your
weight tally, which is obtained by difference, shows a
net weight of 125 lb of NH3 left in the tank at 292
psig. Because the tank is sitting in the sun, the
temperature in the tank is 125 oF. Your boss
complains that his calculations sow that the specific
volume of the gas is 1.2 ft3/lb, and hence that there
are only 100 lb of NH3 in the tank. Could he be
correct?
Basis: 1 lb of NH3
Apparently, your boss used the ideal gas law in
getting his figure of 1.2 ft3/lb of NH3 gas.
R = 10.73
( psia)( ft 3 )
(lbmol )( o R)
P = 292 + 14.7 = 306.7 psia
T = 125 oF + 460 = 585 oR
n=
1lb
17lb / lbmol
1
(10 . 73 )( 585 )
RT
17
ˆ
V =
=
= 1 . 20 ft 3 / lb
P
306 . 7
By Assist. Prof. Dr. Wipada Sanongraj
137
1304 211 Chemical Engineering Principles
and Calculations
However, he should have used the compressibility
factor, because NH3 does not behave as an ideal
gas under the observed conditions of temperature
and pressure. Let us again compute the mass of
gas in the tank this time using
PV = znRT
Tc = 405.5 K = 729.9 R
Pc = 111.3 atm = 1636 psia
Then, since z is the function of Tr and Pr
Tr =
T
585 o R
=
= 0 . 801
T c 729 . 9 o R
Pr =
P
306 . 7 psia
=
= 0 . 187
Pc
1636 psia
From the Nelson and Obert chart, Figure 3 (c), we
can read z = 0.855.
Vˆ = 1 . 2 ( ft 3 / lb ) ideal ( 0 .855 )
= 1 .03 ft 3 / lbNH
3
Mass of NH3 = (120 ft3)(1lb NH3/1.03 ft3)
By Assist. Prof. Dr. Wipada Sanongraj
138
1304 211 Chemical Engineering Principles
and Calculations
Equation of State
Another way to predict p, V, n, and T for real gases
Equations of state are formulated by collecting
experimental data and calculating the coefficients
in a proposed equation by statistical fitting.
Van der Waals is the first one who generated
equation of state considering 2 effects that make
real gas behaves different from ideal gas.
1)Interaction between molecules called Van der
Waals force causing the reduction of pressure.
2) Volume of molecule is not negligible as
compared to volume of gas.
Van der Waals’ Equation can be expressed as:
p =
nRT
n 2a
V - nb
V 2
However, if you want to solve for V (or n), you can see
that the equation becomes cubic in V (or n):
æ
nRT ö 2 n 2 a
n 3 ab
f (V ) = V 3 - ç nb +
V
+
V
=0
÷
p ø
p
p
è
when a and b are the van der Waals constants obtained
from experiments.
By Assist. Prof. Dr. Wipada Sanongraj
139
1304 211 Chemical Engineering Principles
and Calculations
Example 4.8: Application of Van der Waals’ Equation
A cylinder 0.15 m3 in volume containing 22.7 kg of
propane C3H8 stands in the hot sun. A pressure
gauge shows that the pressure is 4790 kPa gauge.
What is the temperature of the propane in the
cylinder? Use van der Waals equation.
Basis : 22.7 kg of propane
The van der Waals constants are
a = 9.24*106 atm(cm3/gmol)2 b = 90.7 cm3/gmol
n 2a
p +
(V - n b ) = n R T
V 2
All the additional information you need is as follows:
P=
(4790 + 101) kPa ´ 1 atm
= 48.3 atm abs
101.3 kPa
R = 8 2 .0 6
n=
( c m 3 )( a tm )
( g m o l )( K )
22.7 kg
= 0.516 kgmol propane
44 kg / kgmol
é
( 0 .5 1 6 ´ 1 0 3 ) 2 ( 9 .2 4 ´ 1 0 6 ) ù
6
ê 4 8 .3 +
ú [ 0 .1 5 0 ´ 1 0
( 0 .1 5 ´ 1 0 6 ) 2
ë
û
- ( 0 .5 1 6 ´ 1 0 3 ) ( 9 0 .7 ) ] = ( 0 .5 1 6 ´ 1 0 3 ) (8 2 .0 6 ) ( T K )
T = 384 K
By Assist. Prof. Dr. Wipada Sanongraj
140
1304 211 Chemical Engineering Principles
and Calculations
Example 4.9: Solution of Van der Waals’ Equation for V
Given the values of
P = 679.7 psia
a = 3.49*104 psia(ft3/lbmol)2
n = 1.136 lbmol
R = 10.73 (psia)(ft3)/(lbmol)(oR)
T = 683 oR
b = 1.45 (ft3/lbmol)
Solve for the volume of the vessel
Write van der Waals’ equation as a cubic equation
in one unknown variable, V.
æ pnb + nRT ö 2 n 2 a
n 3 ab
f (V ) = V 3 - ç
V=0
÷V +
p
p
p
è
ø
Let us apply Newton’s method to obtain the desired
root:
f (V k )
(a)
V
=V k +1
k
f '(V k )
Where f’(Vk) is the derivative of f(V) with respect to V
evaluated at Vk:
æ pnb + nRT
f '(V ) = 3V 2 - 2 ç
p
è
ö
n2a
V
+
÷
p
ø
You can obtain a reasonably close approximation to
V (or n) in many cases from the ideal gas law, useful
at least for the first trial in which k = 0 in Eq. (a)
By Assist. Prof. Dr. Wipada Sanongraj
141
1304 211 Chemical Engineering Principles
and Calculations
nRT
10.73( psia)( ft 3 )
1
Vo =
= 1.136lbmol ´
´ 683o R ´
o
P
(lbmol)( R)
679.7 psia
= 12.26 ft3 at 679.7 psia and 683 oR
The second and subsequent estimates of V will be
calculated using Eq.(a):
V1 = V
o
-
f (V o )
f '( V o )
(679.7)(1.137)(1.45) + (1.137)(10.73)(683)
(12.26)2
679.7
(1.137)2 (3.49 ´104 )
(1.137)3 (3.49 ´104 )(1.45)
+
(12.26) = 738.3
679.7
679.7
f (Vo ) = (12.26)3 -
2[(679.7)(1.137)(1.45) + (1.137)(10.73)(683)]
(12.26)
679.7
(1.137)2 (3.49 ´104 )
+
= 216.7
679.7
f '(Vo ) = 3(12.26)2 -
V1 = 12.26 - 738.3/216.7 = 8.85
On the next iteration
V 2 = V1 -
f (V 1 )
f '(V 1 )
The final solution is 5.0 ft3 at 679.7 psia and 683 oR.
By Assist. Prof. Dr. Wipada Sanongraj
142
1304 211 Chemical Engineering Principles
and Calculations
Gaseous Mixtures
Use Kay’s method of pseudocritical values to
calculate the pseudo-reduced values and predict
P, V, T, and n via the compressibility factor.
In Kay’s method, pseudocritical values for mixtures
of gases are calculated on the assumption that
each component in the mixture contributes to the
pseudocritical value in the same proportion as the
number of moles of that component. Thus, the
pseudocritical values are computed as follows:
Pc' = PcA y A + PcB y B + ...
Tc' = TcA y A + TcB y B + ...
Where yi is the mole fraction
Pc’ is the pseudocritical pressure
Tc’ is the pseudocritical temperature
The pseudoreduced variables can be calculated
as follows:
P
P = '
Pc
'
r
By Assist. Prof. Dr. Wipada Sanongraj
Tr' =
T
Tc'
143
1304 211 Chemical Engineering Principles
and Calculations
Example 4.10: Calculation of P-V-T Properties for
Real Gas Mixture
A gaseous mixture has the following composition (in
mole percent)
Methane
Ethylene
Nitrogen
20
30
50
at 90 atm pressure and 100 oC. Compare the volume
per mole as computed by the method of:
a)The perfect gas law
b) the pseudoreduced technique (Kay’s method)
Basis : 1 gmol of gas mixture
Additional data needed are:
Component
Methane
Ethylene
Nitrogen
Tc (K)
191
283
126
Pc (atm)
45.8
50.9
33.5
R = 82.06 (cm3-atm)/(gmol-K)
By Assist. Prof. Dr. Wipada Sanongraj
144
1304 211 Chemical Engineering Principles
and Calculations
a) Perfect gas law:
nRT 1(82.06)(373)
Vˆ =
=
P
90
= 340 cm 3 at 90 atm and 373 K
b) According to Kay’s method, we first calculate the
pseudocritical values for the mixture
Pc' = PcA y A + PcB y B + PcC yC
= ( 45.8)(0.2) + (50.9)(0.3) + (33.5)(0.5)
= 41.2 atm
Tc' = TcA y A + TcB y B + TcC yC
= (191)(0.2) + (283)(0.3) + (126)(0.5)
= 186 K
Then we calculate the pseudoreduced values for the
mixture
Pr' =
P
90
=
= 2.18
Pc' 41.2
Tr' =
T 373
=
= 2.01
'
Tc 186
By Assist. Prof. Dr. Wipada Sanongraj
145
1304 211 Chemical Engineering Principles
and Calculations
By using these two parameters and the
generalized compressibility chart, we will get z =
0.965. Thus
znRT 0.965(1)(82.06)(373)
Vˆ =
=
P
90
= 328 cm 3 at 90 atm and 373 K
Vapor Pressure and Liquids
Figure 4 Vapor pressure curve for water
(Himmelblau, 1996).
By Assist. Prof. Dr. Wipada Sanongraj
146
1304 211 Chemical Engineering Principles
and Calculations
Change of Vapor Pressure with Temperature
Because the function of vapor pressure (P) vs. T is
not a linear function so we use the Antoine equation
to predict P from T
ln( P*) = A -
B
C +T
Where A, B, C = constants for each substance
T = temperature, K
Example 4.11 : Vaporization of Metals for Thin Film
Deposition
The following figure shows evaporation from a boat
placed in a vacuum chamber. The boat made of
tungsten has a negligible vapor pressure at 972 oC,
the operating temperature for the vaporization of Al
(which melts at 660 oC and fills the boat). The
approximate rate of evaporation is given in g/(m2)s
by
*
1/ 2
m = 0.437
P ( MW )
T 1/ 2
Where P* is the pressure in kPa and T is the
temperature in K.
What is the vaporization rate for Al at 972 oC in
g/(cm2)s?
By Assist. Prof. Dr. Wipada Sanongraj
147
1304 211 Chemical Engineering Principles
and Calculations
Target
Vacuum
chamber
electrode
To pump
electrode
Solution
We have to calculate P* for Al at 972 oC. The Antoine
equation is suitable if data are known for the vapor
pressure of Al. Considerable variation exists in the
data for Al at high temperature, but we will use A =
8.779, B = 1.615*104, and C = 0 with P* in mm Hg
and T in K.
1.615*104
972 + 273
= 0.0154 mmHg (0.00201 kPa)
*
ln P972
= 8.799 o
C
m = 0 .4 3 7
( 0 .0 0 2 0 1) ( 2 6 .9 8 )
(9 7 2 + 2 7 3 )
= 1 .3 * 1 0
By Assist. Prof. Dr. Wipada Sanongraj
-4
g /(c m
2
1
2
1
2
)( s )
148
1304 211 Chemical Engineering Principles
and Calculations
Another way to relate vapor pressure to temperature
is by a graphical technique. The curvature for P* vs.
T can be straightened out by a special plot known as
a Cox chart.
How to make the Cox chart
1. Mark on the horizontal scale values of logP* as to
cover the desired range of P*
2. Next draw a straight line on the plot at a suitable
angle, say 45o, that covers the range of P*
3. To calibrate the vertical axis in common integers
such 25, 50, 100, 200 degrees, and so on, you
use a reference substance, namely water. For the
first integer, say 100 oF, you look up the vapor
pressure of water in the steam tables, or calculate
it from the Antoine equation, to get 0.9487 psia.
Locate this value on the horizontal axis, and
proceed vertically until you hit the straight line.
Then proceed horizontally until you hit the vertical
axis. Mark the scale there as 100 oF.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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4. Pick the next temperature, say 200 oF.
5. Continue as in 3 and 4 until the vertical scale is
established over the desired range for the
temperature.
Figure 5 Cox chart (Himmelblau, 1996).
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Saturation
Saturate : method, ew first calculate the .779, B = 1.615*10
E 1996).•n volume containing .2 7 kg of
Saturated Air : RaaodDr o
When any pure gas (or a gaseous mixture) comes in
contact with a liquid, the gas will acquire molecules
from the liquid. If contact is maintained for a
considerable length of time, vaporization continues
until equilibrium is attained (no more liquid will
vaporize into the gas phase), the gas is then said to
be saturated with the particular vapor at the given
temperature.
PairV nairRT
=
PH2OV nH2O RT
Figure 6 Evaporation of water at constant
pressure and temperature of 65 oC
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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Figure 7 Change of partial and total pressures during
the vaporization of water into air at constant
temperature.
Dew point for the mixture of pure vapor and
noncondensable gas means the temperature at
which the vapor just starts to condense when
cooled at constant pressure.
The dew point is the temperature you must cool
air at constant pressure in order for that air mass
to become saturated. This does not necessarily
mean a cloud or rain drop will form. When water
does condense into a cloud or rain drop, it means
that the air temperature and dew point
temperature are the same, and this would mean
we have 100% relative humidity. High dew points
mean high moisture content of the air, which
often translates to muggy and uncomfortable
conditions.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
The most common confusion over the dew point
is that it is the temperature at which condensation
forms. This is not necessarily true.
Actually condensation is always occurring in our
air. At the dewpoint temperature however, is
when condensation overtakes evaporation, and
this is a step in the process of forming dew,
clouds, rain, fog, basically water droplets.
Example 4.12 : Saturation
What is the minimum number of cubic meters of
dry air at 20 oC and 100 kPa necessary to
evaporate 6.0 kg of ethyl alcohol if the total
pressure remains constant at 100 kPa and the
temperature remains 20 oC? Assume that the air is
blown through the alcohol to evaporate it in such a
way that the exit pressure of the air-alcohol mixture
is at 100 kPa.
Solution
Assuming that the process is isothermal
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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The additional data needed are
MW. of ethyl alcohol (C2H5OH) = 46.07
P*alcohol
ln( P*) = A -
From
B
C +T
A = 18.5242 B = 3578.91
C = -50.50
P*alcohol = 43.647 mm Hg (5.82 kPa)
We will use P*alcohol = 5.93 kPa
20 oC
100 kPa
Air
100 kPa
Saturated
air-alcohol
mixture
Alcohol
6 kg
Basis : 6.0 kg of alcohol
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
The ratio of moles of ethyl alcohol to moles of air
in the final gaseous mixture is the same as the
ratio of the partial pressures of these two
substances. Since we know the moles of alcohol,
we can find the number of moles of air.
*
Palcohol
n
= alcohol
Pair
nair
And once we know the number of moles of air we
can apply the ideal gas law.
Pair = Ptotal – P*alcohol = (100-5.93) kPa = 94.07 kPa
nair =
6 kg alcohol ´ 1 kgmol alcohol ´ 94.07 kgmol air
46.07 kg alcohol ´ 5.93 kgmol alcohol
= 2.07 kgmol air
Using Ideal law
Vair
2.07kgmolair ´ 8.314(kPa ´ m 3 ) ´ 293K
=
(kgmol )( K ) ´ (100kPa)
= 50.3 m3 at 20 oC and 100 kPa
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Another way to solve for Vair at 20 oC and 100 kPa
First…solve for Vair at 20 oC and 94.07 kPa
Vair =
2.07kgmolair ´ 8.314(kPa ´ m 3 ) ´ 293K
( kgmol )( K ) ´ (94.07kPa)
= 53.5 m3 at 20 oC and 94.07 kPa
Vair =
53.5 m 3 ´ 94.07 kPa
100 kPa
= 50.3 m3 at 100 kPa and 20 oC
Example 4.13 : Smokestack Emission and Pollution
A local pollution-solution group has reported the
Simtron co. boiler plant as being an air polluter and
has provided as proof photographs of heavy
smokerstack emissions on 20 different days. As
the chief engineer for the Simtron Co., you know
that your plant is not a source of pollution because
you burn natural gas (essentially methane) and
your boiler plant is operating correctly. Your boss
believes the pollution-solutions group has made an
error in identifying the stack- it must belong to the
company next door that burns coal. Is he correct?
Is the pollution-solutions group correct?
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Solution
Methane contains 2 kg mol of H2 per kgmol of C,
while coal contains 71 kg of C per 5.6 kg of H2 in
100 kg o coal. The coal analysis is equivalent to
71 kg C ´
5.6 kg H 2 ´
1 kgmol C
= 5.92 kgmol C
12 kg C
1 kgmol H 2
= 2.78 kgmol H 2
12 kg H 2
or a ratio of 2.78/5.92 = 0.47 kgmol of H2/kgmol of C.
Suppose that each fuel burns with 40% excess air
and that combustion is completed. We can
compute the mole fraction of water vapor in each
stack gas.
Products
Known Fuel
O2 0.21
N2 0.79
1.00
Basis : 1 kg mol C
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Natural Gas
CH 4 + 2O2 ® CO2 + 2 H 2 O
Composition of combustion gases (kgmol)
Components
kgmol
C
1.0
H2
2.0
CO2 H2O Excess O2
1.0
2.0
Air
Total
1.0
Req. O2: 2
N2
2.0
Excess O2: 2(0.4) = 0.8
0.8
10.5
0.80
10.5
N2: 2.8(79/21) = 10.5
The total kilogram moles of gas produced are 14.3
and the mole fraction H2O is 2/14.3 = 0.14
Coal
1
H 2 + O2 ® H 2O
2
C + O2 ® CO2
Composition of combustion gases (kgmol)
Components
kgmol
C
H2
1.0
CO2 H2O Excess O2
1.0
0.47
0.47
Air
Total
N2
1.0
0.47
0.49
6.5
0.49
6.5
Req. O2: 1+0.47(1/2) = 1.24 Excess O2: (1.24)(0.4) = 0.49
N2: 1.40(79/21)[1+0.47(1/2)] = 6.5
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
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The total kilogram moles of gas produced are 8.46
and the mole fraction H2O is 0.47/8.46 = 0.056
If the barometric pressure is, say, 100 kPa, the
stack gas would become saturated and water
vapor would start condense at P*H2O:
Natural gas
Partial pressure
100(0.14) = 14 kPa
Equivalent temperature 52.5 oC
Coal
100(0.056) = 5.6 kPa
35 oC
Vapor-Liquid Equilibria for Multicomponent Systems
In this section, we will learn how to calculate the
partial pressures and mole fractions of solutes and
solvents in multicomponent mixtures in which the
gas and liquid phases are at equilibrium.
In a two-phase vapor-liquid mixture at equilibrium, a
component in one phase is in equilibrium with the
same component in the other phase. The equilibrium
relationship depends on the temperature, pressure,
and composition of the mixtures.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
For Ideal Solutions
Henry’s law is primarily used to relate the mole
fraction of one component in the vapor phase to the
mole fraction of the same component in the liquid
phase for a component whose mole fraction
approaches zero, such as a dilute gas dissolved in
as liquid:
Pi = Hixi
where pi is the pressure in the gas phase of the
dilute component at equilibrium at some
temperature, and Hi is the Henry’s law constant.
yi =
Pi
H i xi
=
Ptot
Ptot
Raoult’s law is primarily used for a component whose
mole fraction approaches unity or for solutions of
components quite similar in chemical nature, such as
straight chain hydrocarbons.
Pi = Pi * xi
where P*i is the vapor pressure of component i and xi
is the liquid-phase mole fraction. Note that when xi =
1, pi = p*i
An equilibrium constant, Ki, is defined as follows by
assuming that Dalton’s law applies to the gas phase
(Pi = Ptotyi)
yi
Pi*
Ki =
=
xi Ptot
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
The above equation gives reasonable estimates of
Ki values at low pressure for components well
below their critical temperatures, but yields values
too large for components above their critical
temperatures, at high pressures, and/or for polar
compounds. So Ki was modified to nonideal
mixtures by Sandler (if Tc,i/T > 1.2):
[ 7 .7224 - 7 .534 / Tr , i - 2 .598 ln Tr , i ]
Ki =
Pc ,i
Ptot
Typical problems you may be asked to solve that
involve the use of the equilibrium Ki are:
1) Calculate the bubble point temperature of a liquid
mixture given the total pressure and liquid
composition.
2) Calculate the dew point temperature of a vapor
mixture given the total pressure and vapor
composition.
3) Calculate the related equilibrium vapor-liquid
compositions over the range of mole fractions
from 0 to 1 as a function of temperature given the
total pressure.
4) Calculate the composition of the vapor and liquid
streams, and their respective quantities, when a
liquid of given composition is partially vaporized at
a given temperature and pressure.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
To calculate the bubble point temperature (given the
total pressure and liquid composition), we can write
the following equation as yi = Kixi and we know that
∑yi = 1 in the vapor phase.
1=
n
å
K i xi
i =1
Where the Ki’s are functions of solely temperature and
n is the number of component. For an ideal solution the
above equation becomes:
Ptot =
n
å
i =1
Pi * x i
And we might use Antoine’s equation for P*i. Once
the bubble point temperature is determined, the
vapor composition can be calculated from
Pi * x i
yi =
P tot
To calculate the dew point temperature (given the
total pressure and vapor composition, we can write
the following equation as xi = yi/Ki, and we know ∑xi
= 1 in the liquid phase.
n
y
1= å i
i =1 K i
in which the K’s are function of temperature as
explained for the bubble point temperature calculation.
n
yi
For an ideal solution. 1 = P
tot å
*
i = 1 Pi
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
To calculate the amount of the respective vapor and
liquid phases that evolve at equilibrium when a
liquid of known composition flashes (flash
vaporization) at a known temperature and pressure,
we must use the following equation:
FxFi = Lxi + Vyi
Where F is the moles of liquid to be flashed, L is the
moles of liquid at equilibrium, and V is the moles of
vapor at equilibrium. Introduction of yi = Kixi into the
above equation gives:
æy ö
FxFi = Lçç i ÷÷ + Vyi
è Ki ø
yi =
FxFi
=
xFi
L
L
1
+ (F - L) 1- (1- )
Ki
F
Ki
where L/F is the fraction of liquid formed on
vaporization. Consequently, since ∑yi = 1 , after
summing the yi’s we want to solve the following
equation:
1=
n
å
i =1
By Assist. Prof. Dr. Wipada Sanongraj
x Fi
1-
L
1
(1 )
F
Ki
163
1304 211 Chemical Engineering Principles
and Calculations
Example 4.14 : Vapor-Liquid Equilibrium Calculation
Suppose that a liquid mixture of 4.0% n-hexane in
n-octane is vaporized. What is the composition of
the first vapor formed if the total pressure is 1.00
atm?
Solution
The mixture can be treated as an ideal mixture
because the components are quite similar. As an
intermediate step, we must calculate the bubble
point temperature.
First we have to calculate the vapor pressures of the
two components:
ln( P*) = A -
A
B
C +T
B
C
n-hexane (C6)
15.8737 2697.55
-48.784
n-octane (C8)
15.9798 3127.60
-63.633
Basis : 1 kgmol of liquid
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
We need to solve the following equation to get the
bubble point temperature:
2697 . 55 ö
3127 . 60 ö
æ
æ
760 = exp ç 15 . 8737 ÷ 0 . 040 + exp ç 15 .9787 ÷ 0 . 960
- 48 .784 + T ø
- 63 . 633 + T ø
è
è
The solution is T = 393.3 K, where the vapor
pressure of hexane is 3114 mm Hg and the vapor
pressure of octane is 661 mm Hg.
yC6 =
PC*6
Ptot
xC6 =
3114
(0.040) = 0.164
760
yC8 = 1- 0.164= 0.836
Example 4.15 : Flash Calculation
Calculate the fraction of liquid that will exist at
equilibrium at 150 oF and 50 psia when the liquid
concentrations of the solution to be vaporized are as
follows:
Component
Initial liquid
K
mole fraction
C2
C3
i-C4
n-C4
i-C5
n-C5
C6
Total
By Assist. Prof. Dr. Wipada Sanongraj
0.0079
0.1321
0.0849
0.2690
0.0589
0.1321
0.3151
1.0000
16.20
5.2
2.6
1.98
0.91
0.72
0.28
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1304 211 Chemical Engineering Principles
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The K values come from the Engineering Data Book
of the Gas Processors Supply Association (1980).
Solution
We want to solve for L/F. Start with an initial guess
of L/F = 1.0.
Stage
1
2
3
4
5
6
L/F
1.0
0.8565
0.6567
0.5102
0.4573
0.4511
Gibb’s phase rule is a useful guide in establishing
how many properties, such as pressure and
temperature, have to be specified to definitely fix all
the remaining properties and number of phases that
can coexist for any physical system. The rule can
be applied only to systems in equilibrium. The
Gibbs phase rule states that
F=C–P+2
F = number of degrees of freedom
C = number of components in the system
P = number of phases that can exist in the system
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Phase-rule variables are variables of the kind with
which the phase rule is concerned and they are
intensive properties of the system.
Intensive properties are the ones that do not
depend on the quantity of the material present. The
specific (per unit mass) values are intensive
properties.
Extensive properties are the ones that depend on
how much material we have. The total quantities
are extensive properties.
An example of the use of the phase rule is the ideal
gas law, PV = nRT. In order to be able to determine
the remaining one unknown, you might conclude
that F = 3. However, if we apply the phase rule for
a single phase P = 1 and for a pure gas C = 1, so
that
F=C–P+2=1–1+2=2
How can we explain this apparent paradox with our
previous statement? Since the phase rule is
concerned with intensive properties only, the
following are phase-rule variables in the ideal gas
law:
P, T, and Vˆ
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Thus the ideal gas law would be written as
pVˆ = RT
We can see that by specifying two intensive variables
(F = 2), the third can be calculated.
Example 4.16 Application of the phase rule
Calculate the number of degrees of freedom from
the phase rule for the following materials at
equilibrium:
a)Pure liquid benzene
b) A mixture of ice and water only
c) A mixture of liquid benzene, benzene vapor, and
helium gas
d)A mixture of salt and water designed to achieve
a vapor pressure.
What variables might be specified in each case?
By Assist. Prof. Dr. Wipada Sanongraj
168
1304 211 Chemical Engineering Principles
and Calculations
Solution
F=C–P+2
a) C = 1, P = 1, hence F = 1 – 1 + 2 = 2. The
temperature and pressure might be specified in the
range in which benzene remains a liquid.
b) C = 1, P = 2, hence F = 1 – 2 + 2 = 1. Once either
the temperature or the pressure is specified, the
other intensive variables are fixed.
c) C = 2, P = 2, hence F = 2 – 2 + 2 = 2. A pair from
temperature, pressure, or mole fraction can be
specified.
d) C = 2, P = 2, hence F = 2 – 2 + 2 = 2. Since a
particular pressure is to be achieved, you would
adjust the salt concentration and temperature of
the solution.
Partial Saturation and Humidity
Often, the contact time required in a process for
equilibrium (or saturation) to be attained between the
gas and liquid is too long, and the gas is not
completely saturated with the vapor. Then the vapor
is not in equilibrium with a liquid phase, and the
partial pressure of the vapor is less than the vapor
pressure of the liquid at the given temperature. This
condition is called partial saturation.
When the vapor is water vapor and the gas is air, the
special term humidity applies. For other gases or
vapors, the term saturation is used.
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
Relative saturation is defined as
RS =
Pvapor
Psatd
= relative saturation
where Pvapor = partial pressure of the vapor in the
gas mixture
Psatd = partial pressure of the vapor in the
gas mixture if the gas were saturated
at the given temperature of the
mixture
If the subscript l denotes the vapor,
RS =
Pl
Pl / Pt
Vl / Vt
nt
massi
=
=
=
=
Pl* Pl* / Pt Vsatd / Vt nsat d masssatd
We can see that relative saturation, in effect,
represents the fractional approach to total saturation.
If you listen to the radio or TV and hear the
announcer say that temperature is 25 oC and the
relative humidity is 60%, he or she implies that
PH 2 O
PH* 2 O
(100 ) = % R H = 60 %
with both the PH2O and the P*H2O being measured at
25 oC.
By Assist. Prof. Dr. Wipada Sanongraj
170
1304 211 Chemical Engineering Principles
and Calculations
Example 4.17 Application of Relative Humidity
The weather report on the radio this morning was that
the temperature this afternoon would reach 94 oF, the
relative humidity would be 43%, the barometer 29.67
in. Hg. partly cloudy to clear, with the wind from SSE at
8 mi/hr. How many pounds of water vapor would be in
1 mi3 of afternoon air? What would be the dew point of
this air?
Solution
The vapor pressure of water at 94 oF is 16.1 in Hg.
We can calculate the partial pressure of the water
vapor in the air from the given percent relative
humidity.
Pw = (1.61 in. Hg)(0.43) = 0.692 in. Hg
(Pair = Pt –Pw = 29.67 – 0.692 = 28.98 in. Hg
Basis : 1 mi3 water vapor at 94 oF and 0.692 in. Hg
3
æ 5280 ft ö 492 o R 0.692 in.Hg 1 lbmol 18 lbH 2O
1 mi ´ ç
´
´
´
÷ ´
o
3
1 lbmol
è 1 mi ø 555 R 29.92 in.Hg 359 ft
= 1.52 ´108 lb H 2O
3
Now the dew point is the temperature at which the
water vapor in the air will first condense on cooling at
constant pressure and composition. As the gas is
cooled the relative humidity increases because the
partial pressure of the water vapor is constant while the
vapor pressure of water decreases with temperature.
By Assist. Prof. Dr. Wipada Sanongraj
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and Calculations
When the percent relative humidity reaches 100%
100
PH 2 O
PH* 2 O
= 100%
or
PH 2 O = PH* 2 O
The water vapor will start to condense. This mean that
at the dew point the vapor pressure of water will be 0.692
in. Hg. The corresponding temperature is 68-69 oF.
Molal Saturation is another way to express vapor
concentration in a gas. It is the ratio of the moles of
vapor to the moles of vapor-free gas.
nvapor
n vapor - free gas
= molal
saturation
If subscripts 1 and 2 represent the vapor and the dry
gas, respectively, then for a binary system,
P1+ P2 = Ptot
n1+ n2 = ntot
n1 P1 V1
n1
P1
V1
=
=
=
=
=
n2 P2 V2 ntot - n1 Ptot - P1 Vtot - V1
By Assist. Prof. Dr. Wipada Sanongraj
172
1304 211 Chemical Engineering Principles
and Calculations
Humidity refers to the mass of water vapor per
mass of bone-dry air.
(nvapor )(mol.wt .vapor )
mass vapor
H = humidity =
=
(ndrygas )(mol.wt .drygas ) mass drygas
Absolute Saturation (humidity) is defined as the
ratio of the moles of vapor per mole of vapor-free
gas to the moles of vapor that would be present per
mole of vapor-free gas if the mixture were
completely saturated at the existing temperature
æ
ö
and total pressure:
moles vapor
ç
÷
AS =
ç moles
è
vapor - free gas
æ
moles vapor
ç
ç moles vapor
- free
è
gas
÷
ø actual
ö
÷
÷
ø saturated
Using the subscripts 1 for vapor and 2 for vaporfree gas
æ n1 ö
æ P1 ö
çç
÷÷
çç
÷÷
è n 2 ø actual
è P2 ø actual
% AS =
100 =
100
æ n1 ö
æ P1 ö
çç
÷÷
çç
÷÷
è n 2 ø saturated
è P2 ø saturated
Since P1 saturated = P*1 and Ptot = P1 + P2
æ
ö
P1
çç
÷÷
P
P
1 ø
è total
P1 æ Ptotal - P1* ö
÷100
% AS =
100 = * çç
P1 è Ptotal - P1 ÷ø
æ
ö
P1*
çç
÷
* ÷
P
P
1 ø
è total
By Assist. Prof. Dr. Wipada Sanongraj
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1304 211 Chemical Engineering Principles
and Calculations
From P1/P1* = relative saturation, therefore,
æP
- P1*
% AS = ( relative saturation ) çç total
è Ptotal - P1
ö
÷÷100
ø
Percent absolute saturation is always less than
relative saturation except at saturated conditions (or
at zero percent saturation) when percent absolute
saturation = percent relative saturation.
Example 4.18 Partial Saturation
The percent absolute humidity of air at 30 oC (86 oF)
and a total pressure of 750 mm Hg (100 kPa) is 20%.
Calculate a) the percent relative humidity, b) the
humidity, c) the partial pressure of the water vapor in
the air. What is the dew point of the air?
Solution
Data from the steam tables are
P*H2O at 86 oF = 1.253 in Hg = 31.8 mm Hg = 4.242 kPa
By Assist. Prof. Dr. Wipada Sanongraj
174
1304 211 Chemical Engineering Principles
and Calculations
To get the relative humidity, PH2O/P*H2O, we need to
find the partial pressure of the water vapor in the air.
This may be obtained from
æ
ö
PH 2 O
PH 2 O
ç
÷
ç Ptotal - PH O ÷
750 - PH 2 O
2
è
ø
100 ( AH ) = 20 =
100
=
100
31 . 8
æ
ö
PH* 2 O
ç
÷
750 - 31 . 8
ç Ptotal - PH* O ÷
2
è
ø
c)
PH2O = 6.58 mm Hg
a)
%RH = 100(6.58/31.8) = 20.7%
b)
H = [(MWH2O)(nH2O)]/[(MWair)(nair)]
= [18(PH2O)]/[29(Pair)] = 18(6.58)/[(29(750-6.58)] = 0.0055
The dew point is the temperature at which the water
vapor in the air would first commence to condense,
when cooled at constant total pressure, because the
gas becomes completely saturated. This would be
at the vapor pressure of 6.58 mm Hg, or about 5.1
oC.
By Assist. Prof. Dr. Wipada Sanongraj
175