INDUSTRIAL PLANT DESIGN
Edsel G. Anyayahan, PME
VISION
1
Laguna University shall be a socially responsive
educational institution of choice providing holistically
developed individuals in the Asia-Pacific Region.
MISSION
Laguna University is committed to produce academically
prepared and technically skilled individuals who are
socially and morally upright citizens.
Department of Mechanical Engineering
MISSION
The Department of Mechanical Engineering of Laguna
University is committed to produce academically prepared
and technically skilled mechanical engineers who are
socially and morally upright citizens.
VISION
The Department of Mechanical Engineering of Laguna
University is envisioned to be the provincial college of
choice producing well-equipped mechanical engineers who
specializes on energy management.
2
Table of Contents
Module 1:Fuels and Combustion
Introduction
Learning Objectives
Lesson 1. Classification of Fuels
Lesson 2. Hydrocarbons
Lesson 3. Properties of Diesel Engine
Lesson 4. Solid fuels
Lesson 5. Coal Analysis
Lesson 6. Combustion Chemistry
Lesson 7. Air-fuel Ratio in terms of the products of Combustion
(Volumetric Analysis)
Assessment Task 1
Summary
Reference
1
1
1
4
6
10
15
22
27
31
32
32
Module 2: Steam Generating Unit
Introduction
Learning Objectives
Lesson 1. Types of Boiler
Lesson 2. Boiler Ratings and Performance
Lesson 3. Steam Generator Heat Balance
Assessment Task 2
Summary
Reference
33
33
34
35
37
41
42
42
Module 3: Mechanical Dryers
Introduction
Learning Objectives
Lesson 1. Mechanical Dryer
Assessment Task 3
Summary
Reference
43
43
44
54
55
55
3
Course Code:
ME 204
Course Description: This course is a study of mechanical engineering
theories, equipment and systems that are needed in the operation of an
industrial/manufacturing plant.
Course Intended Learning Outcomes (CILO):
At the end of the course, students should be able to:
1. Practice leadership skills in decision-making during group discussion
and group activity
2. Modify and apply basic design concepts of industrial plants systems and
equipment
3. Differentiate and design the basic components of industrial equipment
in accordance with Codes and Standards
4. Build skills in selecting system components and equipment in industrial
plant design
Course Requirements:
Assessment Tasks - 60%
Major Exams -40%
_________
Periodic Grade
100%
Computation of Grades:
PRELIM GRADE
= 60% (Assessment Task 1-3) + 40% (Prelim exam)
MIDTERM GRADE
=
30%(Prelim Grade) + 70 %[60% Assessment Task 1-3) + 40% (Midterm exam)]
FINAL GRADE
=
30%(Midterm Grade) + 70 %[60% (Assessment Task 1-3) + 40% (Final exam)]
4
MODULE 1
FUELS AND COMBUSTION
Introduction
Fuel is any substance or combustible material that by rapid oxidation or burning produces
heat and light. An example of a fuel is coal or gasoline. A fuel is composed of chemical elements
which, in rapid chemical union with oxygen, produce combustion. Fossil fuels are fuels that originate
from the earth as a result of the slow decomposition and chemical conversion of organic material.
The basic forms of fossils are solid (coal), liquid (oil), and natural gas. Synthetic fuels or synfuels, the
new combustible-fuel-options, are liquid or gaseous fuels derived largely from coal, oil shale, and tar
sands (Francisco, 2014).
Learning Outcomes
At the end of this module, students should be able to:
1. Define what fuels is, and determine the different classification of Fuels;
2. Solve problems involving properties of Diesel Engine;
3. Review and learn the different Coal Analysis; and
4. Learn the Combustion Chemistry.
1
Lesson 1. Classification of Fuels
According to Francisco (2014) the different types of fuels are:
1. Solid Fuels
a. Natural solid fuels. Ex. Coal, wood, bagasse
b. Prepared or manmade. Ex. Charcoal, coke, briquette fuels
2. Liquid Fuels
a. Hydrocarbon (Cn Hm )
i. Gasoline – Octane, C8 H18
ii. Diesel – Hexadecadene, C16 H32
iii. Fuel oil – Dodecane, C12 H26
b. Alcohol (Cx Hy Oz )
i. Ethyl alcohol
ii. Methyl alcohol
c. Alcogas (70% alcohol and 30% ethanol)
d. Methanol (Liquid Coal)
3. Gaseous Fuels
a. Natural Gas Derivatives
i. Methane, CH4
ii. Propane, C3 H8
iii. Ethane, C2 H6
iv. Acetylene, C2 H2
b. Liquid Petroleum Gas (LPG)
c. Biogas (Animal waste gas)
Crude Oil

Crude oil is the term used from the raw petroleum as it comes from the oil wells (Francisco,
2014).
Classification of Crude oil from Francisco (2014)
1. Paraffin Base
2. Napthenic Base
2
3. Mixed Base
Products Resulting from the Refinement of Petroleum (Francisco, 2014):
1. Natural gas

The gaseous hydrocarbons are usually associated with liquid petroleum, either
standing above the liquid in the earth or dissolved in it.
2. Gasoline

Liquid petroleum fuel intended for use in spar-ignition engines. Specific gravity is 0.70
to 0.78
3. Kerosene

The next fraction heavier than gasoline

Intended for use in lamps, heaters, stoves, and similar appliances

Excellent fuel for compression-ignition engines and for aircraft gas turbines

Specific gravity is 0.78 to 0.85
4. Distillate

Slightly heavier than kerosene

Obtained from some Western United States crude by distillation at atmospheric
pressure

Substantially the same uses as a kerosene
5. Diesel Fuel

Petroleum fractions that lie between kerosene and lubricating oils

It covers a wide range of specific gravity

Composition is controlled to make them suitable for use in various types of CI engines.
6. Fuel oils

Covers a wide range of specific gravity

Distillation is similar to that of Diesel oils

Its compositions do not require such accurate control as in Diesel oils for it is used in
continuous burners.
7. Lubricating oils

Made up in part from heavy distillates of petroleum and in part from residual oil, that is,
oils remaining after distillation. Tar and asphalt are solid or semi-solid products that
remain undistilled.
3
Compositions of Petroleum Products from Francisco (2014)

The common or main compositions of petroleum products are carbon and hydrogen

Hydrocarbon is the combination of carbon and hydrogen

The general formula is Cn Hm , where n = 1 to 26 and m = 2 to 54
Lesson 2. Hydrocarbons
According to Francisco (2014) hydrocarbons are:

Hydrocarbons are further categorized into sub-families such as:
 Alkynes (Cn H2n−2 ) → Ex. Acetylene, C2 H2
 Alkenes (Cn H2n ) → Ex. Ethylene, C2 H4
 Alkanes (Cn H2n+2 ) → Ex. Octane, C8 H8

Alkynes and alkenes are referred to as unsaturated hydrocarbons.

Alkanes are referred to as saturated hydrocarbons. Alkanes are also known as the paraffin
series and methane series.

Alkenes are subdivided into the chain-structures olefin series and the ring-structured
naphthalene series.

Aromatic hydrocarbons (𝐶𝑛 𝐻2𝑛−6 ) constitute another subfamily. Ex. Benzene, 𝐶6 𝐻6 .
Types of Hydrocarbons from Francisco (2014)
1. Paraffin (Cn H2n+2 )
2. Olefins (Cn H2n )
Ex. Heptane, C7 H16
Ex. Octane, C8 H16
Cracking of Hydrocarbons from Francisco (2014)

Cracking is the process of splitting hydrocarbon molecules into smaller molecules. It is used
to obtain lighter hydrocarbons (such as used in gasoline) from heavy hydrocarbons (say,
Crude oil)
o
Ex. Alkane molecules crack into a smaller member of the alkane subfamily and a
member of the alkene subfamily.

Cracking can proceed under the influence of high temperatures (thermal cracking) or
catalysts (catalytic cracking or “cat cracking”).

Catalytic Cracking also produces gasoline with better antiknock properties than does
thermal cracking.
4
Gaseous Fuels
According to Francisco 2014 the different types of Gaseous Fuels are:
1. Natural Gas

Obtained from oil wells

It is called a casing-head gas

Usually treated for gasoline recovery

It is delivered into the pipeline system to be used as fuel
2. Coke-oven Gas

Obtained as a by-product when making coke

Its analysis depends into the pipeline system to be used as fuel
 Coke; typically used in blast furnaces, is produced by heating coal in the
absence of oxygen. The heavy hydrocarbons crack, leaving only a
carbonaceous residue containing ash and sulfur. Coke burns smokelessly.
Breeze is coke smaller than 5/8 inch (16 mm). it is not suitable for use in blast
furnaces, but steam boilers can be adapted to use it. Char is produced from coal
in 900 °F (500°C) carbonization process.
3. Blast-Furnace Gas

A by-product of melting iron ore
4. Produces Gas
5. Sewage-Sludge Gas

Obtained from sewage disposal plants
Four Main Methods in Making Gasoline from Francisco (2014)
1. Distillation or refining from crude oil
2. Cracking residue oil
3. Polymerization of gases produced by the cracking process, which produce a high-octane
gasoline that can be used either directly or blended with gasoline produced by other methods.
4. Extraction from natural gas by absorption and distillation and blending with some heavier
hydrocarbons.
5
Lesson 3. Properties of Diesel Engine
The Different Properties of Diesel Engine from Francisco (2014):

Density – the mass per unit volume of the fuel, in kg/li, kg/m3 , or lb/Ft 3
𝝆=
𝒎
𝑽
The density of fuel is measure at 60 °F (15.6 °C)
Density at 15.6 °C = SG at 15.6 °C / 15.6 °C, kg/li

Hydrometer- an instrument used for determination of SG (15.6 °C / 15.6 °C) in °API
(American Petroleum Institute) and °Baume.
°𝐁𝐚𝐮𝐦𝐞 =
°𝐀𝐏𝐈 =

𝟏𝟒𝟎
𝐒𝐆 𝐚𝐭 (
𝟏𝟓.𝟔 °𝐂
𝟏𝟓.𝟔 °𝐂
𝟏𝟒𝟏. 𝟓
𝟏𝟓.𝟔 °𝐂
𝐒𝐆 𝐚𝐭 (𝟏𝟓.𝟔 °𝐂)
)
− 𝟏𝟑𝟎
− 𝟏𝟑𝟏. 𝟓
Coefficient of Volumetric Expansion – a coefficient used in oil purchasing and storage, since
oil is bought by volume usually expressed in terms of 42-gallons barrels.
𝐕𝐭 = 𝐕𝐨 [𝟏 + 𝟎. 𝟎𝟎𝟎𝟒(𝐭 − 𝐭 𝐨 )] → 𝐄𝐧𝐠𝐥𝐢𝐬𝐡 𝐔𝐧𝐢𝐭
Where, Vt = volume at a temperature t, Ft 3
t = temperature, °F
Vo = volume at a temperature t o , (usually 60 °F), Ft 3
Coefficient = 0.0004/°F
𝐕𝐭 = 𝐕𝐨 [𝟏 + 𝟎. 𝟎𝟎𝟎𝟕(𝐭 − 𝐭 𝐨 )] → 𝐒𝐈 𝐔𝐧𝐢𝐭
Where, Vt = volume at a temperature t, m3 , li
6
t = temperature, °F
Vo = volume at a temperature t o , (usually 15.6 °F), m3 , li
Coefficient = 0.0007/°C
SAMPLE PROBLEMS from Francisco (2014)
1.
A 28 °API oil has a temperature 40 °F. what is its density?
Solution:
Considering the English System
For SG at 60°F, °API =
𝟏𝟒𝟏.𝟓
𝐒𝐆 𝐚𝐭 (
SG at (
𝟏𝟓.𝟔 °𝐂
)
𝟏𝟓.𝟔 °𝐂
− 131.5
60 °F
141.5
141.5
)=
=
= 𝟎. 𝟖𝟖𝟕
60 °F
°API + 131.5 28 + 131.5
For the density at 60 °F,
60 °F
ρ60°F = ρwater@60°F (Sg @ 60 °F) = 62.4(0.887) = 𝟓𝟓. 𝟑𝟓 𝐥𝐛/ f𝐭 𝟑
For the given volume at a given temperature, t = 40 °F
Vt = Vo [1 + 0.0004(t − t o )] = 1 ft 3 [1 + 0.0004(40 − 60)] = 𝟎. 𝟗𝟗𝟐 𝐟𝐭 𝟑
For the density at 40 °F
55.35
ρ40°F = 0.992 = 𝟓𝟓. 𝟖 𝐥𝐛/f𝐭 𝟑
Considering the SI units
t = 40 °F = 4.44°C
SG (
15.6 °C
141.5
141.5
)=
=
= 𝟎. 𝟖𝟖𝟕
15.6 °C
°API + 131.5 28 + 131.5
ρ15.6°C = SG(at 15.6℃) (1.0) = 𝟎. 𝟖𝟖𝟕 𝐤𝐠/li
15.6℃
Vt = Vo [1 + 0.0007(t − t o )] →
ρt =
m m
= [1 + 0.0007(t − t o )]
ρt ρo
ρo
0.887
𝐤𝐠
=
= 𝟎. 𝟖𝟗𝟒
[1 + 0.0007(t − t o )] [1 + 0.0007(4.44 − 15.6)]
𝐥𝐢
7

Viscosity – property of Diesel fuel used to measure its resistance to flow. It is a fair indication
of how the oil will atomize and how it will affect the injection pump.
o
Absolute viscosity, µ
SI units → 1 poise − 1 Dyne-sec/cm2
English Units → 1 lbf -sec/ft 2 = 1 slug/ft-sec
o
Kinematic Viscosity – the ratio of absolute viscosity to that of the density, v =
µ
ρ
SI units → 1 stoke = 1 cm2 /sec
English Units → Ft 2/sec

Saybolt Viscosimeter – an instrument used to measure viscosity of oil in Saybolt Universal
(SSU) or Saybolt Seconds Furol (SSF).
Centistoke = 0.308(SSU − 26)
62 SSF = 600 SSU

1 SSF = 9.68 SSU
Heating Value – refers to the heat content of the fuel in a unit mass. The heating value of a
liquid fuel is ranging from 18 000 to 19 500 Btu/lb. An instrument or apparatus used to
determine the heating value of a fuel is known as Bomb or Sergent Calorimeter. Heating
value is also called as Calorific Value.

Higher Heating Value (HHV) is the heating value obtained when the moisture or water
in the product of combustion is in the liquid condition.

Lower Heating Value (LHV) is the heating value obtained when the moisture or water
in the product of combustion is in the vapor condition.
ASME formula:
HHV = 41130 + 139.6(°API) → kJ/kg
HHV = 17680 + 60(°API) → Btu/lb
8
Sherman and Knoff Formula:
HHV = 18 250 + 40(°Baume −10) → Btu/lb
HHV = 42 450 + 93(°Baume −10) → kJ/kg
Bureau of Standard Formula:
HHV = 51 716 − 8 793.8(SG)2 → kJ/kg
Lower Heating Value, LHV = HHV −9H2 (2442) → kJ/kg
Where, H = 26 -15(SG), in decimal

Flash Point and Fire Point – the temperature at which oil vapor flashes or burns steadily;
used in relation to ignition and storage hazards of oil.

Pour point – the lowest temperature at which oil flows; used in pumping and flow of oils. It is
more important to the fuel for mobile engines whose fuel tanks are generally exposed and
unheated that for stationary units.

Color is sometime specified on the basis that good color indicates clean fresh stock of
satisfactory volatility.

Purity. Although oils is a relatively pure fuel it may contain some sediment and water, ash, or
sulfur, all of which are undesirable impurities and are limited to low permissible amounts in
specifications. Sediment and water are determined by eluting the fuel with benzol and
centrifuging. They are reported together, sometimes by the number B.S. and W., meaning
bottom sediment and water.

Sulfur Content – in weight percentage; useful with reference to corrosion of plant equipment.

Moisture and Sediment – in weight or volume percentage; useful in firing of oil.

Specific Heat – varies with temperature but for usual values is about 0.4 to 0.5 Btu/lb-F; used
in heating problems.

Carbon Residue – test indicating carbon forming characteristics of oil; used in oil burning.
Lesson 4.Solid Fuels
9
Coal from Francisco (2014)

Coal – is a solid fuel which is a mixture of carbon, hydrogen, oxygen, nitrogen, sulfur, ash,
and moisture. It is believed that the coal once existed primarily as vegetable matter.

Coal is a general term that encompasses a large number of solid organic minerals with widely
differing compositions of properties, although all are essentially rich in amorphous (without
regular structure) elemental carbon.
Coal Compositions from Francisco (2014)

Coal has these components distributed throughout its mass: a) coal substance, b) mineral
matter, and c) moisture.

Coal substance consists of many organic compounds by carbon, hydrogen, and oxygen from
original vegetable matter. The exact composition varies but does not affect practical coal
burning.

The mineral matter to some extent was in the original vegetable fibers. Water, which
percolated through the peat, bogs and coal seams, contained dissolved salts and deposited
most of the mineral content of the coal.

The moisture content of coals, of ten called “mechanical moisture”, means water retained by
coal.
Classification of Coals from Francisco (2014)

There are many ways of classifying coal according to its chemical and physical properties.
The most accepted system is that one used by the American Society for Testing and
Materials (ASTM), which classifies coals by grade or rank according to the degree of
metamorphism (change in form and structure under the influences of heat, pressure, and
water), ranging from the lowest state (lignite) to the highest (anthracite)

These classifications are briefly described below in descending order:
1. Anthracite coal

The highest grade of coal
10

Contains a high content, 86% to 98% by mass, of fixed carbon on a dry, mineralmatter-free basis and a low content of volatile matter less than 2% to 14 % by
mass (chiefly methane, CH4 )

Very hard coal having a shiny black luster

A brittle coal that borders on graphite at upper end of fixed carbon.

Non-coking

With high percentage fixed carbon

Less than 8% volatile matter

Requires strong draft

Burns without flame or very short bluish flame

Desirable when smokeless combustion is essential.
Note:
o
Graphite is a moderately soft allotropic form of carbon. Carbon crystallizes perfectly
into diamond, imperfectly into graphite and is amorphous (having no regular structure,
non-crystalline) in anthracite and charcoal.
o
The anthracite rank of coal is subdivided into three groupings, in descending of fixedcarbon percent, as follows:

Metha-anthracite, greater than 90%

Anthracite, 92 % to 98%

Semi-anthracite, 86% to 92%
2. Semi-anthracite coal

Not used commercially as steam coal

With less fixed carbon

With 8 to 14 % volatile matter

Less luster

Burns with larger and more luminous flames
11
3. Semi-bituminous coal

Highest grade of bituminous coal

Burns with very small amount of smoke

Softer than anthracites coals

Contains 14 to 22 % volatile matter

Has a tendency to break into small sizes during storage or transportation.
4. Bituminous coal

Soft and with high percentage of volatile matter

Burns with long yellow and Smokey flames

Cary greatly in percentage of volatile matter, moisture, ash and sulfur.

Classified as free-burning and caking or coking.
5. Sub-bituminous coal

Sometimes known as black lignites

Low grade bituminous coal which has lost the woody, structural appearance of
lignites

Disintegrates when exposed to the air and requires careful attention during
storage

Contains 36 to 45 % volatile matter

Contains 17 to 20 % moisture
6. Lignites coal

The transition state between peat and the sub-bituminous grade

Have a woody or offer a claylike appearance

With low heating value

Contains 30 to 45 % moisture

High ash content

Unless carefully stored, lignites are subject to spontaneous combustion
PEAT
o
It is not an ASTM rank of coal
o
It is the first geological step in coal’s formation
12
o
It is a heterogeneous material consisting of decomposed plant matter
and in organic minerals
o
It contains up to 90% moisture.
Components of Coals from Francisco (2014)
1. Coal substance – part of coal consisting of organic compounds of carbon, hydrogen, and
oxygen that are derived from the original vegetable.
2. Mineral Matter- composed of inorganic compounds which to some extent with in the original
wood.
3. Moisture Content – referred to as mechanical moisture and means water as such retained by
the coal.
Properties of Solid Fuels (Coal) from Francisco (2014)
1. Heating Value or Calorific Value – the equivalent heat content in a unit mass fuel; determined
either by calorimetric measurement or by Dulong’s formula.
o
Dulong’s Formula (Solid Fuels) for higher heating value
HHV = 33 820C + 144212 (H2 −
HHV = 14544C + 62028 (H2 −
O2
O2
8
8
) + 9304S → kJ/kg
) + 4050S → Btu/lb
Where C, H, O, & S are element in percentage by weight
o
Lower Heating Value
LHV = HHV – QL
where: QL = latent heat of water content,
QL = mw hfg = 9mH2 hfg
mw = mass of water vapor in products of combustion per unit mass of fuel
(due to the combustion of H2 In the fuel, but not including initial H2 O in fuel.
13
mH2 = mass of original hydrogen per unit mass of fuel, known from ultimate
analysis.
hfg = latent heat of vaporization of water vapor at its partial pressure in the
combustion products, Btu/lbm H2 O or kJ/kg H2 O
QL = (W + 9H)(2493 + 1.926t g − 4.187t a ) → for t g ≤ 302 °C
QL = (W + 9H)(2442 + 2.0935t g − 4.187t a ) → for t g > 302°𝐶
W = free moisture in fuel, kg/kg fuel
t g = flue gas temperature, °C
t a = combustible air temperature, °C
2. Ash-fusion temperature – refers to fluid temperature
3. Grindability – indicates the case with which a coal maybe pulverized.
4. Coking – refers to the ability of the coal, when heated, to evolve volatile matter
5. Caking
6. Free-burning
7. Friability
8. Clinkering
Note:
o
Coke is the solid substance remaining after the partial burning of coal in an even or
after distillation properties.
14
Lesson 5. Coal Analysis
According to Francisco (2014) there are two types of coal analysis: proximate and ultimate, both
done on a mass-percent basis.
 Both methods may be used on:
o
“As- received Basis”, useful for combustion calculations
o
“Moisture-Free-Basis”, avoids variation of the moisture content even in the same
shipment and certainly in the different stages of pulverization.
o
“Dry Mineral-Matter-Free Basis”, circumvents the problem of the ash contents not
being the same as the mineral matter in coal.
Proximate Analysis from Francisco (2014)

This is the easier of the two types of coal analysis and the one which, supplied readily
meaningful information for coal’s use in the steam generators.

It is an analysis that gives the gravimetric fraction of moisture, volatile matter, fixed carbon,
and ash.

It determines the mass percentages of fixed carbon, volatile matter, moisture, and ash.

Sulfur maybe combined with ash or may be specified separately.
Fixed Carbon from Francisco (2014)
o
It is the elemental carbon that exists in coal.
o
Its determination is approximated by assuming it to be the difference between the
original sample and the sum of volatile matter, moisture, and ash.
Volatile Matter from Francisco (2014)
o
It is that portion of coal, other than water vapor, which is driven off when the sample is
heated in the pre absence of oxygen in a standard test (up to 1750 °F for 7 minutes)
15
o
It consists of hydrocarbon and other gases that result from distillation and
decomposition.
Moisture from Francisco (2014)
o
it is determined by a standard procedure of drying in an oven
o
this does not account for all the water present, which includes combined water and
water of hydration
o
there are several other terms for moisture in coal. One, inherent moisture, is that,
existing in the natural state of coal and considered to be part of the deposit, excluding
surface water.
Ash from Francisco (2014)
o
it is the inorganic salts contained in coal.
o
It is determined in practice as the non-combustible residue after the combustion of
dried coal in a standard test (at 1380 °F).
Sulfur from Francisco (2014)
o
It is determined separately in a standard test
o
Being combustible, it contributes to the heating value of the coal.
o
It forms oxides which combine with water to form acids. These cause corrosion
problem in the back end of steam generators if the gases are cooled below the dew
point, as well as environmental problems.
Ultimate Analysis from Francisco (2014)

It is a special type of gravimetric analysis in which the constituents are reported by atomic
species rather than by compound.

In this analysis, combined hydrogen from moisture in the fuel is added to hydrogen from
the combustive compounds.
16

It is an analysis showing the chemical elements of coals such as carbon, hydrogen,
oxygen, nitrogen, sulfur, ash, and moisture. Elements are presented in percentage by
weight.
C – Carbon
H – Hydrogen
O – Oxygen
N – Nitrogen
S - Sulfur
A – Ash
W – Moisture
According to Francisco (2014) bases of Reporting the Ultimate Analysis are:
a. “As-received” or “As Fired” basis
C + H + O + N + S + A + W = 100%
b. Dry or moisture-free basis
C+ H + O + N + S + A = 100 %
c. Moisture and Ash Free or Combustible Basis
C+ H + O + N + S = 100 %
d. Moisture, Ash, and Sulfur-free basis
C+ H + O + N = 100 %
Note:
o
The first basis, as fired, is of most use to the power-plant operator because it shows
the constituents of the fuel in the same condition as it was weighed and supplied to the
furnace.
o
The remaining methods are used primarily for comparing coals from various sources.
o
To transfer an analysis from as received to dry, it is only necessary to deduct the
weight of hydrogen and oxygen in the water from these terms and to divide each of the
17
remaining constituents by one minus the decimal equivalent of the moisture. A similar
procedure may be used for converting to any the methods.
SAMPLE PROBLEMS from Francisco (2014)
1. A certain type of coal has the following “as-received” analysis:
C = 80.50 %
S = 1.20 %
H = 4.10 %
A = 5.30 %
O = 3.00 %
W = 4.40 %
N = 1.50 %
Calculate the analysis on the dry basis and combustible basis.
Solution:

Consider the “Dry-Basis" Analysis
Solving for the Factor for the dry-basis analysis,
𝑊
4.40
𝐹𝑎𝑐𝑡𝑜𝑟 = 1 − 100 = 1 − 100 = 𝟎. 𝟗𝟓𝟔
Dry-basis Analysis:
C=
O=
80.50
0.956
3.00
= 84.20 %
H=
= 3.10 %
N=
0.956
1.20
S = 0.956 = 1.30 %
4.10
0.956
1.50
0.956
5.30
= 4.30 %
= 1.60 %
A = 0.956 = 5.50 %
Then the sum of the components:
C+ H + O + N + S + A = 100 %

Consider the Combustible Basis
Solving for the Factor for the Combustible Basis
𝐹𝑎𝑐𝑡𝑜𝑟 = [1 −
(𝑊 + 𝐴)
(5.30 + 4.40)
] = [1 −
] = 𝟎. 𝟗𝟎𝟑
100
100
Solving for the Combustible Basis
18
80.50
𝐶 = 0.903 = 89.20 %
3.00
𝑂 = 0.903 = 3.30 %
1.20
𝑆=
= 1.30 %
0.903
4.10
𝐻 = 0.903 = 4.50 %
𝑁=
1.50
0.903
= 1.70 %
Then the sum of the components:
C+ H + O + N + S = 100%
2. Convert the ultimate analysis of coal shown in the table below to a) an ultimate analysis showing
the amount of moisture (“as-received”), b) moisture-free basis, c) moisture and ash-basis, and d)
moisture, ash, and sulfur-free basis. Determine the heating value in each case.
a. Consider the “As-received” analysis

Deduct 1/9 of W from H2 to obtain the ‘’as-received’’ percent H2 , and deduct 8/9 of W from O2
to obtain the ‘’as-received’’ percent of O2 .
Total = 100 %
Solving for the heating value of the “as-received” fuel
HHV = 14600C + 62000(𝐻2 −
𝑂2
8
) + 4050𝑆
HHV = 14600(0.6623)+6200(0.0425 −
0.0886
8
) + 4050(0.0075)
19
HHV = 11648.305 Btu/lb
b. Consider the Dry-Basis or Moisture-free basis of the analysis

To obtain the moisture-free analysis, divide each component of the as received analysis,
except W, by the following factor:
𝐹𝑎𝑐𝑡𝑜𝑟 = 1 − (
9.77
) = 0.9023
100
66.23
0.9023
4.25
H2 =
0.9023
8.86
O2 =
0.9023
1.51
N2 =
0.9023
0.75
S=
0.9023
8.63
A=
0.9023
73.40 %
C=
4.71 %
9.82 %
1.67 %
0.83 %
9.57 %
TOTAL

100.00 %
Solving for the Heating value
HV = 14600𝐶 + 62000 (𝐻2 −
𝑂2
8
) + 4050𝑆
HV = 14600(0.7340) + 62000 (0.0471 −
0.0982
8
) + 4050(0.0083)
HV = 12 909.165 Btu/lb
c. Consider the Ash and Moisture-free (combustible) Analysis

To obtain the combustible analysis, divide each component, except Ash and Moisture, of the
as received analysis by the following factor:
𝐹𝑎𝑐𝑡𝑜𝑟 = 1 − (
0.0863 + 0.977
) = 0.8160
100
66.23
0.816
4.25
H2 =
0.816
8.86
O2 =
0.816
1.51
N2 =
0.816
C=
81.16 %
5.21 %
10.86 %
1.85 %
20
S=
0.75
0.816
0.92 %
TOTAL

100.00 %
Solving for the Heating Value
HV = 14600C + 62000 (H2 −
O2
8
) + 4050S
HV = 14600(0.8116) + 62000 (0.0521 −
0.1086
8
) + 4050(0.0092)
HV = 14 275.17 Btu/lb
d. Consider the moisture, ash, and sulfur-free basis of analysis

To obtain this analysis, divide each component except moisture, ash and sulfur of asreceived analysis by the following factor:
Factor = 1 − (
0.75 + 0.0863 + 0.977
) = 0.8085
100
66.23
0.8085
4.25
H2 =
0.8085
8.86
O2 =
0.8085
1.51
N2 =
0.8085
81.92 %
C=
5.26 %
10.96 %
1.87 %
TOTAL

100.00 %
Solving for the heating value
HV = 14600C + 62000 (H2 −
O2
8
) + 4050S
HV = 14600 (0.8192) + 62000 (0.0526 −
0.1096
8
) + 4050 (0)
HV = 14 372.12 Btu/lb
21
Lesson6.Combustion
According to Francisco (2014) Combustion is:

Combustion is the rapid chemical union with oxygen of an element in which the exothermic
heat of reaction is sufficiently great and the rate of reaction is sufficiently fast that useful
quantities of heat are liberated at elevated temperature.

Combustion is synonymous with oxidation and is the union of oxygen with a combustible
material.
Standard Air Composition and Standard from Francisco (2014)
1. Components By Weight
𝐍𝟐 = 𝟕𝟔. 𝟗 % 𝐚𝐧𝐝 𝐎𝟐 = 𝟐𝟑. 𝟏 %
2. Components By Volume
𝐍𝟐 = 𝟕𝟗 % 𝐚𝐧𝐝 𝐎𝟐 = 𝟐𝟏 %
3. Air Gas Constant
R = 0.28708 kJ/kg-°K
4. Molecular Weight or Mass
MW = 28.95 kg/kgmol = 28.95 lb/pmol
5. At p = 14.7 psi pr 101.325 kPa and t = 32 °F or 0 °C
Volume of a mole of any gas, 𝐕𝐨 = 𝟑𝟓𝟗 𝐅𝐭 𝟑 /𝐩𝐦𝐨𝐥 = 22.43 𝐦𝟑/kgmol
Combustion Chemistry from Francisco (2014)

The following equations are the combustion chemistry of combustible components of the fuel.
1. Combustion of Carbon, C + O2 = CO2

Analysis by weight

12 kg C + 32 kg O2 = 44 kg CO2
8
11
1 kg C + ( ) kg O2 = ( ) kg CO2
3
3
Oxygen required for 1 kg C
8
Woc = 3 kg C/ kgO2

Molal Analysis
1 mole C + 1 mole O2 = 1 mole CO2
22
2. Combustion of Hydrogen, 2H2 + O2 = 2H2 O

Analysis by weight
4 kg H2 + 32 kg O2 = 36 kg H2 O
1 kg H2 + 8 kg O2 = 9 kg H2 O

Oxygen required for 1 kg H2
WOH2 = 8 kg H2 /kg O2

Molal Analysis
2 moles H2 + 1 mole O2 = 2 moles H2 O
1
1 mole H2 + mole O2 = 1 mole H2 O
2
3. Combustion of Sulfur, S + O2 = SO2

Analysis by Weight
32 kg S + 32 kg O2 = 64 kg SO2
1 kg S + 1 kg O2 = 2 kg SO2

Oxygen required for 1kg sulfur
WOS =1 kg S/ kg O2

Molal Analysis
1 mole S + 1 mole O2 = 1 mole SO2
4. Incomplete Combustion of Carbon, 2 C + O2 = 2 CO

Analysis by Weight
24 kg C + 32 kg O2 = 56 kg CO
4
7
1 kg C + ( ) kg O2 = ( ) kg CO
3
4
4
WOC = kg C/kg O2
3

Molal Analysis
2 moles C + 1 mole O2 = 2 moles CO
1 mole C + ½ mole O2 = 1 mole CO
Theoretical Oxygen Required to Combust Fuel from Francisco (2014)
8
Wo = 3 C + 8 (H2 −
O2
8
)+S
→ kg O2 / kg fuel
Where,
Wo = theoretical oxygen required, kg/kg fuel
23
C = fuel carbon content, kg/kg fuel
O2 = oxygen content, kg/kg fuel
S = sulfur content of the fuel, kg/kg fuel
Theoretical Air Required from Francisco (2014)
𝐖
𝐎
𝐂+𝟖(𝐇𝟐 − 𝟐 )+𝐒
𝟖
𝐨
𝐖𝐚 = 𝟎.𝟐𝟑𝟏
=𝟑

𝟖
𝟎.𝟐𝟑𝟏
Equation by Vopat
Wa = 11.5 C + 34.5 (H2 −

O2
) + 4.32S
8
Equation by Potter
Wa = 11.53 C + 34.36 (H2 −

O2
) + 4.32S
8
Equation Used by Subaran
Wa = 11.44 C + 34.32 (H2 −
O2
) + 4.29S
8
Where, Wa = theoretical air required, kg/kg fuel, lb/lb fuel
Complete Combustion of Carbon with Air from Francisco (2014)
C + Air = CO2 + Nitrogen
C + O2 + 3.76N2 = CO2 + 3.76N2
By Weight:
12 kg C + 32 kg O2 + 3.76(28) kg N2 = 44 kg CO2 + 3.76(28) N2
Relative Weight:
8
1 kg + 3 kg O2 + 8.77 kg N2 =
Theoretical Air/ Fuel Ratio:
11
3
kg CO2 + 8.77 kg N2
8
A : F = + 8.77 = 11.45 kg air/ kg fuel
3
SAMPLE PROBLEMS from Francisco (2014)
24
1 Deduce the combustion equations for a) complete combustion of 1 kg and 1 m3 ethylene, C2 H4 , in
air, and b) combustion with 25 % excess air.
Solution:
a) Consider the complete combustion
C2 H4 + Air = CO2 + H2 O + N2
C2 H4 + aO2 + 3.76aN2 = bCO2 + 3.76aN2 + cH2 O
For the coefficients (atom balance):
Carbon, 2 = b
Oxygen, 2a = 2b + c; a = b + c/2 = 2 + 1 = 3
Hydrogen, 4 = 2c; c = 2
For the combustion equation,
By Volume,
1 m3 C2 H4 + 3 m3 O2 + 11.28m3 N2 = 2 m3 CO2 + 2m3 H2 O + 11.28m3 N2
Analysis by Weight,
[12(2) + 4(1)] + 3(32) + 11.28(28) = 2(44) + 2(18) + 11.28(28)]
28 kg C2 H4 + 96 kgO2 + 315.84 kgN2 = 88kg CO2 + 36 kgH2 O + 315.84 kg N2
Relative Weight,
1 kgC2 H4 + 3.43 kgO2 + 111.28kgN2 = 3.14kgCO2 + 1.285kgH2 O + 11.28kgN2
Theoretical Air/Fuel Ratio,
A : F = 3.43 + 11.28 = 14.71 kg air per kg fuel (ans.)
25
b. Consider the Actual Combustion with e = 25 %
Weight Analysis,
28 kg Fuel + 120 kg O2 + 394.8 kg = 88 kg CO2 + 36 kg H2 O + 24 kg O2 + 394.8 kg N2
1 kgC2 H4 + 4.286 kgO2 + 14.1kgN2 = 3.14CO2 + 1.285kgH2 O + 0.857 kgO2 + 14.1kgN2
Therefore, the air : Fuel Ratio,
𝑊𝑎𝑎 = 4.286 + 14.1 = 18.386 kg/kg fuel
(ans.)
Combustion Products from Francisco (2014)

The composition of the products depends on the kind of fuel used, the A/F ratio, and the
conditions of combustion. The products usually contain carbon dioxide (CO2 ), carbon
monoxide (CO), oxygen, nitrogen, free carbon, fly ash, steam, sulfur dioxide, and unburned
carbons.
Considering the Cn Hm
Cn Hm + aO2 + 3.76aN2 = bCO2 + cH2 O + 3.76aN2
Atom balance,
Carbon: n = b; b = n
Hydrogen: m = 2c; c = 0.5m
Oxygen: 2a = 2b + c; a = n + 0.25m
26
Substituting the coefficients,
Cn Hm + (n + 0.25m)O2 + 3.76(1 + e)(n + 0.25m)N2 = nCO2 + 0.5mH2 O + e(n + 0.25m) +
3.76(1 + e)(n + 0.25m)N2
For the Air-Fuel Ratio with e,
Waa =
Waa =
32(1+e)(n+0.25m)+(1+e)(3.76)(n+0.25m)(28)
12n+m
[32+3.76(28)](1+e)(n+0.25m)
12n+m
𝐀
𝟏𝟑𝟕. 𝟔(𝐧 + 𝟎. 𝟐𝟓𝐦)(𝟏 + 𝐞)
= 𝐖𝐚𝐚 =
𝐅
𝟏𝟐𝐧 + 𝐦
Lesson 7.Air-Fuel Ratio In Terms of The Products Of Combustion (Volumetric
Analysis)
According to Francisco (2014) air-fuel ration of the product of combustion are:
Considering the Dry Gas:
CO2 =
100n
n + e(n + 0.25m) + 3.76(1 + e)(n + 0.25m)
Dividing both the denominator and numerator by n,
𝐶𝑂2 =
100
𝑚
𝑚
1 + 𝑒 (1 + 0.25 𝑛 ) + 3.76(1 + 𝑒) (1 + 0.25 𝑛 )
From the stoichiometric (gravimetric) analysis of fuel,
H2
m
=
;
C
12n
m 12H2
=
n
C
Where, H2 = percent hydrogen by weight
C = percent carbon by weight
Substituting the value of m/n,
27
CO2 =
CO2 =
CO2 =
100
1 + [1 + 0.25 (
12H2
C
)] + 3.76(1 + e) [1 + 0.25 (
100
1 + [e + 3.76(1 + e)] [1 + 0.25 (
12H2
C
)]
100
1 + 3.76 (1 +
3H2
C
) + 4.76e (1 +
3H2
C
)
=
=
12H2
C
)]
100
1 + (3.76 + 4.76e) (1 +
3H2
C
)
100
1 + (3.76 + 11.28
H2
C
) + 4.76e (1 +
3H2
C
)
Solving for the value of e,
1 + [3.76 + 11.28 (
4.76e (1 +
e=
e=
H2
3H2
100
)] + 4.76e (1 +
)=
C
C
CO2
3H2
100
H2
)=
− [3.76 + 11.28 ( )] − 1
C
CO2
C
100 − CO2 − (3.76 + 11.28
4.76e (1 +
3H2
C
100 − (4.76 + 11.28
e=
(4.76 + 14.28
H2
C
) CO2
C
100 − [1 + [3.76 + 11.28
=
(4.76 + 14.28
H2
C
H2
C
]] CO2
) CO2
) CO2
) CO2
H2
C
C
) CO2
H2
100 − (4.76 + 11.28
(4.76 + 14.28
H2
H2
C
) CO2
→ By Maleev
) CO2
A : F Ratio in Terms of H2 and C,
H2
Waa
137.6(n + 0.25m)(1 + e) 137.6 (1 + 0.25 C ) (1 + e)
=
=
H
12n + m
12 + 2
C
But,
m
n
Waa =
=
12H2
C
137.6 [1 + 0.25 (
12 +
12H2
C
12H2
C
)] (1 + e)
=
137.6 (1 +
3H2
C
12 (1 +
) (1 + e)
H2
C
)
28
𝐖𝐚𝐚 =
𝟏𝟏. 𝟓𝟑 (𝟏 +
𝟑𝐇𝟐
(𝟏 +
𝐂
𝐇𝟐
𝐂
) (𝟏 + 𝐞)
)
Determination of The Values Of N And M In 𝐂𝐧 𝐇𝐦 (Sample Problems) from Francisco (2014)
1. An analysis of exhaust gases shows the products to consist of the following percentages vy
volume: 𝐶𝑂2 = 12.1, 𝑂2 = 0.30, CO = 3.3, 𝐻2 = 1.3, 𝐶𝐻4 = 0.30, 𝑁2 = 82.7. Assuming the fuel to be a
pure hydrocarbon of the form Cn Hm , a) find the values of n and m; b) Balance the combustion
equation; and c) Determine the air-fuel ratio.
Solution:
a) Consider 100 moles of the products
Cn Hm + aO2 + 82.7N2 = 12.1CO2 + 0.3 O2 + 3.3 CO + 1.3 H2 + 0.3 CH4 + 82.7 N2 + bH2 O
Where, a = moles of oxygen in air supplied; b = moles of water vapor in the products
#2. A natural gas has the following percentage volumetric compositions:
CH4
= 59.8 %
C2 H 6
= 37.6 %
N2
= 2.2 %
CO2
= 0.4 %
Calculate:
a. The molecular mass of weight of the natural gas
b. The gravimetric analysis of the natural gas
c. The gravimetric analysis per element
d. The amount of air required per kg fuel for complete combustion and an excess air 50 %
e. The volumetric rate of air required at a pressure of 101.325 kPaa and temperature of 15.56 °
C for a fuel mass flow rate of 50 kg/min.
f.
The amount of air required per 𝑚 3of fuel at 116 kPaa and 27 °C
Solution:
a) Solving for the molecular mass or weight
29
MW = 16CH4 + 30C2 H6 + 28N2 + 44CO2
MW = 16(0.598) + 30(0.376) + 28(0.022) + 44(0.004) = 21.64 kg/kgmol fuel
b) Gravimetric analysis by component
16(CH4 )
MW
30(C2 H6 )
MW
28(N2 )
MW
44(CO2 )
MW
CH4
C2 H 6
N2
CO2
16(0.598)
[
] (100%)
21.64
30(0.376)
[
] (100%)
21.64
28(0.022)
[
] (100%)
21.64
44(0.004)
[
] (100%)
21.64
TOTAL
44.21 %
52.13 %
2.85 %
0.81 %
100 %
c) Computation of the gravimetric analysis by weight
C
12CH4 + 24C2 H6 + 12CO2
(100%)
MW
12(0.598) + 24(0.376) + 12(0.004)
(100%)
21.64
75.083 %
H2
4CH4 + 6C2 H6
(100%)
MW
4(0.598) + 6(0.376)
(100%)
21.64
21.479 %
N2
28N2
(100%)
MW
28(0.022)
(100%)
21.64
2.847 %
O2
32CO2
(100%)
MW
31(0.004)
(100%)
21.64
0.591 %
TOTAL
d)
100
Solving for the actual air required per kg fuel
(1 + e)Wa (1 + e) 8
O2
[ C + 8 (H2 − ) + S]
Waa =
=
0.231
0.231 3
8
Waa =
(1.50)Wa (1.50) 8
0.0059
[ (0.7508) + 8 (0.2148 −
=
) + 0]
0.231
0.231 3
8
𝐖𝐚𝐚 = 𝟐𝟏. 𝟏𝟐 kg air per kg fuel
e) Determination of the volumetric rate of air required
Solving for the mass flow rate of air required,
ma = (Waa )mF = (24.12)(50) = 1206
kg
min
30
Solving for the volumetric rate of air required
Va =
ma R a Ta 1206(0.28708)(15.56 + 273)
=
= 985.71 m3 /min
Pa
101.325
For the given Flue Gas Analysis (Orsat Analysis)
Excess Air, e =
O2 − 0.5CO
(100%)
0.26N2 − (O2 − 0.5CO)
Mass of Flue Gas
Wdg =
4CO2 + O2 + 700
3
5
(Cab + S) + S
3(CO2 + CO)
8
8
where, Cab = actually burned carbon, kg/kgfuel
Cab = C − Cub ;
Cub = unburned carbon
Assessment Task 1
Answer the following; write your solution in engineering lettering.
1. A diesel power plant uses fuel with a heating value of 43,000 kJ/kg. What is the density of
the fuel at 25 °C?
2. A typical industrial fuel oil, C16 H32 , has 20 % excess air by weight. Assuming complete
oxidation of the fuel, calculate the actual air-fuel ratio by weight.
3. A gaseous fuel mixture has a molal analysis: H2 = 14 % ; CH4 = 3 % ; CO = 27 % ; O2 =
0.6 % ; CO2 = 4.5 % ; N2 = 50.9%. Determine the air-fuel ratio for complete combustion on molal
basis.
4. There are 20 kg of flue gas formed per kg of fuel oil burned in the combustion of fuel 𝐶12 𝐻32 .
What is the excess air percent?
5. A certain coal has the following ultimate analysis by weight : C =67 %, A = 5 %, M = 8%, S =
7 %, 𝑁2 = 6 %, 𝑂2 = 4 %, 𝐻2 = 3 %. a) Calculate the heating value of the fuel, kJ/kg, b) What is
the lower heating value, kJ/kg, c) Determine the air-fuel ratio of this coal is burned with 50 %
excess air.
6. At a certain temperature of 25 °C, what is the specific gravity of a fuel at 20 °API?
31
Summary

Combustion is the rapidchemical union with oxygen of an element whose heat of reaction is
sufficiently great and rate of reaction is fast enough that quantities of heat are liberated at
elevated temperature.

Straight oils are oils produced entirely from the crude chosen entirely through elimination of
undesired constituents by refining process.

Additive oils are oils produced by adding to straight mineral oils; certain oil are soluble
compounds that enhance the lubricating oil properties for use in Diesel engines.

Proximate analysis – composition of fuel that gives on mass basis the relative amount of
volatile matter, fixed carbon, moisture and ash.

Fuel is any substance or combustible material that by rapid oxidation or burning produces
heat and light. An example of a fuel is coal or gasoline. A fuel is composed of chemical
elements which, in rapid chemical union with oxygen, produce combustion. Fossil fuels are
fuels that originate from the earth as a result of the slow decomposition and chemical
conversion of organic material. The basic forms of fossils are solid (coal), liquid (oil), and
natural gas. Synthetic fuels or synfuels, the new combustible-fuel-options, are liquid or
gaseous fuels derived largely from coal, oil shale, and tar sands (Fossil Fuel, n.d).
Reference
Francisco, Jose R. (2014). Lecture Notes in Industrial Plant Design, Lesson 8: Fuels and
Combustion, p.g 125 to p.g 151.
32
MODULE 2
STEAM GENERATING UNIT
Introduction
Steam Generator is a combination of apparatus for producing, furnishing, or recovering heat,
together with apparatus for transforming to a working fluid the heat thus available (Francisco,
2014).
Learning Objectives
At the end of this module, students should be able to:
1. Learn the different Types of Boiler;
2. Solve problems involving Boiler ratings and performance; and
3. Know and solve Steam Generator Heat Balance problems.
33
Lesson 1.Types of Boiler
According to Francisco (2014) the different types of boiler are:
1. Fire-tube boilers – are those having the heat gases of combustion inside the tubes. Limited
only to a working pressure of 1.7 Mpag.
a. Horizontal-Return-Tubular (HRT) Boiler
b. Packaged Fire-Tube Boiler
c. Vertical Tubular Boiler
2. Water-Tube-Boilers – are those having the hot gases of combustion outside the tubes.
a. Horizontal Straight-Tube Boiler
b. Central Station Steam Generators
c. Marine Steam Generators
d. Forced-circulation Steam Boilers
Water-Tube Boiler Circulation Principle from Francisco (2014)
Figure 2.1. Water-Tube Boiler Circulation Principle from Francisco (2014)
o
Dryness Factor (or fraction) or Quality of steam – refers to the amount of steam in the
mixture of water and steam at the riser.
o
Top Dryness Factor – applies to the mixture leaving the tubes.
34
Lesson 2. Boiler Ratings and Performance
According to Francisco (2014) Boiler rating and performance can be evaluated by:

Boiler Hp = equivalent to the generation of 34.5 lb/hr (15.44 kg/hr) from water at 212 °F (100 °C)
to saturated steam also at 212 °F (100 °C).

Energy of Evaporation
lb
Energy of Evaporation = (34.5)(hfg @ 212 °F) = (34.5 hr) (970.3
kg
Btu
lb
) ≈ 33 500 Btu/hr
kJ
Energy of Evaporation = (15.46)(hfg @ 100 °C) = (15.646 hr ) (2257 kg ) ≈ 35 314 kJ/hr
Where, hfg = enthalpy of evaportion at 212 °F (100 °C) = 970.3
Btu
lb
= 2257 kJ/kg
Therefore, 1 Bo Hp = 33 500 Btu/hr = 35 314 kJ/hr

Boiler Horsepower, Bo Hp
Bo. Hp =
ms (h2 − h1 ) ms (h2 − h1 )
=
35 314
33 500
Where, ms = mass flow rate of steam, kg/hr, lb/hr
h1 = enthalpy of feed water, kJ/kg, Btu/lb
h2 = enthalpy of steam leaving the boiler, kJ/kg, Btu/lb

Factor of Evaporation, FE
Boiler Ouput,
FE =

,
2257 or 970.3
lb
=
h2 − h1 h2 − h1
=
2257
970.3
Equivalent Evaporation, Eq. Evap.
Eq. Evap. = (F. E. )ms =

kJ Btu
kg
Boiler Output, kJ/hr Boiler Output, Btu/ Hr
=
2257
970.3
Rated Boiler Horsepower
Rated Bo. Hp =
Heat Transfer Area or Surface, m2 , or ft 2
k
Where, k = 0.91 m2 /Bo. Hp = 10ft 2 /Bo. Hp → For water-tube boiler
m2
K = 1.10 Bo . Hp = 12ft 2 /Bo. Hp → For fire-tube boiler

Developed Boiler Horsepower
∑ ms (∆h)
Boiler Output, kJ/hr Boiler Output, Btu/hr
Dev. Bo. Hp =
=
=
35 314 or 33 500
35 314
33 500
35
Dev. Bo. Hp =
(F. E. )(ms ) Eq. Evap. (F. E. )(ms ) Eq. Evap.
=
=
=
15.646
15.646
34.5
34.5
Where, ∑ ms (∆h) = total heat transfer to cycle fluid in steam generator (boiler, superheater,
heater, Economizer, reheater), kJ/hr, Btu/hr.

Percent Rating
Percent Rating =

Dev. Bo. Hp
Rated Bo. Hp
Rating for Modern Steam Generators
kW Rating =
Steam Generated at Rated Capacity, kg/hr
Over − all Station Steam Rate, kg/kw ∙ hr
where, Steam Generated = kg/hr of steam at a given pressure and temperature and feed water
temperature.

Boiler Performance – the measure of boiler heat output, QB , in kJ/hr or Btu/hr
a) QB = Based on the flow diagram
b) QB = mF (HHV)EB
c) QB = (Dev. Bo. Hp)( 35 314)
d) QB = (Eq. Evap.)(2257)
e) QB = (Rated Bo. Hp. ) (
% Rating
100 %
or
or
QB = (Dev. Bo. Hp)(33 500)
QB = (Eq. Evap.)(970.3)
) (35 314) or QB = (Rating Bo. HP) (
% Rating
100%
) (33 500)
Where, QB = Boiler Output, kJ/hr, Btu/hr
HHV =higher heating value of the fuel, kJ/kg, Btu/lb
EB = boiler efficiency
mF = mass flow rate of fuel, kg/hr, lb/hr

Boiler Over-all Efficiency
QB
) (100 %)
QA
Where, QB = Boiler output, kJ/hr, kJ/kg, Btu/hr, Btu/lb
EB = (
QA = heat added or heat from the fuel, kJ/hr, Btu/hr = mF (HHV)

New Over-all Boiler Efficiency
36
ƞB =
∑ ms (∆h)
QB ′
(100%) =
(100%)
mF (HHV)
mF (HHV)
Where, QB ′ = heat absorbed by the boiler fluid less the amount of heat used for the boiler
auxiliaries.

Boiler and Furnace Efficiency
EBF
QB
⁄(1 − M − A − C )
ub
] (100 %)
=[
HHV⁄
(1 − M − A)
Where, M = moisture content of the fuel
A = ash content of the fuel
Cub = unburned carbon

Grate Efficiency
EG =
kg combustible burned / kg fuel fired
1 − M − A − Cub
] (100 %)
(100 %) = [
kg combustible / kg fuel fired
1−M−A
Lesson 3.Steam Generator Heat Balance
The fuel supplied to a furnace when completely burned releases its heating value. The
energy primarily changes the feedwater pumped to the boiler into steam. All the heating value does
not go to this useful purpose. These are some losses in the form of incomplete combustion, of flue
gases leaving at high temperature, and of radiated and convected heat from the outside of the steam
generator. An energy balance shows the distribution of the heating value of the fuel to the formation
of steam and to the various losses (Francisco, 2014).
Figure 2.2 Boiler Heat Balance
37
Useful Energy from Francisco (2014)
Energy Absorbed by the steam generator fluid (or Useful Energy)
Q1 = Ws (hg2 − hf1 ) + Wr (hg4 − hg3 ) + WB (hf2 − hf1 )
Where, Q1 = energy absorbed by the boiler fluid or the useful energy, kJ/kg fuel, Btu/lb fuel
hg2 = enthalpy of steam leaving the boiler, kJ/kg, Btu/lb
hf1 = enthalpy of feed water, kJ,kg, Btu/lb
hf2 = enthalpy of water at boiler pressure, kJ/kg, Btu/lb
hg4 = enthalpy of reheat steam leaving the steam generator, kJ/kg, Btu/lb
hg3 = enthalpy of reheat steam entering the steam generator, kJ/kg, Btu/lb
Ws = mass of steam from feedwater per unit mass fuel, kg/kg fuel, lb/lb fuel
Wr = mass of reheated steam per unit mass fuel, kg/kg fuel, lb/lb fuel
WB = mass of blowdown per unit mass fuel (often negligible), kg/kg fuel, lb/lb fuel
Energy Losses from Francisco (2014)
1. Energy Loss due to mechanical moisture in fuel (or energy loss due to evaporating and
superheating moisture in fuel)
English Units:
Q2 = W(1066 + 0.5t g − t a )
Q2 = W(1089 + 0.46t g − t a )
when t g > 575 °𝐹
when t g < 575 °𝐹
SI Units:
Q2 = W(2479.81 + 20.935t g − 4.187t a )
Q2 = W(2533.31 + 1.92602t g − 4.187t a )
when t g > 302 °𝐶
when t g < 302 °𝐶
Where, Q2 = energy loss due to moisture content of the fuel as fired, kJ/kg fuel, Btu/lb fuel
W = moisture content of the fuel, kg/kg fuel, lb/lb fuel
t g = temperature of flue gas after passing all heat-transfer surfaces, °C, °F
t a = temperature of air entering for combustion to the furnace or air heater, if the latter
is used, °C, °F
2. Energy loss due to moisture from hydrogen in the fuel (or energy loss due to evaporating and
superheating moisture formed by combustion of hydrogen)
38
English Units:
Q3 = 9H2 (1066 + 0.5t g − t a )
Q3 = 9H2 (1089 + 0.46t g − t a )
when t g > 575 °𝐹
when t g < 575 °𝐹
SI Units:
Q3 = 9H2 (2479.81 + 20.935t g − 4.187t a )
Q3 = 9H2 (2533.31 + 1.92602t g − 4.187t a )
when t g > 302 °𝐶
when t g < 302 °𝐶
3. Energy Loss due to moisture from air
English Units:
Q4 = 0.47 WAV (t g − t a )
SI Units:
Q4 = 1.96789WAV (t g − t a )
Where, WAV = mass of moisture content n supp air per unit mass fuel, kg/kg fuel, lb/lb fuel
WAV = (1 + e)Wa (g)
Wa = theoretical air-fuel ratio, kg/kg fuel, lb/lb fuel
g = humidity ratio of supply air, kg/kg d.a. ,lb/lbd.a.
3. Energy Loss due to dry flue-gas sensible heat
English Units:
Q5 = WDG cp (t g − t a ) = 0.24WDG (t g − t a )
SI Units:
Q5 = WDG cp (t g − t a ) = 1.005WDG (t g − t a )
Where, WDG = mass of dry flue gas per unit mass fuel, kg/kg fuel, lb/lb fuel
4. Energy Loss due to incomplete combustion
English Units:
39
Q6 = (
28 CO
) (10187)
44CO2 + 28CO + 28N2 + 32O2
Q6 = 10160Ci = 10160Cab (
CO
)
CO2 + CO
SI Units:
Q6 = (
28 CO
) (4380)
44CO2 + 28CO + 28N2 + 32O2
Q6 = 23 631Ci = 23 631Cab (
CO
)
CO2 + CO
Note: CO and CO2 in the second equation must be an Orsat analysis (% by Volume)
5. Heat Loss due to Unburned Carbon
English Units:
Q7 = 14 600 (C − Cab )
Q7 = Wr (HHV)r
SI Units:
Q7 = 33 964 (C − Cab )
Q7 = Wr (HHV)r
Where, Wr = mass of refuse, kg/kg fuel, lb/lb fuel
(HHV)r= heating value of refuse
6. Heat Loss Due to radiation and Unaccounted-for
Q8 = HHV − ∑ Q1−7 = HHV(Q1 + Q2 + Q3 + Q4 + Q5 + Q6 + Q7 )
40
Assessment Task 2
Answer the following problems write your solution in engineering lettering.
1. In a water-tube boiler with 186 m2 of boiler heating surface and operating at 1.75
MPaa, water is fed at the rate of 2 268 kg/hr. Water enters the economizer at 56 °C
and leave therefrom at 112 °C, while steam with 5 % moisture enters the superheater
and leaves therefrom as superheated at the same pressure. Coal with a heating value
of 28 000 kJ/kg is burned at a rate of 250 kg/hr. The gross over-all boiler efficiency is
80 %. The generated steam flows to a steam turbine through a well-insulated pipe
along which the pressure drop is 0.10 Mpa and the steam temperature becomes 250 °
C. in the turbine, steam expands to a dry and saturated at 0.065 MPaa.
Determine:
a) The temperature of steam leaving the superheater
b) The boiler heat losses per hour
c) The equivalent evaporation
d) The percent rating
e) The weight of fuel saved in using the economizer per hour
f)
The weight of additional fuel used per hour in the superheater
g) The energy loss per hour in the piping line
h) The power plant generator output if the turbine generator combined efficiency is 88 %
i)
The power plant gross thermal efficiency.
2. A steam boiler has an actual evaporation rate of 8.23 kg/kg coal fired. Coal as fired
contains 2 % moisture. Dry coal contains 5 % ash and has a heating value of 29 776 kJ/kg.
During the test 12 % of coal fired is delivered from the ashpit as refuse. The steam leaving the
boiler is saturated at a pressure of 0.70 MPaa and the temperature of feed water is 49 °C.
Determine a) the boiler efficiency; b) the boiler-furnace efficiency; c) the grate efficiency.
3. A water-tube boiler having a heating surface of 325.23 m2 evaporates 6349.21 kg of water
in an hour from a feed temperature of 66 °C. Boiler pressure is at 1.04 MPaa and the steam
quality at the boiler outlet is 99 %. What percent of its rated Bo. Hp was the boiler developing?
41
4. A condenser receives 11 338 kg of steam per hour at 89 % dryness factor. Steam
temperature in the condenser is 33 °C and free air at 101.325 kPaa and 15.56 °C leaks into
the condenser at the rate of 0.51 𝑚 3/min. Determine the pressure in the condenser.
Summary

Steam Generator is a combination of apparatus for producing, furnishing, or recovering heat,
together with apparatus for transforming to a working fluid the heat thus available (Francisco,
2014).

The two types of boilers are fire tube boiler and water-tube boiler

The fuel supplied to a furnace when completely burned releases its heating value. The energy
primarily changes the feedwater pumped to the boiler into steam. All the heating value does not
go to this useful purpose. These are some losses in the form of incomplete combustion, of flue
gases leaving at high temperature, and of radiated and convected heat from the outside of the
steam generator. An energy balance shows the distribution of the heating value of the fuel to the
formation of steam and to the various losses (Francisco, 2014).
Reference
Francisco, Jose R. (2014). Lecture Notes in Industrial Plant Design, Lesson 9: Steam Generators
p.g 152 to p.g 161.
42
MODULE 3
MECHANICAL DRYERS
Introduction
Drying is the process of removing moisture in varying amounts from solid or semi-fluid materials.
This process may be accomplished by pressure, suction, decantation, or evaporation (Francisco,
2014).
Learning Objectives
At the end of this module, students should be able to:
1. Learn the different processes of drying;
2. Determine the different types of Dryer; and
3. Solve problems involving Mechanical Dryers.
43
Lesson 1. Mechanical Dryer
Important Information for Mechanical Dryer from Francisco (2014):

Dryer is machine or equipment used for drying process.

Dewaterer is a machine or equipment used for dewatering process.

Dewatering – is the common term for processes using only pressure, suction, or decantation.
This process used to remove or reduce only a portion of the surface moisture of the materials.

Evaporation is the main principle of the term drying process. In drying process, not only the
surface liquid is removed or reduced but also removes internal moisture and in many cases
water in chemical combination.

Inherent moisture and bed moisture are the terms used for describing the moisture in the mass
that is not on the surface of the material.

Chemically combined water is that water occurring when a chemical component of the material
changes its chemical composition by heat or other means.
Parts of Continuous Drying Process from Francisco (2014)
1. Application of heat to the material and/or to the air mixture which carries away the moisture.
2. Means for removing the water vapor, steam, or mixture
3. Conveying the material in its wet, semi-dried, dried condition into, through, and out of the
apparatus, allowing the material the proper time for contact with the heating and moistureremoving elements.
Types of Dryer from Francisco (2014)
1. Direct-heat type dryer
o
This may have the flame from combustion impinging on the material being dried, or the
gases of combustion may be mixed with additional air so that mixture in contact with
the material is reduced in temperature.
2. Indirect-heat type dryer.
o
In this type, the gases of combustion pass through the spaces surrounding, or in other
ways heating, the drying chamber, but the gases are not allowed in contact with the
material being dried.
3. Steam-heated typed dryer
o
In this type, material is in contact with steam piped or the air is passed over steam
heaters and then over or through the material being dried.
44
Six Commercial Types of Dryers and Three Materials Suitable for Each Dryers from Francisco (2014):
1. Rotary Dryer

Materials: copra, sand, wood chips
2. Compartment Batch Dryer

Materials: wood, enamel wares, foodstuffs
3. Centrifugal Dryer

Materials: sugar, fertilizer, salt
4. Hearth Dryer

Materials: copra, enamel wares, chalk
5. Tower Dryer

Materials: palay, wheat, grains
6. Infrared Ray Dryer

Materials: air conditioners, cars, refrigerators
Hygroscopic Materials are those substances which are particularly variable in the moisture content
that they can possess at different times (Francisco, 2014).
o
Example: wood, leather, foodstuff, paper, tobacco, cloth, hair
Bone-Dry Weight (BDW) or Dry Bone Weight is the final constant weight reached by a hygroscopic
substance after being dried out (Francisco, 2014).
Regain is the hygroscopic moisture content of a substance expressed as a percentage of the dry-bone
-weight of the material (Francisco, 2014).
𝐑𝐞𝐠𝐚𝐢𝐧 =
𝐖𝐞𝐢𝐠𝐡𝐭 𝐨𝐟 𝐌𝐨𝐢𝐬𝐭𝐮𝐫𝐞
𝐁𝐨𝐧𝐞 − 𝐝𝐫𝐲 − 𝐖𝐞𝐢𝐠𝐡𝐭
Moisture Content is usually expressed as a percentage of the gross weight of the body, and may refer
to both hygroscopic and purely substance moisture (Francisco, 2014).
45
𝐌𝐨𝐢𝐬𝐭𝐮𝐫𝐞 𝐂𝐨𝐧𝐭𝐞𝐧𝐭 =
𝐖𝐞𝐢𝐠𝐡𝐭 𝐨𝐟 𝐌𝐨𝐢𝐬𝐭𝐮𝐫𝐞
𝐆𝐫𝐨𝐬𝐬 𝐖𝐞𝐢𝐠𝐡𝐭
Gross Weight:
According to Francisco (2014):
Gross Weight = Bone-Dry-Weight + Weight of Moisture
𝐦𝐆 = 𝐦𝐁𝐃𝐖 + 𝐦𝐌
Dryer Schematic Diagram
Figure 3.1 below illustrates the schematic diagram of a dryer with conditioned air that passed through
a CDA and re-heater (Francisco, 2014).
Figure3.1 Dryer Schematic Diagram from Francisco (2014)
Psychrometric Chart Of Drying Process
Figure 3.2 below illustrates the different drying processes in the psychrometric chart (Francisco, 2014).
46
Figure 3.2. Psychrometric Chart of Drying Processes from Francisco (2014)
SAMPLE PROBLEMS from Francisco (2014)
1. Copra enters a dryer, containing 60 % water and 40 % solids, and leaves with 5 % water and 95 %
solids. Find the mass of water removed on each: a) kilogram of original product; b) kilogram of final
product; and c) kilogram of Bone-dry material.
Given: Copra Dryer
Figure 3.3 Copra Dryer from Francisco (2014)
Required:
The weight of water removed based on each
a) Kilogram of original product
b) Kilogram of final product
c) Kilogram of bone-dry material
Solution:
Consider the following notations
MBD1 = mass of bone-dry material of the original or wet product (copra)
MBD2 = mass of bone-dry material of the final or dried product (copra)
47
MC1 = moisture content of the original or wet product
MC2 = moisture content of the final or dried product
MG1 = gross mass of the original or wet product
MG2 = gross mass of the final or dried product
a. Consider 1 kg of original product or wet feed
Solving for the gross weight of the final product,
MBD = MG (1 − MC) = MG (Solid Part)MBD1 = MBD2
MG1 (1 − MC1 ) = MG2 (1 − MC2 )MG2 =
MG1 (1−MC1 )
(1−MC2 )
=
1.0(0.40)
0.95
= 0.421 kg
For the mass of water removed, ∆GW,
MV = MG1 − MG2 = 1.0 − 0.421 = 𝟎. 𝟓𝟕𝟗 𝐤𝐠
b. Consider 1 kg of final or dried product
MBD1 = MBD2 MG1 (1 − MC1 ) = MG2 (1 − MC2 )
MG1 =
MG2 (1 − MC2 ) 1.0(0.95)
=
= 2.375 kg
(1 − MC1 )
0.40
For the mass of water removed, MV = MG2 − MG1 = 2.375 − 1.00 = 𝟏. 𝟑𝟕𝟓 𝐤𝐠
c. Consider 1.0 kg of the bone-dry material, MBD1 = MBD2 = 1.0
1.0
1.0
MG1 =
=
= 2.5 kg
1 − MC1 0.40
1.0
1.0
=
= 1.053 kg
1 − MC2 0.95
For the weight of water removed, MV = MG2 − MG1 = 2.5 − 1.053 = 𝟏. 𝟒𝟒𝟕 𝐤𝐠
MG2 =
2. A tower-type moisture dryer is to deliver 1000 kg/hr of cassava flour with 2 % residual moisture of
20 % in the feed. The air to be heated in the heating chamber is a mixture of fresh air at 33 °C DB and
60 % RH, and hot humid air from dryer at 49 °C DB and 60 % RH. The mixture at 38 °C and 65 % RH
is heated by a steam coil to 93 °C. The dryer is properly insulated so that moisture absorption can be
considered adiabatic. Compute:
48
a. Required flow of heated air mixture to dryer;
b. Capacity of forced draft fan on dryer
c. Heat in kcal/hr for heating the air mixture in the heating chamber
d. Percentage by weight of fresh air in the mixture.
Given: Drying System Shown
Figure 3.4 Drying System from Francisco (2014)
Required:
a. The 𝑚𝑎3 = 𝑚𝑎4
b. The fan capacity, 𝑚 3 /𝑠
c. The heat capacity of the heating chamber, kcal/hr
d. The % outdoor air required
49
Figure 3.5 Temperature from Francisco (2014)
Solution:
a. Determination of the mass flow rate of the heated air
Figure 3.6 Drying System mass flow rate from Francisco (2014)
Consider the dryer and using mass balance,
ma + ma (W3 ) + m5 = ma + ma (W4 ) + m6
ma (W3 ) + m5 = ma (W4 ) + m6
→ ma (W4 − W3 ) = m5 − m6 ma =
m2 − m6
W4 − W3
Where, m5 = gross mass of material entering the dryer
m6 = gross mass of material leaving the dryer
Solving for the bone-dry-mass of the material,
Gross mass = bone-dry-mass + mass of moisture
m6 = BDM + 0.02m6
50
Then, BDM = m6 − 0.02 m6 = m6 (1 − 0.02) = 0.98 m6 = 0.98 (1000)
BDM = 980 kg/hr
Solving for the gross mass of material entering.
BDM = m5 − 0.20 m5 = m5 (1 − 0.20) = 0.80 m5
m5 = BDM + 0.20 m5
ma =
BDM 980
=
= 1225 kg/hr
0.80 0.80
From the Psychrometric Chart (Carrier),
At point 1: 33 °C DB and 60 % RH,
h1 = 82.3 − 0.32 = 81.98 kJ/kg da; &W1 = 0.0192 kg/kg da
At point 2: 38 °C DB and 65 % RH,
h2 = 109 − 0.36 = 108.64 kJ/kg da; &W2 = 0.0275 kg/kg da
At point 3: 93 °C DB and W3 = W2 = 0.0275 kg/kg da
h3 = 1.007(93) − 0.026 + 0.0275[2501 + 1.84(93)] = 167.01 kJ/kg da
At point 4: 49 °C DB and 60 % RH, pv4 = RH(pd )
Where, pd = 11.749 kPaa (from Steam Tables by Keenan at 49 °C)
pv4 = 0.60 (11.749) = 7.0494 kPaa
Then, W4 = (
pv4
7.0494
) = (0.622) (
)
pt − pv4
101.325 − 7.0494
= 0.0465 kg/kg da
h4 = 1.007 t 4 − 0.026 + W4 (2501 + 1.84 t 4 )
h4 = 1.007 (49) – 0.026 + 0.0465 [2501 + 1.84 (49)] = 169.715 kJ/kg da
therefore, ma =
m5 − m6
1225 − 1000
=
= 11842.11 kg/hr
W4 − W3 0.0465 − 0.0275
ma = 3.29 kg/s
51
b. For the capacity of the forced-draft fan, Q3 = ma (v3 )
W3 (pt )
0.0275 (101.325)
where, pv3 =
=
= 4.2901 kPaa
W3 + 0.622
0.0275 + 0.622
v3 =
RT6
(0.28708)(93 + 273)
=
= 1.0828 m3 / kg da
pt − pv3
101.325 − 4.3901
c. For the heat added to the heating chamber, consider the heating chamber
Figure 3.7 Drying System heating chamber from Francisco (2014)
qh = ma (h3 − h2 ) − (3.29)(167.01 − 101.64) = 192.04 kW
But, 1 kcal = 4.187 kJ
1 kcal
qh = 192.04 kW (
) (3 600 sec/ hr) = 165 116.79 kCal/hr
4.187 kJ
d. For the % by weight of fresh air in the mixture, consider junction 1-4-2
Figure 3.8 Drying System from Francisco (2014)
By energy balance, m1 h1 + m4 h4 = m2 h2
52
Where, m4 = m2 − m1
m1 h1 + (m2 − m1 )h4 = m2 h2
→ m1 h1 + m2 h4 − m1 h4 = m2 h2
m1 (h1 − h4 ) = m2 (h2 − h4 )
Then, % by weight of m1 =
m1
m2
(100 %) =
% by weight of m1 =
h2−h4
h1−h4
(100 %)
108.64 − 169.715
(100 %) = 69.61 %
81.98 − 169715
#3. Wet material containing 215 % moisture (dry basis) is to be dried at the rate of 1.5 kg/s in a
continuous dryer to give a product containing 5 % moisture (wet basis). The drying medium consist
of air heated to 373 K and containing water vapor equivalent to a partial pressure of 1.40 kPaa. The
air leaves the dryer at 310 K and 70 % RH. Calculate how much air will be required to remove the
moisture.
Given: Dryer shown
Figure 3.8 Dryer from Francisco (2014)
Required:
Find the required air to remove the moisture.
Solution:
Consider the dryer and solving for the bone-dry-mass
m3 = BDM + 2.15 BDM = 3.15 BDM
BDM =
m3
1.5
=
= 0.476 kg/s
3.15 3.15
Solving for the mass of the material leaving the dryer, m4 = BDM + 0.05 m3
BDM = (1 − 0.05) m4 = 0.95 m4
53
m4 =
BDM 0.476
=
= kg/s
0.95
0.95
Solving for the mass of air required, using mass balance
ma + ma (W1 ) + m3 = ma + ma (W3 ) + m4
ma (W2 − W1 ) = m3 − m4 ; → ma =
m3 − m4
W2 − W1
NOTE: Student is advised to complete the solution
Assessment Task 3
1. A dryer is to deliver 0.30 kg/s of cassava with 2 % moisture and 20 % moisture in the feed.
Determine the mass of air required if the change in humidity ratio is 0.0165 kg/kg da.
2. A certain material enters dryer containing 60 % water and leaves with 5 % water. Find the
mass of the final product if the original product is 1 kg/s.
3. Copra enters a dryer containing 60 % water and 40 % solids and leaves with 5 % water and
95 % solids. Find the weight of water removed based on a kg of original product.
4. A rotary dryer fired with bunker oil of 10 000 kcal/kg higher heating value is to produce 20
metric tons/hr of dried sand with 0.5 % moisture from a wet feed containing 7 % moisture,
specific heat of sand is 0.21 Btu/lb-°F. Temperature of wet sand is 30 °C and temperature of
dried product is 115 °C.
a. Calculate the weight of the wet feed
b. Calculate the weight of water to be removed in kg/hr
c. Calculate the heat required
d. Calculate the liters of bunker oil per hour if specific gravity of bunker oils is 0.90 and
dryer efficiency is 60 %.
Summary
54

Drying is the process of removing moisture in varying amounts from solid or semi-fluid
materials. This process may be accomplished by pressure, suction, decantation, or
evaporation (Francisco, 2014).

Dryer is machine or equipment used for drying process.

Evaporation is the main principle of the term drying process. In drying process, not only
the surface liquid is removed or reduced but also removes internal moisture and in many
cases water in chemical combination.

Inherent moisture and bed moisture are the terms used for describing the moisture in the
mass that is not on the surface of the material.

Chemically combined water is that water occurring when a chemical component of the
material changes its chemical composition by heat or other means.
Reference
Francisco, Jose R. (2014). Lecture Notes in Industrial Plant Design, Lesson 6: Mechanical
Dryers, p.g 90 to p.g 100.
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