U21CH301

Process Calculations
L T P J C Course Category:
Professional Core
3 1 0 0 4 Course (PCC)

Dr. S. Balasubramanian PhD

Department of Chemical Engineering

KPR Institute of Engineering and Technology
Coimbatore 641407
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COMBUSTION

- Combustion Process
- Classification of fuel
- Calorific Value of fuel
- Proximate Analysis and
Ultimate Analysis
- Theoretical Oxygen and
Theoretical Air
 INTRODUCTION COMBUSTION

Ultimate Composition of elements Energy Complete

Incomplete
Analysis (C, H, O, N, S) and ash (Heat) Combustion
Combustion
and Work (CO2 + H2O,
(CO + SO)
SO2)

Proximate 1. Moisture 1. Dry basis (excluding

Fuel Combustion Product Gas Orsat
Analysis 2. Volatile Combustibles Analysis of water or moisture)
(Solid, Liquid (Unit (Stack or Flue Analysis
matter Gas 2. Wet Basis (including
and Gas) Process) Gas)
3. Ash water or moisture)
4. Fixed Carbon
Air O2 = 21%
Calorific Value Burn N2 = 79%
of the fuel Refuse or residue

Amount of Amount of
theoretical air excess air
(oxygen) (oxygen)
required required

Schematic Representation of Combustion

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INTRODUCTION COMBUSTION

Combustion

- Combustion is a unit process in which oxidation reaction takes place.

- It is the rapid oxidation of combustible substances accompanied by the release
of energy
- However, all the oxidation reactions are not termed as combustion.
- For example, oxidation of to to benzaldehyde, oxidation of hydrogen chloride to
chlorine and so on are not normally termed as combustion.
- Broadly with the union of C, H, and S with O2 is termed as combustion.

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INTRODUCTION COMBUSTION

Complete Combustion

- If the products of combustion are carbon dioxide, water and sulfur dioxide from
the elements C, H, and S, the combustion is termed as complete combustion

In complete Combustion

- If carbon monoxide (CO) appears in the product gas, then the combustion is termed
as incomplete or partial combustion because of presence of carbon monoxide.
- Carbon monoxide can further combine with oxygen to produce carbon dioxide (CO2).
- Although, Sulfur dioxide (SO2) can further oxidize to produce Sulfur trioxide (SO3).

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INTRODUCTION COMBUSTION
Classification of fuel Fuels can be broadly classified as follows

Fuels

Solids Liquids Gases

Manmade or Manmade or Manmade or

Natural Natural Natural
Synthetic Synthetic Synthetic

Coal and Natural

Coke, Crude Oil Gasoline,
Wood Gas Coke Oven Gas
Rice Husk, Petrol,
Producer Gas
Saw dust, Diesel,
Kerosene, LPG Gas,
Bagasse etc. Dimethyl ether
Alcohol,
etc.,
fuel oil etc.,

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel

- Calorific value of a fuel is the heat of combustion of the fuel

- It is defined as the heat produced when a unit mass of fuel is burnt
completely with pure oxygen
- It is also said to be the heating value of the fuel

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel

The types of heating value of fuels

1. Low Heating value (LHV) or Net Calorific Value (NCV)
or Net Heating Value (NHV)
2. High Heating Value (HHV) or Gross Calorific value (GCV)
or Gross Heating Value (GCV)

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel

! (Energy/Heat)
The types of heating value Water Vapor (H2O)
Results from
of fuels, LHV or HVL LCV/LHV/NCV
Fuel Combustion Products
depends on the state of the Chamber
Liquid Water (H2O)
water exiting from the Results from
Air HCV/HHV/GCV
combustion process O2 = 21%
N2 = 79%

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel

Low Heating Value or Net Calorific Value High Heating Value or Gross Calorific Value

When the calorific value of the If the water is condensed, the

fuel is determined by considering latent heat of water vapor is
the water present in the form of made available for useful
vapor, it is said to be Low Heating purposes. Therefore if this part of
Value (LHV) or Net Calorific Value heat (latent heat) is added to NCV
(NCV) then GCV is obtained

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel

The formula assumes complete

𝑊𝑒𝑖𝑔ℎ𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝐻! × 9×𝜆 combustion of the fuel. It applies to fuels
𝐺𝐶𝑉 = 𝑁𝐶𝑉 +
100 with significant hydrogen content, such as
hydrocarbons
𝐺𝐶𝑉 = 𝑁𝐶𝑉 + 𝑚𝜆

Where,
𝐺𝐶𝑉 - Grass colorific value or high heating value, kJ/kg
𝑁𝐶𝑉 - Net colorific value or net heating value, kJ/kg
𝑚 - Mass of water produced, kg
𝜆 - late in heat of water vapor, kJ/kg

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel

Problem

A hydrocarbon fuel has, Net Calorific Value (NCV): 42,000 kJ/kg; Weight % of Hydrogen
(wt% of H2): 10% Latent heat of water vapor (λ): 2442 kJ/kg.
Find: The Gross Calorific Value (GCV)

Calculation

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel – Crude Oil

Problem : Calorific value of a liquid fuel (crude oil) is analyzed to contain 87% carbon, 12.5% hydrogen
and 0.5% sulfur by weight. Consider the stoichiometric oxidation of hydrogen. Calculate the net
colorific value of the fuel at 25°C (298 K). Also verify your answer with the relation 𝐺𝐶𝑉 = 𝑁𝐶𝑉 + 𝑚𝜆.
Take gross calorific value of crude oil = 45071 kJ per kg oil and latent heat of water vapor is equal to
2442.5 kJ per kg at 25°C. Take the basis as one kg of crude oil.

Calculation
This is valid if:
- 𝑚 accurately represents the mass of water
produced per unit fuel.
- The relationship between 𝑚 and the fuel's
hydrogen content is correctly defined as per
stoichiometric equation

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel – Crude Oil

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel – Crude Oil

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel – Crude Oil

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel – Crude Oil

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel – Crude Oil

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel – Coal

Coal gasification consists of the chemical transformation of solid coal

into a combustible gas. For many years before the widespread
adoption of natural gas, gas generated from coal served as a fuel
(and also as an illuminant). The heating values of coals differ, but the
higher the heating value, the higher the value of the gas produced.

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INTRODUCTION COMBUSTION

Calorific Value of Fuel (or) Heating value of fuel – Coal

Problem: The analysis of a coal has a reported heating value of 29, 770 kJ per kg as
received. Assume that this grass heating value at atm and 25°C obtained in open
system. Use Dulong formula to check the validity of the reported value.
Dulong Formula
Data 𝑂
𝐻𝐻𝑉 = 14,544 𝐶 + 62,028 𝐻 − + 4050 𝑆
8

Where, HHV – Higher Heating Value or Gross

Calorific value, BTU/lb. C, H and S are the
mass fraction of carbon, hydrogen and sulfur.
The coefficients for C, H and S represents the
energy contributions of carbon, hydrogen,
and sulfur, respectively.

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INTRODUCTION COMBUSTION

Proximate Analysis and Ultimate Analysis

Proximate Analysis and Ultimate Analysis are standard methods for analyzing the composition of coal and
other solid fuels, primarily used in industries such as power generation, metallurgy, and cement
production. These analyses helps us to evaluate the quality of fuel and its suitability

Proximate Analysis
1. Moisture Content
Proximate analysis provides a - The amount of water present in the coal, expressed as a
quick assessment of the percentage.
following four components of - Measured by heating the sample at 105°C until a constant
coal: weight is achieved.
1. Moisture Content 2. Volatile Matter
2. Volatile Matter - The materials, other than moisture, that are released as
3. Ash Content gases upon heating the coal to about 950°C in the absence
4. Fixed Carbon of oxygen.
- Includes hydrocarbons, hydrogen, and other gases.

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INTRODUCTION COMBUSTION

Proximate Analysis
Problem
A coal sample is analyzed, and the following results are
obtained:
Moisture Content (M) : 8%
Volatile Matter (VM) : 28%
Ash Content (A) : 15%
Calculate the Fixed Carbon (FC) in the coal sample.

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INTRODUCTION COMBUSTION

Proximate Analysis
Problem
Calculations
A coal sample is analyzed, and the following results
are obtained:
Moisture Content (M) : 8%
Volatile Matter (VM) : 28%
Ash Content (A) : 15%
Calculate the Fixed Carbon (FC) in the coal sample.

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 INTRODUCTION COMBUSTION

Ultimate Analysis
Ultimate Analysis determines the elemental composition of coal:

1. Carbon (C) - Indicates the fuel’s energy content, usually a major component.
2. Hydrogen (H) - Present in hydrocarbons and water; contributes to the energy released during combustion.
3. Oxygen (O) - Combined with carbon and hydrogen in coal, usually present in lower amounts.
4. Nitrogen (N) - Present in trace amounts; contributes minimally to calorific value.
5. Sulfur (S) - can produce harmful emissions (SO₂) during combustion.
6. Ash - Inorganic matter, residue after combustion similar to proximate analysis.

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 INTRODUCTION COMBUSTION

Theoretical Oxygen and Air Requirement

Problem

100 mol/h of butane (C4H10) and 5000 mol/h of air is fed into a combustion
reactor. Calculate the percent excess air and percent excess oxygen.
Note: The molar percent of oxygen and nitrogen in 1 mole of air is 21% O2 and
79% N2 respectively.

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 INTRODUCTION COMBUSTION

Theoretical Oxygen and Air Requirement

C4H10 + O2 → CO2 + H 2O
Problem (Butane) (Oxygen) (Carbon dioxide) (Water)

100 mol/h of butane (C4H10) and 5000

100 kmol/h Combustion CO2
mol/h of air is fed into a combustion
C4H10 Reactor
reactor. Calculate the percent excess
air and percent excess oxygen. 5000 kmol/h
Note: The molar percent of oxygen 21% O2 H 2O

and nitrogen in 1 mole of air is 21% O2 79% N2

and 79% N2 respectively.

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 INTRODUCTION COMBUSTION
C4H10 + O2 → CO2 + H2O
Theoretical Oxygen and Air Requirement (Butane) (Oxygen) (Carbon dioxide) (Water)

100 kmol/h Combustion CO2

C4H10 Reactor
Problem
5000 kmol/h
H2 O
100 mol/h of butane (C4H10) and 5000 21% O2
79% N2

mol/h of air is fed into a combustion

A (C4H10) + B (O2) → C (CO2) + D (H2O) … (i)
reactor. Calculate the percent excess (Butane) (Oxygen) (Carbon dioxide) (Water)
air and percent excess oxygen. Balance on C Balance on O Balance on H

Note: The molar percent of oxygen 4A = C (1) 2B = 2C+D (2) 10A = 2D (3)
and nitrogen in 1 mole of air is 21% O2
Substitute, A = 1 in (1) and (3), we have (1) ⟹ 4(1) = C ⟹ C = 4
and 79% N2 respectively.
!"
(3) ⟹ 10(1) = 2D ⟹ D = = 2 ⟹D = 5
#

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 INTRODUCTION COMBUSTION

Theoretical Oxygen and Air Requirement

Substitute, D = 2, C = 4 in (2), we have
Problem (2) ⟹ 2B = 2C +D
2B = 2(4) + (5)
100 mol/h of butane (C4H10) and 5000
2B = 8 + 5
mol/h of air is fed into a combustion
13
reactor. Calculate the percent excess B=
2
air and percent excess oxygen. Substitute, A = 1, B = 5, C = 4 and D = 2 in (i), we have
Note: The molar percent of oxygen 13
1 (C4H10) + (O2) → 4 (CO2) + 5 (H2O)
and nitrogen in 1 mole of air is 21% O2 2

and 79% N2 respectively. 13

(C4H10) + (O2) → 4 (CO2) + 5 (H2O)
2

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 INTRODUCTION COMBUSTION

Theoretical Oxygen and Air Requirement

Now, the balanced stoichiometric equation is
Problem (C4H10) +
13
(O2) → 4 (CO2) + 5 (H2O)
2
100 mol/h of butane (C4H10) and 5000
Looking at the above balanced stoichiometric
mol/h of air is fed into a combustion equation,
reactor. Calculate the percent excess 13
1 mole of C4H10 requires = moles of O2
2
air and percent excess oxygen. 100 moles of C4H10 requires = ? moles of O2
Note: The molar percent of oxygen
100 moles of C4H10 × 13 moles ofO2
and nitrogen in 1 mole of air is 21% O2 = 2
1 mole of C4H10
and 79% N2 respectively.
= 650 moles of O2

100 moles of C4H10 requires =650 moles of O2

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 INTRODUCTION COMBUSTION

Theoretical Oxygen and Air Requirement

Problem Actual supply of oxygen through air into the

reactor for 100 mol/h of feed is 5000 mol/h
100 mol/h of butane (C4H10) and 5000
meaning that,
mol/h of air is fed into a combustion 5000 x 0.21 = 1050 mol/h of O2 fed into the reactor
for 100 moles of C4H10
reactor. Calculate the percent excess
air and percent excess oxygen. % Excess of O2 =
Actual O2 in the feed - Theoritically required O2
Theoritically required O2
Note: The molar percent of oxygen
1050 - 650
and nitrogen in 1 mole of air is 21% O2 % Excess of O2 = 1050 = 61.53%
and 79% N2 respectively. % Excess of O2 = 61.53%

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 INTRODUCTION COMBUSTION

Theoretical Oxygen and Air Requirement

Problem % Excess Air Calculation

100 mol/h of butane (C4H10) and 5000 0.21 mole of O2 supplied = 1 mol of air
mol/h of air is fed into a combustion
650 moles of O2 supply = ? moles of air
reactor. Calculate the percent excess
650 moles of O2× 1moles of air
air and percent excess oxygen. =
0.21 mole of O2
Note: The molar percent of oxygen
= 3095.23 moles of air theoretically required
and nitrogen in 1 mole of air is 21% O2
Actual air in the feed − Theoritically required air2
% Excess of air =
and 79% N2 respectively. Theoritically required air
5000 − 3095
% Excess of air = ×100 = 61.53%
3095

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 INTRODUCTION COMBUSTION

Theoretical Oxygen and Air Requirement

Problem
100 mol/h of butane (C4H10) and 5000
Solution
mol/h of air is fed into a combustion
reactor. Calculate the percent excess % Excess of O2 = 61.53%

air and percent excess oxygen. % Excess of air = 61.53%

Note: The molar percent of oxygen
and nitrogen in 1 mole of air is 21% O2
and 79% N2 respectively.

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INTRODUCTION COMBUSTION

Confucius was a Chinese philosopher and politician

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COMBUSTION Text Book References

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