Module 1 Unit 3 EnggChem PDF

Unit 3 Fuels UNIT LEARNING OUTCOMES TLO 3: Demonstrate appropriate calculations in stoichiometry particularly in the combustion of fuels. ENGAGE The fuel is a material whic...

Unit 3 Fuels UNIT LEARNING OUTCOMES TLO 3: Demonstrate appropriate calculations in stoichiometry particularly in the combustion of fuels. ENGAGE The fuel is a material which when once raised to its ignition temperature continues to burn if sufficient oxygen or air is available. The principal constituents of any fuel are carbon and hydrogen. The materials which evolve heat after burning, are called combustibles. Carbon and hydrogen are combustibles. Sulphur is also a combustible material. When anything slowly combines chemically with oxygen, the process is called oxidation. When the same process occurs with a considerable swiftness and exotherm chemical reaction, it is called combustion; whereas such a process with almost instantaneous action is called detonation. An familiar application of fuels is a coal power plant. A coal-fired power station or coal power plant is a thermal power station which burns coal to generate electricity. Coal-fired power stations generate over a third of the world's electricity but cause hundreds of thousands of early deaths each year, mainly from air pollution. A coal-fired power station is a type of fossil fuel power station. The coal is usually pulverized and then burned in a pulverized coal-fired boiler. The furnace heat converts boiler water to steam, which is then used to spin turbines that turn generators. Thus chemical energy stored in coal is converted successively into thermal energy, mechanical energy and, finally, electrical energy. EXPLORE Classification of Fuels: 1. Solid fuel 2. Liquid fuel 3. Gaseous fuel Major forms of fuel being used today 1. Petroleum 2. Natural gas extract 3. Coal – is generally black rock that is composed of C, H2, O2, N2 and S Classification of coal: 1. Anthracite – is the hardest type of coal and has more carbon and more energy. 2. Lignite – has low carbon content but high in hydrogen and oxygen contents. 3. Bituminous – falls in between the range of anthracite and lignite coals for its hardness and energy output. Prepared by: Engr. N. L. Escalante 39 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  In early China, coal was thought to be a stone that could be burned and used for cooking. Later, throughout the world, the used of coal exploded: Transportation, used in factories, fuels for power plants to produce electricity. Oil or petroleum is another type of fuel. Oil has been used for 5,000 to 6,000 years. In ancient times, the Babylonians, the Assyrians, the Samarians used crude oil as a fuel. The crude oil was collected from the Euphrates River. The ancient Egyptians used liquid oil as a medicine to treat wounds and to provide light in lands. In 1859 Edwin Drake discovered crude oil in Titusville, Pennsylvania. He found crude oil underground and pumped it out to the surface and stored it in wood barrels. One of the world’s oil supply is derived from the oil wells of California and the middle east. Crude oil undergoes fractional distillation:  Petrol contains 5-7 carbons in the hydrocarbon chain. Petrol vaporizes at low temperatures and are easily ignited making it useful for cars, bikes, etc.  Kerosene is useful for jets engines, air craft, rockets, lamps.  Diesel is common for vans, trucks, and heavy equipment.  Fuel oil from residue is produce after diesel and is used for ships, power plants etc.  Lubricating oil from residue is produce also after diesel. Natural gas was discovered 2000 years ago. The natural gas seats were first discovered in Iran. The ancient Persians discovered the natural gas. Natural gas is made up of methane and is lighter than air. Natural gas is usually found near petroleum wells pumped out, purified and stored. Two Classification of Natural Gas 1. Dry – Contains generally methane 2. Wet – Aside from methane it contains ethane, propane, butane, pentane Uses of Natural Gas: Domestic, industrial, power generation, fuel for vehicles, feed stocks for fertilizers, hydrogen fuel cells. Other forms of natural gas such as biogas are used in the countryside for domestic purposes. Stoichiometry - is the branch of chemistry which deals with mass and volume relations in unit operations and unit processes. The basic principle of stoichiometry is the Law of Definite and Multiple Proportions. This states that a pure chemical substance always contains its elements in the same proportions by weight, and that, when two elements unite to form a series of compounds, the weight of one element combining with a fixed weight of the other are simple integral multiples of each other. Industrial Stoichiometry - is the application of the laws of conservation of matter, of elements and of energy, and of the chemical laws of combining weights to the processes and operations of industrial chemistry. Prepared by: Engr. N. L. Escalante 40 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Fuels and Combustion The application of stoichiometric principles to problems involving fuels and combustion is important in modern industrial work. Fuel combustion constitutes an important step in almost any chemical process industry where heat generation is desired. Thus, the study of combustion is most important especially in the field of chemical engineering, which encompasses the chemical process industry. Combustion ▪ Commonly employed industrial process for heat generation ▪ It is the chemical reaction of a free reactant (oxygen, O 2 in air) and commonly available chemical called fuel. ▪ It is accompanied by the evolution of light and heat, thus, it is generally used in the generation of heat to supply energy to the process industries. Classes of Fuels Fuels can be divided into three general classes: gaseous, liquids and solids. The stoichiometric treatments of problems involving different classes of fuels are similar. Consequently, in the study of the methods for manipulation of the data supplied, the actual form of the fuel is not of great importance. A. Gaseous Fuels ▪ Industrial fuels that usually contain CO, light or low molecular weight hydrocarbons and sometimes nitrogen and oxygen Examples: 1. Natural Gas - combustible gas that occurs in porous rocks of the earth’s crust (methane (CH4), ethane (C2H6), propane (C3H8), butane (C4H10), CO, H2, N2) 2. Liquefied Natural gas (LNG) - natural gas in liquid form 3. Liquefied Petroleum Gas (LPG) - can be liquefied under moderate pressure at normal temperature but are gaseous under normal atmospheric pressure 4. Re-formed Gas - applied to lower thermal value gases obtained by the pyrolysis and steam decomposition of high thermal value gases 5. Oil Gases - made from thermal decomposition of oils 6. Producer Gas - generated by blasting a deep, hot bed of coal of coke continuously with a mixture of air and steam 7. Blue Water Gas - a blast of air is forced through a fuel bed and steam is passed through forming blue water gas 8. Carbureted Water Gas - water gas enriched with oil 9. Blast Furnace Gas - a byproduct of the manufacture of pig iron in blast surfaces B. Liquid Fuels ▪ Light and heavy oils obtained in the refining of petroleum oil Examples: 1. Non-Petroleum Liquefied Fuels a) Tar Sands - strip-mined and extracted with hot water to recover heavy oil b) Oil Shale - non-porous rocks containing organic kerogen; extracted by pyrolysis after mining Prepared by: Engr. N. L. Escalante 41 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  2. Light Oils a) Alcohols - obtained by synthesis or fermentation process (methyl or ethyl) b) Alcogas - mixtures of methyl or ethyl alcohol and gasoline c) Benzole - obtained by distillation of coal tar or by extraction from coal gas d) Gasoline e) Paraffin f) Kerosene g) Diesel fuel 3. Heavy Oils - fuel oil that contains residual oil left over from distillation a) Bunker oil C. Solid Fuels Examples: coal, wood, charcoal Combustion Analysis: Empirical and Molecular Formula The molecular formula is an expression of the number and type of atoms that are present in a single molecule of a substance. It represents the actual formula of a molecule. Subscripts after element symbols represent the number of atoms. If there is no subscript, it means one atom is present in the compound. The empirical formula is also known as the simplest formula. The empirical formula is the ratio of elements present in the compound. The subscripts in the formula are the numbers of atoms, leading to a whole number ratio between them. ▪ The molecular formula of glucose is C 6H12O6. One molecule of glucose contains 6 atoms of carbon, 12 atoms of hydrogen and 6 atoms of oxygen. ▪ If you can divide all of the numbers in a molecular formula by some value to simplify them further, then the empirical or simple formula will be different from the molecular formula. The empirical formula for glucose is CH 2O. Glucose has 2 moles of hydrogen for every mole of carbon and oxygen. Examples 1. Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO 2 and 0.1159 g of H2O. What is the empirical formula for menthol? Given: weight of sample = 0.1005 g CO2 produced = 0.2829 g H2O produced = 0.1159 g Required: Empirical Formula Solution: ▪ Calculate the amount of C in the sample through the CO 2 produced: 1mol, CO2 1mol, C 12g , C g , C  0.2829g , CO2     0.0772g , C 44g , CO2 1mol, CO2 1mol, C ▪ Calculate the amount of H in the sample though the H 2O produced: 1mol, H 2O 2mol, H 1g , H g , H  0.1159g , H 2O     0.0129g , H 18g , H 2O 1mol, H 2O 1mol, H Prepared by: Engr. N. L. Escalante 42 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  ▪ Calculate the amount of O in the sample by subtracting the calculated amount of C and H from the weight of the sample: g , O  0.1005g  0.0772g  0.0129g  0.0104g , O ▪ Determine the number of moles for each element: 1mol, C mols, C  0.0772g , C   6.4333103 mol, C 12g , C 1mol, H mols, H  0.0129g , H   0.0129mol, H 1g , H 1mol, O mols, O  0.0104g , O   6.5 104 mol, O 16g , O ▪ Find the mole ratio by dividing by the smallest quantity: 6.4333103 mol C  9.8974  10 6.5 10 4 mol 0.0129mol H  19.8462  20 6.5  10 4 mol 6.5 104 mol O 1 6.5 10 4 mol ▪ Empirical Formula: C10H20O 2. A 0.2417 g sample of a compound composed of C, H, O, Cl only, is burned in oxygen yielding 0.4964 g of CO2 and 0.0846 g of H2O. A separate 0.1696 g sample of the compound is fused with sodium metal, the products dissolved in water and the chloride quantitatively precipitated with AgNO 3 to yield 0.1891 g of AgCl. What is the simplest empirical formula for the compound. Given: sample 1 sample 2 weight of sample = 0.2417 g weight of sample = 0.1696 g CO2 produced = 0.4964 g AgCl produced = 0.1891 g H2O produced = 0.0846 Solution: ▪ Calculate the amount of C in sample 1 through the CO 2 produced: 1mol, CO2 1mol, C 12g , C g , C  0.4964g , CO2     0.1354g , C 44g , CO2 1mol, CO2 1mol, C ▪ Calculate the amount of H in sample 1 though the H2O produced: 1mol, H 2O 2mol, H 1g , H g , H  0.0.0846g , H 2O     9.4 103 g , H 18g , H 2O 1mol, H 2O 1mol, H ▪ Calculate the amount of Cl in sample 2 through the AgCl produced: 1mol, AgCl 1mol, Cl 35.45g , Cl g , Cl  0.1891g , AgCl     0.0468g , Cl 143.32g , AgCl 1mol, AgCl 1mol, Cl ▪ Determine the amount of Cl in sample 1 by ratio and proportion: 0.0468g , Cl g , Cl  ; g , Cl  0.0667g , Cl 0.1696g 0.2417g Prepared by: Engr. N. L. Escalante 43 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  ▪ Calculate the amount of O in the sample by subtracting the calculated amount of C, H, and Cl from the weight of sample 1: g , O  0.2417g  0.1354g  9.4 103 g  0.0667g  0.0302g , O ▪ Determine the number of moles for each element: 1mol, C mols, C  0.1354g , C   0.0113mol, C 12g , C 1mol, H mols, H  9.4 103 g , H   9.4 103 mol, H 1g , H 1mol, O mols, O  0.0302g , O   1.8875103 mol, O 16g , O 1mol, Cl mols, Cl  0.0667g , Cl   1.8815103 mol, Cl 35.45g , Cl ▪ Find the mole ratio by dividing by the smallest quantity: 0.0113mol 1.8875103 mol C  6.0055  6 O  1.0032  1 1.8815103 mol 1.8815103 mol 9.4 103 mol 1.8815103 mol H  4.9960  5 Cl  1 1.8815103 mol 1.8815103 mol ▪ Empirical Formula: C6H5OCl 3. Vitamin C (ascorbic acid) contains 40.92% C, 4.58% H, and 54.50% O, by mass. The experimentally determined molecular mass is 176 g/mol. What is the empirical and molecular formula for ascorbic acid? Atomic Weights: C = 12.011 g/mol, H = 1.008 g/mol, O = 15.999 g/mol Given: 40.92% C 4.58% H 54.50% O MW = 176 g/mol Solution: Assume a basis of 100 g sample of ascorbic acid ▪ Determine the number of moles for each element: 1mol, C mols, C  100g  40.92%   3.41mol, C 12.011g , C 1mol, H mols, H  100g  4.58%   4.54mol, H 1.008g , H 1mol, O mols, O  100g  54.50%   3.41mol, O 15.999g , O ▪ Find the mole ratio by dividing by the smallest quantity: 3.41mol 3.41mol C 1 O 1 3.41mol 3.41mol 4.54mol H  1.33 3.41mol  Round your ratio of moles to the nearest whole number as long as it is close to a whole number. In other words, you can round 1.992 up to 2, but you can't Prepared by: Engr. N. L. Escalante 44 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  round 1.33 to 1. You'll need to recognize common ratios, such as 1.333 being 4/3. For some compounds, the lowest number of atoms of an element might not be 1! If the lowest number of moles is four-thirds, you will need to multiply all ratios by 3 to get rid of the fraction. 3.41mol 3.41mol C  1 3  3 O  1 3  3 3.41mol 3.41mol 4.54mol H  1.33 3  3.99  4 3.41mol ▪ Empirical Formula: C3H4O3 ▪ Solve for the empirical weight: EW  3 12.011g / mol  4 1.008g / mol  3 15.999g / mol  88.062g / mol ▪ Determine the formula weight ratio by dividing the molecular weight by the empirical weight: MW 176g / mol R   1.9986  2 EW 88.062g / mol ▪ Multiply the ratio of formula weights to the mole ratio of each element: C  3 2  6 H  4 2  8 O  3 2  6 ▪ Molecular Formula: C6H8O6 Activity 1: Determination of Empirical and Molecular Formula Self-Assessment No. 1 1. What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO2 and 0.306 grams of H2O? 2. Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO 2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene. 3. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO 2 and 14.53 mg of H2O. Determine the empirical and molecular formula of xylene. Molecular weight of xylene is 106.16 g/mol. To be submitted in Google classroom on: Complete and Incomplete Combustion Complete Combustion ▪ All combustible materials are completely gasified, all carbon, C, in the fuel is burned to CO2, all hydrogen is converted to H2O, and all sulfur to sulfur dioxide. Prepared by: Engr. N. L. Escalante 45 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  ▪ Reactions: C  O2 CO2  heat 1 H 2  O2 H 2O  heat 2 S  O2 SO2  heat ▪ The complete combustion of any hydrocarbon is represented by:  n n Cm H n   m  O2 mCO2  H 2O  4 2 Incomplete Combustion ▪ Indicated by the presence of CO and unburned H2 in the product and unburned combustibles in the refuse (in solid fuels only) ▪ It represents a heat loss since this should have been given off for additional power use had the fuel been completely burned. Presence of soot in exhaust also indicated incomplete combustion. ▪ Reaction: 1 C  O2 CO  heat 2 Flue Gas - term used for the gaseous products of combustion Theoretical and Excess Oxygen A. Theoretical Oxygen - oxygen required for complete combustion Methods of Determining Theoretical Oxygen: ▪ Method I. The balanced equation for complete combustion is written and the theoretical O2 is the sum of all oxygen consumed for complete combustion of the fuel. Theoretical O2 = total oxygen required (Method I) - O2 free in the fuel ▪ Method II. The components of the fuel are broken down into its corresponding atoms of carbon, sulfur, hydrogen, and oxygen. One C atom needs 1 mole of O 2, one S atom need 1 mole of O2, one H atom need 1/4 mole of O2. Theoretical O2 = total oxygen required (Method II) - total O2 in the fuel Theoretical O2 = O2 required by C + O2 required by H + O2 required by S - O2 in fuel Note: N2 is non-combustible B. Theoretical Air for Combustion - air that contains the exact amount of theoretical O 2. Air is assumed to contain 21% O 2 and 79% N2 by volume. N2 in air is non-combustible and acts only as diluent to the O2 in air. The theoretical air therefore is determined from theoretical O2. Note: Dry air contains 79% by volume N2 and 21% by volume O2 Prepared by: Engr. N. L. Escalante 46 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Example: 1. 200 kg of pure carbon is burned. Determine the theoretical amount of O 2 and theoretical air needed for complete combustion. Given: Flue Gas composition CO2 (theo O2 only) CO2, N2 (theo air) 200 kg C theo O2 / theo air Required: theo O2, theo air Solution: 1kmol, C 1kmol, O2 theo, O2  200kg, C    16.6667kmol, O2 12kg, C 1kmol, C theo, O2  16.6667kmol, O2 100kmol, air theo, air  16.6667kmol, O2  21kmol, O2 theo, air  79.3652kmol, air 2. Determine the amount of O 2 and air theoretically required for the combustion of 100 kmols of blast furnace gas analyzing 25% CO, 10% CO 2, 5% H2, 10% CH4, 45% N2, and 5% O2. Given: 100 kmols BFG Flue Gas composition 25% CO CO2, H2O, N2 10% CO2 5% H2 10% CH4 45% N2 5% O2 theo O2 / theo air Required: theo O2, theo air Solution: Method I: theo O2 = O2 to burn CO + O2 to burn H2 + O2 to burn CH4 - O2 in fuel 1 2 kmol, O2 1 2 kmol, O2 2kmol, O2 theo, O2  25kmol, CO   5kmol, H 2   10kmol, CH 4   5kmol, O2 1kmol, CO 1kmol, H 2 1kmol, CH 4 theo, O2  30kmol, O2 100kmol, air theo, air  30kmol, O2  21kmol, O2 Prepared by: Engr. N. L. Escalante 47 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  theo, air  142.8571kmol, air Method II: theo O2 = O2 to burn total C + O2 to burn total H - O2 in fuel 1kmol, C 1kmol, C 1kmol, C total, C  25kmol, CO   10kmol, CO2   10kmol, CH 4  1kmol, CO 1kmol, CO2 1kmol, CH 4 total, C  45kmol, C 2kmol, H 4kmol, H total, H  5kmol, H 2   10kmol, CH 4  1kmol, H 2 1kmol, CH 4 total, H  50kmol, H 1 2 kmol, O2 1kmol, O2 O2in, fuel  25kmol, CO   10kmol, CO2   5kmol, O2 1kmol, CO 1kmol, CO2 O2in, fuel  27.5kmol, O2 1kmol, O2 1 4 kmol, O2 theo, O2  45kmol, C   50kmol, H   27.5kmol, O2 1kmol, C 1kmol, H theo, O2  30kmol, O2 100kmol, air theo, air  30kmol, O2  21kmol, O2 theo, air  142.8571kmol, air C. Excess Oxygen or Excess Air - is the amount of O2 or air supplied above that of theoretically required for complete combustion/oxidation of the combustibles in the fuel. This is because in actual practice, theoretical O 2 or theoretical air is not sufficient to get complete combustion. Supplying air in excess will therefore improve combustion efficiency. There are several alternate ways to determine the percent excess air as indicated below: x' s, O2 x' s, O2 % x' s, air  100  100 theo, O2 total, O2 sup plied  x' s, O2 total, O2 sup plied  theo, O2 % x' s, air  100 theo, O2 total, O2 sup plied  theo, O2  x' s, O2 For complete combustion: x’s O2 = O2 free in the flue gas For incomplete combustion: x’s O2 = O2 free in the flue gas - O2 needed to burn unburned combustibles Note: Gaseous fuels require very little excess O 2, liquid fuels require more and solid fuels require the maximum excess O2. Prepared by: Engr. N. L. Escalante 48 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Examples: 3. Combustion of Pure Carbon. 100 kg of pure C is burned. For each of the following cases, calculate the composition of the combustion gases. a) Theoretical amount of O2 is used; complete combustion b) Theoretical amount of air is used; complete combustion c) 25% excess air is used; complete combustion Flue Gas composition theo O2: CO2 Given: theo air: CO2, N2 100 kg C x’s air: CO2, N2, O2-free theo O2 / theo air / 25% x’s air Required: flue gas composition Solution: a) 1kmol, C 1kmol, O2 theo, O2  100kg, C    8.3333kmol, O2 12kg, C 1kmol, C in the flue gas: 1kmol, C 1kmol, CO2 CO2  100kg, C    8.3333kmol, CO2 12kg, C 1kmol, C flue gas composition: kmol percentage CO2 8.3333 100% Total 8.3333 100% b) in the flue gas: CO2  8.3333kmol, CO2 79kmol, N 2 N 2  8.3333kmol, O2   31.3491kmol, N 2 21kmol, O2 flue gas composition: kmol percentage CO2 8.3333 21% N2 31.3491 79% Total 39.6824 100% c) % x' s, O2  theo, O2 25  8.3333kmol, O2 x' s, O2    2.0833kmol, O2 100 100 O2 , sup plied  8.3333kmol, O2  2.0833kmol, O2  10.4166kmol, O2 in the flue gas: Prepared by: Engr. N. L. Escalante 49 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  CO2  8.3333kmol, CO2 79kmol, N 2 N 2  10.4166kmol, O2   39.1863kmol, N 2 21kmol, O2 O2  free  2.0833kmol, O2 flue gas composition: kmol percentage CO2 8.3333 16.8000% N2 39.1863 79.0000% O2-free 2.0833 4.2000% Total 49.6029 100 4. Combustion of Pure Hydrocarbons. The following pure compounds are burned with 40% excess air. Find the composition of the flue gas if the combustion is complete. a) C2H6 b) C6H6 c) C2H4 Given: Flue Gas composition CO2, N2, O2-free, H2O a) C2H6 b) C6H6 c) C2H4 40% x’s air Required: flue gas composition  n n Cm H n   m  O2 mCO2  H 2O  4 2 a) assume 100 kmol of ethane, C2H6 7 2 kmol, O2 theo, O2  100kmol, C2 H 6   350kmol, O2 1kmol, C2 H 6 % x' s, O2  theo, O2 40  350kmol, O2 x' s, O2    140kmol, O2 100 100 O2 , sup plied  350kmol, O2  140kmol, O2  490kmol, O2 in the flue gas: 2kmol, CO2 CO2  100kmol, C2 H 6   200kmol, CO2 1kmol, C2 H 6 79kmol, N 2 N 2  490kmol, O2   1843.3333kmol, N 2 21kmol, O2 O2  free  140kmol, O2 3kmol, H 2O H 2O  100kmol, C2 H 6   300kmol, H 2O 1kmol, C2 H 6 Prepared by: Engr. N. L. Escalante 50 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  flue gas composition: kmol percentage CO2 200 8.0537% N2 1843.3333 74.2282% O2-free 140 5.6376% H2O 300 12.0805% Total 2483.3333 100% *Repeat for parts b and c 4. Pure formaldehyde gas (CH2O) is burned completely in air, using 50% excess air. a) What should be the Orsat analysis of the combustion gas? b) What should be the Orsat analysis of the combustion gas if instead of excess air, only enough were used to burn all the hydrogen to water, 25% of the carbon to CO2 and the remaining 75% to CO, the combustion being conducted that no oxygen gas whatever is left remaining in the combustion products. Note: Orsat analysis - fuel gas composition excluding H2O and SO3 Given: Flue Gas composition a) CO2, N2, O2-free, H2O b) CO, CO2, N2, H2O CH2O a) 50% x’s air b) enough air CH 2O  O2 CO2  H 2O Required: Orsat analysis Solution: assume a basis of 100 kmol formaldehyde a) using the mole ratios provided by the given balanced chemical equation 1kmol, O2 theo, O2  100kmol, CH 2O   100kmol, O2 1kmol, CH 2O % x' s, O2  theo, O2 50  100kmol, O2 x' s, O2    50kmol, O2 100 100 O2 , sup plied  100kmol, O2  50kmol, O2  150kmol, O2 in the flue gas: 1kmol, CO2 CO2  100kmol, CH 2O   100kmol, CO2 1kmol, C2 H 6 79kmol, N 2 N 2  150kmol, O2   564.2857kmol, N 2 21kmol, O2 O2  free  50kmol, O2 Prepared by: Engr. N. L. Escalante 51 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  1kmol, H 2O H 2O  100kmol, CH 2O   100kmol, H 2O (although not required) 1kmol, CH 2O Orsat analysis: kmol percentage CO2 100 14.0000% N2 564.2857 79.0000% O2-free 50 7.0000% Total 714.2857 100% b) using Method II since it is a more appropriate approach in the flue gas: 1kmol, C 1kmol, CO2 CO2  100kmol, CH 2O   25%   25kmol, CO2 1kmol, C2 H 6 1kmol, C 1kmol, C 1kmol, CO CO  100kmol, CH 2O   75%   75kmol, CO 1kmol, C2 H 6 1kmol, C 2kmol, H 1kmol, H 2O H 2O  100kmol, CH 2O    100kmol, H 2O 1kmol, CH 2O 2kmol, H in the air supplied 1kmol, O2 1 2 kmol, O2 O2 , sup plied  25kmol, CO2   75kmol, CO  1kmol, CO2 1kmol, CO 1 2 kmol, O2 1 2 kmol, O2  100kmol, H 2O   100kmol, CH 2O  1kmol, H 2O 1kmol, CH 2O O2 , sup plied  62.5kmol, O2 79kmol, N 2 N 2  62.5kmol, O2   235.1190kmol, N 2 21kmol, O2 Orsat analysis: kmol percentage CO2 25 7.4600% CO 75 22.3801% N2 235.1190 70.1599% Total 335.1190 100% Activity 2: Calculations based on fuel analysis Self-Assessment No. 2 1. 100 kmols of moist hydrogen gas containing 4% water by mole is burnt completely in a furnace with 32% excess dry air. a) Complete analysis b) Orsat analysis 2. A furnace is charged with 100 kmols of fuel gas, the composition of which is: 50% CO, 30% H2, 5% CH4, 5% C2H6 and 10% O2. Dry air is supplied 50% in excess for combustion. Assuming complete combustion. Prepared by: Engr. N. L. Escalante 52 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  a) Complete analysis b) Orsat analysis To be submitted in Google classroom on: EXPLAIN To be able to translate your understanding of fuels, do the following activity. Activity 3: Check Your Understanding on Fuels and Combustion Self-Assessment No. 3 1. Differentiate combustion from oxidation. 2. Is combustion exothermic or endothermic? Explain. 3. What happens during incomplete combustion? To be submitted in Google classroom on: ELABORATE & EVALUATE Activity 4: Fuels and Combustion Self-Assessment No. 4 Combustion encompasses a great variety of phenomena with wide application in industry, the sciences, professions, and the home, and the application is based on knowledge of physics, chemistry, and mechanics; their interrelationship becomes particularly evident in treating flame propagation. Think of an example where fuels and combustion is applied and discuss the whole process of it. To be submitted in Google classroom on: Prepared by: Engr. N. L. Escalante 53 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.