N5 Power Machines COMBUSTION BY MASS

ialegodi@gmail.com 082 396 7717 COMBUSTION BY: (MASS, HYDROCARBONS AND HEAT VALUES) I. BY MASS Combustion is a chemical process where heat is liberated and the combustible elements is changed into another substance. Note: Air consists of 23% oxygen and 77% nitrogen by mass Elements Symbol Atomic mass Molecular mass 1. Carbon C 12 12 2. Hydrogen H2 1 2 3. Oxygen O2 16 32 4. Nitrogen N2 14 28 5. Sulphur S 32 32 Formula: 100 8 The theoretical mass of air = M air  ( C  8 H 2  S  O2 ) 23 3 M total  M air The excess percentage =  100 M air The total mass of air = (1+excess) M air 8 The mass of oxygen = M O2 = ( C  8 H 2  S  O2 ) 3 Products of combustion Symbol Formula 1. Carbon dioxide CO2 11 C 3 2. Water H 2O 9H 2 3. Sulphur dioxide SO2 2.S 4. Nitrogen N2 77%M total  % N 2 5. Excess of oxygen Excess  O2 %Excess M O2 THE POSSIBLE QUESTIONS: Downloaded from hlayiso.com

ialegodi@gmail.com 082 396 7717 1. A sample of coal consists of 84% carbon; 5% hydrogen; 7% sulphur and 4% oxygen. There is a 30% needed to be supplied for the complete combustion of 1 kg of fuel. Calculate the following: 1.1. The theoretical mass of air required for the complete combustion of 1 kg of fuel 1.2. The actual mass of air required to the complete combustion of 1 kg of fuel 1.3. The products of combustion of fuel 1.4. Express the mass’s in question 1.3 as percentages 2. A fuel consists of 87% carbon, 9% hydrogen, 1% sulphur, 1,5% oxygen a remainder is incombustibles. The actual air fuel ratio is 18,5 : 1. Calculate the following: 2.1. The mass of oxygen 2.2. The theoretical mass of air 2.3. The mass of air 2.4. The excess percentage 3. A turbine fuel has the following analysis according to its mass: Carbon 87% Hydrogen 12% Oxygen 2% Sulphur 1% Nitrogen 3% There is 48% excess air provided to ensure the complete combustion of the fuel. Calculate the following: 3.1. The theoretical mass of air required for the complete combustion of the fuel. 3.2. The total mass of air required for the complete combustion of 1 kg of fuel 3.3. The masses of the products of combustion 3.4. Express the masses in QUESTION 3.3 as percentages Downloaded from hlayiso.com

ialegodi@gmail.com 082 396 7717 4. A sample of coal has the following analysis by mass: carbon 87%; hydrogen 5%; oxygen 6%; sulphur 1% and ash 1%. The composition of air is 23% oxygen by mass and the atomic masses are: carbon 12; oxygen 16; sulphur 32; and hydrogen 1. 30% excess air is supplied for the complete combustion of the coal. Calculate the following: 4.1. The mass of oxygen required for the complete combustion of 1 kg of coal from first principle 4.2. The theoretical mass of air required for the complete combustion of 1 kg of coal 4.3. The actual mass of air supplied for the complete combustion of 3, 5kg of coal 5. The chemical analysis by mass of a fuel is 78% carbon, 14% hydrogen and 8% oxygen. 35% of excess air is required for the complete combustion and it is supplied at 101 kPa and 26c. The gas constant R for air is 0, 287 kj/kg.K Calculate the volume of air required for complete combustion of the fuel. 6. A certain fuel has the following composition: 77,7% carbon; 6, 8% hydrogen; 1, 2% nitrogen; 2, 2% sulphur; 8, 8% oxygen; 3,3% ash Calculate the following from the first principles: 6.1. The mass of oxygen required for the combustion of carbon 6.2. The mass of oxygen required for the combustion of hydrogen 6.3. The mass of oxygen required for the combustion of sulphur 6.4. The total mass of oxygen required 6.5. The mass of air required 6.6. The mass of nitrogen 7. A fuel consists of 3% nitrogen; 1% water, 3% hydrogen; 86% carbon; 1% sulphur; 2,5% oxygen and the remainder ash according to mass. 22% excess air is supplied to ensure complete combustion. Calculate the following using the formula method: 7.1. The theoretical mass of oxygen required for the complete combustion of 1 kg of fuel. Downloaded from hlayiso.com

ialegodi@gmail.com 082 396 7717 7.2. The theoretical mass of air required for the complete combustion of 1 kg of fuel. 7.3. Theoretical air fuel ratio 7.4. The actual mass of air supplied for the complete combustion of 1 kg of fuel 7.5. The actual air fuel ratio 8. The chemical analysis of 1 kg of coal is as follows: Carbon = 86%, hydrogen = 3%, sulphur = 0, 5% and the rest is no-combustible. 16, 5 kg of air is provided for the complete combustion of 1 kg of coal Calculate the following: 8.1. The theoretical mass of air required for the complete combustion of 1 kg of the coal 8.2. The percentage of excess air supplied. 8.3. The masses of the products of combustion 8.4. Express the masses in QUESTION 8.3 as percentages 9. 18, 04 kg of air is used for the complete combustion of a kilogram of fuel. The fuel has the following composition: Carbon = 87% Hydrogen = 2, 5% Sulphur = 1% The balance of the fuel is non-combustible. Calculate the percentages of the products of combustion 10. A sample of coal has the following analysis by mass: Sulphur 3% Oxygen 6% Carbon 84% Hydrogen 7% Air is supplied at 27, 8% excess Calculate the following 10.1. The theoretical mass of oxygen required for the complete combustion of 1, 32 kg of the coal from the first principle. Downloaded from hlayiso.com

ialegodi@gmail.com 082 396 7717 10.2. The theoretical mass of air required for the complete combustion of 3, 12 kg of the coal 10.3. The actual mass of air required for the complete combustion of 2, 13 kg of the coal II. BY CALORIFIC VALUE OR HEAT VALUE Formula: H.C.V = Higher Heat Value of fuel or Higher calorific Value of fuel L.C.V = Lower Heat Value of fuel or Lower Calorific Value of fuel mw  mass of water used in calorimeter ; me  Eguivalentmass of water; m f  mass of coal used; t water  risein temperature of water; Cp water  specificheat capacityof water 4,187 kJ / kg.K  (mwater  mequivalent )Cpt water H .C.V  mf O2 H .C.V  CVc .C  CV H 2 ( H 2 )  CVs .S where CVc  calorific/ heat valueof carbon; 8 CVs  calorific/ Heat valueof fuel ; CV H 2  calorificvalueof hydrogen L.C.V  H .C.V  9(hfg) H 2 Possible questions: 1. A bomb calorimeter was used to determine the calorific value of coal. The following readings were obtained during the test: Specific heat capacity of water 4 187 J/kg.K Water equivalent of the calorimeter 0, 6 kg Maximum temperature of water 24, 6 C Temperature of water before ignition 17, 8 C Mass of water used 2, 2 kg Mass of coal used 1, 36 kg Calculate the calorific value of the coal in MJ/kg 2. A mass of coal was tested and the following information was noted: - Hydrogen 6, 7% Downloaded from hlayiso.com

ialegodi@gmail.com 082 396 7717 - Oxygen 2, 8% - Sulphur 7, 3% - Carbon 83, 4% The calorific value of sulphur, carbon and hydrogen are 9 MJ/kg, 34 MJ/kg and 142 MJ/kg respectively. Calculate the following: 2.1. The mass of air required for the complete combustion of 3 kg of coal when there is 35% excess air supplied 2.2. The higher calorific value of coal 2.3. The lower calorific value of coal if the partial pressure is 4, 5 kPa 3. The percentage analysis of fuel according to mass is as follows: Carbon : 80 Hydrogen : 8 Sulphur : 6 Oxygen : 6 The heat value of carbon, hydrogen and sulphur is 35 MJ/kg, 145 MJ/kg and 8 MJ/kg respectively. The air has a temperature of 35and C the dry exhaust gas has the temperature of 206 C . The partial pressure of the stream formed in the exhaust gas is 5 kPa, whilst the specific heat capacity of the exhaust gas is 1, 005 kJ/kg.K Calculate the following: 3.1. The higher heat value of the fuel in MJ/kg 3.2. The lower heat value of the fuel in MJ/kg 3.3. The available heat value of the fuel in MJ/kg 4. The analysis of a fuel according to the mass, is as follows: Carbon = 84% energy of heat = 34 MJ/kg Hydrogen = 6% energy of heat = 142 MJ/kg Sulphur = 7% energy of heat = 9,8 MJ/kg Oxygen = 3% Take the heat of steam from the latent vapour as 2 432 kJ/kg Downloaded from hlayiso.com

ialegodi@gmail.com 082 396 7717 Calculate the following: 4.1. The higher heat value 4.2. The lower heat value 4.3. The mass of air required for the complete combustion of fuel is 23% of excess is supplied 5. A bomb calorimeter was used to determine the heat value of the fuel. The information on hand is as follows; Mass of fuel tested = 0, 7 g Mass of water in the calorimeter = 2 464 g Water equivalent of the calorimeter = 327 g Temperature of the water before the test = 18, 5 C Temperature of water after the test = 20, 69 C Latent heat of the vapour = 2 453 kJ/kg Calculate the following: 5.1. The higher heat value of the fuel in MJ/kg 5.2. The lower heat value of the fuel in MJ/kg if the lower heat value is 14, 5% lower than the higher heat value of the fuel 5.3. The percentage of hydrogen content in the fuel 6. The bomb calorimeter water was used to determine the heat value of the fuel. The following information results were recorded: - Mass of fuel burnt 0, 497 g - Mass of water used in the test 2, 4 kg - Water equivalent of the calorimeter 1 kg - Rise in temperature of water 1, 082 K - Specific heat capacity of water 4, 187 kJ/kg.K - The partial pressure of fuel 6 kPa Calculate the following: 6.1. The higher calorific value of fuel in MJ/kg 6.2. The lower calorific value of fuel in MJ/kg FOR MORE ACTIVITIES (ACTIVITY 6.1 PAGE 96 AND ACTIVITY 6.2 PAGE 99, V.A.A HAAGNER) Downloaded from hlayiso.com

ialegodi@gmail.com 082 396 7717 III. Combustion of hydro-carbon fuels Consider the general case of the hydrocarbon fuel represented by the formula: C x H y e.g C3 H 6 which means x  3 and y  6 The general formula : y y C x H y  ( x  )O2  xCO2  H 2 O 4 2 M air  100 23   M O2 Where M O2  Co  efficient of oxygen fromthe formula M total  (1  excess)  M air The product of combustion Formula Carbon dioxide CO2 From general formula Water H 2O From general formula Excess of oxygen Excess  O2 %Excess M O2 Nitrogen N2 77%M total POSSIBLE QUESTION: 1. 1 kg of a liquid fuel with a formula C15H32 is combusted with 30% excess air The molecular masses are as follows: Hydrogen 1 Carbon 12 Oxygen 16 Calculate the following: 1.1. The theoretical mass of air required for the complete combustion of this liquid 1.2. The actual mass of air required for the complete combustion of 1 kg of this liquid 1.3. The mass of each product of combustion 1.4. The percentages of each product of combustion 2. 1kg of fuel oil has the formula C12 H 4 completely burnt with the aid of 40% excess air. The atomic masses of the carbon, oxygen, and hydrogen are 12, 16, and 1 respectively Calculate the following: Downloaded from hlayiso.com

ialegodi@gmail.com 082 396 7717 2.1. The theoretical mass of air required to burn the fuel 2.2. The actual mass of air used to burn the fuel 2.3. The mass of each product of combustion 2.4. The percentage analysis by mass of the combustion products 3. 7, 5 kg of ethylene which has a formula C3H4 is completely burnt. The atomic mass of hydrogen, carbon and oxygen are 1, 12 and 16 respectively, whilst the air contains 23% oxygen and 77% nitrogen by mass. Calculate the following: 3.1. The mass of air completely burn 1 kg of fuel 3.2. The mass of air completely burn 5, 8 kg of fuel 3.3. The products of combustion for 5, 8 kg of fuel Downloaded from hlayiso.com