Introduction To Combustion Questions Example

Introduction To Combustion Questions Example Introduction To Combustion Questions â€" Assignment Example > Question 1There are three basic states of matter namely solid, liquid and gas. The molecular difference of these matter states are as follows; Solid- This is the state in which the intermolecular attraction forces that exist between the molecules keep the molecules in a spatial relationship which is basically fixed since there is no free movement of molecules. Liquid- This is the state in which intermolecular forces of attraction are responsible for keeping close together but not fixed together as in solids. This means that the molecules in liquids are freer to move around as compared to solids. Gas- In gases, intermolecular forces of attraction in gases are considered the least since molecules are free to move away from each other as compared to both liquids and solids (Kuhl 2003, p. 251). Question 2Free radicals results when a fuel such as a hydrocarbon is exposed to heat which creates a bond breaking energy. This energy breaks the hydrocarbon bond creating free radicals of hydr ogen (H. ) and carbon (C. ). The breaking of this bond releases energy which leads to more fire spread as these free radicals reacts with other gases in the process of combustion (Peters 2000). Free radicals are also reactive and thus leading to more reaction which increases combustion. Question 3Heat combustion occurs when a compound undergoes absolute combustion with the presence of oxygen releasing energy as heat under normal conditions (Peters 2000). Question 4i. 46.5010C = 273 KTherefore add 46.50 to 273K Hence; 273K + 46.50= 319.5Kii. 1740F to KelvinSubtract 32 from 1740F= 142Therefore 142 divided by 1.8 yields; 142/1.8 = 78.89 Then add 273 to 78.89 Yields= 351.89K =351.89Kiii. Converting 7050C 10C = 273KThen add 273 to 7050CYields= 978Kiv. 212 0F to KelvinSubtract 32 from 2120F= 180Therefore 180 divided by 1.8 yields; 180/1.8 = 100 Then add 273 to 100 Yields= 373K =373K(Peters 2000)Question 5H3PO4 + KOH ? K3PO4+ H2OMultiply H3PO4 by 1 and the KOHby 3, in order to balance wit h the other side, multiply K3PO4 by 1 and multiply H2O by 3 which yield; H3PO4 + 3KOH ? K3PO4+ 3H2Ob. H3PO4 + Mg (OH)2 ? Mg3(PO4)2 + H2OMultiply H3PO4 by 2 and the Mg (OH)2 by 3, in order to balance with the other side, multiply H2O by 6 which yields; 2H3PO4 + 3Mg (OH)2 ? Mg3(PO4)2 + 6H2Oc. C2H6 + O2 ? CO2 + H2OMultiply C2H64 by 2 and the O2 by 7, in order to balance with the other side, multiply CO2 by 4 and multiply H2O by 6 which yield; 2C2H6 + 7O2 ? 4CO2 + 6H2Od. Ca3 (PO4)2+ SiO2 + C ? CaSiO3 + CO + PMultiply Ca3 (PO4)2by 1and the SiO2 by 3,and C by 5, in order to balance with the other side, multiply CaSiO3by 3 and multiply CO by 5 and P by 2 which yield; Ca3 (PO4)2+ 3SiO2 + 5C ? 3 CaSiO3 + 5CO + 2P(Siegenthaler 2003)Question 6The factors that determine the movement of gases and flames in an upward direction includes; Convectional current as a result of increased turbulence which is caused by combustion. The chemical type and the amount of fuel used also affect the upward mov ement of flames. The ventilation of the room contributes to the direction of the gases and flame movement since combustion is highly dependent on oxygen presence (Miyanishi 2001)