Introduction to Combustion Questions - Assignment Example

Heading: Introduction to combustion questions Your name: Course name: Professors’ name: Date: Question 1 There 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 2 Free 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 hydrogen (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 3 Heat combustion occurs when a compound undergoes absolute combustion with the presence of oxygen releasing energy as heat under normal conditions (Peters 2000). Question 4 i. 46.50 10C = 273 K Therefore add 46.50 to 273K Hence; 273K + 46.50 = 319.5K ii. 1740F to Kelvin Subtract 32 from 1740F = 142 Therefore 142 divided by 1.8 yields; 142/1.8 = 78.89 Then add 273 to 78.89 Yields= 351.89K =351.89K iii. Converting 7050C 10C = 273K Then add 273 to 7050C Yields= 978K iv. 212 0F to Kelvin Subtract 32 from 2120F = 180 Therefore 180 divided by 1.8 yields; 180/1.8 = 100 Then add 273 to 100 Yields= 373K =373K (Peters 2000) Question 5 a. H3PO4 + KOH → K3PO4+ H2O Multiply H3PO4 by 1 and the KOH by 3, in order to balance with the other side, multiply K3PO4 by 1 and multiply H2O by 3 which yield; H3PO4 + 3KOH → K3PO4+ 3H2O b. H3PO4 + Mg (OH)2 → Mg3(PO4)2 + H2O Multiply 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 + 6H2O c.C2H6 + O2 → CO2 + H2O Multiply 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 + 6H2O d. Ca3 (PO4)2 + SiO2 + C → CaSiO3 + CO + P Multiply Ca3 (PO4)2 by 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 6 The factors that determine the movement of gases and flames in an upward direction includes; 1. Convectional current as a result of increased turbulence which is caused by combustion. 2. The chemical type and the amount of fuel used also affect the upward movement of flames. 3. 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) Question 7 H= MCφ M= C= for gypsum = 1.09KJ/KJ/K Φ= 1800C (+ 273)K-300C(+273)K= 160K H= MCφ Φ= 160K 1*1.09*160= 174.4Joules H= 174.4Joules (Choo & van der 1998) Question 8 When the volume of the space that gases occupy is decreased, the pressure will consequently increase since volume is inversely proportional to pressure. The increase in pressure will be as a result of decreased space of molecules to occupy and hence knocking each other repeatedly causing increase in their kinetic energy which then results to increased rate of reaction (Miyanishi 2001). Question 9 Temperature can be described as a physical property of a matter which shows states of cold and hot situation, in summary temperature quantitatively shows the state of hot or cold. On the other hand, heat is basically energy transfer from one body to another as a result of thermal contact at different temperatures (Siegenthaler 2003). Question 10 Empirical formula= ClCH2 Molecular weight = 98.96 g/mol Masses of the elements Cl= 35.5g C= 12 H2=1*2=2 Let the unknown constant be n; Hence; n(ClCH2) = 98.96 g/mol 35.5n+12n+2n= 98.96 g/mol Then; 49.5n= 98.96 g/mol n= 1.999 approximately 2 n=2 2(ClCH2) = Cl2C2H4 Molecular formula= Cl2C2H4 Cl2C2H4 (Siegenthaler 2003) Question 11 Thermal explosion occurs as a result of the balance between the rates of generation of heat which is essentially a chemical reaction which releases heat to the atmosphere. Thermal explosion usually occurs at the initial stage of self heating process. If assumed to be under isothermal conditions, thermal explosion occur at temperature T in chemical type TD. The origin of the explosion is attributed to self-heating process at constant temperature as shown below (White 2010). The pre-explosion is usually described by;T1=Tc +(RT2c) / E with simplified form as; ϪTc=T1-Tc=(RT2c) / E. In the course of the heating process which is usually under isothermal condition, if the thermal condition is far greater than the atmosphere (u Read More