Appendices A study on the sintering of fine-grained diamond-SiC composites 94 APPENDICES Appendix 1: Calculation of the amount of powder required to achieve a 1?m TiC coating on 2?m diamond powder. Molar mass of Ti = 48g, Molar mass of C = 12g, Molar mass of O2 = 32g, Molar mass of NH4Cl =53g, Density of TiC = 4.93g/cm3, Density of Diamond = 3.54g/cm3. NH4Cl + Ti + C TiC + NH3 + HCl Unit volume of diamond (U) = density g1 = 54.3 1 = 0.282cm3 Volume of diamond grain = 3 3 4 rpi = 3 2 0002.0 3 4 ?? ??? ? pi = 4.19 x 10-12 cm3 Number of particles per g of diamond (n) = ofgrainVol UnitVol . . = 6.74 x 1010 Therefore: Specific surface area of diamond (SSA) = 24 rpi x No. of particles per gram (n) = 8.47 x 103cm2/g Vol. of coating per g of diamond powder (V) = SSA x thickness of coating on diamond (t) = 8.47 x 103cm2/g x 1x 10-4cm = 8.47 x 10-1cm3 Mass of TiC per g of diamond = density of coating compound x vol. of coating per g of diamond = 4.18g Appendices A study on the sintering of fine-grained diamond-SiC composites 95 Vol. fraction of TiC = UV V + = 0.75 Therefore: Vol. fraction of diamond (ignoring porosity) (Vp) = 1-0.75 = 0.25 Assuming 50% porosity of the preform: Vol. fraction of diamond prior to infiltration (porosity included) = Vp x (1-0.5) = 0.125 Estimated vol. of sample (preform) = hr 2pi = pi (0.9)2 x 0.5 = 1.272cm3 Therefore: Vol. of diamond in the sample = 0.125 x 1.272 = 0.159cm3 Therefore: Mass of diamond in the sample = Vol. x Density = 0.159 x 3.52 = 0.560g Final vol. of TiC in the sample (porosity included) = Vol. fraction of TiC x Porosity = 0.75 x 0.5 = 0.375 Therefore: Vol. of TiC in sample = vol. frac. of TiC x vol. of sample = 0.375 x 1.272cm3 = 0.477cm3 Hence: Mass of TiC in the sample = vol. of TiC x density of TiC = 0.477cm3 x 4.93g/cm3 = 2.352g Therefore: Mass of Ti in the sample = 2.352 x 60 48 = 1.882g Hence: Balance C from TiC = 2.352 ? 1.882 = 0.470g Appendices A study on the sintering of fine-grained diamond-SiC composites 96 Therefore: Total mass of C (diamond) in the sample = 0.470g + 0.560g = 1.030g Ratio of reacting moles from the equation = Ti : NH4Cl = 1:1 Moles of Ti that reacted = Moles of NH4Cl that reacted = 48 882.1 = 0.0392 moles Hence: Mass of NH4Cl reacted = moles x Mr = 0.0392 x 53g = 2.078g Therefore the required quantities of the powders are: Diamond = 1.030g : Ti =1.882g : NH4Cl = 2.078g Appendix 2: Calculation of the amount of powder required to achieve a 1?m SiC coating on 2?m diamond powder. Molar mass of Si = 28g, Molar mass of C = 12g, Molar mass of O2 = 32g, Molar mass of NH4Cl =53g, Density of SiC = 3.16g/cm3, Density of Diamond = 3.54g/cm3. NH4Cl + Si + C SiC + NH3 + HCl Unit volume of diamond (U) = density g1 = 54.3 1 = 0.282cm3 Appendices A study on the sintering of fine-grained diamond-SiC composites 97 Volume of diamond grain = 3 3 4 rpi = 3 2 0002.0 3 4 ?? ??? ? pi = 4.19 x 10-12 cm3 Number of particles per g of diamond (n) = ofgrainVol UnitVol . . = 6.74 x 1010 Therefore: Specific surface area of diamond (SSA) = 24 rpi x No. of particles per gram (n) = 8.47 x 103cm2/g Vol. of coating per g of diamond powder (V) = SSA x thickness of coating on diamond (t) = 8.47 x 103cm2/g x 1x 10-4cm = 8.47 x 10-1cm3 Mass of SiC per g of diamond = density of coating compound x vol. of coating per g of diamond = 2.68g Vol. fraction of SiC = UV V + = 0.75 Vol. fraction of diamond (ignoring porosity) (Vp) = 1-0.75 = 0.25 Assuming 50% porosity of the preform: Vol. fraction of diamond prior to infiltration (porosity included) = Vp x (1-0.5) = 0.125 Estimated vol. of sample (preform) = hr 2pi = pi (0.9)2 x 0.5 = 1.272cm3 Therefore: Vol. of diamond in the sample = 0.125 x 1.272 = 0.159cm3 Therefore: Mass of diamond in the sample = Vol. x Density = 0.159 x 3.52 = 0.560g Final vol. of SiC in the sample (porosity included) = Vol. fraction of SiC x Porosity Appendices A study on the sintering of fine-grained diamond-SiC composites 98 = 0.75 x 0.5 = 0.375 Therefore: Vol. of SiC in sample = vol. frac. of SiC x vol. of sample = 0.375 x 1.272cm3 = 0.477cm3 Hence: Mass of SiC in the sample = vol. of SiC x density of SiC = 0.477cm3 x 3.16g/cm3 = 1.507g Therefore: Mass of Si in the sample = 1.507 x 40 28 = 1.055g Hence: Balance C from SiC = 1.507 ? 1.055 = 0.452g Therefore: Total mass of C (diamond) in the sample = 0.452g + 0.560g = 1.012g Ratio of reacting moles from the equation = Si : NH4Cl = 1:1 Moles of Si that reacted = Moles of NH4Cl that reacted = 28 055.1 = 0.0377 moles Hence: Mass of NH4Cl reacted = moles x Mr = 0.0377 x 53g = 1.998g Therefore the required quantities of the powders are: Diamond = 1.012g : Si =1.055g : NH4Cl = 1.998g