Please provide the solution for Part 1, 2 & 3 .

A bag of catalyst has recently been delivered to ACE Pte. Ltd.. The Certificate of Analysis of the catalyst provides the particle size distribution by dynamic light distribution, as laid out in Table 1. Fv is the cumulative volume distribution (normalized) of the particles, and x is the diameter in micrometer. Part 1 As the catalyst will be used in fluidized bed reactors, it is necessary to determine its surface volume mean for further calculations. Using the cumulative volume distribution given, determine the (a) the harmonic mean of volume distribution, (b) the arithmetic mean of surface distribution, and (c) the surface volume mean. All values should be calculated using appropriate conversions from the cumulative volume distribution values given. An Excel spreadsheet should be used for the calculations. Hint: Make use of the general equation and table from the textbook, given in the next page. Part 2 Determine the minimum fluidizing velocity of the catalyst for 1500kg of the catalyst in a tubular fluid bed reactor with a radius of 0.5m, given that the optimum conversion reaction occurs when the bed volume is 1.64 the volume of the catalyst particles. The fluid density and viscosity are 1000kg/m3 and 3.9mPas, respectively, while the particle density is 3000kg/m3. Part 3 For the same particles, what would happen if an upward flowing fluid (=950kg/m3,=0.0015 Pa.s) at 5.5mm/s is passed through the bed? If the catalysts are fluidized, what would be the mass fraction of the catalyst particles that is entrained (i.e. moving upwards and out of the reactor with the fluid flow)? The catalyst sphericity is 1. Hint: Volume distribution in Table 1 can be converted into mass distribution. Tahla 1 Den fore tha ratahuet nartinlas. From the textbook: g(x)=01dF01g(x)dFbut01dF=1andsog(x)=01g(x)dF