Date: 22nd May, 2024 Semester: 4th Faculty Member: Nouman Ahmed Course & Section: CHE-15A School of Chemical and Materials Engineering (SCME) CHE-224 Fluid Mechanics-II Complex Engineering Problem Name Muhammad Usama Bin Asif Ahmed Hassan Haisum Tahir Hamzah Kamran CMS 407178 408785 420292 424447 The Problem: Our Complex Engineering problem will deal with the following problem: You are tasked with designing a fluidized bed reactor for coating small particles with a polymer solution. The particles have an average diameter of 0.5 mm and a density of 2,500 kg/m³. The polymer solution is to be sprayed evenly over the particles as they are fluidized by air. The objective is to achieve a uniform and complete coating of all particles in the reactor. Calculate the minimum air velocity required to fluidize the particles and propose a design for the distribution of air to ensure effective fluidization. Known Parameters: ο Particle Equivalent Diameter or (π π): 0.5mm ππ ο Particle Density (ππ ): 2500 π3 ππ ο Air Density at Operating Conditions (ππ ): 1.2 π3 ο Dynamic Viscosity of air at operating Conditions (ππ ): ππ. π ∗ ππ−π Pa s Glossary: Archimedes Number: Dimensionless Number that is used to related the external forces to interna viscous forces resulting in a difference in the motion of fluids due to density differences. Modified Reynolds Number: The Reynolds number equation that is the ratio of inertial forces to viscous forces, for a given empirical case, in this case, fluidization. Minimum Fluidization Velocity: The minimum velocity required by the fluid phase for the drag forces to be equivalent to the gravitational forces acting on the particles inside for them to be suspended. ππ = −ππ Sphericity Factor: The degree to which a particular object is close to being a perfect sphere with 1 being considered to be an object completely equivalent to a perfect sphere. A perfect sphere is defined as one with no edges. Our Solution: Step 1: Determine the Archimedes Number (π¨π) The Archimedes number is given by: ππ3 ∗ (ππ − ππ ) ∗ π2 π΄π = ππ2 We begin calculations as such: ππ − ππ = 2500 ππ ππ − 1.2 = 2498.8 ππ/π3 3 3 π π Plug in the values into the formula: (0.0005)3 π ∗ (2498.8) ∗ 9.812 π΄π = π π 2 (18.2 ∗ 10−6 )2 ππ π π3 ππ π΄π = 1.85 ∗ 10 π 2 8 Step 2: Determine the Modified Reynolds Number (πΉπππ ) The modified Reynolds Number for fluidization is given by: 1 π πππ = (33.72 + 0.0408 ∗ π΄π)2 − 33.7 Using the Archimedes number value, we obtained before: 1 π πππ = (33.72 + 0.0408 ∗ (1.85 ∗ 108 )2 ) − 33.7 π πππ = 2687.72 Step 3: Determine the Minimum Fluidization Velocity (πΌππ ) The minimum fluidization velocity can be obtained by: πππ = π πππ ∗ ππ ππ ∗ π· Where: π· = ππ ∗ Ο s Diameter (D): To be determined Sphericity Factor (ππ¬ ): 1 As Sphericity factor is 1, the value of D is equal to ππ . Thus, by inputting our values: 2687.72 ∗ (18.2 ∗ 10−6 ππ π ) πππ = ππ 1.2 3 ∗ (0.0005) π π π πππ = 81,53 π Thus, the minimum velocity required for this particular fluidization process is π¦ ππ. ππ . π¬ Step 3: Air Distribution Method To ensure that air is distributed in the reactor without any channeling or voids, we must make sure that the fluid is distribute uniformly. This can be ensured as such: Distributor Plate Design: Use a perforated plate with a high open area percentage (20-30%). Hole Size/Spacing: Holes should be ensured to be small and spaced evenly for effective uniform fluidization.
