August 2014 Mission Statement Background
Download 11.38 Mb.
|
Appendix CCalculation pressure flow rate Details of notations on page 51 A perfect sphere has a surface area of The volume per spherical particle is This means that surface area – volume ratio for a sphere is 1 because: 
A nearly spherical shape is desired, so Ergun equation:  (C-3)
Length of bed assumed to be 0.5 m is approximation of how deep into the sand in the 0.915 m high water butt the anchor leg will fluidise. Density is for the water, 998 kg/m3, and superficial velocity is: Since the water butt is 0.56 m in diameter, maximum fluidised cross-sectional area is: 
Maximum flow rate is the flow rate directly from the water source. Measuring this resulted in a bucket of volume 5.767*10-3 m3 filled in 35 seconds. This gives a maximum flow rate of: 
Maximum possible superficial velocity for the experiment would therefore be: 
Viscosity is 0.001 Ns/m2 for water in room temperature [Eng143]33, so if void fraction of 0.4 is used, then pressure flow rate for a 0.1mm diameter particle is: 
Finding out what part dominates in the Ergun equation: First part with viscosity: 
Second part with density: 
Directory: shs shs -> Stonelaw Mathematics Department Green Course Revision Sheets Block E shs -> Core Classes Offered by Department Indicates ncaa clearinghouse Approved Core Class english shs -> Identify the prepositional phrases in the following sentences. Write them on a separate piece of paper shs -> Smart Phone Class Security and Safety shs -> Bordon Luciano shs -> 1 Week before Scheduled Contact shs -> Informative Research Essay Assignment shs -> 10th Daily “Fix It” Sentences shs -> The story of Perseus as told by Diodorus Download 11.38 Mb. Share with your friends:
|
(C-1) 
(C-2) 
will be assumed to be 0.83. For this, void fraction will vary from 0.4-0.45. Particle diameters evaluated: 0.05, 0.1, 0.2, 0.3, 0.4 and 0.5 mm.
(C-4)