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Unformatted text preview: Exercise 17.1 Subject: Estimation of sphericities Given: (a) a cylindrical needle with height = 5 times the diameter. (b) a rectangular prism of sides a , 2 a , and 3 a . Find: The sphericities Analysis: The sphericity, , is defined by (171) and (172): surface area of a sphere of same volume as the particle surface area of particle = = particle particle 6 p V D S & (1) (a) Let the diameter of the needle be D and the height be H = 5 D The volume of the needle = needle V = ( ) 2 2 3 5 5 4 4 4 D D D H D = = The surface area of the needle = needle S = ( ) 2 2 2 11 2 5 4 2 2 D D DH D D D & + = + = Substitution into (1), gives: 3 2 5 4 6 15 11 11 2 p p D D D D D & = = & (2) From the definition of the sphericity, needle V = sphere V or 3 3 5 4 6 p D D = Therefore, 1/3 30 1.96 4 p D D & = = and from (2), = 15 1 11 1.96 & = 0.696 (b) Let the sides of the prism be a , 2 a , and 3 a . The volume of the prism = prism V = a (2 a )(3 a ) = 6 a 3 The surface area of the prism = ( )( ) ( )( ) ( )( ) 2 prism 2 2 3 2 3 22 S a a a a a a a = + + = Substitution into (1) gives: 3 2 6 6 36 22 22 p p a a D a D & & = = From the definition of the sphericity, prism V = sphere V or 3 3 6 6 p D a = Therefore, 1/3 36 2.55 3.14 p D a & = = and from (2), = 36 1 22 2.25 & = 0.727 Exercise 17.2 Subject: Sphericity of a thin circular plate Given: Circular plate of thickness, t , and diameter, D , with sphericity, , equal to 0.594 Find: The ratio of t to D . Analysis: The sphericity, , is defined by (171) and (172): surface area of a sphere of same volume as the particle surface area of particle = = particle particle 6 p V D S & (1) The volume of the plate = plate V = 2 4 D t The surface area of the plate = plate S = 2 2 2 4 2 D D Dt Dt & + = + Substitution into (1), gives: 2 2 4 6 6 2 4 2 p p D t tD D D D t D Dt & & = = + & + (2) From the definition of the sphericity, sphere V = plate V or 3 2 6 4 p D D t = Therefore, 3 2 3 2 p D D t = and from (2), = 1/3 2 6 0.594 2 4 3 2 tD D t D t & = + & (3) Let R = t / D Therefore, t = RD Substitution into (3) gives: ( ) ( ) 2 1/3 1/3 3 6 0.594 5.24 2 4 2 4 3 2 RD R D RD R R D R & = = + + & which can be rewritten as, 1/3 4/3 0.297 0.594 1.31 R R + = (4) Solving (4), a nonlinear equation, with Polymath, for an initial guess of 0.3, R = t / D = 0.166 Exercise 17.3 Subject: Differential and cumulative undersize plots of a screen analysis for crystals of sodium thiosulfate....
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This note was uploaded on 04/27/2011 for the course CHEM 101 taught by Professor Suh during the Spring '11 term at University of Toronto Toronto.
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