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# ch17 - Exercise 17.1 Subject Estimation of sphericities...

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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, ψ, is defined by (17-1) and (17-2): surface area of a sphere of same volume as the particle surface area of particle ψ = = particle particle 6 p V D S ° ± ² ³ ² ³ ´ µ (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 π π π = = The surface area of the needle = needle S = ( ) 2 2 2 11 2 5 4 2 2 D D DH D D D ° ± π π π + = π + = π ² ³ ´ µ Substitution into (1), gives: 3 2 5 4 6 15 11 11 2 p p D D D D D ° ± π ² ³ ´ µ ψ = = ° ± π ² ³ ´ µ (2) From the definition of the sphericity, needle V = sphere V or 3 3 5 4 6 p D D π π = Therefore, 1/3 30 1.96 4 p D D ° ± = = ² ³ ´ µ and from (2), ψ = 15 1 11 1.96 ° ± ² ³ ´ µ = 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 = + + = · ¸ ¹ Substitution into (1) gives: 3 2 6 6 36 22 22 p p a a D a D ° ± ° ± ψ = = ² ³ ² ³ ² ³ ´ µ ´ µ From the definition of the sphericity, prism V = sphere V or 3 3 6 6 p D a π = Therefore, 1/3 36 2.55 3.14 p D a ° ± = = ² ³ ´ µ and from (2), ψ = 36 1 22 2.25 ° ± ² ³ ´ µ = 0.727

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Exercise 17.2 Subject: Sphericity of a thin circular plate Given: Circular plate of thickness, t , and diameter, D , with sphericity, ψ , equal to 0.594 Find: The ratio of t to D . Analysis: The sphericity, ψ, is defined by (17-1) and (17-2): surface area of a sphere of same volume as the particle surface area of particle ψ = = particle particle 6 p V D S ° ± ² ³ ² ³ ´ µ (1) The volume of the plate = plate V = 2 4 D t π The surface area of the plate = plate S = 2 2 2 4 2 D D Dt Dt ° ± π π + π = + π ² ³ ´ µ Substitution into (1), gives: 2 2 4 6 6 2 4 2 p p D t tD D D D t D Dt ° ± π ² ³ ° ± ´ µ ψ = = ² ³ + ° ± π ´ µ + π ² ³ ´ µ (2) From the definition of the sphericity, sphere V = plate V or 3 2 6 4 p D D t π π = Therefore, 3 2 3 2 p D D t = and from (2), ψ = 1/3 2 6 0.594 2 4 3 2 tD D t D t ° ± = ² ³ + ´ µ ° ± ² ³ ´ µ (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 ° ± = = ² ³ ² ³ + + ° ± ´ µ ² ³ ´ µ which can be rewritten as, 1/3 4/3 0.297 0.594 1.31 0 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. Given: U.S. Screen analysis. Assumptions: particles approximate spheres Find: Differential and cumulative undersize plots of the data Analysis: Use a spreadsheet with apertures for U.S. Mesh from Table 17.4. Results are as follows with the two plots on the next page. Arithmetic plots are preferred here.

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ch17 - Exercise 17.1 Subject Estimation of sphericities...

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