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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 (17-1) and (17-2): 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 (17-1) and (17-2): 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....
View Full Document

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.

Page1 / 63

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online