Crushing and Classification number 2 solution (Particle Technology by Holdich)
Download as DOCX, PDF1 like908 views
This is my solution to Problem Number 2 from Chapter 11 of Fundamentals of Particle Technology by Holdich. Hope it helped. E-mail me if you have questions or corrections at vic_zhou14@yahoo.com.ph. :) -Detective Freddie
More Related Content
Similar to Crushing and Classification number 2 solution (Particle Technology by Holdich) (14)
Recently uploaded (20)
Crushing and Classification number 2 solution (Particle Technology by Holdich)
- 1. Canja, Bobby T. ChE 65 A BS ChE 5 February 28, 2013 Chapter 11 Crushing and Classification below k f p q C k-q M
- 2. By doing material balance in certain sections of the diagram, the following equations are obtained. (1) (2) (3) (4) Manipulating (4): (5) Substituting (3) to (1): (6) Substituting (6) to (5): (7) Manipulating (7): (8) Given from the problem: C= 0.1 0 0 0 0 0.3 0 0 0 0 0.5 0 0 0 0 0.7 For the size distribution in the feed rate, 4 grades are assumed for the classifications of the particle sizes, namely: Grade 1 2 3 4 Size (亮m) 1000 to 750 750 to 500 500 to 250 <250
- 3. The size distribution of the particles in the feed with respect to each grade is obtained by subtracting the number of undersize in a corresponding particle diameter to the undersize percentage of the next smaller particle size diameter. The feed matrix in terms of mass flow rate is therefore: f= 0.15 0.61 0.14 0.10 f= 0.75 3.05 0.7 0.5 X 5 tons/hour The mill matrix is derived with the use of the breakage function unction where y is the lower boundary and x is the mean particle size within a certain grade. M= 0.176 0.358 0.267 0.199 0 0.287 0.427 0.286 0 0 0.448 0.552 0 0 0 1 There is also an identity function which when multiplied by a certain matrix will yield the value of the matrix multiplied to this type of matrix given by: I= 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 The I C matrix is just the difference of each elements in matrix I and C. IC= 0.9 0 0 0 0 0.7 0 0 0 0 0.5 0 0 0 0 0.3
- 4. Multiplying the matrix above the matrix of the feed yields: 0.675 2.135 0.350 0.150 (I C) {f} = Performing operations of matrix for the rest of the terms given in the equation, yields the following matrices. (I C)M = 0.158 0.251 0.134 0.060 0 0.201 0.214 0.086 0 0 0.224 0.165 0 0 0 0.300 I - (I C)M = 0.842 -0.251 -0.134 -0.06 0 0.799 -0.2135 -0.086 0 0 0.776 -0.165 0 0 0 0.7 Let so that ; where The cofactor matrix of H is obtained by using the formula for each of the element of matrix H where m = the minor of an element, which is the determinant of the matrix that excludes the row and column where the element is located. The cofactor matrix, as obtained by applying the formula in each element is: e= 0.434 0 0 0 0.136 0.457 0 0 0.086 0.126 0.471 0 0.08 0.086 0.111 0.522 0 0.457 0.126 0.086 0 0 0.471 0.111 0 0 0 0.522 The transpose of this matrix is: et = 0.434 0.136 0.086 0.08
- 5. From the original equation of inverse matrix, H-1 = = 1.189 0.373 0.236 0.219 0 1.252 0.345 0.236 0 0 1.29 0.304 0 0 0 1.43 1.189 0.373 0.236 0.219 0 1.252 0.345 0.236 0 0 1.29 0.304 0 0 0 1.43 From equation (8): {k-q} = 0.675 2.135 0.350 0.150 Thus, {k-q} = 0.803 2.925 1.347 0.973 tons per hour The size distribution in percentage in the coarse cut from the classifier is: Grade Percentage 1 13.28 % 2 48.36 % 3 22.27 % 4 16.09 % The total flow rate is 6.048 tons/hour.