Wills' Examples to end Chapter 3

Got there at 17:15 on Wednesday 22nd May 2024.

I seem to have started on doing the calculations in "Wills" on about the 24th of April. On-and-off - has been a backdrop to the time, doing these calculations following the "Examples" in Wills' book.
Having cover-to-cover 'ed the Wills' "Mineral Processing Technology" by 09 April 2024.

Going through "Examples" to end of Chapter 3.
So that's been 2 months and 1 day since the book arrived.

So topics covered:

Chapter 1

metallic fraction in minerals
Recovery
Separation Efficiency
Net Smelter Return
Economic Efficiency

Chapter 3

sample size for good accuracy
slurry calculations - volume-based; mass-based
"n-product formula" as introduction to mass-balance

then had to change to "just" re-reading

the mass-balance approaches able to handle circuits

because these involve "sophisticated" mathematical techniques which most (?) would use a computer "application" (a suite of programs forming a "package") to solve. My estimate - these mathematics would take months to a year or two to learn, if as a practical person you wanted to master these techniques (relating to my Finite Element Analysis self-teaching).

The point and driving-force?

These are generalities in minerals processing.
Subsequent Chapters in the book are specific processes used in mineral processing - eg comminution; froth-flotation; etc.

Applying analysis of these generalities I show

interest in minerals processing and minerals in general which is indisputable
that I can "get to work" on anything mathematical in minerals processing - it's believable that what I don't currently know I can soon acquire
that I've grasped many minerals processing concepts

An example from Chapter 3 : slurries (finely ground ore in water - otherwise known as "pulps"):
initially a "brain-buster" for me - in Example 3.5 (in 8th Edition) which is the second slurries Example - where this time there is a volume flow-rate of slurry, yet the slurry is described by a mass % solids. Volume<=>mass...
The most valuable approach is to develop a "model" in your mind which is easy to recall and from which you could re-derive the maths if you come back to the same situation again some distant time later. Given some archane equation alone is hard to remember.
Sketching, I came up with a model of a "Unit mass of slurry" (so in SI-units that would be 1 kg) which has a "corresponding volume" - which can now have a % solids and a % liquids...
You have converted to working in volumes - which is useful given the flow-rate of the slurry is given as a volume per unit time...

For what it's worth - given this may be incomprehensible.
This is my text-file of notes and developing the calculations.
The maths is expressed in Lisp (the "emacs Lisp" of the text-processor I use), with the answer generated by the text-editor's interpretter interpretting that expression "inserted" into the text.
Anything to the right-hand-side of a ";" is disregarded by the interpretter.

Stream 1 5 m^3 hour^-1; 40% solids by mass
Stream 2 3.4 m^3 hour^-1; 55% solids by mass
Density of solids = quartz : 3000 kg m^-3
Density of liquid - assume is water at 1000 kg m^-3
Tonnage of dry solids (equivalent) pumped per hour?

E3.5
====

Stream 1 :

Unit mass

   1 ---
  liq.|
      |
    __|__
 solid|   0.4
      |
   0 ---

goes to a "corresponding volume to unit mass"
which is an entire volume conceptually comprising a volume of solid
and a volume of liquids.

rho_w = 1e3
rho_m = 3e3


Water component
V_c-v_l =
(/ (- 1 0.4) 1e3) ;; 0.0006

Solid component
V_c-v_m =
(/ 0.4 3e3) ;; 0.00013333333333333334

Corresponding volume
V_c-v =
(+ (/ (- 1 0.4) 1e3) (/ 0.4 3e3)) ;; 0.0007333333333333332

(* (+ (/ (- 1 0.4) 1e3) (/ 0.4 3e3)) 1e3) ;; 0.7333333333333333 ;; Litres
Stream 1 slurry has 0.733L per kg mass

Density of the slurry 1
becomes accessible

(/
 1e0 ;; kg ;; unit mass
 (+ (/ (- 1 0.4) 1e3) (/ 0.4 3e3))
 ) ;; 1363.6363636363637 ;; kg m^-3

rho_sl_1 = 1363.6363636363637

--------------------------------

Stream 2 :

   1 ---
  liq.|
    __|__
      | 0.55
 solid|
      |
   0 ---

Water component
V_c-v_l =
(/ (- 1 0.55) 1e3) ;; 0.00044999999999999993 ;; m^3

Solid component
V_c-v_m =
(/ 0.55 3e3) ;; 0.00018333333333333334 ;; m^3

Corresponding volume
V_c-v =
(+ (/ (- 1 0.55) 1e3) (/ 0.55 3e3)) ;; 0.0006333333333333333
0.633L per kg of slurry

Density of slurry 2
(/ 1e0 (+ (/ (- 1 0.55) 1e3) (/ 0.55 3e3))) ;; 1578.9473684210527 ;; kg m^-3

rho_sl_2 = 1578.9473684210527

--------------------------------

With the slurry densities, now going to what is being pumped - the
combined streams...

v-dot = 5
rho_sl_1 = 1363.6363636363637
f_m_min_1 = 0.4

v-dot = 3.4
rho_sl_2 = 1578.9473684210527
f_m_min_2 = 0.55

(+ ;; the two streams

 ;; stream1
 (*
  5 ;; m^3 h_1
  1363.6363636363637 ;; kg m^-3
  ;; ;; 6818.181818181819 ;; kg h^-1
  0.4 ;; f_m_solids
  ) ;; 2727.272727272728

 ;; stream2
 (*
  3.4 ;; m^3 h-1
  1578.9473684210527 ;; kg m^-3
  ;; ;; 5368.421052631579 kg h^-1
  0.55 ;; f_m_solid
  ) ;; 2952.631578947369

 ) ;; 5679.904306220096 ;; kg h^-1 of solids pumped

5.7 Tonnes h^-1 of solids in pumped stream(s).

My answer agrees with Wills, but uses different expressions - mine are
derived from my "unit mass => corresponding volume" model

(R. Smith, 23May2024, 25May2024 (eds), 10Jun2024 (ed. diag1st))