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

- the mass-balance approaches able to handle circuits

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))