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MPT721S - Mineral Processing Techniques and Applications 324 - 1st Opp - Nov 2022 |
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n Am I BI A u n IV ERs ITY
OFSCIEnCEAno TECHnOLOGY
FACULTY OF ENGINEERING
DEPARTMENT OF MINING AND PROCESS ENGINEERING
QUALIFICATION : BACHELOR OF ENGINEERING IN METALLURGY, CHEMICAL
ENGINEERING & MINING ENGINEERING
QUALIFICATION CODE: BSc.
COURSE CODE: MPT721S
LEVEL: 7
COURSE NAME: MINERAL PROCESSING
TECHNIQUES AND APPLICATIONS 324
SESSION: OCTOBER, 2022
DURATION: 3 HOURS
PAPER: THEORY
MARKS:100
EXAMINER(S)
MODERATOR:
FIRST OPPORTUNITY QUESTION PAPER
Dr. Clement K. Owusu
Mr. Thomas Moongo
Prof. Godfrey Dzinomwa
INSTRUCTIONS
1. Answer all questions.
2. Read all the questions carefully before answering.
-------- --
3. Marks for each questions are indicated at the end of each question.
4. Please ensure that your writing is legible, neat and presi]}tab_[. -- . ~--~ ~-:--=-_-
PERMISSIBLE MATERIALS
«::-:-~-L---=- ,~-• :.:..-••-. =-•-•••
-:_•-
1. Examination paper.
THIS QUESTION PAPER CONSISTS OF 3 PAGES (Including this front-page)
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Question 1
a) Discuss the stages involved in the manufacturing process of unattractive rough
diamond to final glittering polished.
(15 marks)
b) Explain the relevance of process mineralogy in the extraction of value minerals
from ores.
(lOmarks)
c) Discuss the impact of carbonaceous matter on gold cyanidation process and
mention two possible control and management techniques used by mining
companies to ensure improved gold recovery
(9 marks)
Question 2
a. Explain the effect of pulp pH on collector adsorption (Remember to support with
chemical equations)
(10 marks)
b. A CIL gold processing plant receives it feed from the thickener after thickening of
the hydrocyclone overflow. After thickening, 95 % of the solids in the thickener
feed reported to the underflow while 80% of the water in the pulp reported to
overflow. The overflow pulp volumetric flow rate is 452 m 3/hr and the pulp
density is 1.05 t/m 3. The solids density is 2.57 t/m 3
i. Sketch and label the circuit.
(5 marks)
ii. Estimate the mass of solids (on dry basis) entering the thickener and mass of
water entering the thickener
(15 marks)
Question 3
a) Discuss briefly 3 possible causes for the failures of the tailings dam. (6 marks)
a) 100 tph of zinc ore containing 8.5% sphalerite and 91.5% gangue is treated in a
processing plant. Recovery of the sphalerite and the gangue in the concentrate are
- 92o/o~dJ:s-%.~Calculate flowrates- of ·co~cenir-ate_~-a tailing, assay ·oP-'"spnai~Ht~=-
b,- . --·----- - ---- .. - ------·---
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---:=..-=: A iingl_e-§.e!li_n ~--giy~:nb~_of-flot_~tjQn_~e.!J.§_giy~sGopper !e~overy of 55_%_J9.I.:._a-_:- . _:-:-_-_-_
p.~o~r~o(sP.IUlai: •..• :.
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\\J -
n Am I BI A u n IVER s ITY
OFSCIEnCE Ano TECHn OLOGY
FACULTY OF ENGINEERING
DEPARTMENT OF MINING AND PROCESS ENGINEERING
QUALIFICATION : BACHELOR OF ENGINEERING IN METALLURGY, CHEMICAL
ENGINEERING & MINING ENGINEERING
QUALIFICATION CODE: BSc.
COURSE CODE: MPT721S
LEVEL: 7
COURSE NAME: MINERAL PROCESSING
TECHNIQUES AND APPLICATIONS 324
SESSION: OCTOBER, 2022
DURATION: 3 HOURS
PAPER: THEORY
MARKS:100
FIRST OPPORTUNITY MEMORANDUM PAPER
EXAMINER(S)
Dr. Clement K. Owusu
Mr. Thomas Moongo
MODERATOR:
Prof. Godfrey Dzinomwa
INSTRUCTIONS
_ _ __ l. __:{\\nswerall questions.
- -r-~_
--
2. ~Read all the questions carefully before answering.
3~ Marks for each questions are indicated at the end of each question.
- _-~~-=~~~?:~-::~-=;_::-~~--i;P~l-ea:-se-ensure-that your writing is iegible, ~"neat=-anpdresentable:-- - -
- - --- "-~~:c,.:~~-
·~~=-=-c~-= - -~00 ~--< .,,PERMISSIBLE-MATERIALS
--~--=---1. Examination paper.
THIS QUESTION PAPER CONSISTS OF 3 PAGES (Including this front page)
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')
Question 1
a) Discuss the stages involved in the manufacturing process of unattractive rough
diamond to final glittering polished.
(15 marks)
b) Explain the relevance of process mineralogy in the extraction of value minerals
from ores.
(10 marks)
c) Discuss the impact of carbonaceous matter on gold cyanidation process and
mention two possible control and management techniques used by mining
companies to ensure improved gold recovery
(9 marks)
Solution
illfh
Manufacturingas appliedin diamondprocessinglooksat processesinvolvedin the transformation
of unattractive rough diamonds to glittering polished diamonds. The process involves acid
cleaning,sortingcutting and polishing.
• Acid cleaning: it involves the removal/cleaningof dirt from the rough diamondswfaces
th.roughthe use of an acid and washing. Th.isis done to ensure effectivesorting process.
(3 marks)
• Sorting: the cleanedrough diamondsare sortedbasedon size (caratweight shape,colour
and clarihJ-At th.isstage, the gem qualihJdiamondsare separatedfrom the industrial
diamonds.
(3 marks)
• Cutting and polishing entail processessuch as marking,sawing,bntting and girdling
and blocking
(3 marks)
Marking: Themarkersets roughdiamondon.system's samplestageand selectsthe
facetingproportionsth.atthe diamondshouldbecut into.
(1.5 marks)
Sawing: is the separationof a rough diamondinto differentpieces,to befinished
as separategems. The markedroughdiamondis then placedon a sawing spindle,
with a blademadefrom copperlayeredwith a mixture of oil and diamondpowder.
The rough diamond is th.enloweredto the bladeand cut where the marker has
markedthe stone. It is the diamondpowderth.atphysicallycuts the diamond,not
the copperblade- diamondsare the hardestknown mineral to man, and only a
diamondwill cut anotherdiamond.
(1.5 marks)
B~!f!ing_gtqes~!1_1d1i!a11~~10i1tsE_._b~ic sh.ap<T!_h..is is acl~ieveq_.by rotai:J!~_g?1_ie
diamondagainstanotherdiamondthat may alsoberotatingor stationan;,cihdthe
ttv9diamondsareprogressivelyJroundaway by niutua.Z-abrasion. (1.5 111.ark_s)_-
Blocking is thefinal stage wfrerebydiamondsare polis/redinto tirestone using
cast-irondisc surfacecoatedwith oil and diamonddust.
(1.5 marks)_
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Qlb.
Processmineralogi;:bridges the gap between mineral processingand traditionalmineralogi;.It
provides mineralsprocessingengineerswith informationon the characteristicsof mineralsin an
ore to optimize the process of recovering valuable minerals from the waste rock or gangue.
Information obtainedfrom processmineralogiJis appliedin mineralsexploration,determine the
best processingroutefor treating an ore and how processcan be optimised.Also, out of process
mineralogi;,one can preempt the benefits that can be harnessed,or limitations that need to be
cateredduring ore processing.Furthermore,it provides hint about sides reactionslikely to take
place during processing, qualihJ of concentrate and by-product to be expected, and possible
environmental impacts.
(10 marks)
Qlc.
Carbonaceousmaterialcan reducethe recoven;of gold by restricting the releaseof goldfrom the
carbonaceousmatrix, or by adsorbingdissolvedgoldfrom the leachliquor(preg-robbing)B. ecause
of their soft nature, they can causesmearing/coatingof Au particlesduring grinding process.The
treatmentof theseoresinvolves;
(4 marks)
• Roasting-where the carbonaceousmatter is burnt offfrom the ore
(1 mark)
• Isolating throughflotation using depressants
(1 mark)
• CIL approach - to competewith carbonaceousmatterfor Au. CILis howevernot effective
for highly preg-robbingores
(1 mark)
• Chlorination
(1 mark)
• Application of Blinding/blanking agents (e.g., Diesel, Keroseneand Heavy machine
oils)in combinationwith CIL -
(1 mark)
Question2
a. Explain the effect of pulp pH on collector adsorption (Remember to support with
chemical equations)
(10 marks)
b. A CIL gold processing plant receives it feed from thickener after thickening of the
hydrocyclone overflow. After thickening, 95% of the solids in the thickener feed
reported to the underflow while 80% of the water in the pulp reported to overflow.
The overflow pulp volumetric flow rate is 452 m3/hr and the pulp density is
1.05 t/m 3. The solids density is 2.57 t/m 3
i. Sketch and label the circuit.
-~(5marks) _ ____
ii. Estimate the mass of solids (on dry basis) entering the thiGkener and mass of
water entering the thickener
__(15 marks)
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Solution
Q2a
Theadsorptionmechanismof collectoron to :mineraslwface is infiueizcedby the pulp pH because
it controls the adsorptionof a collectorby the electron transfer mechanism.Generally,during
collectoradsorption,oxidationof thecollectoroccursat theanodeleadingto thereleaseofelectrons.
Forexample,in terms of xanthate the reactionshown in equation 1 occurs. The corresponding
cathodicreactioninvolves the reductionof oxygen which leadsto the productionof hydroxyl ions
as shown in equation2.
2x- H x2 + 2e
(1)
0 2 + 2H2 0 + 4e H 40H-
(2)
Raising tlzepH increasesthe concentrationof OH- ions will slow or even stops tlzereactionof a
collectorwith the mineraldue to the shift in chemicalequilibrium.When this occurs,the collector
no longeradsorbsonto the mineralsurface.Also, at toohigha pH, metalhydroxidescanprecipitate
andform coatingson the swface of a mineral,therebyblockingthemineralsurfaceand making the
surfacehydrophilic.This subsequentlypreventingcollectorinteractionwith themineralsurface.If
thecollectorhJpeconsideredis xanthate,at acidicpH, it becomesunstable.Thus, it breaksdown to
xanthic acid.
(10 marks)
Q2b
Feed (F)
(Mass of Solid=?)
(Mass of Water=?)
Overflow (0)
(%Solid= 5%)
(%Water= 80%)
(5 marks)
Underflow (U)
(%Solid= 95%)
(%Water= 20%)
.,,(].marks) ··· -- · ---
wherepsppand pwrepresentrespectivelysoliddensihJ,pulp densihJand densih)·oiyoater
= 2·57 <1.o5-l) X 100 7.79So/c
1.05(2.57-1)
O
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Mass of pulp
Pulp density = Volume of pupl
Mass of pulp = pulp density
Mass of pulp = 1.05 t/m3 x 452 1131/h = 474.6 t/h
x volume of pulp
(3 marks)
Mass of solids in overflow =percentsolids x Mass of pulp
=7.795% x474.6 t/h
= 36.995 t/h
(2 marks)
Mass of solidsin overflow=5% x massoffeed
Therefore;mass of feed= massof solidsin overflow/5%
= 36.995 t/h/5%=739.898t/h
( 2 marks)
Mass of overflow pulp = massof solidsin overflow+ massof water in overflow
Therefore,massof overflowwater = massof overflowpulp - massof solidsin overflow
= 474.6 t/h - 36.995 tjh = 437.6 t/h
(2 marks)
Remember,massof water in overflow=82%x massof water infeed
Thus;Mass of water infeed = massof water in overflow/82%
=437.6 t/h/0.82 = 538.92 t/h
( 2 marks)
Question 3
a) Discuss briefly 3 possible causes for the failures of the tailings dam. (6 marks)
b) 100 tph of zinc ore containing 8.5% sphalerite and 91.5% gangue is treated in a
processing plant. Recovery of the sphalerite and the gangue in the concentrate are
92% and 2.5%. Calculate flowrates of concentrate and tailing, assay of sphalerite
and gangue in concentrate and tailing.
(25 marks)
c) A single cell in a given bank of flotation cells gives copper recovery of 55% for a
residence time of 6 minutes. What is the number of similar sized cells in a
continuous flotation bank required to achieve a total recovery of 97%? (5 marks)
Solution
Q3a. (Any 3 for 6 marks)
Liquefaction; - Earthquakes are associated with the release of §>~i~!vllViCaveswhich
cause increase in shear stress on the embankment; and Pore pressure in saturated
-
tailings. Tailings in the impoundment may be liquefied during seismic events
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Liquefaction may also be caused by mine blasting or nearby motion and vibrations
of heavy equipment.
(2 marks)
Rapid increase in dam wall height; If an upstream dam is raised too quickly, .
very high internal pore pressure are produced within the tailings. High pore
pressures decrease the dam stability and may lead to dam failure. (2 marks)
Foundation failure; - If the base below the impoundment is too weak to support
the dam, movement along a failure plane will occur.
(2 marks)
Excessive water levels; - The beach width between the decant pond and dam crest
becomes too small. Flood inflow, high rainfall and improper water management
of the mill operators may cause excessive water level within the impoundment.
This may lead to overlapping and collapse of the embankment.
(2 marks)
Excessive seepage; -Seepage within or beneath the dam causes erosion along the
seepage flow path. Excessive seepage may result in failure of the embankment.
(2 marks)
Q3b.
Given
Feed flow rate
= F = 100 tph
% sphalerite in feed = f = 8.5%
% gangue in feed
= 91.5%
% recovery of sphalerite in concentrate = 92 %
% recovery of gangue in concentrate = 2.5%
Sphalerite recovery=~; x 100 = 92
C: 100 8_5 x 100 = 92
Cc=782
(3 marks)
C(100 - c)
Gangue recovery= F(l00 _ /) x 100 = 2.5
lF0{01C0-0-_.::C:1Yc_-X_--10-0-=-2-.-~5_---
100C- 782 x 100=2 S
100 X (100 - 8.5)
•
= C
10.1
---·-
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Concentrate flow rate= C = 10.1 tph
Solids balance
F = C+ T
Tailing flow rate
Cc=782
= T = 89.9tph
C
=
782
10.1
= 77.4%
100 = 10.1 + T (6 marks)
T=89.9tph
(4marks)
Assay of sphalerite in concentrate= c = 77.4%
Assay of gangue in concentrate = 100 - 77.4 = 22. 6%
(2 marks)
Sphalerite balance Ff= Cc+ Tt => 100 x 8.5 = 782 + 89.9t
=> t=0.76%
Assay of sphalerite in tailing= t = 0.76%
= Assay of gangue in tailing
100 -0.76 = 99.24%
(2 marks)
(2 marks)
Q3c
Data
Recoveryfor a singlecell(R1)= 55%
T11eresidencetime (r) = 6 min
R1 (1 + k-r) = kT 0.55(1 + 6k) = 6k biJmakingk the subject;
k = 0. 204/min
R = 1 - (1 + kT)-N
thereforeat 98% recovery
0.97 = 1 - (1 + 0.205 x 6rN bi; simplifi;ing;
2.zz-N= 0.03 applyinglogon bothsideand solving
N = 3.82691 ~ 4, tlms the numberof cellsrequiredis 5.
(3 marks)
(2 marks)
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