NUMERICALS FROM DAY I-7

(Principles of Extractive Metallurgy)

PRESENTED BY-

PRATIK KURLY(2019UGMM032)

ARYAN RAJ(2019UGMM063)

SAURAV SUMIT (2019UGMM098)

Topics-

❑ L i b e r a t i o n & C o m m i n u t i o n

❑ S i e v e A n a l y s i s

❑ S t o k e s ’ L a w

❑ P i p e t t e A n a l y s i s

❑ R e l a t i v e D e n s i t y

❑ I m p o r t a n c e O f M a t e r i a l S t r e n g t h A t B r e a k a g e

❑ D y n a m i c S t r e s s e s G e n e r a t e d B y I m p a c t L o a d s

I n B r i t t l e n e s s Te s t

❑ S c r e e n i n g

❑ F r e e S e t t l i n g Ve l o c i t y

❑ H y d r o C y c l o n e s

❑ J i g g i n g

❑ F r o t h F l o t a t i o n

❑ F l o t a t i o n

❑ Ta i l i n g

MINERAL PROCESSING

There are two fundamental operations in

mineral processing: namely the release, or

liberation, of the valuable minerals from their

waste gangue minerals, and separation of

these values from the gangue, this latter

process being known as concentration.

Liberation of the valuable minerals from the

gangue is accomplished by comminution,

which involves crushing, and, if necessary,

grinding, to such a particle size that the

product is a mixture of relatively clean particles

of mineral and gangue.

CRUSHING

GRINDING

EQUIPMENT AND STEPS INVOLVED

LIBERATION,COMMINUTION

❑ LIBERATION AND COMMINUTION

Question.)What equipment is suitable for comminution of chromite so that it produces a proper liberation size?Answer: For Comminution of Chromite, various equipment are available like

Grinding mill vertical roller mill roller press etc.

We can also use selfrag option. Using this technology as a part of Comminution process can be economically beneficial, saving on the wear and tears of traditional plant.

Crushing :- C r u s h i n g i s t h e f i r s t m e c h a n i c a l s t a g e i n t h e p r o c e s s o f c o m m i n u t i o n i n w h i c h t h e m a i n o b j e c t i v e i s t h e l i b e r a t i o n o f t h e v a l u a b l e m i n e r a l s f r o m t h e g a n g u e . I t i s g e n e r a l l y a d r y o p e r a t i o n a n d i s u s u a l l y p e r f o r m e d i n t w o o r t h r e e s t a g e s .

T h e m o s t c o m m o n a p p l i c a t i o n i s t o p r o d u c e a c e r t a i n p r o d u c t s i z e d i s t r i b u t i o n o r u p p e r s i z e f o r r e q u i r e m e n t s o f a s p e c i a l m a r k e t o r s u b s e q u e n t m a c h i n e s . C r u s h i n g i s a m e c h a n i c a l p r o c e s s w h e r e l a r g e r s o l i d p a r t i c l e s o f b r i t t l e m a t e r i a l a r e s u b j e c t e d t o s u f f i c i e n t l y h i g h f o r c e s a n d e n e r g y w i t h t h e c o n s e q u e n c e t h a t t h e m a t e r i a l o f t h e p a r t i c l e s f a i l s a n d f r a g m e n t s a r e g e n e r a t e d

• Grinding:- G r i n d i n g i s o f t e n t h e g r e a t e s t e n e r g y c o n s u m e r , a c c o u n t i n g f o r u p t o 5 0 % o f a c o n c e n t r a t o r ' s e n e r g y c o n s u m p t i o n .

• A s i t i s t h i s p r o c e s s w h i c h a c h i e v e s l i b e r a t i o n o f v a l u e s f r o m g a n g u e , i t i s a l s o t h e p r o c e s s w h i c h i s e s s e n t i a l f o r e f f i c i e n t s e p a r a t i o n o f t h e m i n e r a l s , a n d i t i s o f t e n s a i d t o b e t h e k e y t o g o o d m i n e r a l p r o c e s s i n g .

• I n o r d e r t o p r o d u c e c l e a n c o n c e n t r a t e s w i t h l i t t l e c o n t a m i n a t i o n w i t h g a n g u e m i n e r a l s , i t i s n e c e s s a r y t o g r i n d t h e o r e f i n e l y e n o u g h t o l i b e r a t e t h e a s s o c i a t e d m e t a l s .

The grinding operations in mineral processing are carried out in

what are called are the tumbling mills. The general term

tumbling or revolving mill includes the rod mill, the ball mill,

the pebble mill, the autogenous mill, and the tube mill.

GRINDABILITY :-O r e g r i n d a b i l i t y r e f e r s t o t h e e a s e w i t h w h i c h m a t e r i a l s c a n b e c o m m i n u t e d , a n d d a t a f r o m g r i n d a b i l i t y t e s t s a r e u s e d t o e v a l u a t e c r u s h i n g a n d g r i n d i n g e f f i c i e n c yP r o b a b l y t h e m o s t w i d e l y u s e d p a r a m e t e r t o m e a s u r e o r e g r i n d a b i l i t y i s t h e B o n d w o r k i n d e x W ~ .G r i n d a b i l i t y i s b a s e d u p o n p e r f o r m a n c e i n a c a r e f u l l y d e f i n e d p i e c e o f e q u i p m e n t a c c o r d i n g t o a s t r i c t p r o c e d u r e . T h e B o n d s t a n d a r d g r i n d a b i l i t y t e s t h a s b e e n d e s c r i b e d i n d e t a i l b y D e i s t e r ( 1 9 8 7 ) , a n d L e v i n ( 1 9 8 9 ) h a s p r o p o s e d a m e t h o d f o r d e t e r m i n i n g t h e g r i n d a b i l i t y o f f i n e m a t e r i a l s . T a b l e 5 . 1 l i s t s s t a n d a r d B o n d w o r k i n d i c e s f o r a s e l e c t i o n o f m a t e r i a l s .

EVALUATION OF PARTICLE SIZE & SIZE DISTRIBUTION CURVES

Sieve Analysis

Question.)500gm of dry soil was subjected to a sieve analysis the weight of soil retained on each sieve is as follows.Plot the grain size distribution curve and determine the following.•Percentage of gravel, coarse sand, medium sand, fine sand and silt clay fraction in the soil as per I.S.•Effective size•Uniformity coefficient•The gradation of the soil•Soil Classification

I.S. Sieve Size Mass of Soil(gm)4.75mm 10

2mm 165

1mm 100

425µ 85

212µ 60

150µ 20

75µ 40

I.S. Sieve Diameter

of grain

(mm)

Mass

retained

(gm)

Cumulati

ve mass

(gm)

Percentage

(%) retainedPercentage

Finer(%)

4.75mm 4.75 10 10 2 982mm 2 165 175 35 651mm 1 100 275 55 45425µ 0.425 85 360 72 28212µ 0.212 60 420 84 16150µ 0.150 20 440 88 1275µ 0.075 40 480 96 4

ANSWER:-

Soil Classification:≤2µ — Clay

2µ-75µ — Silt75µ-0.425mm — Fine Sand

0.425mm-2mm — Medium Sand2mm-4.75mm — Coarse Sand

4.75mm-80mm — Gravel80mm-300mm — Coble

≥300mm — Boulder

Percentage retained = Mass retained /total mass

* 100

a) % of Gravel = 100-98 = 2%% of Coarse Sand = 98-65 = 33%% of Medium Sand = 65-28 = 37%% of Fine Sand = 28-4 =24%% of Silt Clay = 4%

b) Effective Size (from Graph),D10 = 0.12mm

c) Uniformity Coefficient (Cu),Cu =

= 14.16

d) Gradation of the Soil (Cc),

Cc = = 1.12

e) Soil Classification,Soil is classified as Well Graded Sand (SW) since Cu is greater than 6and Cc is in between 1-3.

=

=

Size Distribution Curve:

Question: Assume that a spherical body is flowing through water. The velocity of the body at particular instant is 2 m/s. Find the drag force on the body due to the fluid. Assume that Stokes law is valid. (Given: viscosity of water = 0.001 Kgm−1s−1, radius of spherical object = 2 mm)

Answer: W e k n o w t h a t t h e d r a g f o r c e o n t h e b o d y w i l l b e

F=6πμ r V

p u t t i n g t h e v a l u e s i n t h e e x p r e s s i o n ,

F = 6 × 3 . 1 4 × 0 . 0 0 1 m s K g × 2 × 1 0−3× 2 s m

h e n c e F = 7 5 . 3 6 × 1 0−6N

STOKES’ LAW _:

Pipette Analysis

Question.)The following observations were taken during a pipette analysis for the determination of particle size distribution of a soil sample.

•Depth below the water surface at which the sample was taken = 100 mm•Capacity of pipette = 10 ml•Mass of sample when dried = 0.3 gm•Time of taking sample = 7 minutes after the start•Volume of soil suspension in the sedimentation tube = 500 ml •Dry mass of soil used in making suspension = 25 gmDetermine the coordinate of the point on the particle size distribution curve corresponding to aboveobservations.Take G = 2.70, η = 10.09 milli poise and Df = 1 gm/ml.Answer : As We know that,

D=

D=

And N= X 100 = X 100 =60%

So,The coordinates of the point on the size distribution curve are (0.0161 mm, 60%).

RELATIVE DENSITY:-

Q u e s t i o n . ) A s o i l h a s a d r y d e n s i t y o f 1 . 8 1 6 g m / m l i n t h e n a t u r a l c o n d i t i o n . W h e n 4 1 0 g m o f t h e s o i l w a s p o u r e d i n av e s s e l i n a v e r y l o o s e s t a t e , i t s v o l u m e w a s 2 9 0 m l . T h e s a m e s o i l w h e n v i b r a t e d a n d c o m p a c t e d

w a s f o u n dt o h a v e a v o l u m e o f 2 1 5 m l . D e t e r m i n e t h e r e l a t i v e d e n s i t y .

gm/ml

gm/ml

ANSWER:

Relative Density,

MECHANISM OF BREAKAGE OF MATERIALS

I m p o r t a n c e o f m a t e r i a l s t r e n g t h a t b r e a k a g e

Question.)The uniaxial compressive strength of limestone is σc = 55 M Pa and its elastic modulus is E = 37 G Pa then

find the stored energy of limestone at failure.

Ans:Stored energy of limestone,

.. The stored energy of limestone at failure is 40.9 kJ.

D y n a m i c s t r e s s e s g e n e r a t e d b y i m p a c t l o a d s i n b r i t t l e n e s s t e s t

Question.)When h = 0, for a suddenly applied load gives a value σmax= 2σ. The stress produced by a suddenly

applied load is twice the static stress. In the case of the brittleness test, the weight (14 kg) is dropped

from a height of 25 cm then find the dynamic load factor.

Answer : G i v e n ,

h = 2 5 c m , A = 1 0 c m 2 ( c h o s e n a r b i t r a r i l y ) , E = 7 0 0 0 0 M P a , W = 1 4 k g , L = 5 c m

D y n a m i c l o a d f a c t o r ,

The dynamic load factor of applied load is 2260.

Question.)A stainless-steel woven wire screen with a square aperture had an aperture 3.18 mm square.

The diameter of the wire was 1.2 mm. Determine

a) the percent open area when the screen was operated in an horizontal position, b) the percent open area when the screen was operated at a slope of 20°,

Answer:As the entire screen is fabricated with wire of diameter of 1.2 mm,

a) for a horizontal screen:

percent open area

= 52.7 %

b) For a screen inclined at 20°:

Effective percent open area A= = 49.5%

INDUSTRIAL SCREENING

Screening

FREE AND HINDERED SETTLING

Free Settling Velocity

Question.)Derive Free Settling Velocity.

Answer:

Forces acting on particles,

Gravitational Force Downward,

Buoyancy force due to displacement of fluid in a pressure gradient,

Drag force due to relative movement of particles and fluid ,

Equation of particle notion,

where,

Fluid density

Particle density

Drag Coefficient

Projected surface area

At terminal settling condition acceleration is zero,

For spherical particle,

Terminal velocity,

THICKNESS, HYDROCYCLONES, FILTRATION, AGITATION AND MIXING

Hydro cyclones

Question.) A hydro cyclone produces two products i.e. underflow and the overflow from the feed. The overflow is treated further for concentration of valuable mineral; whereas underflow is re-circulated. In a hydro cyclone a slurry of density 1140 kg/m3 is fed for separations. It produces underflow of slurry density 1290 kg/m3 and overflow of slurry density 1030 kg/m3. Determine mass flow rate of feed to the cyclone, when 3-litre sample of underflow takes 4s. Density of dry solid is 3000 kg/m3.

Answer: % solids in the feed =

= 18.42 %

%solids in Underflow =

= 33.72 %

% solids in Overflow =

= 4.37%

=Density of solid=Density of slurry

Dilution ratio of feed = 4.42

Dilution ratio of Underflow = 1.97

Dilution ratio of Overflow = 21.88

Mass flow rate of Underflow =

=

= 1172 kg/hr

Question.)In a hydro cyclone a slurry containing 30% solid is fed for separation of coarse and

fines. The underflow has 15% solids. If the feed enters at 20 tons/hr in the hydro cyclone,

calculate the tonnage of solid/hr in underflow.

Answer : Given ,

Feed – 20tons/hr, 30% solid

Overflow – 15% solid

Underflow – 50% solid, Tonnage of solid?

20 = MU + MO

Water balance,

(where are dilution ratios of feed , UF&OF)

On solving we get,

Therefore, Tonnage of dry solids in underflow is 14.3 T/hr.

Question.) If the pulsation frequency is 60 strokes/minute. What is the stroke length?

Ans: T = 1 , s = 1 a n d f r o m e q u a t i o n

I f t h e p u l s a t i o n r a t e i s t o b e h a l v e d , t h e n T = 2 s ( 3 0

p u l s a t i o n s / m i n ) a n d t h e s t r o k e l e n g t h s h o u l d b e i n c r e a s e d

a c c o r d i n g t o

JIGGING

I f t h i s p r i n c i p l e w e r e n o t f o l l o w e d t h e r e s u l t w o u l d b e e x c e s s i v e s t r e s s o n t h e w a l l s o f t h e j i g a n dt u r b u l e n c e w i t h i n t h e b e d t h a t w o u l d d i s r u p t t h e s e p a r a t i o n a n d

l o w e r p o w e r e f f i c i e n c y .

Question.) For a jig treating 5 mm coal and shale of density 1800 and 2500 kg/m3 respectively,

calculate the jig stroke amplitude required to expand the bed. The fluid is water at a density of

1000 kg/m3 and viscosity 0.001 Pa s and the Newtonian drag force is 0.44.

Answer: The pulsation must produce a water velocity capable of raising the largest

pieces of 5 mm scale. The terminal velocity of the largest particle is:

VT = [(4 x 9.81 x (2500-1000) x 0.005)/ 3 x 0.44 x 1000] = 0.4722 m/s

Assume the pulsation rate,

N = 60 pulses/min

Amplitude,

Thus, a surge of about 150 mm would be required in the jig.

SURFACE BEHAVIOUR AND FLOTATION PRINCIPLES

Froth flotation

Question.)Consider the treatment of a very simple ore that contains 5 t/h of pyrite and 95 t/h of mineral quartz,and that the ore is fed to a bank of cells so that the retention time in each cell is two minutes. If we samplethe feed, concentrate and tailings from any cell we may calculate the recovery of the two minerals in the cell from the feed to that cell. Answer:

RPyrite= 37.5% and Rquartz =3.8%

Substituting in Eq,

The rate constant is calculated as,

On solving,

Kpyrite = 0.30 min-1

and similarly, Kquartz = 0.02 min-1

The products of a bank of any number of cells can be calculated.

For example, from six cells in series the recoveries from the feed will be,

RPyrite = l - (l + 0.30x2)-6 = 0.9404 or 94.0% and

Rquartz = l - (l + 0.02x2)-6 = 0.2097 or 21.0%

T h e t o n n a g e s o f t h e t w o m i n e r a l s i n . t h e c o n c e n t r a t e w i l l b e :

Q P y r i t e = Q f e e d ( p y r i t e ) x R P y r i t e = 5 . 0 x 9 4 . 0 % = 4 . 7 0 t / h

Q q u a r t z = Q f e e d ( q u a r t z ) x R q u a r t z = 9 5 . 0 x 2 1 . 0 % = 1 9 . 9 2 t / h

T o t a l c o n c e n t r a t e m a s s = 4 . 7 0 + 1 9 . 9 2 = 2 4 . 6 2 t / h

G r a d e o f t h e c o n c e n t r a t e w i l l b e ,

pyrite by mass

For the circuit in figure calculate:a)the mass flow of A and B,b) the grade and recovery of A to cleaner concentrate for the following conditions:Mass flow (F)=100 t/h; mineral feed grades fA= 5% (0.05) and fB = 95% (0.95).Stage recoveries: RcA=0.75, RcB=0.10, RcsA=0.90, RcsB=0.15.Ans:

❑ Mass flows,

From the mass balance,

F = X [1-Rcs(1-Rc)] [i.e. scaling factor (to determine mass flow ofstream)]

for the individual minerals F = F × f where f = fA or fB.,

Question

Stream Formula Mass flow (th-1)

A B A B

Cleaner Concentrate F fRc/[1-Rcs(1-Rc)] 4.84 10.98

Cleaner Tails F (1-Rc)/ [1-Rcs(1-Rc)] 1.61 98.84

CleanerCirculating

Load

FfRcs(1-Rc)/[1-Rcs(1-

Rc)]

1.45 14.83

Cleaner Feed F f/[1-Rcs(1-Rc)] 6.45 109.83

The table below summarizes the calculations,

where,

Cleaner feed = fresh feed + circulating load

For A, 6.45 = 5 + 1.45 and for B, 109.83 = 95+ 14.83

❑ Grades of and recoveries to cleaner concentrate.

From the table these are readily calculated,

Grade A ==0.9677 or 96.77%

Recovery A = = 0.3058 or 30.58%

Question) Estimate the size of cell for the following conditions related to the data for the Pb-ore feed rate

24,000 t d-1 at 35% solids with ore.

Answer:F = 1,000 t h-1 (assuming 100% availability),

otherwise Multiply by 100% availability

T = 12.5 min (5 min from, giving ~ 95% recovery) × 2.5 (scale-up factor)

E=100/(100 - %GH) =1/85 (default, i.e., 15% GH assumed)

P= pulp volume per tons solid (m^3t^-1)

=1/ (solid density) + 100/(pulp % solids) -1 ( with solid density in t m^-3 which is numerically

equally to specific gravity)

N Veff = 537 m3.

Taking seven cells as the bank length, then Veff= 77 m3 and Vtotal= 80 m3 (rounding).

This cell size may be available (check with cell suppliers) otherwise another cell size is selected and N

recalculated. The objective is to use the least number of cells to meet the minimum bank length

requirements.

QUESTION) The streams in the Zn flotation plant are also assayed for their content in Cu and Fe. The laboratory

reports the measured values in. Calculate the recoveries of zinc and copper and provide the expression of

their variances as functions of the variances of Zn and Cu assays on all 3 streams.

Answer: The recoveries of Zn and Cu are calculated using,

For Zn,

For Cu,

Variances are provided by,

For Zn,

VR = 0.059547 × [172.257Vf + 0.004365Vc + 9652.06Vt]

= 10.25733Vf + 0.00026Vc + 574.7472Vt

For Cu,

VR = 45613.21[10.5625Vf + 0.000918Vc + 12.7551Vt]

= 481789.5Vf + 41.88541Vc + 581801.1Vt

Question )The feed to a Zn flotation plant assays 3.93% Zn. The concentrate

produced assays 52.07% Zn and the tailings 0.49% Zn. Calculate the solid split, ratio of concentration,

enrichment ratio, and Zn recovery.Ans:

The solid split is,

The ratio of concentration is the inverse of the solid split,

The enrichment ratio is,

The Zn recovery is,

FLOTATION MACHINES

Flotation

Question:) A flotation plant treats feed whose grade is 0.8% Cu. The plant produces concentrate and tailings. The

copper grade of concentrate is 26% and that of tailings is 0.16%.Calculate,

•Cu recovery in concentrate

•Fraction of feed in concentrate

•Enrichment ratio

Answer : We know that mass balance,

❑ Cu recovery in concentrate,

Plant recovery

❑ Fraction of feed in concentrate,

❑ Enrichment Ratio

Tailing

QUESTION.) In the circuit shown below, the dry solids of density 3000 kg/m3 are fed at the rate of 25 tons/hr. The

feed to the cyclone contains 36% solids by weight. It is found that 250 µm size in the rod mill discharge,

ball mill discharge and cyclone feed is 27%, 5 % and 14% respectively. Determine the volumetric flow rate

of feed (solid + water) to the cyclone.

Answer: Cyclone feed (CF) = Ball mill feed (B) + Rod mill feed

CF = B + 25

Perform balance on 250 µm size,

CF×14 = 5B + 25×27

B = 36.1 T/hr

Cyclone feed of dry solids = 61.1 tons/hr

Volumetric flow rate of dry solids = m3/hr

Volumetric flow rate of water = m3/hr

Total volumetric flow rate = 20.36 + 108.6 = 128.96 m3/hr