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NAMA : DICKY AFFANDI

NIM : 22115026

ADVANCED MATHEMATHIC TA-5101

1. Sketch the 3D Geometry of coal mining

2. Model the spatial distribution of topography or the ground surface

z = 0.000954118651301749x 0.00430359937402192y + 100.1666168729549

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3. Model the spatial distribution of coal thickness

determine dummy thickness by vertical drilling by : (top of coal bottom of coal)

determine Dip of coal (from plane surface top of coal with horizontal plane)

Plane surface top of coal:

= 0.173084186939821

0.0033884976525822

+ 9.9617690994452

The angle between planes is the same as the angle between their 2 normal vector

Plane 1 : z = 0

Plane 2 : z = 0.173084186939821

0.0033884976525822

+ 99.9617690994452

=[0, 0,1]

= 0.173084186939821,0.0033884976525822,1 ]

cos

= 0.985343810808396

= 9.82154967840292

Determine true thickness

True thickness = thickness of seam (dummy thickness) x cos dip

DH-ID Thickness of

seam

True Thickness

of seam

(m) (m)

1 2.2 2.1678

2 2.6 2.5619

3 2.4 2.36484 2.1 2.0692

5 2.3 2.2663

6 2.2 2.1678

7 2.3 2.2663

8 2.3 2.2663

9 1.9 1.8722

10 2.3 2.2663

11 2.4 2.3648

12 2.3 2.2663

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13 2.4 2.3648

14 2.5 2.4634

15 2.3 2.2663

16 2.1 2.0692

17 2 1.9707

18 2 1.9707

19 2 1.9707

Model spatial distribution of coal thickness :

z = 0.00041308320 8777875

0.000274863120933628

+ 2.19318201099221

4. Model the low wall

Low wall assumed as plane surface of bottom of coal

= 0.17350341442595 0.00310954616588421 + 97.7359652866696

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5. Model the high wall

Built high wall model base on 3 point

2 point in surface and 1 point in plane of high wall

x y z

p1 400 0 103.9826

p2 400 500 101.8308

p3 0 0 -296.017

1x+-0.0043035993740219y+-1z+-

f(x,y,z)= 296.017356665244

=

0.0043035993740219y

296.017356665244

6. Model the mine floor (the intersection between low wall and high wall)

Is an intersection of low wall (bottom of coal) vs high wall

z = -0.17350341442595x 0.324658130601793y+ 97.7359652866696

vs

z = x 0.0043035993740219y 296.017356665244

Mine floor/Intersection line is :

y = 982. 789884427523x + 329761.95639224

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7. Mineable coal resource

Because dip is around 9o

its mean top of coal surface is nearly flat, so we can :

determine volume with double integral usethickness function

z= 0.000413083208777875x 0.000274863120933628y+ 2.19318201099221

X boundary is 0 to mine floor (line) to

Mine floor (line) in f(y) format =

x = 0.00101751149034516y + 335.536579707805

y boundary is 0 to 500

Volume = 368329.71 m3

Tonnage = Volume .p = (368329.72) x(1.35)

Tonnage = 497241.072 ton

8. Volume of overburden rock

To compute volume of overburden rock we can separated by 2 boundary:

By thickness (between surface and top of coal)

By thickness (between surface and high wall)

Volume OB1

Thickness OB1 = Zsurface Ztop of coal

Thickness OB1

500

Sand layer

Highwall

400east

OB1 OB 2

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= (0.00954118651301749x 0.00430359937402192y +

100.166168729549)

Thickness OB1

Z = 0.182625373452838x 0.000915101721439715 + 0.204399630103794

X boundary is 0 to mine floor(line)

Mine floor (line) in f(y) format =

x = 0.00101751149034516y + 335.536579707805

y boundary is 0 to 500

Volume OB1 = 5341910 m3

Volume 0B2

Thickness OB2 = Zsurface Zhighwall

Thickness OB2

= (0.00954118651301749x 0.00430359937402192y + 100.166168729549)

(x-0.0043035993740219y 296.017356665244)

Thickness OB2

Z = 0.990458813486983x 1.56125112837913E17

+ 396.183525394793

X boundary is mine floor(line) to crest of high wall (x=400)

Mine floor (line) in f(y) format =

x = 0.00101751149034516y + 335.536579707805

y boundary is 0 to 500

Volume OB2= 1020870 m3

Total volume of overburden rock = Volume OB1 + Volume OB2= m3

= Volume OB1+ Volume OB2

= 5341910 + 1020870

= 6362780 m3

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9. Stripping ratio

Stripping ratio in coal mining is ratio of volume overburden to be stripping to get 1

tonnage of coal.

Stripping ratio =

=

.

Stripping ratio = 12.79 : 1

10.Estimate ground seepage into the open pit after the coal is fully excavated

Mathematical model of ground water table :

z = 0.000557084933845486x 0.000772691705790307y + 95.390372030160

After coal is fully excavated, we can determine the intersection line between GW

table surface with coal bottom surface.

z = 0.000557084933845486x 0.000772691705790307y + 95.390372030160

vs

z = 0.17350341442595x 0.324658130601793y + 97.7359652866696

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Intersection line is

y = 74.4849550248844x + 1003.73955527146

In f(y) format :

x = 0.0134255300236929y + 13.4757355352652

Then determine area of ground water bearing layer. Because the coal was fully excavated so

plane surface bottom of coal also as a plan surface for sand layer.

z = 0.17350341442595x 0.00310954616588421y + 97.7359652866696

Area= 165269.3634 m2

Elevation of GW Table around 95 masl. we can assumed that Ground water table surface is

flat. We can determine hydraulic gradient by difference elevation between intersection

point divide by length of bearing layer.

P1

P2

Ground

water

table High wall

Bearing layerarea

Seepagedirection

(throughsand layer)

Sand Layer

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All point located at y = 0

Point 1

x = 0.0134255300236929y + 13.4757355352652

x = 13.4757355352652

z= -0.17350341442595.(13.4757355352652)-0.00310954616588421(0)+

97.7359652866696

z = 95.39788

Point 1 = (13.4757355352652,0,95.39788)

Point 2

x = 0.00101751149034516y + 335.536579707805

x = 335.536579707805

z = -0.17350341442595.(335.536579707805)-0.00310954616588421(0) +

97.7359652866696

z = 39.51922

Point2 = (335.536579707805,0,39.51922)

h = 95.39788 39.51922 = 55.8787 m

Length of bearing layer is l = 500 m

Q = K.A.

Q = 104

. (165269.3634 ). ( .

)

Q = 1.84700598481552 m3

/ det

Seepage velocity after the coal is fully excavated = 1.84700598481552 m3 / det