Design a pumping system to deal with the mine water for copper mine with 1.5 million tonnes/annum capacity for next 10-15 years.

Mine drainage design

Where required it may be assumed that g = 9.81 m/s2, Density of water = 1000 kg/m3,

The mine is a medium sized operation producing 1.5 mtpa of coper ore at 3% Cu. The mine was originally designed using a bench height of 15m. Footwall benches were 15 m wide to give a slope angle of 45°, but on the hanging wall side alternate benches are only 10m wide to produce an angle of about 50°.

Design a drainage and pumping system to deal with the mine water for the next 10-15 years based on the data in the following sections. As the mine grade is low and power costs are high every effort should be made to minimise the operating costs of the pumping system. Describe your proposed drainage and pumping system with the aid of diagrams where required and explain the general philosophy behind the system. Detail and assumptions made and make comments and conclusions as required.

Data

Mine water

Ground water: currently averages 1,000,000 litres/day and is expected to increase at a rate of 5% per year as the mine deepens.

Mine water: averages 50,000 l/day with small variations in use on an hour by hour basis.

Water Quality: Run of mine water is expected to be slightly acidic due to the formation of sulphuric acid from sulphides within the rock

Rock properties: SG of 2.9 for all rocks. Hanging wall and ore have moderate strength, footwall is very strong.

Mine Section

Rainfall: the average annual rainfall s 740 mm, the average monthly figures are as follows
January	10mm
February	10mm
March	40mm
April	60mm
May	80mm
June	100mm
July	110mm
August	120mm
September	90mm
October	60mm
November	40mm
December	20mm
Based on measurements in the locality the maximum rainfall intensity recorded over the last 50 years is as indicated below and needs to be accounted for: Storm duration Hours	Intensity (mm/hr)
2	40
4	25
8	15
12	12
16	10
20	9
24	8
36	6
48	5

Solution:

Introduction:

• Here for given mine we have to design drainage and pumping system such that total amount of water inflows from all sources should be handled.
• Water inflow to given mine is mainly from four sources

i)ground water

ii)mine water

iii)rainfall inflows

iv)stormwater inflow

• Here while designing pumping system for mine proposed hydraulic circuit is given as follows

Various component of system are

1)reservoir tank=where all water inflows are supposed to be stored

2)strainer of metal grid circular type with 75% clear area:to filter water to remove higher size particle in flow

3)hydraulic pump driven by motor=to lift water to required head

4)check valve or non return valve =to prevent backward flow of water in emergence shut down of pump to protect strainer and pump.

5)delivery line of length=Ld=165 m=to carry water from pump to outlet

6)suction line=Ls=15 m=to carry water from reservoir to pump

7)regular sized elbow with 90 degree bend,number of elbows are 13.

4)first of all we have to estimate total water inflows in mine from all sources

i)ground water inflow: ground water inflow is fixed for given year of Qgo= 1000000 l/day and expected rise at 5% per annum for next 15 years

so by uniform rise approach we have total rise in ground water in one year is 5% of present ground water,it gives us

Qg2=.05*1000000=50000 l/day

Total rise at the end of 15 years is

Qg1=15*Qg2=750000 l/day

Hence total ground water at the end of 15 year span is

Qg=Qgo+Qg1=1750000 l/day

ii)mine water inflow: mine water inflow will remains constant and it is equal to

Qm=50000 l/day

iii) Rainfall inflows: it is water entering mine from surface due to rainfall in that area

Here catchment area of mine as shown in figure is of diameter D=140 m

A= (pi/4)*D^2=15393.804 m^2

Here average rainfall per annum=h=740 mm

Rainfall inflows=Qra= A*h*10^-3=11391.4 m63/annum

Each annum has 365 days so average rainfall inflow per day is

Qrd=Qra/365=31.209356 m3/day or 31209.356 l/day

But when we look at monthly average rainfall then it observed than peak rainfall observed in the month of August and it is peak rainfall inflow and while designing we must design for peak discharge load.

Rainfall inflow for August=Qrm=A*h*10^-3=15393.804*120*10^-3=1847256.48 l/month

Where august has 31 days in one month so average rainfall per day is

Qrd =1847256.48/31=59588.91 l/day

1. iv) Storm water inflows:

Storm water inflow is estimated based on storm water intensity by analytical method named as rational method equation.

as per mine safety and health organization (MSHA), for low hazard potential application pumping system should handled storm water collected from storm intensity which last for 6 hrs.

From intensity chart given we get polynomial equation for storm intensity as function of time as follows

Time(t)	Intensity mm/hr(I)
2	40
4	25
8	15
12	12
16	10
20	9
24	8
36	6
48	5

so 8 th degree polynomial is given by nine data points as follows

I=-4.896*10^-10*t^8+2.6875*10^-8*t^7+3.068*10^-6*t^6-3.4620*10^-4*t^5+1.3940*10^-2*t^4-2.8972*10^-1*t^3+3.333*t^2-20.8946*t+70.5626

As we want storm intensity for 6 hrs duration then by above equation for t=6 hrs we get I as

I=18.1275 mm/hr

Then by rational method, stormwater discharge is given by

Qs1=c*I*A

C=run off coefficient=for mine we assumed it equal to cultivated land=.4

As most of mining study hold c=.14(dry season) and c=.55(wet season)

Putting value we get

Qs1=.4*18.1275*15393.804=111620.47 l/hr

As we have to calculate for 6 hrs so we get

Qs=Qs1*6=669722.83 l/day

v)total water inflow is sum of this four source inflows as follows

Q=Qg+Qm+Qrd+Qs

Q=2529311.73 l/day

For day there are 1 day=86400 sec

Q=2529311.73*86400=29.2744 l/sec

Q=.0292744 m3/s

5)As per API code for fluid flow having small amount of solid particle flow velocity should be within range of 3 ft/s to 15 ft/s or .9144 m/s to 4.572 m/s

Here we choose maximum velocity of fluid flow in pipe=15 ft/s or 4.572 m/s

By continuity equation we get that

Q=A*V

A=(pi/4)*d^2=(.0292744/4.572)

d=90.2893 mm or 3.5547 in

6)while selecting pipe material and its dimensions we must consider nature of fluid flow

Nature of flow=acidic

For acidic flow,to minimize cost of installation and maintainance we must select plastic pipe like high density polyethylene as pipe material as friction is negligible for fluid flow compare to other material like cast iron or copper tubing

Pipe material=HDPE(high density polyethylene)

From ISO standard for HDPE pipe and fitting, that is ISO-4427 we can select pipe nominal outside diameter and thickness of pipe.

From experience we can select HDPE pipe with following specification

PE100/PN16/DN110

MRS=10 MPa (PE100 interpret as MRS=10 MPa )

Design hoop stress=S=MRS/1.25=8 MPa

Operating pressure=16 bar(we check for operation pressure later on which is present in system)

Outside diameter=110 mm

As lower available class of nominal diameter is 90 mm and it is less than our requirement as we required inner diameter minimum=90 mm

Outside diameter=110 mm

Thickness from 10 mm to 11.1 mm

Inside diameter=d=ID=OD-thickness=110-10=100 mm

For this dimension flow velocity is given by

V=Q/A=3.7272 m/s

And this velocity lies in API limits(.9144<V<4.572)

7)manometric head for operation of pump is given by

Manometric head=suction head+delivery head+frction loss in suction line+friction loss in delivery line+minor head loss in strainer+head loss in check valve

Here for operation of pump from diagram, we must mount pump on first bench from bottom side

And pipeline should in contact with ground for all of its length,which gives us

i)Suction head=Hs=5 m

ii)Delivery head=Hd=90 m

iii)Friction loss in suction pipe by Hazen William method we get head loss as

Hf=.002083*Length*(100/C)61.85*Q^1.85/d^4.8655

All units are imperial and on solving it gives head loss in feet and on converting to m of H2O we get following answer for head loss in suction and delivery line

d=diameter in inch=3.937 in or 100 mm

C=friction coefficient for Hazen Williams method for HDPE=150

Q=386.36 gpm

Ls=15 m or 49.21 ft and Ld=165 m or541.33 ft

Hfs=head loss in suction pipe by above formula=1.62 m

Hfd=head loss in delivery pipe=17.83 m

iv)Head loss in single elbow by K value method and multiplied by number of elbows we get that

Hfe=head loss in elbows=number of elbows*K*V^2/2g

For regular sized elbow for 90 degree bend,K=.3 and V=3.7272 m/s

Hfe=13*.3*3.7272^2/(2*9.81)=2.7614 m

v)Head loss in strainer of metal grid circular type with 75% clear area is given by K value method as

Hfss=K*V^2/2*g

K=.8

Hfss=.61 m

vi)Head loss in check valve of swing type which has inside diameter equal to pipe inside diameter

By K value method as

K=1.89

Hfc=1.33 m

vii)Total Manometric head=suction head+delivery head+frction loss in suction line+friction loss in delivery line+head loss in elbow+minor head loss in strainer+head loss in check valve

H=Hs+Hd+Hfs+Hfd+Hfe+Hfss+Hfc

Putting value we get that

H=129.15 m ofH2o

8)pressure inside pipeline is given by

P=density*g*H=9810*129.1514=12.6697 bar

As nominal pressure PN=16 bar> operating pressure,hence system will operate properly.

Where from operating pressure pipeline thickness is given by

Minimum Thickness=(P*OD/(2S+P))+.1=8.17mm

Maximum thickness=1.1*minimum thickness+.2=9.18 mm

As our standard thickness 10 mm>maximum thickness for operation

And hence our design is safe

9)power required for operation of pump for with 100% overall efficiency is given as folllows(assumption,generally ranges from 65 to 80%)

Power=(density*g*Q*H)/overall efficiency=(1000*9.81*.0292744*129.15)/1=37089.93 watt

Power=37089.93/746=49.71 hp

10)conclusion:in this way as per requirement we have design piping,drainage and pumping system to deal with all mine water inflows and select HDPE as piping material as it has good resistance to acidic attack.