Excavation Costs - Otuonye - 2000

8/13/2019 Excavation Costs - Otuonye - 2000

1/63

[_ X _INDEX ll To chapter 2 ll prof. oruonve Home paqCHAPIDR 1 Prt 3

COST ESTIMATION FOR AN EXCAVATORA. Ownership cost1. Depreciation: same as a truck, except that no tires are involved.Purchase price is estimated at $l10,000/yd3.Include: erection costs, ballast costs, trailing cable costs, extras.Freight: estimate shovel weight as 54,000/yd3 (dead load)2. fnterest: same as a truckB. Operating costsL. No tire replacement2. No tire repair cost3. Repairs and maintenance and supplies (such as cables and parts): same as truck.4. Power:.Assume 55 hp/yd3 to 66hp/yd3.I hp: 746watts: .746kWPower cost : (Power use)(load factor)($/kW-hr): (kWX$/kW-hr): $/hr5. Fuel consumption : (gaUhp-hr)(load factor)(hp)6. Labor cost: same as truck, and include oiler.Follow same procedure for estimating truck haulage.

Page I of63

http ://www.mg.mtu. edu/-frotuony/m 933 6l3mg33 6.htm 5/20/00


2/63

Page2 of 63

EXAMPLEFor a l5-yd3 (t t m3) power shovel. Estimate the total hour cost and unit cost if the desired output is10,000 tons/shift. Assume operating conditions average. Power cost: 7,lkW-hr,lubrication oil:$4.00/gal ($1.06/L), operator wage : $16.00/hr, and oiler wage : $12.001hr. Assume an 8-hr shift, 50minute per hour; and material weighs 2.0 tons/yd3 b.nk measure.SolutionRefer to following table. Follow the procedure outlined for truck, Calculations requiring explanation

http ://www.mg. mtu. edtr/-frotuony/mg33 6 13mg336. htm 5120100


3/63

are summarized below.A. Ownerqhip costsI a. Purchase price : l5 yd3 x $l10,000 /yd3 : $1,650,000.b. Salvage value:0.15 x $1,650,000 : $247,500.c. Weight : 15 yd3 x 54,000 lb/yd3: 810,000 lb.Freight : 810,000 lb x ($4.00/100lb) : $32,400.d. Unloading and moving cost:0.10 x $32,400 : $3240.e. Delivered price : a - b + c + d : $1,438,140, say $1,438,000.f. For operating period of I shift/day, use 2000 hr/yr.g. For average conditions, use economic life of 18,000 h + 2000 : 9 yrs.2. Nl calculations are self explanatory.B. Operating Costs3. Select repair factor :70Yo (average conditions).Repair cost:0.70 x $79.89: $55.92lhr.4. Power rating : 15 yd3 x 55 hp/yd3 :825 hp x0.746: 615 kW.Select load factor :60Yo (avg conditions). Power use: 15 kW x 0.60 : 369 kw.Power cost:369 kw x $0.07lkW-hr: $25.85/tr.5. If diesel, fuel oil consumption : 0.059 gaVhp-hr x 0.0 x 825 hp:29 gal.Lubrication oil cost : 29/lO0 gaVhr x $4.00/gal : $1.16/hr.C. Overall ownership and operating costs : A * B : $272.5811r.D. Unit Cost: hour production: 10,000 tons/shift + I h :7250 tonsr;unit cost : (527258[hr)K1250 ton/hr) : $0.22/ton

Page 3 of63

http ://www. mg. mtu. edu/-frotuony/m g33 6 l3mg33 6. htm sl20l00


4/63

Page 4 of 63

Cost Estimation FormffSsyel Mining Equipment Unit

A. Ownership Costs1. DEPRECIATIONa. Purchase price: $1.650,000b. Salvage value (15%) : - 247.500c. Freight 810.000 lb @$4.oD/cwt:32.400d. Unloading and moving cost: 3.240e. Delivered price: 1.438,140 say $1.438.000f Operating period 2000hrlyrg. Economic live 18.000 hr:-1yr(n)h. Depreciation: $1.438.000 (del. Prices less tire cost) :$ 79.89 lttr18.000 hr2. INTEREST, TAXES, INSURANCE, AND STORAGEa. Rate = interest 14% + taxes 2Yo * other 2o/o = lSYo

http://www.mg.mtu.edu/-frotuony/mg336/3mg336.htm 5120/00


5/63

b. Average annual investment rate: (n+l)/2n: l0l18 :$ 55.6 %c. Average annual investment = $1.438.000 x 55.6Yo: $779,529d. Annual fixed charge: $799,528 xlSYo: $143.915e. Fixed charge = $143.915 :$ Z-96 /hr2000tr/yrTOTAL OWNERSHIP COSTS:$ 151.85 /trB. Operating CostsI, TIRE REPLACEMENT COSTPurchase price I set oftires: $Tire life hrTire cost: $ :$ - /hrhr2. TIRE REPAIR COST: %X$ tire cost:$ - /hr3. REPAIRS, MAIN'TENAI.{CE:70% X $79.89 deprec. :$ 55.92\tv4. FUEL: gaVhr @$_/ealOR POWER: 369 kW @ $ 0.07lkW-hr:$ 25.85 /hr5. LUBRICATION: 29/100 gallhr @ $ 4.00 /gal:$J=l /hr6. ALDilLIARY FUEL: :$ - /hr7. LABoR:_operator @ $ 16.00 /hr: $ 16.00 /hroiler @ $ 12.00 /lr: $ 12.00 lhrhelper @g_lhr: $ /hrTotal $ 28.00 lhr+ 35Yo benefits $ 9.80 /lu:$ 37.80 /hrTOTAL OPERATING COSTS :$ 120.73 ITV

Page 5 of63

http ://www.mg.mtu. edu/-frotuony/mg336/3mg3 3 6.htm 5120/00


6/63

Page 6 of63

TOTAL OWNERSHIP AND OPERATING COSTS :$ 272,58 lhTLINIT COST S 272.58 ltu + 1250 ton/hr: $ 0.22 ltonMATCHING LOADING AND HAULING UNITSIt is crucial that a loading unit is matched to haulage units to avoid either over trucking (where morethan the adequate number of trucks is assigned to a loader) or under trucking where inadequatenumber of trucks is assigned to a loader.Truck production & fleet requirements are affected by:Mine planHaulage roadsMine production requirementsLoading equipmentTruck performance and cycle timeOperating method and practicesMatching of load equipment and trucksEquipment availabilThe total cycle time t is given ast. : t, * tl + tf,f + t * tr, + the + tw + tdewhere: t, is the spotting time t_ is the maneuvering timet" is the loading time tn. is the time to haul emptyt* is the time to haul loaded t* is the waiting timetu is the dumping time tu" is the delay timeThe theoretical number of trucks that can setre a loader is given as:

N=5= to--ta.h tr+thttp ://www, mg. mtu. edu/-frotuony/mg33 6 13mg336. htm sl20l00


7/63

PageT of 63

where t,.: total theoretical truck cycle time excluding wait time.t*: truck spot, and load time.N: theoretical number oftrucksif t. n(t, + ts), truck waits for shovelWait time is n(t, + rJ - ttc,where n: # of trucks in fleet.For loading methods where an arriving truck is spotted while another truck is loaded, truck load andspot times are replaced by loader or shovel loading time.If n is less than \ loader is under-truckedIf n is greater than \ loader is over-trucked

SYNCHRONTZATION OF LOADING AND HAULAGE UNITSFour to six swings of the shovel to fill a truck are considered optimum for normal truck sizes (22-130tons). Five to eight swings for giant truck sizes (150-350 tons).One final check remains to be made before the choice of loading and haulage units is completed. Thesynchronization of the loading portion of the cycle must be investigated to be certain that:a. shovel does not wait for the trucksb. truck wait is not excessive (more than a few minutes)If the synchronization is unsatisfactory, then the selection of equipment must be modified or the cycle(travel, spot, or dump time) changed.Load time can be calculated for the equipment selected. Spot and dump times are usually estimatedbased in operating conditionsOnce these elements have been determined and knowing the cycle time, we can check the

http : I / www. mg. mtu. edu/-frotuony/mg 3 3 6 I 3 mg3 36. htm 5/20100


8/63

Page I of63synchronization by the following eq:t n(tl + tJwhere n: # oftrucks in fleett": load timetr: spot timet: cycle time for one truckFor proper cycle balance or synchroniztionto exist the cycle time for one. truok must be less than thetime required to spot and lo.ad truck fleet: n(t" + tr).

Truck waiting time : n(t" + ts) - t.

http ://www.mg.mtu, edu/-frotuony/mg3 36/3mg33 6.htm


9/63

EXAMPTEGiven the following data on a shovel-truck open pit mine:Required production : 10,000 tonVshiftOperating period : I shifl/day (7 hrlshift)Material: Well blasted rock (2.0 tons/b or 1.3 tonsll,cyOperating conditions: favorable (80%)Job efficiency:83.3YoLoad time/truck:3.4 minSpot time:0.5 minCycle time/truck: 15 minDetermine size and number of shovels required.Desired output: l0'0,Ih.fl . ,ffi . oso = tTtstons/hr= l?r5 tonsitq = gs? yd3/h bank masup2.tl torsiy'd 3Ideal output :857/.8: l07l yd3/nrSelect a 15 yd3 dipper on a shovelAssume: Bucket factor: 90YoTrucks available in75,80, 85 ton sizes only

Page 9 of63

Bucket capac: (ts yd3)(0.9x1.3 tons/yd3) : 17.55 tons/swing4 swings 17.6 x 4 : 70.4 tons5 swings 77.6 x 5 : 88 tons (vs. 85 tons) + 5Yo of calculated6 swings 17.6 x 6 = 105.6 tons (vs, 100 tons) value or overload is okay.

http : //www. mg. mtu. edu/-frotuony/mg33 6 I 3 m9336. htm sl20l00


10/63

Number of trips/shift: ffi = 2gs rrs/sttfrOutput per shift: (88 tons)(23.3 trips):2050 tonsNumberoftrucks: ffi - saysrnd{sSelect 85 ton trucks.Load + spot:3.9 mint n (t" + ts), 15 5(3.9) 19.5Wait time: 19.5 - 15 :4.5 minCycle is satisfactory (trucks wait for shovel)

ALTERNATTVE APPROACHcycle time for shovel :load time/truck# of loads/truck

:0.68 min.

Production ofshovel per hour:

Production with 5 trucks = 5 x 293.3 tons/hr:1466.7 tons/hr

Page l0 of63

3.4 min/tuck5 loads/nuck

88 tons* cJule *50 mincycle 15 min hmr

http ://www, mg. mtu. edu/-frotuony/m 933 6 l3mg33 6. htm sl20l00


11/63

Because the production of 5 trucks is larger than the production ofone shovel, then trucks wait forshovel. This result is consistent with that using the equation,

http ://www. mg. mtu. edu/-frotuony/mg3 3 6|3mg33 6.htm


12/63

Page 12 of 63

and Bucket Wheel ExcavatorsMachine Advantages Disadvantages

Shovel 1. Lower capital cost per yd3 l. More coal damage can result1m3 of bucket capacity, in lower coal recoveryAlthough when boom length 2. Susceptible to spoil slides andor machine weight is consi- pit floodingdered, the capital costs are 3. Cannot easily handle spoilroughly equivalent having poor stability2. Digs poor blasts and tougher 4. Cannot dig deep box cutsmaterials better easily3. Can handle partings well 5. Reduced cover depth capabil-4.Large output/hr. ity compared with a draglineof comparable cost6. Difficult to move

Dragline 1. Flexible operation; easy to move l. Requires bench preparation2.Large digging depth capability 2. Does not dig poor blasts well3. Can handle and stack overburden 3. Higher capital cost per yd3 (m3)having poor stability of bucket capac, although4. Completely safe from spoil capital costs are roughlyPile slides or pit flooding equivalent

http ://www. mg. mtu. edu/-frotuon y / mg33 6 I 3 mg3 3 6. htm 5120100


13/63

Page 13 of63

during normal operation 4. Reduced capacity (70 to g0%)5. High percentage of coal of capacity of shovelrecovery; less coal damage6. Will dig a deeper box cut7. Low maintenance cost8. Can handle partings well9. Is not affected by an unevenor rolling coal seam top surface10. Can move in any direction

Bucket wheel 1. continuous operation; no l. will not dig hard materialsswinging necessary 2. Some surface preparation2.Long discharge range required3. Can be operated on a 3. Lower availabilityhighwall bench or on the 4.Large maintenance crewcoal seam required4. Can easily handle spoil with 5. High capital cost comparedpoor stacking characteristic with outputand poor stabil 6. Can be susceptible to spoil5. Can extend range of shovel slides and floodingor dragline when operated in 7. Can cause coal damage withtandem resulting lower coal recovery6. Can facilitate land reclamation 8. Poor mobilityas it dumps surface material

http :/iwww. mg. mtu. edu/-frotuony/mg33 6 13mg336. htm s/20100


14/63

Page 14 of63

back on top ofthe spoil pile

Source: Anon.,l976a.

Machine Advantages I)isdvantages

Dozer l. Flexible l. Limited to short haul2. Good gradeability 2. Discontinuous3. Negotiates rough terrain 3. Low output, slowTruck, trailer 1. Flexible and maneuverable l Requires good haul roads2. Handles coarse, bloc rock2. Slowed by bad weather3. Moderate gradeabil 3. High operating costScraper 1. Flexible and maneuverable l. May require push loading(rubber-tired) 2. Good gradeability 2. Limited to soil, small fragmentsRail 1. High output, low cost l. Track maintenance costly2. Unlimited haul distance 2. Poor gradeability3. Handles coarse, bloc rock 3. High investment costBelt conveyor l. High output, continuous l. lnflexible2. Yery good gradeability 2. Limited to small or crushed rock3. Low operating cost 3. High investment cost

Sources: Modified from Pfleider, l9'13a, 1973b; Martin et a1.,1982

http ://www. mg. mtu. edu/-frotuony/mg33 6 13m9336. htm 5120100


15/63

Page 15 of63

http ://www. mg. mtu. edu/-frotuony/m 933 6 I 3m933 6. htm sl20l00


16/63

Page 16 of63

EQUIPMENT SELECTION AND PIT DESIGNStrip Mine ShovelIn designing a strip mine operation, the reach and capac of the machine are the two most importantfactors to be considered.The reach determines how high an overburden can be mined and how far and high the spoil can beemplaced to establish the technological strip ratio.

SR= cu Frds of merbudemtons of recmerb minml

The bucket capac determines the productivity of the machine and thus determines the economicfeasibility of an operation.The product of reach and capacity represents an overhanging load which must be adequatelysupported by the main structural frame of the machine.For a given main structural frame ofthe machine, the reach and capacity are inversely proportional i.e.one can easily be increased by an appropriate reduction ofthe other.The maximum usefulness factor (MUF) is the term used to express the relationship between machinestructural weight and its ability to do work (capacity x reach)MUF: (nominal dipper, yd3) or (bucket size, yd3) x (dumping reach, ft)If the gross machine weight for various stripping shovels (draglines) is plotted against maximumusefulness factor for each shovel or dragline, the curve appears to follow a generally straight lineregardless of manufacture or size range of machine.The MUF is usefill in the preliminary evaluation of stripping projects under consideration andapproved machine size other than current designs. Using the attached figure, for a machine grossweight of 27 x 106 lb, usefulness would be36250 ft-yd3 bucket, the dumping reach would be 201 ft.Designing the PitIn order to design a pit width and find the mineral production in the last cut the following 8 stepscould be followed:1. Determine the minimum pit width (or minimum clean up radius)

http ://www. mg. mtu. edu/-frotuony/mg33 6 13m9336. htm sl20l00


17/63

Wp^in : r/z width over both crawlers * clearance radius(or rear end radius at lower edge of frame)2. Determine the maximum pit width, (or maximum clean up radius)Wn-o : t/zwidth over both crawlers * radius of clean up.3. Determine the digging reech: Vzvttdth over both crawlers * m


18/63

Page 18 of63

('trtgli:t'e pl

$G tr.r *l ; l,,,ilf *ili*ntJ*rJ as a fincfivn olrr:hne r,ihg tu-r 'sk+rnii llii'j tJrllrra'*' = 1 ' fi, lclt. sh


19/63

nage ie ofoi

Refening to the figure, the area of the spoil, As can be written asAr=ar+A.,The height of the spoil, l can be written ash=hr+h,Where h, is the height of area Ar, and h, is the height of area Ar.Ar:w/2xh,

http://www.mg.mtu.edu/-froruony/mg336/3mg336.htm S/ZO1}}


20/63

A,r= w/2 xwl2 tanWhere: the angle of repose of the spoil pile or slope of spoil.ar:wxh,A,=w(h-h1): w(h - wl2 tn)+=#*o+wh-#*u

= $rh - *ttur.u4

Spoil volume: volume of overburden + ddilisnal volume due to swell.In terms of bank cut area:Spoil area, Ar: area of overburden + ddilisnal area due to swell(or volume per unit depth into the paper)\: A" + (s/100)*4.4: (1 + s/loo) * A.where, s: o/o swellAa: *HA, : (1 + s/100) x wH or,

Page}} of 63

http ://www. mg.mtu. edu/-frotuony/mg33 6 13 mg336. htm sl20l00


21/63

*t'-$*,u=(t.=r*)*"

o- *.['. *

r' -It-ru=fr. =HcotE 4 t r00,

*=[(t.=r*J n.f*u].o.u


22/63

Page2? of63

Looking at the figure for the dragline, the dragline reach ru will be equal to the shovel reach +additional amount equal to the horizontal projection of the highwall slope.rd=r*+Hcot

n = [[t . #J H . *u].o,u + H cotg

Where is the angle the highwall makes with the horizontal.Having tentatively selected a stripping shovel or machine based on reach requirements, appropriatecapacity must be attained.Dumping radius ofthe shovel: shovel's dumping reach + 0.5 width over crawlers.Dumping radius of dragline : dumping reach of the dragline + 0.75 (tub diameter)Theoretical capac of the dragline: bucket capac x fill factor x efficiency x loose weight/yd3 x #of passes/tr.

http ://www. mg. mtu. edu/-frotuony/mg33 6 13mg336. htm 5120100


23/63

Page 23 of 63

The determination ofthe dumping reach in the previous derivation did not take into consideration thethickness of the mineral deposit. If we considered the thickness ofthe seam and referring to the figurebelow, we obtain the following derivation for the dumping reach.

Insert "surface extraction" figure here

http :1/www.mg.mtu. edu/-frotuony/m g33 6l3mg33 6.htm 5/20/00


24/63

ACB :41 + 42 + 43:whl+wh2+wT= .i*o+w(r- ,^,0) .*r: #*,o + wh - r*0 + wr

ACB: *n- *21"ro+wT4

Page24 of 63

ACS: wH

http : //www. mg. mtu. edu/-frotuony/mg 3 3 6 / 3 mg3 36. htm sna/00


25/63

acB: (r . -)*n

ACB: (r . ;) ,rcs

Page25 of63

Combining equations and

*r, - St"r.o + wr = (t . #)*t

rr-ftano+r=(t.rfu)"

But h: r tan, so

http ://www. mg, mtu. edu/-frotuony/mg33 61 3mg36.trtm


26/63

rtano-*u+r.(t.r*a)"

rtaro = *u . (t . rL*) r - t

'= a*e['-.u.n['.#) -tl

q=6|u[*'.n['.#) 'l

Page26 of 63

Where r:\For Shovel:

http ; //www, mg, mtu. edu/-frotuony/mg 3 3 6 I 3 mg3 36. htm s/20/00


27/63

Page27 of 63

Dumping radius: dumping reach + 0.5 (width of crawlers)\ = .. + 0.5 width over crawlersr. = l(t . =rrJ " . f ,",,o l.otu

{ O = ((t . #) n . f *u) .oru + u.Swidth mer crau,lerFor dragline:Dumping radius: dumping reach + 0.75 width of tubRd: rd + 0.75 width of tubrd:r.+Hcot

-+ * = ((t. r*=J H.l*u) ,o,u +Hcot +o.?Swidthorrrh

NOTE:Maximum Usefulness Factor (MUF): nominal dipper or bucket size x dumping reachSuppose a BE 480w draglineDumping radius: 154 ft

http://www,mg,mtu,edu/-frotuonylmg336l3mg336,htm 5120100


28/63

Bucket Size: 18 yd3Dumping reach: 154 - (0.75)36: 127 ft.Wish to determine the reach of a 15 yd3 dragline bucket"rgBtg = MUF: r1sB15rra:rrrBre/Brs: (127)(L8) / ls: 152.4 ft5. Determine the average highwall depth of last cut, H, or the highwall timit of mining.

Page29 of 63

http ://www. mg. mtu. edu/-frotuony/m933 6/3mg336. htm sl20l00


29/63

$'yH xtft+2ift34d3two x p+ 2000 D/ton

Page 29 of 63

r, /cot0 - E tar e1H= -3I+- 100

Thus for a given machine size, considering the material-pit parameters, it is possible to determine thehighwall limit of mining.6. Determine the number of cuts that can be taken. (from 2 and 5)

Limit is when Hi: H

7. Determine strip ratio per linear ft.

- rmlume of merbrnden ()d1tonnage of mlnnal (tru)

http ://www. mg. mtu. edu/-frotuony/mg33 6/3mg336. htm 5120100


30/63

'Where: t: thickness of mineral deposit: average density ofmineral deposit(co"l: lS5 lb/ft3)8. Find monthly production of mineral in last cut.

volunn of mu:burlur fudl

-

- utnsormrrat

http ://www.mg.mtu. edu/-frotuony/mg3 36/3mg336.htm


31/63

EXAMPLEIn a level terraiq determine the mo


32/63

Page32 of 63

127 = 0.287 H+ 1.326 (1.3 H + 9.425 - 4)2.01 H = 119.8H:59.6 ft.Without including t in the formula,H = 56.96 ft. + 4 ft. of coal= 60.96 ft.

Remember: Cycle times for these and other equipment can be estimated by conducting time studies.

CHAPTER 2

RArL (LOCOMOTTVE) HAULAGEWhenever large tonnages have to be transported over relatively long distances on a relatively flatterrain, locomotive haulage is often the most efficient system to use.

http ://www. mg.mtu. edu/-frotuony/mg3 3 6 l3mg336.htm 5/20100


33/63

Page 33 of63

Rail haulage today in mining stands at the cross roads. Radical departures must be made from pastpractices and new innovations made in order for rail haulage to remain a viable alternative in the faceof new developments for the transport of materials in mines.Advantages of Ril Haulage1. Very low production costs where production requirements and volumes justify their installation.2. Small labor force required.3. Very versatile as to the type of materials handled.4. I\gh tonnage capacity5. Spillage is minimal6. Minimum maintenance requirements7. High loading and unloading costs can be justified by economy of haul and high tonnage.8. Efficient and cheap over long distances. Increased haul distance does not increase costsproportionally.9. Traffic control is simple using two-way radios.10. Can provide a certain amount of regeneration on downhill runs resulting in energy savings.11. Ability to utilize automatic traffic control systems (CTC) and to operate the locomotives remotely.This increases efficiency and reduces manpower requirements.12. Good availabilityDisadvantages of Rail Haulagel. Locomotives are limited to fairly shallow gradients (mainline track grades of 2 to 3 percent adverseto the load and 3.5 to 4 percent favorable to the load are generally used).2. Frictional grip between the driving wheels and the rails also limits the size of train which can behauled.3. Mines considering rail haulage must have large reserves to support long life and large production.Capital intensive and on short hauls can't compete with trucks economically.4. High initial installation costs5. System is inflexible6. Special unloading devices required.

http ://www. mg. mtu. edu/-frotuony/mg33 6 13mg336. htm


34/63

7. Mostly used in ore transport due to difficu in spreading waste.

Types of locomotives used in mines today1. Diesel-electric2. Trolley-electric

Page 34 of63

http : //www. mg. mtu. edu/-frotuony/mg 3 3 6 / 3 mg3 36. htm sl20l00


35/63

Tracliw effutsupto Tl, theryeed can t'e up to Vl, fc tactiveeffortsh/wTr & Tr, ma:c qpecdisVr

I Gcar2II--------.t---i i Gcar3

Page 35 of63

Thrce-gcer Dir*l Lmcnoti'c Chncbristics

1. Diesel Electric - most prevalent because they can be adapted to any track gauge and are availablein all sizes from small switch engines to large freight haulers.Stationary trolley wire is not required since the diesel engine is the primary power source, and themoving of track and adjacent mining operational conditions are simplifiedDiesel engines have a traction generator directly connected to them which furnishes po\/er to axle-mounted traction motors. This also provides dynamic braking through the electrical system effectivelyassisting braking capability.Characteristic of the diesel engine is to provide a steady torque over a wide range of speed, the valueof the torque depending on the amount of fuel injected per engine cycle whioh is controlled by thedrivers throttle oontrol.The graph of available tractive effort against speed is shown above. The tractive effort depending on

http ://www. mg. mtu. edu/-frotuony/mg33 6 I 3 mg336. htm


36/63

Page 36 of63the engine torque and the gear in use. The shaded portions ofthe graph represent the working areas ofthe locomotive.Points d B, and C represent the mrimum power the locomotive can develop usually called thetraction rating and are the points normally specified in manufacturer's details of locomotives.a. Size (standard) 2-45 tonsb. Must meet schedule24 (u.s.B.M.)must have scrubberratio of air-fuel specifications; 20-25 parts airl I part fuel.

2. Trolley Electric - have a top mounted pantograph or a side mounted collector to receive powerprovided by an electrified wire.can provide a certain of power regeneration on downhill runs. Results in energy savings.give offno exhaust fumes, an advantage in tunnels and in providing a pollution free environment.fewer problems of cooling motors than diesels. Have speed advantage over diesels because of thetrolley wire's large power sources.major disadvantages of electric locomotives is the interference to the mining operation that trolleywire, poles, and structures that carry the wire and close tolerance requirements for these structuresgenerate.dump settlement makes the maintenance of trolley wire alignment difficult and the operationcomplicated.

http ://www. mg. mtu. edu/-frotuony/mg33 6 /3 mg336. htm s/20100


37/63

Page37 of 63

final selection of a locomotive type may depend on the availability of electric power as opposed todiesel fuel in certain parts ofthe world.Trolley locomotives (trolley-wire supplied electric)a. two types: coal mine tlpemetal mining type (looks like a square box)b. voltages: 250 V and 500 Vc. Size range: 2 - 50 tons. In ril. Va. the max is 300 tons.d. Motors: Series wound traction motor with characteristics as shown below with low speeds at highcurrent or high tractive effort

http ://www. mg. mtu. edu/-frotuony/mg33 6 /3 m9336. htm sl20l00


38/63

Page 38 of63

The position of the line depends on the supply voltage and the field connections and by effectivelyvarying these, different characteristics can be obtained as shown below.At tractive effort T, the speed may be Vl , V2 , or V, depending on the electrical arrangement.Intermediate characteristics are sometimes provided by including resistors in the electric circuit; butthese are often short term rated and so cannot be used for continuous operation.

http ://www. mg. mtu. edu/-frotuony/mg33 6 13m9336. htm 5/20/00


39/63

e. Advantages ofD.C. motors: - HiSh starting torque- No chargingf. Disadvantages: - Limited to those headings where the trolley has been strung- hazards of exposed line- rails must be well bonded for efficient operationTypes of Rail Carsa. Side dump

Page 39 of63

more versatile and usually with capacities that range from 30 yd3 to 50 yd3.employed in hard rock mines and metal mines.can dump on either side.b. Bottom dump

http ://www. mg. mtu. edu/-frotuony/mg33 6 l3mg336,htm sl20l00


40/63

Page 40 of63

more prevalent in coal minesfast unloadinghigher maintenance costs.c. Rotary dumping machineswhich turn over the rail cars for quick dumping of ore.d. Rocker typeLocomotives are however limited to fair shallow gradients, and the frictional grip between thedriving wheels and the rails also limits the size of train which can be hauled.Factors affecting the selection of a locomotiveThe selection of a locomotive to handle a specified daily tonnage depends upon the following:1. Track - Profile: grade & lengthCondition: wet, dry, sanded.2. Resistance3. Distance to be traveled per trip4. Time oftrip5. Tonnage to be handled per trip or shift,

1. TrackA profile of the track over which the locomotive is expected to operate is necessary for selecting bothdiesel and trolley locomotives. In the case of either type, we must know the steepest grade, both plusand minus, so that the proper weight of locomotive is chosen for starting and stopping the load.In addition, for the trolley locomotive, the various grades with the distances must be known todetermine the heating capacity of the motor(s) in the locomotive.2, ResistanceResistance may be considered as coming from six sources.

http ://www. mg. mtu. edu/-frotuony/mg33 6 13mg336. htm sl20/00


41/63

age 4i of63a. Resistance of Trailing Load (bearing & track), R" - The force that the load must overcome tostart it on a level track.The amount of resistance offered by the trailing load is expressed in pounds per ton of load hauled.For mine work it varies between I and2Yo of the gross load. Roller bearings are considered as havinga resistance of lYo (20 b/ton). Plain bearings are considered as having a resistance of 1.5% (30lb/ton).For example, assun'dng a train of cars with plain bearings and weighing 60 tons, the rolling resistancewould be 60 x 30 =' 1800 lb which is the force necessary to start the load on a level track.Resistance of traihqg [oad, R" = Iad (tr) x r resbtarre x 2fl00lb/ton100RL = L lristance (%) x 20 lb=LxRx20:20LR lb/tonwhere: L is the weight of the trailing load in tons.R is the resistance in oZb. Rolling resistance of Locomotive, R*Amount of drawbar pull or tractive effort required to overcome the retarding effect between thewheels and track.Resistance offered by the locomotive when the motor pinion is removed. Usually taken as 20 lb perton. Some manufacturers claim l5 lbs/ton to be sufficient.R":20 lb/ton.c. Curue Resistance, R" (effort required to overcome the resistance of curves)Resistance due to curves, usually disregarded if the minimum radius of curvature for the curves hasbeen used. Curve resistance depends on a variety of factors:l. track gage on the curve 4. speed2, wheel base 5. radius of curve3. wheel diameter 6. number of cars (load) on the curve

http: lwww. mg. mtu. edu/-frotuony/m g33 6/3m933 6. htm s/20/00


42/63

Page 42 of 63

A 1 curve is one in which 100 ft of track is l/300 of a complete circle. The radius of I : 5730 ft.s=r100 = (r) (t.zas x tO-2)r: 5730 ft.Rc: 0.8 (lb/ton) / degree (Normally used value)Degree of.Curve : 5730 / Radius in feet.Alternative equation for determining curve resistance is given as:ps: 225(B+H) bfionrwhere: B is the wheelbase ofthe wagons, ft, K is track gauge in ft, r is the radius of curve in ft.d. Grade Resistancer RaThe grade resistance is the drawbar pull or tractive effort required to overcome gravity in propelling alocomotive up an incline. Expressed in degrees 0, and percentages (%).Grade resistance, Rr: Golo W tons:20 GW lb:20 G lb/tonWhere W is the weight of the locomotive (tons) and g is the grade in o/o.Grade resistance must be considered for the trailing load and the locomotive. Grades are normally 5%.e. Level of Drawbar Pull (DP)Defined as the force exerted on the coupled load by a locomotive through its drawbar or coupling andis the sum of the tractive resistances of the coupled load.The drawbar pull which a locomotive is capable of developing is determined by subtracting from thetractive effort, the sum of the tractive resistances of the locomotive.DP: Tractive effort - sum oftractive resistance of locomotive.Adhesion between the wheel tread of the locomotive and the steel rails is what permits the load to be

http ://www. mg. mtu. edu/-frotuony/mg33 6 /3m9336. htm sl20l00


43/63

Page 43 o 63started and kept in motion.Adhesive force depends onmaterial of which the tread is madecondition of railsFor estimating purposes for level track use20Yo for cast iron treads25Yo for steelTractive Effort - total force delivered by the motive power of the locomotive through the gearing atthe wheel treads. When the tractive effort force is greater than the product of the locomotive weightand the coefficient of adhesion between the wheels and rails, the wheels will slip.Tractive effort : Yo of adhesion x W : A x 't x 2000 = 20AH lb/ton100Where: W: the weight of locomotive (tons)A: adhesioninYo.Conditions % AdhesionDry rails2iYoDry rails (sanded rails) 33%Wet rails 5-15%Braking dry rails20Yof Force ofacceleration and decelerationForce of acceleration is the force required to overcome the inertia of the body.R"=Ma=\algwhere R": accelerating force, lb.W: weight, tons

http : //www. mg. mtu. edu/-frotuony/m 933 6 I 3m933 6.htm s/20100


44/63

Page 44 of 63

a : acceleration, ff/sec2g : acceleration of gravity :32.2 ff/sec

lrphps=,ffi = t.4?ft/s2Express force in terms of pounds per ton with an acceleration of 1 mphps, we get:

R. = g x l? ft/sec2 = 913 h/ton of rehicle pr I trElps' 32.2Therefore, the force required to impart an acceleration of I mphps to a weight of one ton is 91.3 lb.In actual calculations, 100 lb/ton instead of 9l.l lblton are used as the force necessary to give I ton anacceleration of I mphps because rotational acceleration of the wheels, motors, gears requiresapproximately 8.7Ib. Total: (91.3 + 8.7) lbs or l00lb/ton.If acceleration or deceleration is considered, they must be applied to both the trailing load and thelocomotive;Ra=1o0aWlbor R": 100a lb/tonwhere a is the acceleration in miles per hour per second.Braking EffortThe weight of the locomotive selected for starting and hauling the load should be investigated forstopping the train.When the power is shut offand the train allowed to coast down grade, it is aided by gravity to theextent of 20 lb/ton for each percent of grade. Plain bearings retard the train at 30 lb/ton.Tractive effort :20Yo for cast iron thread

http ://www.mg.mtu. edu/-frotuony/mg3 3 6/3mg3 3 6. htm sl20l00


45/63

Page 45 of63

:25Yo for steel threadBreaking effort: 80 to 85% of the tractive effort.3. Distance to be traveled and tonnege per shiftAffect the size ofthe locomotive and the kilowatt-hr capacity ofthe locomotive selected.4. Time per tripA consideration of this feature is necessary to determine the tonnage that will be handled per shift andthus the number of locomotives required for the work.

RAILROAD HAIILAGE CALCULATIONSA. Against Loads (Eauting)1. Drawbar pull required by load on gradesagainst lods

Jean Le.Rond D'Alembert (French Mathematician)D'Alembert's Principle: A mass develops an inertial force proportional to its acceleration and opposingit, Principle permits the equations of motion to be expressed as equations of dynamic equilibrium,By D'Alembert's Principle, this can then be considered as a system in equilibrium and the balanced

--u*..-Ulsin = 20b /t /ahgr.

http ://www. mg. mtu. edu/-frotuony/mg33 6/3m9336. htm sl20/00


46/63

Page 46 of63force equation can be written down as:Drawbar Pull, (DP) : Grade resistance + Rolling resistance * Accn. resistance * curve resistance (ifany)DP:RsL+RRL+RaL+RLDP : 20 GL + RRL + l00al, + RcLDP : LQG + RR + 100 a * Rc) lb (hauling)

2. Tractive Effort - required for the total train is thesum of the various tractive resistances of both thelocomotive and coupled load including the acceleration resistanceloco * cars train

a. Tractive effort required for hauling on grades against loadsT. : \ + ReL .'- LRa + LRC + WRs + RWW + rilRWa + \ryRWC: DP + sum of resistances:20 AWtherefore, 20AW : 20GL+ LR, + 100aL + RCL + 20GW + \ryRw + l00aW + RcWW(204 -20G -Rw- l00a-Rc)=L(20G+RL+ l00a+Rg)b, The drawbar pull developed by locomotive while hauling on grades against loads is L(20G + RL +

http ://www. mg. mtu. edu/-frotuony/mg33 6 I 3 mg336. htm 5120100


47/63

l00a + Rc) or W(204 - 20c- \ - l00a - R")c. The weight of locomotive required for handling coupled load against loads is

\4 - L f20G+Rr+llllla+R.l2OA - 20G - Rw- ltllta - Rc3. Brakinga. Drawbar Pull Required by Load while Braking on grades against loads

,5

Page 47 of 63

http ://www. mg. mtu. edu/-frotuony/mg33 6/ 3m9336. htm sl20/00


48/63

DP + L\ + LR*: LR"DP:LR"-LRs-LRRDP: L(Ra - Rs - RR)

Page 48 of63

b. Tractive effort required while braking ongrades against loads

T. + LRr + LRc + WRg a WRy: LR" 1WRT, : LRu - LRL - LRC a WR - WR\M - WRg= l00al - LRl, - 20 GL + 100 aW - tilR* - 20 GL

http ://www. mg. mtu. edu/-frotuony/m933 6 I 3 mg336. htm sl20l00


49/63

Page 49 of63

: L(100 u - RL - 20 G) +W(100 u - Rw(0) - 20 G)WhenBrakingT. and DP are opposite to the direction of the motionR" is in the direction of motion or opposite to the direction of positive acceleration or opposite thedrawbar pull.B. fn Favor of Loads1. Drawbar pull required by load on grades in fevor of loeds

,#..f'.$u"- ""tt$"-$-

http ://www.mg.mtu, edu/-frotuony/mg33 613mg336, htm sl20l00


50/63

Page 50 of63

DP +Rr=R *Ra+RCDP=RL+Ra-Rg*RC: LRL + l00al - 20GL + RLDP : L(\ + 100 a - 20G * \) (lb) hauling

2. Trctive Effort required for hauling whengrades are in favor of loads

T. * Rlg + Rrg : RW* RWu * RL + Rl" + RWC+ RrCT. : L(\ - Rre * RL"* RLc)+ (Rw * Rwe * Rw"* Rwc)\ryDrawbar Pull developed by locomotive on grades in favor of loads20 AW : (LRr - 20 GL+ l00al, + LRs) a WRr - 20GW + 100aW + WRcW(204 - Rw * 20G - l00a - \) : rr, - 20G + 100a + \)Weight of locomotive required when hauling on grades in favor of loads

http://www.mg.mtu.edu/-frotuony/mg336l3mg336.htm 5/20100


51/63

= L fRr-20G+lltra+Rcl20.4' - R$+ 20G - lflla - R3. Braking on grades in favor of loeds:a. Drawbar Pull

(lb) hauling

Page 51 of63

,#rf

t'$

http : //www. mg. mtu. edu/-frotuony/ mg33 6/3m93 3 6. htm 5/20/00


52/63

DP + LRL: LRo + LRODP:LRo-LRr,aLRI: 20GL - LRr + l00al,DP:L(2OG-Rr*100a)

Page 52 of 63

b. Tractive EffortI

T. + VR* + LRL: WR, a WR + LRa * LRsT" : - WRW - LRr + WRg a WR + LRa * LRsT" : - WRw - LRr + 20GW + l00aW + l00al, + 20GLT. : W(206 - Rw( o + 100a) + L(20G - Rl * l00a)R" is zero when braking & R*:0For hauling and braking the following cases should be considered

http ://www. mg. mtu. edu/-frotuony/mg33 6/3m9336. htm

Itlt --.7,FrS't

5/20100


53/63

Page 53 of63

(total of 12 cases)a) Te required for hauling on grades against loadsb) TB required for hauling on grades in favor of loadsc) Drawbar pull developed by locomotive on grades against loadsd) Drawbar pull developed by locomotive on grades in favor of loadse) Drawbar pull required by load on grades against loadsf) Drawbar pull required by load on grades in favor of loadsDUTY CYCLESufficient data pertaining to duty cycle and haulageway dimensions, curyes and grades are necessaryto estimate adequate weight, horsepower and speed of a locomotive to perform satisfactorily underspecified conditions.MineJocomotive manufacturers have established by tests and experience certain practicalvalues for:a. the coefficients of adhesion between the threads ofthe locomotive driving wheels and the rail.b. factors that influence train resistance such as tractive resistance and includel. frictional resistance of bearings in locomotives and cars, track resistance to rolling2. grade resistance3. track curvature resistance4. acceleration resistanceWhen locomotives are powered by electric motors, it is also vital that they do not overheat during theduty cycle and cause insulation and brush failures.In selecting a locomotive for haulage purposes, it is important that the duty cycle for the proposedconditions be calculated. The horsepower rating for the locomotive will depend on the duty it mustperform.Assuming that the locomotive weight is sufficient to handle the load under the most adverseconditions of starting, accelerating, stopping etc. it must in addition have a continuous motor capacitywhich will not overheat during the period in which it is in service.The motor capacity required is determined by calculating a duty-cycle for a round trip under operatingconditions. For this purpose there must be available the profile of the route showing the grades curves,



54/63

Page 54 of63and level stretches of track, and the characteristic curves ofthe motors in the proposed locomotive.

CHARACTERISTIC CURVESThe characteristic curves of the motors used in locomotive may be obtained from the manufacturers.These curves are determined for sea-level rating and a temperature rise of 75C from an ambienttemperature of 25C. Ifthe locomotive is to be used under conditions other than these, correctionsmust be applied to the temperature rating at seaJevel.Effect of Altitude and Tempereture on Motor CharacteristicIncreased altitude has the effect of increasing the temperature rise of some types of machinery.It is recommended that when a machine is intended for service at altitudes above 3300ft, thepermissible temperature rise at sea level shall be reduced by l% for each 330ft by which the altitudeexceeds 3300ft.Permissible temperature : ?s - ?S f nhation - gfOO )l 33oxloo I

Suppose we have an elevation of 5280'

Permissible temperature rise: ?s - is f szgo - ss00l = ?5 - 4.s = ?0.s- c[330xr00J

If the temperature under which the locomotive operates is greater than 25 a similar correction must beapplied to the ambient temperature, otherwise the estimated temperature will exceed 100.If the locomotive is to operate under a temperature of 35C which is l0 (35-25) higher than normal,then the temperature under which the machine would be rated is70.5 - (35 - 25) : 70.5 - l0 : 60.5

CONTINUOUS CURRENT RATING

http ://www. mg. mtu. edu/-frotuony/mg33 6 l3mg3 3 6, htm sl20l00


55/63

Page 55 of63

The principal source of heat in locomotive haulage is the resistance of the conductor and it isproportional to the product ofthe square ofthe current flowing and the time during which the specificcurrent flows.It is therefore necessa/ to estimate the average "continuous current" demands of the entire duty cycleand compare them to the "continuous current" rating of the motor chosen.Estimate of "continuous-current" demand of duty cycle is obtained by the rms (root-mean-square)method by:a. dividing duty cycle into main sections requiring different tractive effort valuesb. selecting the corresponding amps and speed for each tractive effort value from the characteristiccurve of the motor.c. calculating from the specific mph and length of the section, the specific time for each section.d. adding the products of the squares of the specific amperes and the respective values of time (I2 x t)e. dividing the sum by the effective time, t.t. is usually calculated by locomotive manufacturers as total time + total idle or terminal time whenmotors are cooled by forced ventilation or total running time + one-half idle or terminal time when selfventilating motors and specified.

r.m.s. currentI"

portl,ffi s\g =@tltrfrce

http ://www. mg. mtu. edu/-frotuony mg336/3mg3 3 6. htm s/20100


56/63

L*- Iftr * lfu * 13 +..-.. + Itt*te*b*---.+t1+ I/2 trle tirpfor self cooled (or)motors

Irrrr. should be less than the continuous current rating for efficient service.

irlle timpmoton cooled hyfored rcntilation

age 5o ot'o3

http ://www. mg. mtu. edu/-frotuony/mg 33 6 /3 mg336. htm 5120100


57/63

Page57 of63

Motor Performance Curves or Characteristic Curves

ESTIMATING CONTINUOUS CURRENT RATINGSuppose a haulage level is at 6600ft and the ambient temperature is 30C.Correction for elevation is f cooo - rsoo * t oo = l0 e/o( 33oo ) --

Temperatre : 7 5C - (10%)(7 5C) : 67 .5CCorrection for ambient temperature is 30C - 25C: 5C.The temperature rise becomes 67.5C - 5C: 62.5For practical applications, the heating varies directly as the square of the current ever the time it isapplied. A temperature rise curve must be constructed and the continuous rating of the motor for thenew conditions must be determined from this.The curve may be constructed as follows:From the 75C curve points are chosen on the curve. For example, at 0.37hr, the current is 250 amps;at 0.6 hr, it is 200 amps; at I hr, it is 155 amps; at 1.5 hr, it is 128 amps; at2vs, it is 113 amps; at2.5hr, it is 102 amps; at 3hrs, it is 96 amps; at 3.5hr, it is 90 amps; at 8 hrs, it is 60 amps.By use ofthe formula , we obtain values that are plotted against 75 in

http : //www, mg. mtu. edu/-frotuony/m933 61 3m9336, htm sl20l00


58/63

Page 58 of63

vrlr * amn-fig 166. From Fig 166 we find the r 5 values for the 62.5C curve. Thesevalues with the actual motor amps may be solved for the time. A continuous rating oroonstant temp is assumed to occur after the motor has run continuously for 8 hrs.The amperes for the 75C curve are plotted against their equivalent time in the 62.5C curve. We findthat the thr rating has decreased to about 52 amps.If the duty cycle requires a rating of 61.5 amp., then the locomotive selected is not large enough toperform the work satisfactorily unless auliary ventilation is used.

Time to rise to 75"



59/63

Continuous current rating = [+J"

continrous crrent mting at 75' - continrors curert rating at y'-

Assume continous current rating at seaJevel is the vertical line tangent to 75 curve.MOTOR HORSEPOWER

http : //www. mg. mtu. edu/-frotuony/mg 33 6 I 3 mg336. htm


60/63

Locomotive horsepower at any given instant

3?5 x efreruy

Efficiency is that of Transmission between motive polver and driving wheels.For spur gears, efficiency:0.95 or 95Yosingle reduction, eff : 0.95double reduction, eff : 0.90won gears, eff:0.90 - 0.96Kilowatt input to motor: HP x ?4 x I00df xlfl00

SPEED TIME & DISTANCE (for a constant rate of acceleration)t:(vl-"J/aIv:u+at]D : 0.5 ((rr - v ) [ s : ut * % at2l1.467 is the conversion factor from mph to s.Rr: (% resistance)(20), lb/ton\:20, lbltonR.:0.8 lb/ton / degree (applied to load and locomotive)Rr:20G lb/ton (applied to load and locomotive)Ru: 100a, lb/ton (applied to load and locomotive)T":204W, ton

Page 60 of63

http ://www. mg. mtu. edu/-frotuony/mg33 6/3m9336. htm s/20/00


61/63

MINE STORAGE BATTERIESTtvo Typesa. Edison or nickel-iron-alkaline combinationb. Lead or acid typeCapacity ofBatterylkwH: 33000 ft lh /min/HP x 0 mln/hr - 2Lll x ftD.?4krH/HP

In practice, it is customary to consider an efficiency of 67YolkwH : .67(2.654x tO6) ftJb: 1,769,500 ft-lbThe battery capacity for each section of the haul is found fromKw_hr _ TxD0*20G)r,76ftpoo

Where T : total tons hauled (including locomotive)D: distance of each sectionG: gradeYoSelection of Batteryl. Determine number of platesa. Lead Battery : voltagel2V cellsb. Edison Battery: voltage/I.2V cells2. Determine number ofplates

Page 61 of63

http ://www.mg. mtu. edu/-frotuony/mg33 613m9336. htm sl20l00


62/63

Page 62 of 63

amD-hours: 1ItH _ M'vvNumber of plates : -anP -hoTs -@Duty CycleDuty cycle is important in selecting a locomotive for haulage purposes.The h.p. rating for the locomotive will depend on the duty it must perform.Assuming locomotive weight is sufficient to handle the load under the most adverse conditions ofstarting accelerating, stopping etc, it must have in addition a continuous motor capacity which will notheat during the period in which it is in servi ce.The motor capacity required is determined by calculating a duty cycle for a round trip under operatingconditions.

REGENERATIONLocomotives like belt conveyors could be driven by industrial internal combustion engines whereelectric power is not available. They could be driven,..a. from a line shaft where steam power is availableb. by hydraulic motors or air turbines in gassy or explosive environmentsc. almost universally by electric motors.If an induction motor is driven by its load in the same direction as the rotation of the flux, its speedrises above synchronous speed. The motor acts as an induction generator, taking magnetizing currentfrom the line and absorbing mechanical power through its shaft, Electric power is then fed back intothe power system.The motor will restrain the load with little rise in speed above synchronous as long as the load torquedoes not exceed the maximum torque of the motor. If maximum torque is exceeded, the motorbecomes unstable and runs away.As the motor drains magnetizing current from the line, it follows that dynamic braking is possiblewhen the connection between the motor starter and the power line is intemrpted. The motor serves as

http ://www. mg. mtu. edu/-frotuony/mg33 6 I 3 m9336. htm sl20l00


63/63

a generator feeding power back into the system without special control.In this case retardation must be effected by other braking methods.

Page 63 of63

Excavation Costs - Otuonye - 2000

Documents

Transcript of Excavation Costs - Otuonye - 2000