Calculations on Materials Handling: Surface Coal Mining By Dailyn Nivero
Calculation All the production calculations for loading and haulage equipment that treat the material as discrete units can be characterized by a cycle. Capacity Capacity ² refers efers to the volume of material that a loading or haulage unit can hold at any point in time. e.g. volume of loading machine bucket or truck bed y
y
Classification of Capacity y
y
Struck Capacity ² Capacity ² volume olume of of mater material ial that that a loading or haulge unit when it is filled to the top, but with no material material above the sides side s or carried on any external attechments such as bucket teeth. Capacity ² maxim maximum um volume olume of Heaped Capacity ² material that a loading or haulage unit can handle when the material is heaped above the sides. While the struck stru ck capacity capaci ty is is a constant for any unit, u nit, the heaped capacity cap acity is a function of the material properties and the shape of the unit.
Calculate capacity The general relationship relationship between production rate, rat e, cycle time, time , and capacity capacity is quite simple and can be stated as: production production rate = capacity × (no. (no. of cycles/unit time) When all efficiency factors are considered, produc productivity tivity = produc production tion rate × efficiency efficiency factors y
y
Loading
and Haulage Production Calculations: Bucket Payload An excavator·s excavator·s bucket bucket payload payload (actual amount of material in the bucket bucket on each digging cycle) is dependent on bucket size, shape, shape , curl force, and certain certain soil characteristics. Average Average Bucket Payload Payload = Heaped Bucket Bucket Capacity x Bucket Bucket Fill Factor y
e. g. Select the bucket size for a fleet on mining minin g shovels at a coal ore operation given the following assumptions about the operation: Daily Required Capacity/Machine (3 Machines)=32,000 Machines)=32,000 tpd Estimated Daily Operating Time=17.02 hours Diggability Rating = very hard digging Estimated Work Cycle =37 seconds Material Bulk Weight = 6,000 lb/yrd3 Swell Factor = 0.60 Dipper Fill Factor = 0.80 y
y y y y y y
Solution: Conventional double backup loading planned; therefore, a standard boom length is satisfactory. Shovel Cycles per Day = y
= = 1,656 cycles/day
y
Required Required material.
Tonnage per Cycle = = = 19.8 tons or 39,600 lbs Dipper Size = =
=13.8, say 14 yd3
Production Capacity of Continuous Flow Loaders y
If I is nominal bucket bucket capacity in yd3, z is number of buckets per wheel in ft, then:
Ss = Vi x Z/ D Where Ss = no. of bucket bucket discharges per second Qt = I x Ss x 3,600 And Qt = theoretical theoretical capacity of the excavator in yd3/hr (m3/h)
M x g = M x Vi2/R M = mass of material, lb (kg) R = radius of the wheel in ft (m) g = acceleration due to gravity Vi = = Vmax Practical values of speed lie between 0.4 to 0.6 Vmax.
Yet
another factor that affects the output of BWE is the bucket bucket filling capacity. capacity. Actual capacity of the BWE in any soil is: Qa = I x Bf x Ss x 3,600 e.g. A BWE has eight buckets with a nominal capacity of 2 m3/bucket. /bucket. The wheel has a diameter of 15 m and operates at a speed of the 0.4 m/s. In a material with a digging dig ging resistance of 25 kg/cm, the BWE is producing 400 m3/hr. If the speed of the bucket does not change cha nge,, what would be b e the bucket-fill factor for the buck bu ckets ets when cutting a material with a digging resistance of 45 kg/cm?
y
Solution:
Q2 = Q
= 400
= 123.5 m3/hr
The production rate can also be expressed as: Q = I x Bf x Ss x 3,600 Solving for Bf, Bf = but Ss= , therefore
Bf =
=
=0.252 or 25.2%
Carying Capacity of Discrete Unit Carying Haulers y
y
Theoretical Productivity The tons/kg or cubic yards/cubic yards/cubic meters per hour produced by an operating unit if no delays were encount encountered. ered. This This indicates 100% potential, which is rarely, if ever achieved.
Tons =
Bank yd3 per hr, bcy/hr =
Loading Time
There are several methods for determining the no. of shovel, draglin dragline e, or loader loader passes per load and the resulting load time. Weight per yd3 of material (loose) = weight in bank x swell swell factor
No. of tons per pass = bucket capacity(yd 3 or m3) x fill factor x loose weight per yd3 or m3 No. of passes per load = rate capacity of haulage unit (tons or kg) per pass Load time (min) = no. pf passes x excavator cycle time (min) Travel time can be calculated by: Travel Time(m Time(min in)) = or Travel Time(m Time(min in)) =
e.g. An operator is determining which of the two possible haul roads should be used: 1. The first alternative is a level roadway, 6000 ft long with a rolling resistance factor of 120 lb/ton and a coefficient of traction of 0.45. 2. The second alternative is a 3,000 ft roadway roadway with a 5% adverse grade on the haul, and a rolling resistance factor of 200 lb/ton and coefficient traction of 0.40. The hauling unit is an off-highway truck with a gross vehicle weight of 130,000 lb; with 81,000 lb on the rear wheels when loaded to rated capacity.
Solution: Alternative Alternative 1: Rolling Resistance Factor = 120 lb/ton = 6%(20 lb/ton =1 %) Grade resistance in Haul =0% Grade resistance on return = 0% Effective Effective Grade on Haul = 6% 6 % + 0% = 6% Effective Effective Grade on Return = 6% 6 % + 0% = 6% a.) Gross vehicle weight = 130,000 lb Power Power Required Required = 130,000 130,00 0 lb x 6% = 7,800lb Power Power Available Available = 8,000lb at 16 mph Power Power Usable = 81,000 x 0.45 = 36,450lb Vehicle will be able to haul under the given conditions: b.) haul time 4.70 min return time + 2.40 min TOTAL 7.10 min
Alternative Alternative 1: Rolling Resistance Factor = 200 lb/ton = 10%(20 lb/ton =1 %) Grade resistance in Haul =5% Grade resistance on return = -5% Effective Effective Grade on Haul = 10% 1 0% + 5% = 15% Effectiv Effective e Grade Grade on Return Return = 10% - 5% = 6% a.) Gross vehicle weight = 130,000 lb Power Power Required Required = 130,000 130,00 0 lb x 15% = 19,500lb 1 9,500lb Power Power Available Available = 20,000lb at 7 mph Power Power Usable = 81,000 x 0.40 = 32,400lb Vehicle will be able to haul under the given conditions: b.) haul time 5.50 min return time + 1.30 min TOTAL 7.10 min
Discrete Unit Haulers with Fixed Paths Productivity =
Dragline Calculations y
The dragline reach factor is the horizontal distance between the bench crest and the spoil pile peak.
Reach Factor = Stac St acki king ng Heigh Heightt = h ² D
e.g. Consider an operation with the following parameters: =1:3 (highwall slope) = 1.25:1 (spoil pile slope) D = 90 ft (overburden depth) S = 25% (swell factor) W = 120 ft (pit width) T = 5 ft (coal thickness) t = 0 ft (spoil pile touches the coal)
Solution: 1st calculation is to find the dragline reach factor RF= = 140.63 + 30 + 30 RF = 200.63 ft 2nd calculation is to find the stacking height Stac St acki king ng heig height ht = h ² D h= = 131. 50 ft 131. 131.50 50 ² 90.0 90.0 = 41.5 41.5 ft
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