Calculations on Materials Handling: Surface Coal MiningBy Dailyn Nivero

CalculationAll the production calculations for

loading and haulage equipment that treat the material as discrete units can be characterized by a cycle.

Capacity – refers 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

Classification of Capacity

Struck Capacity – volume of material that a loading or haulge unit when it is filled to the top, but with no material above the sides or carried on any external attechments such as bucket teeth.

Heaped Capacity – maximum volume of material that a loading or haulage unit can handle when the material is heaped above the sides. While the struck capacity is a constant for any unit, the heaped capacity is a function of the material properties and the shape of the unit.

Calculate capacityThe general relationship between

production rate, cycle time, and capacity is quite simple and can be stated as:

production rate = capacity × (no. of cycles/unit time)

When all efficiency factors are considered,

productivity = production rate × efficiency factors

Loading and Haulage Production Calculations: Bucket Payload

An excavator’s bucket payload (actual amount of material in the bucket on each digging cycle) is dependent on bucket size, shape, curl force, and certain soil characteristics.

Average Bucket Payload = Heaped Bucket Capacity x Bucket Fill Factor

e. g.Select the bucket size for a fleet on mining

shovels at a coal ore operation given the following assumptions about the operation:

Daily Required Capacity/Machine (3 Machines)=32,000 tpd

Estimated Daily Operating Time=17.02 hours

Diggability Rating = very hard diggingEstimated Work Cycle =37 secondsMaterial Bulk Weight = 6,000 lb/yrd3

Swell Factor = 0.60Dipper Fill Factor = 0.80

Solution:Conventional double backup

loading planned; therefore, a standard boom length is satisfactory.

Shovel Cycles per Day =

=

= 1,656 cycles/day

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

If I is nominal bucket capacity in yd3, z is number of buckets per wheel in ft, then:

Ss = Vi x Z/ DWhere Ss = no. of bucket discharges

per secondQt = I x Ss x 3,600And Qt = theoretical capacity of the

excavator in yd3/hr (m3/h)

M x g = M x Vi2/RM = mass of material, lb (kg)R = radius of the wheel in ft (m)g = acceleration due to gravity Vi = = VmaxPractical values of speed lie

between 0.4 to 0.6 Vmax.

Yet another factor that affects the output of BWE is the bucket filling capacity.

Actual capacity of the BWE in any soil is:Qa = I x Bf x Ss x 3,600e.g.A BWE has eight buckets with a nominal

capacity of 2 m3/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 resistance of 25 kg/cm, the BWE is producing 400 m3/hr. If the speed of the bucket does not change, what would be the bucket-fill factor for the buckets when cutting a material with a digging resistance of 45 kg/cm?

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 HaulersTheoretical ProductivityThe tons/kg or cubic yards/cubic

meters per hour produced by an operating unit if no delays were encountered. 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, dragline, or loader passes per load and the resulting load time.

Weight per yd3 of material (loose) = weight in bank x swell factor

No. of tons per pass = bucket capacity(yd3 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(min) =or

Travel Time(min) =

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 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 1:

Rolling Resistance Factor = 120 lb/ton = 6%(20 lb/ton =1 %)

Grade resistance in Haul =0%Grade resistance on return = 0%Effective Grade on Haul = 6% + 0% = 6%Effective Grade on Return = 6% + 0% = 6%a.) Gross vehicle weight = 130,000 lb

Power Required = 130,000 lb x 6% = 7,800lb Power Available = 8,000lb at 16 mph

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 1:Rolling Resistance Factor = 200 lb/ton

= 10%(20 lb/ton =1 %)Grade resistance in Haul =5%Grade resistance on return = -5%Effective Grade on Haul = 10% + 5% = 15%Effective Grade on Return = 10% - 5% = 6%a.) Gross vehicle weight = 130,000 lb

Power Required = 130,000 lb x 15% = 19,500lb Power Available = 20,000lb at 7 mph

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 CalculationsThe dragline reach factor is the

horizontal distance between the bench crest and the spoil pile peak.

Reach Factor =

Stacking Height = h – 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 factorRF= = 140.63 + 30 + 30RF = 200.63 ft2nd calculation is to find the

stacking heightStacking height = h – Dh = = 131. 50

ft

131.50 – 90.0 = 41.5 ft

Thanks for Listening!!