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OPS 1 Reviewer

PRELIMS

SIZE REDUCTION

1. A material is crushed in a Blake jaw crusher such that the average size of particle is reduced from 50

mm to 10 mm, with the consumption of energy of 13.0 kW/(kg/s). What will be the consumption of

energy needed to crush the same material of average size 75 mm to average size of 25 mm:

(a) assuming Rittingers Law applies,

(b) assuming Kicks Law applies?

Which of these results would be regarded as being more reliable and why?

Soln:

a) Rittingers law.

This is given by: E = KRfc[(1/L2) (1/L1)] (equation 2.3)

Thus: 13.0 KRfc[(1/10) (1/50)]

and: KRfc = (13.0 50/4) = 162.5 kW/(kg-mm)

Thus the energy required to crush 75 mm material to 25 mm is:

E = 162.5[(1/25) (1/75)] = 4.33 kJ/kg

b) Kicks law.

This is given by: E = KKfc ln(L1/L2) (equation 2.4)

Thus: 13.0 = KKfc ln(50/10)

and: KKfc = (13.0/1.609) = 8/08 kW/(kg/s)

Thus the energy required to crush 75 mm material to 25 mm is given by:

E = 8.08 ln(75/25) = 8.88 kJ/kg

The size range involved by be considered as that for coarse crushing and, because Kicks law

more closely relates the energy required to effect elastic deformation before fracture occurs,

this would be taken as given the more reliable result.

***

2. If crushing rolls, 1 m in diameter, are set so that the crushing surfaces are 12.5 mm apart and the

angle of nip is 31, what is the maximum size of particle which should be fed to the rolls?

If the actual capacity of the machine is 12 per cent of the theoretical, calculate the throughput

in kg/s when running at 2.0 Hz if the working face of the rolls is 0.4 m long and the bulk density

of the feed is 2500kg/m3.

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Soln:

The particle size may be obtained from:

cos = (r1 + b)/(r1 + r2)

In this case: 2 = 31 and cos = 0.964, b = (12.5/2) = 6.25 mm or 0.00625 m and:

r1 = (1.0/2) = 0.5 m

Thus: 0.964 = (0.5 + 0.00625)/(0.5 + r2)

and: r2 = 0.025 m or 25 mm

The cross sectional area for flow = (0.0125 0.4) = 0.005 m2

and the volumetric flowrate = (2.0 0.005) = 0.010 m3/s.

Thus, the actual throughput = (0.010 12)/100 = 0.0012 m3/s

or: (0.0012 2500) = 3.0 kg/s

Materials Handling

1. A belt conveyor is required to deliver crushed limestone having a bulk density

of 75 lb/cu ft at the rate of 200 tons/hr. The conveyor is to be 200 ft between

centers of pulleys with a rise of 25ft. The largest lumps are 4 in an andconstitute 15% of the total. The conveyor will discharge over the end. For a

belt speed of 200 fpm, what is the minimum width of belt that can be used?

Calculate the horsepower for the drive motor.

Soln:

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