Question

It is desired to crush 100 ton/hr of phosphate rock from a feed size where 80% is less than 4 into a product where 80% is less than 1/8 in. The work index is 10.13.

(a) Calculate the power required

(b) Calculate the power required to crush the product further where 80% is less than 1000pm

Answer 1

Explanation:

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Answer 2

a) The power required is 147.98kW

b) The power required to crush the product a further where 80% is less than 1000pm is 140.54kW

Given:

a) m = 100ton/hr

   Feed size = dpa = 101.6mm

   Product size = dpb = 1/8 inches = 3.175mm

   work index (wi) = 10.13

b) dpa = 3.175mm

   dpb = 1mm

   work index (wi) = 10.13

To Find:

a) The power required to crush the phosphate rock

b) The power requires to crush the product further where 80% is less than 1000pm

Solution:

Using Bond's Law of size reduction,

$\frac{P}{m} = 0.3162wi(\frac{1}{\sqrt{dpb} } - \frac{1}{\sqrt{dpb}})$

$\frac{P}{100} = 0.3162 * 10.13(\frac{1}{\sqrt{3.175} } - \frac{1}{\sqrt{101.6}})$

P = 147.98kW

Therefore, the power requires is 147.98kW

b) According to Bond's Law

$\frac{P}{m} = 0.3162wi(\frac{1}{\sqrt{dpb} } - \frac{1}{\sqrt{dpb}})$

$\frac{P}{100} = 0.3162 * 10.13(\frac{1}{\sqrt{1} } - \frac{1}{\sqrt{3.175}})$

P = 140.54 kW

Therefore, the power required to crush the product a further 80% is less than rpm is 140.54kW.

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