Question

It is found that the energy required to reduce particle from a mean diameter of 10 mm to 5 mm is 1 kJ/kg. Using Rittinger's law, what is the energy requirement to reduce the same from a diameter of 1 mm to 0.5 mm?

• A. 5 kJ/kg
• B. 100 kJ/kg
• C. 10 kJ/kg
• D. 1 kJ/kg

Hint:

Rittinger's law depends on the difference of reciprocals of the diameters, not on the ratio of the diameters, so halving a smaller particle costs more energy than halving a bigger one.

Answer 1

The Correct Option is C

Step 1: Rittinger's law states that the energy needed for size reduction is proportional to the new surface area created, which gives the relation

\[E = K_R \left( \dfrac{1}{D_2} - \dfrac{1}{D_1} \right)\]

where \(D_1\) is the initial particle diameter, \(D_2\) is the final particle diameter, and \(K_R\) is Rittinger's constant for the material.

Step 2: Use the first case to find \(K_R\). Here \(D_1 = 10\) mm, \(D_2 = 5\) mm, and \(E = 1\) kJ/kg.

\[1 = K_R \left( \dfrac{1}{5} - \dfrac{1}{10} \right) = K_R (0.2 - 0.1) = K_R \times 0.1\]

So \(K_R = 10\) kJ.mm/kg.

Step 3: Now apply the same constant to the second case, reducing from \(D_1 = 1\) mm to \(D_2 = 0.5\) mm.

\[E = 10 \left( \dfrac{1}{0.5} - \dfrac{1}{1} \right) = 10 (2 - 1) = 10 \text{ kJ/kg}\]

Step 4: So the energy required for the second size reduction is 10 kJ/kg, matching option 3.

Why the other options fail: 1 kJ/kg would only apply if the same size ratio and absolute size range as the first case were used, but the particles here are ten times smaller, needing much more energy. 5 kJ/kg and 100 kJ/kg do not come out of the constant found from the first case when the calculation is carried through correctly.