Question

In grinding of black pepper, 80% feed passes through 6 mm screen, and 80% product passes through 0.5 mm screen. The gross energy required is 10 kW.h per ton of material. Assuming Bond's law holds good, and for the same feed conditions, if 80% product passes through 0.2 mm screen, the energy required, in kW.h per ton of material, is nearest to

• A. 18.17
• B. 9.09
• C. 5.50
• D. 11.00

Hint:

Use Bond's law to find the grinding constant from the first size reduction, then reapply it for the finer product size.

Answer 1

The Correct Option is A

Step 1: State Bond's law.
Bond's law gives the specific grinding energy as \( E = K_b\left(\dfrac{1}{\sqrt{D_p}} - \dfrac{1}{\sqrt{D_f}}\right) \), where \(D_f\) and \(D_p\) are the 80% passing sizes of feed and product.

Step 2: Find the Bond constant from the first grind.
Here \(D_f = 6\,mm\) and \(D_{p1} = 0.5\,mm\), with \(E_1 = 10\,kW.h/ton\).
\( \dfrac{1}{\sqrt{0.5}} - \dfrac{1}{\sqrt{6}} = 1.414 - 0.408 = 1.006 \).
So \( K_b = E_1 / 1.006 = 10/1.006 = 9.94 \).

Step 3: Use the same constant for the finer grind.
Now \(D_{p2} = 0.2\,mm\), same feed size \(D_f = 6\,mm\).
\( \dfrac{1}{\sqrt{0.2}} - \dfrac{1}{\sqrt{6}} = 2.236 - 0.408 = 1.828 \).

Step 4: Compute the new energy requirement.
\( E_2 = K_b \times 1.828 = 9.94 \times 1.828 = 18.17\,kW.h/ton \).

Final Answer:
Grinding finer to 0.2 mm needs about 18.17 kW.h per ton, so option (A) is correct. The other options come from using Rittinger's or Kick's law instead of Bond's law. \[ \boxed{E_2 \approx 18.17\ kW.h/ton} \]