Discussion
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
Answer: A
The energy required is about 18.17 kW.h per ton.
Bond's law gives the energy as \(E = K\left(\frac{1}{\sqrt{x_{p}}} - \frac{1}{\sqrt{x_{f}}}\right)\), with \(x_{f}\) and \(x_{p}\) the 80% passing sizes of feed and product. Find \(K\) from the first case:
\[ \begin{aligned} 10 &= K\left(\frac{1}{\sqrt{0.5}} - \frac{1}{\sqrt{6}}\right) \\ &= K(1.4142 - 0.4082) \\ K &= 9.941 \end{aligned} \]
Then the finer product:
\[ \begin{aligned} E &= 9.941\left(\frac{1}{\sqrt{0.2}} - 0.4082\right) \\ &= 9.941 \times 1.8279 = 18.17 \end{aligned} \]
A finer product must need more energy than 10 kW.h, which rules out 9.09 and 5.50 at once. 11.00 understates the effect: under Bond's law energy grows with the inverse square root of product size, so going from 0.5 mm to 0.2 mm nearly doubles it.
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