Question:
The gross energy requirement, to reduce a very large feed of coal to such a size that 80% of the product passes through a 100 \(\mu\text{m}\) screen, is 13 kWh per ton of feed. In a process, 250 tons \(\text{h}^{-1}\) of coal is crushed. The range of feed sizes is such that 80% of the feed passes through an opening of 100 mm. The product size range is to be such that 80% of the product passes through an opening of 25 mm. According to Bond's law, which one of the following is the power consumption (in kW)?
The gross energy requirement, to reduce a very large feed of coal to such a size that 80% of the product passes through a 100 \(\mu\text{m}\) screen, is 13 kWh per ton of feed. In a process, 250 tons \(\text{h}^{-1}\) of coal is crushed. The range of feed sizes is such that 80% of the feed passes through an opening of 100 mm. The product size range is to be such that 80% of the product passes through an opening of 25 mm. According to Bond's law, which one of the following is the power consumption (in kW)?
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First find $W_i$ from the infinite-feed data point ($F_{80}\to\infty$ gives $W_i=13$), then reapply Bond's law with $F_{80}=100$ mm and $P_{80}=25$ mm.
Updated On: Jul 17, 2026
- 12.8
- 25.7
- 51.4
- 102.7
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The Correct Option is D
Solution and Explanation
Step 1: State Bond's law.
\[ E = 10\,W_i\left(\frac{1}{\sqrt{P_{80}}} - \frac{1}{\sqrt{F_{80}}}\right) \]
Step 2: Find work index from calibration data.
With F80 to infinity, P80=100 micron, E=13: 13 = 10 W_i(1/10 - 0), so W_i=13 kWh/ton.
Step 3: Apply Bond's law to the actual process.
F80=100mm=100000 micron, P80=25mm=25000 micron.
\[ E = 10 \times 13 \times (0.006325 - 0.003162) = 0.4112\ \text{kWh/ton} \]
\[ E = 10 \times 13 \times (0.006325 - 0.003162) = 0.4112\ \text{kWh/ton} \]
Step 4: Convert to power for 250 tons/h.
\[ P = 0.4112 \times 250 = 102.8\ \text{kW} \]
\[ \boxed{P \approx 102.7\ \text{kW}} \]Was this answer helpful?
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