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GATE 2026 CH – Question 45

Fluid Mechanics and Mechanical Operations · Particle size, size reduction and classification · 2 marks · Multiple choice

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 µm screen, is 13 kWh per ton of feed. In a process, 250 tons 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 the Bond's law, which one of the following is the power consumption (in kW)?

  1. 12.8
  2. 25.7
  3. 51.4
  4. 102.7

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Correct answer: (D) 102.7

Explanation

Bond's law: $\frac{P}{\dot{m}} = 10 W_i\left(\frac{1}{\sqrt{D_p}} - \frac{1}{\sqrt{D_f}}\right)$ with sizes in micrometres. For a very large feed and a 100 µm product, $13 = 10W_i \times \frac{1}{\sqrt{100}}$, so $W_i = 13$ kWh/ton. For the actual duty, $D_p = 25000$ µm and $D_f = 100000$ µm: $\frac{P}{\dot{m}} = 130\left(\frac{1}{158.1} - \frac{1}{316.2}\right) = 130 \times 0.003162 = 0.411$ kWh/ton. Then $P = 0.411 \times 250 = 102.8 \approx 102.7$ kW.