Ball mill scaling — power requirement vs. mill diameter For a given ball load, the power required to drive a ball mill is proportional to which function of the mill diameter D?

Difficulty: Easy

Correct Answer: D^2.5

Explanation:


Introduction / Context:
Power draw correlations for tumbling mills are crucial for scale-up. A commonly used empirical relation shows a strong dependence of power on the mill diameter for mills with similar loading and speed fraction of critical.


Given Data / Assumptions:

  • Same ball load fraction, liner condition, and speed fraction.
  • Geometric similarity is assumed.


Concept / Approach:
Empirical and semi-theoretical treatments (e.g., Bond/Rowland) indicate that mill power P scales approximately as D^2.5 * L, where L is length. Holding L proportional or fixed, the diameter exponent remains dominant, thus P ∝ D^2.5.


Step-by-Step Solution:
Recognize the standard scaling: P ∝ D^2.5 (for given load and operating conditions).Eliminate inverses 1/D and 1/D^2.5 as contrary to observed behavior.Linear D underpredicts power growth with size; D^2.5 matches practice.


Verification / Alternative check:
Industrial datasets corroborate the super-quadratic dependence of power on D for dynamically similar mills.


Why Other Options Are Wrong:

  • D or inverse powers do not reflect measured power draw trends in scale-up.


Common Pitfalls:

  • Using only cross-sectional area (∝ D^2) and ignoring charge lift and trajectory effects that steepen the exponent.


Final Answer:
D^2.5

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