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?
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AD
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B1/D
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CD^2.5
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D1/D^2.5
Answer
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