# Grinding & Classification Circuits

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## Rod Mill Scale Up (4 replies)

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weethengtan
1 year ago
weethengtan 1 year ago

Hi, there. Currently our open-circuit, wet overflow rod mill 6ft x 12ft, feed size <50mm, product size 90% < 2mm has a 14mt/hr throughtput. We would like to triple throughput capcity to 42Mt/hr per rod mill. What would be the scale up dimension of the new rod mill which has 42Mt/hr throughput capacity, given feed and product sizes remain unchanged?

Looking forward to some advices. Thank you.

SmartDog
1 year ago
SmartDog 1 year ago
1 like by weethengtan

For a good approximation some more information would be needed, such as grindability index.  But using the Mill Calc spreadsheet from here: https://www.smartdogmining.com/software/

An approximation would be 7' x 14'

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weethengtan
1 year ago
weethengtan 1 year ago

Hi, Smart Dog, I am grateful for your prompt guidance and will work through your spreadsheet to establish the preliminary size selection. Thank you Regards

Alex Doll
1 year ago
Alex Doll 1 year ago
1 like by weethengtan

Rod mill power draw is scaled between two mills using this relationship:

kW1 / kW2 = (D1^2.5 × L1) / (D2^2.5 × L2)

{some older models use an exponent of 2.3, but 2.5 is more common}

Where D is the diameter of a mill (inside the liners), and L is the length of the rods in a rod mill (not the length of the mill).  There is a limit to the maximum size of a rod mill that is possible: rod length of 22 ft (7 metres) and mill diameter of 17-18 ft according to Rowland and Kjos (Mineral Processing Plant Design, SME, 1979).  So if you know how much power your current mill draws (kW1) and how much power you want to draw for the higher tonnage (kW2), then you can calculate the (D2^2.5 × L2) term.  Then use another relationship (and the limits above):

L2 = D2

where is between 1.4 and 1.6 (this is not exact because L2 and D2 must be in the set of permissible sizes, usually integer values in feet).  The length of a mill shell is usually 100 to 150 mm longer than the rods, so you can compute that, too.

So assuming that your ore characteristics will not change, you can scale up from your existing mill this way.

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weethengtan
1 year ago
weethengtan 1 year ago

Hi Alex, Thank you for your guidance. I came across a relationship in an article stating Q = k*L*D^2.5 where Q is capacity and k is a constant. It's great to find similarity with your relationship in term of kW. I think it's safe to assume a direct relationship between Q and kW. Thank you