I WANT TO CALCULATE THE MATERIAL PRESENT IN KG IN A BALL MILL , IF THE FRESH FEED IS 100 TPH & 300 TPH IS THE MILL OUTLET PRODUCT.
IT IS A CEMENT MILL
BY THIS I SHALL BE ABLE TO CALCULATE THE THICKNESS OF MATERIAL LAYER INSIDE THE MILL OR IN OTHER WORDS THE FILLING % OF MATERIAL ALONE IN A BALL MILL EXCLUDING GRINDING MEDIA.
IN MY OPENION I REQUIRE THE RETENTION TIME OF MATERIAL INSIDE TRHE MILL TO GET THIS VALUE.
PLEASE DO COMMENT
SIDDU
Dear
The material retaintion time in a ball mill depends on the tons of ball and total material fed to the mill (fresh feed +separator’s rejects). It can be approximated using the following expression:
T(minutes) = 6.7 x (F/P)
Where:
F = total tons of grinding bodies in the mill
P = Mill’s production (total dry feed material) (ton/hr)
It is also possible to calculate the tons of resident material inside the mill. This means, the tons of material that would remain if the mill if mill is suddenly shut down.
M(ton) = P x (T(minutes)/60)
For example, a mill that produces 100 ton/hr with a circulation factor of 2 and 250 tons of ball, retains material for an average period of:
6.7 x (250/100) = 16.75 minutes
The material inside the mill will be
100 x 16.75/60 = 27.91 mt
Practically checking of retaintion time in a mill :
The residence time of material is checked practically by using some fluorescent material . It is added to the feed material at mill inlet. The mill outlet sample is collected in every 2 to 3 minutes. The time when the concentration of the fluorescent is high, the time is noted.
IT IS DISCUSSED EARLIER IN THE FORUM.
The retention time is calculated from time inlet to time outlet.
Dear Chari
In my openion if the circulation factor is 2 with 100 tph product than retention time should be
6.7*250/300 = 5.58 minutes
Corresponding material inside the mill
5.58*300/60 = 27.9 t
It is ok ,matching with the end result of 27.9 t.
Now how much power the mill takes to rotate this 27.9 t?
How can I calculate the material layer thickness inside mill if the density of the material is 1.2 t/m3 & mill dimentions are known with known media tons?
Sorry for the delay in reply, You are right as I have not considered the return load by mistake. General retaintion time in a mill with 70% rpm of critical speed is taken as 5 to 7 minutes for Low L/D mills.
If the total media is 250 mt in the mill and BD of the media taken as 4.55t/m3 , the specific gravity of steel balls is taken as 7.8t/m3 so the volume of voids in the media is
250 x ( 1/4.55 - 1/7.8) = 22.9 m3
if BD of mix is 1.2 than material in the mill without increase in the height of the media level is 22.9x1.2 = 27.5 mt.
if the retaintion material is high then there will be an increase in volume of material inside the mill.
chari