Theoretical Consideration of the Power Required for the Cylindrical Equipments Rotating at Low Speed.

Abstract

The power required for the operation of cylindrical equipments rotating at low speed may be considered as consisting of the following four kinds.
(1) Power absorbed in turning over the charge of granular materials in the cylinder.
-The horse power to turn over the charge of granular materials in the cylinder is given by the following formula:
where D=inside diameter of cylinder, m.
L=length of cylinder, m.
S_p=specific gravity of materials.
n=rate of rotation, rpm.
α=natural angle of repose of materials.
c₁, c₂=coefficient of fullness.
(2) The power required for driving the cylinder. (including gear, tire, etc.)
-The horsepower to drive the cylinder is given by the following formula:
where M=weight of cylinder, ton.
D_m=mean diameter of cylinder, m.
(3) Friction losses at the supporter of the cylinder.
-The horsepower to make up the friction loss caused in the operation of the cylindrical equipments is given by the following formula:
In case the cylinder is supported by horizontal or vertical rollers.
In case the cylinder is supported by turnion hearings.
where X=coefficent depending upon shapes of supporter.
D_T=diameter of tires, m.
D_R=diameter of rollers, m.
D₁=diameter of shaft in bearings, m.
μ=cofficient of friction at surface of bearings.
μ'=coefficient of friction between tires and rollers.
(4) Power absorbed in the reduction stages.
-It is usual, in a case like this, to suppose that a certain percentage of the total horsepowerconsumed for the operation is lost in each stage of reduction.
It can theoretically be said that when we add (1), (2), and (3), we get the approximate horse power required for driving the cylindrical equipment. But, in order to cover the small friction loss and allow some reserve for the motor, it is necessary to allocate some more power in addition to the power calculated above, as follows:
where η=total efficiency of reduction unit.