THE MECHANICAL HANDLING OF MATERIALS 337
THE CALCULATION OF HORSE-POWER
The power necessary to drive the various types of elevators and conveyors has in many cases been arrived at by practical experience and reference to the makers’ records. It is, however, useful to Le able to form aa opinion as to the likely horse-power which. will be required by any one installation, and the following simple but reliable methods will probably prove of some assistance to the designer.
First, in the case of conveyors such as the gravity-bucket and tipping-tray types, practical experience has shown that the tractive force is rather under 5 per cent, of the total weight of the moving parts and load. In addition to this 100 per cent, should be allowed for contingencies, such as neglect of lubrication and general inefficiencies. A gravity-bucket almost invariably performs two functions, first conveying horizontally, and secondly elevating the material.
Actually, four main factors liave to be considered when calculating the power required to drive a conveyor, namely:—
(a) The work done against friction in moving the chain along the conveyor trough and return path.
(s) The work done in dragging or carrying the material along the effektive length of the conveyor.
(t) The frictional losses at the terminal sprocket wheel bearings and otter guide wheels, if any.
(cZ) The frictional losses in the driving gear.
For most purposes the horse-power required. may be arrived at approximately by assuming a coefficient of friction in accordance with the conditions upon which the elevator or conveyor is operated. The various cases would be treated in the following manner:—
(u) Vertical elevator.
Let M = weight of material carriod in pounds.
W = weight of moving parts in pounds.
K = coefficient of friction (see below).
S = speed in feet per minute.
H p __ MS KS (M + W) ’ ~ 33,000 + 33,000 ~
(v) Inclined elevator.
Let a = angle of inclination. with horizontal.
.g p __MS sin a KS (M + W) cos a 33^000 33,000
(w) Horizontal conveyor.
. . p KS (M + W)
33.000
So far as the valne of K is concerned the result will be sufficiently accurate if this is taken as 045 to represent the average coefficient between coke sliding on metal, or metal on metal, i.e. it strikes an average for those conditions where both coke and the conveyor chain are sliding on a metal trough. It should be borne in