THE MECHANICAL HANDLING OF MATERIALS 339
= 37 C.
(b) Work done in conveying coke
= wL tanA S ft. Ib. per minute.
Assume angle of repose of 45°, then tan A = 1.
Capacity in Ib. per minute — wS
_ C x 2240
60 .-. wL tanA S = 37 CL
(c) Frictional loss on terminal sprocket.
This depends upon ratio of diameter (D) of sprocket to the diameter (<Z) of the shaft bearing. For simplicity, this may be assumed to be 10 to 1. Take the co-efficient of friction as 0-25, and assume that the friction is evenly distributed round the axle.
Then, taking the pull (P) to be equal on. both sides of the bearing, the friction will be equal to 2PF.
The rate at which work is done against this friction
=SX-=S
D 10
, ,, , , . , 2PFS
and the work done per minute =—----------
10
p ,, , 2PFS , PS
r = 0-25, ttieretore ------ becomes —
10 20
i.e. 5 per cent, of PS.
Hence, about 5 per cent, of tlie driving power is lost at each sprocket shaft. That is, a loss of 10 per cent, for the two sprocket skafts.
(Æ) Frictional loss in driving gear.
The driving mechanism most usually consists of a double reduction. spur gear, and it is usually assumed that 10 per cent, is lost at each. reduction. This is equivalent to allowing 80 per cent, efliciency for the gear.
Combining the four above results, therefore, tlie horse-power may be obtained as follows:—
H.P. = (2LWFS + 37 CL) x -- X® v 33,000
' 90 80
If tlie coefficient of friction is taken as 0-25, by no means too high, a figure for a conveyor in which coke dust is likely to get between the rubbing surfaces ; and if the weight of the chain is taken at 50 Ib. per foot run, i.e. a good practical average, then the above expression may be reduced to—
Hp _ 1'4 L (25 S +37 C) ’ 33,000
Conveyor, partly inclined and partly horizontal.
If the conveyor is of the inclined type with an incline of l feet in length and rises to a vertical height of h feet. Then the additional power required can be arrived at as follows:—
Two new factors are introduced, namely—