338
MODERN GASWORKS PRACTICE
mind, however, that the above methods do not include the power absorbed by frictional losses at the terminal sprockets and in the driving gear.
In all cases where the cxpression “ foot nm ” of conveyor or elevator has been usod, it rofers to the distance between the terminal sprockets and not to the length of chain. employed. The latter will, of course, be approximately double the foot run length.
The following examples will show the mothod of working :—
(a) Horizontal coke conveyor, capacity 20 tons per hour, speed 40 feet per minute, length between terminal sprockets 100 feet, weight of moving parts of conveyor 50 1b. per foot.
M = 100 X 20 X 2240 = 1867 1b.
40 X 60
W = 100 X 2 x 50 = 10,000 Ib.
H p _KS (M + W) __ 0-45 x 40 (11,867) ‘ ~ 33,000 33,000 ~
= 6-44
(6) Inclined conveyor, of same type and capacity as above, rising at an angle of 30 degrees with the horizontal.
M =■ 1867 Ib. W = 10,000 1b. Sin 30° = 0-5
Cos 30° = 0-866
j l p __MS sin 30° KS (M + W) cos 30° ’ 33,000 + 33,000~
_ 1867 x 40 X 0-5 045 X 40 X 11,867 x 0-866
33,0 00 33,000
= 1-13 + 5-58
= 6-71
The horse-powcr required may be more accurately arrived at by taking into con-sideration tliose items which have been neglected (owing to tlieir comparatively small influence) in the above treatment. An alternative methocl is, therefore, given below.
First assume the conveyor to be horizontal, and let—
L — effective length in feet.
W — weight per foot (in 1b.) of the moving parts which convey the material.
F = coefficicnt of friction between the moving parts and the guides.
A = angle of repose of the material on the conveyor trough—35 deg. for coal, 45 deg. for coke.
C = capacity of conveyor in tons per hour.
w = weight of material per foot carried by conveyor.
S = speed of conveyor in feet per minute.
P = pull on conveyor chain.
Then, for a drag-bar coke conveyor the work done in foot-pounds per minute is as follows:—
(a) Work done in driving moving parts only
— 2 LWFS ft. 1b. per minute.