Calculate Horsepower For Inclined Screw Conveyors

Screw conveyors can be used to convey bulk materials at an inclined angle, but several areas of design should be examined first.

Because of the incline of the screw, several conveying problems begin to occur.
1. The horsepower per unit of material increases
2. The efficiency of moving the material forward decreases.

A U-trough conveyor, because of its shape allows the material to flow back over the top of the screw when the angle of incline increases. This problem is complicated by the presence of the hanger bearing which speeds up this fall back.

One way to avoid this is through the use of a tubular housing, or round tube conveyor. This prevents the material from riding above the screw and slows the turbulence caused by the fall back. Although the type of material being conveyed greatly effects the amount of fall back incurred, a general area where the U-trough conveyor starts to lose significant efficiency is 15 degrees of incline.

Round tube conveyors are available with and without intermediate hanger bearings. One instance would be in the conveying of stringy or fibrous materials where the material will wrap around the bearing and eventually cause plugging. Elimination of the hanger bearing may cause excessive deflection of the central screw pipe and cause a vibration or chatter as a result of the screw hitting the tubing. This can be avoided by the introduction of most materials which will tend to lift and support the screw, provided the material is introduced upon start-up of the conveyor and fills the entire length. Care should be taken to empty the conveyor before shut-down to avoid overload start-up conditions.

Due to the fallback created in inclined screw conveyors by tumbling and agitation, the amount of material being pushed forward through the conveyor is reduced. To offset this loss of capacity, increased screw rotation is applied. This increased rotation creates a greater forward material velocity, the net result being greater capacity.

In general, as the angle of incline of the conveyor increases, the greater the loss in the percentage of fill of the screw.

The flowability of a material affects the percentage fo fill greatly. The greater the flowability, the more quickly the screw pitches become filled before conveying the material along the length of the screw. Also, at some inclines, depending on pitch and diameter, a section of the helical flight is actually at a near horizontal plane. This tends to slice or sling the material outward, rather than to move the material forward.

Several things can be done to maintain the fill percentage of an inclined conveyor.
1. Completely cover the screw intake to prevent spatter and fallback.
2. Increase the length of exposed intake screw.
3. Force feed the incline conveyor by means of a pressure fed boot supplied by a horizontal intake screw.
4. Use a close tolerance between screw and tube.
5. Increase RPM (revolutions per minute).

In general the horsepower and capacities of incline conveyors depends on the characteristics of the material being conveyed and may require consulting the manufacturer.

Horsepower, Inclined Conveyors
To calculate the horsepower of the inclined conveyor, use the method as described for horizontal conveyors. See this post.

HPF =
L X N X Fd X Fb
------------------
1,000,000

HPm =
C X L X W X Fm
------------------
1,000,000

Inclined conveyors require additional calculations to measure the horsepower required to elevate the material.

HPh =
C X W X H
-------------
1,980,000

Where: H = Discharge height in feet
Total Incline HP =
(HPf + HPm + HPh)
---------------------
e

Hanger Bearing Factors (Fb):
Ball Bearing - 1.0
Babbitt - 1.7
Bronze - 1.7
*Graphite bronze - 1.7
*Canvas base phenolic - 1.7
*Oil impregnated bronze - 1.7
*Oil impregnated wood - 1.7
*Plastic - 2.0
*Nylon - 2.0
*Teflon - 2.0
No bearings - 4.0
"*" = Non-lubricated bearings, or bearings not additionally lubricated.
Conveyor Diameter and Diameter Factor (Fd)
4.0 - 10
5.0 - 13
6.0 - 16
8.0 - 24
10.0 - 33
12.0 - 49