Horsepower Requirements, Torsional Ratings for Conveyor Screws, End Thrust.
HORSEPOWER REQUIREMENTS, HORIZONTAL SCREW CONVEYORS
The horsepower required to operate a horizontal screw conveyor is based on proper installation,
uniform and regular feed rate to the conveyor and other design criteria.
The following factors determine the horsepower requirements of a screw conveyor operating under the foregoing conditions.
C = Capacity in cubic feet per hour.
e = Drive efficiency.
Fb = Hanger bearing factor.
Fd = Conveyor diameter factor.
Fm = Material factor
L = Total length of conveyor, feet.
N = Operating speed, RPM (revolutions per minute).
W = Apparent density of the material as conveyed, lbs, per cubic foot.
The
horsepower requirement is the total of the horsepower to overcome conveyor friction (
HPf) and the horsepower to transport the material at the specified rate (
HPm), divided by the total drive efficiency, e, or:
HPf =
LN
Fd Fb------------
1,000,000
HPm =
CLW Fm
------------
1,000,000
Total HP =
(
HPf +
HPm)
---------------
e
Note: Inclined Auger HP Requirements are higher than for horizontal augers.
It is apparent that with conveyor capacity, size, speed, and length, all known that factors Fm,
Fd and
Fb are quite important. Small changes in these factors cause significant changes in the required horsepower. A discussion of these factors follows.
The factor
Fb is related to the friction in the hanger bearing, due to rubbing of the journals in the bearing metal and including, for sleeve type hanger bearings, an allowance for the entry into the bearing of some foreign material. This factor is empirically derived.
Factor
Fd has been computed proportional to the average weight per foot of the heaviest rotating parts and to the coupling shaft diameter.
The factor Fm depends upon the characteristics of the material. It is an entirely empirical factor determined by long experience in designing and operating screw conveyors. It has no
measurable relation to any physical property of the material transported.
While it is good procedure in conveying of bulk materials to run the conveyor until it is empty, prior to a work stoppage, frequently conveyors must of necessity be stopped while fully loaded. In that event, starting the conveyor again may possibly cause a serious overloading of the Drive and motor. The characteristics
of the material have much to do with the restarting of a fully loaded screw conveyor. Some materials will settle and pack or otherwise change their "as conveyed" characteristics. For example, Portland cement may take on the characteristics of a solid. Granulated sugar may pick up moisture from the
atmosphere and form a crust or cake. These situations will require a larger than normal driving motor.
It is important that a conveyor system operate as demanded by its controls. Start-up conditions or temporary overloads should not cause interruptions in service, so all components of the drive, as well as, the motor, should be chosen accordingly.
It is generally accepted practice that most power transmitting elements of a screw conveyor be sized and selected to handle safely the rated motor
horsepower. If, for example, a screw conveyor requires 3.5 horsepower as determined by the horsepower formula, a 5 horsepower motor must be used, and it is
desirable that all power transmitting elements be capable of safely handling the full 5 horsepower. However, on a screw conveyor made up of several lengths of conveyor screw, only the drive shaft has to handle the full motor load. The succeeding screw lengths and couplings only have to handle loads proportionate to the distance these parts are from the drive shaft. For economy, ease of design and maintenance, it is usual to select conveyor couplings, coupling bolts and other rotating parts such that all are of the same size and interchangeable, even if they are a bit larger than necessary.
NOTE: The foregoing load carrying requirements really constitute a minimum shock loading, metal fatigue from 24 hour per day continuous service, etc., must be considered in addition.