This is the formula you use to convert bushels per hour to tons per hour:
BPH x 1.25 = Cubic Feet Per Hour
Cubic Feet Per Hour x Pounds Per Cubic Feet = Pounds Per Hour
Pounds Per Hour / 2000 = Tons Per Hour
Common Assumed Per Cu Ft Weights:
Corn = 45 lbs per cubic ft
Feed = 35 lbs per cubic ft
Wheat = 48 lbs per cubic ft
Pellets = 55 to 60 lbs per cubic ft
(I highly recommend you VERIFY the weight of the material you will be conveying)
The above formula takes the form of an interactive calculator at sudenga.com: Click here to use this calculator.
Friday, July 11, 2008
Wednesday, January 30, 2008
Important Bucket Elevator Formulas
TO FIND LEG CAPACITY:
1. DETERMINE BELT SPEED IN FEET PER MINUTE
a. Motor RPM x motor pulley diameter, divided by driven pulley diameter = Input shaft speed to drive.
b. Divide Input shaft speed by Drive reduction ratio (15:1, 25:1, etc.) This gives the head shaft RPM.
c. Multiply the head shaft RPM x head pulley diameter in feet, x
3.1416. This gives the Belt speed in feet per minute.
2. FIND THE NUMBER OF CUPS FILLED IN ONE MINUTE:
a. Multiply feet per minute of belt speed x 12”, divided by cup spacing in inches.
b. Find cup capacity from Manufacturer’s chart in cubic feet.
Use water level + 10% or 75% of gross cup capacity.
c. Multiply cups filled per minute x cup capacity in cubic ft. This gives capacity of leg in one minute. Multiply result x 60 minutes
for hourly capacity in cubic feet.
d. For Bushels per hour, multiply cubic ft. per hour x .8
HORSEPOWER FORMULA FOR BUCKET ELEVATORS:
1. DISCHARGE HEIGHT IN FEET X BUSHELS PER HOUR,
Divided by 33,000 gives BARE HORSEPOWER
2. Multiply Bare Horsepower x 1.25 (safety factor) to get DESIGN
HORSEPOWER.
3. This calculation is based on grain weighing 60# per bushel.
ALTERNATE HORSEPOWER FORMULA:
1. Multiply DISCHARGE HEIGHT IN FEET X POUNDS OF MATERIAL
RAISED IN 1 MINUTE. Divided by 33,000 gives BARE HORSEPOWER.
2. Multiply Bare Horsepower by 1.25 (Safety Factor) for DESIGN
HORSEPOWER.
Wednesday, January 9, 2008
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
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
Tuesday, January 1, 2008
Horizontal Auger Horsepower Requirements
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.
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.
Intermediate Auger Bearing Recommendations
Selection of bearing material for an intermediate hanger is based on experience, together with a consideration of the characteristics of the material to be conveyed. Normally, the bearing selection will be made from one of the following three bearing types:
A. Babbitted or Bronze Bearings
Lubricated babbitted bearings are very frequently used, but have a maximum operating temperature of 130° F; lubricated bronze bearings may be operated at temperatures up to 225° F. This temperature figure for bronze bearings may be exceeded by the use of appropriate high temperature lubricants. CARE MUST BE EXERCISED IN THE USE OF BABBITTED OR BRONZE BEARINGS WHEN THE CONVEYED MATERIAL MUST NOT BE CONTAMINATED BY THE PRODUCTS OR BEARING WEAR OR THE LUBRICANTS USED.
B. Self Lubricated Bearings
Self lubricated bearings are available in several types.
1. Oil impregnated hard maple wood has a maximum operating temperature of 160° F.
2. Oil impregnated sintered bronze has a maximum operating temperature of 200° F.
3. Plastic and reinforced fibre compounds are available in a wide variety of compositions and constructions, and can be obtained from many sources. They require no grease or oil lubrication and are usually run dry. They are best suited for use in conveyors wetted with water. Maximum operating temperatures vary with the composition and construction of the bearing. When appropriately used, the wear rate is usually low.
C. Ball Bearings
Ball bearings are preferably used when handling granular or pelletized materials not containing any fine powder. Maximum operating temperature is 225° F with petroleum based lubricants, or 270° F with high temperature synthetic lubricants. When appropriately used and sealed against loss of lubricant, ball bearings usually involve no contamination of the material conveyed.
A. Babbitted or Bronze Bearings
Lubricated babbitted bearings are very frequently used, but have a maximum operating temperature of 130° F; lubricated bronze bearings may be operated at temperatures up to 225° F. This temperature figure for bronze bearings may be exceeded by the use of appropriate high temperature lubricants. CARE MUST BE EXERCISED IN THE USE OF BABBITTED OR BRONZE BEARINGS WHEN THE CONVEYED MATERIAL MUST NOT BE CONTAMINATED BY THE PRODUCTS OR BEARING WEAR OR THE LUBRICANTS USED.
B. Self Lubricated Bearings
Self lubricated bearings are available in several types.
1. Oil impregnated hard maple wood has a maximum operating temperature of 160° F.
2. Oil impregnated sintered bronze has a maximum operating temperature of 200° F.
3. Plastic and reinforced fibre compounds are available in a wide variety of compositions and constructions, and can be obtained from many sources. They require no grease or oil lubrication and are usually run dry. They are best suited for use in conveyors wetted with water. Maximum operating temperatures vary with the composition and construction of the bearing. When appropriately used, the wear rate is usually low.
C. Ball Bearings
Ball bearings are preferably used when handling granular or pelletized materials not containing any fine powder. Maximum operating temperature is 225° F with petroleum based lubricants, or 270° F with high temperature synthetic lubricants. When appropriately used and sealed against loss of lubricant, ball bearings usually involve no contamination of the material conveyed.
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