Showing posts with label math. Show all posts
Showing posts with label math. Show all posts
Wednesday, August 15, 2012
Calculating Bucket Elevator Spout Length
Dry grain typically flows in a spout at an angle of 45° or more. High moisture grain, sunflowers and ground feed generally require spouts at a minimum angle of 60°. To calculate spout length, the following calculations apply:
Wednesday, June 13, 2012
Hydraulic Motor Size vs. RPM Calculation:
Some useful formulas are listed below for figuring out how to size a hydraulic motor for your auger or other grain handling equipment.
Gallons per minute of supply x 231
RPM = Cubic inches of motor displacement
Gallons per minute x 231
Motor cubic inches = RPM
RPM x cubic inches of motor displacement
GPM required = 231
PTO Shaft
A common question is, "How long can I expect my PTO shaft to last?" While there are many factors that contribute to that answer, there are a few rules of thumb that should be kept in mind.
Grease! Grease! Did I mention grease? The more often the better. The number one thing you can do to extend the life of your PTO shaft is to grease it often. Want to make sure it gets done? Hang a grease gun on the equipment with a PTO shaft.
Some information regarding how operating angle contributes to the life expectancy of a PTO shaft:
At 540 RPM and transmitting 40 HP, a category 4 PTO shaft at 22 degrees has a 12% decrease in life expectancy when the angle is changed by 3 degrees.
At 540 RPM and transmitting 24 HP, a category 3 PTO shaft at 22 degrees has a 22% decrease in life expectancy when the angle is changed by 6 degrees.
Gallons per minute of supply x 231
RPM = Cubic inches of motor displacement
Gallons per minute x 231
Motor cubic inches = RPM
RPM x cubic inches of motor displacement
GPM required = 231
PTO Shaft
A common question is, "How long can I expect my PTO shaft to last?" While there are many factors that contribute to that answer, there are a few rules of thumb that should be kept in mind.
Grease! Grease! Did I mention grease? The more often the better. The number one thing you can do to extend the life of your PTO shaft is to grease it often. Want to make sure it gets done? Hang a grease gun on the equipment with a PTO shaft.
Some information regarding how operating angle contributes to the life expectancy of a PTO shaft:
At 540 RPM and transmitting 40 HP, a category 4 PTO shaft at 22 degrees has a 12% decrease in life expectancy when the angle is changed by 3 degrees.
At 540 RPM and transmitting 24 HP, a category 3 PTO shaft at 22 degrees has a 22% decrease in life expectancy when the angle is changed by 6 degrees.
Thursday, April 5, 2012
Calculating Conveyor Angle
One could pull out their calculator and mess with sin, cos, tan, but when it comes to figuring out discharge height, or conveyor length, or angle of incline for an auger or conveyor installation, I just pull up this handy web page: http://www.csgnetwork.com/righttricalc.html
Enjoy!
Enjoy!
Thursday, March 15, 2012
Screw Conveyor Lump Size Limitations
The size of a screw conveyor not only depends on the capacity required, but also on the size and proportion of lumps in the material to be handled. The size of a lump is the maximum dimension it has. A closer definition of the lump size would be the diameter of a ring through which the lump would pass. However, if a lump has one dimension, much longer than its transverse cross-section, the long dimension or length would determine the lump size.
The character of the lump also is involved. Some materials have hard lumps that won't break up in transit through a screw conveyor. If that is the case, provision must be made to handle these lumps. Other materials may have lumps that are very hard, but degradable in transit through the screw conveyor, thus really reducing the lump size to be handled. Still other materials have lumps that are easily broken in a screw conveyor and therefore impose no limitations.
Three classes of lump sizes apply as follows:
Class I - A mixture of lumps and fines in which not more than 10% are lumps ranging from maximum size to one half of the maximum; and 90% are lumps smaller than one half of the maximum size.
Class II - A mixture of lumps and fines in which not more than 25% are lumps ranging from the maximum size to one half of the maximum; and 75% are lumps smaller than one half of the maximum size.
Class III - A mixture of lumps only in which 95% or more are lumps ranging from maximum size to one half of the maximum size; and 5% or less are lumps less than one tenth of the maximum size.
Table III shows the recommended maximum lump size for each customary screw diameter and the three lump classes. The ration, R, is included to show the average factor used for the normal screw diameters which then may be used as a guide for special screw sizes and constructions. For example:
Radial Clearance, Inches
Ratio, R = Lump Size, Inches
The allowable size of a lump in a screw conveyor is a function of the radial clearance between the outside diameter of the central pipe and the radius of the inside of the screw trough, as well as the proportion of lumps in the mix. The following illustration shows this relationship:
The character of the lump also is involved. Some materials have hard lumps that won't break up in transit through a screw conveyor. If that is the case, provision must be made to handle these lumps. Other materials may have lumps that are very hard, but degradable in transit through the screw conveyor, thus really reducing the lump size to be handled. Still other materials have lumps that are easily broken in a screw conveyor and therefore impose no limitations.
Three classes of lump sizes apply as follows:
Class I - A mixture of lumps and fines in which not more than 10% are lumps ranging from maximum size to one half of the maximum; and 90% are lumps smaller than one half of the maximum size.
Class II - A mixture of lumps and fines in which not more than 25% are lumps ranging from the maximum size to one half of the maximum; and 75% are lumps smaller than one half of the maximum size.
Class III - A mixture of lumps only in which 95% or more are lumps ranging from maximum size to one half of the maximum size; and 5% or less are lumps less than one tenth of the maximum size.
Table III shows the recommended maximum lump size for each customary screw diameter and the three lump classes. The ration, R, is included to show the average factor used for the normal screw diameters which then may be used as a guide for special screw sizes and constructions. For example:
Radial Clearance, Inches
Ratio, R = Lump Size, Inches
The allowable size of a lump in a screw conveyor is a function of the radial clearance between the outside diameter of the central pipe and the radius of the inside of the screw trough, as well as the proportion of lumps in the mix. The following illustration shows this relationship:
Friday, July 11, 2008
Bushels Per Hour to Tons Per Hour Formula
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.
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.
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.
Monday, November 5, 2007
Figuring Auger RPM
A very common problem. How do I figure my auger's RPM?
Use these formulas to figure auger RPM.
For straight pulley reduction:
Drive pulley x motor RPM = SUM ÷ driven pulley diameter = auger RPM
For auger with a reduction gearbox:
Drive pulley x motor RPM = SUM ÷ driven pulley diameter = SUM ÷ gearbox reduction ratio = auger RPM
To use the auger RPM you've figured to figure capacity, click here.
Use these formulas to figure auger RPM.
For straight pulley reduction:
Drive pulley x motor RPM = SUM ÷ driven pulley diameter = auger RPM
For auger with a reduction gearbox:
Drive pulley x motor RPM = SUM ÷ driven pulley diameter = SUM ÷ gearbox reduction ratio = auger RPM
To use the auger RPM you've figured to figure capacity, click here.
Wednesday, October 10, 2007
Auger Flight Rotation and Material Flow
The above diagrams are a simple means of determining screw rotation. When the material flow is in the direction away from the end being viewed, a RIGHT hand screw will turn counter clockwise and a LEFT hand screw will turn clockwise rotation as shown by the arrows.
How to Identify Right Hand or Left Hand Flighting
A picture is worth a thousand words. Below is a picture of how to tell if your auger has right hand, or left hand flighting.
A conveyor screw is either right hand or left hand depending on the form of the helix. The hand of the screw is easily determined by looking at the end of the screw.
The screw pictured to the left has the flight helix wrapped around the pipe in a counter-clockwise direction, or to your left. Same as left hand threads on a bolt. This is arbitrarily termed a LEFT hand screw.
The screw pictured to the right has the flight helix wrapped around the pipe in a clockwise direction, or to your right. Same as right hand threads on a bolt. This is termed RIGHT hand screw.
A conveyor screw viewed from either end will show the same configuration. If the end of the conveyor screw is not readily visible, then by merely imagining that the flighting has been cut, with the cut end exposed, the hand of the screw may be easily determined.Monday, October 1, 2007
Useful Grain Handling Sites
-Purdue University Post-Harvest Grain Quality & Stored Product Protection Program.
-Mycotoxins in Cereal Grains
-Grain Storage Management (Univ. of Neb. Lincoln)
-Iowa Grain Quality Initiative (Iowa State University)
-Univ. of Minnesota Post-Harvest Handling of Crops
-Oklahoma State Univ. Stored Products Research
-North Dakota State Univ. Grain Drying, Handling and Storage
-USDA ARS Grain Marketing and Production Research Center
-Kansas State University Grain Science Industry Extension Program
-Multistate Project NC-213 (Management of Grain Quality and Security)
-Mycotoxins in Cereal Grains
-Grain Storage Management (Univ. of Neb. Lincoln)
-Iowa Grain Quality Initiative (Iowa State University)
-Univ. of Minnesota Post-Harvest Handling of Crops
-Oklahoma State Univ. Stored Products Research
-North Dakota State Univ. Grain Drying, Handling and Storage
-USDA ARS Grain Marketing and Production Research Center
-Kansas State University Grain Science Industry Extension Program
-Multistate Project NC-213 (Management of Grain Quality and Security)
Friday, September 21, 2007
Thursday, September 20, 2007
Grain Handling Knowledge
It is difficult to find a place where one can look up all the math, and theory, and layout ideas for grain handling systems. So, in response, I am beginning this blog. I will post as much "free domain" info as possible on this site. Check back often, I hope to update frequently. Also, I'd love to hear from those of you out there that have new ideas or experiences that apply to this area of need in the agricultural industry.
Thanks for visiting! I hope you find this site useful.
Thanks for visiting! I hope you find this site useful.
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