Use this useful chart to estimate the required length of your auger when trying to achieve a specific discharge height and angle of auger incline.
Thursday, April 21, 2022
Friday, July 21, 2017
A good source for online grain (& feed) handling calculators
There can be a lot of math associated with planning for a grain or feed handling system. Sudenga Industries, who makes equipment for those applications, posts a lot of their common formulas and interactive calculators on a page of their web site called "Sudenga University". Check it out Sudenga University.
Here is a list of their current offering of calculators as of July 2017. I happen to know that they are always adding more though so bookmarking their page might be a good thing.
Here is a list of their current offering of calculators as of July 2017. I happen to know that they are always adding more though so bookmarking their page might be a good thing.
ONLINE CALCULATORS
- Calculate electric motor horsepower required to drive a hydraulic pump
- Calculate belt length and distance between pulleys
- Calculate Auger RPM
- Convert BPH to TPH (U.S. or Metric)
- Calculate feed density
- Convert cubic feet to tons
- Calculate feed truck capacity needed to obtain given tonnage
Friday, May 13, 2016
How Tall Should Your Bucket Elevator Be?
A bucket elevator is a sound investment if your goal is to streamline throughput and minimize trailer unloading time at your grain bin site. Benefits of an elevator leg can include higher capacity with lower horsepower requirements than a comparable capacity auger system, higher reliability factors and lower long term maintenance requirements.
Maybe you are thinking about putting up a bucket elevator and want to get some rough estimates on costs? Maybe you've gotten bids from several contractors to put up a bucket elevator, but the bids came back with different elevator heights? What should it cost to put up a bucket elevator? To answer these questions, first you need to know the answer to the question, "how tall should my bucket elevator be?"
The height of your bucket elevator is determined by three factors: Required spout angle. How far away do you need to spout? How tall is your highest discharge point?
You'll find that there are a lot of opinions about what is correct, and what you might "get away with" when it comes to spout angle. The safe rule of thumb is, dry grain typically flows reliably 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°.
You might hear some say, "you don't need to be at 45° with your leg spouts. You can get by with 43° or even 40°. Working within these shallower spout angles is a way for some to shorten the height of the leg they are bidding. Buyer beware! While you might save a couple bucks putting up the elevator leg that is the shortest, it is generally not the place to try and save money. Remember, if the bucket elevator you install is too short, it is very difficult (read:expensive) to fix later. If your material can't efficiently flow down your spouts at the rated capacity of the rest of the upstream system, you'll have to slow all the equipment down upstream. This obviously reduces your throughput and increases your unload time. In a particularly bad scenario, overfeeding a spout at too shallow of an angle can cause plugging of the bucket elevator. This is something people strive to only do once!
I think you get my point. Here is one of the easiest ways I know to calculate how tall your bucket elevator should be.
In the illustration below, assume the grain tank shown is the furthest point away from where you'd like to place your bucket elevator. For a 45° spout angle, dimension A and B must equal one another. For a 60° spout angle, dimension B should equal dimension A x 1.75. All other discharge points (closer to the leg) can generally be assumed to be at spout angles steeper than your shallowest spout angle.
Maybe you are thinking about putting up a bucket elevator and want to get some rough estimates on costs? Maybe you've gotten bids from several contractors to put up a bucket elevator, but the bids came back with different elevator heights? What should it cost to put up a bucket elevator? To answer these questions, first you need to know the answer to the question, "how tall should my bucket elevator be?"
The height of your bucket elevator is determined by three factors: Required spout angle. How far away do you need to spout? How tall is your highest discharge point?
You'll find that there are a lot of opinions about what is correct, and what you might "get away with" when it comes to spout angle. The safe rule of thumb is, dry grain typically flows reliably 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°.
You might hear some say, "you don't need to be at 45° with your leg spouts. You can get by with 43° or even 40°. Working within these shallower spout angles is a way for some to shorten the height of the leg they are bidding. Buyer beware! While you might save a couple bucks putting up the elevator leg that is the shortest, it is generally not the place to try and save money. Remember, if the bucket elevator you install is too short, it is very difficult (read:expensive) to fix later. If your material can't efficiently flow down your spouts at the rated capacity of the rest of the upstream system, you'll have to slow all the equipment down upstream. This obviously reduces your throughput and increases your unload time. In a particularly bad scenario, overfeeding a spout at too shallow of an angle can cause plugging of the bucket elevator. This is something people strive to only do once!
I think you get my point. Here is one of the easiest ways I know to calculate how tall your bucket elevator should be.
In the illustration below, assume the grain tank shown is the furthest point away from where you'd like to place your bucket elevator. For a 45° spout angle, dimension A and B must equal one another. For a 60° spout angle, dimension B should equal dimension A x 1.75. All other discharge points (closer to the leg) can generally be assumed to be at spout angles steeper than your shallowest spout angle.
Example for a 45° grain spout: The furthest point you want to spout to is 50 feet from where you want the base of the bucket elevator. The peak (spout discharge point) of your furthest grain tank is 30 feet tall. So, 30 feet (grain tank height) + 50 feet (dim A) = a leg discharge height of 80 feet. Keep in mind that you should plan to make the bucket elevator's discharge height slightly higher than that because typically you will have a grain distributor, or other spout connection accessories that bring the discharge point of the leg down a bit. A good rule of thumb is to plan for an extra 6 to 10 feet of discharge height on the leg, putting the leg in this example at a discharge height of 86 to 90 feet.
I hope this is somewhat clear and that you find this information helpful. Admittedly, this is a very involved topic that often needs to factor in site specific information. Hopefully, I've kept you moving in the planning process of improving your grain handling system. If in doubt, I always encourage people to find a respected millwright or agricultural contractor that 1.can provide solid references and 2. you can get along with. They'll be able to come out to your site, take measurements and draw your system to scale to give you a lot of this information.
Got questions about spout length? I've posted a drawing and the formula for how to calculate that here.
Size matters when it comes to your leg spouts. For a handy chart concerning what to expect for capacity through different sizes and diameters of grain spouts, click here.
Wednesday, May 4, 2016
Free Calculator for Belt Length and Pulley Distance
Calculate belt length for an electric motor/drive belt
A common challenge experienced by many farmers, millwrights or grain handling professionals, you've broken a belt on a piece of grain unloading equipment. Looking closer, you realize that the belt is worn to the point that you can't read the size. You've got to get back up and running fast, so the belt needs to be replaced. How can you figure out what size belt you need? Sudenga Industries has come up with a handy online calculator to do just that. Just plug in your info and you'll know what size belt to order.
Click here to go to the Sudenga online belt length calculator
Monday, March 16, 2015
What Am I Dealing With? Auger Sheaves, Pulleys and Belts!
First of all, what is the difference between a "sheave" and a "pulley"? Check out this description on Wikipedia. In short, it comes down to minor details and the words are often used interchangeably.
Sheaves and Belts
Sheaves are a common means of transmitting power from an electric motor or gas engine to the drive member. Sheaves are available in many diameters which allow a screw conveyor (also known as an auger), to be run at many different RPM's.
"B" section belts carry the load from the motor sheaves to the drive. The number of belts that are required to transmit the load depends on the pitch diameter of the driver (or motor pulley) and the type of belts used.
Two types of "B" section belts are used. The ratings between the two types differs greatly.
The gripnotch or cog belt features a notched section on the inside diameter of the belts. This allows the belt to coil around the smaller motor sheave instead of flexing the fibers of the belts which causes heat, creating a loss of horsepower. Also, the sides of the belt are cut, not molded which creates a raw gripping edge.
Because of high horsepower ratings per belt, gripnotch belts should not be used on aluminum or pressed steel pulleys.
The 5L belt is a FHP, or fractional horsepower belt. This belt is molded and relies mainly on tension of the belt to transmit the load. The 5L belt is usually recommended for small loads where single belt only applications are required. This is because 5L belts are not made in matched sets as are gripnotch belts, which would be required on multiple groove applications. In short, because 5L belts are not matched, exact length cannot be guaranteed from belt to belt making tensioning difficult.
Proper belt tension is important. Belts that are too loose will slip and wear prematurely. If tightened too tight, the belts will stretch or create an overhung load on the bearings of the reducer. Refer to the specifications provided by the manufacturer of your belt for instructions on how to properly tension the belt(s) on your equipment.
Sheaves and Belts
Sheaves are a common means of transmitting power from an electric motor or gas engine to the drive member. Sheaves are available in many diameters which allow a screw conveyor (also known as an auger), to be run at many different RPM's.
"B" section belts carry the load from the motor sheaves to the drive. The number of belts that are required to transmit the load depends on the pitch diameter of the driver (or motor pulley) and the type of belts used.
Two types of "B" section belts are used. The ratings between the two types differs greatly.
The gripnotch or cog belt features a notched section on the inside diameter of the belts. This allows the belt to coil around the smaller motor sheave instead of flexing the fibers of the belts which causes heat, creating a loss of horsepower. Also, the sides of the belt are cut, not molded which creates a raw gripping edge.
Because of high horsepower ratings per belt, gripnotch belts should not be used on aluminum or pressed steel pulleys.
The 5L belt is a FHP, or fractional horsepower belt. This belt is molded and relies mainly on tension of the belt to transmit the load. The 5L belt is usually recommended for small loads where single belt only applications are required. This is because 5L belts are not made in matched sets as are gripnotch belts, which would be required on multiple groove applications. In short, because 5L belts are not matched, exact length cannot be guaranteed from belt to belt making tensioning difficult.
Proper belt tension is important. Belts that are too loose will slip and wear prematurely. If tightened too tight, the belts will stretch or create an overhung load on the bearings of the reducer. Refer to the specifications provided by the manufacturer of your belt for instructions on how to properly tension the belt(s) on your equipment.
Calculating Drive Belt Length and Belt Horsepower Capability For An Auger
To determine the drive belt length required for an auger:
-Measure the center to center distance between the auger input shaft, and the motor shaft. (center to center distance the belt will have to span between the drive pulley and the driven pulley)
-Many auger manufacturers' catalogs often offer this measurement in the specifications section of their information.
(C) = Center to Center Distance
D = Drive Pulley Diameter
d = Driven Pulley Diameter
Note: The pitch length equation gives the required circumference for the belt at the midsection. The pitch diameter must be cross referenced in Chart X to determine the proper length for ordering belts.
Note: If the pitch length determined from the equation falls between the lengths given in chart No. X, choose the next larger belt.
-Measure the center to center distance between the auger input shaft, and the motor shaft. (center to center distance the belt will have to span between the drive pulley and the driven pulley)
-Many auger manufacturers' catalogs often offer this measurement in the specifications section of their information.
(C) = Center to Center Distance
D = Drive Pulley Diameter
d = Driven Pulley Diameter
Note: The pitch length equation gives the required circumference for the belt at the midsection. The pitch diameter must be cross referenced in Chart X to determine the proper length for ordering belts.
Note: If the pitch length determined from the equation falls between the lengths given in chart No. X, choose the next larger belt.
Note: Table XI shows how many horsepower a belt can handle based on pulley size. As the chart shows, belt wrap determines, to a great extent, the horsepower carrying capability of each belt.
Monday, May 6, 2013
Calculating Horsepower For an Incline Drag Conveyor
Use the following formula to estimate required horsepower for an incline drag (en-masse) conveyor.
1. Determine overall ground length of drag conveyor
2. Multiply LENGTH X BUSHELS PER HOUR
3. Divide results by 55,000 - Write this total down.
4. Determine the discharge height in feet and multiply by the bushels per hour desired.
5. Divide the result by 26,500. - Write this total down.
6. If a curve section is used, divide the BPH desired by 1325.
(If no curve section is used, ignore step 6.)
7. Add the sums of the individual horsepowers to find the total horsepower.
The above formula has been found to be adequate for products such as shelled corn and soybeans. HEAVIER PRODUCTS MAY REQUIRE ADDITIONAL HORSEPOWER.
Formula example:
Overall length of the example conveyor is 50 feet. The discharge height of the conveyor is 7 feet from the bottom of the conveyor. There is a curve section used.
50 feet x 5000 bushels per hour = 250,000 divided by 55,000 = 4.5455
7 foot conveyor discharge height x 5000 bushels per hour = 35,000 divided by 26,500 = 1.3208
5000 bushels per hour divided by 1325 (curve section factor) = 3.7736
Add 4.5455 + 1.3208 + 3.7736 = 9.6399 total required horsepower (Round up to the nearest motor means this conveyor requires a 10 HP electric motor.)
1. Determine overall ground length of drag conveyor
2. Multiply LENGTH X BUSHELS PER HOUR
3. Divide results by 55,000 - Write this total down.
4. Determine the discharge height in feet and multiply by the bushels per hour desired.
5. Divide the result by 26,500. - Write this total down.
6. If a curve section is used, divide the BPH desired by 1325.
(If no curve section is used, ignore step 6.)
7. Add the sums of the individual horsepowers to find the total horsepower.
The above formula has been found to be adequate for products such as shelled corn and soybeans. HEAVIER PRODUCTS MAY REQUIRE ADDITIONAL HORSEPOWER.
Formula example:
Overall length of the example conveyor is 50 feet. The discharge height of the conveyor is 7 feet from the bottom of the conveyor. There is a curve section used.
50 feet x 5000 bushels per hour = 250,000 divided by 55,000 = 4.5455
7 foot conveyor discharge height x 5000 bushels per hour = 35,000 divided by 26,500 = 1.3208
5000 bushels per hour divided by 1325 (curve section factor) = 3.7736
Add 4.5455 + 1.3208 + 3.7736 = 9.6399 total required horsepower (Round up to the nearest motor means this conveyor requires a 10 HP electric motor.)
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