Handling

Because of differences in shape, size, and consistency, each product must have a handling system capable of moving that specific product. The designer of a handling system also must consider product perishability and the desired form of the finished product. A harvester designed to harvest tomatoes for the fresh vegetable market will be different from one designed to harvest tomatoes used for catsup. Because of the prevalence of grains across the United States, they will be used to illustrate some of the basic principles of handling biological products.

Grains were one of the first products to be mechanically moved because they flow by gravity, are small, and have a relatively hard outer coat. These characteristics

Minimum distance

FIGURE 20.1. Auger conveyor.

Minimum distance

FIGURE 20.1. Auger conveyor.

allow them to be moved by different mechanical devices. Augers and pneumatic conveyors will be discussed in the following sections.

20.4. Augers

Augers are available in two types, Archimedean screw and belt. Of the two, the screw type auger is the most popular in agriculture and it is the one that will be used in the following discussion and sample problems. An auger is like a bolt, but instead of threads it uses flights that turn inside a tube. Another name for an auger is a screw conveyor. As the auger rotates, the flights move the product through the tube similarly to the way that threads move a nut on a bolt. Augers are available in several diameters and are capable of handling many different types of products. Augers have the advantage of requiring less horsepower per bushel and having fewer mechanical parts than pneumatic systems; but their disadvantages include the danger of the exposed auger at the inlet, and their requirement of more space because the inlet is some distance from the outlet, Figure 20.1.

Augers are selected on the desired capacity (bu/hr) and length requirements. The length is usually predetermined by the distance needed to move the grain, if horizontal, or by the discharge height and angle if the auger is at an angle. The selection criteria are primarily based on the auger diameter and speed because the capacity increases as the speed increases. The following discussion will illustrate how the capacity and energy requirements for augers can be determined.

Table 20.1 and Table 20.2 contain typical values for two sizes of screw augers and two different crops. This type of information can be used to make decisions in managing a grain handling system, such as determining the size of auger required to convey grain at a given rate (bu/hr).

Table 20.1. Screw auger capacity handling dry corn (12 in exposure).

Auger angle of elevation

Table 20.1. Screw auger capacity handling dry corn (12 in exposure).

Size

RPM

bu/hr

hp/10 ft

bu/hr

hp/10 ft

bu/hr

hp/10 ft

4 in

200

150

0.12

120

0.15

60

0.11

400

290

0.29

220

0.29

130

0.24

600

420

0.38

310

0.45

190

0.36

6 in

200

590

0.38

500

0.44

280

0.32

400

1090

0.56

850

0.88

520

0.70

600

1510

0.84

1160

1.28

740

1.05

Note: For total horsepower, 10% must be added for drive train losses.

Source: Structures and Environment Handbook, MWPS-1, Midwest Plan Service, Iowa State University, Ames, Iowa, 1987, Section 534.

Problem: What is the minimum size of auger that can be used to convey dry corn at the rate of 500 bushels per hour when the auger is inclined 45°?

Solution:

Auger angle of elevation

Auger

45°

90°

Size

RPM

bu/hr

hp/10 ft

bu/hr

hp/10 ft

bu/hr

hp/10 ft

4 in

200

150

0.12

120

0.15

60

0.11

400

290

0.29

220

0.29

130

0.24

600

420

0.38

310

0.45

190

0.36

6 in

200

590

0.38

500

0.44

280

0.32

400

1090

0.56

S50

0.SS

520

0.70

600

1510

0.84

1160

1.2S

740

1.05

Using Table 20.1, the minimum size of auger is 6 inches.

Table 20.2. Screw auger capacity handling dry soybeans (12 inch exposure).

Auger angle of elevation

Auger

45°

90°

Size

RPM

bu/hr

hp/10 ft

bu/hr

hp/10 ft

bu/hr

hp/10 ft

4 in

200

140

1.00

125

0.17

70

0.12

400

270

0.21

215

0.35

130

0.26

600

390

0.34

315

0.51

180

0.40

6 in

200

500

0.40

360

0.57

220

0.40

400

990

0.84

690

1.20

390

0.79

600

1350

1.20

930

1.71

500

1.10

Note: For total horsepower, 10% must be added for drive train losses.

Source: Structures and Environment Handbook, MWPS-1, Midwest Plan Service, Iowa State University, Ames, Iowa, 1987, Section 534.

Another use of Table 20.1 and Table 20.2 is to determine the horsepower required to operate the auger.

Problem: How much horsepower (including drive train) is required to operate a 100-ft, 6-inch auger, installed at 45°, conveying 690 bushels of soybeans per hour?

Solution:

Auger angle of elevation

Auger angle of elevation

Size

RPM

bu/hr

hp/10 ft

bu/hr

hp/10 ft bu/hr

hp/10 ft

4 in

200

140

1.00

125

0.17

70

0.12

400

270

0.21

215

0.35

130

0.26

600

390

0.34

315

0.51

180

0.40

6 in

200

500

0.40

360

0.57

220

0.40

400 600

990 1350

0.84 1.20

930

1.20

390

0.79 1.10

1.71

500

Using Table 20.2, the power requirement is 1.2 hp/10 ft. It is important to read the note at the bottom of Table 20.2. 10% must be added for drive train losses. Therefore:

The horsepower required by the auger, including the drive train, is 13 hp.

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