Auger Horsepower and Capacity Reduction

Horsepower Interpolation

Example by Length:

Given: a 34’ horizontal auger, 8” diameter running at 430 rpm, (15% corn) 2,160 bph.

Per the chart, 30’ takes 2.8 hp in 15% corn.

34 / 30 = 1.13

1.13 x 2.8 = 3.16 hp required

Example by Capacity:

Given: The above auger, 8” x 34’ at 430 rpm but instead of 2,160 bph, we need 2,500 bph.

2,500 / 2,160 = 1.16

1.16 x 3.16 = 3.67 hp required

Capacity Reduction Factor

	CAPACITY REDUCTION FACTOR
ANGLE OF INCLINE	CORN & WHEAT	SOYBEANS & SUNFLOWERS
0	1.0	1.0
5	0.98	0.97
10	0.96	0.94
15	0.94	0.90
20	0.92	0.87
25	0.89	0.84
30	0.87	0.80
35	0.85	0.77
40	0.83	0.74
45	0.80	0.70
50	0.77	0.65
55	0.74	0.63
60	0.70	0.60
65	0.67	0.57
70	0.64	0.54
75	0.60	0.50
80	0.57	0.47
85	0.54	0.44
90	0.50	0.40

The horizontal capacity of the equivalent diameter and length of auger can be multiplied by this number to determine the incline capacity at the same horsepower and rpm.

Auger Capacity Per 100 Revolutions

The capacities listed below are based on clean, dry corn at approximately 90% load on a horizontal operation. Grain conditions, grain type, and loading methods will change capacities listed below. (bph = bushels per hour)

Auger Diameter	Capacity Per 100 rpm (90% load)
4”	60 bph
6”	240 bph
8”	480 bph
10”	1200 bph
12”	2000 bph

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.