Alternative conveyor speeds and delivery rates to convey wheat Suppose we want to convey wheat having a density of 30 lb per cubic foot from a delivery truck to the top of a grain silo. From Table 10.26, we see that the maximum angle of incline for grain is 20° . If the vertical distance from the e

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Alternative conveyor speeds and delivery rates to convey wheat

Suppose we want to convey wheat having a density of [latex]30 \mathrm{lb}[/latex] per cubic foot from a delivery truck to the top of a grain silo. From Table [latex]10.26[/latex], we see that the maximum angle of incline for grain is [latex]20^{\circ}[/latex]. If the vertical distance from the entry point to the exit point of the conveyor is [latex]50^{\prime}[/latex], then a trigonometric calculation establishes that the conveyor will need to be [latex]146^{\prime}[/latex] long. Assuming conveyor sections are available in [latex]50^{\prime}[/latex] increments, a [latex]150^{\prime}[/latex] conveyor would be used.

Table 10.26 Maximum Angle of Incline for Various Bulk Materials (in degrees)
Bulk Material	Maximum Angle of Incline	Bulk Material	Maximum Angle of Incline
Ashes, dry	20°	Gravel, bank	18°
Ashes, wet	23°	Gravel, washed	12°
Briquettes	10°	Grain	20°
Coffee, green bean	10°	Sand, damp	20°
Cornmeal	22°	Sand, dry	15°
Coal, bituminous	20°	Salt, dry/coarse	20°
Coal, mine run	18°	Salt, dry/fine	11°
Earth, loose	20°	Steel trimmings	18°

Accepted Answer

If we use a belt that is [latex]18^{\prime \prime}[/latex] wide, the maximum belt speed is [latex]500 \mathrm{fpm}[/latex]. Further, at that speed, from Table [latex]10.27[/latex], the wheat will be conveyed at a rate of [latex](16.0)(500 / 100)[/latex], or 80 tons per hour. On the other hand, if we used a belt that is [latex]30^{\prime \prime}[/latex] wide, the speed could be 700 fpm, and the delivered capacity would be [latex](48.5)(700 / 100)=339.5[/latex] tons per hour. Let us select the [latex]30^{\prime \prime}[/latex] belt and the maximum belt speed to obtain a delivered capacity of approximately 340 tons per hour.

Table 10.27 Delivered Capacity in Tons per Hour(tph) per 100 fpm of Belt Speed for Various Belt Widths (in inches) and Material Densities (in pounds per cubic feet)
Belt Width	30 lb/[latex]ft^{3}[/latex] Density	50 lb/[latex]ft^{3}[/latex] Density	100 lb/[latex]ft^{3}[/latex] Density
18″	16.0	26.5	53.0
24″	31.0	52.0	104.0
30″	48.5	81.0	162.0
36″	68.0	113.5	227.0

Based on a conveyor length of [latex]150^{\prime}[/latex], a [latex]30^{\prime \prime}[/latex] belt width, and a belt speed of [latex]700 \mathrm{fpm}[/latex], from Table [latex]10.28[/latex], the horsepower required to drive the empty conveyor is [latex](0.9)(700 / 100)[/latex], or [latex]6.3 \mathrm{hp}[/latex]. For a delivered capacity of 340 tons per hour, the additional horsepower required to convey the material horizontally over the [latex]150^{\prime}[/latex], length of the conveyor is found in Table [latex]10.29[/latex] by interpolating between 300 tph and 350 tph; the result is [latex]3.1[/latex] hp. Lifting the wheat [latex]50^{\prime}[/latex] at a rate of [latex]340 \mathrm{tph}[/latex], from Table [latex]10.30[/latex], will require [latex]17.2 \mathrm{hp}[/latex]. Hence, the total horsepower requirement is

[latex]H P=(6.3+3.1+17.2)(1.2)=31.92 \mathrm{hp}[/latex]

Table 10.28 Horsepower Required to Drive Empty Conveyor per 100 fpm of Belt Speed for Various Belt Widths (in inches) and Conveyor Lengths (in feet)
Belt Width	50′	100′	150′	200′	250′	300′	400′	500′
18″	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.1
24″	0.5	0.6	0.7	0.8	0.9	1.0	1.3	1.5
30″	0.6	0.7	0.9	1.0	1.2	1.3	1.6	1.9
36″	0.7	0.9	1.1	1.3	1.5	1.6	2.0	2.4

Table 10.29 Additional Horsepower Needed to Convey Material Horizontally at Any Speed Based on Various Conveyor Lengths (in feet) and Delivered Capacities (in tons per hour)
Conveyor Length	50 tph	100 tph	150 tph	200 tph	250 tph	300 tph	350 tph	400 tph	450 tph	500 tph
50′	0.3	0.6	0.9	1.2	1.5	1.8	2.1	2.4	2.7	3.0
100′	0.4	0.8	1.1	1.5	1.9	2.3	2.7	3.0	3.4	3.8
150′	0.5	0.9	1.4	1.8	2.3	2.7	3.2	3.6	4.1	4.5
200′	0.5	1.1	1.6	2.1	2.7	3.2	3.7	4.2	4.8	3.5
250′	0.6	1.2	1.8	2.4	3.0	3.6	4.2	4.8	5.5	6.1
300′	0.7	1.4	2.0	2.7	3.4	4.1	4.8	5.5	6.1	6.8
400′	0.8	1.7	2.5	3.3	4.2	5.0	5.8	6.7	7.5	8.3
500′	1.0	2.0	3.0	3.9	4.9	5.9	6.9	7.9	8.9	9.8

Table 10.30 Additional Horsepower Required to Elevate Any Material at Any Speed for Various Handling Rates (in tons per hour) and Various Lifting Heights (in feet)
Handling Rate (tph)	5′	10′	15′	20′	25′	30′	40′	50′
25	0.2	0.3	0.4	0.5	0.6	0.8	1.0	1.3
50	0.3	0.6	0.8	1.0	1.3	1.5	2.0	2.5
75	0.4	0.8	1.1	1.5	1.9	2.3	3.0	3.8
100	0.5	1.0	1.5	2.0	2.5	3.0	4.0	5.1
125	0.6	1.3	1.9	2.5	3.2	3.8	5.1	6.3
150	0.8	1.5	2.3	3.0	3.8	4.5	6.1	7.6
175	0.9	1.8	2.7	3.5	4.4	5.3	7.1	8.8
200	1.0	2.0	3.0	4.0	5.1	6.1	8.1	10.1
225	1.1	2.3	3.4	4.5	5.7	6.8	9.1	11.4
250	1.3	2.5	3.8	5.1	6.3	7.6	10.1	12.6
300	1.5	3.0	4.5	6.1	7.6	9.1	12.1	15.2
350	1.8	3.5	5.3	7.1	8.8	10.6	14.1	17.7
400	2.0	4.0	6.1	8.1	10.1	12.1	16.1	20.0
450	2.3	4.5	6.8	9.1	11.4	13.6	18.1	23.0
500	2.5	5.1	7.6	10.1	12.6	15.2	20.1	25.0

Note that the horsepower required to drive the empty conveyor is about twice the horsepower required to move the grains horizontally! Also, not surprisingly, by far the largest horsepower requirement is due to lifting the grains vertically by [latex]50^{\prime}[/latex]. If motors come in increments of [latex]5 \mathrm{hp}[/latex], we conclude that a [latex]35 \mathrm{hp}[/latex] motor would be required if the conveyor runs at [latex]700 \mathrm{fpm}[/latex] to deliver 340 tons of wheat per hour.

By slowing down the conveyor, the horsepower requirement can be reduced. However, slowing down the conveyor would also lengthen the time required to unload the delivery truck. Suppose a rate of 300 tons per hour is deemed feasible for transferring the wheat to the silo. With a [latex]30^{\prime \prime}[/latex] belt, a belt speed of [latex](100)(300 / 48.5)=618.56 \mathrm{fpm}[/latex] is required. At that speed, the horsepower requirement is reduced to

[latex]H P=(5.567+2.7+15.2)(1.2)=28.16 \mathrm{hp}[/latex]

or [latex]30 \mathrm{hp}[/latex], if motors come in [latex]5 \mathrm{hp}[/latex] increments.

Question 10.59: Alternative conveyor speeds and delivery rates to convey whe...

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