Technical handbook for the paddy rice postharvest industry in ...

59 CONVEYINGHorsepower requirements may be determined by using the following formulas.The determination does not consider power loss in drive equipment (belts, chains, orgear reducers), imperfect alignment, or the power required for starting under load.Additional power is therefore required for the average installation to overcome drivelosses and imperfect alignment.(DS + QK)(1) H= L1,000,000Where L = overall length in feetD = factor depending on typeof bearings (Table 3.6)S = speed in rpmQ = quantity of paddy inpounds per hourK = material factor, forpaddy = 0.4H×P(2) hp=0.85Where P = 2 when H is less than 1P = 1.5 when H is between 1and 2P = 1.25 when H is between 2and 4P = 1.1 when H is between 4and 5P = 1 when H is greater than5 and 0.85 is estimatedefficiency of the driveA sample problem:Determine conveyor size, speed, and horsepower requirements to move 20 tpaddy/ hour over a distance of 24 m.Solution:From Figure 3.22 (20 t/ h × 1,000 kg/ t) ÷ 576 kg/ m 3 = 34.72 m 3 / h), a 30-cmscrew at 92 rpm would be adequate. Then, assuming self-lubricating bronzebearings from Table 3.6, D = 96, H = (24 × 3.281) × (171 × 150 + 44,080 × 0.4) ÷1,000,000 = 3.41 then:3.41 × 1.25hp = =5.010.85Table 3.6. “D” factors in computing horsepower for screw conveyors.Conveyordiam(cm)Ball orroller“D” factor for type of hanger bearingsWood, babbitt,bronze ormoldedfabricSelflubricatingbronzeWhiteiron ormanganesesteel7.5101523253035401012183238557810615213354669613518624335496114171255336355080130160250350480