Question

- 300 min = 800 kgm = 2.0 m/s = 0.0725 = 2.0 = 80 = 260 m = 260 m = 255 m = 1.0 The following data refers to a horizontal belt conveyor used for conveying a coal in a therm Standard electric motor ratings: 5, 5.5, 7.5, 10, 11, 12.5, 15, 20, 22, 25 kW power station : Capacity of conveyor Density of coal Belt speed Surcharge factor for polyamide belt, Ċ Number of plies for polyamide belt Material factor for plies, K, Belt tension and arc of contact factor for belt, K, Material conveying horizontal length Centre distance between drive pulley and tail pulley Centre distance between snub pulleys Ratio of tail pulley to drive pulley diameters Ratio of snub pulley to drive pulley diameters Mass of each carrying run idler Mass of each return run idler Pitch of carrying run idlers Pitch of return run idlers Friction factor for idlers, fc Snub factor for snub pulleys, €, Snub factor for drive and tail pulleys, €2 Material velocity component along belt line Frictional resistance due to belt cleaner near drive pulley where B = belt width, m Angle of lap on drive pulley Coefficient of friction between belt and drive pulley Ultimate tensile strength for belt per unit length (width)of ply Drive efficiency Standard belt widths : Belt Width, mm 500 600 750 800 900 1000 Mass Per Unit 7.75 9.3 11.6 12.4 Length of Belt, kg/m 14.0 15.5 Electric motor speed = 0.5 = 25 kg = 20 kg = 1 m = 2.5 m = 0.02 = 0.03 = 0.06 = 1 m/s = 100 B.N = 210° = 0.4 = 60 Nm = 939 1200 / 1400 / 1600 18.6 21.7 24.8 = 1440 r.p. Determine the following parameters for belt conveyor : (i) the standard belt width; cs Scanned with CamScanner

Answer 1

5 P3 P4 e P2 li Tail Drive Pulley la pulley of Belt Layout conveyor. 1. Belt width (B) = M = 300 x 103 2 m/s P = 800 kg/m3 C = 0.0725 Υ M = P Q M= PC (0.98 -0.05]? V X 3600 kg/hr 300 x103 - 800 x 0.0725 *[0.98 -0.05]? x2 *3600 :: B = 997. 3 mm B-1000 mm Standard Hence, belt selected as wichth is of drive CD) - 2. Diameter pulley Zp=3 Ki= 2 K2= 80 minimum diam of drive pulley is given a → The au, D = ko. K2. 2P 2x 80 X3 D-480 mm CS Scanned with CamScanner

3. Reduction ratio of gear no 1440 rpm reducer CG)- V= 2 m/s V= TDN 60 2 = TT X0:48X + N 60 N = 79.58 pm 18.096 Gun N 1440 79.58 4. Number of and return nun idlers - camping 4 : 235 m tc - 1 m 12 = 260 m ty : 2.5m Return nun idlers (zr)- tra di (Zr+1) 2.5 = 255 Zrt! Z7= 101 carrying nun idlers (zo) - tca ta Zc+I 1 = 260 Zct Zc: 259 CS Scanned with CamScanner

5. Belt belt- conveyor tensions along i] At initial Fi = Fslack ji] At point 1:- point is a FCL = 100 B N F1 - Fi + FCL fslack + loo B F1 - fslack + 100 x1 Fi = (Fslack + 10o) b iii] At point 2:- Epi = 0.03 F2 - Fi + fpi Fi + Epi Fi C1 +0.03) C Islack Floo) = F2 - 1-03 Fslack + 103 с iv] At point 3 = mai = 20 kg fcs 0:02 los 255 m Zr= lol Mb = B = = looo mm, 15.5 for kg/m F3 = Fat Frr F2 + fc ть + Пhi Zr li Edge (1.03 fslack + 103) + 0.02 (15.5 + 20x 101 ) 9.81 x 255 255 F3 = 1.03 fslack + 1274.8 N d CS Scanned with CamScanner

v] At point 4: Ep2 = 0.03 F4 = f3 +Fp2 F3 + Epz F3 CI+ 0.03) x (1.03 fs lack + 1274.8) f4 = 1.0609 islack + 1313.05 vi] At point 5- EP3 - 0.06 Fs F4 + Fp3 F4 + Ep3 F4 Ci to.06) x (1.0609 Fslack + 1313.05) f 1.1245 54 Fslack + 1391.83 N Fs = vii] At point Vi= 1 m/s M= 300 x103 3600 83.33 kg/s co F6 = Fs + FL fs + M (V-VI) (1.124554 fslack + 1391 · 833) + 83.33 (2-1) F6 - 1. 124554 fslack + 1475.17 N viii] At point 7 - 83.33 41.665 2 mb: 15.5 kg/m 5kg mm = mci: 25 Zc: 259 cs Scanned with CamScanner

77 = FG + for F6 + fc + mci Zc tmb t mm maiz)gta + 1475.17-) + 0.02 (41.665 65 + 15.5+ (1.124554 fslack 25 x 259 X 9.81 X 260 260 h F7s 1.124554 Fslack + 5661.61 N ix] At final point fa EP4 = 0.06 Fright fy t Fp4 F7 + Ep4 Ft C1 +0.06) x (1.124554 Folack + 5661.67) - 1.192 fslack 60 01:37 fright tensions om drive belt pulley 6] Ratio of y effective = u= 0.4 210 XT 3.6652 rad 180 0.4 X 3.6652 e MO ftight Fótack 4.3322 Ftight fslack 7] Effective best tensions drive pulley: 4.3322 fslack Fright: iting (0) substituting ego get 4.3322 Fslack - 1192 fslack + 6001.37 in we Fslack = 1911.14 N CS Scanned with CamScanner

Fright 4-3322 Fslack = 4.3322 x 19.11.14 Ftight 8279.44 N 8] Power required on drive ση (Po)= pulley Pou (Fight - Fslack) v looo (8279:44 - 1911-14) x2 looo Po 12. 7366 Kny C Pida g) Input power to belt Conveyor no 0.93 12-7366 0.93 Pia Po n Pi= 13.7 kW 10] Power rating of Standard electric motor selected 15 kW 11] Available factor of Sut - 60 N/mm CNF) = B = 1000 mm Zp=3 safety in Belt Conveyor Femaz - Fright maximum tension in + Fc fright right t mbuz ftmax 8279.44 + 15.5 x 22 8341.44 N CS Scanned with CamScanner

Breaking Strength of belt is Conveyor Fos Sut B ZP 60 x 1000 X3 180000 N Available Factor of safety 180000 NF Fos Ftmar 8341.44 NF = 21.58 cs Scanned with CamScanner