Question

Determine the number of V-belts required to transmit a power of 60 kW at 750 rpm from an electric motor to a compressor shaft at 300 rpm. The larger pulley diameter is 1500 mm. The centre distance is 1650 mm, and overload factor is to be taken as 1.5. A belt with cross- sectional area of 350 mm², density of 1000 kg/m and having an allowable tensile strength of 2 MPa is available for use. The coefficient of friction between the belt and the pulley may be tak- en as 0.28. The groove angle of the pulley is 35. An open belt configuration is adopted.

Answer 1

/ 2 Given! Power p = 60 kW = 60 x10²w motor speed N, = 750 rpm Compressor shaft speed,Mg 200 rpm, The larger pulley ella loro compressor side pulley odometer da = 1500 mm centre distance x = 1650 mm, overlood factor a 1.5; belt cross section area, a=350 min density of belt material, I = 1000 kg/m3 Allowable fensile strength Co-efferent of Archon, lfs 0.28 Groove angle of the pulley, 2 B = 85' objective ! To find the number of robelt regurred howing open belt configuration Solunon Ta belt, 02 - 2 Mpa, M L - 22 on Ti K 5 ik 2 N x= 1650 mm Mofor pulley compressor pulley Ni= 750 rpm Nga 300 rpm di da i 1500 mm Ni da » d = da XN2 1500 x 300 N2 = 600 mm 750 From the geometry, sind - N2n 이212 x x => sind a 750 - 300 = 0.27.27 1650 => do Stra' (0.2727) 5 15.8269 180° -22 wrap angle of smaller pulley, => 0 = 180 - (2x15.826) = 148.348 = 2,589 voeltorn li 1° T/180 Veella) Ň 2. 6 nol dila Let T Tight side tension of belt To = slack side tension of belt

e 1 28 .. el H s e Sin 1715 - 60 60 Ratio of tensions for v-belt is given by: T, Cello. cosec Tz where 8B = groove angle of pulley > B = 35,= 17:59 (0.28 x 2.6 x csec 17.5') e cosec 17:52 >> ㅍ 11: 256 To The belt velocity is; V V = V = TTCM TTX600X750 Ve 23561.94 mm/s = 23.56 m/s sine v > 10 m/s, there for centrifugal tension will Te = my? where m= Area x length x density = (300 x 106)m x 1m x 1000 Kolm 0.35 bey / meter length Tc: 0135 X (23-5632 194 · 275 N Also the mansmam belt tension is; To Stress x area 2 N/mm x 350 mm 700N Tight side tension Ti = T- Tc = 700- 194.275 N act. mom 2 རབ .. m = 505.72 AN and the slack side tension T2-T 11.256 3 505724 44 93 N 1256 -) T2 -

P 3 W Also power transmitted, P= (to - 72.) xv >> P = (505.724 - 44.93) * 23.56 Watt 10856. 10.856 kW This is the power which ss to be transmitted one belt But the rated power is = 60 KW with over lood factor of 1.5 The designed power GOX 15= 90 kW - The number of vabelts required The designed power Power of one belt n = 90 10.856 8129 ñ 9 (Ars.)