Question

5) [40 pts) Tensions in a flat belt drive A 5-in wide polyamide F-1 flat belt is used to connect a 2-in diameter pulley to drive a larger pulley with an angular velocity ratio of 0.5. The center-to-center distance is 8 ft. The angular speed of the small pulley is 1800 rev/min as it delivers 2 hp. The service is such that a service factor of Ks of 1.25 is appropriate Find F, F, (F1)a, and F2, assuming operation at the maximum tension limit. (You must provide an answer before moving to the next part.) The value of Fis Ibf. The value of F; is Ibf. The value of (F1) is Ibf. The value of F2 is Ibf.

Answer 1

Given deta" Reynoldis number, Rea ved solution : pere, a a m% 6) Voy average velocity in (m/s) d Characteristie dimension in (m) dynamie viscositas o kinematic viscosity in (mals (ow strokes) >> density a) Units 요 V termi Voo Re m/tim m/ m% mas 1 (dimensionley) Su, ant of kinematic discosity ) is V tem represents (kinematic visusity 6 P= 1.25 bylms : de 0.5 m; Velocity (ga 35 ml M= 1.70x too Ns/ma evd Reynolds number Rea 1.29x 35 x 0.5 м 172x co-5 1312500 The flow is turbulent here. (134950> 2000) 0.072 0.012 Su, oefficient of drag, az Me= (Rejas ($12500 ks Co= 4.8x403 Ams! efficient of dag = 4.8*20-3

Od = 178.8068 3.120 rad Iso - v=A danning (n) 9 ep (= Quáp (0.5 = ? (24) Csep FQ) =261.569 4.761 Calculate the velocity of driving pulley, TT XO (ax %) * 1800 V = 942.678 forum Centrifugal tersion ) fca w 0.658 949.6789 92, 2 x 60 x 60 f = 0.8046 lst amo Torque (7) = 63025 Prona Pa Power delivered na speed les Service fuctor ; has design factor Ta 63025 & 2 x 1.25x 1 (800 T2 87.535 lbs.in ۸ 10

Go Pulley correction factor a velocity correction factor Wa = b fa Coc (ha = 5x 354 QIX a = 122.5 lbs) com Differene in force, کادو Sf = 2T 2x 87 1535 13 dodoniring SE= 87.55 lbfit fia ~Af 12215- 87.535 force, fac to = 32.965 16t can force h Chitro f 2 122.5+ 34.986 - 0.8046 2 6 77.928 lbs Mwis 0.8046 Abt fja = 122.5 ka fi $77.928) this fa - $31.966 Mt