Question

Two C180 V-Belts are used in a drive consisting of a 9 inch driving sheave rotating at 1275 rpm, and a 27 inch driven sheave.

a) Find the center-to-center distance.

b) Find the power capacity of the drive assuming a service factor of 1.0 and a design factor of 1.

c) Find the tight and loose side tension in the belts, assuming operation at full power capacity.

d) What is the recommended preload?

Answer 1

XVIII Date: Page: Here, Given data is :- diameter of driving sheave = 9 inches. (Di). - RPM af deiving sheave = Ni = 1275 RPM diametee of driven sheave =D2 = 27 inches We know that for v pulleys, D2= Di (I-s). VR. [a= slip factol = 0.01 -0.03 V = Velocity Ratio 27= 9(1-0-02). VR 27 = 9x 0.98. VR 8.82 Vro 3.061. 27 - VR There, Ni a 3.061 N2. 1275 - N2 > 3.062 5) N22417 rpm (approx) 1. a) | fol Centre to_centre distance (c) We know that c = 1.58 D2 WR123 » c = 1.SX 27 (3.06173 => c=4005 lus

Date: Page: => c = 27.93 2 28 inches Gay). 6) Power capacity of drive We know that D1 = (38-42). M+2) = _9x2.54 x10. z 40 (Mtu) => 9x25.42 473) 3. [mm]. [Mt: Torque]. tincn = 25.4mm] - ya 1 » 5.715 = (MA) cubing both sides A (5:715) P=[(MAJES 311866 = MAL They torque by sheave 1 is We know that 1866 Nm, to Power = 25 N, MA Power transmitted og sheave 1 =) p = 2x3.14x1275x18666 L with service and I to design factor 1 - P = lugu 106.2 Go 2 P= 24901.77 watts 2. p= 24.9 kW. [1kW 21000 w). TETY

Date: Page: Tight and loose side Tensions and I let f and fo be the tensions at tight loose sides respectively We know that Design Power = (fi-f2) xv. - looo V = TDINI = 3,14 x 9x25.4 x 1275 Goxl000 m toto Put v= 15.25 - 15:25 mai in ③ we get 24.9 kw): (Fy-fz) x 15:25 -> (f1-f2) = 16.32-7_N. 0. looo Also - ما هي f2 . I where O= Angle of contact on smallee pulley and le u fu= coeff of friction sin/s/2 ogrove angle] Por calculating o. G = A -a sin D2-D1 26 a 180°– 20in1 27-9- 2x28 = 180°- 2 sint 18 56

O2 180°- 2 sin-1 (0.321) >> or 180°-2x18.74_ => o=180-37-48 - o=142-52° 2) O= 2048 rad free = 0.0175 rad I Me = u sin 8/2 u ee rubber material / t .ß = atom all > ß f atan (0.3). - ß = 2x1607 & B F 34° =) Me= 003 le = sin 34 0.3 bin 17 2 -- U N TUVATTIOSAAN

Date: Page: le = 0.3 0029 2 Me=1.03. Put the values of Me and o in eum, 1.03x2.48 a 280 a fia 1208 F2 put value of fi in ev - 12-8f2 -622 1632-4 N. 1 11-8f2 = 1632.7 N > F2 1632-7 a 138.36 N. 18 Put Value of f2 in . W fi – 138, 36 = 163 2.7. afiz 1771:06 N. ربیباک Tensions on and Tension Tight side = 1771 N. loose side = 138.36 N. on
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Preload is given by fp = ft- fc [ft- Tight side Tension] where fca ciray m ecente le=fi fz 3 1771.06-138.36_ z 1632.7 2 ») fpz 1771.06-1632.7 -) ерт 454 N Hence preload zasun