Question #313144

4. A pulley 600 mm. in diameter transmits 40 kW at 500 rpm. The arc of contact between the belt and pulley is 144° having a center distance of 1100 mm. Determine: (a) diameter of the driven pulley for open belt drive, (b) speed of the driven pulley, and (c) belt length for open belt connection.


1
Expert's answer
2022-03-19T02:46:19-0400

P=2πNT60P =\frac{2\pi NT}{60}

T=40000×602π×500=763.94T =\frac{40000\times 60}{2\pi\times500 } = 763.94 Nm

(a) Diameter of the driven pulley for open belt drive

sinβ=Dd2Csin \beta = \frac{D-d}{2C}

D=Sinβ×2C+d=sin144×2200+600=1893.13D = Sin \beta \times 2C +d =sin 144 \times 2200 + 600 = 1893.13 mm

(b) Speed of the driven pulley

nN=Dd\frac{n}{N}=\frac{D}{d}


n=N×Dd=500×1893.13600=1577.611578n = N\times \frac{D}{d}= 500\times \frac{1893.13}{600}= 1577.61 \approx 1578 rpm


(c) belt length for open belt connection


L=2C+π(D+d)2+(Dd)24C=2×1100+π(1893.13+600)2+(1893.13600)24×1100=6496.24L = 2C+\frac{\pi(D+d)}{2}+\frac{(D-d)^2}{4C} =2\times 1100+\frac{\pi (1893.13+600)}{2}+\frac{(1893.13-600)^2}{4\times1100}=6496.24 mm




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