Presentation on theme: "Frictional Drive in Textile Part 1:Drum drive, belt drive and chain drive."— Presentation transcript:
Frictional Drive in Textile Part 1:Drum drive, belt drive and chain drive
Frictional Drive 1.What is frictional drives 2.Types of frictional drives 1.Drum drive (eg cheese winding) 2.Belt drive (spinning, weaving) 3.Vee-belt drive (motor power ) 4.Chain-and-sprocket drive
1) Frictional Drives Many drives for textile machines rely on the friction between the driving element and the driven element. Examples: –Belt and rope drives; –Drum drives to cone in yarn preparation –Drum drives to warping beams The common feature is one element has a hard surface and the other is a relatively soft surface
Drum Drive Cone winding is a typical example of frictional drive Driver Driven
Drum drive Figure 1 shows the cheese sits on top of the drum (driven). The surface speed of drum (driver) is V=n1 x pi x d1 Assume no slippage between the drum and cheese, the surface speed will be: n1 x pi x d1 =n2 x pi x d2 n1 /n2 = d2 /d1
Drum drive The last equation shows that the ratio of the rotational speeds of the two elements is the inverse ratio of their diameter. It should be noted that the running speed (rev/min) for the empty cheese (smaller diameter) is higher than full cheese (bigger diameter).
Question 1 A drum has 12cm diameter and rotates at 2650 rev/min., find out –A) the actual winding rate (m/min) –B) the rev/min of the cheese at start (empty package) –C) the rev/min of the cheese at 18cm diameter (full package)
Answer 1 1) Actual winding rate: –V= n1 x pi x d1 –V= (2650 x 3.14 x 12)/100 =1000 m/min 2) Rev/min at start: –n1 x d1 x pi = n2 x d2 x pi –2650 x 12 = n2 x 5; n2=6360 rev/min 3) Rev/min at 18 cm: –2650 x 12 =n2 x 18; n2=1767 rev/min It should note that the rev/min for empty package is running faster than full package
2) Belt drive Flat belt drive Driver Driven Figure 2
Flat belt drive If the belt has no thickness, the surface speed: n1 x d1 x pi = n2 x d2 x pi If the thickness is t, the surface speed : n1 x (d1 + t) = n2 x (d2 + t)
Question 2 In Fig. 2, pulley A rotates at 120 rev/min and has diameter of 30 cm. Pulley B has a diameter of 50 cm. If the belt is 0.6 cm thick, estimate the number of rev/min of pulley B, with slippage ignored. Suppose 4% slippage occurred, what is the rev/min of pulley B?
Answer 2 n1 x (d1 + t) = n2 x (d2 + t) 120 x (30+0.6) = n2 x (50+0.6) n2=72.6 rev/min If 4% slippage, the pulley B will be 72.6 x 0.96% =69.7 rev/min
Vee-belt drive The flat belt drive as discussed has the limitation to the power that it can transmit. Owning to the “wedging” effect, the Vee-belt drive can trasmit four times as much power as the flat belt for the same cross-sectional area. Vee-belt drive is a common transmitting power on textile machine, e.g from the electric motor to the main shaft of the machine.
Vee-belt drive High transmitting power of Vee-belt drive
Question 3 Vee-belt The linear velocity of a vee-belt is 1200 m/min. If the pitch diameters of the motor pulley and machine pulleys are 16 and 36 cm, calculate the motor speed and the machine speed
Answer 3 Surface speed V= n1 x pi x d1 Linear speed = speed of pulley (rev/min) x pi x d1 1200 m/min =(n1 x 3.14 x 16)/100; n1=2385 rev/min. (motor speed) 1200 m/min = (n2 x 3.14 x 36)/100; n2=1060 rev/min (machine speed)
Chain and sprocket drive This is a positive drive with a driving sprocket connected to the driven sprocket by a roller chain. The driving chain from bicycle is an example of chain-and-sprocket drive. It is simple and versatile. (Rev/min of driving / rev/min of driven) = (no. of teeth of driven / no of teeth of driver)
Question 4 A sprocket of 24 teeth rotates at 56 rev/min and, by roller-chain drives, causes a sprocket of 40 teeth to rotate. Calculate the number of rev/min of the driven sprocket.
Answer 4 56 x (24/40) = 33.6 rev/min. One advantage of a chain drive is that several sprockets may be driven from one driving sprocket. One of these guide sprockets may be adjustable so that any stretch in the chain may be taken up.