Basics
Sprocket: NT14 (PCD: 142.68)
Sprocket: NT30 (PCD: 303.75)
Speed Reducer (i=1:60)
Motor with brake
Roller Chain: SUPER 1OO
Spocket: NT14 (PCD: 171.22)
Roller Chain: SUPER 120 (Chain Speed=6.2 m/min.)
M=3,000kg {W=3,000kgf}
Sprocket: NT14 (PCD: 171.22)
Figure 4.14 Example of a Hanging Chain Machine
EXAMPLE: You are planning to use a hanging transmission machine like the one shown in Figure 4.14. Determine if you can use SUPER120 for hanging and SUPER100 for the drive chain. The power source is a 3.7-kW motor (with brake). The motor shaft rotational speed is 1,500 rpm. Step 1.Check the motor characteristics.
Rated torque: T n = 0.038 kN • m Starting torque: T s = 0.083 kN • m Braking torque: T b = 0.096 kN • m Motor moment of inertia: I m = 0.015 kN • m Step 2. Calculate the chain tension based on load.
The chain tension Fw = M = 3,000 3 9.80665 3 10 - 3 = 29.4 kN Service factor Ks = 1.3 (with some shock)
The chain speed coefficient Kv = 1.02 (from the chain speed 6.2 m/min.) Coefficient for number of sprocket teeth Kc = 1.28 (14-tooth sprocket) Coefficient of unbalanced load Ku = 0.6 (two sets of chains) Determine the chain design tension F'w = Fw 3 Ks 3 Kv 3 Kc 3 Ku = 29.4 3 1.3 3 1.02 3 1.28 3 0.6 = 29.9 kN Step 3.Calculate the chain tension based on the motor loading. Calculate moment of inertia of motor shaft.
V ( ) 2
(
) 2
6.2
= 0.0013 kg •m 2
I = M 3
= 3,000 3
2 π n 1 2 3 π 3 1,500 Moment of inertia of motor I m = 0.015 kg •m 2 Inertia ratio R = I / I m = 0.087
44
Powered by FlippingBook