Design of Flat Belt Drive System for Conveyor System

2. Design of Flat belt drive system:
Requirements:
Power, P = 50 KW
Application: Conveyor system
Given data:
2 Pulley equal diameters
Mass, m = 0.5 Kg/m
Friction coefficient, μ = 0.4
T0 = 10000 N
θ = 2.5132 degree = 0.0438636148 Rad
To find:
Design of flat belt drive system.
Solution:
Now,
T1
T2
= T t−T c
T s−T c
=eμθ
.................. eqn 1
Where,
Tt is the tension in tight side.

Ts is the tension in slack side.
T t +T s = 2 T0 = 20000 ..................... eqn 2
T c=M V 2 =0.5 V 2 ................................eqn 3
From equations 1
Tt - Tc = (Ts - Tc) (1.017700)
= (20000 −¿ (Tt - Tc)) (1.017)
Tenstion on tight side , T t= 20354−1.0177 T c
2.0177
Tension on Slack side ,T s=20000−T t
To find the centrifugal force:
WKT,
V C= √ T 0
3 m=81.649 m/s
T c=M V 2 =0.5 V 2=3333.333 N
Now,
T t = 8406.43 N
T s = 11593.56 N

Now,
P= ( Tt −T s ) V
V = P/ (T t−T s) = 15.6 m/s
Calculation of Pulley diameter:
The velocity and the diameter ratios is given as (Kulkarni, R., et.al 2016):
V = 3.14 (D) (N) / 60
Now,
D is the diameter of the pulley.
N is the speed in rpm = 720 rpm (Assumed)
V is the velocity in m/s : 15.6 m/s
D = 0.413 m = 413 mm
The pulleys are of equal diameter so that D1 = D2 = 413 mm
The Pulley ratio, i = 1 (as both diameters are same)
Calculation of Design power:
Design power is given as (Han, F., et.al 2018),
D p = Rated power X Load correction factor
For conveyor system the LCF is taken as 1.2