Design Project: 50 KW Flat Belt Drive System for Conveyor System

This project focuses on the design of a flat belt drive system for a conveyor application requiring 50 KW of power, utilizing equal diameter pulleys. The design process includes calculating tensions on the tight and slack sides of the belt, determining centrifugal force, and selecting appropriate belt...Read More

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 =0.5V 2 …………………………..eqn 3
From equations 1
Tt - Tc = (Ts - Tc) (1.017700)
= (20000 −¿ (Tt - Tc)) (1.017)
Tenstion ontight side , Tt = 20354−1.0177T c
2.0177
Tension on Slack side , T s=20000−Tt
To find the centrifugal force:
WKT,
V C= √ T 0
3 m=81.649 m/ s
T c=M V 2 =0.5V 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

Dp = 50 (1.2) = 60 KW
Selection of belt:
From PSG design data book, pg 7.54, Fort Duck belting is selected. The capacity
of the belt is 0.0289 KW/mm/ply.
No. of plies:
Depending upon the pulley diameter and the speed of the belt the number of plies
require is taken from the standard PSG design data Pg 7.52 as 8. (Book, D.D.,
1971)
No of plies required: 8
Width of the belt:
The belt width is given as:
W = DP
Load rating X No. of plies
Load rationg = Load rating at 10 m/s X (V/10) = 0.04358 KW /mm/ply.

Width of the belt = 172 mm
Determination of Pulley width:
The pulley width should be greater than belt width by 25 mm. (Book, D.D., 1971)
Pulley width: 172 + 25 = 197 mm: 200 mm
Belt length:
L = 2C + π
2 ( D+d ) +(D−d)2
4 C
Here,
D: d = 413 mm
C: 3m (Assumed)
Length of the belt required: L = 1303 mm
References:

Book, D.D., 1971. PSG college of Technology. Coimbatore-641, 4.
Han, F., He, R., Yan, H. and Xiong, F., 2017. Lateral motion of the endless flat
belt in a two-pulley belt system. Advances in Mechanical Engineering, 9(4),
p.1687814017695955.
Kulkarni, R., Patel, B., Mulay, C., Musale, P. and Bhaskar, R.V., 2018. Design of
Stairlift for Curved Path.