Selection and installation of TKM48 hyperbolic gear reducer on mining conveyor belt

In the production operation of the mine, the conveyor belt system is a key link in material transportation. As an important component of the conveyor belt drive system, the correct selection and installation of the TKM48 hyperbolic gear reducer is of great significance for ensuring the stable operation of the conveyor belt, improving production efficiency, and reducing maintenance costs. This article will elaborate on the selection and installation process of KM48 hyperbolic gear reducer on mining conveyor belt through a practical case.

2、 Project Background

A large mining site needs to upgrade and renovate one of its main ore conveyor belts. The conveyor belt has a conveying length of 500 meters, a bandwidth of 1.2 meters, and a conveying speed of 2.5 meters
At a speed of meters per second, the conveyed material is ore, with a maximum packing density of 2.5 tons per cubic meter. The working environment is relatively harsh, with dust, moisture, and certain vibrations.

3、 Selection process

Calculate load torque

Firstly, it is necessary to calculate the load torque of the conveyor belt. According to the formula: torque=force x force arm. On the conveyor belt, the force is mainly composed of the gravity and friction of the material.

The gravity of a material=material volume x packing density x gravitational acceleration

Material volume=bandwidth x tape length x material thickness

Assuming a material thickness of 0.2 meters, the material volume is 1.2 meters x 500 meters x 0.2 meters=120 cubic meters

Material gravity=120 cubic meters x 2.5 tons/cubic meter x 9.8 meters/second ²=2940 kilonewtons

Friction force=coefficient of friction x positive pressure

Assuming a coefficient of friction of 0.3 and a positive pressure equal to the gravity of the material, the frictional force is 0.3 × 2940 kN=882 kN

The force arm is the radius of the drum, assuming it is 0.3 meters

Then the load torque=(2940+882) kN x 0.3 meters=1146.6 N · m

Consider starting and braking torque

Due to the significant torque impact generated by the conveyor belt during start-up and braking, a certain safety factor needs to be considered. Usually, the starting torque is 1.5-2 times the load torque, and the braking torque is equal to the load torque
1.2-1.5 times. Assuming a starting torque coefficient of 1.8 and a braking torque coefficient of 1.3, the starting torque=1146.6 N · m x 1.8=2063.88
Nm, braking torque=1146.6 Nm x 1.3=1490.58 Nm