1. Coupling Selection and Things to Pay Attention in Power Transmission Systems
Fittings called COUPLINGS have been used to transmit power and connect drive elements to each other for a long time. Especially in our country, low-cost fittings that do not comply with standards have been used for drive systems for a long time. As complicated, expensive and important power transmission systems have been popular recently, quality couplings that conform to standards have started to be reviewed in more detail. Therefore, selection of the most suitable coupling has been one of the priorities in R&D.

Besides, couplings, which act as a fuse for power transmission elements such as motor, pump or reducers, have made users feel this function more and more day by day and allowed a precise and successful torque-power transmission by compensating axial, angular and radial offsets.

As mentioned above, coupling selection is highly significant for the whole system. Also, they act as a low-cost fuse element compared to the whole system. Things to consider for a correct coupling selection and the coupling selection process are described below.

Coupling is a component that transfers the rotating movement generated in a power source (motor, etc.) to another system (reducer, machine, pump, conveyor band, etc.). Besides rotation, couplings;

• Should not cause loss of power
• Should damp vibrations
• Should not transfer potential knocks to the power supply (motor, etc.)
• Should break in case of overload and act as a fuse that protects the motor or other components

2. Coupling Selection
2.1. Couplings are selected according to two torque values.

TKN= Nominal Torque TKmax= Maximum Torque

HKTM – Mechanical Automation and Sales - Power Transmission Systems (Couplers, Gears)

2.2. Drive is calculated according to nominal torque value.
TN=9550 X P(kW)/n(rpm)

TKN≥ TN X St

HKTM – Mechanical Automation and Sales - Power Transmission Systems (Couplers, Gears)

2.3. For maximum torque

TKmax≥ Ts X Sz X St + TN X St

Drive side torque Ts=TAS X MA X SA

Torque on the driven side Ts=TLS X ML X SL

MA= JL/( JA+ JL) M= JA/( JA+ JL)

3. Example Coupling Selection

Driving Side

Electric motor size = A.C. Motor 315 L P= 160 kW

Speed (n) = 1485 rpm

Inertia Torque on Drive Side (JA) = 2.9 kgm2

Initial frequency (z) = 6 1/h Sz= 1.0

Ambient Temperature = 60°C St=1,4

Driven Side

Screw Compressor Nominal Torque (TLN) = 930 Nm

Inertia Torque on Driven Side (JL) = 6.8 kgm2

Calculation
TAN (Nm)= 9550 X PAN (kW)/nAN (rpm)

TAN= 9550 X 160/1485=1029 Nm

TKN≥TLN X St TKN≥ 930 X 1.4 = 1302 Nm

The coupling that has the closest higher value to the resulting Nominal Torque value can be selected. Another thing to consider here is the TKmax value. Maximum torque value is usually twice as nominal torque.

4. Conclusion

Power transmission systems used today are similar to a human heart for all industries. Elements running on this system must be carefully and meticulously selected, run for long years and must not cause stoppages. Couplings are essential for these systems and at the same time they are affordable insurance tools. Choosing the correct coupling is an issue that increases and affects the efficiency and health of the system.