Example 1 Let's take a non-reversible motor that is supposed to run at a speed of 300 rpm and deliver a torque of 10 Nm. How do you choose the most suitable motor for this application? 1. In this example, the required power P = π × torque × speed 30 = 3.14 × 10 × 300 30 = 314 W (0.314 kW) 2. Now you choose a motor with a suitable power level from the catalog. In this case, we take an M39 (with 0.39 kW). 3. Now let's look at the power curves for each motor variant from the M39 series and choose the one whose maximum power comes closest to the operating point. 4. The model M39-520 is best suited for these requirements because we can then work at a speed above the maximum power point, resulting in a higher starting torque and a more stable speed. Power [kW] 0,4 Torque [Nm] Air consumption [l/s] . . 0,3 0,2 0,1 24 20 16 12 8 4 200 400 M39-520 600 Speed (rpm) Air motor speed (rpm) 5. To check the selection, the operating point should be transferred to the power diagram shown in the Chicago Pneumatic catalog. ■ 1 Find the line with the suitable motor speed. ■ 2 & 3 Transfer the desired torque and speed to the diagram. ■ 4 The two curves should intersect below the speed/ torque curve When the operating point is drawn, it is often found that the motor needs to be adjusted slightly so that the oper- ating point is on the power curve. This can be achieved in two ways: By changing the air supply or changing the air pressure. Table 1 Correction factors Air Pressure (Bar) 7 6 5 4 3 (Psi) 101 87 73 58 44 Performance 1.13 0.94 0.71 0.51 0.33 Speed 1.01 0.99 0.93 0.85 0.75 Power torque 1.09 0.95 0.79 0.63 0.48 Air consumption 1.11 0.96 0.77 0.61 0.44 19000 3000 2400 1180 520 Output speed (rpm) 2 Air motor speed (rpm) 4 3 10 8 6 4 2 1 Chicago Pneumatic air motors 7 6 bar (87psi) 4 bar (58psi) Power - kW , Torque - Nm , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 520 1180 2400 3000 19000
