Question

Figure 1 shows the rotary screen being driven by a 3-phase Brook Crompton DOL motor via a compound gear train having an overall efficiency of 85% and velocity ratio of 4:1. The two gears 1 and 3 have 16 teeth each and the two larger gears 2 and 4 have the same number of teeth (same size). All the four gears have module of 6 mm and pressure angle of 20°. The rotary screen requires a torque of 388.2 Nm at a speed of approximately 184 rpm and it operates for 8 hours per day under steady load conditions. The shaft AB has a diameter of 50 mm. -phase Motor Gear 1 Bearing A Bearing B Gear 3 Coupling Gear Shaft AB Ø50 mm Bearing Bearing D Gear 2 Rotary Screen Required Torque = 388.2 Nm Required Speed = 184 rpm Gear 4 Compound Gear Train Efficiency = 85% V.R. = 4:1 Figure 1 3190 N Gear 2 Activate Windows Go to Settings to activate Win 200mm 5096 N 400mm 200mm

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Answer 1

Given overall efficiency of compound gear train 85% velocity ratio of 4:1

required speed = 184 rpm

we know that velocity ratio = input speed /output speed

so rotational speed required by motor = (4/1 )X required speed

rotational speed required by motor = 736 rpm