Technical Data Reducers Worm Gear Reducers Selection
Selection example
Follow these procedures for sizing a TD series reducer, or when the operating conditions in “Sizing table 3” fall outside sizing table 1 or 2 for a EWJ/EW/SWJ/SW series reducer.
Conditions
- ・Application ：Agitator (pure liquid)
- ・Motor ：15kW、1450r/min
- ・Output shaft speed ：24r/min
- ・Output shaft torque ：4000N・m {408kgf・m}
- ・others ：Vertical, hollow output shaft
- ・Output shaft load ：Axial load only 18000N
- ・Operating hours ：10 hours/day
- ・cycles ：1 time/hour
- ・Ambient Temperature ：30℃
Selection
1.Determine compensation factor
From the table of Load Categories by application, pure liquids have uniform load (U), which gives a service factor (Table1), Sf = 1.0.
2.Determine compensation kW and compensation torque
Use the service factor and load torque to obtain the compensation torque.
Correction torque = 4000N・m × 1.0 = 4000N・m
3.Determination of reduction ratio
Use the motor speed and output speed to obtain the reduction ratio.
Reduction ratio = 1450r/min ÷ 24r/min ≒ 60
4.Determine a size
From the Transfer Capacity table, select the size that satisfy es the compensation torque.
Size ：TD175 Nominal reduction ratio ：60(At input of 1450 r/min, output torque is 4785N・m)
Checking the equivalent thermal capacity for TD Series.
[Equivalent thermal capacity ]
Look up the temperature compensation factor (Table 3) where the ambient temperature of 30°C arrive at f1 = 1.0.
Equivalent thermal capacity = 4000N・m × 1.0 = 4000N・m
5.Check shaft load
Check that the axial load on the output shaft is within the allowable load.
Axial load ：18000N < allowable axial load = 34255 N, and is acceptable (hollow output shaft ：H output 59 r/min or less )
6.Calculate required input kW
Look up the required motor capacity (kW) for TD175-1/60 in the Transfer Capacity table.
Required input kW = 14.2kW×4000N・m 4785N・m × 1.0 = 11.87kWgives 15 kW, and is acceptable.
From vertical mount (V type), hollow output shaft (H), your selection is Model: TD175H60VRF (LF).
