Agitator Gear Box

Ratio = motor rpm ÷ desired agitator rpm.

The given numbers don't make any sense.

So, there's no consideration of the agitator speed and its relationship with the vessel contents and the power they need, then? Just the Mechanical Engineering considerations only, then? No consideration of the motor size and the power drawn from the supply at all at the revised speed, then?

The big question is:

• Will the motor turn the agitator sufficiently at the revised speed to provide the agitation power the process needs?

Until that has been considered, put the Mechanical considerations to one side.

• Has anyone done any pilot-scale trials, for example?
• And if the revised speed is lower, how about installing a variable speed drive instead?
• What are the true process needs?

So, how must i do, can i use auxiliary gearbox as maindrive, without auxiliary gearbox. it's means i just intall one gearbox in agitator

First your numbers didn't make sense; now your words don't make sense. No one will know how to use an auxiliary gearbox without an auxiliary gearbox.

What rpm(s) do you want to turn the agitator, and what rpm(s) of motor do you have? Do you want a choice of two speeds, or just one?

Dear Mr dedek,

your speed ratio appears to have some error.

Pl. inform correct data specifically.

DHAYANANDHAN.S

are you driving the auxiliary gearbox from the main gearbox ? I can imagine this if the ratio required at the output is not obtained with only one box, not enough reduction.

This application requires a little clarification.

Dear Dedek,

I think that the "speed ratio" is messed up, maybe you multiplied the input rpm by the "i" (reduction factor) of the gearbox, instead of divide. Maybe.

The gearbox ratio is given usually as "i=x" or "1:x" being x the reduction factor.

i.e: a gearbox drive of 1000 rpm with i=5, would have an output speed of 200rpm.

If there were one more reduction stage, with i=4, the output speed would be 200/4= 50rpm.

I understand from your information that:

1. You have one agitator.

2. The drive, could be a motoreducer with two stages?. (please clarify)

3. You want to remove one of the stages.

And that raise the questions, as stated by PWS previously:

a. What are the process requirements of speed rate for this agitator? (a minimum required fluid speed at the tank to avoid settling or ensure mixing, a maximum speed allowable to avoid undesirable effects like air , foaming, reactions, or simply unnecesary power consumption)

b. Are you aware that removing one stage of reduction, the speed will increase as per the reduction stage removed, and power demand will increase? (and non linearly!)

c. This increase in speed can result as well, in damage of the mechanical parts if they are not properly sized for the new conditions (increased axial load, more chances for vibration/radial load due to unbalanced impeller spinning faster).

Salam,

Abel