Grain Drag

[ptonsparky]

Ok. Another simple issue. Should be.
Millwrights have the angle of a grain drag to steep.. The corn rolls back down over the top of the paddles filling up the distributor drop tube which causes the grain to fall back down the leg. Motor of drag is just below FLA at this point. IDK what the leg amperage is. We stopped the process before it got that far.

Millwrights just want to speed the drag up saying that usually makes the current go down. Are they FOS or am I missing something?

 

[retirede]

FOS, unless I’m missing something.

It might be the case if the conveyor is fed at a constant rate, you would have less material between each paddle thus less material weight. But I don’t know if that’s a net gain?

 

[wwhitney]

Not familiar with the equipment, but will speeding up the drag somehow reduce the mechanical power output required? E.g. if this is like a bucket conveyor is there some mechanism whereby running it 50% faster makes each bucket less than 2/3s as full?

Since you mention falling grain (not clear what the distributor drop tube is), if speeding up the conveyor would successfully deliver grain at the same rate, and cause the same amount of falling grain, but cause it to fall quicker, i.e. get raised up less before falling, then that decreases the mechanical power output.

Just brainstorming here.

Cheers, Wayne

 

[ptonsparky]

A bucket elevator brings grain to the top of the leg and is dumped into a tube that directs the grain to the drag inlet. The drag is a series of paddles on a chain that drags the grain to an open discharge. The drag needs to be at an angle that does not allow the grain to start falling back to the inlet. A flat drag would move the most grain. An angled one the least. All else being equal.

An angled one with more rpm may move the same amount of grain but will it require a larger HP?
Not sure how to fit all that into W=fd. Too many years away from a physics class.

 

[retirede]

Not familiar with the equipment, but will speeding up the drag somehow reduce the mechanical power output required? E.g. if this is like a bucket conveyor is there some mechanism whereby running it 50% faster makes each bucket less than 2/3s as full?

Since you mention falling grain (not clear what the distributor drop tube is), if speeding up the conveyor would successfully deliver grain at the same rate, and cause the same amount of falling grain, but cause it to fall quicker, i.e. get raised up less before falling, then that decreases the mechanical power output.

Just brainstorming here.

Cheers, Wayne

We’re thinking along the same lines.

If speeding up reduces the amount of material between paddles, it may also stop the spilling over the paddles.

I’m still not sure if that’s a net reduction in required HP. Losses will definitely increase with higher speeds.

 

[synchro]

Can they reduce the speed of the bucket elevator somewhat so that the drag can keep up with the input flow without the spillover on the paddles? Maybe with a little speed increase on the drag as well.

 

[wwhitney]

Not sure how to fit all that into W=fd. Too many years away from a physics class.

In this case, for raising the grain, F = weight of the grain (the only resisting force is gravity) and d = change in height of the grain (the distance moved parallel to F). Of course, for a horizontal conveyor the above is 0, so this in addition to whatever power is required for the horizontal case to overcome the mechanism's internal friction. [Although I would expect this also would be increasing with the weight of grain on the drag.]

Thinking about it a little more, the above mechanical power is just the weight of the grain on the drag * the vertical component of the conveyor velocity (i.e. if it runs 1.4 ft/sec and is at a 45 degree angle, that's 1 ft/sec). For the steady state where the grain falls off the end at the same rate that it enters at the bottom, that would be constant. But with the grain falling down to the bottom, the weight on the drag is large than normal operating conditions, and the mechanical power demand is higher, assuming the motor power is sufficient to prevent the drag from slowly down.

So anything you can do to reduce the amount of grain on the drag while keeping the speed constant will reduce the mechanical power demand. If speeding up the drag reduces the amount of grain on it by a greater factor than the speed increase (by preventing the accumulation), that will also reduce the mechanical power demand. E.g. if 20% faster causes 40% less grain on the drag, then power demand is now 120% * 60% = 72% of previously.

Cheers, Wayne

 

[ptonsparky]

The two existing constants we have are the leg and a drag feeding the leg. We can slow up the feed drag via a VFD, and have, to temporarily ease the problem. The added equipment by the millwrights was not to place a restriction on either. They were to run at full speed with no issues.

I am beginning to think maybe speeding the new drag up will help.

Yesterday we had to run our VFD @ 40hz to prevent back legging. Today with the change in rpm of the new drag we could run about 55hz before it started to hit FLA. No back legging. Running into the SFA might work, but I don't like that practice.

Millwrights said they will make it right whatever it takes.

 

[wwhitney]

Let us know how it gets resolved.

I'm inclined to think now that speeding up can only improve things by decreasing the amount of grain per paddle, and hence decreasing the spill over and overaccumulation on the drag.

When there's no spillover, the mechanical power demand of raising the grain is independent of the drag speed, and only depends on the bucket elevator deposition rate. Since in the steady state, grain exits at the top at the same rate it enters in at the bottom, and the power is just weight flow rate * height difference. That is, with respect to the earlier formulation, speeding up the drag causes less grain to be on the drag, in inverse proportion.

Cheers, Wayne

 

[ptonsparky]

Let us know how it gets resolved.

I'm inclined to think now that speeding up can only improve things by decreasing the amount of grain per paddle, and hence decreasing the spill over and overaccumulation on the drag.

When there's no spillover, the mechanical power demand of raising the grain is independent of the drag speed, and only depends on the bucket elevator deposition rate. Since in the steady state, grain exits at the top at the same rate it enters in at the bottom, and the power is just weight flow rate * height difference. That is, with respect to the earlier formulation, speeding up the drag causes less grain to be on the drag, in inverse proportion.

Cheers, Wayne

This may actually be a partial solution for my SILs employer. they are actually breaking shafts and destroying bearings due to a PP design.