Calculation. Not electrical.

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markstg said:
No formula needed, just a low level switch to turn pump off and high level switch to turn pump on. Am I missing something.
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Other than the original design was to have a continuous flow of product within a 12-15" space at the top, No, and for my part it worked. Well that and it is not a pump. Oh, and there is a lot of steam involved so switches are SS in the $1300 apiece range. We now have two spare 12"ers.

I may still work Steve's formula into the new logic. The closer I can keep the corn to the top of the kettles, the less steam works it way up into grain cleaner and on up to the leg belt.
 
141021-1237 EDT

ptonsparky:

I believe you have two cooking tanks each with a full volume of 486 bushels, and a level control range of 60 bushels at the top. I also believe this 60 bushels corresponded to about 15" of level change near the tank full point. I will assume the low level switch is asserted at this -15" point, and the high level at the 0" level (full tank).

Your new specified minimum flow rate for the mixing auger is 690 bushels per hour. If the output flow rate of both cooking tanks is less than 690 bushels per hour for a sustained time period, then the mixing auger must operate in a start stop mode.

There is a problem with my explanation to follow if both tanks have to time share the mixing auger, and therefore, I will use a one tank illustration.

If the cooking tank outputs more than 890 bushels per hour on a sustained basis, then the mixing auger will not be able to keep up with needed product and the cooking tank level will drop below the low level limit.

Between 890 and 690 bushels per hour there is opportunity to keep the level nearer the top limit. At a cooking tank output flow rate of 690 bushels per hour the tank level changes at the rate:
60/690 = 0.087 hours (5.217 minutes)
15/5.217 = 2.875 inches per minute @ 690 bushels per hour
60/5.217 = 11.5 bushels per minute @ 690 bushels per hour
60/15 = 4 bushels per inch

If you hit the lower limit and want to get back to full in 1 hour, then while the output remains at 690 the input will have to be 690 + 60 = 750 bushels per hour.

Now suppose you hit the high limit and shut off the mixing auger for 1 minute. How much does the level change in 1 minute? 2.875".

If you want to stay within 1" of full, then at an output flow of 690 and an input of 750 bushels per hour your timed off time would be 1/2.875 = 0.348 minutes or 20.9 seconds.

To return to full takes 4/60 hours or 4 minutes at the 750 and 690 for rates.

The choice of operating rates will depend upon how accurate averaged power measurements predict what is the state of the system.

Always check my calculations. I can proofread something and still not see errors.

.
 
I haven't checked your math but feel we are headed in the same direction. The flow rate of my corn is not as predictable as I had thought it would be. Different operators get different #s. Without posting several 15000 plus line spreadsheets, just take my word for it. Right now we don't know why but part of it seems to be the ability to keep grain as high as possible and the attitude in general. That and the grain just does not flow the same at all times.

This is what I have come up with for part of my logic. This timer happens to be in the Stop. Another timer using the same numbers will set the Start or at least that is how I see it at the moment.

View attachment 11262

Essentially what I am attempting to do is compensate for when the operator changes the Empty rate. If my 60 bushel space is at 45% full at rate 1, changing to rate 2, calculates how long it will take to finish the fill from that point on, be it 2 seconds or an addition 400 minutes. I only see a manually set rate, at other than the default when tanks are empty.

Time will tell.
 
Have you considered whether the motor has a limit on the number of starts per hour, even with a VFD?
Especially since the auger will be at least partially full each time it is started.
 
I am at a loss for understanding exactly how the system is set up. First, if all you have to measure tank fill level while running are high-high and high-low level switches, then any calculation for level in between has to use a compounded flow rate calculation for in between levels based on the last trip of either switch. If the flow rates are based on motor watts, I have to wonder how accurate the wattage values are in relation to actual flow rates. I would think motor rpm would more accurately depict flow rate than would watts...???

Isn't there a level range detection scheme available for this application?
 
GoldDigger said:
Have you considered whether the motor has a limit on the number of starts per hour, even with a VFD?
Especially since the auger will be at least partially full each time it is started.
Click to expand...

I had not but even at the Min empty rate it will take the VFD 218 seconds to catch up at 58 HZ. Depending on if I allow it to drop the full 60 bushel before it starts again.
 
Smart $ said:
I am at a loss for understanding exactly how the system is set up. First, if all you have to measure tank fill level while running are high-high and high-low level switches, then any calculation for level in between has to use a compounded flow rate calculation for in between levels based on the last trip of either switch. If the flow rates are based on motor watts, I have to wonder how accurate the wattage values are in relation to actual flow rates. I would think motor rpm would more accurately depict flow rate than would watts...???

Isn't there a level range detection scheme available for this application?
Click to expand...

The mixing auger will stay running once the production process is started. The Holding Bin VFD will be the only motor starting & stopping. Flow rates via motor watts of the MA are critical for the OV. I will let them deal with that. For me, they are interesting and I can set some warnings and/or limits as needed.
 
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ptonsparky said:
The mixing auger will stay running once the production process is started. The Holding Bin VFD will be the only motor starting & stopping. Flow rates via motor watts of the MA are critical for the OV. I will let them deal with that. For me, they are interesting and I can set some warnings and/or limits as needed.
Click to expand...
To me, your reply can be likened to a political candidate's debate response to a major topic yes or no answer question... :slaphead:
 
Asleep at the wheel?

Asleep at the wheel?

gar said:
141021-1237 EDT

ptonsparky:

I believe you have two cooking tanks each with a full volume of 486 bushels, and a level control range of 60 bushels at the top. I also believe this 60 bushels corresponded to about 15" of level change near the tank full point. I will assume the low level switch is asserted at this -15" point, and the high level at the 0" level (full tank).

Your new specified minimum flow rate for the mixing auger is 690 bushels per hour. If the output flow rate of both cooking tanks is less than 690 bushels per hour for a sustained time period, then the mixing auger must operate in a start stop mode.

There is a problem with my explanation to follow if both tanks have to time share the mixing auger, and therefore, I will use a one tank illustration.

If the cooking tank outputs more than 890 bushels per hour on a sustained basis, then the mixing auger will not be able to keep up with needed product and the cooking tank level will drop below the low level limit.

Between 890 and 690 bushels per hour there is opportunity to keep the level nearer the top limit. At a cooking tank output flow rate of 690 bushels per hour the tank level changes at the rate:
60/690 = 0.087 hours (5.217 minutes)
15/5.217 = 2.875 inches per minute @ 690 bushels per hour
60/5.217 = 11.5 bushels per minute @ 690 bushels per hour
60/15 = 4 bushels per inch

If you hit the lower limit and want to get back to full in 1 hour, then while the output remains at 690 the input will have to be 690 + 60 = 750 bushels per hour.

Now suppose you hit the high limit and shut off the mixing auger for 1 minute. How much does the level change in 1 minute? 2.875".

If you want to stay within 1" of full, then at an output flow of 690 and an input of 750 bushels per hour your timed off time would be 1/2.875 = 0.348 minutes or 20.9 seconds.

To return to full takes 4/60 hours or 4 minutes at the 750 and 690 for rates.

The choice of operating rates will depend upon how accurate averaged power measurements predict what is the state of the system.

Always check my calculations. I can proofread something and still not see errors.

.
Click to expand...

Waiting breathlessly, I know.

We still won't have the additional level indicators, to answer Smart$, until the end of this week, so we are cycling with a high level limit where all feeding motors stop, then start is allowed again 97 seconds later. One of the operators came in late Saturday night to make flakes for the next day. Startup takes time. Just the way it is. Once he got things cycling, the top space temperature began coming down, means no steam is being lost to the atmosphere or traveling up to the leg. One of two things happened, he quit screwing with the process intentionally or fell asleep. The start/stop cycled with the top 11" of the tank, or about 18 bushel.

Motor start/hour is between 8 and 12 times.
 
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