Calculation. Not electrical.

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ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
I need a 7th grader. My closest is in bed already.

I have a tank that fills at a set rate. It empties at a variable rate. Start is initiated at a set low level and run time needs to vary with the rate of drain so it stops filling at approximately the same high level.

Who has a formula for me? Maybe I should ask what is the formula for a solution? It has been a long time since Jr High.
 
141014-2340 EDT

With the information you have available there is no way to calculate how long to fill the tank.

What I mean by this is that you have no information on the output flow other than that it is variable. The simple solution is a high level switch added to the circuit.

You could make some assumptions. but that won't prevent overfilling the tank.

.
 
gar said:
141014-2340 EDT

With the information you have available there is no way to calculate how long to fill the tank.

What I mean by this is that you have no information on the output flow other than that it is variable. The simple solution is a high level switch added to the circuit.

You could make some assumptions. but that won't prevent overfilling the tank.

.
Click to expand...
And if ptonsparky wants to have a "smooth" solution, one control procedure would be to have the fixed fill rate be greater than the maximum drain rate (by a factor of two if possible), and to use an analog depth signal (proportional to the distance the water level was down from the top.)
Then run the pump at a constant rate until the tank is nearly full and then go to some form of proportional, on/off, control until the desired fill level was reached.
The critical problem with what pt described is that the tank would never be able to stay full, since is not way to match the inflow to the outflow. Instead the level would have to oscillate between almost full and some lower trigger level.

If you were able to get a signal proportional to the flow rate and a second signal which was the low level float switch, you could subtract the flow signal from a constant current representing the fill rate and then set the on time of the pump control to be inversely proportional to that difference signal. (I.E. when the flow signal and the constant fill signal were equal, the pump would stay on forever in the perfect approximation. If the outflow signal was 1/2 the input signal the time would be twice the time needed to go from low level to full while the outflow was zero. )
You would still need a full level switch as a safety cutoff, since the time algorithm would be approximate at best.
 
Musings from the official in-house plumber..

Musings from the official in-house plumber..

What size line/drain pipe? Easy to find max GPM flow rate, at any particular pressure, of any size pipe, do it all of the time with hydronic pipe sizing.
 
mtfallsmikey said:
What size line/drain pipe? Easy to find max GPM flow rate, at any particular pressure, of any size pipe, do it all of the time with hydronic pipe sizing.
Click to expand...

Not liquid, and operator has control of rate discharge, but it does vary throughout the day. Minute to minute @ startup then pretty consistent for most of the time.


GoldDigger said:
And if ptonsparky wants to have a "smooth" solution, one control procedure would be to have the fixed fill rate be greater than the maximum drain rate (by a factor of two if possible), and to use an analog depth signal (proportional to the distance the water level was down from the top.)
Then run the pump at a constant rate until the tank is nearly full and then go to some form of proportional, on/off, control until the desired fill level was reached.
The critical problem with what pt described is that the tank would never be able to stay full, since is not way to match the inflow to the outflow. Instead the level would have to oscillate between almost full and some lower trigger level.

If you were able to get a signal proportional to the flow rate and a second signal which was the low level float switch, you could subtract the flow signal from a constant current representing the fill rate and then set the on time of the pump control to be inversely proportional to that difference signal. (I.E. when the flow signal and the constant fill signal were equal, the pump would stay on forever in the perfect approximation. If the outflow signal was 1/2 the input signal the time would be twice the time needed to go from low level to full while the outflow was zero. )
You would still need a full level switch as a safety cutoff, since the time algorithm would be approximate at best.
Click to expand...

The fill rate is variable and in "auto" does what I want. Not necessarily what others want.

We do have a full level indicator but when that trips all fill stops leaving everything in a full condition. The motors can handle the restart during normal temperatures. Now if you can guarantee we won't have temperatures below freezing and winds are calm we are good to go.

gar said:
141014-2340 EDT

With the information you have available there is no way to calculate how long to fill the tank.

What I mean by this is that you have no information on the output flow other than that it is variable. The simple solution is a high level switch added to the circuit.

You could make some assumptions. but that won't prevent overfilling the tank.



.
Click to expand...

The owner wants a Manual fill speed override, with automatic adjustments in time of fill. :? Makes it difficult when the whole thing was set up on a variable rate of fill.

Glad to hear there is "no way". I can get close, but it would always be loosing a bit of ground at a constant fill speed.
 
ptonsparky said:
Not liquid, and operator has control of rate discharge, but it does vary throughout the day. Minute to minute @ startup then pretty consistent for most of the time.

The fill rate is variable and in "auto" does what I want. Not necessarily what others want.

... The motors can handle the restart during normal temperatures. Now if you can guarantee we won't have temperatures below freezing and winds are calm we are good to go.
The owner wants a Manual fill speed override, with automatic adjustments in time of fill. :? Makes it difficult when the whole thing was set up on a variable rate of fill.

Glad to hear there is "no way". I can get close, but it would always be loosing a bit of ground at a constant fill speed.
Click to expand...
Who says no way... maybe no way lacking all the details necessary.

Sounds like the objective is to keep the motors running... and the owner wants an override so that the tank is never full while being emptied at the same time, and the motors will run continuously...
 
I don't really understand the process that is going on in the tank other than emptying so that you need to sync product input and output?

But if the problem is that you don't have a large enough tank inside or room for a large enough tank to run the process for a full shift and have to receive product from a larger tank elsewhere.

The solution would be to match your controls on both tanks, and by paralleling the controls with both tanks.

If that is a problem because the outside tank is larger and would empty at a higher rate then you could have another tank outside that is identical to your inside tank being refilled by the larger outside tank.

Sounds expensive though.
 
Smart $ said:
Who says no way... maybe no way lacking all the details necessary.

Sounds like the objective is to keep the motors running... and the owner wants an override so that the tank is never full while being emptied at the same time, and the motors will run continuously...
Click to expand...

That was the original objective & I pretty much had that working. The new objective is to manually set the incoming speed and change the run time. The run time would be determined by the amount of product flowing out in relation to that flowing in.


Pick a fill rate of 690 - 890 units per hour. (Automatic rate would be 92 to 890.) The top capacity we have to work with is about 60 bushel and takes 240 seconds to fill @ max speed with no discharge. Start would be initiated by a low limit sensor, then incoming rate would be at FR

Empty rate is 27 - 416 per hour. The low rate may occur for 30 minutes or more until things get warmed up. Literally. After that they increase speed as quickly as they can.
 
So the tank fills at a set rate, and it empties at a set rate that the operator determines. (Assume both stay the same during a single fill).

Lets assume everything is in gallons/minute. Assume the fill rate = F, and the empty rate = E.

Also, assume there is a fixed level to start filling the tank, and there is a fixed level to stop filling the tank. Lets call the difference in these two the volume "V", which is in gallons. Just as an example, say we start filling the tank when its down to 2 feet, which is 200 gallons, and we want to stop filling the tank at 10 feet, which is 1000 gallons. The difference in volume would be 800 gallons. So in this case our V would be 800 gal.

Finally, assume the time we are going to let the fill run is T.

The equation for T would be T = V/F + E*T/F. It takes some algebra to get the T together on one side, but the result is T=V/(F-E).

Steve
 
You don?t need a seventh grader. You need someone who has, at the very least, taken a high school calculus class. A formula for the level of the tank as a function of time will involve the following variables:

  • The initial level (i.e., at the moment of time at which you begin to monitor the level),
  • The geometry of the tank (i.e., is it a cylinder, does it taper down at the top or bottom, what it its diameter at various levels, etc.),
  • The fill rate (you first said that it was constant, and later said that it varied through the day, so I am confused about this),
  • The discharge rate (if this varies throughout the day, then you can?t create a formula for level without first creating a formula for the discharge rate as a function of time of day).

I don?t think that creating a formula for level is what you need. I also don?t think it is possible to create one. I say that because not only did you not give us enough of the required input information (as listed above), but I also doubt that anyone will be able to provide that information.

Instead, I suspect what you need is a means of control. For this, you need access to the level at every point in time, and you need the drain rate at every point in time, and you need to establish the desired behavior of the tank level, and you need a means of remotely controlling the fill rate. When I say ?behavior,? I mean you need to know if the goal is to maintain a constant level, or to allow the tank level to drop to near the bottom before starting to fill it up again, or to keep the level within a specific band (e.g., between 70% and 95% full).
 
141015-1553 EDT

ptonsparky:

Is this the same project that has the 5 and 7.5 HP motors that move corn with augers and to which you want to monitor power input?

Why does someone want you to go from proportional control to bang-bang (on-off) control?

.
 
gar said:
141015-1553 EDT

ptonsparky:

Is this the same project that has the 5 and 7.5 HP motors that move corn with augers and to which you want to monitor power input?

Why does someone want you to go from proportional control to bang-bang (on-off) control?

.
Click to expand...

Yes, same project. The monitoring of power was an eye opener for me and is working well.

Why, you ask? To drive me nuts.
The other vendors equipment can not deliver, at this moment, chemicals in amounts small enough to treat below 300 lbs per minute. Their field tech is going to pass on the problem to engineering. They may come up with a better solution on their end.
 
steve66 said:
So the tank fills at a set rate, and it empties at a set rate that the operator determines. (Assume both stay the same during a single fill).

Lets assume everything is in gallons/minute. Assume the fill rate = F, and the empty rate = E.

Also, assume there is a fixed level to start filling the tank, and there is a fixed level to stop filling the tank. Lets call the difference in these two the volume "V", which is in gallons. Just as an example, say we start filling the tank when its down to 2 feet, which is 200 gallons, and we want to stop filling the tank at 10 feet, which is 1000 gallons. The difference in volume would be 800 gallons. So in this case our V would be 800 gal.

Finally, assume the time we are going to let the fill run is T.

The equation for T would be T = V/F + E*T/F. It takes some algebra to get the T together on one side, but the result is T=V/(F-E).

Steve
Click to expand...

That works so well on a spreadsheet, I may have to see how it works in practice. The catch will be if they drop the E rate, especially if it is close to full. I could subtract the last time from the new time.
 
141017-0845 EDT

ptonsparky:

I would like to get a clear picture in my mind of this system. So some questions.

Background and questions:

(1) There is a mixing auger. I believe this is driven by the 7.5 HP motor, but I am not sure. Is this correct?

(2) There is a power input monitor that is, I believe, presently monitoring the 7.5 HP motor. Is this correct?

(3) The mixing auger is manually adjusted to output
"Empty rate is 27 - 416 per hour. The low rate may occur for 30 minutes or more until things get warmed up. Literally. After that they increase speed as quickly as they can.".
What are the units for emptying rate? Bushels per hour or some other units?

(4) What are the min and max power inputs to the mixing auger motor at the 27 units per hour rate? Same question for the 416 units per hour?

(5) There is a 60 bushel tank from which the mixing auger extracts corn. How many bushels in this tank at the low level switch point?

(6) There is an input auger that can feed the 60 bushel tank at
"Pick a fill rate of 690 - 890 units per hour. (Automatic rate would be 92 to 890.) The top capacity we have to work with is about 60 bushel and takes 240 seconds to fill @ max speed with no discharge."
I believe this says the maximum input rate is 15 bushels per minute, or 900 bushels per hour, and that implies that the above rates of 890 units per hour are 890 bushels per hour.

The input rate capability appears to be about double the output rate capablity. Is that correct?

Intuitively I would expect the 7.5 HP motor on the input and the 5 HP on the output, but you have previously implied that the 5 HP was heavily loaded, and the 7.5 HP lightly loaded.

What is the input auger motor power input (min max) at 90 bushels per hour, and at 900 bushels per hour?

(7) Is product quality affected by the head (level) pressure of the corn in the 60 bushrel tank?

(8) I also believe that a high level switch has been mentioned for the 60 bushel tank. Is that correct?

.
 
gar said:
141017-0845 EDT

ptonsparky:

I would like to get a clear picture in my mind of this system. So some questions.

Background and questions:

(1) There is a mixing auger. I believe this is driven by the 7.5 HP motor, but I am not sure. Is this correct?

(2) There is a power input monitor that is, I believe, presently monitoring the 7.5 HP motor. Is this correct?

(3) The mixing auger is manually adjusted to output
"Empty rate is 27 - 416 per hour. The low rate may occur for 30 minutes or more until things get warmed up. Literally. After that they increase speed as quickly as they can.".
What are the units for emptying rate? Bushels per hour or some other units?

(4) What are the min and max power inputs to the mixing auger motor at the 27 units per hour rate? Same question for the 416 units per hour?

(5) There is a 60 bushel tank from which the mixing auger extracts corn. How many bushels in this tank at the low level switch point?

(6) There is an input auger that can feed the 60 bushel tank at
"Pick a fill rate of 690 - 890 units per hour. (Automatic rate would be 92 to 890.) The top capacity we have to work with is about 60 bushel and takes 240 seconds to fill @ max speed with no discharge."
I believe this says the maximum input rate is 15 bushels per minute, or 900 bushels per hour, and that implies that the above rates of 890 units per hour are 890 bushels per hour.

The input rate capability appears to be about double the output rate capablity. Is that correct?

Intuitively I would expect the 7.5 HP motor on the input and the 5 HP on the output, but you have previously implied that the 5 HP was heavily loaded, and the 7.5 HP lightly loaded.

What is the input auger motor power input (min max) at 90 bushels per hour, and at 900 bushels per hour?

(7) Is product quality affected by the head (level) pressure of the corn in the 60 bushrel tank?

(8) I also believe that a high level switch has been mentioned for the 60 bushel tank. Is that correct?

.
Click to expand...

1. Yes
2. Yes
3. BPH
4, 5, &6. The VFD delivers 92 to 890 BPH to the Mixing Auger, then the tanks . As requested the min rate has been increased to 690. Each tank is about 486 bushel. The top 60 is where the VFD was to come into play. The High limit in the tank is a Stop. All grain flow must stop. Before that is reached I need to time the fill in order get the last 30 bushel in the tanks to prevent freezing this winter... at whatever rate is being requested by the metering rolls that Empty the tanks. Empty rate is 27 to 416 bush per hour per tank. We have seen the 416 with one mill running. The Flake Drag (5HP) is at its max to handle this flake rate. Realistically I never expected the VFD to run over 45 HZ with both mills in operation.

Mix auger KW. No efficiency applied. Max load: Max 2021, Min 1294, avg 1721; Min load 690 bph: Max 2288, Min 967, avg 1534. (80 sec time frame including start & stop) No load Max 895, Min 606.

Product quality is no bothered by level of corn.
Yes, a high level switch.

Thanks for the interest and FWIW, I had another requested change of operation today.

Sometimes I really do wish for more alcohol in my blood stream. Hey...I have VO and ice. What more can I ask for.
 
141020-0743 EDT
ptonsparky:

I need to unscramble this statement. I do not understand it ---
Mix auger KW. No efficiency applied. Max load: Max 2021, Min 1294, avg 1721; Min load 690 bph: Max 2288, Min 967, avg 1534. (80 sec time frame including start & stop) No load Max 895, Min 606.
Click to expand...

(10) What does "Max load" mean? Is "maximum load" the maximum short time average motor power input when the corn input end of the mixing auger is fully covered with corn (no shortage of input corn), and the auger is moving at its highest RPM that corresponds with the the desired maximum flow rate, 890 bushels per hour, of corn and additives leaving the mixing auger?

(11) If so, then why the huge range of 1290 to 2021 watts?

(12) Minimum load appears to be defined as 690 bushels per hour. I assume this means the auger is running slower to produce the lower flow rate. The motor power input range at this flow rate is 967 to 2288 watts. This is a greater range, lower on the low side and higher on the high side, than at maximum load. Why the greater range?

(13) Would not auger RPM be a better measure of flow rate than motor power input?

(14) The items referred to as tanks have level switches at 486 bushels (high level) and 426 bushels (low level), 60 bushels less than full, and these limit switches are input signals to the VFD control. Is this correct?

(15) There are two 486 bushel tanks. Are these tanks simultaneously filled? Are there high and low level switches in both tanks? Are these cooking tanks?

(16) With no product, corn and additives, in the mixing auger is the mixing auger motor power input 606 ton 895 watts?

.
 
gar said:
141020-0743 EDT
ptonsparky:

I need to unscramble this statement. I do not understand it ---


(10) What does "Max load" mean? Is "maximum load" the maximum short time average motor power input when the corn input end of the mixing auger is fully covered with corn (no shortage of input corn), and the auger is moving at its highest RPM that corresponds with the the desired maximum flow rate, 890 bushels per hour, of corn and additives leaving the mixing auger?

(11) If so, then why the huge range of 1290 to 2021 watts?

(12) Minimum load appears to be defined as 690 bushels per hour. I assume this means the auger is running slower to produce the lower flow rate. The motor power input range at this flow rate is 967 to 2288 watts. This is a greater range, lower on the low side and higher on the high side, than at maximum load. Why the greater range?

(13) Would not auger RPM be a better measure of flow rate than motor power input?

(14) The items referred to as tanks have level switches at 486 bushels (high level) and 426 bushels (low level), 60 bushels less than full, and these limit switches are input signals to the VFD control. Is this correct?

(15) There are two 486 bushel tanks. Are these tanks simultaneously filled? Are there high and low level switches in both tanks? Are these cooking tanks?

(16) With no product, corn and additives, in the mixing auger is the mixing auger motor power input 606 ton 895 watts.

.
Click to expand...

Mixing Auger is single speed. Product is supplied to it via a drag with a VFD. Max load is the maximum amount of corn supplied to the MA. The Mixing auger is 16" diameter with portions of the flighting removed ever few feet to mix the corn and water. The range includes start up with product in the auger, but yes it does jump around. Possibly the way flighting is cut. Not precise in any way shape or form.

The min load of 690 bushel was a short period of time and again it included the start. I have several days of data now with more questions coming up every day.

The High Limit is at about the 481 +-.
The lower is at the 411 area. VFD is timed to allow the Leg and Mixing auger to empty before it gets to the High Limit. About 3 bushel short.
The Tanks are in a row, one fills than the other. We can select via gates on the top of the tanks, which one or both to fill.
 
View attachment 11244

Holding bin is to Rt.
2 HP VFD Drag is at bottom of Holding Bin and moves grain to Leg.
Leg is hidden by HB, but dumps corn via a grain cleaner to Rt end of Mixing Auger.
Water chemical mix is added via blue pipe running up the outside of the building
Red rectangles at top of building indicate 60 bushel top portion of Steam Kettles.
There is a gate at the top of each Kettle to select which one or both fills.
Product is Steamed for approximately 1 hour then feed into rollers that process it into flakes. One each Kettle.
Product the drops onto drag and exits to the left bottom of the building.
 
141020-1352 EDT

ptonsparky:

The mixing auger is driven by a fixed speed 7.5 HP motor with a power monitor. The flow rate thru the mixing auger is determined by the rate that corn is fed into the mixing auger. This input flow is from a conveyor driven by a VFD controlled motor. At times both the mixing auger and the conveyor feeding corn to it may have to be stopped to prevent overflow of the cooking (steaming) tanks. Recently you have had new specifications on the min and max flow rates. The raising of the minimum flow rate is what make cause start-stop operation of the corn feed.

At maximum corn flow, 890 bushels per hour, the mixing auger input power varies from 1290 to 2021 watts with an average of 1721 W.

What is the time roughly between 1290 and 2021 watts? 5 seconds, 50 sec, or 500 sec?

Would a running average from the last 5 minutes with data update every 5 seconds give you a stable signal for control purposes?

The present minimum flow rate is 690 bushels per hour. The mixing auger power input varies from 967 to 2288 watts with an average of 1534 watts. Much more peak-to-peak variation.

The corn flow ratio is 690/890 = 0.78 .
Avg power input ratio is 1534/1721 = 0.89 . But we really need to subtract residual power.

The residual was at no load 895 and 606 or an average of 750 W. Thus, the incremental power load resulting from just the corn is 1721-750 = 971 W at 890 bushels per hour, and 1534-750 = 784 W at 690 bushels per hour. Now the ratio is 784/971 = 0.81 and closer to the corn feed ratio.

The photo is useful.

Any errors in my summary?

.
 
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