Math: If Train A leaves the station traveling...

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GoldDigger

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gadfly56 said:
Your original statement of the problem seemed to indicate that if there were no product outflow that it would take 3 minutes for the tank to fill to some stop point. Is that still correct?
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Another description stated that it took three minutes after the startup of the raw material motor before anything arrives at the processing tank. Is this at initial startup of the process only or does it occur every the raw material motor starts up?

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gar

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EE
180509-1957 EDT

ptonsparky:

Back to rereading your post #12.

Finished product leaves at a continuous rate but raw product starts to enter 3 minutes after the product in the tank has reached a Low Set Point. The raw product motor continues to run until the LSP has been reached, then shuts Off. Raw product will continue to flow for 30 seconds or so, but I'm not concerned with that level.
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Apparently you have only a low level limit.

1. When low level is reached, then supply motor starts.

2. At this start time it will be 3 minutes before any new stock flows into the tank. Is it really true that this 3 minutes is a constant independent of input flow rate? We will assume it is.

3. For 3 minutes there is an out flow of product. Thus, the product in the tank has been reduced from the low level by 3*Fo.

4. The time to get back to the low limit or motor pump up time is
T = 3*Fo / (Fi - Fo) = 3 / (Fi/Fo - 1).
If Fi = 2*Fo, then 3 minutes.
If Fi = 1.1*Fo, then 30 minutes.
If Fi = 4*Fo, then 1 minute.

.
 
ptonsparky

ptonsparky

Tom
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gar said:
180509-1957 EDT

ptonsparky:

Back to rereading your post #12.



Apparently you have only a low level limit. Not to confuse things, but it is the high limit. Incoming product covers it somewhat before things empty out. Once product drops below it there is an additional time delay that ends up to be about 3 minutes before the input motor starts.

1. When low level is reached, then supply motor starts. After timer delay.

2. At this start time it will be 3 minutes before any new stock flows into the tank. Is it really true that this 3 minutes is a constant independent of input flow rate? We will assume it is. the 3 minutes is dependent on the outflow. Assume it is constant for now.

3. For 3 minutes there is an out flow of product. Thus, the product in the tank has been reduced from the low level by 3*Fo. The flow out of the tank is continuous. Only the inflow cycles. Off at the high limit and On after the timer sequence

4. The time to get back to the low limit or motor pump up time is
T = 3*Fo / (Fi - Fo) = 3 / (Fi/Fo - 1).
If Fi = 2*Fo, then 3 minutes.
If Fi = 1.1*Fo, then 30 minutes.
If Fi = 4*Fo, then 1 minute.

.
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(this message is not too short now)
 
kwired

kwired

Electron manager
Location
NE Nebraska
ptonsparky said:
No, the fill motor does not start for three minutes as product is leaving the tank
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From what you have already said before - I think you mean the fill motor does not start until three minutes after the level in tank drops below the high level limit switching device.
 
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gar

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Location
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EE
180509-2353 EDT

ptonsparky:

My equations still apply. But there is a better description of what the operation is.

1. Filling motor is off and tank level is dropping. On tank level going below a set level a 3 minute time delay is initiated.

2. During this 3 minute period the volume outflow from the tank is 3*Fo.

3. At the end of the 3 minute time delay the input motor starts and the in flow rate is Fi while the out flow rate is still Fo. Thus, net flow rate into the tank is Fi - Fo.

4. When the threshold is reached on pump up, then the input motor stops. Doesn't matter if there is some overflow of product above the threshold.

5. The time to refill the lost 3*Fo is T = 3*Fo / (Fi - Fo) or 3 / (Fi/Fo - 1) minutes.

It is self evident that if output flow rate is Fo continuously, then to continue that rate and refill in the same time as draining occurred, then Fi must = 2*Fo without any equations.

.
 
ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
gar said:
180509-2353 EDT

ptonsparky:

My equations still apply. But there is a better description of what the operation is.

1. Filling motor is off and tank level is dropping. On tank level going below a set level a 3 minute time delay is initiated.

2. During this 3 minute period the volume outflow from the tank is 3*Fo.Fo=8.85 Fi=12.2 V=26.54

3. At the end of the 3 minute time delay the input motor starts and the in flow rate is Fi while the out flow rate is still Fo. Thus, net flow rate into the tank is Fi - Fo. Fn=3.35

4. When the threshold is reached on pump up, then the input motor stops. Doesn't matter if there is some overflow of product above the threshold.

5. The time to refill the lost 3*Fo is T = 3*Fo / (Fi - Fo) or 3 / (Fi/Fo - 1) minutes.3/(12.2/8.85-1) or 3/(12.2/7.85)=1.55 ??

It is self evident that if output flow rate is Fo continuously, then to continue that rate and refill in the same time as draining occurred, then Fi must = 2*Fo without any equations. Todays time for fill off was about 3:05 to fill on of 7:45 Fo was a bit slower than the average. Fi is always the same.

.
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I don't think I did the math the way you intended in step 5.
 
G

gar

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EE
180510-1223 EDT

ptonsparky:

It looks like your Fi = 1.38*Fo. If you can go to 3 times instead 0f 1.38 does the equation still hold? It should.

What is your end objective?

From several years ago we had a discussion on power monitoring a motor instead of current. I don't remember if you acquired a power monitor.

Relative to your present application is your interest related to energy consumption vs Fi and/or Fo as well? Totally what are you trying to study?

.
 
ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
gar said:
180510-1223 EDT

ptonsparky:

It looks like your Fi = 1.38*Fo. If you can go to 3 times instead 0f 1.38 does the equation still hold? It should.

What is your end objective?

From several years ago we had a discussion on power monitoring a motor instead of current. I don't remember if you acquired a power monitor.

Relative to your present application is your interest related to energy consumption vs Fi and/or Fo as well? Totally what are you trying to study?

.
Click to expand...

This is related to the power monitoring project. I didn’t supply the equipment and they now use it to monitor a leg motor instead of a U trough auger.

My my goal is to have longer periods of run time for the fill process which will keep the steam in the chests, instead of escaping out the top level of product. Waste of energy. The owners are installing a high limit sensor that will now lower the product by an additional foot. Not a good idea IMO. I may attempt to offset this by timing in the PLC.

it may be a wasted effort because of limiting equipment at the input-to large, and output-to small.
 
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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
180410-1721 EDT

ptonsparky:

Since I have no knowledge of what this machine does, how it is constructed, or how it works my comments may not make sense or relate.

If the volume for storage remains fixed, and the output flow rate is constant, then to lessen the change in level the off time of the timer must be shortened independent of the input feed rate. This assumes overflow is not a factor.

If number of motor starts per given time is important, then other criteria apply to the off timer.

Can you explain your comment
My my goal is to have longer periods of run time for the fill process which will keep the steam in the chests, instead of escaping out the top level of product. Waste of energy. The owners are installing a high limit sensor that will now lower the product by an additional foot. Not a good idea IMO. I may attempt to offset this by timing in the PLC.
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in relation to the structure of the machine and what the process requires?

.
 
G

gar

Senior Member
Location
Ann Arbor, Michigan
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EE
180511-0730 EDT

ptonsparky:

How tall is the tank? What is its diameter?

How tall is the product in the tank?

Is finished product removed from the bottom, and raw input fed in at the top?

What is the maximum peak-to-peak variation across the top surface of the material?

What kind of sensor is used for level measurement?

What is the temperature and environment around the level sensor?

Is it just one sensor or in some fashion do several sensors measure an average of the top surface?

At your present output feed rate and no input feed how much level change occurs in 3 minutes?

How close would you like to hold the level?

.
 
ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
gar said:
180511-0730 EDT

ptonsparky:

How tall is the tank? What is its diameter?

How tall is the product in the tank?

Is finished product removed from the bottom, and raw input fed in at the top?

What is the maximum peak-to-peak variation across the top surface of the material?

What kind of sensor is used for level measurement?

What is the temperature and environment around the level sensor?

Is it just one sensor or in some fashion do several sensors measure an average of the top surface?

At your present output feed rate and no input feed how much level change occurs in 3 minutes?

How close would you like to hold the level?

.
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Wow...how about a simple graph. Note the purple line. That is the speed at which I believe the VFD should run to maintain a constant level. I'm limited to what I have control over. I will PM you with added details if you still want them. My OP has been adequately answered by three of you. Now I need to see how I can make it work for me.
Screenshot (1).jpg

Thank you all, again.
 
kwired

kwired

Electron manager
Location
NE Nebraska
gar said:
180509-2353 EDT

ptonsparky:

My equations still apply. But there is a better description of what the operation is.

1. Filling motor is off and tank level is dropping. On tank level going below a set level a 3 minute time delay is initiated.

2. During this 3 minute period the volume outflow from the tank is 3*Fo.

3. At the end of the 3 minute time delay the input motor starts and the in flow rate is Fi while the out flow rate is still Fo. Thus, net flow rate into the tank is Fi - Fo.

4. When the threshold is reached on pump up, then the input motor stops. Doesn't matter if there is some overflow of product above the threshold.

5. The time to refill the lost 3*Fo is T = 3*Fo / (Fi - Fo) or 3 / (Fi/Fo - 1) minutes.

It is self evident that if output flow rate is Fo continuously, then to continue that rate and refill in the same time as draining occurred, then Fi must = 2*Fo without any equations.

.
Click to expand...
I thought this was simple at first - as long as Fi > Fo you should never run the tank empty, but then you added more objectives:

ptonsparky said:
This is related to the power monitoring project. I didn’t supply the equipment and they now use it to monitor a leg motor instead of a U trough auger.

My my goal is to have longer periods of run time for the fill process which will keep the steam in the chests, instead of escaping out the top level of product. Waste of energy. The owners are installing a high limit sensor that will now lower the product by an additional foot. Not a good idea IMO. I may attempt to offset this by timing in the PLC.

it may be a wasted effort because of limiting equipment at the input-to large, and output-to small.
Click to expand...
:)
 
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