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

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kwired

kwired

Electron manager
Location
NE Nebraska
ptonsparky said:
Raw Product Arrives at a set speed. Finished Product Leaves at a set speed. PA>PL

PL has a 3 minute head start on PA. What is the formula for raw product level to reach the start/stop point again?
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I think there are too many details in the process that are left out to be able to answer.

How do you get a three minute head start on leaving when you haven't been processed yet?

Probably more a matter of asking the right question.
 
gadfly56

gadfly56

Senior Member
Location
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kwired said:
I think there are too many details in the process that are left out to be able to answer.

How do you get a three minute head start on leaving when you haven't been processed yet?

Probably more a matter of asking the right question.
Click to expand...

If the product is binned, at t=0 it will take 3 minutes to fill the bin up from its current level to "full". Since PA>PL, at some point the bin will fill up, but it will be longer than 3 minutes, based on the difference between PA and PL and the size of the bin, which has a total volume VT and a starting volume V0.

I have to think about this for a bit.

(VT - V0)/PA = 3

PA - PL = PN (net accumulation of product)

(VT - V0)/PN = X (time to fill the bin with material flowing in and out)

Since we aren't given VT or V0, rearrange the first equation to:

(VT - V0) = 3PA

or 3PA/(PA - PL) = X
 
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W

wwhitney

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Berkeley, CA
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Retired
ptonsparky said:
Raw Product Arrives at a set speed. Finished Product Leaves at a set speed. PA>PL

PL has a 3 minute head start on PA. What is the formula for raw product level to reach the start/stop point again?
Click to expand...

I'm interpreting what you've written as follows: Raw product is accumulated. Finished product leaves as soon as it is created, and PL is the rate of conversion from raw product to finished product.

At time t = 0, there is some amount I of raw product accumulated. The finished product starts getting made at rate PL.

At time t = 3, raw product has run out yet (I > 3 * PL), and new raw product starts arriving at rate PA > PL.

At what time does the quantity of raw product return to the initial level I? The deficit created was 3 * PL, and the rate of growth is now (PA - PL). So the additional time required is 3 * PL / (PA - PL).

Cheers, Wayne
 
ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
kwired said:
I think there are too many details in the process that are left out to be able to answer.

How do you get a three minute head start on leaving when you haven't been processed yet?
Click to expand...

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.

The Google train sequence only gave me when they would be at the same point on the track.

I want to be able to eventually vary both PL speed and PA speed then have an idea of run time for the raw product motor.
 
G

gar

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Location
Ann Arbor, Michigan
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EE
180509-1343 EDT

ptonsparky:

You need to know the volume between the two level switches, call it V.

Let Fo be the output flow rate, and Fi the input rate. Let Vover be the overfill volume.

Assume Vover is not large, then the approximate time to empty to low level Temptying = (V + Vover)/Fo.

On filling the time motor has to run is Trun = 3 minutes + ( V + 3*Fo )/(Fi - Fo).

So descriptively:

1. You have some change in volume between the high and low limit switches. This is V. You also have some overflow, and you imply this can be ignored, but it does have an effect on the time to get to the low level switch.

2. Once you get to the low level switch the filling motor starts, but it takes 3 minutes before it starts to add product to the tank. That is the 3 minutes in the run time equation.

3. Also during this 3 minutes the tank continues to have product removed. Thus, the volume that has to be added to reach the full limit is V + 3*Fo. That is the volume between the high and low limits plus the additional lost volume during the 3 minutes there was no flow into the tank after low limit was reached.

4. Now we know how much has to be added to reach the high limit switch, V + 3*Fo.

5. The time to add this much product is determined by the net inflow rate. That is Fi - Fo.

6. Thus, time to reach full limit is INITIAL DELAY + TIME to FILL.

TIME to FILL = Change in volume / Net fill rate (this is once filling starts)

See if I made any mistakes.

.
 
gadfly56

gadfly56

Senior Member
Location
New Jersey
Occupation
Professional Engineer, Fire & Life Safety
gar said:
180509-1343 EDT

ptonsparky:

You need to know the volume between the two level switches, call it V.

Let Fo be the output flow rate, and Fi the input rate. Let Vover be the overfill volume.

Assume Vover is not large, then the approximate time to empty to low level Temptying = (V + Vover)/Fo.

On filling the time motor has to run is Trun = 3 minutes + ( V + 3*Fo )/(Fi - Fo).

So descriptively:

1. You have some change in volume between the high and low limit switches. This is V. You also have some overflow, and you imply this can be ignored, but it does have an effect on the time to get to the low level switch.

2. Once you get to the low level switch the filling motor starts, but it takes 3 minutes before it starts to add product to the tank. That is the 3 minutes in the run time equation.

3. Also during this 3 minutes the tank continues to have product removed. Thus, the volume that has to be added to reach the full limit is V + 3*Fo. That is the volume between the high and low limits plus the additional lost volume during the 3 minutes there was no flow into the tank after low limit was reached.

4. Now we know how much has to be added to reach the high limit switch, V + 3*Fo.

5. The time to add this much product is determined by the net inflow rate. That is Fi - Fo.

6. Thus, time to reach full limit is INITIAL DELAY + TIME to FILL.

TIME to FILL = Change in volume / Net fill rate (this is once filling starts)

See if I made any mistakes.

.
Click to expand...

You, wwhitney, and I have made slightly different assumptions regarding system behavior. Ptonsparky needs to provide a little more information to clarify exactly what's going on to determine the actual form of the equation.
 
gadfly56

gadfly56

Senior Member
Location
New Jersey
Occupation
Professional Engineer, Fire & Life Safety
ptonsparky said:
it will be a while before I can fill in the blanks with known values. Thank you all.
Click to expand...

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?
 
W

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
wwhitney said:
At time t = 3, raw product has run out yet (I > 3 * PL), and new raw product starts arriving at rate PA > PL.
Click to expand...
I dropped a "not" in the above, I meant to say that at time t=3, raw product has not run out yet, which is what the inequality signifies.

Cheers, Wayne
 
ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
For a change, I’m busy at this moment.

Fulthrotl, you’re right to some extent, but it’s logged data every 5 seconds and I have 4 years or so of data. I know how it operates and can predict what I’m looking for but I know the formula might be a quicker prediction as well as a check.
 
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