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

  1. #11
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    Quote Originally Posted by ptonsparky View Post
    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?
    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

  2. #12
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    Quote Originally Posted by kwired View Post
    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?
    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.
    Tom
    TBLO

  3. #13
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    Quote Originally Posted by Phil Corso View Post
    You wili probably ask for the name of conductor's sister!
    Phil Corso
    I am close to 70, her dress color means nothing to me.
    Tom
    TBLO

  4. #14
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    it will be a while before I can fill in the blanks with known values. Thank you all.
    Tom
    TBLO

  5. #15
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    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.

    .

  6. #16
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    Quote Originally Posted by gar View Post
    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.

    .
    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.

  7. #17
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    Quote Originally Posted by ptonsparky View Post
    it will be a while before I can fill in the blanks with known values. Thank you all.
    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?

  8. #18
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    Quote Originally Posted by ptonsparky View Post
    it will be a while before I can fill in the blanks with known values. Thank you all.
    i'm getting an image of someone with a clipboard and a whistle and a stopwatch....
    ~New signature under construction.~
    ~~~~Please excuse the mess.~~~~

  9. #19
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    Quote Originally Posted by wwhitney View Post
    At time t = 3, raw product has run out yet (I > 3 * PL), and new raw product starts arriving at rate PA > PL.
    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

  10. #20
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    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.
    Tom
    TBLO

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