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

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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?
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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2. gar
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180509-1957 EDT

ptonsparky:

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.
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.

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Originally Posted by gar
180509-1957 EDT

ptonsparky:

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)

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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?
No, the fill motor does not start for three minutes as product is leaving the tank

5. Originally Posted by ptonsparky
No, the fill motor does not start for three minutes as product is leaving the tank
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.

6. gar
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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.

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If outflow is by gravity, calculus is necessary to arrive at the form of equation.

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Originally Posted by Sahib
If outflow is by gravity, calculus is necessary to arrive at the form of equation.
Outflow is not gravity, but my elevation is 2200’ and if the operators sister looks anything like him, I really don’t care about the color of her dress. Only that she is wearing one.

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Originally Posted by gar
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.

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