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I've come across a scenario where I have a vertical cylindrical tank filled with water, that has an inlet flow valve and outlet flow valve.

Known values:

  • Resistance or gain R = 1.5 m3/min
  • Area of tank = 2 m2
  • Gravity due to acceleration = 9.81 m/s2
  • Density p = 1000 kg/m3

The transfer function for this open-loop control system (I think):

$${H(s)\over Qi(s)} = {k\over ts+1}$$

I'm a bit stuck on how to get the time constant, a lot of the examples I've looked at are two tanks or the tank is horizontal.

Can anyone help explain if I've got the transfer function right and how to get the time constant? I think I'm meant to use Laplace transform but don't know what variables to use.

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  • \$\begingroup\$ Why don't you be less ambiguous and call the valve resistance 1.5 minutes per cubic metre? But I'm still unsure how this can be solved from what info you've supplied. \$\endgroup\$
    – Andy aka
    Commented Jan 5, 2020 at 17:20
  • \$\begingroup\$ I've found online that T=A/Rpg and K=1/Rpg, I plugged that into the transfer function I think I'm meant to use and got 4.077 e-3 / 8.153 e-3 s+1? Using p=1000 kg/m^3 of water and g= 9.81 m/s^2. Does this look right, or have I totally mixed stuff up and gone wrong? \$\endgroup\$
    – Marie87
    Commented Jan 5, 2020 at 17:26
  • \$\begingroup\$ What is your gain if current output = input and gain is measured by fluid volume change ? Gain =0 Your question is unclear for specs. Re-write in the most concise form with all assumptions. \$\endgroup\$
    – D.A.S.
    Commented Jan 5, 2020 at 17:39
  • \$\begingroup\$ What is the resistance to? Flow? What is it of? The outlet? How can the resistance be in \$\mathrm{m^3/minute}\$? Would't it be in flow/head (making it \$\mathrm{m^2/minute}\$) or flow/pressure (making it \$\mathrm{m^3/kPa}\$ or for the pedantic, \$\mathrm{m^5/N}\$)? \$\endgroup\$
    – TimWescott
    Commented Jan 5, 2020 at 21:47
  • \$\begingroup\$ The resistance is of the valve, and the flow is the amount of water that travels through the valve per minute. Those are the values I was given. \$\endgroup\$
    – Marie87
    Commented Jan 6, 2020 at 11:26

1 Answer 1

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If the "current or fluid flow" resistance is equal for both input and output then the currents are equal and the storage remains constant. It never fills. right?

enter image description here

It's a like a battery charger , battery and load. The battery charge does not change if in = out. Then the battery or tank capacity is irrelevant.

schematic

simulate this circuit – Schematic created using CircuitLab

Summary

Your time is empty volume/ max fill rate and to fill up only occurs when the outlet is reduced or closed. If volume fills per unit height based on maximum height velocity=volume flow/Area= 1.5 m^3/min / 2m^2 = 3/4 m/min then depends on volume to fill and flow rate current difference. Nothing fancy.

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  • \$\begingroup\$ The equation given is question is incorrect. But the rate of change in water height is known as (in-out)x0.75m/s \$\endgroup\$
    – D.A.S.
    Commented Jul 22, 2022 at 16:40
  • \$\begingroup\$ You think it’s flow is linear? \$\endgroup\$
    – RemyHx
    Commented Jul 22, 2022 at 18:22
  • \$\begingroup\$ There are no control parameters and if there were, they would likely be nonlinear. The only flow given was fixed. But they could be made quasi-linear with flow sensors \$\endgroup\$
    – D.A.S.
    Commented Jul 22, 2022 at 19:54

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