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 m^3/min
  • Area of tank = 2 m^2
  • Gravity due to acceleration = 9.81 m/s^2
  • Density p = 1000 kg/m^3

The transfer function for this open-loop control system (I think): H(s)/Qi(s) = k/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?

  • \$\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 Jan 5 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 Jan 5 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\$ – Tony Stewart Sunnyskyguy EE75 Jan 5 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 Jan 5 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 Jan 6 at 11:26

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.


simulate this circuit – Schematic created using CircuitLab


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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