[Math] Graph representation of proportionally pumped storage

[DonDegow]

Hey guys, I'm trying to use a graph to represent a storage's capacity in function of time where the storage is filled at a fixed rate but it's also emptied at a proportional rate at the same time (and thus, will reach equilibrium at a point I want to find out).

To illustrate in ONI terms, let's say I have a gas pipe constantly venting at 1kg/s in a box and that box is pumped out at a proportional rate of 10%/s. This system would reach equilibrium when filled with 10kg of gas.

I could always try it out or make the calculation for a specific case but I'm looking for a way to visualize it easily (as I can alter the pumping rates and storage size).

I'm not particularly good at maths so I figured I could ask for help. There probably is a specific math function that does exactly that, I just don't know it...

(Trying to draw my graph on desmos)

Thank you and sorry if this was the wrong place to ask for help.

[MrAnimaniac]

If I'm not mistaken the solution should be

V = (c•t + V_0) exp(-m•t)

Here V_0 is a potential filling of the storage at t=0, m the loss rate aka 10%/s and c the inflow (your 1kg/s).

[DonDegow]

Wow, it's quite close, never thought of putting an exp here, thank you.

However it doesn't reach equilibrium with that formula; it reaches a peak then loses stored material over time until 0 : 

 

If it can help, my Excel formula looks like this : B3 = B2 + 8 - B2 * 0.05

(Where 8 and 0.05 are constants in a cell like $G$1)

[LadenSwallow]

Try:

V(t) = c - (c-V(0))e^(-mt)

c is equilibrium value (10 in our case), V(0) is storage at t=0, m is the proportion pumped out (1/10th in our case)

[MrAnimaniac]

I rechecked, should be:

V = c/m + (V_0 - c/m)•exp(-m•t)

Edit:
My brain is lame but LadenSwallow's solution and mine are more or less the same. If you want to solve an issue like this you're trying to solve the differential equation. Yours should look like

d/dt V = - m V + c (*)

First you solve the homogeneous part: d/dt V = - m V => V = V_0 exp(-m t)
Hence the e-function.
Than you assume that V_0 = V_0(t) and put it into (*)
This will give you sth like

d/dt V_0 = c•exp( m•t)

If you integrate this expression over an interval of [0, t_now] and put this new V_0 into the homogeneous solution you end with the above solution (t_now = t)

[Slvrsrfr]

yay big brain stuff, i cant really follow. but i kno its cool!

[ALCRD]

On 5/2/2022 at 5:36 PM, MrAnimaniac said:

I rechecked, should be:

V = c/m + (V_0 - c/m)•exp(-m•t)

Edit:
My brain is lame but LadenSwallow's solution and mine are more or less the same. If you want to solve an issue like this you're trying to solve the differential equation. Yours should look like

d/dt V = - m V + c (*)

First you solve the homogeneous part: d/dt V = - m V => V = V_0 exp(-m t)
Hence the e-function.
Than you assume that V_0 = V_0(t) and put it into (*)
This will give you sth like

d/dt V_0 = c•exp( m•t)

If you integrate this expression over an interval of [0, t_now] and put this new V_0 into the homogeneous solution you end with the above solution (t_now = t)