How Does Euler's Method Help Solve Dynamic System Changes in a Tank System?

[djspuddy]

Homework Statement

here is picture I've drawn up : http://www.mathhelpforum.com/math-help/attachments/f12/20731d1297284244-five-iterations-euler-s-method-please-help-eulermethod.jpg
It is a tank fed by 2 streams with volumetric flow rates A& B - the tank operates at atmospheric pressure and drains naturally through C, The outlet flow rate C is known to be proportional to the liquid level in the tank.
question 1 = if C is shut and the tank is empty , calculate time taken for the tank to fill.

question 3 = calculate the steady state liquid level in the tank for the given conditions

question 4 = Assume that the system is at steady state with the inlets given, demonstrate how the system would respond dynamically to attain steady state if inlet A is was stopped. Show hand calculations for 5 iterations of the Euler method using step size of 20 minuites.

im just really stuck on question 4 - question 1 -3 i think I am ok on.

Homework Equations

The Attempt at a Solution

managed to answer question 1 - = (6x4) - 0.15 = 23.85 mins for tank to fill

unsure the on question 4

 

[gneill]

Try again for the time to fill. Why are you subtracting the fill rate from the volume? The units are not even then same.

Why does the diagram state a flow rate for the outlet in units of m²/sec? Are you assuming that the areal velocity should be multiplied by the water height in the tank to yield the volumetric flow rate? If so, you should make this clear.

For part 4, you say to use a step size of 20. 20 what? Minutes? Hours? Seconds?

 

[djspuddy]

what i have done for question 1=
A=dh/dt = Qi-0 ?

Qi= 150/1000=0.15
( height x area) take away inlet flows added together 120+30 divided by 1000 =

(6x4) - 150/1000 = 23.85 mins - and gneill yes that's right about the velocity should be multiplied i think -

not sure on the question 4 - don't know where to start

please also check the picture link

 

[gneill]

V = 6m x 4m² = 24m³

Rate = (120L/min + 30L/min)*1m³/1000L = 0.15m³/min

They have different units. You can't just subtract one from the other. It's like subtracting velocity from distance to find time -- it doesn't work that way.

 

[djspuddy]

thanks - I am struggling on question 4

 

[gneill]

Well, I don't know what your approach will be, but if it were me I'd first turn all the units into something manageable that I have a good "feel" for how they relate. For example, I'd express the volume in liters rather than cubic meters. That would tie in nicely with the inflow rates given in liters per minute. The outflow rate would become 30 liters per minute per meter of water height. Oh, and I'd express the area of the tank in terms of liters per meter of height (yes, m² --> L/m)

Using these units you have:

Inflows:
FA = 120 L/min
FB = 30 L/min

Outflow:
VC = 30 L/min/m

Geometry:
Vol = 24000 L
Area = 4000 L/m

So given a volume of water V in the tank, the water height would be

h(V) = V/Area

and the outflow rate would be

Vout = h(V)*VC

Suppose you had a time step of ∆T = 20 minutes, assuming no change in rates, etc., over that timestep, can you write an expression for the new volume in the tank after that time ∆T?