Solving for Time in a Volume Flow Rate DE - Help Needed

[ryancalif]

Hi all,

So I've got a fairly straight forward problem to solve here regarding flow rate out of a tank with uniform cross-sectional area. I am treating this is a volumetric flow problem since there is assumed to be volume flow out of the tank, and volume flow into the tank.

I have two Qout terms (one out of a hole in the bottom of the tank, and one from fluid consumed by an engine), as well as one Qin term (via a pump feeding fluid into the tank at a constant rate).

Qouthole (Qh) = a*C*sqrt(2*g*z)
Qoutengine (Qe) = constant
Qin (Qi) = constant

NOTE: a = exit hole area
C = energy loss coefficient
g = gravity
z = head height in tank

My DE looks like:

-A*(dh/dt) = Qh - Qi + Qe

Separating the terms and solving for time by integrating h from Hi to Hf and t from 0 to T, I get this...

((a*C)/A)*sqrt(2*g)*(2/3)*((Hf^(3/2))-(Hi^(3/2))) = (1/A)q*t

I combined the Qi and Qe terms early on since they are constants (easier to integrate), therefore, q in the above equation equals Qi-Qf.

The problem is, solving for t eliminates the tank cross-sectional area (A) since it is divided out.

I feel confident with my general equation, but may have made a mistake somewhere in solving for t. Any help would be GREATLY appreciated! Thank you

 

[coelho]

Could you outline how did you arrive at the expression you found after integrating?

 

[ryancalif]

Could you outline how did you arrive at the expression you found after integrating?

I found one (stupid) mistake I made in my original equation (divided out the Qleak term rather than adding it to the other side). So ignore that equation... Below is how I arrived at the new equation, and this one still has a problem.

Qleak = a*C*sqrt(2*g*h)
Qi = constant
Qe = constant

integrate dH from Hi to Hf
integrate dt from 0 to T
---------------------

1. -A * (dH/dt) = Qleak - Qi + Qe

2. ... simplify by combining constants... Qi + Qe = Qtot

3. dH = (-Qleak/A + Qtot/A)dt

4. dH = 1/A*(-Qleak + Qtot)dt

5. (A*dH) / (-Qleak + Qtot) = dt

6. int((A*dH) / (-Qleak + Qtot)) = int(dt)

7. ... integrate dt from 0 to T

8. int((A*dH) / (-Qleak + Qtot)) = T

---------

This is where I stopped since I could not find an easy way to integrate the left side. The basic form of the integral would be: dH/(h^1/2 + constant)

Any tips on where I may have gone wrong, or how to solve that integral would be appreciated! Thanks

 

[coelho]

This is where I stopped since I could not find an easy way to integrate the left side. The basic form of the integral would be: dH/(h^1/2 + constant)

Are H and h the same variable, or related in some way?

 

[blue_raver22]

Hello!

It looks like you have a good grasp on the problem and have set up the relevant equations correctly. However, I think the issue lies in your integration and solving for time.

First, let's take a look at your equation for the time (t) term:

((a*C)/A)*sqrt(2*g)*(2/3)*((Hf^(3/2))-(Hi^(3/2))) = (1/A)q*t

In this equation, the term on the left side is a function of the tank cross-sectional area (A), while the term on the right side is a constant (q). This means that as the cross-sectional area changes, the time it takes for the tank to drain will also change. This is not consistent with the problem statement, which assumes a uniform cross-sectional area.

To solve for time, we need to keep the cross-sectional area (A) constant. This can be achieved by using the initial and final heights (Hi and Hf) as limits of integration, rather than integrating h from 0 to T. This will result in a time term that is dependent on the tank cross-sectional area, which makes more sense intuitively.

Also, I noticed that you have a negative sign in front of the (dh/dt) term. This should not be there, as we are looking for the time it takes for the tank to drain, which is a positive value.

Overall, I would recommend double-checking your integration and making sure that the limits and signs are correct. Additionally, it may be helpful to plug in some numbers and see if your equation gives reasonable results. If you're still having trouble, it may be helpful to consult with a colleague or your professor for further guidance.

Best of luck with your problem!