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

In summary, the conversation discusses a problem involving flow rate out of a tank with uniform cross-sectional area. The problem is treated as a volumetric flow problem and involves multiple terms such as Qouthole, Qoutengine, and Qin. The DE equation is given and the individual steps of integrating are outlined. However, there is a mistake in the original equation and the speaker is looking for help in solving the integral.
  • #1
ryancalif
4
0
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
 
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  • #2
Could you outline how did you arrive at the expression you found after integrating?
 
  • #3
coelho said:
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
 
  • #4
ryancalif said:
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?
 
  • #5
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!
 

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

1. What is volume flow rate?

Volume flow rate is the amount of fluid that passes through a given cross-sectional area in a specific amount of time. It is commonly measured in units such as liters per second or cubic meters per hour.

2. How is volume flow rate calculated?

Volume flow rate can be calculated by dividing the volume of fluid that flows through a given cross-sectional area by the time it takes for the fluid to pass through that area. The formula for calculating volume flow rate is Q = V/t, where Q is the volume flow rate, V is the volume, and t is the time.

3. What factors affect volume flow rate?

Volume flow rate can be affected by a number of factors, including the size and shape of the cross-sectional area, the viscosity of the fluid, and the pressure and velocity of the fluid. These factors can impact the rate at which the fluid flows through the area.

4. How is volume flow rate used in real-world applications?

Volume flow rate is an important measurement in many industries, such as in the design of pipelines, pumps, and ventilation systems. It is also commonly used in the measurement of blood flow in medical applications.

5. How is volume flow rate different from mass flow rate?

Volume flow rate measures the amount of fluid passing through a given area in a specific amount of time, while mass flow rate measures the amount of mass passing through the same area in the same amount of time. The two can be related by the density of the fluid, as mass flow rate is equal to volume flow rate multiplied by the density of the fluid.

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