Volume flow rate DE - Help needed


by ryancalif
Tags: flow, rate, volume
ryancalif
ryancalif is offline
#1
Feb9-12, 11:45 PM
P: 4
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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coelho
coelho is offline
#2
Feb11-12, 04:49 AM
P: 52
Could you outline how did you arrive at the expression you found after integrating?
ryancalif
ryancalif is offline
#3
Feb12-12, 10:46 AM
P: 4
Quote Quote by coelho View Post
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
coelho is offline
#4
Feb13-12, 02:16 PM
P: 52

Volume flow rate DE - Help needed


Quote Quote by ryancalif View Post
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?


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