Tank Draining Problem (Related Rates?)

[sowinski]

I'm supposed to find how long it takes for a cylindrical tank (diameter=0.3m) to drain from a certain level (1.0m) to a steady state level (calculated to 0.9m).

The flow into the tank from 20mm pipe with $\displaystyle v_{1}=0.595 \;\ m/s$.

\(\displaystyle V__{in dot}\) is constant and calculated to be $\displaystyle A_{1}\cdot v_{1}={\pi}\cdot (\frac{.02}{2})^{2}\cdot (0.595)=.0001869 m^3/s$

The velocity of the water (thru 10mm pipe) out of the tank is dependent on the water level and given by the equation

$\displaystyle v_{2}=0.85\sqrt{9.81\cdot (H-0.1)}$, so the flow out would be

$\displaystyle V_{out dot}={\pi}(\frac{.01}{2})^{2}\cdot 0.85\sqrt{9.81\cdot (H-0.1)}=.000066759\cdot \sqrt{9.81\cdot H-0.981}$

so flow in - flow out = dV/dt (continuity)

since H and V are related, I replaced H in terms of V... $\displaystyle V=\frac{H{\pi}(0.3 \;\ m/2)^{2}}{4}\Rightarrow H=\frac{4V}{0.3^2{\pi}}=14.15\cdot V$

so $\displaystyle V_{out dot}=.00006676\cdot \sqrt{9.81\cdot 14.15V-0.981}=.00006676\cdot \sqrt{138.8V-0.981}$

ALL THE PREVIOUS STUFF IS SETUP, I MAINLY NEED HELP WITH FOLLOWING PART

then subst into continuity equation... $\displaystyle .0001869 - .00006676\cdot \sqrt{138.8V-0.981} = \frac{dV}{dt}$

not sure here, rearranged to $\displaystyle dt = \frac{1}{0001869 - .00006676\cdot \sqrt{138.8V-0.981}}dV$

will refer to as $\displaystyle dt = \frac{1}{a - b\sqrt{cV-d}}dV$

then integrated both sided (wolfram)... $\displaystyle t = \frac{-2[a\cdot ln(b\sqrt{cV-d}-a)+b\sqrt{cV-d}]}{b^{2}\cdot c}$

height of 1m to 0.9m correspond to volumes of 0.0707 m^3 to 0.0636 m^3, respectively.

imputing boundaries I get t = 1580 seconds = 26.3 minutes, I was told the answer is around 52 minutes. I think I might have done the related rates part incorrectly.

Thanks.

 

[galactus]

I tidied up your calculations with LaTex. If you see any typos let me know. It was difficult to read as it was, especially with that scientific notation using e.

 

[sowinski]

Thanks, looks good, just when back and added "dot" to flow rate- I might have left them out to begin with.

 

[sowinski]

Hmm, seems like it is very sensitive to rounding off numbers. Using more digits I actually get 48.8 minutes which is pretty close to "about 52".

Does the part where I seperate dV/dt and the LHS then integrate look right?

 

[galactus]

Could you please post the exact problem the way it was given. That may help us to determine if you're on the right track or not.

Thanks for showing all your work.