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.
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.