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May 11th, 2017, 03:42 PM  #1 
Member Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1  Time taken for the volume of fluid in a leaky container to rise from Vo/4 to 3Vo/4
Fluid enters a leaky vessel through a valve. The valve admits fluid at a rate proportional to the volume of fluid already in the vessel, and the rate of leakage is proportional to the square of the volume already in the vessel. There is a balance between inflow and outflow when the volume in the vessel is $V_o$. Initially there is a volume $\frac{V_o}{4}$ in the vessel, and the volume increases to $\frac{V_o}{2}$ in time $T$. Find the time taken for the volume to increase from $\frac{V_o}{4}$ to $\frac{3V_o}{4}$. For this problem the following differential equation can be set up: $$\frac{dV}{dT} = aV  bV^2$$ With the information given, what is the best way to find the values of $a$ and $b$ so that the differential equation can be solved (the answer for the volume increase should be $1.39T$). 
May 12th, 2017, 05:25 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,959 Thanks: 802 
We can write the equation as $\displaystyle \frac{dV}{aV bV^2}= dT$. By "partial fractions", $\displaystyle \frac{dV}{V(a bV)}= \frac{\frac{1}{a}dV}{V}\frac{\frac{b}{a}dV}{a bV}$. That can be integrated to give $\displaystyle \frac{1}{a}\lnV+ \frac{b}{a}\lna bV= \ln\left(V^{\frac{1}{a}}(a bV)^{\frac{b}{a}}\right)$. Setting that equal to the integral of $dT$, $T+ C$, we have $\displaystyle \ln\left((V^{\frac{1}{a}})\left(a bV\right)^{\frac{b}{a}}\right)= T+ C$ and taking the exponential of both sides, $\displaystyle V^{\frac{1}{a}}(a bV)^{\frac{b}{a}}= C' e^T$ where $\displaystyle C'= e^C$. Now, for what values of a and b can that NOT be done? Last edited by skipjack; May 12th, 2017 at 06:24 AM. 
May 12th, 2017, 06:10 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493 
Let $t$ be the elapsed time, and $v$ be the volume of fluid in the vessel, then $dv/dt = av  bv^2$. I get $v = V_0e^{at}/(e^{at} + 3)$ and $V_0 = a/b$, with $v = V_0$ for the balance situation. Substituting $t = T$ and $v =V_0/2$ leads to $a = (\ln 3)/T$. However, solving for $t$ when $v = 3V_0/4$ doesn't seem to lead to the answer given. Perhaps I made a slip somewhere. 
May 12th, 2017, 01:02 PM  #4 
Member Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 
For the partial fractions, I get: $$ \frac{1}{V(abV)} = \frac{{1}/{a}}{V} + \frac{\frac{b}{a}}{abV}$$ so that when multiplied out gives $$\frac{1}{aVbV^2}$$ and for the integral of $$\frac{\frac{b}{a}}{abV}$$ I get $$\frac{1}{a} \ln(abV)$$ using the substitution $u=abV$ so $du/dv = b$ and $dV/du = \frac{1}{b}$ so the $b$ cancels. Last edited by skipjack; May 12th, 2017 at 05:16 PM. 
May 12th, 2017, 02:35 PM  #5 
Member Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 
The above gives $$Ce^T = \frac{V^\frac{1}{a}}{(abV)^\frac{1}{a}}$$ but $$V= \frac{V_oe^{aT}}{e^{aT} + 3}$$ is a solution to the differential equation, using the initial condition $V(0) = \frac{V_o}{4}$. Last edited by Statistics132; May 12th, 2017 at 02:55 PM. 
May 12th, 2017, 05:30 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493 
Those equations are equivalent for a suitable value of C.


Tags 
3vo or 4, container, differential equation, fluid, fluid flow, leaky, rise, time, v0 or 4, vo or 4, volume 
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