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May 11th, 2017, 02:42 PM   #1
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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 $1.39T$).
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May 12th, 2017, 04:25 AM   #2
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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 $\displaystyle \frac{1}{a}\ln|V|+ \frac{b}{a}\ln|a- bV|= \ln\left(V^{\frac{1}{a}}(a- bV)^{\frac{b}{a}}\right)$.

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?
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Last edited by skipjack; May 12th, 2017 at 05:24 AM.
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May 12th, 2017, 05:10 AM   #3
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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.
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May 12th, 2017, 12:02 PM   #4
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For the partial fractions, I get:

$$ \frac{1}{V(a-bV)} = \frac{{1}/{a}}{V} + \frac{\frac{b}{a}}{a-bV}$$

so that when multiplied out gives $$\frac{1}{aV-bV^2}$$

and for the integral of $$\frac{\frac{b}{a}}{a-bV}$$ I get $$\frac{-1}{a} \ln(a-bV)$$ using the substitution $u=a-bV$ so $du/dv = -b$ and $dV/du = \frac{-1}{b}$ so the $b$ cancels.

Last edited by skipjack; May 12th, 2017 at 04:16 PM.
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May 12th, 2017, 01:35 PM   #5
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The above gives $$Ce^T = \frac{V^\frac{1}{a}}{(a-bV)^\frac{1}{a}}$$

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 01:55 PM.
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May 12th, 2017, 04:30 PM   #6
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Those equations are equivalent for a suitable value of C.
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3vo or 4, container, differential equation, fluid, fluid flow, leaky, rise, time, v0 or 4, vo or 4, volume



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