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 May 11th, 2017, 02:42 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, 04:25 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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? Thanks from Statistics132 Last edited by skipjack; May 12th, 2017 at 05:24 AM. May 12th, 2017, 05:10 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,029 Thanks: 2259 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. Thanks from Statistics132 May 12th, 2017, 12:02 PM #4 Member   Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 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. May 12th, 2017, 01:35 PM #5 Member   Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 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. May 12th, 2017, 04:30 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,029 Thanks: 2259 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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post civilengineeringstudent Physics 3 January 16th, 2017 12:32 AM Nicodemus Algebra 5 February 23rd, 2016 07:16 PM shunya Elementary Math 1 November 30th, 2015 02:19 AM Luckman1 Calculus 1 May 26th, 2014 05:00 AM frankivalli Geometry 7 April 23rd, 2014 04:54 AM

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