Calculus Calculus Math Forum

 September 20th, 2015, 05:07 AM #1 Newbie   Joined: Sep 2015 From: AU Posts: 5 Thanks: 0 Mixing problem!!!!! I'm struggling with a). Any help would be greatly appreciated!!! I tried setting up a equation and using separable ODE and then using integration by parts,however i'm getting a weird answer. Whisky consists mainly of water and alcohol. Before bottling, the alcohol is diluted with water to the standard bottle concentration of (44−46)%. Suppose a 2500 gallon tank initially contains 600 gallons of pure alcohol. An alcohol-water mix of (1+cos t)/6 % alcohol is fed in at 6 gallons per minute and the mixture stirred; simultaneously the mixture is withdrawn at 3 gallons per minute. (a) Find the amount of alcohol (in %) in the tank at time t. (b) Approximately how many hours will it take for the mixture in the tank to reach 44%? (c) Approximately how many hours will it take for the mixture in the tank to overflow?  September 20th, 2015, 05:22 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 What was the equation you tried? September 20th, 2015, 05:48 AM #3 Newbie   Joined: Sep 2015 From: AU Posts: 5 Thanks: 0 it was input - output ----> dC(%)/dt = (1+cos(t))-(3C(%)/2500) September 20th, 2015, 10:27 PM #4 Member   Joined: Apr 2015 From: USA Posts: 46 Thanks: 32 If you look at the units of $1+\cos t$, they are in alcohol amount (not concentration) per time. That means the left-hand side is not $\frac{dC}{dt}$ but rather $\frac{d(CV)}{dt}$ where $V$ is the volume of mixture. Both $C$ and $V$ are functions of time $t$. From the question, we can see that $V(t)=600+3t$. Therefore, \begin{align}\frac{d(CV)}{dt}&=1+\cos t-3C\\ \frac{V\,dC+C\,dV}{dt}&=1+\cos t-3C\\ V\frac{dC}{dt}+C\frac{dV}{dt}&=1+\cos t-3C\quad\text{Note: }\frac{dV}{dt}\text{ is a constant }6-3=+3\text{ gallons/minute}\\ \left(600+3t\right)\frac{dC}{dt}+3C&=1+\cos t-3C\\ \left(600+3t\right)\frac{dC}{dt}+6C&=1+\cos t\\ \frac{dC}{dt}+\frac{6}{600+3t}C&=\frac{1+\cos t}{600+3t}\quad\text{Note: Integrating factor is }e^{\int\frac{6}{600+3t}dt}=e^{2\ln(600+3t)}=(600+ 3t)^2\\ \frac{dC}{dt}(600+3t)^2+6(600+3t)C&=(1+\cos t)(600+3t)\\ \frac{d}{dt}\left(C(600+3t)^2\right)&=600+3t+600 \cos t+3t\cos t\\ \int d\left(C(600+3t)^2\right)&=\int\left(600+3t+600 \cos t+3t\cos t\right)dt\\ C(600+3t)^2&=600t+\frac{3}{2}t^2+600\sin t+3t\sin t+3\cos t+K\quad\text{Note: solve for arbitrary constant }K\text{ using initial conditions}\\ \end{align} Etc. To reach $44\%$, it should take about ~102 minutes or ~1.7 hours. Thanks from mikehepro September 21st, 2015, 08:41 PM #5 Newbie   Joined: Sep 2015 From: AU Posts: 5 Thanks: 0 what about this approach here? I'm getting different answers, did i do something wrong here.  September 23rd, 2015, 04:26 AM #6 Member   Joined: Apr 2015 From: USA Posts: 46 Thanks: 32 I think $Q(0)=V(0)\cdot C(0)=600\cdot\color{red}{100}\%=60,000$ My solution for $\displaystyle Q(t)=\frac{t^2+400t+2t\sin t+400\sin t+2\cos t+K}{400+2t}$ Use initial conditions to find that $\displaystyle60,000=\frac{2+K}{400} \Longrightarrow23,999,998=K$ Divide $Q(t)$ by $V(t)$ to get $\displaystyle C(t)=\frac{t^2+400t+2t\sin t+400\sin t+2\cos t+23,999,998}{(400+2t)(600+3t)}$ I think this can be massaged into my original equation above. To check, WolframAlpha gives me $t\approx101.836$ minutes to reach $44\%$, the same answer I got before. Tags mixing, problem 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 neko Differential Equations 0 May 3rd, 2014 02:23 AM kobiz Linear Algebra 4 June 15th, 2012 08:33 PM finalight Calculus 11 February 29th, 2012 08:48 PM Singularity Applied Math 2 August 21st, 2010 11:04 AM symmetry Algebra 1 June 7th, 2007 07:13 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      