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 alcoholwater mix of (1+cos t)/6 % alcohol is fed in at 6 gallons per minute and the mixture stirred; simultaneously the mixture is withdrawn at 3 gallons per minute.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 lefthand 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, $\begin{align}\frac{d(CV)}{dt}&=1+\cos t3C\\ \frac{V\,dC+C\,dV}{dt}&=1+\cos t3C\\ V\frac{dC}{dt}+C\frac{dV}{dt}&=1+\cos t3C\quad\text{Note: }\frac{dV}{dt}\text{ is a constant }63=+3\text{ gallons/minute}\\ \left(600+3t\right)\frac{dC}{dt}+3C&=1+\cos t3C\\ \left(600+3t\right)\frac{dC}{dt}+6C&=1+\cos t\\ \frac{dC}{dt}+\frac{6}{600+3t}C&=\frac{1+\cos t}{600+3t}\quad\text{Note: Integrating factor is }e^{\int\frac{6}{600+3t}dt}=e^{2\ln(600+3t)}=(600+ 3t)^2\\ \frac{dC}{dt}(600+3t)^2+6(600+3t)C&=(1+\cos t)(600+3t)\\ \frac{d}{dt}\left(C(600+3t)^2\right)&=600+3t+600 \cos t+3t\cos t\\ \int d\left(C(600+3t)^2\right)&=\int\left(600+3t+600 \cos t+3t\cos t\right)dt\\ C(600+3t)^2&=600t+\frac{3}{2}t^2+600\sin t+3t\sin t+3\cos t+K\quad\text{Note: solve for arbitrary constant }K\text{ using initial conditions}\\ \end{align}$ Etc. To reach $44\%$, it should take about ~102 minutes or ~1.7 hours. 
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. 

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