# input/output mixing problem [closed]

'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 $\frac{1+\cos t}{6}$ % alcohol is fed in at 6 gallons per minute and the mixture stirred; simultaneously the mixture is withdrawn at 3 gallons per minute.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?

## closed as off-topic by Harish Chandra Rajpoot, N. F. Taussig, Mankind, Dan, Tim RaczkowskiSep 20 '15 at 21:14

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• I don't see why this is downvoted. The question seems to be on on-topic. – Calle Sep 20 '15 at 14:41
• Can you post your attempt? Hint: the "nice" quantities are the total volume of the mixture (this is easy to compute) and the total volume of alcohol in the mixture. The percentage should be calculated by dividing these two quantities, rather than trying to manipulate it directly. – Ian Sep 20 '15 at 14:55

Let $V(t)$ be the volume of solution in your tank at time $t$. We have $V(0) = 600$. Let $Q(t)$ be the volume of alcohol in your tank at time $t$, so we have $Q(0) = 600$. The concentration at time $t$ is: $$C(t) = \frac{Q(t)}{V(t)}.$$ We measure the volume and the quantity of alcohol in gallons. The concentration has no unit. We measure time in minutes.
The volume is increased by $R_{in} = 6$ gallons per minute and decreased by $R_{out} = 3$ gallons per minute. Hence we have $$\frac{dV}{dt} = R_{in} - R_{out} = 3.$$ So, this gives us equations for the volume: \begin{align} V(0) &= 600 \\ \frac{dV}{dt} &= 3 \end{align} which has solution $V(t) = 600 + 3t$.
At time $t$, the quantity of alcohol entering the tank is $R_{in} \frac{1 + \cos t}{6}$ (flow rate in times alcohol concentration in the inflow) and the amount of alcohol exiting the tank is $R_{out} C(t) = R_{out}\frac{Q(t)}{600 + 3t}$ (flow rate out times alcohol concentration in the tank). So we can now setup the equations for $Q(t)$: \begin{align} Q(0) &= 600 \\ \frac{dQ}{dt} &= R_{in} \frac{1 + \cos t}{6} - R_{out} C(t) = 1 + \cos t - \frac{Q(t)}{200 + t} \end{align} The solution to this is: $$Q(t) = \frac{239998+400 t+t^2+2 \cos t}{400+2 t}+\sin t$$ and we get that the concentration is: $$C(t) = \frac{Q(t)}{V(t)} = \frac{\frac{239998+400 t+t^2+2 \cos t}{400+2 t}+\sin t}{600+3t} = \frac{239998+400t+t^2+2 \cos t+(400+2t) \sin t}{6 (200+t)^2}.$$
• Yes, you should solve $C (t) = 44\% = 0.44$. Make sure you get the initial condition for $C (0)$ right. – Calle Sep 22 '15 at 11:07