My Math Forum Mixing problem!!!!!

 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mix of (1+cos t)/6 % alcohol is fed in at 6 gallons per minute and the mixture stirred; simultaneously the mixture is withdrawn at 3 gallons per minute.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 Display Modes Linear 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