A tank contains 50 liters of mixture containing equal parts of alcohol and water . A mixture containing 2 parts of water to 1 part alcohol flows into the tank at the rate of 3 liters/min and the mixture flows out at the rate of 2liters/min. how much alcohol will there be in the solution after half an hour?

Question

A tank contains 50 liters of mixture containing  equal parts of alcohol and water . A mixture containing 2 parts of water to 1 part alcohol flows into the tank at the rate of 3 liters/min and the mixture flows out at the rate of 2liters/min. how much alcohol will there be in the solution after half an hour?

simontheengineer · Answer

The tank initially contains 25L of alcohol. 

As the flow in is 3L per min and the flow out is 2L per min the increase in volume of the liquid is 1L per minute. 

30 mintues therefore is 30L and of this 20L are water and 10L are alcohol. 

Thus, 25 initial liters of alcohol plus the new 10 liters brought in by the new fluid is 35 liters.

dumbcow · Answer

@prizzyjade, is this for a calculus class or diff equations?

@SimonTheEngineer, you are assuming the mixture going out is same concentration as going in

dumbcow · Answer

The concentration in the tank is continually changing as new mixture flows in and tank mixture flows out.

Set up equation based on rates of alcohol going in and out
$$\frac{dA}{dt} = Rate_{i n} - Rate_{out}$$
$$\frac{dA}{dt} = 3(\frac{1}{3}) - 2(\frac{A}{t +50})$$
Rearranging you get
$$\frac{dA}{dt} + (\frac{2}{t+50})A = 1$$
Solve the Diff equation to get function A(t) which will tell you the amount of alcohol at time t

dumbcow · Answer

Here is site that will help with solving this type of diff equation: http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx