# Finding differential equation (mixing problem)Watch

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Thread starter 2 years ago
#1
The problem states the following:
A full tank initially contains 100 litres of pure water. A brine solution containing 2 grams of salt per litre is pumped in at 3 litres per minute. The solution in the tank is well mixed and pumped out at 4 litres per minute. What is the volume of fluid in the tank after time t? Write down a differential equation for the amount of salt x in grams in the tank and verify that:
x(t) = 2(100 − t) + A(100 − t)^4

So, for the first part I got that V = 100+(3t-5t) at time t (in here I just calculated the amount of liquid as I was not sure if when they say 'fluid' will that also include salt or not)

As for the differential equation I did the following:
Let x be the amount of salt in the tank at time t, then the concentration at time t is x/100.
Now in the short period of time between delta(t) and delta(t) + t, the amount of salt entering the tank is 2*3*delta(t) = 6*delta(t)
Amount of salt leaving the tank:
(x/100)*4*delta(t) = (x/25)*delta(t)

Hence, increase in salt in the tank will be:
delta(x) = 6delta(t) - (x/25)delta(t)
and so
(dx/dt) = 6 - (x/25)
the solution to which is x = 150 - Ae^(-t/25) where A is the constant of integration

In the question they also give this expression: x(t) = 2(100 − t) + A(100 − t)^4 but I am not sure where they got it from. I thought of using Taylor expansion for the exponential but it did not seem to help as there were huge numbers in denominator and it looked nothing like the suggested expression.
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2 years ago
#2
(Original post by spacewalker)
The problem states the following:
A full tank initially contains 100 litres of pure water. A brine solution containing 2 grams of salt per litre is pumped in at 3 litres per minute. The solution in the tank is well mixed and pumped out at 4 litres per minute. What is the volume of fluid in the tank after time t? Write down a differential equation for the amount of salt x in grams in the tank and verify that:
x(t) = 2(100 − t) + A(100 − t)^4

So, for the first part I got that V = 100+(3t-5t) at time t (in here I just calculated the amount of liquid as I was not sure if when they say 'fluid' will that also include salt or not)

As for the differential equation I did the following:
Let x be the amount of salt in the tank at time t, then the concentration at time t is x/100.
Not followed it all through, but the first thing that strikes me as incorrect is - in red. Volume is not 100 at time t.
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2 years ago
#3
each minute 3 liters go in and 4 go out... so each minute the volume decreases by 1 liter ? so after t minutes ...
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Thread starter 2 years ago
#4
(Original post by ghostwalker)
Not followed it all through, but the first thing that strikes me as incorrect is - in red. Volume is not 100 at time t.
Yes, thank you! I have just realised that! I managed to solve it
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Thread starter 2 years ago
#5
Thanks a million! This helped me a lot!
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