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#1 2012-10-08 10:58:37

Eileen
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DiffEq Mixing Problems

I am having difficulty getting the correct answers to two mixing problems. If anyone would kindly help me out, it'd be much appreciated!

Problem1:
A tank initially holds 25 gal of water. Salt enters at a rate of 2gal/min and the mixture leaves at a rate of 1 gal/min. What will be the concentration of salt when 50 gal of fluid is in the tank?
   dA/dt=2-[A(t)/(t+25)]
   IF=(t+25)
   A(t+25)sadt^2)+50t+c
   A=[(t^2)+50t+c]/(t+25)
   For A(0)=0; c=0
   A=[(t^2)+50t]/(t+25)
   50=[(t^2)+50t]/(t+25)
   t=35.36
   35.36/50=.71

       The answer should be 75%.


Problem 2:
A tank initially contains 100 gal of a solution that holds 30lb of a chemical. A solution containing 2 lb/gal of the chemical runs in at a constant rate and the mixture runs out at the same rate. What should be the rate of flow if the tank is to contain 70lb of the chemical after 40min?
   dA/dt=2r-(A(t)*r/100)
   A(t)=ce^(-rt/100)+200
   70=ce^(-40r/100)+200
   -130=ce^(-2r/5)
   ln(-130)=ln(c)*ln(e^-2r/5)
   ln(-130)=c*2r/5
   (5/2)ln(-130)=r

       The answer should be (5/2)ln(17/13)=.671

#2 2012-10-08 19:20:26

anonimnystefy
Real Member

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Re: DiffEq Mixing Problems

Hi

For 2, you should first find the constant c from the two initial conditions given, and only then find the flow rate.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#3 2012-10-09 02:22:46

bob bundy
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Re: DiffEq Mixing Problems

hi Eileen

Q1.



That looks good to me, but I'm very confused by what you then did.

It's a linear differential equation requiring an 'integrating factor'.

Re-write as



The integrating factor is just (t + 25)





t= 0 A = 0 makes C = 0

Put t = 25



So concentration is



Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

#4 2012-10-09 12:02:35

Eileen
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Re: DiffEq Mixing Problems

Thankyou very much to both of you!

#5 2012-10-09 18:52:31

anonimnystefy
Real Member

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Re: DiffEq Mixing Problems

I also think that in 2, you calculated the function incorrectly in the second step.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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