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Archy [21]
3 years ago
8

A tank contains 350 liters of fluid in which 50 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of

5 L/min; the well-mixed solution is pumped out at the same rate. Let the A(t) be the number of grams of salt in the tank at time t. Find the number A(t).)
Engineering
1 answer:
Phantasy [73]3 years ago
8 0

Answer:

A(t)=50e^{-\frac{t}{70}}

Explanation:

The volume of fluid in the tank =350 liters

Initial Amount of Salt, A(0)=50 grams

<u>Amount of Salt in the Tank </u>

\dfrac{dA}{dt}=R_{in}-R_{out}

Pure water is then pumped into the tank at a rate of 5 L/min. Therefore:

R_{in}=(concentration of salt in inflow)(input rate of brine)

=(0\frac{g}{L})( 5\frac{L}{min})\\\\R_{in}=0\frac{g}{min}

R_{out}=(concentration of salt in outflow)(output rate of brine)

=(\frac{A(t)}{350})( 5\frac{L}{min})\\R_{out}=\frac{A(t)}{70}

Therefore:

\dfrac{dA}{dt}=0-\dfrac{A}{70}

We then solve the resulting differential equation by separation of variables.

\dfrac{dA}{dt}+\dfrac{A}{70}=0\\$The integrating factor: e^{\int \frac{1}{70}dt} =e^{\frac{t}{70}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{70}}+\dfrac{A}{70}e^{\frac{t}{70}}=0\\(Ae^{\frac{t}{70}})'=0

Taking the integral of both sides

\int(Ae^{\frac{t}{70}})'=\int 0 dt\\Ae^{\frac{t}{70}}=C\\$Divide both sides by e^{\frac{t}{70}}\\A(t)=Ce^{-\frac{t}{70}}

Recall that when t=0, A(0)=50 grams (our initial condition)

A(t)=Ce^{-\frac{t}{70}}\\50=Ce^{-\frac{0}{70}}\\C=50

Therefore, the number of grams of salt in the tank at time t is:

A(t)=50e^{-\frac{t}{70}}

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