Question

A tank contains 250 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Answer 1

Answer:

$A(t)=250-230e^{-\frac{t}{50} }$

Step-by-step explanation:

A(t) is the amount of salt in the tank at time t.

dA / dt = rate of salt flowing into the tank - rate of salt going out of the tank

dA / dt = (1 g/L)*(5 L/min) - (A(t)/250 g/L) * (5L/min)

dA / dt = 5 g/min - (A(t) / 50) g/min

$\frac{dA}{dt}+\frac{A(t)}{50} = 5\\\\The\ integrating\ factor(IF)= e^{\int\limits \frac{1}{50}dt }=e^{\frac{t}{50} }\\\\Multiplying\ through\ by\ the\ I.F:\\\\\frac{dA}{dt}*e^{\frac{t}{50} }+\frac{A(t)}{50}*e^{\frac{t}{50} } = 5*e^{\frac{t}{50} }\\\\Integrating \ both \ sides:\\\\\int\limits[ \frac{dA}{dt}*e^{\frac{t}{50} }+\frac{A(t)}{50}*e^{\frac{t}{50} }] dt=\int\limits 5e^{\frac{t}{50} } dt\\\\A(t)e^{\frac{t}{50} } =\int\limits 5e^{\frac{t}{50} } dt$

$A(t)e^{\frac{t}{50} } =250e^{\frac{t}{50} }+C\\\\A(t)=250+Ce^{-\frac{t}{50} }\\\\But\ A(0) = 20\\\\20=250+Ce^{-\frac{0}{50} }\\\\C+250=20\\\\C=-230\\\\A(t)=250-230e^{-\frac{t}{50} }$