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Exercise 10.8 · Q10

Q.A tank initially contains 5050 litres of pure water. Starting at time t=0t=0 a brine containing 22 grams of dissolved salt per litre flows into the tank at the rate of 33 litres per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time t>0t>0.

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Set up the mixture equation IN−-OUT, solve the resulting linear equation via the integrating factor, and apply the "starts as pure water" initial condition.

Step 1. Set up IN and OUT. IN =2 g/L×3 L/min=6=2\ \text{g/L}\times3\ \text{L/min}=6 g/min. Volume stays constant at 5050 L (inflow rate == outflow rate), so OUT =x50×3=3x50=\dfrac{x}{50}\times3=\dfrac{3x}{50} g/min.

Step 2. Write the differential equation. dxdt=6−3x50 ⟹ dxdt+350x=6\dfrac{dx}{dt}=6-\dfrac{3x}{50}\ \Longrightarrow\ \dfrac{dx}{dt}+\dfrac{3}{50}x=6.

Step 3. Integrating factor. P=350⇒I.F.=e3t/50P=\dfrac3{50}\Rightarrow I.F.=e^{3t/50}. …

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