Question

A tank has inflow, outflow, and a stirring mechanism. Initially, the tank holds 500 L of a brine solution of concentration 200 g/L. At \( t = 0 \), an inflow of another brine solution of concentration 100 g/L starts entering the tank at the rate of 15 L/minute. At the same time, the outflow of thoroughly stirred mixture also takes place at the same rate so that the volume of brine in the tank remains constant. The brine concentration \( C \) (g/L) in the tank at any time \( t \) (minute) can be expressed by the following differential equation: \[ \frac{dC}{dt} + 0.03 C = 3 \] The brine concentration in the tank at \( t = 1.5 \) hour is ________ g/L. (rounded off to two decimal places)

Hint:

For solving first-order linear differential equations, use the integrating factor method and apply the initial conditions carefully to find the particular solution.

Answer 1

Step 1: Solve the differential equation. We are given the equation: \[ \frac{dC}{dt} + 0.03 C = 3 \] This is a first-order linear differential equation of the form \( \frac{dC}{dt} + p(t)C = q(t) \). The integrating factor is \( e^{\int p(t) \, dt} \), where \( p(t) = 0.03 \). The integrating factor is: \[ e^{\int 0.03 \, dt} = e^{0.03t} \] Now multiply the entire differential equation by \( e^{0.03t} \): \[ e^{0.03t} \frac{dC}{dt} + 0.03 e^{0.03t} C = 3 e^{0.03t} \] The left-hand side is now the derivative of \( e^{0.03t} C \): \[ \frac{d}{dt} \left( e^{0.03t} C \right) = 3 e^{0.03t} \] Integrating both sides with respect to \( t \): \[ e^{0.03t} C = \int 3 e^{0.03t} \, dt \] The integral of \( 3 e^{0.03t} \) is: \[ \int 3 e^{0.03t} \, dt = \frac{3}{0.03} e^{0.03t} = 100 e^{0.03t} \] So, we have: \[ e^{0.03t} C = 100 e^{0.03t} + C_1 \] Dividing both sides by \( e^{0.03t} \): \[ C = 100 + C_1 e^{-0.03t} \] Step 2: Apply the initial condition. At \( t = 0 \), the concentration is 200 g/L: \[ C(0) = 200 = 100 + C_1 e^{0} \] \[ 200 = 100 + C_1 \] \[ C_1 = 100 \] Thus, the solution is: \[ C = 100 + 100 e^{-0.03t} \] Step 3: Calculate the concentration at \( t = 1.5 \) hours. Since \( t = 1.5 \) hours = 90 minutes: \[ C(90) = 100 + 100 e^{-0.03 \times 90} \] \[ C(90) = 100 + 100 e^{-2.7} \] Using the approximate value \( e^{-2.7} \approx 0.0672 \): \[ C(90) = 100 + 100 \times 0.0672 = 100 + 6.72 = 106.72 \] Step 4: Conclusion. The brine concentration in the tank at \( t = 1.5 \) hours is approximately: \[ \boxed{106.72 { g/L}} \]