Let's go through each part step by step: --- ### **(a) Mathematical model for the mixing process** Let: - \( V(t) \): Volume of liquid in the chamber at time \( t \) (L) - \( Q(t) \): Amount of dye in the chamber at time \( t \) (kg) - \( C(t) \): Concentration of dye in the chamber at time \( t \) (kg/L), so \( Q(t) = V(t) C(t) \) - Inflow rates: clear liquid (\( 10 \) L/min), dye solution (\( r \) L/min, \( 0.25 \) kg/L) - Outflow rate: \( 10 + r \) L/min **1. Volume balance:** \[ \frac{dV}{dt} = \text{Total inflow} - \text{Total outflow} = (10 + r) - (10 + r) = 0 \] So, \( V(t) = 5 \) L (constant). **2. Dye balance:** \[ \frac{dQ}{dt} = \text{Dye in} - \text{Dye out} \] - Dye in: Only the dye solution contributes, at rate \( r \) L/min, concentration \( 0.25 \) kg/L: \[ \text{Dye in} = 0.25r \ \text{kg/min} \] - Dye out: Outflow rate is \( 10 + r \) L/min, concentration in tank is \( C(t) \): \[ \text{Dye out} = (10 + r)C(t) \] So the model is: \[ \frac{dQ}{dt} = 0.25r - (10 + r)C(t) \] But \( Q(t) = V C(t) = 5C(t) \), so: \[ \frac{d}{dt}[5C(t)] = 0.25r - (10 + r)C(t) \] \[ 5 \frac{dC}{dt} = 0.25r - (10 + r)C(t) \] \[ \boxed{ \frac{dC}{dt} + \frac{10 + r}{5} C = \frac{0.25r}{5} } \] --- ### **(b) Required injection rate \( r \) for equilibrium concentration \( 0.2 \) kg/L** At equilibrium (\( \frac{dC}{dt} = 0 \)), \( C = C_{eq} = 0.2 \) kg/L: \[ 0 + \frac{10 + r}{5} (0.2) = \frac{0.25r}{5} \] \[ (10 + r) \cdot 0.2 = 0.25r \] \[ 2 + 0.2r = 0.25r \] \[ 2 = 0.05r \] \[ r = \frac{2}{0.05} = 40 \] **Final Answer:** \[ \boxed{r = 40 \text{ L/min}} \] --- ### **(c) Would the equilibrium \( r \) value be different if initial fluid had dye?** No, the equilibrium value **does not depend on initial conditions**. It depends only on the rates and desired equilibrium concentration. Initial conditions only affect **how long it takes** to reach equilibrium. **Final Answer:** \[ \boxed{\text{No, the equilibrium value of } r \text{ would not be different.}} \] --- ### **(d) Time for outflow concentration to rise to within 4% of 0.2 kg/L** We want \( C(t) \) to reach at least \( 0.2 \times 0.96 = 0.192 \) kg/L (within 4% of 0.2 kg/L). From (a), the ODE: \[ \frac{dC}{dt} + \frac{10 + r}{5} C = \frac{0.25r}{5} \] With initial condition \( C(0) = 0 \). General solution for a first-order linear ODE: \[ C(t) = C_{eq} + (C(0) - C_{eq}) e^{-kt} \] where \( k = \frac{10 + r}{5} \), \( C_{eq} = 0.2 \), \( C(0) = 0 \): \[ C(t) = 0.2 \left[ 1 - e^{-kt} \right] \] Set \( C(t) = 0.192 \): \[ 0.192 = 0.2 \left[ 1 - e^{-kt} \right] \] \[ \frac{0.192}{0.2} = 1 - e^{-kt} \] \[ 0.96 = 1 - e^{-kt} \] \[ e^{-kt} = 0.04 \] \[ -kt = \ln(0.04) \] \[ t = -\frac{\ln(0.04)}{k} \] With \( r = 40 \): \[ k = \frac{10 + 40}{5} = 10 \] \[ t = -\frac{\ln(0.04)}{10} \] Calculate: \[ \ln(0.04) \approx -3.2189 \] \[ t = -\frac{-3.2189}{10} = 0.3219 \text{ min} \] **Final Answer:** \[ \boxed{t \approx 0.322 \text{ min} \approx 19.3 \text{ seconds}} \] --- ## **Summary of Answers** **(a) Model:** \(\frac{dC}{dt} + \frac{10+r}{5}C = \frac{0.25r}{5}\) with \(C(0)\) initial. **(b) Required \( r \):** \(\boxed{40 \text{ L/min}}\) **(c) Equilibrium \( r \) depends only on rates/concentrations, not initial condition:** \(\boxed{\text{No, it would not be different.}}\) **(d) Time to within 4%:** \(\boxed{t \approx 0.322 \text{ min} \approx 19.3 \text{ seconds}}\) Let me know if you want more explanation or the full derivation for part (a)!

The mixing process can be modeled using a first-order linear differential equation that balances dye input and output, assuming perfect mixing and constant flow rates. Equilibrium concentration is determined by the ratio of dye inflow to total outflow.