### (a) Time-dependent BOD Profile and Steady-State Expression Let \( L(t) \) be the BOD concentration in the reservoir at time \( t \). Initial concentration: \( L() = L_ \). At steady state (as \( t \to \infty \)): \( L(t) \to L_\infty \). **Mass balance on the reservoir:** \[ \text{Accumulation} = \text{In} - \text{Out} - \text{Decay} - \text{Settling} \] \[ V \frac{dL}{dt} = Q_{in}L_{in} - Q_{out}L - V k_d L - V k_s L \] Assume \( Q_{in} = Q_{out} = Q \): \[ V \frac{dL}{dt} = Q L_{in} - Q L - V (k_d + k_s)L \] \[ \frac{dL}{dt} = \frac{Q}{V}(L_{in} - L) - (k_d + k_s)L \] \[ \frac{dL}{dt} = \frac{Q}{V}L_{in} - \left(\frac{Q}{V} + k_d + k_s\right)L \] Define the hydraulic residence time \( \tau = \frac{V}{Q} \), and \( K = k_d + k_s \): \[ \frac{dL}{dt} = \frac{1}{\tau}L_{in} - \left(\frac{1}{\tau} + K\right)L \] At steady state (\( \frac{dL}{dt} = \)): \[ = \frac{1}{\tau}L_{in} - \left(\frac{1}{\tau} + K\right)L_\infty \] \[ \left(\frac{1}{\tau} + K\right)L_\infty = \frac{1}{\tau}L_{in} \] \[ L_\infty = \frac{L_{in}}{1 + K\tau} \] So, the BOD concentration \( L(t) \) will exponentially decay from \( L_ \) to \( L_\infty \). --- ### (b) Steady-State BOD Removal Efficiency Removal efficiency is: \[ E = \frac{L_{in} - L_\infty}{L_{in}} \] Substitute \( L_\infty \): \[ E = 1 - \frac{L_\infty}{L_{in}} = 1 - \frac{1}{1 + K\tau} \] \[ E = \frac{K\tau}{1 + K\tau} \] This shows that removal efficiency increases with higher \( K \) (removal rate) and higher \( \tau \) (residence time). --- ### (c) Impact of Temperature on \( K \) and \( L_\infty \) The removal rate constant \( K \) is temperature dependent. The Arrhenius equation for \( K \) is: \[ K(T) = K_{20} \theta^{(T-20)} \] where \( K_{20} \) is the rate constant at \( 20^\circ\text{C} \), \( \theta \) is typically 1.03–1.07, and \( T \) is temperature in °C. As \( T \) increases, \( K(T) \) increases, which leads to a lower steady-state BOD \( L_\infty \): \[ L_\infty = \frac{L_{in}}{1 + K(T)\tau} \] Higher temperature → higher \( K \) → lower \( L_\infty \), meaning better BOD removal at higher temperatures. --- **Summary of Answers:** (a) The BOD concentration \( L(t) \) decays exponentially from \( L_ \) to \( L_\infty = \frac{L_{in}}{1 + K\tau} \). (b) The steady-state BOD removal efficiency is \( E = \frac{K\tau}{1 + K\tau} \). (c) As temperature increases, the removal rate constant \( K \) increases according to \( K(T) = K_{20}\theta^{(T-20)} \), thereby decreasing \( L_\infty \) and improving removal efficiency.