Chapter 16, Problem 17P

TheStreeter-Phelps model can be used to compute the dissolved oxygen concentration in a river below a point discharge of sewage (Fig. P16.17),

$0=0_s-\frac{k_d{L}_0}{k_d+k_s-k_a}\left(e^{-k_{a}t}{e}^{\left(-k_d+k_s\right)t}\right)-\frac{S_b}{k_b}\left(1-e^{-k_{a}t}\right)$

where $0=$ dissolved oxygen concentration (mg/L), $0_s=$ oxygen saturation concentration (mg/L), $t=$ travel time (d), $L_0=$ biochemical oxygen demand (BOD) concentration at the mixing point (mg/L), $k_d=$ rate of decomposition of BOD $\left(d^{-1}\right),$ $k_s=\text{rate of setling of BOD }\left(d^{-1}\right),k_a=\text{reaction rate }\left(d^{-1}\right),\text{ and }{S}_b=\text{sen}\dim \text{ent}$ oxygen demand $\left(\text{mg/L/d}\right)$.

As indicated in Fig. P16.17, Eq. (P16.17) produces an oxygen “sag” that reaches a critical minimum level $o_c$ some travel time $t_c$

FIGURE P16.17

A dissolved oxygen “sag” below a point discharge of sewageinto a river.

below the point discharge. This point is called “critical” because it represents the locationwhere biota that depend on oxygen (like fish) would be the most stressed. Determine the critical travel time and concentration, given the following values:

$\begin{array}{l}{0}_s=10\text{ mg/L }{k}_d=0.1d^{-1}{k}_a=0.6d^{-1}\\ k_s=0.05d^{-1}{L}_0=50\text{ mg/L }{S}_b\text{=1 mg/L/d }\end{array}$