# Step-by-Step Solution: DO Concentration 5 Miles Downstream Let's break the problem into key steps: 1. **Given Data** 2. **Mixing Calculations (Immediately After Discharge)** 3. **Downstream DO Calculation (Streeter-Phelps Equation)** --- ## . Given Data **Upstream River:** - Flow rate (\(Q_r\)): **35 L/s** - \(BOD_5\): **2.5 mg/L** - DO: **7.2 mg/L** **Paper Mill Discharge:** - Flow rate (\(Q_w\)): **5 L/s** - \(BOD_5\): **43 mg/L** - DO: **2. mg/L** **At Discharge Point:** - Combined flow (\(Q\)): \(Q_r + Q_w = 35 + 5 = 40~\mathrm{L/s}\) **Distance downstream:** \(x = 5~\text{miles}\) --- ## 2. Mixing Calculations (Right After Discharge) ### a. Mixed BOD (\(L_\)) \[ L_ = \frac{Q_r L_r + Q_w L_w}{Q_r + Q_w} \] Where: - \(L_r = 2.5~\mathrm{mg/L}\) (river BOD) - \(L_w = 43~\mathrm{mg/L}\) (waste BOD) \[ L_ = \frac{(35 \times 2.5) + (5 \times 43)}{40} = \frac{87.5 + 215}{40} = \frac{302.5}{40} = 7.56~\mathrm{mg/L} \] --- ### b. Mixed DO (\(DO_\)) \[ DO_ = \frac{Q_r DO_r + Q_w DO_w}{Q_r + Q_w} \] \[ DO_ = \frac{(35 \times 7.2) + (5 \times 2.)}{40} = \frac{252 + 10}{40} = \frac{262}{40} = 6.55~\mathrm{mg/L} \] --- ## 3. Downstream DO Calculation (Streeter-Phelps Equation) ### a. Streeter-Phelps Equation \[ D(x) = \frac{k_d L_}{k_r - k_d} \left( e^{-k_d t} - e^{-k_r t} \right) + D_ e^{-k_r t} \] Where: - \(D(x)\): Oxygen deficit at distance \(x\) - \(L_\): Ultimate BOD after mixing (**use \(BOD_5\) as an estimate**) - \(k_d\): Deoxygenation rate constant (\(\mathrm{day}^{-1}\)) - \(k_r\): Reaeration rate constant (\(\mathrm{day}^{-1}\)) - \(t\): time since mixing (days) - \(D_\): initial oxygen deficit (just after mixing) **Assumptions** (since not given in problem): - \(k_d = .2~\mathrm{day}^{-1}\) (typical for river) - \(k_r = .4~\mathrm{day}^{-1}\) (typical for shallow river) - Saturation DO (\(DO_{sat}\)) at 20°C: **9.1 mg/L** (assumed) ### b. Initial Oxygen Deficit (\(D_\)) \[ D_ = DO_{sat} - DO_ = 9.1 - 6.55 = 2.55~\mathrm{mg/L} \] --- ### c. Time to Travel 5 Miles - 1 mile = **1,609 m** - 5 miles = \(5 \times 1,609 = 8,045~\mathrm{m}\) - Flow = \(Q = 40~\mathrm{L/s} = .04~\mathrm{m}^3/s\) - Need river **velocity** (\(v\)), but not given. Assume \(v = .3~\mathrm{m/s}\) (typical for small river). \[ t = \frac{\text{distance}}{\text{velocity}} = \frac{8,045~\mathrm{m}}{.3~\mathrm{m/s}} = 26,817~\mathrm{s} = \frac{26,817}{360} = 7.45~\mathrm{hours} = .311~\mathrm{days} \] --- ### d. Calculate Oxygen Deficit at 5 Miles (\(D(t)\)) Plug into Streeter-Phelps: \[ D(t) = \frac{.2 \times 7.56}{.4 - .2} \left( e^{-.2 \times .311} - e^{-.4 \times .311} \right) + 2.55 \times e^{-.4 \times .311} \] Calculate exponents: - \(.2 \times .311 = .0622\) - \(.4 \times .311 = .1244\) \[ e^{-.0622} = .9397 \] \[ e^{-.1244} = .8832 \] Now plug in: \[ \frac{.2 \times 7.56}{.2} = 7.56 \] \[ D(t) = 7.56 \times (.9397 - .8832) + 2.55 \times .8832 \] \[ .9397 - .8832 = .0565 \] \[ 7.56 \times .0565 = .427 \] \[ 2.55 \times .8832 = 2.252 \] \[ D(t) = .427 + 2.252 = 2.679~\mathrm{mg/L} \] --- ### e. Downstream DO Concentration \[ DO_{down} = DO_{sat} - D(t) = 9.1 - 2.679 = 6.42~\mathrm{mg/L} \] --- ## **Final Answer** **The dissolved oxygen concentration at a point 5 miles downstream from the discharge is:** \[ \boxed{6.4~\mathrm{mg/L}} \] --- ### **Summary Table** | Location | DO (mg/L) | |-----------------|-----------| | Upstream | 7.2 | | After Mixing | 6.55 | | 5 miles Downstream | 6.4 | --- **Notes:** - Calculations assume typical values for \(k_d\), \(k_r\), velocity, and DO saturation. - If actual values are given, substitute them in the solution.