Four Worked Examples with Detailed Formula Applications
📋 Step 1: Given Data
River Discharge (Qₛ)
10 m³/s
Pollution Discharge (Qₚ)
0.1 m³/s
Pollution BOD (Lₚ)
50 mg/L
Pollution DO (DOₚ)
2.0 mg/L
Stream Velocity (v)
0.2 m/s
🔢 Step 2: Calculate Mixed Temperature
Formula:
\[T_{mix} = \frac{Q_s \times T_s + Q_p \times T_p}{Q_s + Q_p}\]
Calculation:
\[T_{mix} = \frac{10 \times 22 + 0.1 \times 22}{10 + 0.1} = \frac{220 + 2.2}{10.1} = \frac{222.2}{10.1} = 22°C\]
Tmix = 22°C
🔢 Step 3: Calculate Mixed BOD (L_mix)
Formula:
\[L_{mix} = \frac{L_s \times Q_s + L_p \times Q_p}{Q_s + Q_p}\]
Calculation:
\[L_{mix} = \frac{3 \times 10 + 50 \times 0.1}{10 + 0.1} = \frac{30 + 5}{10.1} = \frac{35}{10.1} = 3.47 \text{ mg/L}\]
Lmix = 3.47 mg/L
🔢 Step 4: Calculate Mixed DO
Formula:
\[DO_{mix} = \frac{DO_s \times Q_s + DO_p \times Q_p}{Q_s + Q_p}\]
Calculation:
\[DO_{mix} = \frac{8.0 \times 10 + 2.0 \times 0.1}{10 + 0.1} = \frac{80 + 0.2}{10.1} = \frac{80.2}{10.1} = 7.94 \text{ mg/L}\]
DOmix = 7.94 mg/L
🔢 Step 5: Adjust Rate Constants for Temperature
Formula for K at temperature T:
\[K_T = K_{20} \times \theta^{(T-20)}\]
Calculation:
\[K_{22} = 0.20 \times 1.047^{(22-20)} = 0.20 \times 1.047^2 = 0.20 \times 1.096 = 0.219 \text{ /day}\]
K₂₂ = 0.219 /day
Formula for R at temperature T:
\[R_T = R_{20} \times \theta^{(T-20)}\]
Calculation:
\[R_{22} = 0.40 \times 1.047^{(22-20)} = 0.40 \times 1.047^2 = 0.40 \times 1.096 = 0.438 \text{ /day}\]
R₂₂ = 0.438 /day
🔢 Step 6: Calculate Ultimate BOD (L₀)
Formula:
\[L_0 = \frac{L_{mix}}{1 - 10^{-5K_T}}\]
Note: This converts 5-day BOD to Ultimate BOD. The factor accounts for the BOD that remains after 5 days.
Calculation:
\[K_T \text{ (base 10)} = \frac{0.219}{2.303} = 0.0951\]
\[10^{-5K_T} = 10^{-5 \times 0.0951} = 10^{-0.4755} = 0.334\]
\[L_0 = \frac{3.47}{1 - 0.334} = \frac{3.47}{0.666} = 5.21 \text{ mg/L}\]
L₀ = 5.21 mg/L
🔢 Step 7: Get Saturation DO and Calculate Initial Deficit
From saturation table at 22°C:
Cs = 8.83 mg/L
Calculate Initial Deficit (D₀):
\[D_0 = C_s - DO_{mix}\]
\[D_0 = 8.83 - 7.94 = 0.89 \text{ mg/L}\]
D₀ = 0.89 mg/L
🔢 Step 8: Calculate Self-Purification Factor
Formula:
\[F_s = \frac{R}{K}\]
\[F_s = \frac{0.438}{0.219} = 2.00\]
Fs = 2.00 (Good self-purification capacity)
🔢 Step 9: Calculate Critical Time (tc)
Formula:
\[t_c = \frac{1}{K(F_s - 1)} \times \log_{10}\left[F_s \times \left(1 - \frac{(F_s-1) \times D_0}{L_0}\right)\right]\]
Calculation:
\[\text{First calculate the ratio: } \frac{(F_s-1) \times D_0}{L_0} = \frac{(2.0-1) \times 0.89}{5.21} = \frac{0.89}{5.21} = 0.171\]
\[\text{Inner term: } 1 - 0.171 = 0.829\]
\[F_s \times 0.829 = 2.0 \times 0.829 = 1.658\]
\[\log_{10}(1.658) = 0.2196\]
\[t_c = \frac{1}{0.219 \times (2.0-1)} \times 0.2196 = \frac{0.2196}{0.219} = 1.00 \text{ days}\]
tc = 1.00 days
🔢 Step 10: Calculate Critical Distance (Xc)
Formula:
\[X_c = v \times t_c \times \frac{86400 \text{ sec/day}}{1000 \text{ m/km}}\]
\[X_c = 0.2 \times 1.00 \times \frac{86400}{1000} = 0.2 \times 86.4 = 17.28 \text{ km}\]
Xc = 17.28 km
🔢 Step 11: Calculate Critical Deficit (Dc) and Critical DO
Formula:
\[D_c = \frac{L_0}{F_s} \times 10^{-K \times t_c}\]
\[10^{-K \times t_c} = 10^{-0.219 \times 1.00} = 10^{-0.219} = 0.604\]
\[D_c = \frac{5.21}{2.0} \times 0.604 = 2.605 \times 0.604 = 1.57 \text{ mg/L}\]
Dc = 1.57 mg/L
Calculate Critical DO:
\[DO_{critical} = C_s - D_c\]
\[DO_{critical} = 8.83 - 1.57 = 7.26 \text{ mg/L}\]
✅ Critical DO = 7.26 mg/L (EXCELLENT)
📊 Interpretation
Water Quality Status: EXCELLENT
- Critical DO (7.26 mg/L) is well above the 6.5 mg/L threshold for sensitive species.
- Dilution ratio is 100:1, indicating exceptional dilution capacity.
- Self-purification factor of 2.0 indicates reaeration rate is double the deoxygenation rate.
- Oxygen sag occurs at 17.28 km downstream, with rapid recovery.
- River can support diverse aquatic ecosystems including trout and salmon.
- Current pollution load is well within the river's assimilative capacity.
Oxygen Sag Curve - Example 1
📋 Step 1: Given Data
River Discharge (Qₛ)
5 m³/s
Pollution Discharge (Qₚ)
0.5 m³/s
Pollution BOD (Lₚ)
120 mg/L
Pollution DO (DOₚ)
1.5 mg/L
Stream Velocity (v)
0.15 m/s
🔢 Step 2: Calculate Mixed Parameters (T=24°C)
Mixed BOD (Lmix):
\[L_{mix} = \frac{4 \times 5 + 120 \times 0.5}{5 + 0.5} = \frac{20 + 60}{5.5} = \frac{80}{5.5} = 14.55 \text{ mg/L}\]
Lmix = 14.55 mg/L
Mixed DO:
\[DO_{mix} = \frac{7.5 \times 5 + 1.5 \times 0.5}{5.5} = \frac{37.5 + 0.75}{5.5} = \frac{38.25}{5.5} = 6.95 \text{ mg/L}\]
DOmix = 6.95 mg/L
🔢 Step 3: Adjust Rate Constants for 24°C
\[K_{24} = 0.20 \times 1.047^{(24-20)} = 0.20 \times 1.047^4 \approx 0.240 \text{ /day}\]
\[R_{24} = 0.40 \times 1.047^{(24-20)} = 0.40 \times 1.047^4 \approx 0.480 \text{ /day}\]
K₂₄ = 0.240 /day, R₂₄ = 0.480 /day
🔢 Step 4: Calculate Ultimate BOD (L₀)
\[K_T \text{ (base 10)} = \frac{0.240}{2.303} = 0.1042\]
\[10^{-5 \times 0.1042} = 10^{-0.521} = 0.301\]
\[L_0 = \frac{14.55}{1 - 0.301} = \frac{14.55}{0.699} = 20.81 \text{ mg/L}\]
L₀ = 20.81 mg/L
🔢 Step 5: Calculate Initial Deficit and Critical Parameters
Saturation DO at 24°C:
Cs = 8.53 mg/L
Initial Deficit ($D_0$) and Self-Purification Factor ($F_s$):
\[D_0 = 8.53 - 6.95 = 1.58 \text{ mg/L}\]
\[F_s = \frac{R}{K} = \frac{0.480}{0.240} = 2.0\]
D₀ = 1.58 mg/L, Fs = 2.0
Critical Time ($t_c$):
\[\frac{(F_s-1) \times D_0}{L_0} = \frac{(2.0-1) \times 1.58}{20.81} = 0.076\]
\[\log_{10}\left[2.0 \times (1 - 0.076)\right] = \log_{10}(1.848) = 0.2667\]
\[t_c = \frac{1}{0.240 \times 1.0} \times 0.2667 = 1.11 \text{ days}\]
tc = 1.11 days
Critical Distance ($X_c$):
\[X_c = 0.15 \times 1.11 \times 86.4 = 14.39 \text{ km}\]
Xc = 14.39 km
🔢 Step 6: Calculate Critical Deficit (Dc) and Critical DO
Critical Deficit ($D_c$):
\[D_c = \frac{L_0}{F_s} \times 10^{-K \times t_c}\]
\[10^{-K \times t_c} = 10^{-0.240 \times 1.11} = 0.541\]
\[D_c = \frac{20.81}{2.0} \times 0.541 = 10.405 \times 0.541 = 5.63 \text{ mg/L}\]
Dc = 5.63 mg/L
Critical DO:
\[DO_{critical} = 8.53 - 5.63 = 2.90 \text{ mg/L}\]
❌ Critical DO = 2.90 mg/L (CRITICAL)
📊 Interpretation
Water Quality Status: CRITICAL (Severe Stress)
- Critical DO (2.90 mg/L) is below the 4 mg/L survival threshold for most fish.
- High initial BOD (20.81 mg/L) and elevated temperature accelerate oxygen consumption.
- The sag occurs relatively quickly (1.11 days), suggesting severe anoxic conditions will develop nearby.
- Management requires immediate reduction in pollution load ($L_p$) or significant artificial reaeration.
Oxygen Sag Curve - Example 2
🔢 Step 2 & 3: Mixed BOD (L_mix) and Mixed DO
Mixed BOD (Lmix):
\[L_{mix} = \frac{5 \times 3 + 350 \times 0.8}{3.8} = \frac{15 + 280}{3.8} = 77.63 \text{ mg/L}\]
Mixed DO:
\[DO_{mix} = \frac{7.0 \times 3 + 0.5 \times 0.8}{3.8} = \frac{21.4}{3.8} = 5.63 \text{ mg/L}\]
Lmix = 77.63 mg/L, DOmix = 5.63 mg/L
🔢 Step 4: Ultimate BOD (L₀) at 20°C
\[K_{20} \text{ (base 10)} = 0.40 / 2.303 = 0.1737\]
\[L_0 = \frac{77.63}{1 - 10^{-5 \times 0.1737}} = \frac{77.63}{0.865} = 89.75 \text{ mg/L}\]
L₀ = 89.75 mg/L
🔢 Step 5, 7, 8: Initial Deficit ($D_0$) and Factors
Saturation DO at 20°C:
Cs = 9.09 mg/L
Initial Deficit ($D_0$) and Self-Purification Factor ($F_s$):
\[D_0 = 9.09 - 5.63 = 3.46 \text{ mg/L}\]
\[F_s = \frac{R}{K} = \frac{0.20}{0.40} = 0.5 \text{ (Poor reaeration capacity)}\]
D₀ = 3.46 mg/L, Fs = 0.5
🔢 Step 9, 10, 11: Critical Parameters
Because $F_s < 1$ (reaeration rate is less than decay rate), the theoretical minimum DO occurs far downstream, but the river will reach **anoxia ($\text{DO}=0$)** much sooner.
Critical Time ($t_c$):
\[t_c = 3.37 \text{ days (Calculated using natural log formula)}\]
\[X_c = 0.1 \times 3.37 \times 86.4 = 29.1 \text{ km}\]
tc = 3.37 days, Xc = 29.1 km
Critical DO:
The theoretical minimum deficit $D_c$ is $> C_s$ (Deficit $> 9.09$), meaning the water reaches **0 mg/L DO** (anoxic) almost immediately. The critical DO is limited by zero.
💀 Critical DO = 0.00 mg/L (ANOXIC)
📊 Interpretation
Water Quality Status: ANOXIC / BIOLOGICALLY DEAD ZONE
- The self-purification factor is only 0.5, meaning oxygen is consumed twice as fast as it is replaced.
- The combination of extremely high BOD ($L_0=89.75 \text{ mg/L}$) and poor reaeration leads to rapid oxygen depletion.
- The river becomes anoxic ($DO=0$) very close to the discharge point, creating a dead zone.
- **Required Action:** Pre-treatment must be implemented to significantly reduce $L_p$ (e.g., to $\approx 50 \text{ mg/L}$).
Oxygen Sag Curve - Example 3
🔢 Step 2 & 3: Mixed BOD (L_mix) and Mixed DO (T=18°C)
Mixed BOD (Lmix):
\[L_{mix} = \frac{6 \times 6 + 250 \times 1.0}{7.0} = \frac{286}{7.0} = 40.86 \text{ mg/L}\]
Mixed DO:
\[DO_{mix} = \frac{8.5 \times 6 + 0.5 \times 1.0}{7.0} = \frac{51.5}{7.0} = 7.36 \text{ mg/L}\]
Lmix = 40.86 mg/L, DOmix = 7.36 mg/L
🔢 Step 4 & 5: Adjusted Ultimate BOD and Rates at 18°C
Adjusted K and R ($\theta=1.047$):
\[K_{18} = 0.25 \times 1.047^{-2} = 0.228 \text{ /day}\]
\[R_{18} = 0.25 \times 1.047^{-2} = 0.228 \text{ /day}\]
K₁₈ = 0.228 /day, R₁₈ = 0.228 /day. Thus $F_s = 1.0$
Ultimate BOD (L₀):
\[K_{18} \text{ (base 10)} = 0.228 / 2.303 = 0.099\]
\[L_0 = \frac{40.86}{1 - 10^{-5 \times 0.099}} = \frac{40.86}{0.680} = 60.09 \text{ mg/L}\]
L₀ = 60.09 mg/L
🔢 Step 7, 9, 10: Initial Deficit and Critical Time ($F_s=1$)
Initial Deficit ($D_0$) at $C_s=9.50 \text{ mg/L}$:
\[D_0 = 9.50 - 7.36 = 2.14 \text{ mg/L}\]
D₀ = 2.14 mg/L
Critical Time ($t_c$) for $K=R$:
\[t_c = \frac{1}{K} \left(1 - \frac{D_0}{L_0}\right) \quad \text{Using base } e \text{ rates}\]
\[t_c = \frac{1}{0.228} \times \left(1 - \frac{2.14}{60.09}\right) = 4.386 \times 0.9644 = 4.23 \text{ days}\]
tc = 4.23 days
Critical Distance ($X_c$):
\[X_c = 0.25 \times 4.23 \times 86.4 = 91.33 \text{ km}\]
Xc = 91.33 km
🔢 Step 11: Critical Deficit (Dc) and Critical DO
Critical Deficit ($D_c$) for $K=R$:
\[D_c = (L_0 \times K \times t_c + D_0) e^{-K t_c}\]
\[K t_c = 0.228 \times 4.23 = 0.965\]
\[D_c = (60.09 \times 0.228 \times 4.23 + 2.14) e^{-0.965}\]
\[D_c = (57.85 + 2.14) \times 0.381 = 22.86 \text{ mg/L}\]
Dc = 22.86 mg/L
Critical DO:
\[DO_{critical} = 9.50 - 22.86 = -13.36 \text{ mg/L}\]
Practical Interpretation: Since the theoretical deficit ($22.86 \text{ mg/L}$) exceeds $C_s$ ($9.50 \text{ mg/L}$), the DO drops to $\mathbf{0 \text{ mg/L}}$ (anoxic conditions) long before the critical time.
💀 Practical Critical DO = 0.00 mg/L (ANOXIC)
📊 Interpretation
Water Quality Status: CRITICAL / ANOXIC RISK
- $F_s = 1.0$ (decay equals reaeration) means the river has zero spare capacity to absorb BOD.
- High $L_0$ and initial deficit ($D_0=2.14 \text{ mg/L}$) push the DO below zero quickly, leading to a long stretch of anoxia.
- This scenario typically requires significant **wastewater treatment upgrades** (tertiary treatment) before discharge.
Oxygen Sag Curve - Example 4