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:
\[L_0 = \frac{L_{mix}}{1 - e^{-5K_T}}\]
Note: This converts 5-day BOD to Ultimate BOD using natural exponential (base e).
Calculation:
\[e^{-5K_T} = e^{-5 \times 0.219} = e^{-1.096} = 0.334\]
\[L_0 = \frac{3.47}{1 - 0.334} = \frac{3.47}{0.666} = 5.20 \text{ mg/L}\]
L₀ = 5.20 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:
\[t_c = \frac{1}{R - K} \times \ln\left[\frac{R}{K} \times \left(1 - \frac{D_0(R-K)}{K \times L_0}\right)\right]\]
Calculation:
\[\text{Calculate ratio: } \frac{D_0}{L_0} = \frac{0.89}{5.20} = 0.171\]
\[\text{Calculate argument: } \frac{F_s}{F_s-1} \times (1 - 0.171 \times (F_s-1))\]
\[= \frac{2.0}{1.0} \times (1 - 0.171 \times 1.0) = 2.0 \times 0.829 = 1.658\]
\[\ln(1.658) = 0.506\]
\[t_c = \frac{0.506}{R - K} = \frac{0.506}{0.438 - 0.219} = \frac{0.506}{0.219} = 2.31 \text{ days}\]
tc = 2.31 days
🔢 Step 10: Calculate Critical Distance (Xc)
Formula:
\[X_c = v \times t_c \times \frac{86400 \text{ sec/day}}{1000 \text{ m/km}}\]
\[X_c = 0.2 \times 2.31 \times \frac{86400}{1000} = 0.2 \times 2.31 \times 86.4 = 39.86 \text{ km}\]
Xc = 39.86 km
🔢 Step 11: Calculate Critical Deficit (Dc) and Critical DO
Formula (Full Streeter-Phelps):
\[D_c = \frac{K \times L_0}{R - K} \times \left(e^{-K \times t_c} - e^{-R \times t_c}\right) + D_0 \times e^{-R \times t_c}\]
\[\text{Calculate: } e^{-K \times t_c} = e^{-0.219 \times 2.31} = e^{-0.506} = 0.603\]
\[e^{-R \times t_c} = e^{-0.438 \times 2.31} = e^{-1.012} = 0.364\]
\[\frac{K \times L_0}{R - K} = \frac{0.219 \times 5.20}{0.219} = 5.20\]
\[D_c = 5.20 \times (0.603 - 0.364) + 0.89 \times 0.364\]
\[= 5.20 \times 0.239 + 0.324 = 1.24 + 0.32 = 1.57 \text{ mg/L}\]
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 39.86 km downstream, taking 2.31 days.
- 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)
\[e^{-5K_T} = e^{-5 \times 0.240} = e^{-1.200} = 0.301\]
\[L_0 = \frac{14.55}{1 - 0.301} = \frac{14.55}{0.699} = 20.82 \text{ mg/L}\]
L₀ = 20.82 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 (\(t_c\)):
\[\frac{D_0}{L_0} = \frac{1.58}{20.82} = 0.076\]
\[\frac{F_s}{F_s-1} \times (1 - 0.076 \times (F_s-1)) = \frac{2.0}{1.0} \times (1 - 0.076) = 2.0 \times 0.924 = 1.848\]
\[\ln(1.848) = 0.614\]
\[t_c = \frac{0.614}{R - K} = \frac{0.614}{0.480 - 0.240} = \frac{0.614}{0.240} = 2.56 \text{ days}\]
tc = 2.56 days
Critical Distance (\(X_c\)):
\[X_c = 0.15 \times 2.56 \times 86.4 = 33.18 \text{ km}\]
Xc = 33.18 km
🔢 Step 6: Calculate Critical Deficit (Dc) and Critical DO
Critical Deficit (\(D_c\)):
\[D_c = \frac{K \times L_0}{R - K} \times \left(e^{-K \times t_c} - e^{-R \times t_c}\right) + D_0 \times e^{-R \times t_c}\]
\[e^{-K \times t_c} = e^{-0.240 \times 2.56} = e^{-0.614} = 0.541\]
\[e^{-R \times t_c} = e^{-0.480 \times 2.56} = e^{-1.229} = 0.293\]
\[\frac{K \times L_0}{R - K} = \frac{0.240 \times 20.82}{0.240} = 20.82\]
\[D_c = 20.82 \times (0.541 - 0.293) + 1.58 \times 0.293\]
\[= 20.82 \times 0.248 + 0.463 = 5.16 + 0.46 = 5.62 \text{ mg/L}\]
Dc = 5.62 mg/L
Critical DO:
\[DO_{critical} = 8.53 - 5.62 = 2.91 \text{ mg/L}\]
❌ Critical DO = 2.91 mg/L (CRITICAL)
📊 Interpretation
Water Quality Status: CRITICAL (Severe Stress)
- Critical DO (2.91 mg/L) is below the 4 mg/L survival threshold for most fish.
- High initial BOD (20.82 mg/L) and elevated temperature accelerate oxygen consumption.
- The sag occurs at 2.56 days (33.18 km), suggesting severe anoxic conditions will develop.
- 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
\[e^{-5K_{20}} = e^{-5 \times 0.40} = e^{-2.0} = 0.135\]
\[L_0 = \frac{77.63}{1 - 0.135} = \frac{77.63}{0.865} = 89.75 \text{ mg/L}\]
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):
\[e^{-5K_{18}} = e^{-5 \times 0.228} = e^{-1.14} = 0.320\]
\[L_0 = \frac{40.86}{1 - 0.320} = \frac{40.86}{0.680} = 60.09 \text{ mg/L}\]
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