Exploring the Universe Through Equations.

✨ Equation of the Week ✨

Dt=K1L0K2K1(eK1teK2t)+D0eK2tD_t = \frac{K_1 L_0}{K_2 - K_1}(e^{-K_1 t} - e^{-K_2 t}) + D_0 e^{-K_2 t}

This is the Streeter-Phelps equation, which describes the dissolved oxygen sag curve in a water body following the discharge of organic pollutants. It models the dynamic balance between oxygen depletion due to biological oxygen demand (BOD) and oxygen replenishment through reaeration.

  • DtD_t: Dissolved oxygen deficit at time tt (mg/L)
  • K1K_1: Deoxygenation rate constant (day⁻¹) - rate of oxygen consumption by microorganisms
  • K2K_2: Reaeration rate constant (day⁻¹) - rate of oxygen replenishment from atmosphere
  • L0L_0: Initial ultimate BOD (biochemical oxygen demand) concentration (mg/L)
  • D0D_0: Initial dissolved oxygen deficit (mg/L)
  • tt: Time downstream from pollution source (days)

The equation shows how dissolved oxygen levels initially decrease due to microbial consumption of organic matter, reach a critical minimum, then gradually recover as reaeration dominates. This "sag" pattern is crucial for water quality management and environmental protection.

Example Problem

A river receives an organic waste discharge with the following parameters: K1=0.3 day1K_1 = 0.3 \text{ day}^{-1}, K2=0.5 day1K_2 = 0.5 \text{ day}^{-1}, L0=30 mg/LL_0 = 30 \text{ mg/L}, and D0=2 mg/LD_0 = 2 \text{ mg/L}. Find the time and magnitude of the critical dissolved oxygen deficit (maximum oxygen depletion).

Solution

Step 1: Find the critical time

The critical time occurs when the rate of change of deficit equals zero:

dDtdt=0\frac{dD_t}{dt} = 0

Taking the derivative of the Streeter-Phelps equation:

dDtdt=K1L0K2K1(K1eK1t+K2eK2t)K2D0eK2t\frac{dD_t}{dt} = \frac{K_1 L_0}{K_2 - K_1}(-K_1 e^{-K_1 t} + K_2 e^{-K_2 t}) - K_2 D_0 e^{-K_2 t}

Setting this equal to zero and solving for critical time:

tc=1K2K1ln(K2K1(1D0(K2K1)K1L0))t_c = \frac{1}{K_2 - K_1} \ln\left(\frac{K_2}{K_1}\left(1 - \frac{D_0(K_2 - K_1)}{K_1 L_0}\right)\right)

Step 2: Calculate critical time

Substituting the given values:

tc=10.50.3ln(0.50.3(12(0.50.3)0.3×30))t_c = \frac{1}{0.5 - 0.3} \ln\left(\frac{0.5}{0.3}\left(1 - \frac{2(0.5 - 0.3)}{0.3 \times 30}\right)\right)
tc=10.2ln(1.667×(10.044))=5ln(1.594)=2.35 dayst_c = \frac{1}{0.2} \ln\left(1.667 \times (1 - 0.044)\right) = 5 \ln(1.594) = 2.35 \text{ days}

Step 3: Calculate critical deficit

Substituting tc=2.35t_c = 2.35 days into the original equation:

Dc=0.3×300.50.3(e0.3×2.35e0.5×2.35)+2e0.5×2.35D_c = \frac{0.3 \times 30}{0.5 - 0.3}(e^{-0.3 \times 2.35} - e^{-0.5 \times 2.35}) + 2 e^{-0.5 \times 2.35}
Dc=45(e0.705e1.175)+2e1.175=45(0.4940.309)+2(0.309)=8.95 mg/LD_c = 45(e^{-0.705} - e^{-1.175}) + 2 e^{-1.175} = 45(0.494 - 0.309) + 2(0.309) = 8.95 \text{ mg/L}

Step 4: Final result

The critical dissolved oxygen deficit occurs at 2.35 days downstream from the pollution source, with a maximum deficit of 8.95 mg/L. This represents the point of lowest dissolved oxygen concentration in the river, which is crucial for assessing potential fish kills and ecosystem impacts.

Contributor: H.W. Streeter & Earl B. Phelps

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