3.5 Marine Pollution

Step 1: Conservation of Mass for a Dissolved Pollutant

Consider a control volume in the ocean. The concentration C of a pollutant changes due to advective transport by currents, turbulent diffusion, and chemical/biological decay. The mass balance for an infinitesimal volume element gives:

$$\frac{\partial C}{\partial t} = -\nabla \cdot (\mathbf{u} C) + \nabla \cdot (K \nabla C) + S - \lambda C$$

Step 2: Simplify to 1-D Advection-Diffusion with Decay

For a uniform flow field u in the x-direction with constant diffusivity K and first-order decay rate lambda, and no additional sources (S = 0):

$$\frac{\partial C}{\partial t} + u\frac{\partial C}{\partial x} = K\frac{\partial^2 C}{\partial x^2} - \lambda C$$

Step 3: Analytical Solution for an Instantaneous Point Source

For an instantaneous release of mass M at x = 0, t = 0, we seek a solution by substituting a Gaussian ansatz. Transforming to a moving coordinate xi = x - ut reduces to a diffusion equation with decay. The solution is:

$$C(x,t) = \frac{M}{A\sqrt{4\pi K t}} \exp\!\left[-\frac{(x - ut)^2}{4Kt}\right] \exp(-\lambda t)$$

where A is the cross-sectional area of the plume. The Gaussian spreads with standard deviation sigma = sqrt(2Kt), while the center advects at velocity u and the total mass decays exponentially.

Step 4: Derive the Gaussian Spreading Width

The variance of the concentration distribution grows linearly with time. From the second moment of the diffusion equation (Einstein relation):

$$\sigma^2(t) = \langle x^2 \rangle - \langle x \rangle^2 = 2Kt \quad \Rightarrow \quad \sigma = \sqrt{2Kt}$$

Step 5: Environmental Half-Life of a Pollutant

When multiple removal processes act in parallel (degradation k_d, burial k_b, volatilization k_v), the total effective decay rate is their sum. The environmental concentration decays as:

$$C(t) = C_0 \, e^{-(k_d + k_b + k_v)t}, \qquad t_{1/2} = \frac{\ln 2}{k_d + k_b + k_v}$$

Step 1: Single-Organism Uptake-Elimination Model

Consider an organism exposed to a contaminant at water concentration C_w. The organism absorbs the pollutant at uptake rate k_u and eliminates it at rate k_e. The internal concentration C_org obeys:

$$\frac{dC_{\text{org}}}{dt} = k_u \cdot C_w - k_e \cdot C_{\text{org}}$$

Step 2: Solve for Steady-State Bioconcentration Factor

At steady state, dC_org/dt = 0. Solving for the ratio of organism to water concentration gives the bioconcentration factor (BCF):

$$k_u C_w = k_e C_{\text{org}} \quad \Rightarrow \quad \text{BCF} = \frac{C_{\text{org}}}{C_w} = \frac{k_u}{k_e}$$

Step 3: Include Dietary Uptake for Biomagnification

For organisms at trophic level n, contaminant uptake comes from both water and food. Adding dietary assimilation (efficiency α, feeding rate I_R, prey concentration C of the level below):

$$\frac{dC_n}{dt} = k_u C_w + \alpha \cdot I_R \cdot C_{n-1} - k_e \cdot C_n$$

Step 4: Steady-State Biomagnification Factor (BMF)

At steady state, and when dietary uptake dominates over waterborne uptake (true for lipophilic compounds like methylmercury and PCBs):

$$C_n \approx \frac{\alpha \cdot I_R}{k_e} \cdot C_{n-1} \quad \Rightarrow \quad \text{BMF} = \frac{C_n}{C_{n-1}} = \frac{\alpha \cdot I_R}{k_e}$$

Step 5: Concentration at Trophic Level n

If BMF is approximately constant across trophic levels, the concentration at level n relative to the base level (phytoplankton, which bioconcentrate from water) is a geometric progression:

$$C_n = C_w \cdot \text{BCF} \cdot \text{BMF}^{n-1}$$

For methylmercury with BCF ~ 10³ and BMF ~ 5-10, a top predator at trophic level 5 accumulates concentrations 10³ x 5⁴ ~ 6 x 10⁵ times higher than seawater, explaining the ~10⁶-fold enrichment observed in tuna.