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Environmental Plume Mapping (Environmental Engineering) Solution

Formulas & Definitions

Problem Overview

We analyze an expanding circular chemical contaminant plume spreading across an impermeable geological barrier layer with a constant volumetric surface area growth rate of dAdt=12.0 m2/hr\frac{dA}{dt} = 12.0\text{ m}^2/\text{hr}dtdA​=12.0 m2/hr. We calculate the instantaneous radial velocity drdt\frac{dr}{dt}dtdr​ at specific radial snapshots to establish emergency containment boundaries.


Step-by-Step Mathematical Solution

Step 1: Formulate the Geometric Model & Time Derivative

The governing formula is the Area of a Circle:

A=πr2A = \pi r^2A=πr2
(This formula and its time derivative principles are defined in Related Rates.)

Because the contaminant spreads continuously over time, both the area A(t)A(t)A(t) and the radius r(t)r(t)r(t) are dynamic variables depending implicitly on time ttt. Applying the time differential operator ddt\frac{d}{dt}dtd​ to both sides using the Chain Rule:

dAdt=ddt[πr2]=2πrdrdt\frac{dA}{dt} = \frac{d}{dt}[\pi r^2] = 2\pi r \frac{dr}{dt}dtdA​=dtd​[πr2]=2πrdtdr​

Isolating the instantaneous radial expansion rate drdt\frac{dr}{dt}dtdr​:

drdt=12πrdAdt\frac{dr}{dt} = \frac{1}{2\pi r} \frac{dA}{dt}dtdr​=2πr1​dtdA​

Step 2: Calculate Radial Expansion Speed at r=15.0 mr = 15.0\text{ m}r=15.0 m

Substitute the given telemetry values (r=15.0 mr = 15.0\text{ m}r=15.0 m and dAdt=12.0 m2/hr\frac{dA}{dt} = 12.0\text{ m}^2/\text{hr}dtdA​=12.0 m2/hr):

drdt=12.02π(15.0)=12.030.0π=25π m/hr≈0.1273 m/hr\frac{dr}{dt} = \frac{12.0}{2\pi (15.0)} = \frac{12.0}{30.0\pi} = \frac{2}{5\pi}\text{ m/hr} \approx 0.1273\text{ m/hr}dtdr​=2π(15.0)12.0​=30.0π12.0​=5π2​ m/hr≈0.1273 m/hr

Converting to centimeters per hour for field technicians:

drdt≈12.73 cm/hr\frac{dr}{dt} \approx 12.73\text{ cm/hr}dtdr​≈12.73 cm/hr

Step 3: Evaluate Radial Velocity when Area Reaches A=400π m2A = 400\pi\text{ m}^2A=400π m2

  1. Determine the snapshot radius rrr:
    A=πr2  ⟹  400π=πr2  ⟹  r2=400  ⟹  r=20.0 mA = \pi r^2 \implies 400\pi = \pi r^2 \implies r^2 = 400 \implies r = 20.0\text{ m}A=πr2⟹400π=πr2⟹r2=400⟹r=20.0 m
  2. Compute drdt\frac{dr}{dt}dtdr​ at r=20.0 mr = 20.0\text{ m}r=20.0 m:
    drdt=12.02π(20.0)=12.040.0π=310π m/hr≈0.0955 m/hr≈9.55 cm/hr\frac{dr}{dt} = \frac{12.0}{2\pi (20.0)} = \frac{12.0}{40.0\pi} = \frac{3}{10\pi}\text{ m/hr} \approx 0.0955\text{ m/hr} \approx 9.55\text{ cm/hr}dtdr​=2π(20.0)12.0​=40.0π12.0​=10π3​ m/hr≈0.0955 m/hr≈9.55 cm/hr

Engineering Conclusion

In environmental contamination response and remediation engineering, knowing the areal influx rate (dAdt\frac{dA}{dt}dtdA​) is necessary for estimating the total mass of leaked contaminant, but it is insufficient for physical containment deployment. Physical containment equipment—such as absorbent surface booms, trench cut-off walls, and sheet-pile barriers—must be positioned at a specific radial distance rrr from the spill source.

Because drdt=12πrdAdt\frac{dr}{dt} = \frac{1}{2\pi r}\frac{dA}{dt}dtdr​=2πr1​dtdA​, the radial expansion velocity is inversely proportional to the radius. When the plume is small (r=5 mr = 5\text{ m}r=5 m), the boundary advances rapidly (≈38.2 cm/hr\approx 38.2\text{ cm/hr}≈38.2 cm/hr), posing an immediate threat of overrunning field crews. As the plume expands to r=20 mr = 20\text{ m}r=20 m, the radial velocity drops to 9.55 cm/hr9.55\text{ cm/hr}9.55 cm/hr. Calculating drdt\frac{dr}{dt}dtdr​ gives remediation engineers the exact time window available to excavate cutoff trenches safely ahead of the advancing plume boundary.