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Contaminant Dispersion Wellhead Flux (Environmental & Civil Engineering) Solution

Formulas & Definitions

Problem Overview

We evaluate the boundary behavior of contaminant solute flux J(r)=βˆ’D0rln⁑(r/R0)J(r) = -D_0 r \ln(r / R_0)J(r)=βˆ’D0​rln(r/R0​) as radial distance rβ†’0+r \to 0^+rβ†’0+ where D0=1.80Β g/(mβ‹…day)D_0 = 1.80\text{ g}/(\text{m}\cdot\text{day})D0​=1.80Β g/(mβ‹…day) and R0=100.0Β mR_0 = 100.0\text{ m}R0​=100.0Β m. Direct substitution yields the indeterminate product form 0β‹…(βˆ’βˆž)0 \cdot (-\infty)0β‹…(βˆ’βˆž). We convert the expression to a quotient, evaluate the limit using L'HΓ΄pital's Rule, determine peak flux location rβˆ—r^*rβˆ—, and interpret results for groundwater filter pack design.


Step-by-Step Mathematical Solution

Step 1: Algebraic Inversion & L'HΓ΄pital's Rule Evaluation

The governing formulation is L'HΓ΄pital's Rule for Indeterminate Forms:

lim⁑xβ†’af(x)g(x)=lim⁑xβ†’afβ€²(x)gβ€²(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}xβ†’alim​g(x)f(x)​=xβ†’alim​gβ€²(x)fβ€²(x)​
(This formula and transformation techniques are defined in Indeterminate Forms and L'HΓ΄pital's Rule.)

  1. Classify the Indeterminate Form: As r→0+r \to 0^+r→0+:

    rβ†’0,r100β†’0+β€…β€ŠβŸΉβ€…β€Šln⁑(r100)β†’βˆ’βˆžr \to 0, \qquad \frac{r}{100} \to 0^+ \implies \ln\left(\frac{r}{100}\right) \to -\inftyrβ†’0,100r​→0+⟹ln(100r​)β†’βˆ’βˆž
    Thus, the product rln⁑(r/100)r \ln(r / 100)rln(r/100) produces the indeterminate form 0β‹…(βˆ’βˆž)0 \cdot (-\infty)0β‹…(βˆ’βˆž).

  2. Invert to a Fractional Quotient: To apply L'HΓ΄pital's Rule, keep the logarithmic term in the numerator and push the linear factor rrr to the denominator as rβˆ’1=1rr^{-1} = \frac{1}{r}rβˆ’1=r1​:

    lim⁑rβ†’0+J(r)=βˆ’D0lim⁑rβ†’0+ln⁑(r/100)rβˆ’1\lim_{r \to 0^+} J(r) = -D_0 \lim_{r \to 0^+} \frac{\ln(r / 100)}{r^{-1}}rβ†’0+lim​J(r)=βˆ’D0​rβ†’0+lim​rβˆ’1ln(r/100)​
    Direct substitution now yields the form βˆ’βˆžβˆž\frac{-\infty}{\infty}βˆžβˆ’βˆžβ€‹, which grants a green light for L'HΓ΄pital's Rule.

  3. Differentiate Numerator and Denominator Independently: Numerator derivative using the Chain Rule:

    ddr[ln⁑(r100)]=1r100β‹…1100=1r\frac{d}{dr}\left[\ln\left(\frac{r}{100}\right)\right] = \frac{1}{\frac{r}{100}} \cdot \frac{1}{100} = \frac{1}{r}drd​[ln(100r​)]=100r​1​⋅1001​=r1​
    Denominator derivative using the Power Rule:
    ddr[rβˆ’1]=βˆ’rβˆ’2=βˆ’1r2\frac{d}{dr}\left[r^{-1}\right] = -r^{-2} = -\frac{1}{r^2}drd​[rβˆ’1]=βˆ’rβˆ’2=βˆ’r21​

  4. Simplify Algebraically and Evaluate:

    lim⁑rβ†’0+1rβˆ’1r2=lim⁑rβ†’0+[1rβ‹…(βˆ’r2)]=lim⁑rβ†’0+(βˆ’r)=0\lim_{r \to 0^+} \frac{\frac{1}{r}}{-\frac{1}{r^2}} = \lim_{r \to 0^+} \left[\frac{1}{r} \cdot \left(-r^2\right)\right] = \lim_{r \to 0^+} (-r) = 0rβ†’0+limβ€‹βˆ’r21​r1​​=rβ†’0+lim​[r1​⋅(βˆ’r2)]=rβ†’0+lim​(βˆ’r)=0
    Multiplying by the constant βˆ’D0-D_0βˆ’D0​:
    Jboundary=lim⁑rβ†’0+J(r)=βˆ’D0β‹…(0)=0.00Β g/(mβ‹…day)J_{\text{boundary}} = \lim_{r \to 0^+} J(r) = -D_0 \cdot (0) = 0.00\text{ g}/(\text{m}\cdot\text{day})Jboundary​=rβ†’0+lim​J(r)=βˆ’D0​⋅(0)=0.00Β g/(mβ‹…day)

Step 2: Peak Contaminant Flux Calculation (rβˆ—r^*rβˆ—)

To find the location of maximum radial flux, differentiate J(r)=βˆ’D0[rln⁑(r)βˆ’rln⁑(R0)]J(r) = -D_0 [r \ln(r) - r \ln(R_0)]J(r)=βˆ’D0​[rln(r)βˆ’rln(R0​)] using the Product Rule:

dJdr=βˆ’D0[(1)ln⁑(rR0)+rβ‹…1r]=βˆ’D0[ln⁑(rR0)+1]\frac{dJ}{dr} = -D_0 \left[(1)\ln\left(\frac{r}{R_0}\right) + r \cdot \frac{1}{r}\right] = -D_0 \left[\ln\left(\frac{r}{R_0}\right) + 1\right]drdJ​=βˆ’D0​[(1)ln(R0​r​)+rβ‹…r1​]=βˆ’D0​[ln(R0​r​)+1]
Setting dJdr=0\frac{dJ}{dr} = 0drdJ​=0:
ln⁑(rR0)+1=0β€…β€ŠβŸΉβ€…β€Šln⁑(rR0)=βˆ’1β€…β€ŠβŸΉβ€…β€ŠrR0=eβˆ’1=1e\ln\left(\frac{r}{R_0}\right) + 1 = 0 \implies \ln\left(\frac{r}{R_0}\right) = -1 \implies \frac{r}{R_0} = e^{-1} = \frac{1}{e}ln(R0​r​)+1=0⟹ln(R0​r​)=βˆ’1⟹R0​r​=eβˆ’1=e1​
rβˆ—=R0e=100.02.71828β‰ˆ36.79Β metersr^* = \frac{R_0}{e} = \frac{100.0}{2.71828} \approx 36.79\text{ meters}rβˆ—=eR0​​=2.71828100.0β€‹β‰ˆ36.79Β meters

Evaluating the maximum flux at rβˆ—r^*rβˆ—:

Jmax=J(R0e)=βˆ’D0(R0e)ln⁑(eβˆ’1)=βˆ’D0(R0e)(βˆ’1)=D0R0eJ_{\text{max}} = J\left(\frac{R_0}{e}\right) = -D_0 \left(\frac{R_0}{e}\right) \ln\left(e^{-1}\right) = -D_0 \left(\frac{R_0}{e}\right) (-1) = D_0 \frac{R_0}{e}Jmax​=J(eR0​​)=βˆ’D0​(eR0​​)ln(eβˆ’1)=βˆ’D0​(eR0​​)(βˆ’1)=D0​eR0​​
Substituting D0=1.80Β g/(mβ‹…day)D_0 = 1.80\text{ g}/(\text{m}\cdot\text{day})D0​=1.80Β g/(mβ‹…day) and R0=100.0Β mR_0 = 100.0\text{ m}R0​=100.0Β m:
Jmax=1.80β‹…100.0e=180.0eβ‰ˆ66.22Β g/(mβ‹…day)J_{\text{max}} = 1.80 \cdot \frac{100.0}{e} = \frac{180.0}{e} \approx 66.22\text{ g}/(\text{m}\cdot\text{day})Jmax​=1.80β‹…e100.0​=e180.0β€‹β‰ˆ66.22Β g/(mβ‹…day)

Step 3: Engineering Analysis & Filter Pack Sizing

  • Boundary Behavior vs. Peak Flux: At the aquifer capture perimeter (r=R0=100Β mr = R_0 = 100\text{ m}r=R0​=100Β m), ln⁑(1)=0β€…β€ŠβŸΉβ€…β€ŠJ(100)=0\ln(1) = 0 \implies J(100) = 0ln(1)=0⟹J(100)=0. As groundwater travels inward, flux increases, reaching a maximum of 66.22Β g/(mβ‹…day)66.22\text{ g}/(\text{m}\cdot\text{day})66.22Β g/(mβ‹…day) at rβ‰ˆ36.8Β mr \approx 36.8\text{ m}rβ‰ˆ36.8Β m, and then decreases smoothly back to zero as rβ†’0+r \to 0^+rβ†’0+.
  • Why lim⁑rβ†’0+J(r)=0\lim_{r \to 0^+} J(r) = 0limrβ†’0+​J(r)=0 Matters: If J(r)J(r)J(r) diverged to ∞\infty∞ at the well screen (rβ†’0r \to 0rβ†’0), the massive concentration gradient would cause local supersaturation of heavy metals and dissolved solids, cementing the sand pack with mineral scale. Because the flux mathematically approaches 000, the cylindrical surface area of the well screen (A=2Ο€rLA = 2\pi r LA=2Ο€rL) is sufficient to absorb incoming groundwater smoothly. Environmental engineers can use standard silica sand filter packs and schedule periodic mild surfactant flushes without fear of localized structural clogging.

Engineering Conclusion

In environmental hydrology, products of decaying distances and logarithmic driving potentials frequently generate 0β‹…βˆž0 \cdot \infty0β‹…βˆž indeterminate forms near point sinks and source boundaries. Relying on intuition might suggest the singularity ln⁑(r)β†’βˆ’βˆž\ln(r) \to -\inftyln(r)β†’βˆ’βˆž dominates, predicting runaway solute buildup.

By inverting the linear factor rrr to rβˆ’1r^{-1}rβˆ’1 and applying L'HΓ΄pital's Rule, engineers prove that the linear convergence of distance to zero (rβ†’0r \to 0rβ†’0) completely overpowers the weak logarithmic divergence (ln⁑rβ†’βˆ’βˆž\ln r \to -\inftylnrβ†’βˆ’βˆž). This confirms that the contaminant flux vanishes at the borehole axis (Jboundary=0J_{\text{boundary}} = 0Jboundary​=0). This guarantees that extraction well screens will not experience catastrophic chemical encrustation, providing rigorous mathematical validation for long-term municipal groundwater remediation designs.