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Math problem set Grade 12 Math

Quillbeck Solvent Plume: Darcy Velocity and Travel Time

A grade 12 problem set applying Darcy's law to a fictional spill site, finding seepage velocity, sorption-slowed plume travel time to a well, and solvent mass flux.

Objectives

  1. Students will compute hydraulic gradient, Darcy velocity, and seepage velocity from field measurements.
  2. Students will solve distance-rate-time equations for contaminant arrival at a downgradient well.
  3. Students will apply the pounds formula to find the mass of solvent moving past a cross-section each day.

Vocabulary

Hydraulic conductivity
Hydraulic conductivity is a number that tells you how easily water can flow through a particular soil or rock. Gravel has high hydraulic conductivity because water flows through it easily, while dense clay has very low hydraulic conductivity.

Also: hydraulic gradient, Darcy velocity, seepage velocity, effective porosity, retardation factor

Teacher overview

Groundwater moves slowly enough to measure in feet per day, which is why a spill can sit undetected for years and still be on its way to a well. This set uses the fictional village of Quillbeck, where a closed machine shop released a chlorinated solvent into a sand aquifer upgradient of the municipal well. Students work from two monitoring well readings to a gradient, then to Darcy velocity, seepage velocity, and an arrival time. The error to correct here is reporting Darcy velocity as the speed of the water: the water in the pores moves faster than that figure by a factor of one over the effective porosity, and a sorbing solvent moves slower than the water by the retardation factor.

Quillbeck Solvent Plume: Darcy Velocity and Travel Time — Grade 12 Math
Name: ____________________________ Date: ______________

Problems

All values below describe the fictional village of Quillbeck and its aquifer.

Two monitoring wells sit 600 feet apart on a straight line running from the spill site toward the municipal well. The upgradient well reads a water level elevation of 412.6 feet, and the downgradient well reads 408.4 feet. The sand aquifer has a hydraulic conductivity of 85 feet per day and an effective porosity of 0.28. The aquifer is 30 feet thick, and the plume is 220 feet wide where it crosses the monitoring line. The municipal well stands 2,400 feet downgradient of the spill site. The dissolved solvent sorbs onto the sand grains, giving it a retardation factor of 3.4. The solvent concentration at the monitoring line is 0.18 mg/L. Use 7.48 gallons per cubic foot, 1,440 minutes per day, and pounds per day = concentration in mg/L x flow in MGD x 8.34.

  1. Find the hydraulic gradient between the two monitoring wells, in feet of head loss per foot of distance. Report three decimal places.

  2. Find the Darcy velocity, where Darcy velocity = hydraulic conductivity x hydraulic gradient. Report feet per day to three decimal places.

  3. Find the seepage velocity of the water, where seepage velocity = Darcy velocity divided by effective porosity. Report feet per day to three decimal places.

  4. Water leaving the spill site travels 2,400 feet to the municipal well. Write and solve a distance-rate-time equation for the number of days that trip takes at the seepage velocity. Round to the nearest whole day.

  5. Convert your problem 4 answer to years, using 365 days per year. Round to the nearest hundredth.

  6. The solvent moves more slowly than the water by the retardation factor. Find the solvent's velocity, where solvent velocity = seepage velocity divided by 3.4. Report feet per day to three decimal places.

  7. Using the problem 6 velocity, how many years will the solvent take to travel the 2,400 feet from the spill site to the municipal well? Round to the nearest tenth.

  8. How far has the solvent front traveled 6 years after the release, at the problem 6 velocity? Round to the nearest foot.

  9. Find the groundwater flow moving through the plume's cross-section, in gallons per minute. The cross-section is 30 feet thick and 220 feet wide, and flow through it equals Darcy velocity x area. Round to the nearest tenth.

  10. Convert the problem 9 flow to million gallons per day, then use the pounds formula to find the pounds of solvent moving past the monitoring line each day at 0.18 mg/L. Round to three decimal places.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: 0.007 ft/ft.
    Head loss = 412.6 - 408.4 = 4.2 feet over 600 feet. Gradient = 4.2 / 600 = 0.007 feet per foot.

  2. Answer: 0.595 ft/day.
    Darcy velocity = 85 feet per day x 0.007 = 0.595 feet per day. This figure describes flow per unit of total aquifer area, not the speed of a water particle.

  3. Answer: 2.125 ft/day.
    Seepage velocity = 0.595 / 0.28 = 2.125 feet per day. Water moves only through the pore space, so dividing by the effective porosity of 0.28 raises the speed by a factor of about 3.6.

  4. Answer: 1,129 days.
    Equation: 2,400 = 2.125t, so t = 2,400 / 2.125 = 1,129.4, which rounds to 1,129 days.

  5. Answer: 3.09 years.
    1,129.4 / 365 = 3.094, which rounds to 3.09 years for water to make the trip.

  6. Answer: 0.625 ft/day.
    Solvent velocity = 2.125 / 3.4 = 0.625 feet per day. Sorption onto the sand grains holds the solvent back, so it trails the water that carries it.

  7. Answer: 10.5 years.
    Time = 2,400 / 0.625 = 3,840 days. In years: 3,840 / 365 = 10.52, which rounds to 10.5 years. The solvent needs about 3.4 times as long as the water, matching the retardation factor.

  8. Answer: 1,369 ft.
    6 years x 365 days = 2,190 days. Distance = 0.625 x 2,190 = 1,368.75, which rounds to 1,369 feet, so the front has covered a little over half the distance to the well.

  9. Answer: 20.4 gpm.
    Cross-section area = 30 x 220 = 6,600 square feet. Flow = 0.595 x 6,600 = 3,927 cubic feet per day. In gallons: 3,927 x 7.48 = 29,374 gallons per day. Per minute: 29,374 / 1,440 = 20.4 gallons per minute.

  10. Answer: 0.044 lb/day.
    Flow in million gallons per day = 29,374 / 1,000,000 = 0.029374 MGD. Pounds per day = 0.18 x 0.029374 x 8.34 = 0.044 pounds per day. Small as that sounds, it is about 16 pounds of solvent per year crossing the monitoring line, and only a few pounds spread through a well field can push a supply above a drinking water limit.

Differentiation

Support

Give students a labeled cross-section of the fictional Quillbeck aquifer showing the two monitoring wells, their water levels, the plume, and the municipal well, with the 600-foot spacing and 2,400-foot distance marked. Provide problems 1 and 2 with the formulas already written and only the substitution left to complete, and supply a worked example of the same three-step chain using different numbers.

Extension

Ask students to find what pumping schedule would be required to intercept the plume at the fictional site: if a recovery well must capture the full 20.4 gallons per minute crossing the monitoring line plus a 40 percent safety margin, what pumping rate is needed, and how many pounds of solvent would that well remove in a year at the stated concentration?

ELL supports

Pre-teach "gradient," "conductivity," "porosity," and "retardation" with a labeled diagram of sand grains, pore space, and a solvent molecule attaching to a grain. Provide a step chart for the fictional scenario listing the four quantities in order, gradient, Darcy velocity, seepage velocity, and solvent velocity, with a box for each answer so students can see how each one builds on the previous step.

Extensions

Have students test how sensitive the arrival time is to the porosity estimate. Using the fictional Quillbeck data, they recompute problems 3 and 7 with effective porosity values of 0.22 and 0.34 while holding everything else constant, then describe how many years the estimated arrival at the municipal well shifts in each case. The result gives a concrete argument for why a remediation plan states a range of arrival times rather than a single date.