WR/Env. Six Minute Solutions Prob. 56

[DanHalen]

Problem 56 states:

A small pest-control business routinely discharges pesticide contaminated water to a drainage ditch when washing their equipment. The ditch infiltrates to a shallow aquifer with a bulk groundwater velocity of 172 cm/day. The concentration of the pesticide in the aquifer just below the ditch is 0.182 mg/L. How much time will be needed for the pesticide to reach a drinking water well located in a direct line 1600 m downgradient of the source at its maximum containmanent level (MCL) of 1.0 microgram/L?

A) 0.8 d

B) 120 d

C) 860 d

D) 5300 d

The answer is B, 120 days.

They use a strange equation to solve this problem and I'm having a difficult time trying to figure out how they got their answer. I think the solution is skipping a few key steps that would otherwise clarify how they solved the problem. They start out by solving for alphaL and DL.

AlphaL=0.0175L1.46 = 0.0175*16001.46 = 834 m

DL=alphaLvx = 834 m * 172 cm/day * 1 m/1000 cm = 1434 m2/day

(2C)/(Co) = (2*1.0 microgram/L*1mg/1000 micrograms)/(0.182 mg/L) = 0.01099

(L-vxt)/(2*sqrt(DLt)) = (1600m-(172cm/day)(1m/100cm)*t)/(2*sqrt(1434 m2/day * t))

(2C)/(Co)=erfc((L-vxt)/(2*sqrt(DLt)))

0.01099 = erfc((1600m-(172cm/day)(1m/100cm)*t)/(2*sqrt(1434 m2/day * t)))

1.66 = (1600-(172*1/100*t))/(2*sqrt(1434t))

t = 122.2 days ~ 120 days

How did they go from 0.01099 to 1.66? I have not been able to find this equation anywhere. I thought for sure Metcalf & Eddy would have it but it doesn't.

 

[IlPadrino]

I don't have any books with me, but this is a diffusion problem. erfc is the "complementary error function". Using wolframalpha, (http://www.wolframalpha.com/input/?i=erfc^%28-1%29%280.01099%29) for the inverse erfc (0.01099) gives a result of 1.80. I'm guessing there are inverse erfc tables somewhere and 1.66 is the result of linear interpolation.

Does this help at all?

 

[DanHalen]

Thanks for the reply IIPadrino. You're right, erfc is the complementary error function. I tried using Google to see if I could find a chart or table but was unsuccessful. I went to the link you posted and that helps some. Unfortunately I can't use WolframAlpha on the exam. I need to find a way to do this by hand using a chart or table. Is this one of those problems that has a very low probability of being on the exam? If so I will toss this one out and move on.

 

[DanHalen]

I think this table will help but I still don't see where the 1.66 comes from.

http://www.geophysik.uni-muenchen.de/~malservisi/GlobaleGeophysik2/erf_tables.pdf

 

[K19]

Is this one of those problems that has a very low probability of being on the exam? If so I will toss this one out and move on.

Absolutely, this type of problem (and many from this 6-min solutions book, so I've heard) is off the charts compared to the type and level of difficulty you'll actually see on the exam. The groundwater problems in the 2011 and 2008 ncees practice exams are much more representative. If you have the 2008 note that there was errata issued addressing an error in one such problem.

 

[IlPadrino]

I think this table will help but I still don't see where the 1.66 comes from.

http://www.geophysik.uni-muenchen.de/~malservisi/GlobaleGeophysik2/erf_tables.pdf

That's a table for the erf function, not the erfc function. Try this table: http://www.coe.montana.edu/ee/tjkaiser/ee407/notes/comperrfnc.pdf

I'd guess 1.66 comes from a different table that was interpolated.