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Homework Help: Flow rate in unconfined aquifier?

  1. Dec 26, 2017 #1
    1. The problem statement, all variables and given/known data

    A 40cm well was drilled penetrating straight into unconfined aquifier at depth of 90m . The other wells were drilled at a distance of 20m and 100m from initial well have decrease in water level of 12m and 8m from groundwater level . If the aquifier has a permeability value of 200m / day , what’s the pumping rate of well ?

    2. Relevant equations


    3. The attempt at a solution
    Permeability in m^3 / s = 200/24/3600
    H = 90 – 8 = 82
    H = 90- 12 = 78

    My working is Q = πk( H^2 -h^2) / ln (R /r)= π( (200/24/3600)x (82^2 - 78^2)) / ln(100/200 = 2.89m^3 / s , But the ans provided is 1.06m^3 / s , what's wrong with my answer ? The ans provided is in m^3 /s
     

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  3. Dec 26, 2017 #2

    haruspex

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  4. Dec 26, 2017 #3
    what do you mean ? Can you explain further ?
     
  5. Dec 26, 2017 #4

    haruspex

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    If the water level of an aquifer decreases, is it shallower or deeper?
     
  6. Dec 26, 2017 #5
    Shallower , am i right ?
     
  7. Dec 26, 2017 #6

    haruspex

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    Just realised my question was ambiguous.
    I meant shallower in the sense of nearer the ground surface. If the water level falls it will be further from the ground surface, right?
     
  8. Dec 26, 2017 #7
    yes , any part of my working is wrong ? I couldnt get the ans ...
     
  9. Dec 26, 2017 #8

    haruspex

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    You are not getting my hints.
    The first well is at a depth of 90m, i.e. its water's surface is at 90m below ground level.
    The next well has a water surface 12m lower, so how far below ground level is that?
     
  10. Dec 26, 2017 #9
    102 m ...
     
  11. Dec 26, 2017 #10
    H = 90+ 8 = 98
    H = 90 + 12 = 102

    My working is Q = πk( H^2 -h^2) / ln (R /r)= π( (200/24/3600)x (102^2 - 98^2)) / ln(100/200 = 3.61m^3 / s , but the ans provided is 1.06m^3 / s , what's wrong with my answer ? Which is still wrong
     
  12. Dec 26, 2017 #11

    haruspex

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    Sorry, I should have looked at the diagram sooner. I now see that the wording is misleading. Your original calculation of 82m and 78m was correct. The only error I see in your working is that you have ln(100/200) instead of ln(200/100), but that would have simply flipped the sign.
    That said, if I put your numbers through my calculator I get 6.71m3/s.
    The equation puzzles me. I do not understand where that π comes from, since it is not getting multiplied by a radius anywhere. The given 40cm does not seem to feature in the equation (is that radius or diameter?), but I can see why it might be irrelevant.

    Edit, just noticed you had 200 instead of 20.., just a typo I guess. Now I get your 2.89.

    Edit 2: I tried dividing 2.89 by 1.06. It gives a number very close to e. Perhaps a result of fat-fingering a calculator?
     
    Last edited: Dec 26, 2017
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