Flow rate in unconfined aquifier?

AI Thread Summary
The discussion revolves around calculating the pumping rate of a well in an unconfined aquifer, given its depth and the decrease in water levels from nearby wells. Initial calculations yielded a flow rate of 2.89 m³/s, but the provided answer was 1.06 m³/s, prompting confusion. Key errors identified include miscalculating the depth of water levels and using the incorrect logarithmic ratio in the equation. The participants clarified that a decrease in water level indicates it is further from the ground surface, and the relevance of the well's diameter was questioned. Ultimately, the conversation highlights the importance of precise calculations and understanding the parameters involved in aquifer flow rate equations.
tzx9633

Homework Statement



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 ?

Homework Equations

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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tzx9633 said:
H = 90 – 8 = 82
H = 90- 12 = 78
tzx9633 said:
depth of 90m ... decrease in water level
 
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what do you mean ? Can you explain further ?
 
tzx9633 said:
what do you mean ? Can you explain further ?
If the water level of an aquifer decreases, is it shallower or deeper?
 
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haruspex said:
If the water level of an aquifer decreases, is it shallower or deeper?
Shallower , am i right ?
 
tzx9633 said:
Shallower , am i right ?
Just realized 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?
 
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haruspex said:
Just realized 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?
yes , any part of my working is wrong ? I couldn't get the ans ...
 
tzx9633 said:
yes , any part of my working is wrong ? I couldn't get the ans ...
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?
 
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haruspex said:
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?
102 m ...
 
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haruspex said:
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?
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
 
  • #11
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?
 
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