<p>Based on the Theis equation, increasing the transmissivity will increase the radius of influence and decrease the total drawdown.  Drawdown is directly proportional to the pumping rate and inversely proportional to aquifer transmissivity and storativity.</p>

<p>$$s = \frac{Q}{4T\pi}\int_u^\infty \frac{e^{-y}}y dy$$
and $$u = \frac{r^{2}s}{4Tt}$$
Where s is drawdown, r is radius from well, t is time, T is transmissivity, Q is pumping rate, and u is the well function.</p>

<p>Applying this equation, you can produce plots like this:<a href="https://i.sstatic.net/UUajW.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UUajW.jpg" alt="cone of depression with well at 0; note the log-log scales"></a><a href="https://i.sstatic.net/tsJn5.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/tsJn5.jpg" alt="radius of influence as a function of T"></a></p>

<p>for these plots, I used Q = 500, S = 0.01, and t = 4</p>

CC BY-SA 3.0
2017-07-31T06:52:51.463Z
Wonderful and well explained answer. Thanks. A few more questions: is the radius of influence in the second graph defined by drawdown equals 0? Do you use the taylor expansion of the well function to find W(u) or do you use the table or any other way? What do you mean by "total drawdown"? and lastly :), I find it counter intuitive that larger transmissivity has a larger hydraulic radius of influence. Is there a simple physical explanation for this? or is my intuition just off?
pls igonre the first comment.
asher
CC BY-SA 3.0
2017-07-31T18:23:48.827Z
For this, I think I used s = 0.001. I used the python scripts here to calculate W(u): http://nbviewer.jupyter.org/github/Applied-Groundwater-Modeling-2nd-Ed/Chapter_3_problems-1/blob/master/P3.4_Theis_pumping.ipynb. By 'total drawdown,' I mean drawdown at the pumping well.  Better connected porous media (higher T) will allow for a larger area to be drained, dissipating the drawdown over a larger area.
Inkenbrandt
CC BY-SA 3.0
2017-08-01T09:17:17.143Z
Again thanks. Is it correct to say that the volume of depression (or the area if we take the 2 dimensional case) has to be equal between the 2 situations? and if so, is the V=S*deltaH*A the right equation to prove this? (deltaH being the integrated change of head in the depression cone.
asher
CC BY-SA 3.0
2017-08-01T19:41:49.837Z
If you pump for at a specified rate over a specified amount of time, you will pump the same volume independent of the transmissivity, assuming drawdown is not greater than the depth of the pump. Volume of water extracted will remain constant.
Inkenbrandt
CC BY-SA 3.0
2017-07-30T20:21:09.917Z
2017-07-30T20:21:09.917Z
Inkenbrandt
<p>My question concerns the depression cone of a pumping well in a confined aquifer: Say the transmissivity of the aquifer is 1200 m^2/day how will the cone change if the transmissivity is 600 m^2/day? All the aquifers parameters stay constant in both scenarios and so is the pumping rate (say- 300 cm/hr).</p>

<p>The main interest is what happens to the radius of influence (where there will be no change in head from the aquifers regional head)? and also what happens to the head at the well it self?</p>

<p>Thanks, Asher</p>

CC BY-SA 3.0
2017-08-03T00:32:27.553Z
If you are interested in the head at the well, then you need to consider well-loss in addition to the aquifer response.
haresfur
<hydrology><groundwater>
Hydrology - Depression cone VS transmissivity
asher
# Hydrology - Depression cone VS transmissivity

(Asked by: username_0 on 2017-07-30)

My question concerns the depression cone of a pumping well in a confined aquifer: Say the transmissivity of the aquifer is 1200 m^2/day how will the cone change if the transmissivity is 600 m^2/day? All the aquifers parameters stay constant in both scenarios and so is the pumping rate (say- 300 cm/hr).

The main interest is what happens to the radius of influence (where there will be no change in head from the aquifers regional head)? and also what happens to the head at the well it self?

Thanks, username_0

**Tags:** hydrology, groundwater

> **Comments:**
> **username_1**: If you are interested in the head at the well, then you need to consider well-loss in addition to the aquifer response.  

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## 1 answers

### Answer by username_2 on 2017-07-30. Score: 3 | Accepted answer

Based on the Theis equation, increasing the transmissivity will increase the radius of influence and decrease the total drawdown. Drawdown is directly proportional to the pumping rate and inversely proportional to aquifer transmissivity and storativity.

$$s = \frac{Q}{4T\pi}\int\_u^\infty \frac{e^{-y}}y dy$$
and $$u = \frac{r^{2}s}{4Tt}$$
Where s is drawdown, r is radius from well, t is time, T is transmissivity, Q is pumping rate, and u is the well function.

Applying this equation, you can produce plots like this:[![cone of depression with well at 0; note the log-log scales](https://i.sstatic.net/UUajW.jpg)](https://i.sstatic.net/UUajW.jpg)[![radius of influence as a function of T](https://i.sstatic.net/tsJn5.jpg)](https://i.sstatic.net/tsJn5.jpg)

for these plots, I used Q = 500, S = 0.01, and t = 4

> **Comments:**
> **username_0**: Wonderful and well explained answer. Thanks. A few more questions: is the radius of influence in the second graph defined by drawdown equals 0? Do you use the taylor expansion of the well function to find W(u) or do you use the table or any other way? What do you mean by "total drawdown"? and lastly :), I find it counter intuitive that larger transmissivity has a larger hydraulic radius of influence. Is there a simple physical explanation for this? or is my intuition just off? pls igonre the first comment.  
> **username_2**: For this, I think I used s = 0.001. I used the python scripts here to calculate W(u): http://nbviewer.jupyter.org/github/Applied-Groundwater-Modeling-2nd-Ed/Chapter_3_problems-1/blob/master/P3.4_Theis_pumping.ipynb. By 'total drawdown,' I mean drawdown at the pumping well.  Better connected porous media (higher T) will allow for a larger area to be drained, dissipating the drawdown over a larger area.  
> **username_0**: Again thanks. Is it correct to say that the volume of depression (or the area if we take the 2 dimensional case) has to be equal between the 2 situations? and if so, is the V=S*deltaH*A the right equation to prove this? (deltaH being the integrated change of head in the depression cone.  
> **username_2**: If you pump for at a specified rate over a specified amount of time, you will pump the same volume independent of the transmissivity, assuming drawdown is not greater than the depth of the pump. Volume of water extracted will remain constant.  

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