# How to find the drawdown of a well in an unconfined aquifer?

I tried to calculate the dropdown of a well of an unconfined aquifer but the drawdown of an observation well is not given. Here's the full question:

A 10 cm diameter well in an unconfined aquifer was pumped at a uniform rate of 500 litres/min, while observations of drawdown were made in an observation well at a distance of 50 m from the well. The original head of water, measured from the top of the impervious layer, was 25 m. The hydraulic conductivity of the aquifer was 1.83 x 10-3 m/min. Determine the drawdown at the face of the well, assuming that flow at the well is steady.

The answer provided is 20 cm. I tried using the Dupuit-Thiem equation to arrive at the answer using the well and observation well but all to no avail. With the dropdown of the observation well not provided, is it still possible to calculate the dropdown of the well? A prompt response would be highly appreciated.

• Why the edit if I may ask? @fred Commented Jul 2 at 10:17
• The italics make clearer indication of what text the question you were provided with has, the spacing with the units is both proper and consistency with the rest of the question, and thanks messages tend to be taken out on SE as superfluous... Commented Jul 2 at 10:32
• @JeopardyTempest: thanks for your comment, you've nicely summed up why I made the edits.
– Fred
Commented Jul 3 at 10:14
• Using Dupuit-Thiem (R=50 m/r=0,05 m/h₀=25 m and given k/Q) I get roughly 16 m of drawdown! Where do the 0.2 m come from? Commented Jul 4 at 18:04
• @klanomath what assumption did you make to arrive at this solution? Well the answer you got wrong show your assumptions could be wrong too. Using the dupuit-thiem equation without natural logarithms, does an equation like that exist? Commented Jul 5 at 14:48

To find the drawdown of a well with a steady discharge in an unconfined aquifer you have to use the Dupuit-Thiem equation which looks like this:

$$Q = \pi \cdot k_f \cdot \frac{\left(h_2^2-h_1^2\right)}{\ln \left( \tfrac{r_2}{r_1}\right)}$$

with:

$$Q$$: discharge
$$k_f$$: hydraulic conductivity
$$r_1$$: distance to observation well 1
$$r_2$$: distance to observation well 2
$$h_1$$: groundwater level at observation well 1
$$h_2$$: groundwater level at observation well 2

An older variant of the Dupuit-Thiem equation (Dupuit’s work) is similar but uses slightly different parameters:

$$Q = \pi \cdot k_f \cdot \frac{\left(h_0^2-h^2\right)}{\ln \left( \tfrac{R}{r}\right)}$$

with:

$$Q$$: discharge
$$k_f$$: hydraulic conductivity
$$R$$: radius of influence of the well
$$r$$: radius of the well
$$h_0$$: undisturbed groundwater level
$$h$$: water level in the well

To answer the question you have to use the latter equation and one has to assume that the drawdown at the observation well is marginal and this well is therefore situated at the radius of influence of the well.

$$Q = \pi \cdot k_f \cdot \frac{\left(h_0^2-h^2\right)}{\ln \left( \tfrac{R}{r}\right)}$$

$${h^2} = h_0^2 - \frac{Q \ln \left( \tfrac{R}{r}\right)}{\pi \cdot k_f \cdot}$$

$${h} = \sqrt {\left( h_0^2 - \frac{Q \ln \left( \tfrac{R}{r}\right)}{\pi \cdot k_f \cdot} \right)}$$ or drawdown $${s} = h_0 - h$$

With the provided values of $$R$$, $$r$$, $$h₀$$, $$k_f$$ and $$Q$$: $$h = 4.92 m$$ and the drawdown $$s = 20.08 m$$

• thank you very much. Your submission sums it all up for me. I'm glad to have benefitted from your body of knowledge. Cheers! Commented Jul 6 at 19:38