# Hydrogeology

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## Goals of this unit

• Understand the derivation of one-dimensional groundwater flow
• Apply integrals to our our modeling techniques

## Handouts for out class discussion

Handout - application of integration: arc length.

Handout: 1D groundwater flow. Note: there are some sample calculations you can do here.

Handout - Darcy's Law and Well Production. Note: there are some sample calculations you can do here.

Reference Figure: Unconfined aquifer during pumping. How would we describe the water flow?

Reference Figure: Confined aquifer during pumping. How would we describe the water flow?

Well Intake (surface) Area

## Deriving the flow equation for the confined aquifer case

Here is a sketch of the confined aquifer case:

FIGURE A (above)

Assumptions we will make:

• Aquifer well-screen exists throughout the entire thickness, B, of the aquifer.
• The aquifer is very long laterally (i.e. ~ infinitely long). Note: think about how this assumption is implicitly used below.
• Piezometric surface is initially horizontal
• Groundwater flow is horizontal towards the well (see figure below)

FIGURE B (above)

HERE WE GO

Darcy's Law says:

$Q=K\cdot A\cdot \frac{{h}_{A}-{h}_{B}}{L}$$Q=K\cdot A\cdot \frac{h_{A} - h_{B}}{L}$. Using Figure A, we can rewrite this as $Q=K\cdot A\cdot \frac{\mathrm{\Delta }h}{\mathrm{\Delta }r}$$Q=K \cdot A \cdot \frac{\Delta h}{\Delta r}$, or replacing the $\mathrm{\Delta }$$\Delta$ with differentials (as we set up for integration below) we have: $Q=K\cdot A\cdot \frac{dh}{dr}$$Q=K \cdot A \cdot \frac{dh}{dr}$

The area $A$$A$ of the well-screen is the cylinder area: $A=2\pi rB$$A=2 \pi r B$. Substituting this for $A$$A$ in we can how get: $Q=K\cdot 2\cdot \pi \cdot r\cdot B\cdot \frac{dh}{dr}$$Q=K \cdot 2 \cdot \pi \cdot r \cdot B \cdot \frac{dh}{dr}$. This is a nice little differential equation.

Looking at Figure A and our last equation, the flow rate $Q$$Q$ is made up of contributions across the radii and corresponding head, $h$$h$ changes. So we want to add all of that together -- i.e. integrate.

But in what form will we integrate the last equation? Let's rearrange to separate the differentials:

$Q\cdot \frac{dr}{r}=2\pi BKdh$$Q \cdot \frac{dr}{r}=2 \pi B K dh$.

We can now integrate each side - but what are the limits? On the left, ${r}_{1}$$r_1$ to ${r}_{2}$$r_2$, and on the right the corresponding ${h}_{1}$$h_1$ to ${h}_{2}$$h_2$.

${\int }_{{r}_{1}}^{{r}_{2}}\frac{Q}{r}\mathrm{d}r={\int }_{{h}_{1}}^{{h}_{2}}2\pi KB\mathrm{d}h$$\int_{r_1}^{r_2} \frac{Q}{r} \mathrm{d}r = \int_{h_1}^{h_2} 2 \pi K B \mathrm{d}h$

This gives us, after integrating and subbing in the limits of integration:

$Q \cdot \ln \frac{r_1}{r_2} = 2 \pi K B (h_2 - h_1)$

Finally, solving for $Q$$Q$ gives us:

$Q=\frac{2\pi BK\left({h}_{2}-{h}_{1}\right)}{\mathrm{ln}\left(\frac{{r}_{1}}{{r}_{2}}\right)}$$Q = \frac{2 \pi B K (h_2 - h_1)}{\ln( \frac{r_1}{r_2})}$

Remember this is for the confined aquifer case.

For the unconfined case, there is a modification because now the drawdown curve physically occurs within the aquifer (unlike the hypothetical potentiometric surface of the confined aquifer), so the height of the intake screen on the well is not constantly equal to $B$$B$. In other words, the height of the aquifer is variable, so $B$$B$ gets replaced with $h$$h$.

This will then integrate out to: $Q=\frac{\pi K\left({h}_{2}^{2}-{h}_{1}^{2}\right)}{\mathrm{ln}\left(\frac{{r}_{1}}{{r}_{2}}\right)}$$Q = \frac{\pi K (h_2^2 - h_1^2)}{\ln( \frac{r_1}{r_2})}$

The above material is motivated by this from the book Geology for Engineers & Environmental Scientists by Alan E. Kehew (3rd edition).