Vertical Flow Through Three Layered Formations: 18-Geol-A2 Hydrogeology, December 2016, Q3

Question 3: Vertical Flow Through Three Layered Formations (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three horizontal formations, each $L=25\ \text{m}$ thick, stacked with hydraulic conductivities $K_1=0.001\ \text{m/s}$ (top), $K_2=0.005\ \text{m/s}$ (middle), $K_3=0.001\ \text{m/s}$ (bottom). Steady, one-dimensional vertical flow with total head $H_{top}=200\ \text{m}$ at the top surface and $H_{bot}=80\ \text{m}$ at the base (75 m below the top).

Layer	K	Thickness
1 (top)	0.001 m/s	25 m
2 (middle)	0.005 m/s	25 m
3 (bottom)	0.001 m/s	25 m

Find. (a) The total hydraulic heads at the two internal boundaries (1/2 and 2/3); (b) the corresponding water pressure heads at those same boundaries.

Figure 3. Vertical flow column through the three formations, with total heads at the top, base, and two internal boundaries.

Approach. Continuity requires the same specific discharge $q$ through all three layers in series; treat the head loss across each layer as $q(L_i/K_i)$ (the layer's hydraulic resistance) so that the total head drop is $\sum q(L_i/K_i)=H_{top}-H_{bot}$, solve for $q$, then step the head down layer by layer. Pressure heads follow from $h_p = H - z$, taking the base as the elevation datum ($z=0$).

Effective vertical resistance and flux. For flow in series across layers, the same $q$ passes through each, so the resistances $L_i/K_i$ add: $$\sum \frac{L_i}{K_i} = \frac{25}{0.001}+\frac{25}{0.005}+\frac{25}{0.001} = 25{,}000+5{,}000+25{,}000=55{,}000\ \text{s},$$ $$q = \frac{H_{top}-H_{bot}}{\sum L_i/K_i} = \frac{200-80}{55{,}000}=\boxed{2.182\times10^{-3}\ \text{m/s}}.$$
(a) Step the head down through Formation 1 to the 1/2 boundary. $$\Delta h_1 = q\left(\frac{L_1}{K_1}\right) = (2.182\times10^{-3})(25{,}000)=54.55\ \text{m}\;\Rightarrow\; H_{1/2}=200-54.55=\boxed{145.45\ \text{m}}.$$
Step through Formation 2 to the 2/3 boundary. $$\Delta h_2 = q\left(\frac{L_2}{K_2}\right) = (2.182\times10^{-3})(5{,}000)=10.91\ \text{m}\;\Rightarrow\; H_{2/3}=145.45-10.91=\boxed{134.55\ \text{m}}.$$ (Check: stepping through Formation 3, $\Delta h_3 = q(25{,}000)=54.55\ \text{m}$, giving $134.55-54.55=80.0\ \text{m}$ — matches the given base head exactly.)
(b) Convert to pressure heads. Taking the base of the stack as the elevation datum ($z=0$), the 2/3 boundary sits at $z=25\ \text{m}$ and the 1/2 boundary at $z=50\ \text{m}$; pressure head is $h_p = H - z$: $$h_{p,1/2} = 145.45-50=\boxed{95.45\ \text{m}},\qquad h_{p,2/3} = 134.55-25=\boxed{109.55\ \text{m}}.$$

Location	Total head H	Elevation z (above base)	Pressure head $h_p$
Top surface	200.00 m	75 m	125.00 m
1/2 boundary	145.45 m	50 m	95.45 m
2/3 boundary	134.55 m	25 m	109.55 m
Base	80.00 m	0 m	80.00 m

Concept & Theory

Topic. One-dimensional steady vertical flow across layered media in series (the hydraulic analogue of electrical resistors in series).

Key relations. Continuity of specific discharge $q$ across all layers; per-layer head loss $\Delta h_i = q(L_i/K_i)$; effective vertical conductivity $K_v = \sum L_i/\sum(L_i/K_i)$; total head vs. pressure head via $H = z + h_p$ (elevation head + pressure head).

Why this works. Because the flow is strictly vertical and steady, mass balance forces the same $q$ through every layer regardless of its $K$ — a low-$K$ layer must therefore carry a disproportionately large share of the total head loss to pass that same $q$, exactly as a large electrical resistor drops a disproportionate share of voltage in a series circuit. The two identical 0.001 m/s layers here each take 54.55 m of the 120 m total drop, while the 5× more conductive middle layer takes only 10.9 m despite being the same thickness. Splitting total head into elevation head + pressure head is just Bernoulli's equation with velocity head neglected (valid for the very slow, laminar groundwater regime), so once $H$ and $z$ are both known at a point, $h_p$ follows immediately.

Common pitfall. Averaging the three $K$ values arithmetically instead of using the series (harmonic-mean-style) resistance formula — vertical flow through layered media is controlled by the harmonic mean of $K$, heavily weighted toward the least-permeable layers, not the arithmetic mean. Also, forgetting to fix an elevation datum before separating $H$ into $z$ and $h_p$ — only differences in $z$ matter, but the datum must be stated and applied consistently.

Source. Freeze & Cherry, Groundwater, Ch. 2.10 (heterogeneity, layered media and equivalent vertical/horizontal conductivity) and Ch. 2.2 (total, pressure and elevation head).

Question 3 of 5: Vertical Flow Through Three Layered Formations

Question 3: Vertical Flow Through Three Layered Formations (equal value)