Question 4, Part 3: Landfill Groundwater – Advection and Piezometric Pressure: 04-BS-14, May 2014, Q7

Question 4, Part 3: Landfill Groundwater – Advection and Piezometric Pressure (10 marks)

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.

Quantity	Value
Head under landfill, $h_1$	210 m
Head at stream, $h_2$	203 m
Flow-path length, $L$	1000 m
Hydraulic conductivity, $K$	$3\times10^{-5}$ m/s
Porosity, $n$	0.27
Piezometer intake elevation, $z$	205 m

Find. (a) Advective travel time from the landfill to the stream; (b) the pressure read at the piezometer's intake.

Approach. Get the Darcy (specific-discharge) velocity from the gradient and $K$, convert to the true seepage velocity via porosity, then divide the flow-path length by seepage velocity for travel time. For the piezometer, use $h=z+u/\gamma_w$ with the head at that location.

(a) Hydraulic gradient. $i=\Delta h/L=(210-203)/1000=\boxed{0.007}$ (dimensionless).
(a) Darcy (specific-discharge) velocity. $v=Ki=(3\times10^{-5})(0.007)=\boxed{2.1\times10^{-7}\text{ m/s}}$ – the apparent bulk velocity, not the true speed of water in the pores.
(a) Seepage (true average linear) velocity and travel time. $v_s=v/n=2.1\times10^{-7}/0.27=7.78\times10^{-7}\text{ m/s}$. Travel time $t=L/v_s=1000/7.78\times10^{-7}=1.286\times10^{9}\text{ s}$, which converts to $t\approx \boxed{40.7\text{ years}}$ (dividing by $3.156\times10^{7}$ s/yr) – using $v$ instead of $v_s$ would have overstated the travel time by a factor of $1/n\approx3.7$.
(b) Piezometer pressure. The piezometer is installed under the landfill, where the regional water-table head is $h=210$ m; for essentially horizontal groundwater flow the total head is very nearly constant with depth (Dupuit approximation), so the head AT the intake (z = 205 m) is still $h\approx210$ m. From $h=z+u/\gamma_w$: $u/\gamma_w=h-z=210-205=5$ m of pressure head, so $u=\gamma_w\times5\text{ m}=9.81\times5=\boxed{49.05\text{ kPa}}$.

Landfill-to-stream flow path: water-table head drops 210→203 m over 1 km; the piezometer intake sits 5 m below the local 210 m water-table head, giving 49.05 kPa of pressure.

Item	Result
Gradient, $i$	0.007
Darcy velocity, $v$	$2.1\times10^{-7}$ m/s
Seepage velocity, $v_s$	$7.78\times10^{-7}$ m/s
4.3(a) Travel time	≈ 40.7 years
4.3(b) Piezometer pressure	49.05 kPa

Topic: Darcy's law, seepage velocity, and piezometric pressure.

Key relations: $i=\Delta h/L$; $v=Ki$; $v_s=v/n$; $h=z+u/\gamma_w$.

Why this works: Contaminant particles travel with the true pore-water (seepage) velocity, not the bulk Darcy velocity, because they only move through the connected void space; a piezometer reads the pressure head needed to bring the water level up to the LOCAL total head at its intake depth.

Common pitfall: Reporting the travel time using $v$ instead of $v_s$ (understating the true seepage velocity by a factor of $n\approx0.27$, i.e. overstating the travel time by 1/0.27); assuming the piezometer reads atmospheric/zero pressure just because its intake is below the ground surface, instead of using the actual local head.

Source: Freeze & Cherry, Groundwater, Ch. 2 (Darcy's law, piezometers).

Question 7 of 10: Question 4, Part 3: Landfill Groundwater – Advection and Piezometric Pressure

Question 4, Part 3: Landfill Groundwater – Advection and Piezometric Pressure (10 marks)