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Definition of Terms

Permeability is the property of soil that permits water or other liquids to flow through its interconnected voids. The coefficient of permeability, also called hydraulic conductivity, indicates how easily water moves through soil.

Transmissivity or transmissibility is the ability of an aquifer to transmit water through its full saturated thickness. Discharge velocity is flow per unit gross area, while seepage velocity is the actual average velocity through the void spaces.

Darcy's Law and Seepage Velocity

$$i = \frac{h}{L}$$
$$v = ki$$
$$q = kiA$$
$$v_s = \frac{v}{n}$$

Darcy's law uses the gross cross-sectional area. Because water actually travels only through voids, seepage velocity is larger than discharge velocity.

Hydraulic Conductivity and Intrinsic Permeability

Hydraulic conductivity depends on both the soil skeleton and the fluid. Intrinsic permeability is a property of the porous medium alone.

$$k = \frac{K\gamma_w}{\eta}$$

Here $K$ is intrinsic permeability, $\gamma_w$ is unit weight of water, and $\eta$ is dynamic viscosity.

Constant Head Test

The constant head test is commonly used for coarse-grained soils, where enough water flows through the specimen to measure accurately.

$$k = \frac{QL}{Aht}$$

$Q$ is collected water volume, $L$ is specimen length, $A$ is specimen area, $h$ is constant head difference, and $t$ is collection time.

Falling Head Test

The falling head test is commonly used for fine-grained soils, where flow is slow and the head drops with time.

$$k = \frac{aL}{At}\ln\left(\frac{h_1}{h_2}\right)$$
$$k = \frac{2.303aL}{At}\log_{10}\left(\frac{h_1}{h_2}\right)$$

$a$ is standpipe area, $A$ is soil specimen area, $h_1$ and $h_2$ are initial and final heads, and $t$ is elapsed time.

Empirical Conductivity of Sands

For sands with small uniformity coefficient, empirical equations may estimate hydraulic conductivity from effective particle size.

$$k = C(D_{10})^2 \quad \text{for uniform loose sand}$$
$$k = 0.35(D_{15})^2 \quad \text{for dense or compacted sand}$$

In these relations, $k$ is in cm/sec when particle size is in cm and the empirical constant is used consistently.

Equivalent Conductivity of Layered Soils

For horizontal flow through horizontal layers, the same hydraulic gradient acts through the layers and flow adds by layer thickness.

$$k_h = \frac{k_1H_1+k_2H_2+\cdots+k_nH_n}{H_1+H_2+\cdots+H_n}$$

For vertical flow across horizontal layers, the flow rate is the same through each layer and head losses add.

$$k_v = \frac{H_1+H_2+\cdots+H_n}{\frac{H_1}{k_1}+\frac{H_2}{k_2}+\cdots+\frac{H_n}{k_n}}$$

Pumping Tests

In a pumping test, observation wells measure drawdown at known radial distances from the pumping well. Use consistent distance and discharge units.

$$k = \frac{Q\ln(r_1/r_2)}{2\pi t(h_1-h_2)} \quad \text{confined aquifer}$$
$$k = \frac{Q\ln(r_1/r_2)}{\pi(h_1^2-h_2^2)} \quad \text{unconfined aquifer}$$
$$T = kt$$
$$T = \frac{Q\ln(r_1/r_2)}{2\pi(s_2-s_1)}$$

Problem: Constant Head Test

For a constant head laboratory permeability test on fine sand: specimen length = 17 cm, specimen diameter = 5.5 cm, constant head difference = 40 cm, collected water = 50 g, and duration = 12 sec. Find hydraulic conductivity.

$$A = \frac{\pi(5.5)^2}{4}=23.76 \text{ cm}^2$$
$$Q = 50 \text{ cm}^3$$
$$k = \frac{50(17)}{23.76(40)(12)}$$
$$k = 0.0745 \text{ cm/sec}$$

Answer: $k = 0.0745$ cm/sec.

Problem: Falling Head Test

A falling head test uses a soil sample 50 mm in diameter and 200 mm high. The head in a 10 mm diameter standpipe drops from 900 mm to 600 mm in one minute. Evaluate $k$ in cm/sec.

$$a = \frac{\pi(1)^2}{4}=0.7854 \text{ cm}^2$$
$$A = \frac{\pi(5)^2}{4}=19.635 \text{ cm}^2$$
$$k = \frac{0.7854(20)}{19.635(60)}\ln\left(\frac{90}{60}\right)$$
$$k = 0.00541 \text{ cm/sec}$$

Answer: $k = 5.41\times10^{-3}$ cm/sec.

Problem: Falling Head on Silty Soil

A falling-head test has sample length 8 cm, sample area 10 cm2, standpipe area 1.5 cm2, $h_1=100$ cm, $h_2=90$ cm, and elapsed time 60 min. Find $k$ in cm/min.

$$k = \frac{1.5(8)}{10(60)}\ln\left(\frac{100}{90}\right)$$
$$k = 0.00211 \text{ cm/min}$$

Answer: $k = 2.11\times10^{-3}$ cm/min.

Problem: Sloping Permeable Layer

A permeable soil layer underlain by an impervious layer slopes at 5 degrees and is 4 m thick measured vertically. If $k=0.005$ cm/sec, determine seepage rate per meter width in liters per hour.

$$k = 0.005 \text{ cm/sec}=5.0\times10^{-5}\text{ m/sec}$$
$$i = \sin5^\circ = 0.0872$$
$$q = kiA = (5.0\times10^{-5})(0.0872)(4)(1)$$
$$q = 1.743\times10^{-5}\text{ m}^3/\text{sec}$$
$$q = 62.8 \text{ L/hr}$$

Answer: About 62.8 L/hr per meter width.

Problem: Confined Aquifer Flow

A confined aquifer has thickness 25 m, width 4 km, hydraulic conductivity 40 m/day, and porosity 0.25. Piezometer heads in two wells 1.325 km apart are 65 m and 60 m. Find flow rate, seepage velocity, and travel time for 4 km.

$$i = \frac{65-60}{1325}=0.00377$$
$$A = 25(4000)=100000 \text{ m}^2$$
$$q = kiA = 40(0.00377)(100000)=15094 \text{ m}^3/\text{day}$$
$$v_s = \frac{ki}{n}=\frac{40(0.00377)}{0.25}=0.604 \text{ m/day}$$
$$t = \frac{4000}{0.604}=6625 \text{ days}$$

Answer: $q=15094$ m3/day, $v_s=0.604$ m/day, and travel time is about 6625 days.

Problem: Layered Soil Method

For layered deposits, evaluate equivalent horizontal or vertical coefficient of permeability depending on the flow direction. Some reference examples require table or figure values embedded in the PDF image.

For horizontal flow through layers, substitute each layer thickness and coefficient into:

$$k_h = \frac{\sum k_iH_i}{\sum H_i}$$

For vertical flow through layers, use:

$$k_v = \frac{\sum H_i}{\sum(H_i/k_i)}$$

Then compute flow by Darcy's law.

$$q = k_{eq}iA$$

Problem: Unconfined Pumping Test

A 300 mm diameter test well penetrates 27 m below the static water table. Pumping is 69 L/sec. At 95 m, drawdown is 0.5 m; at 35 m, drawdown is 1.1 m. Find discharge in m3/day, coefficient of permeability, and transmissibility.

$$Q = 0.069(86400)=5961.6 \text{ m}^3/\text{day}$$
$$h_1 = 27-0.5=26.5 \text{ m}$$
$$h_2 = 27-1.1=25.9 \text{ m}$$
$$k = \frac{5961.6\ln(95/35)}{\pi(26.5^2-25.9^2)}$$
$$k = 60.27 \text{ m/day}$$
$$T = \frac{5961.6\ln(95/35)}{2\pi(1.1-0.5)}=1579 \text{ m}^2/\text{day}$$

Answer: $Q=5961.6$ m3/day, $k=60.27$ m/day, and $T=1579$ m2/day.

Problem: Confined Pumping Test

A well in a confined aquifer pumps at 13 L/sec. Aquifer thickness is 15 m. Observation wells at 10 m and 30 m have drawdowns of 3.7 m and 2.4 m. Find permeability and transmissibility.

$$Q = 0.013(86400)=1123.2 \text{ m}^3/\text{day}$$
$$k = \frac{1123.2\ln(30/10)}{2\pi(15)(3.7-2.4)}$$
$$k = 10.07 \text{ m/day}$$
$$T = kt = 10.07(15)=151.1 \text{ m}^2/\text{day}$$

Answer: $k=10.07$ m/day and $T=151.1$ m2/day.

Concept: Validity and Limits of Darcy's Law

Darcy's law is valid only for laminar flow, which occurs in most fine-grained and medium-grained soils under typical hydraulic gradients. The Reynolds number for porous media uses the effective particle size and discharge velocity.

$$Re = \frac{v D_{10}}{\nu} < 1 \quad \text{for laminar (Darcy valid)}$$

In coarse gravels and fractured rock, turbulent flow may develop at realistic gradients and Darcy's law overestimates the discharge. In very fine clays, a threshold gradient may need to be exceeded before flow begins. Within the laminar range, the coefficient of permeability is essentially constant and independent of gradient, which makes Darcy's law extremely useful for engineering calculations.

Problem: Discharge Velocity and Seepage Velocity

A sandy soil has a hydraulic conductivity of $4\times10^{-3}$ cm/sec, a porosity of 32 percent, and is subjected to a hydraulic gradient of 0.015. Compute the discharge velocity and the seepage velocity.

$$v = ki=4\times10^{-3}(0.015)=6\times10^{-5}\text{ cm/sec}$$
$$v_s = \frac{v}{n}=\frac{6\times10^{-5}}{0.32}=1.875\times10^{-4}\text{ cm/sec}$$

The seepage velocity is always greater than the discharge velocity because flow only passes through the void space, not the entire cross-sectional area.

Answer: $v=6\times10^{-5}$ cm/sec and $v_s=1.875\times10^{-4}$ cm/sec.

Problem: Equivalent Horizontal Permeability of Layered Soil

A soil deposit consists of three horizontal layers: Layer 1 is 1.5 m thick with $k_1=5\times10^{-4}$ cm/sec; Layer 2 is 2.0 m thick with $k_2=3\times10^{-3}$ cm/sec; Layer 3 is 1.0 m thick with $k_3=8\times10^{-4}$ cm/sec. Compute the equivalent horizontal permeability.

$$k_h = \frac{k_1H_1+k_2H_2+k_3H_3}{H_1+H_2+H_3}$$
$$k_h = \frac{5\times10^{-4}(1.5)+3\times10^{-3}(2.0)+8\times10^{-4}(1.0)}{1.5+2.0+1.0}$$
$$k_h = \frac{7.5\times10^{-4}+6.0\times10^{-3}+8.0\times10^{-4}}{4.5}$$
$$k_h = \frac{7.55\times10^{-3}}{4.5}=1.678\times10^{-3}\text{ cm/sec}$$

Answer: $k_h=1.678\times10^{-3}$ cm/sec.

Problem: Equivalent Vertical Permeability and Seepage Rate

Using the same three-layer deposit as the previous problem (Layer 1: 1.5 m, $k_1=5\times10^{-4}$ cm/sec; Layer 2: 2.0 m, $k_2=3\times10^{-3}$ cm/sec; Layer 3: 1.0 m, $k_3=8\times10^{-4}$ cm/sec), compute the equivalent vertical permeability. Then find the vertical seepage velocity if the total head loss across all three layers is 0.30 m.

$$k_v = \frac{H_1+H_2+H_3}{\dfrac{H_1}{k_1}+\dfrac{H_2}{k_2}+\dfrac{H_3}{k_3}}$$
$$k_v = \frac{4.5}{\dfrac{1.5}{5\times10^{-4}}+\dfrac{2.0}{3\times10^{-3}}+\dfrac{1.0}{8\times10^{-4}}}$$
$$k_v = \frac{4.5}{3000+666.7+1250}=\frac{4.5}{4916.7}=9.15\times10^{-4}\text{ cm/sec}$$
$$i_{v}=\frac{h}{H}=\frac{30\text{ cm}}{450\text{ cm}}=0.0667$$
$$v = k_v i_v=9.15\times10^{-4}(0.0667)=6.10\times10^{-5}\text{ cm/sec}$$

Answer: $k_v=9.15\times10^{-4}$ cm/sec and vertical seepage velocity $v=6.10\times10^{-5}$ cm/sec.

Problem: Time of Travel through a Confined Aquifer

A confined aquifer is 8 m thick, 500 m wide, and 2 km long. The hydraulic conductivity is 25 m/day, the porosity is 28 percent, and the hydraulic gradient is 0.004. Compute the total flow through the aquifer and the time for a contaminant particle to travel the 2 km length via seepage velocity.

$$A = 8(500)=4000\text{ m}^2$$
$$q = kiA=25(0.004)(4000)=400\text{ m}^3/\text{day}$$
$$v = ki=25(0.004)=0.1\text{ m/day}$$
$$v_s = \frac{v}{n}=\frac{0.1}{0.28}=0.357\text{ m/day}$$
$$t = \frac{L}{v_s}=\frac{2000}{0.357}=5602\text{ days}\approx15.3\text{ years}$$

Answer: $q=400$ m3/day and travel time $\approx 5602$ days or about 15.3 years.