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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.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: v7

HGE - Geotechnical Engineering / Permeability / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A constant-head permeameter uses a soil sample 230 mm in diameter and 360 mm long. The head is held at 440 mm. In 6 minutes, 320 cm3 of water is collected. Determine the coefficient of permeability in cm/hr.

  1. 4.66 cm/hr
  2. 8.44 cm/hr
  3. 6.30 cm/hr
  4. 10.52 cm/hr

Constant-head test: $Q=k\,i\,A$ with hydraulic gradient $i=\dfrac{h}{L}$, area $A=\dfrac{\pi D^{2}}{4}$, and flow $Q=\dfrac{V}{t}$ (work in cm and minutes):

$$k=\frac{Q}{iA}=\frac{V/t}{(h/L)\,A}.$$

Multiply by 60 to convert from cm/min to cm/hr.


Computed answer: 6.30 cm/hr

Question Bank: v8

HGE - Geotechnical Engineering / Permeability / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A cylindrical soil sample is 45 mm in diameter and 240 mm high. The constant head is 340 mm. In 2 minute, 1.3 liters of water is collected. Evaluate the coefficient of permeability in cm/sec.

  1. 0.63 cm/sec
  2. 0.38 cm/sec
  3. 0.48 cm/sec
  4. 0.01 cm/sec

Constant-head test: $Q=k\,i\,A$ with $i=\dfrac{h}{L}$, $A=\dfrac{\pi D^{2}}{4}$, and $Q=\dfrac{V}{t}$ (cm and minutes):

$$k=\frac{V/t}{(h/L)\,A}\ \text{(cm/min)}.$$

Divide by 60 to express $k$ in cm/sec.


Computed answer: 0.48 cm/sec

Question Bank: v76

HGE - Geotechnical Engineering / Permeability / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A permeable layer 3.5 m thick dips at 10°. Two holes are 45 m apart horizontally and the hydraulic grade drops 2 m. If permeability is 0.00045 cm/s, find flow per meter width in m3/hr.

  1. 0.005866 m³/hr
  2. 0.002957 m³/hr
  3. 0.006794 m³/hr
  4. 0.002444 m³/hr
Darcy law is $q=kiA$. Along the dipping layer, $i=\Delta h/(L_h/\cos\alpha)$ and $A=t\cos\alpha$ per meter width. After converting $k$ from cm/s to m/s, $$q=k i A(3600)=0.00244401270219\text{ m}^3/\text{hr}.$$
Computed answer: 0.002444 m³/hr

Question Bank: v286

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A confined aquifer has k=36 m/day, porosity 0.22, thickness 3.5 m, average width 4000 m, and head drop 5.5 m over length 5500 m.

Find the daily flow.

  1. 504.0 m^3/day
  2. 393.6 m^3/day
  3. 573.0 m^3/day
  4. 635.5 m^3/day

Find seepage velocity.

  1. 0.164 m/day
  2. 0.128 m/day
  3. 0.186 m/day
  4. 0.206 m/day

Estimate travel time across the aquifer.

  1. 33611 days
  2. 26250 days
  3. 38216 days
  4. 42384 days
Apply Darcy's law and divide discharge velocity by porosity.
Computed answers: 504, 0.163636363636, 33611.1111111.

Question Bank: v300

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A layered soil has thicknesses 2, 3.5, and 5.5 m with permeabilities 0.00019, 0.00025, and 0.000035 m/s.

Find equivalent horizontal permeability.

  1. 0.0001316 m/s
  2. 0.0001028 m/s
  3. 0.0001496 m/s
  4. 0.0001659 m/s

Find transmissivity for 100 m width.

  1. 523148.7 L/day
  2. 408579.2 L/day
  3. 594820.1 L/day
  4. 659690.6 L/day

Find vertical discharge per 100 m^2 under unit gradient.

  1. 0.0000605 m/s
  2. 0.0000473 m/s
  3. 0.0000688 m/s
  4. 0.0000764 m/s
Use the arithmetic thickness average for horizontal flow and harmonic average for vertical flow.
Computed answers: 0.000131590909091, 523148.745965, 0.0000605496233755.

Question Bank: v301

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A constant-head test uses a specimen length 5 cm, area 54 cm2, head 36 cm, collects 370 mL in 55 s, and has Gs=2.65 with dry mass 550 g.

Find permeability.

  1. 0.0173 cm/s
  2. 0.0135 cm/s
  3. 0.0197 cm/s
  4. 0.0218 cm/s

Find discharge velocity.

  1. 0.0173 cm/s
  2. 0.0135 cm/s
  3. 0.0197 cm/s
  4. 0.0218 cm/s

Find seepage velocity.

  1. 0.5386 cm/s
  2. 0.4206 cm/s
  3. 0.6124 cm/s
  4. 0.6792 cm/s
Use the constant-head equation, compute void ratio from specimen geometry, and divide Darcy velocity by porosity.
Computed answers: 0.0173026561915, 0.0173026561915, 0.538588299918.

Question Bank: v309

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Adding coarser sand generally causes what?

  1. Lower LL and higher PI
  2. Lower both
  3. Higher permeability
  4. Higher LL and PI
Coarser pores transmit water more readily.

Question Bank: v311

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Which property describes a soil's ability to transmit water?

  1. Moisture content
  2. Permeability
  3. Capillarity
  4. None
Permeability measures flow through connected voids.

Question Bank: v312

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Which permeability test best suits coarse soil?

  1. Constant-head
  2. Falling-head
  3. Both
  4. Neither
Constant-head testing suits high-permeability soils.

Question Bank: v315

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

In stratified soil, horizontal permeability is usually what relative to vertical permeability?

  1. Greater
  2. Smaller
  3. Equal
  4. Unrelated
Layering favors flow parallel to bedding.

Question Bank: v316

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Darcy discharge is proportional to which?

  1. Permeability
  2. Head loss
  3. Neither
  4. Both A and B
Q = kAΔh/L.

Question Bank: v319

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A soil has void ratio 0.8. Compare discharge velocity v with seepage velocity vs.

Find v/vs.

  1. 0.444
  2. 0.347
  3. 0.505
  4. 0.560
Because vs=v/n, the ratio v/vs equals porosity n=$e$/(1+$e$).
Computed answers: 0.444444444444.

Question Bank: v320

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

In a falling-head test, head decreases from 70 cm to 45 cm in 10 minutes. How long from 45 cm to 30 cm?

Find the additional time.

  1. 9.18 min
  2. 7.17 min
  3. 10.43 min
  4. 11.57 min
Falling-head time is proportional to ln(hinitial/hfinal).
Computed answers: 9.17689116564.

Question Bank: v322

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A two-layer unconfined aquifer is 1900 m long and 45 m wide. Total upstream and downstream heads are 45 m and 28 m; layers have thicknesses , m and permeabilities 3.75, 3.5 m/day.

Find equivalent permeability.

  1. 3.62 m/day
  2. 2.83 m/day
  3. 4.12 m/day
  4. 4.57 m/day

Find hydraulic gradient.

  1. 0.00895
  2. 0.00699
  3. 0.01017
  4. 0.01128

Find flow rate.

  1. 65.63 m^3/day
  2. 51.26 m^3/day
  3. 74.62 m^3/day
  4. 82.76 m^3/day
Thickness-average the horizontal permeability and use Darcy's law.
Computed answers: 3.62222222222, 0.00894736842105, 65.6289473684.

Question Bank: v334

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A well of diameter 0.25 m penetrates 27 m below a static water table. After pumping at 55 L/s, drawdown is 0.4 m at radius 85 m. A second observation radius is 30 m.

Find pumped volume per day.

  1. 4752.0 m^3/day
  2. 3711.3 m^3/day
  3. 5403.0 m^3/day
  4. 5992.3 m^3/day

Estimate permeability.

  1. 460.139 m/day
  2. 359.368 m/day
  3. 523.178 m/day
  4. 580.235 m/day

Estimate drawdown at the second radius.

  1. 0.336 m
  2. 0.262 m
  3. 0.382 m
  4. 0.423 m
Use the Dupuit-Thiem relation for an unconfined aquifer.
Computed answers: 4752, 460.138690322, 0.335725038134.

Question Bank: v373

HGE - Geotechnical Engineering / Permeability / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A layered deposit has horizontal-flow layers and vertical-flow segments with thicknesses 4,9,2 cm and permeabilities 0.04,0.08,0.03 cm/s.

Find equivalent horizontal permeability.

  1. 0.0627 cm/s
  2. 0.0489 cm/s
  3. 0.0713 cm/s
  4. 0.0790 cm/s

Find equivalent vertical permeability.

  1. 0.0537 cm/s
  2. 0.0420 cm/s
  3. 0.0611 cm/s
  4. 0.0678 cm/s

Find flow through a 2x4 cm section at unit gradient.

  1. 0.430 cm^3/s
  2. 0.336 cm^3/s
  3. 0.489 cm^3/s
  4. 0.542 cm^3/s
Use arithmetic and harmonic thickness averages.
Computed answers: 0.0626666666667, 0.0537313432836, 0.429850746269.