CELE Geotechnical Engineering — Permeability and SeepageExam Answer Templates

Marks

1

Topic

Darcy's Law

Difficulty

easy

Template Id

T1

Examiner Tip

For a 1-mark question, the equation alone with correct symbols is sufficient. A full verbal statement without the formula is risky — always include the equation.

Model Answer

Darcy's Law states that the discharge velocity of flow through a saturated soil is directly proportional to the hydraulic gradient: v = ki, and the volumetric flow rate is Q = kiA, where k is the coefficient of permeability (m/s), i = h/L is the hydraulic gradient, and A is the gross cross-sectional area of flow.

Question Type

very_short_answer

Answer Structure

• Line 1: State the proportionality principle and write the equation v = ki, Q = kiA [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing v = kh instead of v = ki (confusing head h with gradient i = h/L).
• Omitting the equation entirely and only writing a verbal statement without the formula.
• Using Q = kia instead of Q = kiA (lowercase a is the standpipe area in falling-head test, not the flow area).

Key Phrases To Include

• discharge velocity
• hydraulic gradient
• v = ki
• Q = kiA
• coefficient of permeability

Marks

2

Topic

Darcy's Law

Difficulty

easy

Template Id

T2

Examiner Tip

Examiners specifically look for the equation vs = v/n and the inequality vs > v. Stating both earns the second mark reliably.

Model Answer

Discharge velocity (v) is the volumetric flow rate per unit gross cross-sectional area of the soil: v = ki. It is a fictitious velocity because flow actually occurs only through the void spaces, not through the entire cross-section. Seepage velocity (vs) is the actual average velocity of water moving through the pore spaces: vs = v/n, where n is the porosity of the soil. Since porosity n < 1, the seepage velocity is always greater than the discharge velocity (vs > v).

Question Type

short_answer

Answer Structure

• Line 1–2: Define discharge velocity v = ki and explain it is based on gross area [1 mark]
• Line 3–4: Define seepage velocity vs = v/n and state that vs > v because n < 1 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating vs = v × n instead of vs = v/n — a very common reversal error.
• Failing to explain why vs > v (because n < 1).
• Confusing seepage velocity with Darcy velocity without noting the difference in area basis.

Key Phrases To Include

• gross cross-sectional area
• void spaces
• porosity
• vs = v/n
• seepage velocity always greater

Marks

3

Topic

Laboratory Permeability Tests — Constant-Head

Difficulty

medium

Template Id

T3

Examiner Tip

Always convert all units to a consistent system at the top of your solution before substituting. Show the unit conversion explicitly — it can earn a mark even if the arithmetic is slightly off.

Model Answer

Given: Length L = 200 mm Diameter D = 100 mm → A = π(100)²/4 = 7854 mm² Head h = 400 mm Volume collected V = 240 cm³ = 240 000 mm³ Time t = 3 min = 180 s Formula (Constant-Head Test): k = VL / (A · h · t) Substitution: k = (240 000 × 200) / (7854 × 400 × 180) k = 48 000 000 / 565 488 000 k = 8.49 × 10⁻² mm/s ∴ k = 8.49 × 10⁻⁵ m/s This value is typical of a clean sand or gravel, consistent with the use of a constant-head test.

Question Type

numerical

Answer Structure

• Step 1: List all given data with consistent units (mm, mm², mm³, s) [0.5 mark]
• Step 2: Write the constant-head formula k = VL/(Aht) [1 mark]
• Step 3: Compute A = πD²/4 correctly [0.5 mark]
• Step 4: Substitute and evaluate k with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the falling-head formula instead of constant-head formula — identify the test type first.
• Forgetting to convert volume from cm³ to mm³ (multiply by 1000) or time from minutes to seconds.
• Using diameter as the area instead of computing A = πD²/4.
• Reporting k in mm/s without converting to m/s for the final answer.

Key Phrases To Include

• constant-head test
• k = VL/(Aht)
• A = πD²/4
• unit conversion
• 8.49 × 10⁻⁵ m/s

Marks

3

Topic

Laboratory Permeability Tests — Falling-Head

Difficulty

medium

Template Id

T4

Examiner Tip

Memorize which area is which: lowercase a = standpipe (small tube), uppercase A = soil sample (large cross-section). The factor 2.303 is there because the formula is derived using natural logarithms converted to base-10 logs.

Model Answer

Given: Standpipe area a = 100 mm² Sample area A = 2 500 mm² Sample length L = 120 mm Initial head h₁ = 600 mm, Final head h₂ = 250 mm Time t = 75 s Formula (Falling-Head Test): k = (2.303 · a · L) / (A · t) × log₁₀(h₁/h₂) Compute log ratio: log₁₀(600/250) = log₁₀(2.40) = 0.3802 Substitute: k = (2.303 × 100 × 120) / (2 500 × 75) × 0.3802 k = 27 636 / 187 500 × 0.3802 k = 0.14739 × 0.3802 k = 0.05604 mm/s ∴ k = 5.60 × 10⁻² mm/s = 5.60 × 10⁻⁵ m/s This value is characteristic of a silty sand or fine sand, consistent with use of a falling-head test.

Question Type

numerical

Answer Structure

• Step 1: Identify test type as falling-head and list a, A, L, h₁, h₂, t [0.5 mark]
• Step 2: Write falling-head formula k = (2.303·a·L)/(A·t) × log₁₀(h₁/h₂) [1 mark]
• Step 3: Evaluate log₁₀(h₁/h₂) = log₁₀(2.40) = 0.3802 [0.5 mark]
• Step 4: Substitute and compute k with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using natural log (ln) instead of log₁₀ — the factor 2.303 converts ln to log₁₀, so using ln without 2.303 is redundant.
• Swapping a (standpipe area) and A (sample area) in the formula.
• Inverting the log ratio to log(h₂/h₁) giving a negative k — always use h₁/h₂ where h₁ > h₂.
• Using the constant-head formula instead of falling-head formula.

Key Phrases To Include

• falling-head test
• standpipe area a
• 2.303
• log₁₀(h₁/h₂)
• 5.60 × 10⁻⁵ m/s

Marks

2

Topic

Quick Condition and Piping

Difficulty

easy

Template Id

T5

Examiner Tip

Even in a 2-mark question, show the force-balance derivation in one line. It demonstrates understanding and protects you from losing the mark if you mis-remember the final formula.

Model Answer

The critical hydraulic gradient (icr) is the upward hydraulic gradient at which the effective stress in a soil mass becomes zero, causing the soil to lose all shear strength and behave like a liquid — a condition known as quicksand or boiling. Derivation: At the critical condition, upward seepage force equals the submerged weight of soil: i_cr · γ_w = γ' = γ_sat − γ_w i_cr = γ'/γ_w = (G_s − 1)/(1 + e) For a typical soil with G_s = 2.65 and e = 0.65: i_cr ≈ 1.0.

Question Type

short_answer

Answer Structure

• Line 1–2: Define icr as the gradient at which effective stress = 0 (quicksand condition) [1 mark]
• Line 3–5: Derive i_cr = (Gs − 1)/(1 + e) by equating upward seepage force to submerged unit weight [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing i_cr = Gs/(1+e) — omitting the '−1' in the numerator.
• Stating the condition as 'total stress equals zero' instead of 'effective stress equals zero'.
• Not linking the formula to the balance of forces.

Key Phrases To Include

• effective stress equals zero
• upward seepage force
• submerged unit weight γ'
• i_cr = (Gs − 1)/(1 + e)
• quicksand or boiling

Marks

3

Topic

Quick Condition and Piping

Difficulty

medium

Template Id

T6

Examiner Tip

Board exams frequently ask for FS interpretation. A one-sentence conclusion ('FS = 2.17 > 1.5, therefore safe') often carries the final mark. Never leave interpretation blank.

Model Answer

Given: G_s = 2.68, e = 0.72, i_exit = 0.45 (a) Critical Hydraulic Gradient: i_cr = (G_s − 1) / (1 + e) i_cr = (2.68 − 1) / (1 + 0.72) i_cr = 1.68 / 1.72 i_cr = 0.977 ≈ 0.98 (b) Factor of Safety Against Piping: FS = i_cr / i_exit FS = 0.977 / 0.45 FS = 2.17 ∴ FS = 2.17 > 1.5 (generally acceptable minimum); the soil is safe against piping under the given conditions.

Question Type

numerical

Answer Structure

• Step 1: Write i_cr = (Gs − 1)/(1 + e) and substitute [1 mark]
• Step 2: Compute i_cr = 0.977 correctly [0.5 mark]
• Step 3: Write FS = i_cr / i_exit and substitute [0.5 mark]
• Step 4: Compute FS = 2.17 and state engineering interpretation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using i_cr = Gs/(1+e) without subtracting 1 from Gs.
• Inverting the FS formula as FS = i_exit / i_cr — FS must be ≥ 1 for safe condition, so i_cr goes on top.
• Not providing an engineering judgment on the adequacy of the FS.

Key Phrases To Include

• i_cr = (Gs − 1)/(1 + e)
• FS = i_cr / i_exit
• 0.977
• 2.17
• safe against piping

Marks

5

Topic

Layered Soils — Equivalent Permeability

Difficulty

hard

Template Id

T7

Examiner Tip

A memory trick: 'Parallel = Powerful (arithmetic)' because the high-k layer drives the flow; 'Perpendicular = Pinched (harmonic)' because the tight layer controls the flow. This avoids formula swap errors.

Model Answer

Given: k₁ = 4 × 10⁻⁴ m/s, H₁ = 1.5 m k₂ = 6 × 10⁻⁶ m/s, H₂ = 2.5 m Total thickness H = H₁ + H₂ = 1.5 + 2.5 = 4.0 m (a) Flow PARALLEL to Layers (horizontal flow; high-k layer dominates): Formula: k_eq(H) = (k₁H₁ + k₂H₂) / (H₁ + H₂) k_eq(H) = [(4×10⁻⁴)(1.5) + (6×10⁻⁶)(2.5)] / 4.0 = [6×10⁻⁴ + 1.5×10⁻⁵] / 4.0 = [6.00×10⁻⁴ + 0.015×10⁻⁴] / 4.0 = 6.015×10⁻⁴ / 4.0 ∴ k_eq(H) = 1.50 × 10⁻⁴ m/s Note: Dominated by the more permeable Layer 1 (k₁ >> k₂). (b) Flow PERPENDICULAR to Layers (vertical flow; low-k layer dominates): Formula: k_eq(V) = (H₁ + H₂) / (H₁/k₁ + H₂/k₂) Compute denominators: H₁/k₁ = 1.5 / (4×10⁻⁴) = 3 750 s H₂/k₂ = 2.5 / (6×10⁻⁶) = 416 667 s k_eq(V) = 4.0 / (3 750 + 416 667) = 4.0 / 420 417 ∴ k_eq(V) = 9.51 × 10⁻⁶ m/s Note: Dominated by the less permeable Layer 2 (k₂ << k₁). Summary: k_eq(H) = 1.50 × 10⁻⁴ m/s >> k_eq(V) = 9.51 × 10⁻⁶ m/s — horizontal permeability is about 16× greater than vertical, a common observation in natural sedimentary deposits.

Question Type

numerical

Answer Structure

• Step 1: Identify total thickness H = H₁ + H₂ = 4.0 m [0.5 mark]
• Step 2: State parallel-flow formula k_eq(H) = Σ(kᵢHᵢ)/ΣHᵢ [1 mark]
• Step 3: Compute k_eq(H) = 1.50 × 10⁻⁴ m/s correctly [1 mark]
• Step 4: State perpendicular-flow formula k_eq(V) = ΣHᵢ / Σ(Hᵢ/kᵢ) [1 mark]
• Step 5: Compute H₁/k₁ and H₂/k₂, then k_eq(V) = 9.51 × 10⁻⁶ m/s [1 mark]
• Step 6: State engineering observation — parallel > perpendicular; dominant layer interpretation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping the formulas — using the harmonic formula for parallel flow and arithmetic for perpendicular flow.
• Not computing H₁/k₁ and H₂/k₂ separately and making arithmetic errors in the denominator.
• Omitting units or reporting both answers in the same unit without checking order of magnitude.
• Not commenting on which layer dominates, losing the interpretation mark.

Key Phrases To Include

• parallel flow
• perpendicular flow
• k_eq(H) = Σ(kᵢHᵢ)/ΣHᵢ
• k_eq(V) = ΣHᵢ/Σ(Hᵢ/kᵢ)
• high-k dominates
• low-k dominates
• 1.50 × 10⁻⁴ m/s
• 9.51 × 10⁻⁶ m/s

Marks

2

Topic

Seepage and Flow Nets

Difficulty

easy

Template Id

T8

Examiner Tip

The most common error in the flow-net formula is inverting Nf/Nd. Remember: you want more flow channels (Nf in numerator) to get more seepage, so Nf goes on top.

Model Answer

A flow net is a graphical solution of Laplace's equation for two-dimensional seepage flow. It consists of two families of curves drawn orthogonally: flow lines (or streamlines) that trace the path of water particles, and equipotential lines that connect points of equal total head. Together they form a network of curvilinear squares. Seepage per unit width of dam: Q = k · H · (N_f / N_d) where k = coefficient of permeability, H = total head difference across the structure, N_f = number of flow channels (spaces between adjacent flow lines), and N_d = number of equipotential drops (spaces between adjacent equipotential lines).

Question Type

short_answer

Answer Structure

• Line 1–3: Define flow net as orthogonal grid of flow lines and equipotential lines forming curvilinear squares [1 mark]
• Line 4–6: Write Q = kH(Nf/Nd) and define all symbols correctly [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating Nf = number of flow lines and Nd = number of equipotential lines — these are the spaces (channels and drops), not the lines themselves.
• Omitting 'per unit width' qualifier on the seepage formula.
• Writing Q = kH(Nd/Nf) — inverting the ratio.

Key Phrases To Include

• flow lines
• equipotential lines
• orthogonal
• curvilinear squares
• Q = kH(Nf/Nd)
• flow channels
• equipotential drops

Marks

3

Topic

Seepage and Flow Nets

Difficulty

medium

Template Id

T9

Examiner Tip

Always double-check: Nf goes on top because it represents how many flow paths carry water. Increasing Nf increases seepage — this physical check prevents the inversion error.

Model Answer

Given: N_f = 5 (flow channels) N_d = 15 (equipotential drops) H = 9 m k = 3.5 × 10⁻⁵ m/s Formula: Q = k · H · (N_f / N_d) Substitute: Q = (3.5 × 10⁻⁵)(9)(5/15) Q = (3.5 × 10⁻⁵)(9)(0.3333) Q = (3.5 × 10⁻⁵)(3.0) Q = 1.05 × 10⁻⁴ m³/s per metre of dam ∴ Q = 1.05 × 10⁻⁴ m³/s·m (seepage loss per unit width of dam)

Question Type

numerical

Answer Structure

• Step 1: List N_f, N_d, H, k and identify these as flow channels and equipotential drops [0.5 mark]
• Step 2: Write Q = kH(Nf/Nd) [1 mark]
• Step 3: Substitute and compute correctly [1 mark]
• Step 4: State unit as m³/s per metre of dam width [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using Nf/Nd = 15/5 = 3 instead of 5/15 — inverting the ratio.
• Omitting 'per metre width' from the unit of Q.
• Confusing Nf (number of flow tubes/channels) with the number of flow lines (which is Nf + 1).

Key Phrases To Include

• Q = kH(Nf/Nd)
• flow channels
• equipotential drops
• 1.05 × 10⁻⁴ m³/s
• per metre width

Marks

2

Topic

Laboratory Permeability Tests

Difficulty

easy

Template Id

T10

Examiner Tip

This is a common 2-mark board question. Answer in two clearly labeled parts (a) and (b) and always include the reason. The reason typically earns the mark, not just the test name.

Model Answer

(a) Clean gravel → Constant-Head Test. Reason: Gravel has high permeability (k > 10⁻³ m/s), so water flows quickly through the sample. The constant-head test maintains a fixed head difference, allowing steady discharge to be measured over a short period without the head dropping uncontrollably. (b) Silty clay → Falling-Head Test. Reason: Silty clay has very low permeability (k < 10⁻⁶ m/s), meaning discharge rates are extremely small. The falling-head test measures the rate of head drop in a small-diameter standpipe, which is sensitive enough to detect the low flow rates without requiring an impractically long collection period.

Question Type

short_answer

Answer Structure

• Part (a): Constant-head for coarse soils; reason — high k, measurable discharge [1 mark]
• Part (b): Falling-head for fine soils; reason — low k, flow rate too small for constant-head measurement [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the tests — stating constant-head for fine soils and falling-head for coarse soils.
• Naming the correct test but giving no reason, which forfeits the mark.
• Stating 'any test can be used' without qualification.

Key Phrases To Include

• constant-head
• falling-head
• high permeability
• low permeability
• standpipe
• measurable discharge

Marks

1

Topic

Darcy's Law — Hydraulic Gradient

Difficulty

easy

Template Id

T11

Examiner Tip

Hydraulic gradient is dimensionless. If you end up with units, check that h and L are in the same unit before dividing.

Model Answer

Hydraulic gradient: i = h/L i = 350 mm / 250 mm = 1.4 (dimensionless) Note: A hydraulic gradient greater than 1.0 is possible in the laboratory and in some field conditions (e.g., beneath sheet piles), but sustained values approaching icr ≈ 1 in upward seepage indicate risk of boiling.

Question Type

very_short_answer

Answer Structure

• Line 1: Write i = h/L and substitute values [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing i = L/h — inverting the ratio.
• Attaching a unit to i (e.g., writing i = 1.4 mm/mm instead of just 1.4) — gradient is dimensionless.
• Using different units for h and L without cancellation (e.g., h in m and L in mm).

Key Phrases To Include

• i = h/L
• hydraulic gradient
• dimensionless
• 1.4

Marks

3

Topic

Darcy's Law — Seepage Velocity

Difficulty

medium

Template Id

T12

Examiner Tip

Seepage velocity problems almost always ask you to confirm vs > v. A one-line explanation ('since n = 0.40 < 1, dividing by n gives a larger value') earns the interpretation mark.

Model Answer

Given: A = 0.30 m × 0.30 m = 0.09 m² k = 5 × 10⁻⁴ m/s h = 180 mm = 0.18 m L = 600 mm = 0.60 m n = 0.40 (a) Hydraulic Gradient: i = h/L = 0.18/0.60 = 0.30 Discharge Velocity: v = ki = (5 × 10⁻⁴)(0.30) v = 1.50 × 10⁻⁴ m/s (b) Seepage Velocity: v_s = v/n = (1.50 × 10⁻⁴)/0.40 v_s = 3.75 × 10⁻⁴ m/s ∴ v = 1.50 × 10⁻⁴ m/s (discharge velocity) v_s = 3.75 × 10⁻⁴ m/s (seepage velocity; vs > v as expected since n < 1)

Question Type

numerical

Answer Structure

• Step 1: Compute i = h/L = 0.30 [0.5 mark]
• Step 2: Compute v = ki = 1.50 × 10⁻⁴ m/s [1 mark]
• Step 3: Compute vs = v/n = 3.75 × 10⁻⁴ m/s [1 mark]
• Step 4: Confirm vs > v and state physical meaning [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using vs = v × n instead of vs = v/n.
• Not converting h and L to the same units before computing i.
• Forgetting to state the physical reason why vs > v.

Key Phrases To Include

• i = h/L
• v = ki
• vs = v/n
• 1.50 × 10⁻⁴ m/s
• 3.75 × 10⁻⁴ m/s
• vs > v because n < 1

Marks

5

Topic

Quick Condition and Piping

Difficulty

hard

Template Id

T13

Examiner Tip

Five-mark questions reward engineering judgment. Always include a conclusion in part (c) that references an accepted minimum FS. Even if you are unsure of the exact minimum, stating 'FS = 2.06 > 1.5, therefore safe for a permanent structure' demonstrates professional reasoning.

Model Answer

Given: G_s = 2.70, e = 0.65 γ_sat = 19.8 kN/m³, γ_w = 9.81 kN/m³ i_exit = 0.50 (a) Critical Hydraulic Gradient: i_cr = (G_s − 1) / (1 + e) i_cr = (2.70 − 1) / (1 + 0.65) i_cr = 1.70 / 1.65 i_cr = 1.030 Verification using submerged unit weight: γ' = γ_sat − γ_w = 19.8 − 9.81 = 9.99 kN/m³ i_cr = γ'/γ_w = 9.99/9.81 = 1.018 ≈ 1.03 ✓ (consistent) (b) Factor of Safety Against Heave (Piping): FS = i_cr / i_exit FS = 1.030 / 0.50 FS = 2.06 (c) Engineering Comment: A factor of safety of 2.06 exceeds the commonly adopted minimum FS = 1.5 for piping under temporary structures and approaches the recommended FS ≥ 2.0 for permanent structures. Therefore, the structure is adequately safe against upward heave under the stated conditions. However, the designer should verify that i_exit was computed at the critical exit point (immediately downstream of the sheet pile toe) and that conservative k values were used, as recommended in general geotechnical practice.

Question Type

numerical

Answer Structure

• Step 1: Write i_cr = (Gs − 1)/(1 + e) and substitute [1 mark]
• Step 2: Compute i_cr = 1.030 [1 mark]
• Step 3: Verify using γ' = γsat − γw (optional but earns full credit) [0.5 mark]
• Step 4: Write FS = i_cr / i_exit and substitute [1 mark]
• Step 5: Compute FS = 2.06 [0.5 mark]
• Step 6: Engineering comment comparing FS to minimum recommended values for permanent vs. temporary structures [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not computing the verification step using γ' — while optional, skipping it loses the opportunity for an extra-credit impression.
• Comparing FS only to 1.0 (instead of 1.5 or 2.0) — examiners expect you to know minimum acceptable FS values.
• Omitting the engineering comment (part c) entirely, losing 1 of 5 marks.
• Using i_exit in numerator and i_cr in denominator (inverted FS).

Key Phrases To Include

• i_cr = (Gs − 1)/(1 + e)
• 1.030
• FS = i_cr / i_exit
• 2.06
• permanent structure
• FS ≥ 2.0
• exit hydraulic gradient

Marks

1

Topic

Layered Soils — Equivalent Permeability

Difficulty

easy

Template Id

T14

Examiner Tip

For a 1-mark question, name both conditions and identify which average each uses. A two-sentence answer is ideal.

Model Answer

Equivalent permeability (k_eq) is a single representative permeability value that replaces a multi-layered soil system and produces the same total seepage rate. It is computed under two conditions: (1) flow parallel to the layers, where k_eq = Σ(kᵢHᵢ)/ΣHᵢ (weighted arithmetic average — high-k layer dominates); and (2) flow perpendicular to the layers, where k_eq = ΣHᵢ/Σ(Hᵢ/kᵢ) (harmonic-weighted average — low-k layer dominates).

Question Type

very_short_answer

Answer Structure

• Line 1: Define k_eq as a single representative value for a layered system [0.5 mark]
• Line 2: State the two conditions with formula identifiers [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Mentioning only one condition instead of both.
• Reversing which average applies to which flow direction.

Key Phrases To Include

• equivalent permeability
• parallel to layers
• perpendicular to layers
• arithmetic average
• harmonic average

Marks

5

Topic

Layered Soils — Equivalent Permeability

Difficulty

hard

Template Id

T15

Examiner Tip

Show each kᵢHᵢ product on a separate line. Examiners check intermediate steps, and partial marks are awarded for correct products even if the final sum has an arithmetic error.

Model Answer

Given (horizontal flow → parallel to layers): k₁ = 2×10⁻³ m/s, H₁ = 1.0 m k₂ = 5×10⁻⁴ m/s, H₂ = 2.0 m k₃ = 8×10⁻⁵ m/s, H₃ = 1.5 m i = 0.006 (horizontal) Total H = 1.0 + 2.0 + 1.5 = 4.5 m Step 1 — Equivalent Permeability (Parallel Flow): k_eq = (k₁H₁ + k₂H₂ + k₃H₃) / (H₁ + H₂ + H₃) Numerator: k₁H₁ = (2×10⁻³)(1.0) = 2.000×10⁻³ k₂H₂ = (5×10⁻⁴)(2.0) = 1.000×10⁻³ k₃H₃ = (8×10⁻⁵)(1.5) = 0.120×10⁻³ Σ(kᵢHᵢ) = 3.120×10⁻³ m²/s k_eq = 3.120×10⁻³ / 4.5 = 6.933×10⁻⁴ m/s Step 2 — Discharge Velocity: v = k_eq · i = (6.933×10⁻⁴)(0.006) = 4.16×10⁻⁶ m/s Step 3 — Total Seepage Flow per unit width: Q = v · A = v · (H × 1 m) Q = (4.16×10⁻⁶)(4.5) Q = 1.872×10⁻⁵ m³/s per metre width Alternatively using Q = k_eq · i · A directly: Q = (6.933×10⁻⁴)(0.006)(4.5 × 1) Q = 1.872×10⁻⁵ m³/s per metre width ∴ Q = 1.87 × 10⁻⁵ m³/s per metre width of deposit

Question Type

numerical

Answer Structure

• Step 1: Identify horizontal flow → parallel → use arithmetic-weighted formula [0.5 mark]
• Step 2: Write k_eq = Σ(kᵢHᵢ)/ΣHᵢ and compute each kᵢHᵢ term [1.5 marks]
• Step 3: Compute k_eq = 6.933 × 10⁻⁴ m/s [1 mark]
• Step 4: Apply Q = k_eq · i · A with A = H × 1 m [1 mark]
• Step 5: Final Q = 1.87 × 10⁻⁵ m³/s per metre width, correctly stated [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the perpendicular (harmonic) formula for horizontal flow.
• Forgetting to multiply by the total thickness H when computing A = H × 1 m.
• Arithmetic errors in summing the kᵢHᵢ products — show each product separately.
• Not specifying 'per metre width' in the final answer unit.

Key Phrases To Include

• parallel to layers
• k_eq = Σ(kᵢHᵢ)/ΣHᵢ
• 3.120 × 10⁻³
• 6.933 × 10⁻⁴ m/s
• Q = k_eq · i · A
• 1.87 × 10⁻⁵ m³/s per metre