Fluid Mechanics and Soil Mechanics for Soil and Water Conservation

The first of four chapters for Section 4, Soil and Water Conservation Engineering, of the GATE Agricultural Engineering (AG) paper. The section has six sub-headings that are really three disciplines — the engineering sciences of water and soil, hydrology and surveying, and conservation proper — so it is written as four chapters, and this one covers the two engineering sciences the others rest on. Fluid Mechanics: ideal and real fluids and their properties, hydrostatic pressure and its measurement, continuity, kinematics and dynamics, Bernoulli’s theorem, laminar and turbulent pipe flow with the Darcy–Weisbach and Hazen–Williams equations, orifices, weirs and notches, open channels, and dimensional analysis. Soil Mechanics: engineering and index properties and the phase relationships, permeability and seepage, compaction and the Proctor test, shear strength and Mohr’s circle, active and passive earth pressure, slope stability, and Terzaghi’s one-dimensional consolidation.

1. Fluid properties, hydrostatic pressure and its measurement

An ideal fluid is incompressible and inviscid; a real fluid has viscosity, and its layers resist sliding past each other with shear stress τ = μ du/dy (a Newtonian fluid). The properties used throughout are density ρ, specific weight γ = ρg, dynamic viscosity μ, kinematic viscosity ν = μ/ρ, and surface tension σ, which raises water in a narrow tube by the capillary rise h = 4σ cos θ/(ρgd) — inversely proportional to the tube diameter, which is why fine-textured soils hold water higher above a water table than coarse ones.

Pressure in a fluid at rest increases linearly with depth, p = p₀ + ρgh, and acts equally in all directions. It is measured by piezometers, U-tube manometers (a heavier gauge liquid balancing the pressure difference: Δp = (ρ_m − ρ)gh for a differential manometer), and Bourdon gauges. On a plane surface of area A whose centroid is at depth h̄, the resultant force is F = ρgAh̄, and it acts at the centre of pressure h_cp = h̄ + I_G/(Ah̄), always below the centroid. For a vertical gate 2 m wide and 3 m high with its top at the water surface, F = 9.81 × 6 × 1.5 = 88.3 kN acting 2 m below the surface — at two-thirds of the height.

2. Continuity, kinematics and dynamics of flow, Bernoulli’s theorem and pipe flow

Flow is steady or unsteady, uniform or non-uniform, and is pictured by streamlines. Mass conservation for incompressible flow in a conduit is the continuity equation A₁V₁ = A₂V₂ = Q. Along a streamline of steady, incompressible, frictionless flow, energy per unit weight is conserved: p/(ρg) + V²/(2g) + z = constant — Bernoulli’s theorem, whose three terms are the pressure, velocity and elevation heads. For a real fluid a head-loss term h_L is added between two sections; the venturimeter and orifice meter use Bernoulli with a discharge coefficient to measure pipe flow, and the Pitot tube reads the velocity head directly.

Whether pipe flow is laminar or turbulent is decided by the Reynolds number Re = ρVD/μ = VD/ν: laminar below about 2000, turbulent above about 4000, transitional between. The head lost to friction in a pipe of length L and diameter D is given by the Darcy–Weisbach equation, h_f = fLV²/(2gD), where the friction factor is f = 64/Re in laminar flow and, in turbulent flow, depends on Re and the relative roughness (Moody diagram). For f = 0.02, L = 100 m, D = 0.1 m and V = 2 m/s, h_f = 0.02 × 1000 × 4/19.62 = 4.08 m. The empirical Hazen–Williams equation, V = 0.849 C R^0.63 S^0.54 in SI units (R the hydraulic radius, S the friction slope, C a roughness coefficient that is higher for smoother pipes), is used for water in pressure pipelines and is the usual basis for designing irrigation mains.

⚠️ Bernoulli’s theorem has conditions
Steady, incompressible, frictionless, along a streamline (or anywhere in irrotational flow), with no pump or turbine between the sections. Applying it across a hydraulic jump or a partly closed valve without a loss term is the usual mistake; "the flow must be laminar" is not one of its conditions.

3. Orifices, weirs and notches, open channels and dimensional analysis

Discharge through measuring structures
StructureDischargeHead dependence
Small orifice under head HQ = C_d a √(2gH)H^(1/2)
Rectangular weir of crest length LQ = (2/3) C_d L √(2g) H^(3/2)H^(3/2)
Triangular (V) notch of angle θQ = (8/15) C_d tan(θ/2) √(2g) H^(5/2)H^(5/2)
Cipolletti (trapezoidal, side slopes 1 horizontal : 4 vertical)behaves like a rectangular weir without end contractionsH^(3/2)

The V-notch is preferred for small field flows because its discharge rises as H^(5/2), so a small flow still gives a readable head. For a 90° notch with C_d = 0.6 and H = 0.2 m, Q = (8/15)(0.6)(1)(4.429)(0.2)^2.5 = 0.0254 m³/s = 25.4 L/s. In open channels under uniform flow, Manning’s equation V = (1/n)R^(2/3)S^(1/2) (and Chezy’s V = C√(RS)) relates velocity to the hydraulic radius R = A/P, the bed slope S and the roughness n; for n = 0.025, R = 0.5 m and S = 0.0004, V = 40 × 0.630 × 0.02 = 0.504 m/s. The Froude number Fr = V/√(gy) separates subcritical (Fr < 1) from supercritical (Fr > 1) flow.

Dimensional analysis (Buckingham’s π theorem: n variables in m fundamental dimensions form n − m independent dimensionless groups) and similitude let a model stand for a prototype. Similarity is geometric (all lengths in one ratio), kinematic (velocities in one ratio) and dynamic (force ratios equal), and dynamic similarity is achieved by matching the governing dimensionless number — the Reynolds number where viscosity dominates (pipes), the Froude number where gravity dominates (spillways, channels), and the Weber, Euler and Mach numbers where surface tension, pressure or compressibility do.

4. Engineering and index properties of soils, permeability and seepage

Soil is a three-phase system of solids, water and air. The definitions: void ratio e = V_v/V_s; porosity n = V_v/V = e/(1 + e); water content w = W_w/W_s (on the dry weight); degree of saturation S = V_w/V_v; specific gravity of solids G. They are tied by Se = wG and by the unit weights γ_d = Gγ_w/(1 + e) = γ/(1 + w) and γ_sat = (G + e)γ_w/(1 + e). A saturated soil at w = 20 per cent with G = 2.7 has e = 0.54; a soil of bulk unit weight 19.2 kN/m³ at w = 20 per cent has γ_d = 16.0 kN/m³.

Index properties classify a soil: grain-size distribution (sieve and hydrometer analysis; the coefficients of uniformity and curvature), and for fine soils the Atterberg limits — liquid limit LL, plastic limit PL, shrinkage limit — with the plasticity index PI = LL − PL. Water moves through soil by Darcy’s law, v = ki (k the coefficient of permeability, i the hydraulic gradient), and k is measured in the laboratory by the constant-head test k = QL/(Aht) for coarse soils and the falling-head test k = (aL/(At)) ln(h₁/h₂) for fine soils. Seepage under a structure is solved with a flow net: q = kH N_f/N_d per unit width, N_f flow channels and N_d equipotential drops. Where seepage flows upward, the effective stress vanishes at the critical hydraulic gradient i_c = (G − 1)/(1 + e), about 1 for most soils — the quick (boiling) condition that undermines the toe of dams and drop structures.

🧠 The falling-head test in one line
Standpipe area a = 1 cm², sample area A = 50 cm², length L = 10 cm, head falling from 60 cm to 30 cm in 600 s: k = (1 × 10)/(50 × 600) × ln 2 = 2.31 × 10⁻⁴ cm/s. The logarithm is natural, and the ratio h₁/h₂ is always greater than 1.

5. Compaction, shear strength, earth pressure, slope stability and consolidation

Compaction densifies soil by expelling air. The Proctor test compacts soil in a standard mould with a standard rammer and number of blows at a series of water contents; dry density rises to a peak, the maximum dry density, at the optimum moisture content (OMC), then falls as water takes the place of solids. The curve lies below the zero-air-voids line γ_d = Gγ_w/(1 + wG), which it can approach but never cross. Greater compactive effort (the modified test) raises the maximum dry density and lowers the OMC. Earthen dams and farm-pond embankments are specified as a percentage of Proctor maximum dry density at a moisture near optimum.

Shear strength follows the Mohr–Coulomb law τ_f = c + σ tan φ (in effective-stress terms c′ and φ′). On Mohr’s circle of stress the failure envelope is tangent to the circle at failure, and the failure plane is inclined at 45° + φ/2 to the major principal plane; for a cohesionless soil σ₁ = σ₃ tan²(45° + φ/2). Rankine’s earth pressure coefficients are K_a = (1 − sin φ)/(1 + sin φ) = tan²(45° − φ/2) for the active case (wall moving away from the soil) and K_p = 1/K_a for the passive case; the active thrust on a smooth vertical wall of height H in dry cohesionless soil is P_a = ½K_aγH², acting at H/3 above the base. For φ = 30°, K_a = 1/3, and a 6 m wall retaining soil of 18 kN/m³ carries P_a = 108 kN per metre. An infinite slope of dry cohesionless soil has a factor of safety FS = tan φ/tan β, independent of depth.

Terzaghi’s one-dimensional consolidation describes a saturated clay squeezing out water under load. The time factor is T_v = c_v t/d², with d the drainage path — half the layer thickness for double drainage, the full thickness for single drainage — and the average degree of consolidation U relates to T_v by T_v = (π/4)U² for U up to about 60 per cent. A 4 m clay layer drained at both faces (d = 2 m) with c_v = 1.0 m²/year reaches 50 per cent consolidation at t = (π/4)(0.25)(4)/1.0 = 0.785 year. The final settlement of a normally consolidated clay is S = C_cH/(1 + e₀) log₁₀((σ₀′ + Δσ)/σ₀′).

⚠️ The drainage path is half the layer when both faces drain
Time goes as d², so using the full 4 m thickness for a doubly drained layer makes the predicted time four times too long (3.14 years instead of 0.785).

Key takeaways

  • Force on a plane surface is ρgAh̄ and acts at the centre of pressure h̄ + I_G/(Ah̄), below the centroid.
  • Darcy–Weisbach h_f = fLV²/(2gD), with f = 64/Re in laminar flow; Hazen–Williams is the empirical design formula for water mains.
  • Orifice Q ∝ H^(1/2), rectangular weir ∝ H^(3/2), V-notch ∝ H^(5/2); Manning V = (1/n)R^(2/3)S^(1/2).
  • Se = wG and γ_d = γ/(1 + w) carry most phase-relation problems; the falling-head k uses the natural logarithm.
  • K_a = (1 − sin φ)/(1 + sin φ), P_a = ½K_aγH²; infinite dry slope FS = tan φ/tan β; T_v = c_v t/d² with d the drainage path.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Water flows at 2 m/s through a pipe 0.1 m in diameter and 100 m long with a Darcy friction factor of 0.02. What is the head loss due to friction, in m (to two decimal places)? Take g = 9.81 m/s².

    Numerical answer — type the value.

    Show answer

    Answer: 4.08

    h_f = fLV²/(2gD) = 0.02 × 100 × 2²/(2 × 9.81 × 0.1) = 8/1.962 = 4.08 m. Leaving out the 2g gives 80 m; using the Fanning factor convention (4f) with f = 0.02 would give four times the loss.
  2. What is the Darcy friction factor for laminar flow in a pipe at a Reynolds number of 1600?

    Numerical answer — type the value.

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    Answer: 0.04

    In laminar (Hagen–Poiseuille) flow f = 64/Re = 64/1600 = 0.04, independent of roughness. 16/Re = 0.01 is the Fanning factor, a quarter of the Darcy value, and is the usual trap.
  3. A 90° V-notch with a discharge coefficient of 0.6 runs under a head of 0.2 m. What is the discharge, in L/s (to one decimal place)? Take g = 9.81 m/s².

    Numerical answer — type the value.

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    Answer: 25.4

    Q = (8/15)C_d tan(θ/2)√(2g)H^(5/2) = 0.5333 × 0.6 × 1 × 4.429 × 0.2^2.5. With 0.2^2.5 = 0.01789, Q = 0.02536 m³/s = 25.4 L/s. Using the rectangular-weir exponent 3/2 gives a very different number, and tan θ in place of tan(θ/2) is undefined at 90°.
  4. A channel has Manning’s n = 0.025, a hydraulic radius of 0.5 m and a bed slope of 0.0004. What is the mean velocity under uniform flow, in m/s (to three decimal places)?

    Numerical answer — type the value.

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    Answer: 0.504

    V = (1/n)R^(2/3)S^(1/2) = 40 × 0.5^(2/3) × 0.02 = 40 × 0.6300 × 0.02 = 0.504 m/s. Swapping the exponents (R^(1/2)S^(2/3)) gives 0.154 m/s, and forgetting the square root of the slope gives 0.0101 m/s.
  5. A saturated soil has a water content of 20 per cent and a specific gravity of solids of 2.7. What is its void ratio?

    Numerical answer — type the value.

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    Answer: 0.54

    Se = wG with S = 1 gives e = 0.20 × 2.7 = 0.54. Using the porosity relation instead, n = e/(1 + e) = 0.35, answers a different question; putting w = 20 (not 0.20) gives 54.
  6. A soil has a bulk unit weight of 19.2 kN/m³ at a water content of 20 per cent. What is its dry unit weight, in kN/m³?

    Numerical answer — type the value.

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    Answer: 16

    γ_d = γ/(1 + w) = 19.2/1.20 = 16.0 kN/m³. Subtracting 20 per cent (19.2 × 0.8 = 15.36) treats the water content as a fraction of the total weight, but w is measured on the dry weight.
  7. In a falling-head permeability test, the standpipe area is 1 cm², the sample area 50 cm² and the sample length 10 cm. The head falls from 60 cm to 30 cm in 600 s. What is the coefficient of permeability, in units of 10⁻⁴ cm/s (to two decimal places)?

    Numerical answer — type the value.

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    Answer: 2.31

    k = (aL/(At)) ln(h₁/h₂) = (1 × 10/(50 × 600)) × ln 2 = 3.333 × 10⁻⁴ × 0.6931 = 2.31 × 10⁻⁴ cm/s. Using log₁₀ without the factor 2.303 gives 1.00 × 10⁻⁴, the classic slip.
  8. A smooth vertical retaining wall 6 m high holds dry cohesionless soil with a unit weight of 18 kN/m³ and an angle of internal friction of 30°. What is the total active thrust per metre length of wall, by Rankine’s theory, in kN?

    Numerical answer — type the value.

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    Answer: 108

    K_a = (1 − sin 30°)/(1 + sin 30°) = 0.5/1.5 = 1/3. P_a = ½K_aγH² = 0.5 × (1/3) × 18 × 36 = 108 kN/m, acting 2 m above the base. Using K_p = 3 gives the passive resistance, 972 kN/m, nine times as large.
  9. A clay layer 4 m thick is drained at both its top and bottom and has a coefficient of consolidation of 1.0 m²/year. Using T_v = (π/4)U², how long does it take to reach 50 per cent average consolidation, in years (to three decimal places)?

    Numerical answer — type the value.

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    Answer: 0.785

    T_v = (π/4)(0.5)² = 0.1963. With double drainage d = 4/2 = 2 m, so t = T_v d²/c_v = 0.1963 × 4/1.0 = 0.785 year. Taking d as the full 4 m gives 3.14 years, and using U = 50 rather than 0.5 gives nonsense.
  10. An infinite slope of dry cohesionless soil with an angle of internal friction of 35° stands at 25° to the horizontal. What is its factor of safety against sliding (to one decimal place)?

    Numerical answer — type the value.

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    Answer: 1.5

    FS = tan φ/tan β = tan 35°/tan 25° = 0.7002/0.4663 = 1.50, the same at every depth. Taking the ratio of the angles themselves (35/25 = 1.4) is the tempting shortcut and is wrong.
  11. A vertical rectangular gate 2 m wide and 3 m high holds water, with its top edge at the water surface. At what depth below the surface does the resultant hydrostatic force act, in m?

    Numerical answer — type the value.

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    Answer: 2

    h̄ = 1.5 m, A = 6 m², I_G = bd³/12 = 2 × 27/12 = 4.5 m⁴. h_cp = h̄ + I_G/(Ah̄) = 1.5 + 4.5/(6 × 1.5) = 1.5 + 0.5 = 2.0 m, i.e. two-thirds of the height. Answering 1.5 m places the force at the centroid, ignoring that pressure grows with depth.
  12. Which of the following are conditions for applying Bernoulli’s equation between two points in the simple form p/(ρg) + V²/(2g) + z = constant?

    1. The flow is steady
    2. The fluid is incompressible
    3. Friction losses are negligible
    4. The flow is laminar, with Re below 2000
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    Answer: A — The flow is steady; B — The fluid is incompressible; C — Friction losses are negligible

    The simple Bernoulli equation assumes steady, incompressible, frictionless flow along a streamline with no machine in between. Laminar flow is not required — in fact laminar flow is dominated by viscosity, the very thing the equation neglects, so the fourth option is false.
  13. The capillary rise of water in a clean glass tube is:

    1. inversely proportional to the tube diameter
    2. directly proportional to the tube diameter
    3. independent of the tube diameter
    4. proportional to the square of the tube diameter
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    Answer: A — inversely proportional to the tube diameter

    h = 4σ cos θ/(ρgd): halving the diameter doubles the rise, which is why capillary water rises highest in fine-textured soils with the narrowest pores. The "directly proportional" option reverses the dependence.