(a) The velocity profile in a laminar boundary layer on a flat plate is modelled by the cubic expression μ/μ₁ = a₀ + a₁y + a₂y²…

(a) Cubic boundary‑layer profile, curves and significance

The velocity profile is given by u/U = a₀ + a₁y + a₂y² + a₃y³, where U = μ₁ is the main‑stream velocity and y is measured normal to the plate. Four boundary conditions determine the four constants. First, the no‑slip condition at the wall, y = 0, u = 0, gives a₀ = 0. Second, the wall momentum equation for a flat plate in zero pressure gradient, (1/ρ)(∂p/∂x) = ν(∂²u/∂y²) at y = 0, with ∂p/∂x = 0, requires ∂²u/∂y² = 0 at y = 0; differentiating twice gives 2a₂ = 0, so a₂ = 0. Third, at the edge of the layer, y = δ, the velocity equals the free stream, u = U, so a₁δ + a₃δ³ = 1. Fourth, at the edge the gradient vanishes, ∂u/∂y = 0, giving a₁ + 3a₃δ² = 0. Solving these two equations yields a₃ = −1/(2δ³) and a₁ = 3/(2δ). The resulting profile is u/U = (3/2)(y/δ) − (1/2)(y/δ)³, a curve that starts at the origin with its maximum slope 3U/(2δ), has its only inflection at the wall (y = 0, where ∂²u/∂y² = 0), and flattens to U at y = δ. The shear stress τ = μ(∂u/∂y) = (3μU/2δ)[1 − (y/δ)²] is parabolic, maximum τ_w = 3μU/(2δ) at the wall and zero at y = δ. The boundary layer is significant because it confines all skin‑friction drag, flow separation, transition to turbulence and convective heat transfer to a thin region, allowing the outer flow to be treated as inviscid; it is therefore fundamental to the design of aerofoils, ship hulls, pipes and heat exchangers. Moving plate: the no-slip condition now requires u = 2 m/s at y = 0, while u must still reach U at y = δ. Measured relative to the plate, the profile keeps the same cubic shape with the relative stream velocity (U − 2), so in the fixed frame u = 2 + (U − 2)[(3/2)(y/δ) − (1/2)(y/δ)³]. The curve starts at 2 m/s on the wall, the wall shear falls to 3μ(U − 2)/(2δ), and if U < 2 m/s the plate drags the fluid, so the profile and the shear stress reverse in sign.

(b) Total, neutral and effective stress variation

First compute the unit weights. With porosity n = 0.40, the void ratio is e = n/(1−n) = 0.667. The dry unit weight is γ_d = Gγ_w/(1+e) = 2.7×10/1.667 = 16.2 kN/m³, and the saturated unit weight is γ_sat = (G+e)γ_w/(1+e) = (2.7+0.667)×10/1.667 = 20.2 kN/m³. For the moist zone (0–4 m, degree of saturation S = 0.10), γ_moist = γ_d + (Seγ_w)/(1+e) = 16.2 + (0.1×0.667×10)/1.667 = 16.6 kN/m³. The capillary fringe (4–5 m) is saturated (γ_sat) but carries negative pore pressure, varying linearly from −γ_w×1 = −10 kN/m² at 4 m to 0 at the water table (5 m). The computed values are:

• z = 0 m: σ = 0, u = 0, σ′ = 0
• z = 4 m (top of capillary fringe): σ = 4 × 16.6 = 66.4 kN/m², u = −10 kN/m², σ′ = 76.4 kN/m² (66.4 just above, in the moist zone where u = 0)
• z = 5 m (water table): σ = 66.4 + 20.2 = 86.6 kN/m², u = 0, σ′ = 86.6 kN/m²
• z = 10 m: σ = 86.6 + 5 × 20.2 = 187.6 kN/m², u = 5 × 10 = 50 kN/m², σ′ = 137.6 kN/m²

The total‑stress line is linear with slope 16.6 from 0 to 4 m, then slope 20.2 from 4 to 10 m, giving a single kink at 4 m. The neutral‑stress line is zero from 0 to 4 m, drops to −10 at 4 m, rises to 0 at 5 m, then rises linearly to +50 at 10 m. The effective‑stress line rises with slope 16.6 to 66.4 at 4 m, jumps by 10 to 76.4 there (the suction effect), then rises with the buoyant slope 10.2 through the fringe and below to 137.6 at 10 m. The jump at 4 m, rather than a gradual change, is the signature of the capillary fringe and must be shown on the plot.

(c) Pump power and HGL/EGL

Take g = 9.81 m/s² (2g = 19.62) and γ_w = 9.81 kN/m³. In pipe J–C (L = 1000 m, d = 0.20 m, f = 0.025), A = 0.03142 m², V_C = 0.10/0.03142 = 3.183 m/s, V_C²/2g = 0.516 m, so h_f,C = 0.025(1000/0.20)(0.516) = 64.55 m. Energy from the junction to C gives H_J = 90 + 64.55 = 154.55 m. The flow from A to the junction must equal the flow to B plus the flow to C. In pipe J–B (L = 300 m, d = 0.15 m, f = 0.015), A = 0.01767 m² and V_B²/2g = (Q_B/0.01767)²/19.62 = 163.2 Q_B², so h_f,B = 0.015(300/0.15)(163.2 Q_B²) = 4896 Q_B². Energy from the junction to B gives 154.55 = 110 + 4896 Q_B², so Q_B² = 0.009099 and Q_B = 0.0954 m³/s. Hence the total discharge from the pump is Q_A = Q_B + Q_C = 0.0954 + 0.10 = 0.1954 m³/s. In pipe A–J (L = 400 m, d = 0.30 m, f = 0.02), A = 0.07069 m², V_A = 0.1954/0.07069 = 2.764 m/s, V_A²/2g = 0.389 m, and h_f,AJ = 0.02(400/0.30)(0.389) = 10.38 m. The pump head is H_p = H_J − 65 + h_f,AJ = 154.55 − 65 + 10.38 = 99.93 m. The required power is P = γ_w Q_A H_p / η = 9.81×0.1954×99.93/0.70 = 273.6 kW, about 274 kW. HGL and EGL. The pump sits at the junction. The EGL starts at 65 m at reservoir A and falls linearly by the suction-pipe loss of 10.38 m to 54.62 m just upstream of the pump. It rises vertically by the pump head of 99.93 m to 154.55 m, then falls linearly by 64.55 m to 90 m at C and by 44.55 m to 110 m at B. The HGL runs parallel to the EGL, lower by the velocity head of each pipe (0.39 m in A–J, 0.52 m in J–C, 1.49 m in J–B), and the two meet at the reservoir surfaces once the exit velocity head is lost.