(a) A rigid body having dimensions of 0·6 m wide, 0·9 m high and 1·2 m long weighs 10 kN when submerged in water. What will be…

(a) Volume of the rigid body = 0·6 m × 0·9 m × 1·2 m = 0·648 m³. By Archimedes’ principle, the buoyant force when fully submerged is F_B = γ_water × V = 9790 N/m³ × 0·648 m³ = 6343·92 N = 6·34392 kN. The submerged weight is the apparent weight, so W_air − F_B = 10 kN. Therefore, W_air = 10 + 6·34392 = 16·34392 kN ≈ 16·344 kN. Mass of the body = W_air/g = 16343·92 N / 9·81 m/s² = 1666·047 kg. Density = mass/volume = 1666·047 kg / 0·648 m³ = 2571·06 kg/m³ ≈ 2571 kg/m³. Its specific weight in air = 16343·92 N / 0·648 m³ = 25222·1 N/m³ = 25·22 kN/m³. Thus, weight in air = 16·344 kN, and density = 2571 kg/m³.

(b) For a plate or surface in a turbulent stream, a very thin laminar sublayer of thickness δ′ exists immediately adjacent to the wall. The nature of the surface is decided by comparing the roughness height k_s with δ′.

Hydrodynamically smooth surface: If k_s < δ′, the roughness elements are completely submerged inside the viscous sublayer. The turbulent eddies do not touch the roughness projections, so the wall friction is governed mainly by Reynolds number and not by roughness. In the sketch, the wall is shown with small roughness bumps lying below the dashed line of the laminar sublayer; the velocity profile remains smooth near the wall, and the shear stress is essentially viscous in the sublayer. The friction factor depends on Re only.

Hydrodynamically rough surface: If k_s > δ′, the roughness elements protrude through the viscous sublayer into the turbulent zone. Each projection creates its own small wake and pressure drag. The skin friction then depends mainly on relative roughness k_s/D and becomes nearly independent of Reynolds number in the fully rough regime. In the sketch, the roughness bumps project above the laminar sublayer, eddies form behind them, and the near-wall velocity profile is disturbed.

Boundary layer separation: When a fluid flows over a surface under an adverse pressure gradient, the pressure increases in the direction of flow. The slow-moving fluid near the wall loses momentum and cannot overcome this adverse pressure gradient. The velocity gradient at the wall first becomes zero and then negative. The point where (∂u/∂y)_y=0 = 0 is the separation point. Downstream of it, reverse flow and a wake or eddying region are formed. In the sketch, the velocity profiles before separation are forward; at separation the wall velocity gradient is zero; after separation a backflow region appears near the wall.

Effects of separation: It increases pressure drag or form drag. It causes large energy losses, unsteady wakes, vibration and noise. In airfoils it reduces lift and causes stall. In diffusers, bends, pumps and turbines it reduces efficiency and may cause cavitation. It also makes flow unstable and increases pumping or power requirements. Hence, separation should be delayed or avoided by streamlining, proper pressure-gradient control, suction or blowing, and suitable design of flow passages.

(c) Given: P = 20,000 kW = 20,000,000 W, H = 40 m, boss diameter d = 0·4D, speed ratio φ = 2·5, flow ratio ψ = 0·80, overall efficiency η = 0·80. Take g = 9·81 m/s² and ρg = 9810 N/m³. First, √(2gH) = √(2 × 9·81 × 40) = √784·8 = 28·014 m/s. Flow velocity V_f = ψ√(2gH) = 0·80 × 28·014 = 22·411 m/s. Peripheral velocity u = φ√(2gH) = 2·5 × 28·014 = 70·036 m/s.

Discharge through the turbine is obtained from power: P = ηρgQH, so Q = P/(ηρgH) = 20,000,000 / (0·80 × 9810 × 40) = 63·710 m³/s. The flow area through the Kaplan runner is annular: A = π/4 (D² − d²) = π/4 (D² − 0·16D²) = 0·21πD² = 0·659734D². Also A = Q/V_f = 63·710 / 22·411 = 2·842 m². Hence D² = 2·842 / 0·659734 = 4·309, giving D = √4·309 = 2·076 m.

(i) Diameter of runner = D ≈ 2·076 m.

(ii) Rotational speed: u = πDN/60, so N = 60u/(πD) = 60 × 70·036 / (π × 2·076) = 644·4 rpm ≈ 644 rpm.

(iii) Specific speed in metric units is N_s = N√P / H^(5/4), with P in kW. H^(5/4) = 40^1·25 = 100·595. N_s = 644·4 × √20,000 / 100·595 = 644·4 × 141·421 / 100·595 = 905·9 ≈ 906. So, D = 2·076 m, N = 644 rpm, and N_s = 906 (metric).

(d) The test points are (N, w) = (4, 68%), (9, 58%), (28, 50%), (46, 45%). For the flow curve, plot moisture content w on the linear y-axis and number of blows N on the logarithmic x-axis. A best-fit straight line is drawn through the points. Using least squares on w versus log₁₀ N gives approximately w = 79·54 − 20·82 log₁₀ N. At the liquid limit, N = 25 blows. Since log₁₀ 25 = 1·39794, w_LL = 79·54 − 20·82 × 1·39794 = 50·43%. Thus, liquid limit LL ≈ 50·4%.

The flow index is the magnitude of the slope of the flow curve: I_f = −(Δw / Δlog₁₀ N) = 20·82% ≈ 20·8%. Plastic limit PL = 27%. Plasticity index PI = LL − PL = 50·4 − 27 = 23·4%. For classification, use the plasticity chart. The A-line equation is PI = 0·73(LL − 20). At LL = 50·4%, A-line PI = 0·73(50·4 − 20) = 22·2%. Since the soil has LL > 50% and PI = 23·4% > 22·2%, its point lies above the A-line. Therefore, the soil is an inorganic clay of high plasticity, classified as CH in the USCS/IS system.

(e) For a saturated clay, undrained cohesion is obtained from unconfined compressive strength: c = q_u/2 = 90/2 = 45 kN/m². Since the clay is saturated and undrained, take φ = 0. Using Skempton’s method for net ultimate bearing capacity of a footing on saturated clay: q_nu = c N_c, where N_c = 5·14 (1 + 0·2 D_f/B) (1 + 0·2 B/L). For a square footing, B = L = 0·75 m, so B/L = 1. The footing rests on the surface, so D_f = 0. Thus, N_c = 5·14 × (1 + 0) × (1 + 0·2 × 1) = 5·14 × 1·2 = 6·168. Therefore, q_nu = 45 × 6·168 = 277·56 kN/m². Safe net bearing capacity = q_nu / FOS = 277·56 / 2·5 = 111·024 kN/m² ≈ 111·02 kN/m². Since D_f = 0, the gross safe bearing capacity is also 111·02 kN/m². Thus, net ultimate bearing capacity = 277·56 kN/m², and safe bearing capacity = 111·02 kN/m².