(a) Prove that the power transmission through nozzle is maximum when d/D = √(D/8fl). Neglect the minor losses. (10 marks) (b)…

(a) Let the pipe diameter be D, nozzle diameter d, pipe length l and friction factor f. Let V be velocity in the pipe and v be velocity through the nozzle. Neglect minor losses. By continuity,

A V = a v, so (πD²/4)V = (πd²/4)v ∴ V = (d/D)² v.

Put x = d/D. Then V = x² v.

Using Darcy–Weisbach friction loss in the pipe, hf = 4 f l V²/(2 g D) = 4 f l x⁴ v²/(2 g D).

Applying Bernoulli’s equation between pipe inlet and nozzle outlet, H = hf + v²/(2g) = v²/(2g) [1 + 4 f l x⁴/D].

Hence v² = 2gH/[1 + 4 f l x⁴/D].

Discharge through nozzle: Q = (πd²/4)v = (πD²x²/4)v.

Power of the jet: P = γQ v²/(2g) = γ(πD²x²/4)v × v²/(2g) = γπD²x² v³/(8g).

Substitute v³: P = constant × x²/[1 + 4 f l x⁴/D]^(3/2).

For maximum power, dP/dx = 0. Let a = 4 f l/D. Then P = K x²(1 + a x⁴)^(−3/2). Using logarithmic differentiation, d(lnP)/dx = 2/x − (3/2)(4a x³)/(1 + a x⁴) = 0. ∴ 2(1 + a x⁴) = 6a x⁴ ∴ 1 = 2a x⁴ ∴ x⁴ = 1/(2a) = D/(8 f l).

Therefore, x² = √(D/(8 f l)). Since x = d/D, the maximum-power condition is (d/D)² = √(D/(8 f l)). If d/D in the printed statement is interpreted as the nozzle-to-pipe area ratio, the stated form d/D = √(D/(8 f l)) follows. For the diameter ratio, it is d/D = (D/(8 f l))^(1/4). Validity: minor losses neglected and f constant.

(b) Given: γd,max = 18 kN/m³, w = 16% = 0.16, G = 2.65, γw = 9.81 kN/m³.

Using γd = Gγw/(1 + e), e = Gγw/γd − 1 = (2.65 × 9.81)/18 − 1 = 25.9965/18 − 1 = 1.44425 − 1 = 0.44425.

For degree of saturation, Se = wG ∴ S = wG/e = (0.16 × 2.65)/0.44425 = 0.424/0.44425 = 0.9544 = 95.44%.

The soil can be further compacted at the same moisture content only up to zero air voids, i.e. S = 1. Then γd,zav = Gγw/(1 + wG) = 25.9965/(1 + 0.16 × 2.65) = 25.9965/1.424 = 18.26 kN/m³ approximately.

Thus the maximum dry unit weight it can be further compacted to is 18.26 kN/m³, an increase of about 0.26 kN/m³, with no change in moisture content.

(c) For a critical slope, S0 = Sc, and therefore normal depth yn equals critical depth yc. The normal-depth line and critical-depth line coincide; zone 2 has zero thickness. The possible gradually varied flow profiles are C1, C2 and C3.

Schematic: `` Depth y ↑ | ______ C1: y > yc = yn | __/ yc=yn |---------------- C2: y = yc = yn ---------------- | __/ | __ C3: y < yc |______________________________________________→ x ``

Boundary conditions:

• C1 profile: y > yc = yn. Since y > yc, flow is subcritical. The depth increases downstream and approaches yc = yn asymptotically at far upstream. Downstream boundary is a control such as a dam, gate or confluence where y > yc. It is controlled downstream.

• C2 profile: y = yc = yn everywhere. This is uniform critical flow. It occurs only when the slope is exactly critical and no downstream or upstream control changes the depth. The boundary condition is simply y = yc at all sections.

• C3 profile: y < yc. Flow is supercritical. The depth increases downstream and approaches yc asymptotically at far downstream. The upstream boundary is a control such as a sluice gate or steep chute giving y < yc. It is controlled upstream. If the tailwater is raised, it may terminate in a hydraulic jump.

No GVF profile can cross yc smoothly; the GVF equation becomes indeterminate at critical depth.

(d) Using Buckingham’s π theorem. Variables: T = torque, D = diameter, N = speed, ρ = density, μ = dynamic viscosity.

Dimensions: T = M L² T⁻² D = L N = T⁻¹ ρ = M L⁻³ μ = M L⁻¹ T⁻¹

Number of variables n = 5. Number of fundamental dimensions m = 3. Therefore number of π terms = n − m = 2.

Choose repeating variables: D, N, ρ.

First π term: π1 = T ρ^a D^b N^c. Equating dimensions: M: 1 + a = 0 ⇒ a = −1 T: −2 − c = 0 ⇒ c = −2 L: 2 − 3a + b = 0 ⇒ 2 + 3 + b = 0 ⇒ b = −5.

Thus π1 = T/(ρ N² D⁵).

Second π term: π2 = μ ρ^a D^b N^c. M: 1 + a = 0 ⇒ a = −1 T: −1 − c = 0 ⇒ c = −1 L: −1 − 3a + b = 0 ⇒ −1 + 3 + b = 0 ⇒ b = −2.

Thus π2 = μ/(ρ N D²) = 1/Re, where Re = ρ N D²/μ is the Reynolds number.

Therefore the dimensionless relationship is T/(ρ N² D⁵) = φ(μ/(ρ N D²)) or T = ρ N² D⁵ φ(Re).

Use of non-dimensional numbers in design: they allow model testing and scale-up, ensure dynamic similarity, reduce the number of variables, predict torque/power for different speeds, diameters and fluids, and help select or design pumps, turbines, compressors and other fluid machines.

(e) Terzaghi’s bearing capacity equation for a square footing with c = 0 is qu = qNq + 0.4γBNγ.

Here: q = γDf = 19 × 0.8 = 15.2 kN/m² B = 2 m γ = 19 kN/m³ Nq = 65, Nγ = 80.

Then qu = 15.2 × 65 + 0.4 × 19 × 2 × 80 = 988 + 1216 = 2204 kN/m².

Net ultimate bearing capacity: qnu = qu − q = 2204 − 15.2 = 2188.8 kN/m².

Net safe bearing capacity with FS = 3: qns = qnu/3 = 2188.8/3 = 729.6 kN/m².

Area of footing: A = 2 × 2 = 4 m².

Safe net load carried by the footing: Q_safe = qns × A = 729.6 × 4 = 2918.4 kN ≈ 2918 kN.

If gross allowable bearing capacity is defined as qu/3, then q_all = 2204/3 = 734.67 kN/m² and gross allowable load = 734.67 × 4 = 2938.7 kN. The net load carried by the footing above the existing surcharge is 2918 kN. Water table at great depth, so no water-table correction is required.