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

(a) Condition for maximum power through a nozzle

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)). In terms of diameters this is d/D = (D/8fl)^(1/4), the standard result d = (D⁵/8fl)^(1/4). The printed form √(D/8fl) is obtained if d/D is read as the area ratio a/A = (d/D)². At this condition the friction loss is h_f = H/3, so the maximum efficiency at maximum power is 2/3. Validity: minor losses neglected and f constant.

(b) Degree of saturation and maximum dry unit weight

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) Gradually varied flow profiles on a critical slope

On a critical slope S₀ = S_c, so the normal depth equals the critical depth (y_n = y_c). The normal-depth line (NDL) and critical-depth line (CDL) coincide, zone 2 has zero thickness, and only the C1 and C3 profiles exist (C2 degenerates into uniform critical flow, y = y_n = y_c).

The GVF equation, dy/dx = S₀[1 − (y_n/y)³]/[1 − (y_c/y)³], gives dy/dx = S₀ when y_n = y_c (wide channel, Chezy). The water surface is therefore practically horizontal in both profiles, and the depth grows downstream at the rate S₀.

• C1 (zone 1, y > y_n = y_c): subcritical flow, so it is controlled from downstream. It occurs behind a dam, weir or gate placed on a critical-slope channel, or where a critical-slope channel enters a reservoir. Boundary conditions: the downstream control fixes y > y_c, and upstream the profile meets the coincident NDL/CDL.
• C3 (zone 3, y < y_c): supercritical flow, so it is controlled from upstream. It occurs below a sluice gate or at the foot of a steep chute discharging onto a critical-slope channel. Boundary conditions: the upstream control fixes y < y_c, and downstream the depth rises towards y_c, or a hydraulic jump forms if the tailwater is high enough.
• C2: y = y_n = y_c everywhere, a uniform critical flow line that is unstable and wavy in practice.

A GVF profile cannot pass smoothly through y_c, because dy/dx becomes indeterminate at critical depth.

(d) Dimensionless torque relation by Buckingham π

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) Safe load on the square footing

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.