(a) A long cylindrical rod of radius 10 cm consists of a nuclear reacting material (k = 0·5 W/m-K) generating 24000 W/m³…

(a)

Given: inner rod r₁ = 0.10 m, k₁ = 0.5 W/m·K, q̇ = 24 000 W/m³; outer cylinder r₂ = 0.20 m, k₂ = 4 W/m·K; h = 20 W/m²·K, T∞ = 100 °C.

Assumptions: steady, one-dimensional radial conduction in a long cylinder; constant properties; perfect contact at the interface; no generation in the outer cylinder. All the heat generated in the rod passes by conduction through the shell and then by convection to the fluid.

Heat generated per metre length: Q′ = q̇ πr₁² = 24 000 × π × 0.1² = 753.98 W/m

Outer surface temperature (convection): Q′ = h(2πr₂)(T_s – T∞), so T_s – T∞ = 753.98/(20 × 2π × 0.2) = 753.98/25.133 = 30.0 K T_s = 130.0 °C

Interface temperature (conduction through the shell): R_cond = ln(r₂/r₁)/(2πk₂) = ln 2/(2π × 4) = 0.02758 K·m/W T_i – T_s = Q′R_cond = 753.98 × 0.02758 = 20.79 K T_i = 150.8 °C

Maximum temperature (at the rod axis). The rod equation (1/r)d/dr(r dT/dr) + q̇/k₁ = 0 with dT/dr = 0 at r = 0 gives T(r) = T_i + (q̇/4k₁)(r₁² – r²). Hence T_max – T_i = q̇r₁²/(4k₁) = 24 000 × 0.01/(4 × 0.5) = 120 K T_max = 270.8 °C, at r = 0.

Check. At the interface, –k₁ dT/dr = q̇r₁/2 = 1200 W/m², and 1200 × 2π × 0.1 = 754 W/m, which matches Q′. Most of the temperature rise occurs inside the poorly conducting fuel (120 K), compared with 21 K across the shell and 30 K across the film.

(b)

Mass flow rate. Eye annulus area A₁ = (π/4)(0.30² – 0.15²) = 0.05301 m² There is no prewhirl, so C₁ = C_a = 143 m/s. T₁ = T₀₁ – C₁²/2c_p = 295 – 143²/2010 = 284.83 K p₁ = p₀₁(T₁/T₀₁)^(3.5) = 1.1 × (0.96551)^(3.5) = 0.9729 bar ρ₁ = p₁/RT₁ = 97 290/(287 × 284.83) = 1.190 kg/m³ ṁ = ρ₁A₁C_a = 1.190 × 0.05301 × 143 = 9.02 kg/s

Work input and power. Impeller tip speed U₂ = πD₂N = π × 0.5 × 290 = 455.5 m/s Work per kg = ψσU₂² = 1.04 × 0.9 × 455.5² = 194.23 kJ/kg T₀₃ – T₀₁ = 194 230/1005 = 193.3 K Power = ṁ × ψσU₂² = 9.02 × 194.23 = 1752 kW ≈ 1.75 MW

Overall stagnation pressure ratio. η_c = (T₀₃ₛ – T₀₁)/(T₀₃ – T₀₁), so the isentropic rise is 0.78 × 193.3 = 150.7 K. p₀₃/p₀₁ = (1 + 0.78 × 193.3/295)^(3.5) = (1.5110)^(3.5) = 4.24, giving p₀₃ = 4.66 bar.

Axial depth of the impeller channel at the periphery. The total loss corresponds to (1 – 0.78) × 193.3 K. Half of it occurs in the impeller, so the effective impeller efficiency is η_imp = 1 – 0.11 = 0.89. p₀₂/p₀₁ = (1 + 0.89 × 193.3/295)^(3.5) = 4.99, so p₀₂ = 5.49 bar. The diffuser does no work, so T₀₂ = T₀₃ = 488.3 K. Whirl velocity C_w2 = σU₂ = 0.9 × 455.5 = 410.0 m/s; radial velocity C_r2 = 143 m/s C₂²/2c_p = (410.0² + 143²)/2010 = 93.8 K, so T₂ = 488.3 – 93.8 = 394.5 K p₂ = p₀₂(T₂/T₀₂)^(3.5) = 5.49 × (0.8079)^(3.5) = 2.60 bar ρ₂ = 260 200/(287 × 394.5) = 2.30 kg/m³ Radial flow area = ṁ/(ρ₂C_r2) = 9.02/(2.30 × 143) = 0.02744 m² Axial depth b₂ = 0.02744/(π × 0.5) = 0.0175 m ≈ 17.5 mm

T–s diagram. The stagnation temperature rises only in the impeller (01 → 02) and stays constant through the diffuser (02 → 03). Entropy rises in both components because of losses, so p₀₃ is lower than p₀₂. The static state rises from 1 to 2 in the impeller, and the diffuser then converts kinetic energy into a further static-pressure rise from 2 to 3.

Summary: ṁ = 9.02 kg/s, p₀₃/p₀₁ = 4.24, power = 1.75 MW, and the impeller channel depth at the tip is about 17.5 mm.

(c)

The mean (bulk) temperature is the enthalpy-flow-weighted average over the cross-section. For constant ρ and c_p:

T_m = ∫₀^r₀ u T 2πr dr / ∫₀^r₀ u 2πr dr

Let ξ = r/r₀, so 2πr dr = 2πr₀² ξ dξ. The constants cancel, leaving

T_m = ∫₀¹ (1 – ξ²)(344.8 + 75ξ² – 18.8ξ⁴) ξ dξ / ∫₀¹ (1 – ξ²) ξ dξ

Denominator: ∫₀¹(ξ – ξ³)dξ = 1/2 – 1/4 = 0.25

Numerator: expanding, (1 – ξ²)(344.8 + 75ξ² – 18.8ξ⁴) = 344.8 – 269.8ξ² – 93.8ξ⁴ + 18.8ξ⁶. Multiplying by ξ and integrating term by term: = 344.8/2 – 269.8/4 – 93.8/6 + 18.8/8 = 172.4 – 67.45 – 15.633 + 2.35 = 91.667

T_m = 91.667/0.25 = 366.7 K

Check. The wall temperature is 344.8 + 75 – 18.8 = 401.0 K and the centreline temperature is 344.8 K, so T_m lies between them as it must. It is lower than the simple area average (376.0 K) because the velocity weighting favours the cooler, faster core flow. Since the wall is hotter than T_m, heat is flowing from the wall into the fluid.