(a)(i) Show that the effective conductance, (A_1F̄₁₂) for two black, parallel plates of equal area connected by re-radiating…

(a) Re-radiating walls and radiation shields

(i) Let surfaces 1 and 2 be black with A₁ = A₂, joined by re-radiating (adiabatic) walls R. Then F₁R = 1 − F₁₂ and, by symmetry, F₂R = 1 − F₂₁ = 1 − F₁₂. Heat flows from 1 to 2 by two parallel routes: directly, with conductance A₁F₁₂, and through the wall node, where resistances 1/(A₁F₁R) and 1/(A₂F₂R) are in series. The series route has conductance A₁(1 − F₁₂)/2. Adding the parallel conductances,

A₁F̄₁₂ = A₁F₁₂ + A₁(1 − F₁₂)/2 = A₁(1 + F₁₂)/2.

Check: the wall node floats at J_R = (E_b1 + E_b2)/2.

(ii) Two thin shields (3 and 4) lie between planes 1 and 2. Each of the three gaps has the same resistance per unit area, 1/ε + 1/ε − 1 = 1/0.8 + 1/0.8 − 1 = 1.5. In steady state the same flux crosses each gap, so the drops in σT⁴ are equal:

T₁⁴ − T₃⁴ = T₃⁴ − T₄⁴ = T₄⁴ − T₂⁴ = (T₁⁴ − T₂⁴)/3.

With T₁⁴ = 9.488 × 10¹⁰ and T₂⁴ = 5.973 × 10⁹ K⁴:

T₃⁴ = (2T₁⁴ + T₂⁴)/3, so T₃ = 505.4 K; T₄⁴ = (T₁⁴ + 2T₂⁴)/3, so T₄ = 434.4 K.

Check: q = σ(T₁⁴ − T₂⁴)/(3 × 1.5) = 1120 W/m², one-third of the 3361 W/m² without shields.

(b) Laminar sublayer thickness

Assume a smooth tube, the core profile u/u_c = (y/r₀)^(1/7) with y measured from the wall, a linear sublayer, and the Blasius friction factor f = 0.316 Re^(−1/4) with Re = u_m d/ν.

Mean velocity: u_m = 2u_c∫₀¹(1 − η)^(1/7) η dη = 2u_c(49/120), so u_m = (49/60)u_c and u_c = 1.2245u_m.

Wall shear: τ_w = fρu_m²/8 = 0.0395ρu_m²Re^(−1/4).

In the sublayer, u = (τ_w/μ)y. Matching at y = δ with the core profile,

(τ_w/μ)δ = u_c(δ/r₀)^(1/7), so δ^(6/7) = μu_c/(τ_w r₀^(1/7)).

Substituting u_c and τ_w, and using ν/u_m = 2r₀/Re,

(δ/r₀)^(6/7) = 62.0 Re^(−3/4).

Therefore δ/r₀ = 123 Re^(−7/8), or δ/d = 61.7 Re^(−7/8).

Check: at Re = 10⁵, δ/d = 0.0026 and y⁺ = δu*/ν ≈ 12, close to the usual edge of the sublayer.

(c) Reheating in a gas turbine

In reheating, the expansion is split between a high-pressure and a low-pressure turbine, and the gas leaving the HP turbine is reheated in a second combustor, back to about the maximum temperature, before the LP expansion.

Effects on performance:

• Specific work output rises: the temperature drop for a given pressure ratio is proportional to the inlet temperature, so expanding from the reheated state gives more work, while compressor work is unchanged. Work is greatest when the intermediate pressure is √(p₁p₂), and the plant is smaller for a given power.
• Work ratio improves, which lowers sensitivity to component losses.
• Thermal efficiency by itself falls slightly, because the extra heat is added at a lower mean temperature and the exhaust temperature is higher.
• With a regenerator, the hot exhaust preheats the compressed air, so efficiency rises; reheat is therefore paired with regeneration.

Costs are an extra combustor, more fuel per kg of air and higher LP turbine blade temperatures.