(a) A 50 kg block of iron at 500 K is placed into open atmosphere which is at a temperature of 285 K. The iron block eventually…

(a) Entropy change and irreversibility of a cooling iron block

For a solid of constant specific heat, ΔS = mc ln(T_f/T_i). Here mc = 50 × 0.45 = 22.5 kJ/K, T_i = 500 K and T_f = T₀ = 285 K.

(i) Iron block: ΔS_iron = 22.5 ln(285/500) = 22.5 × (−0.5621) = −12.65 kJ/K. Heat rejected to the atmosphere: Q = mc(T_i − T_f) = 22.5 × 215 = 4837.5 kJ. The atmosphere is a reservoir at 285 K, so ΔS_atm = Q/T₀ = 4837.5/285 = +16.97 kJ/K. Total entropy generation: S_gen = −12.65 + 16.97 = 4.33 kJ/K (positive, so the process is irreversible).

(ii) Irreversibility (exergy destroyed): I = T₀S_gen = 285 × 4.326 = 1233 kJ.

This assumes constant specific heat and no phase change.

**(b) Proof that M*ₓ M*ᵧ = 1 for a normal shock**

Define the critical Mach number M* = V/a*, where a*² = 2γRT₀/(γ + 1) is fixed by the stagnation temperature. T₀ is the same on both sides of an adiabatic shock, so a* is the same on both sides. Using V² = M²γRT and T₀/T = 1 + [(γ − 1)/2]M²: M*² = (γ + 1)M²/[2 + (γ − 1)M²].

Upstream, with u = Mₓ²: M*ₓ² = (γ + 1)u/[(γ − 1)u + 2]. Downstream, the normal-shock relation gives Mᵧ² = [(γ − 1)u + 2]/[2γu − (γ − 1)]. Let D = 2γu − (γ − 1). Then M*ᵧ² = (γ + 1)[(γ − 1)u + 2]/D ÷ {2 + (γ − 1)[(γ − 1)u + 2]/D}. The denominator is {2D + (γ − 1)[(γ − 1)u + 2]}/D = (γ + 1)²u/D, so M*ᵧ² = [(γ − 1)u + 2]/[(γ + 1)u]. Multiplying, M*ₓ² M*ᵧ² = 1, and taking the positive root, **M*ₓ M*ᵧ = 1**. This holds for a steady, adiabatic normal shock in a perfect gas of constant γ. Since a shock makes Mₓ > 1, it follows that M*ᵧ < 1: the flow becomes subsonic.

(c) Why 50% reaction blading is called symmetrical

Degree of reaction R = (static enthalpy rise in rotor)/(static enthalpy rise in stage). For R = 1/2, rotor and stator share the rise equally. Rotor: h₂ − h₁ = (W₁² − W₂²)/2. Stator: h₃ − h₂ = (V₂² − V₃²)/2, with V₃ = V₁ for a repeating stage. Equal rises give W₁² − W₂² = V₂² − V₁².

With constant axial velocity C_a, W² = C_a² + (V_u − U)², so (V_u1 − U)² − (V_u2 − U)² = V_u2² − V_u1². Expanding, (V_u1 − V_u2)[2(V_u1 + V_u2) − 2U] = 0, so U = V_u1 + V_u2. Hence W_u1 = V_u1 − U = −V_u2 and W_u2 = V_u2 − U = −V_u1.

Therefore tanβ₁ = −tanα₂ and tanβ₂ = −tanα₁: in magnitude the relative inlet angle equals the absolute exit angle (β₁ = α₂) and the relative exit angle equals the absolute inlet angle (β₂ = α₁). The rotor velocity triangle is the mirror image of the stator one, so rotor and stator blades have the same shape, the same turning and the same diffusion. This mirror symmetry of the triangles is why the design is called symmetrical blading.

(d) Blackbody furnace emission at 2700°C

T = 2700 + 273 = 2973 K.

(i) At λ = 1.2 μm: λT = 3567.6 μm·K, C₂/λT = 14 380/3567.6 = 4.031, and e^4.031 − 1 = 55.30. E_bλ = 3.742 × 10⁸/(1.2⁵ × 55.30) = 3.742 × 10⁸/(2.488 × 55.30) = 2.72 × 10⁶ W/m²·μm.

(ii) Wien's law: λ_max = 2898/2973 = 0.975 μm.

(iii) At λ_max: C₂/λT = 14 380/2898 = 4.962, e^4.962 − 1 = 141.9, and λ⁵ = 0.975⁵ = 0.880. E_bλ,max = 3.742 × 10⁸/(0.880 × 141.9) = 3.0 × 10⁶ W/m²·μm.

(iv) Total emissive power: E_b = σT⁴ = 5.67 × 10⁻⁸ × (2973)⁴ = 5.67 × 10⁻⁸ × 7.812 × 10¹³ = 4.43 × 10⁶ W/m².

(e)(i) Basic assumptions of the LMTD method

The heat exchanger operates at steady state with constant mass flow rates of both fluids. Specific heats and the overall heat transfer coefficient U are constant over the whole surface. The exchanger is adiabatic to the surroundings, with no internal heat generation and negligible kinetic and potential energy changes. Axial conduction along the tubes is negligible, and a phase change, if present, occurs at constant temperature. The basic formula ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) is for parallel or counter flow; for cross-flow and multipass exchangers it is multiplied by a correction factor F.

(e)(ii) Where LMTD and NTU methods apply

The LMTD method suits design (sizing) problems in which the inlet and outlet temperatures of both fluids are known, so that the area follows directly from Q = UAΔT_lm. If outlet temperatures are unknown it needs trial and error.

The ε-NTU method suits rating or performance problems in which the inlet temperatures, flow rates and exchanger size (UA) are known but the outlet temperatures and heat duty are not. It needs no iteration, using ε = f(NTU, C_min/C_max).