(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)(i) For a solid with average specific heat c, entropy change is obtained from dS = m c dT/T.

ΔS_iron = ∫ from T_i to T_f of m c dT/T = m c ln(T_f/T_i)

m c = 50 × 0·45 = 22·5 kJ/K, T_i = 500 K, T_f = 285 K.

ΔS_iron = 22·5 ln(285/500) = 22·5 ln(0·57) = 22·5 × (−0·5621189) = −12·65 kJ/K

Heat given by iron to atmosphere:

Q = m c (T_i − T_f) = 50 × 0·45 × (500 − 285) = 22·5 × 215 = 4837·5 kJ

Atmosphere behaves as a thermal reservoir at T_atm = 285 K.

ΔS_atm = Q/T_atm = 4837·5/285 = +16·974 kJ/K

Thus total entropy generation:

S_gen = ΔS_iron + ΔS_atm = −12·6477 + 16·9737 = 4·326 kJ/K

(a)(ii) Irreversibility is the exergy destroyed:

I = T₀ S_gen = T_atm S_gen

I = 285 × 4·326 = 1232·9 kJ

Condition: average specific heat is constant and no phase change occurs.

(b) For a perfect gas across a normal shock, define the critical Mach number as M* = V/a*, where

a* = √(2γRT₀/(γ+1))

Using T₀/T = 1 + ((γ−1)/2)M²,

M*² = V²/a*² = γRTM² / [2γRT₀/(γ+1)]

M*² = (γ+1)M² / [2 + (γ−1)M²]

For upstream x:

M*ₓ² = (γ+1)Mₓ² / [2 + (γ−1)Mₓ²]

The normal-shock relation for downstream y is

Mᵧ² = [(γ−1)Mₓ² + 2] / [2γMₓ² − (γ−1)]

Let u = Mₓ². Then

Mᵧ² = [(γ−1)u + 2] / [2γu − (γ−1)]

Now

M*ᵧ² = (γ+1)Mᵧ² / [2 + (γ−1)Mᵧ²]

Substitute Mᵧ² and simplify. Let D = 2γu − (γ−1).

M*ᵧ² = [(γ+1)((γ−1)u + 2)/D] / [2 + (γ−1)((γ−1)u + 2)/D]

Denominator simplifies to

[2D + (γ−1)((γ−1)u + 2)]/D = [(γ+1)²u]/D

Hence

M*ᵧ² = [(γ−1)u + 2] / [(γ+1)u]

But

M*ₓ² = (γ+1)u / [(γ−1)u + 2]

Therefore

M*ₓ² M*ᵧ² = 1

Taking positive square roots,

**M*ₓ M*ᵧ = 1**

This result holds for a steady, adiabatic normal shock in a perfect gas with constant γ.

(c) For an axial compressor stage, reaction R is

R = static enthalpy rise in rotor / stage static enthalpy rise

R = (h₂ − h₁)/(h₃ − h₁)

For 50% reaction, R = 1/2, so the rotor and stator have equal static enthalpy rise.

For the rotor, using the relative-flow energy equation with constant blade speed at the mean radius,

h₂ − h₁ = (W₁² − W₂²)/2

For the stator,

h₃ − h₂ = (V₂² − V₃²)/2

For a repeating stage, V₃ = V₁. Hence for R = 1/2,

W₁² − W₂² = V₂² − V₁²

Assume constant axial velocity C_a. Let U be blade speed and V_u be whirl component. Then

V² = C_a² + V_u²

W² = C_a² + (V_u − U)²

Substituting:

(V_u1 − U)² − (V_u2 − U)² = V_u2² − V_u1²

Expanding and simplifying gives

U = V_u1 + V_u2

Therefore,

W_u1 = V_u1 − U = −V_u2

W_u2 = V_u2 − U = −V_u1

So the magnitude of the rotor-inlet relative whirl equals the rotor-exit absolute whirl, and the magnitude of the rotor-exit relative whirl equals the rotor-inlet absolute whirl. Thus

|β₁| = |α₂| and |β₂| = |α₁|

or, with the usual opposite-angle convention,

β₁ = α₂ and β₂ = α₁

The inlet and outlet velocity triangles are therefore mirror images. The rotor and stator blade angles are identical in magnitude, so the rotor and stator blade passages have the same shape and same angles. This is why 50% reaction blading is called symmetrical blading. It also gives equal diffusion and equal static pressure rise in rotor and stator.

(d) Furnace temperature:

T = 2700 + 273 = 2973 K

(d)(i) Planck’s law:

E_bλ = C₁ / [λ⁵(exp(C₂/(λT)) − 1)]

λ = 1·2 μm, λT = 1·2 × 2973 = 3567·6 μm-K

C₂/(λT) = 14380/3567·6 = 4·0307

exp(4·0307) − 1 = 55·3015

λ⁵ = 1·2⁵ = 2·48832

E_bλ = 3·742 × 10⁸ / [2·48832 × 55·3015]

E_bλ = 3·742 × 10⁸ / 137·608 = 2·72 × 10⁶ W/(m²·μm)

(d)(ii) Wien’s displacement law:

λ_max T = 2898 μm-K

λ_max = 2898/2973 = 0·9748 μm ≈ 0·975 μm

(d)(iii) At λ_max, using Planck’s law:

C₂/(λ_max T) = C₂/2898 = 14380/2898 = 4·9620

exp(4·9620) − 1 = 141·884

λ_max⁵ ≈ (0·9748)⁵ = 0·8802

E_bλ,max = 3·742 × 10⁸ / [0·8802 × 141·884]

E_bλ,max = 2·997 × 10⁶ W/(m²·μm) ≈ 3·00 × 10⁶ W/(m²·μm)

(d)(iv) Stefan-Boltzmann law:

E_b = σT⁴

E_b = 5·67 × 10⁻⁸ × (2973)⁴

(2973)⁴ = 7·8123 × 10¹³ K⁴

E_b = 5·67 × 10⁻⁸ × 7·8123 × 10¹³

E_b = 4·43 × 10⁶ W/m²

(e)(i) Basic assumptions for LMTD method:

• Steady-state operation of the heat exchanger.
• Constant mass flow rates of both fluids.
• Constant specific heats and fluid properties.
• Constant overall heat transfer coefficient U over the entire heat transfer surface.
• No heat loss to the surroundings; heat exchanger is adiabatic.
• Negligible kinetic and potential energy changes.
• No internal heat generation.
• For the basic LMTD formula, flow arrangement is usually parallel-flow or counter-flow.
• For cross-flow or multipass exchangers, LMTD is multiplied by a correction factor F.
• If phase change occurs, it is assumed to occur at constant temperature, or the fluid is treated as having very large heat capacity.

(e)(ii) LMTD method is applicable when the inlet and outlet temperatures of both fluids are known or can be determined directly. It is mainly used for design or sizing problems, e.g. to find the required heat transfer area from Q = U A ΔT_lm. If outlet temperatures are unknown, LMTD requires trial-and-error iteration.

NTU method is applicable when inlet temperatures and mass flow rates or fluid properties are known, but outlet temperatures are unknown. It is mainly used for performance or rating problems, e.g. to find heat duty and outlet temperatures from known exchanger size. The ε-NTU method avoids iteration and is especially convenient when one fluid undergoes phase change or when C_min/C_max is known.