(a) A piston-cylinder contains 2 kg of argon at T₁ = 300 K and V₁ = 1 m³ (state 1). Argon is heated by an electric heater of 10…

(a) Index of expansion for argon (10 marks)

Data: m = 2 kg, T₁ = 300 K, V₁ = 1 m³, R = 0.2081 kJ/kg·K and C_v = 0.312 kJ/kg·K.

State 1. P₁ = mRT₁/V₁ = 2 × 0.2081 × 300/1 = 124.9 kPa.

Process 1–2 (constant volume heating). The electrical energy supplied is W = VIt = 220 × 10 × 10 = 22 000 J = 22 kJ. With 20% lost as heat, the energy reaching the gas is 0.8 × 22 = 17.6 kJ. Since the volume is constant, no boundary work is done and

Q = m C_v (T₂ − T₁): 17.6 = 2 × 0.312 × (T₂ − 300)

T₂ = 300 + 28.2 = 328.2 K.

P₂ = P₁ T₂/T₁ = 124.86 × 328.2/300 = 136.6 kPa.

State 3. T₃ = 40 °C = 313.15 K and P₃ = 0.5 P₂ = 68.3 kPa.

Process 2–3 (PVⁿ = constant). For a polytropic process, T₃/T₂ = (P₃/P₂)^((n−1)/n):

313.15/328.2 = (0.5)^((n−1)/n)

ln(0.9541) = ((n − 1)/n) ln 0.5, so −0.04696 = ((n − 1)/n)(−0.6931)

(n − 1)/n = 0.06774, so n = 1/(1 − 0.06774) = 1.073.

The index is close to 1, that is, the expansion is nearly isothermal (heat is added to the gas during it), and it is well below γ = 1.667 for argon.

(b) Maximum work of an ideal Brayton cycle (10 marks)

Data: T₁ = T_min = 300 K, T₃ = T_max = 1200 K and C_p = 1.005 kJ/kg·K.

Condition for maximum work. The net work per kg is w = C_p (T₃ − T₄) − C_p (T₂ − T₁). With T₂ = T₁ r^((γ−1)/γ) and T₄ = T₃/r^((γ−1)/γ), the work is greatest when d w/d r = 0, which gives

T₂ = T₄ = √(T_min T_max) = √(300 × 1200) = 600 K.

The corresponding pressure ratio is r_p = (T₂/T₁)^(γ/(γ−1)) = 2^3.5 = 11.3 (with γ = 1.4).

Maximum net work.

w_max = C_p (T₃ − T₄ − T₂ + T₁) = 1.005 (1200 − 600 − 600 + 300) = 301.5 kJ/kg

This is the same as C_p (√T_max − √T_min)² = 1.005 × (34.64 − 17.32)² = 301.5 kJ/kg.

Efficiency at that condition.

η = 1 − T₁/T₂ = 1 − 300/600 = 50%

This is the ideal Brayton efficiency 1 − 1/r_p^((γ−1)/γ) at the optimum pressure ratio, and it is well below the Carnot efficiency between the same limits, 1 − 300/1200 = 75%.

(c) Air jet at sonic velocity (10 marks)

Data: γ = 1.4, R = 287 J/kg·K, and the temperature at sonic velocity (M = 1) is T* = 425 K.

At M = 1, T*/T₀ = 2/(γ + 1) = 1/1.2, so T₀ = 1.2 × 425 = 510 K.

(i) Speed of sound at the stagnation state.

a₀ = √(γ R T₀) = √(1.4 × 287 × 510) = 452.7 m/s

(ii) Maximum velocity. Isentropic expansion to absolute zero temperature converts all the enthalpy into kinetic energy. With C_p = γR/(γ − 1) = 1004.5 J/kg·K:

V_max = √(2 C_p T₀) = √(2 × 1004.5 × 510) = 1012 m/s

(Also V_max = a₀ √(2/(γ − 1)) = 452.7 × √5 = 1012 m/s.)

(iii) Crocco number. Ca = V/V_max. At sonic condition, V = a* = √(γ R T*) = √(1.4 × 287 × 425) = 413.2 m/s, so

Ca = 413.2/1012.2 = 0.408

This equals √((γ − 1)/(γ + 1)) = √(0.4/2.4) = 0.408, the value at M = 1.

(d) Heat transfer through the composite wall (10 marks)

Thermal circuit. Heat flows through A, then through B and C in parallel, then through D, then through E, so the resistances A, (B ∥ C), D and E are in series. Take the one-dimensional flow, area 1 m² overall, with B and C each taking half the height (0.5 m² each), as drawn. R = L/(kA):

• R_A = 0.05/(50 × 1) = 0.0010 K/W
• R_B = 0.10/(10 × 0.5) = 0.0200 K/W
• R_C = 0.10/(8 × 0.5) = 0.0250 K/W
• R_D = 0.10/(20 × 1) = 0.0050 K/W
• R_E = 0.05/(25 × 1) = 0.0020 K/W

B and C in parallel: 1/R_BC = 1/0.02 + 1/0.025 = 50 + 40 = 90, so R_BC = 0.01111 K/W.

Total resistance: R_total = 0.0010 + 0.01111 + 0.0050 + 0.0020 = 0.01911 K/W.

Heat transfer.

Q = (T₁ − T₂)/R_total = (800 − 200)/0.01911 = 31 400 W ≈ 31.4 kW

(e) Steady one-dimensional conduction in a plane wall (10 marks)

Data: L = 0.25 m, k = 3 W/m·°C, area 10 m², heat flux at the left face q₀ = 750 W/m² into the wall, and left surface temperature T(0) = 90 °C.

Governing equation. For steady one-dimensional conduction with constant k and no generation, the heat equation d/dx(k dT/dx) + ė = 0 becomes

d²T/dx² = 0

Boundary conditions.

• At x = 0 (specified heat flux): −k dT/dx |ₓ₌₀ = q₀ = 750 W/m².
• At x = 0 (specified temperature): T(0) = 90 °C.

Solution. Integrating twice, T(x) = C₁x + C₂. The flux condition gives −k C₁ = 750, so C₁ = −750/3 = −250 °C/m. The temperature condition gives C₂ = 90 °C. Hence

T(x) = 90 − 250 x (°C, with x in m)

The temperature falls linearly in the direction of the heat flow.

Right side. At x = L = 0.25 m: T(L) = 90 − 250 × 0.25 = 27.5 °C.

The total heat transferred is 750 × 10 = 7500 W, and the temperature difference across the wall is 62.5 °C, which agrees with q L/k = 750 × 0.25/3 = 62.5 °C.