(a) A convergent-divergent nozzle receives steam at 5 bar, 250 °C and expands it isentropically into a space at 1 bar. Neglecting…

(a) Exit area of a steam nozzle for equilibrium and supersaturated expansion

Inlet: 5 bar, 250°C, v₁ = 0.4744 m³/kg, s₁ = 7.2709 kJ/kg·K, h₁ = 2960.7 kJ/kg; exit pressure 1 bar; ṁ = 0.5 kg/s; inlet velocity neglected. Then C₂ = √[2(h₁ − h₂)] and A₂ = ṁv₂/C₂.

(i) Equilibrium (isentropic) flow. Since s₂ = s₁ = 7.2709 is less than s_g = 7.3549 at 1 bar, the exit steam is wet. x₂ = (7.2709 − 1.3069)/(7.3549 − 1.3069) = 0.9861. h₂ = 419.04 + 0.9861 × 2257.06 = 2644.75 kJ/kg, so the enthalpy drop is 315.95 kJ/kg. C₂ = √(2 × 315 950) = 794.9 m/s. v₂ = 0.001044 + 0.9861 × (1.6729 − 0.001044) = 1.6497 m³/kg. A₂ = 0.5 × 1.6497/794.9 = 1.038 × 10⁻³ m² = 1038 mm².

(ii) Supersaturated (metastable) flow with pv^1.3 = constant. Condensation lags, so the vapour expands as superheated steam. v₂ = v₁(p₁/p₂)^(1/1.3) = 0.4744 × 5^0.7692 = 1.6361 m³/kg. Enthalpy drop = [n/(n − 1)](p₁v₁ − p₂v₂) = (1.3/0.3)(500 × 0.4744 − 100 × 1.6361) = 318.9 kJ/kg. C₂ = √(2 × 318 900) = 798.6 m/s. A₂ = 0.5 × 1.6361/798.6 = 1.024 × 10⁻³ m² = 1024 mm².

Supersaturation gives a smaller specific volume and a slightly higher velocity, so the area needed is about 1.3% smaller. The metastable exit temperature is 523.15 × (1/5)^(0.3/1.3) = 360.8 K (87.7°C), about 12 K below the 1 bar saturation temperature (99.6°C).

(b) R134a refrigeration cycle with superheat horn and throttling loss

Data: at −10°C, h_f = 186.78, h_g = 392.75 kJ/kg, s_f = 0.951, s_g = 1.734 kJ/kg·K, v_g = 0.09963 m³/kg. At 44°C, h_f = 262.38, h_g = 421.28 kJ/kg, s_f = 1.209, s_g = 1.710 kJ/kg·K. 1 TR = 3.5167 kW.

State 1 is saturated vapour at −10°C: h₁ = 392.75, s₁ = 1.734. State 2 is after isentropic compression to 11.3 bar. Since s₁ exceeds s_g at 44°C, the vapour is superheated: T₂ = T_k exp[(s₁ − s_g,k)/c_p] = 317.15 × exp(0.024/1.26) = 323.25 K, and h₂ = 421.28 + 1.26 × (323.25 − 317.15) = 428.96 kJ/kg. State 3 is saturated liquid at 44°C, h₃ = 262.38, and throttling gives h₄ = h₃.

Refrigerating effect q₀ = 392.75 − 262.38 = 130.37 kJ/kg. Compressor work w = 428.96 − 392.75 = 36.21 kJ/kg. (i) Mass flow: ṁ = 3.5167/130.37 = 0.0270 kg/s. (ii) Compressor power = 0.0270 × 36.21 = 0.977 kW. (iii) Volumetric cooling capacity = q₀/v₁ = 130.37/0.09963 = 1309 kJ/m³. (iv) COP = 130.37/36.21 = 3.60 (Carnot COP = 263.15/54 = 4.87).

Reversed Carnot cycle 1–2″–3–4′: work = (T_k − T₀)(s₁ − s₃) = 54 × (1.734 − 1.209) = 28.35 kJ/kg.

(v) Superheat horn. Heat is rejected above T_k along 2–2′, and the extra work is the horn area: Δw_sh = c_p(T₂ − T_k) − T_k(s₁ − s_g,k) = 1.26 × 6.10 − 317.15 × 0.024 = 0.07 kJ/kg. This is very small because the discharge is superheated by only about 6 K. (vi) Throttling loss. For isentropic expansion from state 3, x₄′ = (1.209 − 0.951)/(1.734 − 0.951) = 0.3295 and h₄′ = 186.78 + 0.3295 × 205.97 = 254.65 kJ/kg, so Δw_th = 262.38 − 254.65 = 7.73 kJ/kg. Check: 28.35 + 0.07 + 7.73 = 36.15 kJ/kg, which matches w = 36.21 within table round-off. Throttling is the dominant loss.

(c) Lubricating oil characteristics and IC engine operation

A lubricating oil must keep a load-carrying film between moving parts from cold start to full-load temperature. Viscosity sets the film thickness in bearings, rings and cams. Too low, especially when hot, and the film breaks down, causing wear, scuffing and blow-by. Too high, and fluid friction and pumping loss raise fuel consumption and the oil reaches bearings slowly on starting. Viscosity index measures how little viscosity changes with temperature. A high-VI (multigrade) oil is thin enough for cold cranking and still thick enough when hot. A low-VI oil either starves a cold engine or thins out when hot. Pour point is the lowest temperature at which the oil still flows. It must lie well below the lowest ambient temperature, or wax crystals stop the oil being pumped at a cold start and cause dry-start wear. Flash point is the lowest temperature at which the oil vapour ignites momentarily under a flame, and fire point is the higher temperature at which it keeps burning. Both must be well above cylinder-wall and crankcase temperatures, so the oil does not evaporate, burn off and raise oil consumption and deposits, or create a crankcase fire hazard.