(a) Steam at 40 °C is condensed in a power plant using cooling water available at 12 °C. The flow rate of steam is 0·15 kg/s. The…

(a) Steam condenser cooled by water in tubes (20 marks)

Data: ṁ_steam = 0.15 kg/s, h_fg = 2406 kJ/kg, water at 12 °C in and 20 °C out, tube wall at 40 °C, tube length 6 m, internal diameter 1.25 cm, mean water velocity 3.5 m/s, ρ = 998.8 kg/m³ and c_p = 4184.5 J/kg·°C.

Condenser duty. Q = ṁ h_fg = 0.15 × 2406 = 360.9 kW.

Water flow in one tube. Flow area = π/4 × 0.0125² = 1.227 × 10⁻⁴ m². ṁ_tube = ρ V A = 998.8 × 3.5 × 1.227 × 10⁻⁴ = 0.4290 kg/s.

Heat taken by one tube. Q_tube = ṁ c_p (T_out − T_in) = 0.4290 × 4184.5 × (20 − 12) = 14 361 W.

Average heat transfer coefficient. The wall is nearly isothermal at 40 °C, so the temperature differences at the ends are ΔT₁ = 40 − 12 = 28 °C and ΔT₂ = 40 − 20 = 20 °C, and the log-mean temperature difference is

ΔT_lm = (28 − 20)/ln(28/20) = 8/0.3365 = 23.78 °C

The inside surface area of one tube is A_s = π D L = π × 0.0125 × 6 = 0.2356 m². From Q_tube = h A_s ΔT_lm:

h = 14 361/(0.2356 × 23.78) = 2564 W/m²·°C

Number of tubes. N = Q/Q_tube = 360 900/14 361 = 25.1, so 26 tubes are needed (rounding up so that the full duty is achieved).

(b) Axial-flow compressor stage (20 marks)

Data: D = 0.5 m, N = 15 000 rpm, axial velocity V_a = 220 m/s (constant), inlet whirl V_w1 = 80 m/s, T₁ = 300 K, P₁ = 1 bar, η_stage = 0.88, pressure ratio 1.5, γ = 1.4 and c_p = 1.005 kJ/kg·K.

Blade speed. U = πDN/60 = π × 0.5 × 15 000/60 = 392.7 m/s.

Temperature rise and work. The isentropic temperature rise is

ΔT_s = T₁ [(P₂/P₁)^((γ−1)/γ) − 1] = 300 × (1.5^0.2857 − 1) = 300 × 0.1228 = 36.85 K

The actual rise is ΔT = ΔT_s/η = 36.85/0.88 = 41.87 K, and the work input per kg is

w = c_p ΔT = 1.005 × 41.87 = 42.08 kJ/kg

Outlet whirl. From Euler's equation w = U (V_w2 − V_w1): 42 081 = 392.7 (V_w2 − 80), so V_w2 − 80 = 107.2 and V_w2 = 187.2 m/s.

Velocity triangles. All angles are measured from the axial direction, and the relative velocity direction is taken as the blade direction (zero incidence and deviation).

(i) Blade angles.

• Inlet: the relative whirl is U − V_w1 = 392.7 − 80 = 312.7 m/s, so tan β₁ = 312.7/220 = 1.421 and β₁ = 54.9°. (The absolute flow angle is α₁ = tan⁻¹(80/220) = 20.0°.)
• Outlet: the relative whirl is U − V_w2 = 392.7 − 187.2 = 205.5 m/s, so tan β₂ = 205.5/220 = 0.934 and β₂ = 43.1°. (The absolute flow angle is α₂ = tan⁻¹(187.2/220) = 40.4°.)

(ii) Power input. The specific work is 42.08 kJ/kg, that is, 42.1 kW per kg/s of air. For a mass flow ṁ, the power is P = 42.08 ṁ kW (the mass flow is not given).

(iii) Degree of reaction. For constant axial velocity,

R = (V_a/2U)(tan β₁ + tan β₂) = (220/785.4)(1.421 + 0.934) = 0.66

which agrees with R = 1 − (V_w1 + V_w2)/(2U) = 1 − 267.2/785.4 = 0.66. About 66% of the pressure rise in the stage occurs in the rotor.

(c) Air velocity above which natural convection is negligible (10 marks)

Criterion. For flow parallel to a heated plate, the natural-convection contribution is less than the stated fraction when Gr_L/Re_L² is below that fraction (Gr/Re² < 1 means forced convection dominates, and below 0.1 natural convection is negligible). Setting the limit for 12%:

Gr_L/Re_L² = 0.12

Properties. The film temperature is (280 + 20)/2 = 150 °C = 423 K, so β = 1/T_f = 1/423.15 = 2.363 × 10⁻³ K⁻¹, and ν = 2.859 × 10⁻⁵ m²/s (given). The characteristic length is L = 2.5 m.

Grashof number.

Gr_L = g β (T_s − T_∞) L³/ν² = 9.81 × 2.363 × 10⁻³ × (280 − 20) × 2.5³/(2.859 × 10⁻⁵)²

Gr_L = 94.18/(8.174 × 10⁻¹⁰) = 1.152 × 10¹¹

Reynolds number and velocity.

Re_L² = Gr_L/0.12 = 9.60 × 10¹¹, so Re_L = 9.80 × 10⁵

V = Re_L ν/L = 9.80 × 10⁵ × 2.859 × 10⁻⁵/2.5 = 11.2 m/s

Natural convection effects are below 12% for an air velocity above about 11.2 m/s.