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Air nozzle: exit temperature and exit area

Short answer

Exit temperature T₂ = 336 K (63 °C), obtained from energy equation with heat loss q = (h₂-h₁)+(V₂²-V₁²)/2. Mass flow is 1.045 kg/s; exit area A₂ = 33.6 cm².

Air enters a nozzle steadily at 300 KPa, 127°C and with a velocity of 40 m/s and leaves at 100 KPa and with a velocity of 300 m/s. The heat lost from nozzle surface is 20 KJ/Kg of the air. The inlet area of nozzle is 100 cm². Determine: i) Exit temperature of the air and ii) Exit area of the nozzle [Take R = 287J/kgK and Cp = 1005 J/kgK]

Given: air, steady flow: P₁ = 300 kPa, T₁ = 127 °C = 400 K, V₁ = 40 m/s, A₁ = 100 cm² = 0.01 m²; P₂ = 100 kPa, V₂ = 300 m/s; heat lost q = 20 kJ/kg. W = 0, ΔPE = 0.

Energy equation per kg:
−q_loss = (h₂ − h₁) + (V₂² − V₁²)/2

c_p (T₂ − T₁) = −20 − (300² − 40²)/2000 = −20 − 44.2 = −64.2 kJ/kg

i) Exit temperature

T₂ = T₁ − 64.2/1.005 = 400 − 63.88 = 336.1 K (≈ 63 °C)

ii) Exit area

Inlet specific volume: v₁ = RT₁/P₁ = 0.287 × 400/300 = 0.3827 m³/kg
Mass flow: ṁ = A₁V₁/v₁ = 0.01 × 40/0.3827 = 1.045 kg/s

Exit: v₂ = RT₂/P₂ = 0.287 × 336.1/100 = 0.9647 m³/kg
A₂ = ṁ v₂/V₂ = 1.045 × 0.9647/300 = 0.003361 m² = 33.6 cm²

Answer: T₂ = 336 K (63 °C); A₂ = 33.6 cm².

✓ Answer checked by Claude Opus 5.5 (high effort)

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