- 2070 Magh · 8 marks
Steam nozzle 4 MPa: mass flow, exit velocity, area
Short answer
Using steam table data, mass flow rate ṁ = A₁V₁/v₁ = 5.0 kg/s. Exit velocity from energy equation is 759 m/s. Exit area A₂ = 8.26 cm².
Steam at 4 MPa, 450°C enters a nozzle operating at steady state with a velocity of 50 m/s. Steam leaves the nozzle at 2 MPa and 300°C. The inlet area of the nozzle is 80 cm² and heat loss from the nozzle surface occurs at the rate of 100 kW. Determine: i) The mass flow rate of steam ii) The exit velocity of the steam and iii) The exit area of the nozzle [Refer the attached table for the properties of steam]
Given: steam nozzle, steady flow: P₁ = 4 MPa, T₁ = 450 °C, V₁ = 50 m/s, A₁ = 80 cm² = 0.008 m²; exit 2 MPa, 300 °C; heat loss Q̇ = −100 kW; W = 0, ΔPE = 0.
Steam table values:
- State 1 (4 MPa, 450 °C): v₁ = 0.08004 m³/kg, h₁ = 3331.0 kJ/kg
- State 2 (2 MPa, 300 °C): v₂ = 0.12550 m³/kg, h₂ = 3024.2 kJ/kg
i) Mass flow rate
ṁ = A₁V₁/v₁ = 0.008 × 50/0.08004 = 4.997 kg/s ≈ 5.0 kg/s
ii) Exit velocity
Energy equation: Q̇ = ṁ [ (h₂ − h₁) + (V₂² − V₁²)/2 ]
V₂²/2 = V₁²/2 + (h₁ − h₂) + Q̇/ṁ
= 50²/2 + (3331.0 − 3024.2) × 1000 + (−100 × 1000)/4.997
= 1250 + 306 800 − 20 012 = 288 038 J/kg
V₂ = √(2 × 288 038) = 759 m/s
iii) Exit area
A₂ = ṁ v₂/V₂ = 4.997 × 0.1255/759 = 8.26 × 10⁻⁴ m² = 8.26 cm²
Answer: ṁ = 5.0 kg/s; V₂ = 759 m/s; A₂ = 8.26 cm².
✓ Answer checked by Claude Opus 5.5 (high effort)
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