Nitrogen, modeled as an ideal gas, flows at a rate of 3 kg/s through a well-insulated horizontal nozzle operating at steady state. The nitrogen enters the nozzle with a velocity of 20 m/s at 340 K, 400 kPa and exits the nozzle at 100 kPa. To achieve an exit velocity of 478.8 m/s, determine (a) the exit temperature, in K. (b) the exit area, in m2 .

The answers to the questions are

a). T₂ exit = 228.8 K

b). Area, A₂ exit = 15.66 cm²

Explanation:

a). γ/(γ-1)×P1/p1(1-(P₂/P₁)^((γ-1)/γ)) = V₁²/2-V₁²/2

Solving with P₁/p₁ = RT₁=290×340=98600

T₂/T₁=(P₂/P₁)^((γ-1)/γ) = (100/400)^((1.4-1)/1.4)

T₂/340 = 0.673 or T₂ = 340 × 0.673 = 228.8 K

T₂ = 228.8 K

b). Mass flow rate = 3 kg/s

Density ρ₁ = P₁/98600 = 400000/98600 = 4 kg/m^3

Volume flow rate = Mass flow rate/Density = 0.75 m³/s

Area = Volume flow rate/velocity = 0.75 m³/s/20 m/s = 0.0375 m²

A₁ ×v₂ =A₂×v₂

∴A₂ = 1.566×10^(-3) m³ =⁻15.66 cm²

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