- 2067 Mangsir · 8 marks
Adiabatic diffuser: exit temperature and pressure
Short answer
Steady flow energy: T2 = T1 + (V1²−V2²)/(2cp) = 300 + (200²−20²)/(2×1005) = 319.7 K. ρ1=P1/RT1=1.1614 kg/m³, ṁ=ρ1A1V1=0.02323 kg/s. Exit ρ2=ṁ/(A2V2)=1.3505 kg/m³, then P2=ρ2 R T2 = 124 kPa.
An adiabatic diffuser has air entering at 100kPa, 300K, with a velocity of 200m/s. The inlet cross sectional area of the diffuser is 100mm². At the exit, the area is 860mm², and the exit velocity is 20m/s. Determine the exit temperature and pressure of the air. [Take Cp = 1005 J/kg K, R = 287J/kg K].
A diffuser is a steady-flow device with no work and (here) no heat transfer. Use the steady-flow energy equation and continuity.
Given: P₁ = 100 kPa, T₁ = 300 K, V₁ = 200 m/s, A₁ = 100 mm² = 100×10⁻⁶ m², A₂ = 860 mm², V₂ = 20 m/s, cp = 1005 J/kg·K, R = 287 J/kg·K.
Exit temperature
Energy equation (Q = W = 0, Δz = 0): h₁ + V₁²/2 = h₂ + V₂²/2
cp (T₂ − T₁) = (V₁² − V₂²)/2
T₂ = 300 + (200² − 20²)/(2 × 1005) T₂ = 300 + 39600/2010 = 300 + 19.70
T₂ = 319.7 K (46.7 °C)
Exit pressure
Inlet density: ρ₁ = P₁/(RT₁) = 100000/(287 × 300) = 1.1614 kg/m³
Mass flow rate: ṁ = ρ₁A₁V₁ = 1.1614 × 100×10⁻⁶ × 200 = 0.02323 kg/s
Continuity at the exit: ρ₂ = ṁ/(A₂V₂) = 0.02323/(860×10⁻⁶ × 20) = 1.3505 kg/m³
P₂ = ρ₂RT₂ = 1.3505 × 287 × 319.7 = 123 915 Pa
P₂ ≈ 124 kPa
The pressure rises from 100 to 124 kPa, which is the purpose of a diffuser (kinetic energy converted to pressure). (If the flow were reversible too, P₂ = P₁(T₂/T₁)^(k/(k−1)) = 124.9 kPa. The actual value is slightly lower because of irreversibility.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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