Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the air stream

0 thoughts on “Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the air stream”

1. Answer:

   a) The exit temperature is 430 K

   b) The inlet and exit areas are 0.0096 m² and 0.051 m²

   Explanation:

   a) Given:

   T₁ = 127°C = 400 K

   At 400 K, h₁ = 400.98 kJ/kg (ideal gas properties table)

   The energy equation is:

   [tex]q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +delta-p[/tex]

   For a diffuser, w = Δp = 0

   The diffuser is adiabatic, q = 0

   Replacing:

   [tex]0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +0[/tex]

   Where

   V₁ = 250 m/s

   V₂ = 40 m/s

   Replacing:

   [tex]0-0=h_{2} -400×10^{3} +\frac{40^{2}-250^{2} }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg[/tex]

   Using tables, at 431.43 kJ/kg the temperature is 430 K

   b) The inlet area is:

   [tex]m=\frac{P_{1} }{RT_{1} } A_{1} V_{1} \\\frac{6000}{3600} =\frac{80}{0.287*400} A_{1} *250\\A_{1} =0.0096m^{2}[/tex]

   The exit area is:

   [tex]m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}[/tex]

   Reply