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

Question

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 is decreased from 250 m/s to 40 m/s as it passes through the diffuser. Determine

(a) the exit temperature of the air, and
(b) the inlet and exit areas of the diffuser.

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Ngọc Diệp 3 years 2021-07-19T13:53:56+00:00 1 Answers 867 views 0

Answers ( )

  1. Eirian
    0
    2021-07-19T13:55:33+00:00
    Reply

    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:

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

    For a diffuser, w = Δp = 0

    The diffuser is adiabatic, q = 0

    Replacing:

    0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +0

    Where

    V₁ = 250 m/s

    V₂ = 40 m/s

    Replacing:

    0-0=h_{2} -400x10^{3} +\frac{40^{2}-250^{2} }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg

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

    b) The inlet area is:

    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}

    The exit area is:

    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}

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