The compressor of a large gas turbine receives air from the ambient surroundings at 95 kPa and 20 ºC with a low velocity. At the compressor

Question

The compressor of a large gas turbine receives air from the ambient surroundings at 95 kPa and 20 ºC with a low velocity. At the compressor discharge, air exits at 1.25 MPa and 430 ºC with a velocity of 90 m/s. The power input to the compressor is 5000 kW. Determine the mass flow rate of air through the unit

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Amity 4 years 2021-08-27T19:01:51+00:00 1 Answers 64 views 0

Answers ( )

  1. Ben
    0
    2021-08-27T19:03:47+00:00
    Reply

    Answer:

    The mass flow rate is m = 12.0 kg/s

    Explanation:

    From the question we are given the following parameters

              inlet temperature , T_1 = 20°C = 293 K

             inlet pressure , P_1 = 95KPa        

              Outlet pressure P_2 = 1.25*10^3 kPa

              Outlet temperature , T_2 = 430°C

                                                      =703K

              Final velocity , V_2 = 90m/s

               Power Input , W_c =5000kW

    Considering the energy equation we have

            h_1 +\frac{V_1^2}{2} +q = h_2 +\frac{V_2^2}{2} +w

                          q is the net heat transferred

                         w is the net workdone  

    Lets assume that q = 0 and V_1 = 0 Hence in this question the specific heat capacity is constant

    = >         -w =h_2 - h_1 +\frac{V_2^2}{2}          

                =(C_P)_0 (T_2 -T_1) + \frac{V_2^2}{2}

                = (1.004)(703-293) + \frac{90^2}{2(1000)}

    The division by 1000 is to convert the kinetic energy to KJ

    Note the specific heat of air is 1.004  kJ/kg⋅K

               = 415.5 KJ/kg

    The mass flow rate is given as  m = \frac{Z}{-w}

    Where Z is The power input to the compressor which is given as  5000 kW

                            m = \frac{5000}{415.5}

                                12kg/s

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