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

Ben · Answer

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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