The mixing chamber below sees air entering 79 N2 and 21 O2 b

The mixing chamber below sees air entering (79% N_2 and 21% O_2 by volume) at the bottom at a temperature of 537 degree R, carbon dioxide entering from the top at a temperature of 640 degree R, and equal parts of water and O_2 (by volume) entering at the left at a temperature of 580 degree R. The mixing chamber is also heated from a reservoir at 1300 degree R. All flows are at a pressure of 1.1 bar and the total mass flow rates at States 1, 2, and 3 are 5 Ibm/hr, 4 Ibm/hr, and 3 Ibm/hr, respectively. Compute: The mass and mole fractions exiting the mixing chamber at State 4. Show all calculations and unit conversions for full points.

Solution

a) Mass flow rate out = mass flow rate in     = 5 + 4 + 3 = 12 lbm/hr

     Mole fraction of N2 out   = 0.79 / 3   = 0.2633

     Mole fraction of CO2 out   = 1 / 3   = 0.3333

     Mole fraction of O2 out   = (0.21 + .5) / 3   = 0.2367

     Mole fraction of H2O out   = ( .5) / 3   = 0.1667

    

     Mass fraction of N2 out   = 0.79*5 / 12 = 0.32917

     Mass fraction of CO2 out   = 4 / 12 = 0.3333

     Mass fraction of O2 out   = (0.21*5 + .5 * 3) / 12 = 0.2125

     Mole fraction of H2O out   = ( .5 * 3) / 12 = 0.125