Question

Practice -TEST#3 4-8-2016 Student Name: Make any necessary Assumptions Problem (D-30 Points pressure (300 kPa), Ethane is burned and the following are the products of combustion at Number of Moles Gas CO2 H20 13.16 If the gasses were at 2200 K and then cooled to 298 K Determine: Deter The mole fraction of each component of the mixture, a) The mole e mass fraction of each component of the mixture b) The mass frac e) The molar mass of the mixture, d) The amount of heat rejected in kJ/kg (Table A-2, using co )Th )be in the liquid or vapor phase

Please answer as a) b)....e)

And using Applied thermodynamics equations. Thanks

Answer 1

mole fraction of co_2 = 2/2 + 3 + 13.16 = 0.11 mole fraction of H_2O = 3/ 2 + 3 middot 13.16 = 0.166 mole fraction of N-2 = 13.16/2 + 3 + 13.16 = 0.725 (b) mass fraction of CO_2 = 0.11 times 44gm = 4.84gm mass fraction of H_2 = 0.166 times 18gm = 2.99gm mass fraction of N_2 = 0.725 times 28gm = 20.3gm (c) mass fraction of mixture (molar mass) = Total mass fraction/Total no of moles = (4.84 + 2.99 + 20.3)gm/(0.11 + 0.166 + 0.725)moles = 28.27gm.mol = 0.02827kg/mol = 35.37mol /kg (d) Q = mc_p Delta T = 0.02827kg/mol times 0.0525KJ/mol.K times (2200 - 298)K = kg/mol times (KJ/mol) times (KJ/mol times 0.0525) times 1902 = 35.37mol/kg times 0.0525KJ/mol times 1902 Q = 3531.87KJ/kg. (e) 293K = 25 degree c water will be in liquid