Question

For a high-intensity oxidation process, a stream of pure O2 (10.0 mol/sec) at 100oC and 1.0 atm and a stream of pure N2 (10.0 mol/sec) at 150oC and 1.0 atm are combined in a mixer that has a heat loss to the surroundings equal to 209(T – 25) J/sec, where T is the outlet gas mixture temperature in oC. Calculate the temperature of the exiting gas stream from the mixer. Use the following heat capacity equations:

O2: Cp = 6.15 + 3.1 x 10-3 T

N2: Cp = 6.50 + 1.25 x 10-3 T

where T is in K and Cp is in cal/mol/K

Answer 1

heat loss by N2 =heat gained by O2

=>n_N2*Cp_N2*dT =n_o2*Cp_O2*dT

=>10*[(6.15 + 3.1 x 10-3 T)dT ]T^N2_T =10*[(6.5 + 1.25 x 10-3 T)dT ]TO²_T

Integrating

=>(6.15*(423-T)+3.1*10^-3*(423^2-T^2)/2) = (6.5*(T-373)+1.25*10^-3*(T^2-373^3)/2)

=>2601.45 - 6.15*T +277.34 - 1.55*10^-3*T^2 = 6.5*T -2424.5 + 0.625*10^-3*T^2 -86.95

=>2.175*10^-3*T^2 +12.65*T -5390.24 =0

solving quadratic equation

=.T =398.76K =125.76 oC

temperature of the exiting gas stream from the mixer =398.76K =125.76 oC