Question 14.9: With initial temperature and pressure being 1 atm and 298 K,......

Given: (1) Combustion reactants: Air and Methane, (2) Initial condition: 1 atm pressure and 298 K temperature.
Find out: Adiabatic flame temperature at constant volume.
Properties: Molar mass of C,  H_2 and O_2 and air are: 12 kg/kmol, 2 kg/kmol, 32 kg/kmol and 29 kg/kmol
respectively.
Assumption: (1) Combustion is complete, (2) Air and combustion gases are ideal (3) Kinetic and potential energies are negligible (4) There is no work interaction (5) Combustion chamber is isolated.
Chemical reaction for the combustion is:

CH_4 + 2(O_2 + 3.78  N_2) \longrightarrow 1  CO_2 + 2H_2O + 7.56  N_2

Here, N_{CH_4} = 1,  N_{O_2} = 2,  N_{N_2} = 7.56,  N_{CO_2} = 1,  N_{H_2O} = 2.

Adiabatic flame temperature for constant volume given by the relation:

ΔU = 0
U_{products} = U_{reactants}

Since,

H_{R} (T_{in} , p_{in}) – H_p (T_{adi} , p_{out}) – R (N_R(T_{in}, p_{in})T_i – N_p (T_{adi} , p_{out})T_{adi}) = 0                           (14.7)
H_R (T_{in}, p_{in})= \sum{N_p} (H^0_f + H – H_0)_p = N_{CH_4}(H^0_f +H – H_0)_{CH_4} + N_{O_2}(H^0_f + H – H_0)_{O_2} + N_{N_2}(H^0_f + H – H_0)_{N_2}
= 1 × (-74,831) + 2 × 0 + 7.56 × 0 = – 74,831 kJ/kmol                  (14.8)

H_p(T_{adi}, p_{out}) = \sum{N_i} \int\limits_{T_{ref}}^{T}{C_{p,i}} dT + Q_L

Assuming c_p to be constant,

H_{products} = \sum{Ni} (H^0_f + C_{p,i} (T_{adi} – 298))_{N_2}
= 1 \times (-3,93,546 + 56.21 (T_{adi} – 298))+ 2 \times (-2,41,845 +43.87 (T_{adi} – 298)) + 7.52 \times (0+33.71)(T_{adi} – 298))kJ/kmol                               (14.9)

Now,

R(N_R(T_{in}, p_{in}) T_{i} – N_p(T_{adi}, p_{out})T_{adi})
N_R (T_{in}, p_{in})T_i = 1 + 2 + 3.78 = 10.78  kmol  ,  N_p(T_{adi}, p_{out}) = 1 + 2 + 7.56 = 10.56  kmol
R(N_R (T_{in} , p_{in})T_i – N_p(T_{adi} , p_{out})T_{adi}) = 8.315(10.56)(298 – T_{adi})  kJ/kmol              (14.10)

Substituting (14.8), (14.9) and (14.10) in expression (14.7):
– 74,831 – \left[1 \times (-3,93,546 + 56.21(T_{adi}- 298))+2\times  (-2,41,845 + 43.87(T_{adi}- 298))+ 7.52 \times (0+33.71(T_{adi}-298))\right] + 8.315 (10.56)(298 – T_{adi})= 0
Solving it for T_{adi},  T_{adi} = 2,889  K