Find the adiabatic flame temperature for a stoichiometric me

Find the adiabatic flame temperature for a stoichiometric methane/air flame if the initial temperature of the fuel and air is 10. Take the net calorific value of methane as 50.14 MJ/kg.

Clear steps please

Solution

Given data

Fuel - stoichiometric methane/air flame

Initial temperature of the fuel and air is 10.

Net calorific value of methane -50.14 x 10^6 j/kg

To find

the adiabatic flame temperature

Solution:
The first step is to evaluate, for 1 kg fuel, the mass of each of the reactants and products. Reactants            Products
1 kg CH_{4 --------->}    2.75 kg CO₂
4 kg O₂ ------>    2.25 kg H₂O
13.17 kg N_{2 ------>}               13.17 kg N₂

The initial temperature of the reactants is 10 and the value of H_R is the sensible heat in the reactants (ref. 25):
       H_R = (10-25){1C_PCH₄ + (4C_PO₂) + (13.17C_PN₂)}

The mean temperature of the reactants over the interval is 17.5. At this temperature the values of the specific heats are:
               CH₄                       2.23 kJ/kg/K
               O₂                          0.92
               N₂                          1.04
so           H_R = -293.9 kJ/kg methane

Giving H_P = CV + H_R =50,144 – 293.9 = 49,850 kJ/kg
This value of H_P represents the sensible heat in the combustion products above the reference temperature of 25, i.e.
49,850 = (t_f - 25) {(13.17C_PN₂) + (2.25C_PH₂O) + (2.75C_PCO₂)}

To solve this equation by trial and error requires an initial guess for the flame temperature.
If we assume 1,575, the mean temperature of the products above 25 is (1,575 + 25)/2, i.e. 800

The specific heats of nitrogen, water vapor and carbon dioxide at this temperature are
        N₂                          1.18 kJ/kg/K
        H₂O                       2.33
        CO₂                       1.26
so the energy equation is
49,850 = (t_f-25){(13.17×1.18)+(2.25×2.33)+(2.75×1.26)}
t_f=2,081

The new mean temperature of the products is (2,081+25)/2=1,053

The specific heats are now:
               N₂                          1.22 kJ/kg/K
               H₂O                       2.50
               CO₂                       1.30
solving the energy equation for the third time gives:
               t_f = 1,998

The mean temperature of the products now becomes (1,998+25)/2=1,012 giving
               N₂                          1.22 kJ/kg/K
               H₂O                       2.48
               CO₂                       1.30
A further evaluation of the energy equation gives
               t_f = 2,001