Question 6.11.3: The adiabatic flame temperature is that temperature reached ...

The adiabatic flame temperature is that temperature reached when all of the sensible heat released by the combustion of a fuel is used to raise the temperature of the gaseous products of combustion. Consider the adiabatic flame temperature reached when acetylene, C_2H_2 , is combusted at 298 K with (1) the stoichiometric amount of oxygen and (2) the number of moles of air containing the stoichiometric number of moles of oxygen. Air is, by mole or volume percent, 21% O_2 and 79% N_2.

The combustion reaction with stoichiometric oxygen is

C_2H_2+2.5  O_2=2  CO_2+H_2O                                   (i)

For C_2H_2: \Delta H_{298}=+226,700 J

For CO_2: \Delta H_{298}=-393,500 J

For H_2O: \Delta H_{298}=-241,800 J

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Thus, for the reaction given by Equation (i):

\Delta H_{(i)_{298} }=(-2\times 393,500)-241,800-226,700=-1,255,500 J

The constant-pressure molar heat capacities of the products of reaction are

For H_2O:   c_{p},H_2O =30.00+10.71\times 10^{-3} T+0.33\times 10^{5}T^2 J/K.mole

For CO_2:   c_{p},CO_2 =44.14+9.04\times 10^{-3} T-8.54\times 10^{5}T^2 J/K.mole

The adiabatic flame temperature, T , is then obtained from the requirement

\Delta H_{(i)_{298} }+\int_{298}^{T}{(2c_p,CO_2+c_p,H_2O)}dT=0

or

-1,255,500+118.28\times (T-298)+14.40\times 10^{-3}\times (T^2-298^2)

 

+16.75\times 10^5(1/T-1/298) =0

which has the solution T=6,236 K.

For combustion with the stoichiometric amount of air, the reaction is written as

C_2H_2+2.5  O_2+2.5\times (79/21)N_2=2CO_2+H_2O+9.41N_2

For N_2:   c_p,N_2=27.87+4.27\times 10^{-3}T J/mole.K

and the adiabatic flame temperature, T , is obtained from

\Delta H_{(i)_{298} }+\int_{298}^{T}{(2c_p,CO_2+c_p,H_2O+9.41c_p,N_2)}dT=0

or

-1,255,500+380.1\times (T-298)+40.16\times 10^{-3}\times (T^2-298^2)

 

+16.75\times 10^5(1/T-1/298)=0

as T=2,797 K

This high adiabatic flame temperature facilitates the use of acetylene for welding metals with high melting temperatures.

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