A gas that contains 20 percent carbon monoxide and 80
percent nitrogen is burnt with 100 percent excess air at 1000 degrees C. What
is the flame temperature if no heat is lost?
Solution:
Combustion reaction for carbon monoxide is given, where one
mole of carbon monoxide reacts with ½ mole of oxygen to produce one mole of
carbon dioxide. Note that heat of
reaction is negative 67,636 calories per mol.
Negative sign indicates that heat is coming out of the system. Obviously, combustion reactions are
exothermic.
The weighted value of aF is given as sum of
product (ai nFi) for all feed components. This value is calculated to be 62.823
calories per mole per degree K. These
weighted values of a, b, and c for both feed and product are given in the
table. (See Problem 3.1.1). Using these constants, weighted specific
heat values are in calories per degree C.
Let us consider one mole of carbon monoxide as a basis for
calculations.
Feed contains 1 mole of Carbon monoxide, and 4 moles of
nitrogen, as it is present 80 percent versus 20 percent for carbon
monoxide. For complete combustion, ½
mole of oxygen is required for every mole of carbon dioxide. But, with 100 percent excess, we would
provide another ½ mole of oxygen, thus 1 mole of oxygen would be required. As oxygen is derived from air, nitrogen
would also be fed to the system.
Nitrogen in the air is present as 78 percent on volume basis, where
oxygen makes 21 percent. The ratio of
nitrogen to oxygen in the air is 3.76.
Total nitrogen fed in the system is 4 plus 3.76 equaling 7.76 moles.
Feed temperature is 1000 degrees C or 1273.15 K. Enthalpy of both feed and products at 298.15
K is zero. In order to express,
enthalpy as a function of temperature, we need to calculate the integration
constants. Integration constant, delta
HFo can be calculated using corresponding values of a, b,
and c for feed and is found to be equal to negative 1.93 times 10 to the power
4. Similarly, integration constant,
delta HPo is found to be equal to negative 1.87 times 10
to the power 4. Now, enthalpy of feed
can be calculated using this integration constant and temperature 1273.15 K to
be 7.09 times 10 to the power 4 calories per mole.
Using energy balance, delta HP is equal to sum of
delta HF minus q minus delta HR. This value of delta HP is calculated to be
1.385 times 10 to the power 5. Now for
products, we can find the temperature of the products. It can be seen that equation for enthalpy of
the products is a cubic in temperature, which can be solved either by trial and
error or by using an approach like Newton’s method. Should we decide to solve this equation by trial error, it is
fine. Otherwise, we can differentiate
the main function with respect to temperature and apply Newton’s method. New temperature value, T_new , is equal to
T_old minus function of T divided by derivative of the function at old T. We can pick any reasonable point, but be
warned that a totally unreasonable value could lead to a non-converging
situation. Final converged value is
2059 degrees K.