A gas that contains 20 percent carbon monoxide and 80
percent nitrogen is burnt with 100 percent excess air at 25 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.
Specific heat data is given as a function of temperature as
cp is equal to a plus bT plus cT squared.
Constants a, b, and c are given for carbon monoxide, oxygen, carbon
dioxide and nitrogen. Using these
constants, specific heat values are in calories per mole 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.
In the product stream, no carbon monoxide, only excess
amount of oxygen, one mole of carbon dioxide and all nitrogen would be
present. In order to calculate the
enthalpy of feed or product, we need to calculate heat capacity coefficients
for both feed and product. 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. This can be converted to joules per mole per
degree K by multiplying calories by 4.184.
These weighted values of a, b, and c for both feed and product are given
in the table.
Feed temperature is 298.15 K, and considering this as a base
temperature, enthalpy of the feed is zero.
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 2.83 times 10 raised to the power 5. Integration constant, delta Hpo
can be calculated using corresponding values of a, b, and c for products. Please note that enthalpy of products at
298.15 K is again zero. 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. However, in case of Newton’s
method, we can differentiate the main function with respect to
temperature. The new iteration value
for temperature is equal to T_new is equal to T_old minus function of T divided
by derivative of the function and 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 1216 degrees K.