An oil having empirical formula C12H26
is burnt with 20 percent excess air.
The burnt gases at enter a superheater at 1550 degrees Fahrenheit and
exit at 1150 degrees Fahrenheit. Steam
enters the superheater at 400 and is 92 percent saturated. It is required that steam from superheater
leaves at 650 degrees Fahrenheit.
Pressure drop in the superheater is negligible.
Specific heat capacity of the product gases is given as 0.25
BTU per pound per degree F.
Solution:
We need to draw a block diagram to understand the process
and then do a material balance to find out the amount and distribution of
individual product components. Let us
take one lb mole of oil as the basis of calculations. Write the combustion reaction for oil. We can see that one mole of the oil requires 18.5 moles of
oxygen, thus producing 12 moles of carbon dioxide and 13 moles of water. The problem statement desires us to use 20
percent excess air. Remember that
oxygen is derived from ambient air that has a fixed ratio of 3.764 moles of
nitrogen to one mole of oxygen. Now, we
are in a position to calculate the amount on weight basis of the product
components.
Thus carbon dioxide produced is 1.2 times 44 equals 528
pounds.
Water produced is 13 times 18
equals 234 pounds.
Oxygen in the product would only be the excess and is equal
to 0.2 times 18.5 times 32 equals 118.4 pounds.
Associated nitrogen is 3.764 times amount of oxygen fed to
the system. This value is found to be
3.764 times 1.2 times 18.5 times 32 equals 2340 pounds.
Total mass flow rate of combustion gases equals 528 plus 234
plus 118.4 plus 3220.4 pounds. Or 18.94
pound per pound of oil burnt.
As these gases are cooled from 1550 to 1150 degrees F, heat
provided by combustion gases equals mass times specific heat times temperature
difference. This translates to 18.94
times 0.25 times 400 equals 1894 BTU.
Please note that the temperature difference is 1550 minus 1150 equals
400 degrees F.
Enthalpy of superheated steam at 400psi and 600 degrees F cane
be obtained from steam tables. It is
interpolated as 1335.4 BTU.
Now let us calculate how heat is required to superheat,
steam entering into the superheater.
Enthalpy of the entering steam is to be based upon its conditions, that
is, 400 psi and 92 percent quality.
Enthalpy of liquid water is 424.17 BTU and latent heat of vaporization
is 1204.6 minus 424.17 equals 780.43 BTU.
Thus enthalpy of entering steam is 424.17 plus 0.9 times 780.43 equals
1142.14 BTU.
Enthalpy that needs to be provided to steam is 1335.4 minus
1142.14 equals 193.26 BTU per lb.
One pound of oil can provide heat that is sufficient to
superheat 1894 over 193.26 equals 9.78 pounds.