Combustion Calculation?

Question

In a 600-MWel-coal-fired power plant, the following coal shall be used for combustion. C = 76 %, H = 5 %, O = 14 %, S = 3.5 %, N = 1.5 % (waf); w = 11 %, a = 14 %, calorific value: Hu = 23,4 MJ/kg. The efficiency of the power plant is to be assumed as 38 %
a) Calculate the mass flow of dry air for full load for a by 70 % overstochiometric combustion (fluidised bed firing system).
b) Calculate the mass flows of the constituents of the flue gas.
c) The desulphurisation is done with a rather cheap mineral called dolomite (CaMg(CO3)2) which is given to the fluidised bed reactor and spent in large quantities (four times stochiometric quantity). How much dolomite is consumed? How much gypsum is created?

Answer 1

First calculate the answers per unit mass of coal and then multiply to get the total.

For complete combustion, the carbon becomes carbon dioxide, the hydrogen becomes water, and the the sulfur becomes sulfur dioxide. Thus for each atom of carbon, and sulfur, we need two of oxygen, while for each of hydrogen we need only 1/2.

Since hydrogen has an atomic weight of 1 as compared with oxygen's 16, for each gram of hydrogen we need 8 grams of oxygen. Similarly for carbon and sulfur.

(It isn't clear what to do about the nitrogen. This site suggests that we give each nitrogen atom 1/5 of an oxygen atom)
http://en.wikipedia.org/wiki/Nitrogen_oxide

In any case, given the relative weights of the various components, we can determine the amount of oxygen needed per kg of coal for a stochiometric mixture. But since we want to run 70% over, scale the needed oxygen accordingly.

Since the coal contains some oxygen already, we have to subtract that out to get a net additional oxygen per kg.

Then we use the proportion of dry air that is oxygen to compute the mass of air needed:
http://www.engineeringtoolbox.com/air-composition-d_212.html

The flue gas will contain all the gases mentioned above plus the gases (including the excess oxygen) brought in by the dry air.

Once we have the per kg values, we have to scale. Given the efficiency of the power plant and its total output, we can determine the total energy input per unit time.

Given the total energy needed per unti time and energy per kg of coal, we can determine the total number of kilograms of coal needed per unit time.

Multiply the per kg quantities to get the total masses.

All of this is a gross simplification, but it seems to be what is called for.