One of your fellow Energy Engineers comments that the flue gas leaving the natur

Question

One of your fellow Energy Engineers comments that the flue gas leaving the natural gas-fired boiler that you are designing should be at a high temperature because of dew point. Your boss asks you to examine that argument by performing some calculations. The natural gas boiler that you are designing uses a gas with a composition of CH4-95% and C2H6 – 4% and other inert gases 1%. You have a coal fired facility next door and they use a coal with the following analysis C-73.4% H-4.3%, S-1.1%, N-1.3%. The fuel’s proximate analysis is given as follows: Moisture 6.8%, Ash-8.6%, FC- 65.4 and VM 19.2%. The HHV of the Fuel is 13,380 BTUs/lb. The air that is used is ambient at 78 F and 40% relative humidity. This plant is designed to use 22% excess air. Evaluate both cases and show to your boss with calculations whether the dew point is different? If so by how much? And what is the impact?

If 90% of the sulfur dioxide produced has to be captured from the coal fired power plant in problem 3 (500 MW Thermal) by using lime (CaO). What is the flow rate of the CaCO3 that would be required if CaO is produced by calcining CaCO3.The efficiency of sulfur capture is 65%.

Explanation / Answer

We know that the dew point it the temperature at which first drop of a gas condenses

In case of mixture of gases it can be calcualted as

1) first we will calcualte the vapour pressure of each gas by antoine equation

Boiling point of methane = -161.5 °C

Boiling point of ethane = -890C

We will ignore the ideal gas here

so the dew point at atmospheric pressure will lie within this temperature range

a) Antoine equation constant methane

b) ethane

A = 6.83452   B = 663.700   C = 256.470

Antoine equation is

logP = A - ( B/ C+T)

MEthane

logP = 6.6438 - (395.74 / 266.681 - 161.5) = 2.881

So P = 760.32 mmHg

Ethane

logP = 6.8345 - (663.7 / 256.47 - 89) = 2.871

P = 743.02 mmHg

Assuming ideal behaviour, the liquid mole fractions at the dew point can be calculated using:

xmethane = 0.95 X 760 / 760.32 = 0.95

x ethane = 0.04 X 760 / 743.02 = 0.04

Adding the liquid mole fractions together gives: 0.95 + 0.04 = 0.99

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