Specific volume of a gas:  It is very important to realize that measurement of the gas is reported at conditions, which are quite different from the standard conditions.  Standard conditions are 29.92 in Hg and 32 °F.  The measurement temperature is 60 °F and measurement pressure is 30" Hg.  If the gas behaves ideally, then its specific volume can be calculated using ideal gas law and the results are presented for various selected conditions.

 

One pound-mole of a gas at 32 °F and 29.92 in Hg occupies 359.05 ft3. 

One pound-mole of a gas at 60 °F and 30 in Hg occupies 378.48 ft3. 

 

The vapor pressure of water at 60 °F = 0.52" Hg.

Mole fraction of dry gas = (30 - 0.52)/30 = 0.9827

 

If the gas is completely saturated at 60 °F and 30 in Hg, then mole fraction of dry gas is 0.9827:

The volume of 1 mole of dry gas is 378.48/0.9827 = 385.16 ft3.

 

Combustion: To perform combustion calculations it is necessary to apply the chemistry of the reaction.  Good combustion requires that excess air be present.  The percent of excess air may be as low as 5% in a good combustor.  Excess air may also be used to control the temperature when burning a high heating value waste.  The combustion reactions for some hydrocarbon gases are given as follows:

 

 

If one mole of a combustible material A is reacted with nO2 moles of oxygen, then nCO2 moles of carbon dioxide and nH2O moles of liquid water will be produced.  The following table provides heat of formation for various components.  This table also lists moles of oxygen required for complete combustion and moles of combustion products.

 

 

 

 

CH4

 

C2H6

 

C3H8

 

C4H10

 

H2

 

CO

 

CO2

 

H2O (l)

 

Hof, cal/mol

 

-17889

 

-20236

 

-24820

 

-29812

 

0

 

-26416

 

-94052

 

-68317.4

 

nO2

 

2

 

3.5

 

5

 

6.5

 

0.5

 

0.5

 

0

 

0

 

nCO2

 

1

 

2

 

3

 

4

 

0

 

1

 

1

 

0

 

nH2O

 

2

 

3

 

4

 

5

 

1

 

0

 

0

 

1

 

Heating Value:

 

Example 3.2:  Calculate heating value of a natural gas given the following composition in volume percent:

CH4 = 94.44, C2H6 = 3.4, C3H8 = 0.55, C4H10 = 0.5,

CO = 0.01, CO2 = 0.57, H2 = 0.03, N2 = 0.5

 

Solution 

C                  CO2 produced = 0.944(1) + 0.068(2) + 0.017(3) + 0.020(4) + 0.0001(1) = 1.049

C                  H2O(l) produced = 0.944(2) + 0.068(3) + 0.017(4) + 0.020(5) + 0.0001(0) = 2.038

C                  Enthalpy of products, SH Products = 1.049(-94052) + 2.038(-68317.4) = -237891.4 cal

C                  Enthalpy of feed, SH Feed = 0.944(-17889) + 0.068(-20236) + 0.017(-24820)

+ 0.020(-29812) + 0.0001(-26416) = -18407 cal

$                 Heat of reaction, ΔH298 = SH Products - SH Feed

= -237891.4 - (-18407) = -219492 cal/mol = -395086 Btu/lbmol

$                 High heating value, HHV =  -ΔH298 = 3.951 ´ 105 Btu = 1044 Btu/ft3