Dry Gas Analysis

5. Dry Gas Analysis

The products of combustion are analyzed on a volumetric or molar basis in order to determine if combustion is complete. In the context of Eq. (5b) the percentage of CO 2 in the products equals (1 ×100÷(1+2+1+11.28) = 7%. This is an example of a wet analysis, since the

products also include water vapor. If the combustion products are analyzed with water re- moved as a constituent, this is called a dry gas analysis. Again, applying Eq. (5b), the CO 2 percentage on a dry basis is 1 ×100/(1+1+11.28) = 7.53%. The volume percentages of the con-

stituents of an ideal gas mixture are identical to their molar percentages. The measured dry or wet gas compositions can be used to determine (A:F) as illustrated in the following example.

b. Example 2

A dry gas analysis of the gas exhaled by a human lung is as follows– O 2 :16.5% and CO 2 :3.1%. Assume the “fuel” burned by humans is characterized by the chemical formula CH x and is completely burned. Determine the values of “x” and (A:F).

Solution: If the reactant CH x and is completely consumed, the exhaled products consist of only CO 2 H 2 O, N 2 and O 2 . The stoichiometric relation for this chemical activity is

(A) There are seven unknowns, namely x, a, b, c, d, e, f. From an atom balance of the

CH x +aO 2 +bN 2 → c CO 2 +dH 2 O+eO 2 +fN 2 .

elements C, H, O, and N we obtain the following four equations, i.e.,

C: 1 = c, (B)

H: ×= 2 d, (C) O: 2 a = 2 c + d + 2 e, and

(D) N: 2 b = 2f, where

(E) b/a = 3.76.

(F) From the dry gas analysis:

Percentage of CO 2 =c × 100 ÷ (c + e + f)

(G)

Percentage of O 2 =e × 100 ÷ (c + e + f)

(H)

The seven equations are obtained using the seven equations Eqs. (B)–(G), i.e.,

c = 1, (I)

d = x/2, (J)

e = (2 a – 2 c – d)/2 = a – 1 – x/4, (K)

f = b = 3.76 a, (L) (Percentage of CO 2 )/(Percentage of O 2 ) = c/e = 3.1 ÷16.5 = 0.19,

(M) c+e+f=c ÷0.031 = 32.26,

(N) Hence,

e=1 ÷0.19 = 5.26,

f = b = 32.26 – 1– 5.26 = 26,

a = 6.91,

5.26 = 6.91 – 1– x/4, i.e., × = 2.6, and

d = 1.3. Consequently, the chemical relation assumes the form

CH 2.6 + 6.91 O 2 + 26 N 2 → CO 2 + 1.3 H 2 O + 5.26 O 2 + 26 N 2 . (A) The air fuel ratio

A:F = (6.91 + 26) ÷1 = 32.91, and (A:F) mass = 65. For a stoichiometric reaction, the corresponding relation is

CH 2.6 + 1.65 O 2 + 1.65 ×3.76 N 2 → 1 CO 2 + 1.3 H 2 O+O 2 + 1.65 ×3.76 N 2 , and (A:F) mass = 16.

Remarks Once the exhaust gas composition is known, the fuel used and the A:F ratio for the combustion process can be determined. Modern gas analyzers that incorporate the ap- propriate software for determining (A:F) are widely used. For the sake of illustration, consider the oxidation of the following fuel

CH m O n N p S q + air → (1-e) CO 2 , e CO, H 2 O, SO 2 ,O 2 ,N 2 , Ar.

For stoichiometric oxidation, a stoich =(1+q+m/4-n/2).

If the percentages CO 2 and CO are measured, then

N dry , number of dry moles in products = 1/(x CO2 +x CO ).

e=x CO ×N dry . a= (N dry – (e/2 - m/4 + n/2 +p/2)) ×x O2,a , φ=a stoich /a= (A:F) stoich /(A:F). Here, x O2,a denotes the ambient oxygen concentration (mole fraction), and a the actual

oxygen content supplied.

If the percentages O 2 and CO are measured, then

B =(x O2 –x CO /2)/(1-x CO /2). a= (1+q+m/4-n/2 + B(n/2+p/2-m/4))/(1-B/x O2,a ), φ=a stoich /a== (A:F) stoich /A:F.

N dry = (a /x O2,a - m/4 + n/2 +p/2)/(1 - x CO /2).

A generic wet gas analysis of human exhalation is as follows – N 2 : 78%, O 2 : 16%,

CO 2 : 3%, and H 2 O: 3%.

Figure 1: Schematic illustration of the determination of the dew point.

The wet exhaust from a diesel engine generally comprises N 2 : 77%, O 2 : 13.54%, CO 2 : 5%, and H 2 O: 4%.

c. Example 3 Consider the combustion of natural gas (which is assumed to have the same properties as CH 4 ). Determine (A:F) if the products contain 3% O 2 (on a dry basis), and the composition on both a wet and dry basis. Solution: The reaction equation is

CH 4 + a (O 2 + 3.76 N 2 ) → CO 2 + 2H 2 O+bN 2 +dO 2 (A) From an O atom balance,

2 a = 2 + 2 + 2 d, or a = 2 + d. (B)

The exhaust contains 3% O 2 on a dry basis, i.e.,

d/(1 + 3.76 a + d) = 0.03. (C) Therefore,

d = 0.2982, and (D)

a = 2.2982. (E) Also, from an N atom balance

b = 3.76 a. (F) Therefore, (A:F) = 2.2982 × 4.76 ÷ 1 = 10.9, and

(G) Eq. (A) assumes the form CH 4 + 2.3 (O 2 + 3.76 N 2 ) → CO 2 +2H 2 O + 8.6 N 2 + 0.3 O 2 .

(H) The composition is as follows:

CO 2 % in exhaust (wet) = 100 ×(1÷(1+2+3.76×2.2982+0.2982)) = 8.4%, CO 2 % in exhaust (dry) = 100 ×(1÷(1+3.76×2.2982+0.2982)) = 10.1%,

N 2 % in exhaust (wet) = 100 ×(3.76×2.2982÷(1+2+3.76×2.2982+0.2982)) = 72.4%, N 2 % in exhaust (dry) = 100 ×(3.76×2.2982÷(1+3.76×2.2982+0.2982)) = 86.9%.

Remark Using Eq. (H), the mole fraction of water on wet basis has a value equal to

2 ÷(1+2+8.6+0.3) = 0.17. Consequently, the partial pressure of water is 0.17 bar if the mixture pressure is atmospheric. At this pressure T sat = T DP = 56ºC, where T DP de- notes the dew point temperature. Inversely, if T DP is known, the water percentage and

(A:F) can be determined (cf. Figure 1 ).