Question 4.10: A boiler is fired with a high-grade fuel oil (consisting onl...

Take as a basis 100 mol dry flue gases, consisting of:

CO_{2}                 11.2 mol
CO                0.4 mol
O_{2}                        6.2 mol
N_{2}                        82.2 mol

Total            100.0 mol

This analysis, on a dry basis, does not take into account the H_{2}O vapor present in the flue gases. The amount of H_{2} O formed by the combustion reaction is found from an oxygen balance. The O_{2} supplied in the air represents 21 mol-% of the air stream. The remaining 79% is N_{2} , which goes through the combustion process unchanged. Thus the 82.2 mol N_{2} appearing in 100 mol dry flue gases is supplied with the air, and the O_{2}  accompanying  this  N_{2} is:

Moles O_{2} entering in air = (82.2)(21 / 79) = 21.85

and

Total moles O_{2} in the dry flue gases  =  11.2 + 0.4 / 2 + 6.2 = 17.60

The difference between these figures is the moles of O_{2} that react to form H_{2} O.

Therefore on the basis of 100 mol dry flue gases,

Moles H_{2} O formed = (21.85 − 17.60) (2) = 8.50

Moles H_{2} in the fuel = moles of water formed = 8.50

The amount of C in the fuel is given by a carbon balance:

Moles C in flue gases = moles C in fuel = 11.2 + 0.4 = 11.60

These amounts of C and H_{2} together give:

Mass of fuel burned = ​( 8.50 ) ( 2 )  + ​( 11.6 ) ( 12 )  = 156.2 g

If this amount of fuel is burned completely to CO_{2}(g)  and  H_{2}O(l) at 25°C, the heat of combustion is

\Delta H_{298}^{°} = (− 43,515)(156.2) = − 6,797,040  J

However, the reaction actually occurring does not represent complete combustion, and the H_{2}O is formed as vapor rather than as liquid. The 156.2 g of fuel, consisting
of 11.6 mol of C and 8.5 mol of H_{2} , is represented by the empirical formula
C _{11.6} H_{17}. Omit the 6.2 mol O_{2} and 82.2 mol N_{2} which enter and leave the reactor unchanged, and write the reaction:

C _{11.6} H_{17} (l) + 15.65 O_{2} (g) → 11.2 CO_{2} (g) + 0.4 CO(g) + 8.5 H_{2} O(g)

This result is obtained by addition of the following reactions, for each of which the standard heat of reaction at 25°C is known:

C _{11.6} H_{17} (l) + 15.85 O_{2} (g)  →  11.6 CO_{2} (g) + 8.5 H_{2} O(l)

8 .5 H_{2}O(l) → 8.5 H_{2} O(g)

0.4 CO_{2}(g) → 0.4CO(g) + 0.2 O_{2} (g)

The sum of these reactions yields the actual reaction, and the sum of the \Delta H_{298}^{°} values gives the standard heat of the reaction occurring at 25°C:

\Delta H_{298}^{°} = −6,797,040 + (44,012)(8.5) + (282,984)(0.4) = − 6,309,740  J

The actual process leading from reactants at 25°C to products at 300°C is represented by the dashed line in the accompanying diagram. For purposes of calculating ΔH for this process, we may use any convenient path. The one drawn with solid lines is a logical one: \Delta H_{298}^{°} has already been calculated and \Delta H_{298}^{°}  is easily evaluated.

The enthalpy change associated with heating the products of reaction from 25 to
300°C is:

\Delta H_P^{\circ}=\left(\sum_i n_i\left\langle C_{P_i}^{\circ}\right\rangle_H\right)(573.15-298.15)

where subscript i denotes products. The \left\langle C_{P_i}^{\circ}\right\rangle_H / R values are:

Therefore,

\Delta H_P^{\circ}=(8.314)[(11.2)(5.2352)+(0.4)(3.6005)+(8.5)(4.1725)+(6.2)(3.7267)+(82.2)(3.5618)](573.15 − 298.15) = 940,660 J

and

\Delta H=\Delta H_{298}^{\circ}+\Delta H_P^{\circ}=-6,309,740+940,660=-5,369,080 J

Because the process is one of steady flow for which the shaft work and kineticand potential-energy terms in the energy balance [Eq. (2.32)] are zero or negligible, ΔH = Q. Thus, Q = −5369 kJ, and this amount of heat is transferred to the boiler for every 100 mol dry flue gases formed. This represents

ΔH = Q + ​W_{s}     (2.32)

\frac{5,369,080}{6,797,040}(100)=79.0 \%

of the higher heat of combustion of the fuel.