A boiler is fired with a high-grade fuel oil (consisting only of hydrocarbons) having a standard heat of combustion of −43,515 J·g^−1 at 25°C with CO2(g) and H2O(l) as products. The temperature of the fuel and air entering the combustion chamber is 25°C. The air is assumed dry. The flue gases

Question

A boiler is fired with a high-grade fuel oil (consisting only of hydrocarbons) having a standard heat of combustion of [latex]−43,515 J·g^{−1} at 25°C with CO_{2}(g) and H_{2}O(l) [/latex] as products. The temperature of the fuel and air entering the combustion chamber is 25°C. The air is assumed dry. The flue gases leave at 300°C, and their average analysis (on a dry basis) is 11.2% [latex] CO_{2}[/latex], 0.4% CO, 6.2% [latex]O_{2}, and [/latex] 82.2 % [latex] N_{2} [/latex]. Calculate the fraction of the heat of combustion of the oil that is transferred as heat to the boiler.

Accepted Answer

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

[latex] CO_{2}           11.2 mol [/latex]
[latex] CO 0.4 mol [/latex]
[latex] O_{2}    6.2 mol [/latex]
[latex] N_{2}    82.2 mol [/latex]

Total 100.0 mol

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

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

and

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

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

Therefore on the basis of 100 mol dry flue gases,

Moles [latex]H_{2} O formed = (21.85 − 17.60) (2) = 8.50 [/latex]

Moles [latex]H_{2}[/latex] 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 [latex]H_{2}[/latex] together give:

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

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

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

However, the reaction actually occurring does not represent complete combustion, and the [latex]H_{2}O[/latex] 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 [latex]H_{2}[/latex] , is represented by the empirical formula
[latex]C _{11.6} H_{17}[/latex]. Omit the 6.2 mol [latex]O_{2}[/latex] and 82.2 mol [latex]N_{2}[/latex] which enter and leave the reactor unchanged, and write the reaction:

[latex]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)[/latex]

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

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

[latex]8 .5 H_{2}O(l) → 8.5 H_{2} O(g)[/latex]

[latex]0.4 CO_{2}(g) → 0.4CO(g) + 0.2 O_{2} (g)[/latex]

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

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

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: [latex]\Delta H_{298}^{°}[/latex] has already been calculated and [latex]\Delta H_{298}^{°}[/latex] is easily evaluated.

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

[latex] \Delta H_P^{\circ}=\left(\sum_i n_i\left\langle C_{P_i}^{\circ} ight angle_H ight)(573.15-298.15) [/latex]

where subscript i denotes products. The [latex] \left\langle C_{P_i}^{\circ} ight angle_H / R [/latex] values are:

[latex] CO _2: \quad \operatorname{MCPH}\left(298.15,573.15 ; 5.457,1.045 imes 10^{-3}, 0.0,-1.157 imes 10^5 ight)=5.2352 [/latex]

[latex] ext { CO: } \quad ext { MCPH }\left(298.15,573.15 ; 3.376,0.557 imes 10^{-3}, 0.0,-0.031 imes 10^5 ight)=3.6005 [/latex]

[latex] H _2 O : \quad MCPH \left(298.15,573.15 ; 3.470,1.450 imes 10^{-3}, 0.0,0.121 imes 10^5 ight)=4.1725 [/latex]

[latex] O _2: \quad MCPH \left(298 . \mid 15,573.15 ; 3.639,0.506 imes 10^{-3}, 0.0,-0.227 imes 10^5 ight)=3.7267 [/latex]

[latex] N _2: \quad \operatorname{MCPH}\left(298.15,573.15 ; 3.280,0.593 imes 10^{-3}, 0.0,0.040 imes 10^5 ight)=3.5618 [/latex]

Therefore,

[latex] \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 [/latex]

and

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

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

[latex]ΔH = Q + W_{s}[/latex] (2.32)

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

of the higher heat of combustion of the fuel.

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

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