Calculating Enthalpy of Combustion of Methane
Calculate the enthalpy of combustion of gaseous methane, in kJ per kg of fuel, (a) at 25°C, 1 atm with liquid water in the products, (b) at 25°C, 1 atm with water vapor in the products. (c) Repeat part (b) at 1000 K, 1 atm.
Known The fuel is gaseous methane.
Find Determine the enthalpy of combustion, in kJ per kg of fuel, (a) at 25°C, 1 atm with liquid water in the products, (b) at 25°C, 1 atm with water vapor in the products, (c) at 1000 K, 1 atm with water vapor in the products.
Engineering Model
1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert.
2. Combustion is complete, and both reactants and products are at the same temperature and pressure.
3. The ideal gas model applies for methane, the combustion air, and the gaseous products of combustion.
Analysis The combustion equation is obtained from Eq. 13.4
CH _{4}+2\left( O _{2}+3.76 N _{2}\right) \rightarrow CO _{2}+2 H _{2} O +7.52 N _{2} (13.4)
CH _{4}+2 O _{2}+7.52 N _{2} \rightarrow CO _{2}+2 H _{2} O +7.52 N _{2}
Using Eq. 13.9 in Eq. 13.18, the enthalpy of combustion is
\bar{h}(T, p)=\bar{h}_{ f }^{\circ}+\left[\bar{h}(T, p)-\bar{h}\left(T_{ ref }, p_{ ref }\right)\right]=\bar{h}_{ f }^{\circ}+\Delta \bar{h} (13.9)
\bar{h}_{ RP }=\sum_{ P } n_{e} \bar{h}_{e}-\sum_{ R } n_{i} \bar{h}_{i} (13.18)
\bar{h}_{ RP }=\sum_{ P } n_{e}\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{e}-\sum_{ R } n_{i}\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{i}
Introducing the coefficients of the combustion equation and evaluating the specific enthalpies in terms of the respective enthalpies of formation
\begin{aligned}\bar{h}_{ RP }=& \bar{h}_{ CO _{2}}+2 \bar{h}_{ H _{2} O }-\bar{h}_{ CH _{4}( g )}-2 \bar{h}_{ O _{2}} \\=&\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{ CO _{2}}+2\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{ H _{2} O } \\&-\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{ CH _{4}( g )}-2\left(h_{ f }^{\circ \nearrow0}+\Delta \bar{h}\right)_{ O _{2}}\end{aligned}
For nitrogen, the enthalpy terms of the reactants and products cancel. Also, the enthalpy of formation of oxygen is zero by definition. On rearrangement, the enthalpy of combustion expression becomes
\begin{aligned}\bar{h}_{ RP }=&\left(\bar{h}_{ f }^{\circ}\right)_{ CO _{2}}+2\left(\bar{h}_{ f }^{\circ}\right)_{ H _{2} O }-\left(\bar{h}_{ f }^{\circ}\right)_{ CH _{4}( g )} \\&+\left[(\Delta \bar{h})_{ CO _{2}}+2(\Delta \bar{h})_{ H _{2} O }-(\Delta \bar{h})_{ CH _{4}( g )}-2(\Delta \bar{h})_{ O _{2}}\right]\end{aligned}
=\bar{h}_{ RP }^{\circ}+\left[(\Delta \bar{h})_{ CO _{2}}+2(\Delta \bar{h})_{ H _{2} O }-(\Delta \bar{h})_{ CH _{4}( g )}-2(\Delta \bar{h})_{ O _{2}}\right] (1)
The values for \bar{h}_{ RP }^{\circ} \text { and }(\Delta \bar{h})_{ H _{2} O } depend on whether the water in the products is a liquid or a vapor.
a. Since the reactants and products are at 25°C, 1 atm in this case, the \Delta \bar{h} terms drop out of Eq. (1) giving the expression for \bar{h}_{ RP }. Thus, for liquid water in the products, the enthalpy of combustion is
\bar{h}_{ RP }^{\circ}=\left(\bar{h}_{ f }^{\circ}\right)_{ CO _{2}}+2\left(\bar{h}_{ f }^{\circ}\right)_{ H _{2} O (1)}-\left(\bar{h}_{ f }^{\circ}\right)_{ CH _{4}( g )}
With enthalpy of formation values from Table A-25
\begin{aligned}\bar{h}_{ RP }^{\circ} &=-393,520+2(-285,830)-(-74,850) \\&=-890,330 kJ / kmol (\text { fuel })\end{aligned}
Dividing by the molecular weight of methane places this result on a unit mass of fuel basis
h_{ RP }^{\circ}=\frac{-890,330 kJ / kmol (\text { fuel })}{16.04 kg (\text { fuel }) / kmol (\text { fuel })}=-55,507 kJ / kg \text { (fuel) }
The magnitude of this value agrees with the higher heating value of methane given in Table A-25.
b. As in part (a), the \Delta \bar{h} terms drop out of the expression for \bar{h}_{ RP }, Eq. (1), which for water vapor in the products reduces to h_{ RP }^{\circ}, where
\bar{h}_{ RP }^{\circ}=\left(\bar{h}_{ f }^{\circ}\right)_{ CO _{2}}+2\left(\bar{h}_{ f }^{\circ}\right)_{ H _{2} O ( g )}-\left(\bar{h}_{ f }^{\circ}\right)_{ CH _{4}( g )}
With enthalpy of formation values from Table A-25
\begin{aligned}\bar{h}_{ RP }^{\circ} &=-393,520+2(-241,820)-(-74,850) \\&=-802,310 kJ / kmol (\text { fuel })\end{aligned}
On a unit of mass of fuel basis, the enthalpy of combustion for this case is
h_{ RP }^{\circ}=\frac{-802,310}{16.04}=-50,019 kJ / kg (\text { fuel })
The magnitude of this value agrees with the lower heating value of methane given in Table A-25.
c. For the case where the reactants and products are at 1000 K, 1 atm, the term h_{ RP }^{\circ} in Eq. (1) giving the expression for h_{ RP } has the value determined in part (b): \bar{h}_{ RP }^{\circ}=-802,310 kJ/kmol (fuel), and the \Delta \bar{h} \text { terms for } O _{2}, H _{2} O ( g ), \text { and } CO _{2} can be evaluated using specific enthalpies at 298 and 1000 K from Table A-23. The results are
\begin{array}{r}(\Delta \bar{h})_{ O _{2}}=31,389-8682=22,707 kJ / kmol \\(\Delta \bar{h})_{ H _{2} O ( g )}=35,882-9904=25,978 kJ / kmol \\(\Delta \bar{h})_{ CO _{2}}=42,769-9364=33,405 kJ / kmol\end{array}
For methane, the \bar{c}_{p} expression of Table A-21 can be used to obtain
(\Delta \bar{h})_{ CH _{4}( g )}=\int_{298}^{1000} \bar{c}_{p} d T
=\bar{R}\left(3.826 T-\frac{3.979}{10^{3}} \frac{T^{2}}{2}+\frac{24.558}{10^{6}} \frac{T^{3}}{3}\right.
\left.-\frac{22.733}{10^{9}} \frac{T^{4}}{4}+\frac{6.963}{10^{12}} \frac{T^{5}}{5}\right)_{298}^{1000}
= 38,189 kJ/kmol (fuel)
1.
Substituting values into the expression for the enthalpy of combustion, Eq. (1), we get
\bar{h}_{ RP }=-802,310+[33,405+2(25,978)-38,189-2(22,707)]
= -800,522 kJ/kmol (fuel)
On a unit mass basis
2. h_{ RP }=\frac{-800,552}{16.04}=-49,910 kJ / kg (\text { fuel })
1 Using Interactive Thermodynamics: IT, we get 38,180 kJ/kmol (fuel).
2 Comparing the values of parts (b) and (c), the enthalpy of combustion of methane is seen to vary little with temperature. The same is true for many hydrocarbon fuels. This fact is sometimes used to simplify combustion calculations.
Skills Developed
Ability to…
• calculate the enthalpy of combustion at standard temperature and pressure.
• calculate the enthalpy of combustion at an elevated temperature and standard pressure.
Quick Quiz
What is the lower heating value of methane, in kJ/kg (fuel) at 25°C, 1 atm? Ans. 50,020 kJ/kg (Table A-25).