Known Combustion products exit a combustor at a specified state and a specified mass flow rate.
Find Determine the total flow exergy rate of the combustion products, in MW.
Schematic and Given Data:
Engineering Model
1. The combustor is analyzed as a control volume at steady state.
2. The effects of motion and gravity are ignored.
3. The combustion products form an ideal gas mixture.
4. The exergy reference environment of Table A-26 (Model II) applies.
Analysis The total flow exergy on a unit of mass basis is given by Eq. 13.47 as
e _{ f }=\underline{h-h_{0}-T_{0}\left(s-s_{0}\right)+\frac{ V ^{2}}{2}+g z}+ e ^{ ch } (13.47)
e _{ f }=\underline{\left(h_{4}-h_{0}\right)-T_{0}\left(s_{4}-s_{0}\right)}+e^{ ch } (a)
where the underlined term is the thermomechanical contribution to flow exergy, Eq. 7.14, subject to assumption 2, and e ^{ ch } is the chemical contribution, each on a unit of mass basis. These contributions will now be considered in turn.
e _{ f }=h-h_{0}-T_{0}\left(s-s_{0}\right)+\frac{ V ^{2}}{2}+g z (7.14)
Thermomechanical Contribution The thermomechanical contribution to Eq. (a) requires values for h_{4}, s_{4}, h_{0}, \text { and } s_{0}. The table provided with the problem statement gives h_{4} \text { and } s_{4}. They are mixture values determined using Eqs. 13.9 and 13.23, respectively. While the evaluation of h_{4} \text { and } s_{4} s left as an exercise, the evaluation of h_{0} \text { and } s_{0} is detailed next.
\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)
\overline{s_{i}}\left(T, p_{i}\right)=\bar{s}_{i}^{\circ}(T)-\bar{R} \ln \frac{y_{i} p}{p_{\text {ref }}} (component of an ideal gas mixture) (13.23)
When determining thermomechanical exergy at a state, we think of bringing the substance from the given state to the dead state where temperature is T_{0} \text { and pressure is } p_{0}. In applications dealing with gas mixtures involving water vapor, some condensation of water vapor to liquid may occur in such a process, which is the case in the present application. See note 1 .
As shown in Fig. E13.15b, a 1-kmol sample of the combustion products at the dead state where T_{0}=298.15 K , p_{0}=1 atm consists of a gas phase, including the “dry” products N _{2}, O _{2}, \text { and } CO _{2} together with water vapor in equilibrium with the condensed water. In kmol, the analysis of the gas phase is N _{2},0.7507 ; O _{2}, 0.1372 ; CO _{2}, 0.0314 ; H _{2} O ( g ), 0.0297, while the liquid water phase contains the rest of the water formed on combustion: 0.0510 kmol. Within the gas phase N _{2}, O _{2}, CO _{2}, and H _{2} O ( g ) have the following mole fractions, denoted by y^{\prime}:
2 y_{ N _{2}}^{\prime}=0.7910, y_{ O _{2}}^{\prime}=0.1446, y_{ CO _{2}}^{\prime}=0.0331,{ y}_{ H _{2} O ( g )}^{\prime}=0.0313 (b)
The mixture enthalpy h_{0} is determined using Eq. 13.9 to obtain the specific enthalpy of each mixture component and then summing those values using the known molar analysis of the mixture. For each component, the \Delta \bar{h} term of Eq. 13.9 vanishes, leaving just the enthalpy of formation. Accordingly, with data from Table A-25,
The first four terms of this calculation correspond to the gas phase whereas the last term is the contribution of the liquid water phase. Using the mixture molecular weight, we get the mixture enthalpy on a unit mass basis:
The mixture entropy s_{0} is also determined by summing the entropies of the gas and liquid water phases. For each substance in the gas phase the following special form of Eq. 13.23 is used:
where y′ is the mole fraction of that substance in the gas phase, as given by Eqs. (b).
With absolute entropy data from Table A-23, we get the following values, each in kJ/kmol, for the gas phase
The liquid water value is obtained from Table A-25 as
Summing as in the calculation of h_{0} above
When expressed on a unit mass basis
The thermomechanical contribution to the total flow exergy is then
Chemical Contribution
At the dead state a sample of the combustion products consists of a gas phase and a liquid water phase. The chemical exergy is determined by adding the chemical exergies of these phases.
For the gas phase, Eq. 13.41b is applied in the form
where y′ is the mole fraction of component i of the gas phase, as given by Eqs. (b). Accordingly, with chemical exergy values from Table A-26 (Model II),
For the liquid phase, Table A-26 (Model II) gives 900 kJ per kmol of liquid.
On the basis of 1 kmol of combustion products at T_{0}, p_{0}, the gas phase accounts for 0.949 kmol and the liquid phase accounts for 0.0510 kmol. The chemical exergy is then
When expressed on a unit mass basis,
Collecting results, Eq. (a) gives the total flow exergy as the sum
On a time-rate basis
1 At the dead state where T_{0} = 25°C, p_{0} = 1 atm, a 1-kmol sample of the combustion products consists of 0.9193 kmol of “dry products” (N_{2}, O_{2}, CO_{2}) plus 0.0807 kmol of water. Of the water, n kmol is water vapor and the rest is liquid. Considering the gas phase, the partial pressure of the water vapor is the saturation pressure at 25°C: 0.0317 bar. The partial pressure also is the product of the water vapor mole fraction and the mixture pressure, 1.01325 bar. Collecting results,
Solving, the amount of water vapor is n = 0.0297 kmol. The amount of liquid is (0.0807 − n) = 0.0510 kmol. These values appear in Fig. E13.15b. See Example 13.2(d) for a closely similar analysis.
2 On the basis of 1 kmol of combustion products the gas phase accounts for 0.949 kmol. Then, for N_{2}, y′ = 0.7507/0.949 = 0.7910; for O_{2}, y′ = 0.1372/0.949 = 0.1446; and so on.
3 In the present application, T = T_{0}, p_{i} = y′_{i} p_{0}, where p_{0} = p_{ref} =1 atm.
4 In keeping with expectations for high-temperature combustion products, the thermomechanical exergy contribution dominates; chemical exergy contributes only 1.4% to the total.
5 In this application, evaluation of the total flow exergy at state
4 can be simplified by assuming a hypothetical dead state
where all water formed by combustion is in vapor form only. With this assumption, the thermomechanical and chemical exergy contributions to the total flow exergy are 1086 kJ/kg and 17 kJ/kg, respectively, as can be verified (Problem 13.107). The total flow exergy is then 1103 kJ/kg, which differs negligibly from the value determined in the solution: 1104 kJ/kg.
Skills Developed
Ability to…
• determine the flow exergy, including the chemical exergy contribution, of combustion products modeled as an ideal gas mixture.
Quick Quiz
If the total flow exergy of the preheated compressed air entering the combustor at state 3 is 41.9 MW and heat transfer from the combustor is ignored, evaluate the rate exergy is destroyed within the combustor, in MW. Ans. 25.
