Determination of Syngas Premixed Gasoline and ... - ACS Publications

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Determination of Syngas Premixed Gasoline and Methanol Combustion Products at Chemical Equilibrium via Lagrange Multipliers Method Osman Sinan Süslü*,† and Ipek Becerik†,‡ Energy Institute and ‡Chemistry Department, Istanbul Technical University, Maslak, Iṡ tanbul 34469, Turkey

†

ABSTRACT: This article investigates theoretically the products from adiabatic combustion of synthesis gas reformed out of aqueous methanol with a variable gasoline−methanol mixture assuming chemical equilibrium. Previous research has generally focused on the combustion of a given fixed fuel composition, whereas in this article the composition of the fuel is varied by three different parameters: the amount of water in the aqueous methanol used as the synthesis gas feedstock, the amount of methanol in the gasoline−methanol mixture, and the ratio of synthesis gas energy to total fuel energy. The effects of these parameters on the adiabatic flame temperature and combustion product distribution are determined. The method used in this article allows the above parameters and the air/fuel ratio parameter to vary in equation sets derived using Lagrange undetermined multipliers in order to determine the combustion temperature and products with a computer solution. Stoichiometric combustion of synthesis gas reformed out of an equimolar water−methanol mixture causes a decrease in the NO ratio by 20% compared with methanol and 40% compared with gasoline under equilibrium conditions, whereas the adiabatic flame temperature decreases by only 50 and 100 K, respectively. Stoichiometric combustion of synthesis gas reformed out of neat methanol causes a higher adiabatic flame temperature and higher NOx emissions than stoichiometric combustion of gasoline or methanol. However, neat-methanolbased synthesis gas has a higher power density and extends the lean flammability limit of the engine more efficiently than aqueous-methanol-based synthesis gas because the reforming enthalpy of neat methanol is higher than that of aqueous methanol. Hydrogen-rich synthesis gas fuels should be combusted with lean air/fuel ratios at part load to increase the thermal efficiency, combustion stability, and control emissions simultaneously. If the road load increases, the engine charge’s energy density and motor octane number will be increased by substitution of synthesis gas with liquid fuel while decreasing the air/fuel ratio simultaneously.

1. INTRODUCTION An alternative to three-way catalytic converters, the current standard for emission reduction in spark ignition (SI) engines, is the modification of the initial combustion process by using lean mixtures.1 Moreover, stable combustion of leaner mixtures decreases the engine charge’s energy density for an efficient operation at part load without throttling the flow of engine charge. Hydrogen’s wide flammability limit and high flame speed allow ultralean combustion, reducing exhaust emissions. The exhaust emissions of an SI engine were shown to be reduced by addition of pure hydrogen to gasoline.2 However, on a commercial level, hydrogen stored on board poses a number of problems, such as hydrogen infrastructure, mechanical problems involved with refueling, limited driving range, high weight involved with the size of the storage cylinder, and safety concerns related to carrying a high-pressure cylinder of hydrogen.3 Instead of pure hydrogen storage on board, partial oxidation of gasoline with air generates H2-rich synthesis gas (syngas), which when added to gasoline reduces engine exhaust emissions as well.4 However, partial oxidation is not considered to be attractive in terms of efficiency because the product gas has a lower calorific value than the original feedstock as a result of the exothermic reaction.1 The exhaust gases of SI engines for automotive propulsion leave the engine at a temperature between 900 and 1000 K. © 2014 American Chemical Society

Although these exhaust gases possess substantial availability, this energy is not recovered for further use in state-of-the-art automotive technology. The resulting exhaust heat could increase the energy level of a fuel by reforming the fuel to a hydrogen-rich synthesis gas. The prerequisite is that the fuel’s reforming reaction take place at a temperature lower than the exhaust gas temperature. The lower the temperature of the reforming reaction, the higher will be the reactor’s heat recovery and syngas output. Compared with alkanes and higher alcohols, methanol has a higher oxygen-to-carbon ratio, allowing the catalytic decomposition of the neat fuel vapor to CO and H2 at relatively low temperatures. Methanol’s steam reforming reaction causes a less reducing reaction environment that retards coke formation compared with other commercial hydrocarbons. The resulting lower propensity to form coke preserves the catalyst from deactivation due to solid carbon formation and thereby prevents a premature reactor shutdown.5 Moreover, steam reforming of methanol consumes less energy to produce 1 mole of hydrogen compared with commercial fuels such as compressed natural gas (CNG), liquefied petroleum gas (LPG), gasoline, and ethanol. Given the same exhaust gas amount and temperature, the lower energy requirement of a Received: November 7, 2013 Revised: January 19, 2014 Published: February 11, 2014 2076

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not stored in a separate tank onboard, such an ethanol solution cannot be combusted with air in the engine efficiently. When the temperature difference between the SI engine exhaust gas temperature and the ethanol reforming temperature is considered, the reaction enthalpy for complete ethanol steam reforming to H2 and CO2 (ΔHr = 173.46 kJ/mol) is too high to be recovered out of exhaust gas.6 Low-temperature steam reforming of an equimolar ethanol solution to a mixture of H2, CH4, and CO2 is a promising reaction pathway for ethanol steam reforming via exhaust gas recovery. A copper-plated Raney nickel catalyst is stable and active enough to reform ethanol below 300 °C.17 Neat ethanol or gasoline-blended ethanol (E85) can be decomposed to CO, H2, and CH4 without water in the fuel using the same catalyst, and the liquid fuel containing no water can be used as a direct fuel too.18 Although the H2 selectivity of low-temperature ethanol decomposition is not as high as that in methanol decomposition, its reaction enthalpy (ΔHr = 49.1 kJ/mol) is lower. The main obstacle to the use of ethanol as a sustainable engine fuel is its higher cost of commercial production compared with methanol. The opportunity cost of renewable ethanol production is higher since it requires input such as corn or sugar cane, which are in high demand as human food reserves. On the other hand, methanol can be produced from widely available agricultural, forest, and municipal waste as a renewable feedstock that can be mixed with lower-cost inputs such as coal and natural gas as fossil feedstocks.19 If the water-to-methanol molar ratio (w) in the methanol solution is less than unity, one can assume that all of the steam is reduced to hydrogen by methanol according to eq 4. The rest of the methanol not oxidized by steam decomposes to CO and H2 according to eq 1 if there is enough heat for 100% conversion. The dilution of methanol with water increases the hydrogen selectivity and catalyst stability while decreasing the overall reaction enthalpy, product gas energy density, and methanol dehydration selectivity to higher hydrocarbons. If reactions other than steam reforming and decomposition are neglected,20 the product gas composition as a function of w (for 0 < w < 1) can be calculated as w·(eq 4) + (1 − w)·(eq 1), leading to eq 5a:

reformer increases the output of a reactor in contact with an exhaust gas heat exchanger.6 Besides steam reforming, methanol can also be decomposed to syngas for use as a motor fuel in order to reduce automobile emissions.7 The endothermic decomposition of methanol (eq 1) has been demonstrated to produce H2-rich syngas as an engine fuel.8,9 The decomposition products of methanol enable an SI engine to run with an increased compression ratio (14:1), increasing the thermal efficiency.10 Several patents for generating hydrogen-rich syngas out of methanol to combust the syngas in an automotive engine have been published.11,12 CH3OH ⇄ CO + 2H 2

ΔHr1° = 90.36 kJ/mol

(1)

Cu/Cr-based catalysts are stable in methanol decomposition, but they are still gradually deactivated because the methanol decomposition products constitute a reducing reaction environment.13 The higher the amount of CO in the product gas, the higher is the propensity for carbon deposition to deactivate the catalyst because CO in the product gas can be reduced to carbon by hydrogen (eq 2) or even by itself (eq 3). Furthermore, these reductions to carbon are exothermic reactions, the selectivity for which is increased by the comparably lower decomposition temperature of methanol. CO + H 2 ⇄ C(s) + H 2O

ΔHr2° = −131.3 kJ/mol (2)

2CO ⇄ C(s) + CO2

ΔHr3° = −172.5 kJ/mol

(3)

Reforming of aqueous methanol solutions preserves the catalyst better than the decomposition of neat methanol because steam evaporated out of the solution can oxidize methanol directly to CO2 (eq 4). The presence of H2O and CO2 in the reaction environment shifts the thermodynamic equilibrium from carbon toward CO since both are products of the above reduction reactions. On the other hand, the potential for heat recovery is decreased because of the lower reaction enthalpy of steam reforming compared with the reaction enthalpy of the decomposition reaction. CH3OH + H 2O ⇄ CO2 + 3H 2

ΔHr4° = 49.19 kJ/mol (4)

Methanol can be reformed with steam to give a H2-rich gas at a temperature higher than 160 °C with an Al2O3-supported CuO/ZnO catalyst.14 If an equimolar steam−methanol mixture is reformed by a Cu/Cr catalyst (80 mol % Cu) at 235 °C and atmospheric pressure, the CO2 selectivity reaches 90%.3 This selectivity corresponds to molar concentrations of around 74% H2 and 1% CO, while the rest of the product gas is mainly CO2. H2-rich syngas can also be generated from aqueous ethanol with a Co catalyst supported on Al2O3 or SiO2 at 400 °C.15 MgAl2O4 is an alternative support to suppress ethanol dehydration to ethylene, which leads to catalyst deactivation due to coke deposition.16 Noble metals such as Ru, Pt, and Ir are added as promoters to increase catalyst stability and H2 selectivity. Complete conversion of ethanol requires reactor temperatures as high as 500 °C. Such a high reaction temperature challenges a reforming reactor heated with engine exhaust gas to produce the amount of syngas required to run the engine by syngas alone. The stoichiometry of ethanol steam reforming to H2 and CO2 requires 3 moles of water per mole of ethanol in the solution. This requirement decreases the energy density of the energy stored on board. If water required to reform ethanol is

CH3OH + w H 2O ⇄ wCO2 + (1 − w)CO + (2 + w)H 2 (5a)

Likewise, the overall reaction enthalpy corresponding to eq 5a can be calculated as w·ΔH°r4 + (1 − w)·ΔH°r1, leading to eq 5b: ° = (90.36 − 41.17 ·w) kJ/mol ΔHr5a

(5b)

The current work will illustrate the equilibrium products of the combustion of an evaporated liquid fuel mixture with aqueous-methanol-based syngas produced by the overall reaction in eq 5a. The liquid fuel consists of a variable methanol−gasoline mixture. The amount of water in the aqueous methanol solution varies the syngas composition according to eq 5a. The amount of hydrogen in the syngas increases with dilution of the aqueous methanol until aqueous methanol becomes stoichiometric (w = 1), where the hydrogen output is maximized. However, before the determination of the combustion products, the impact of syngas generation on the system efficiency and the energy density of the fresh charge must be estimated. Methanol’s calorific value and heat of vaporization are 676.2 and 37.9 kJ/mol, respectively. Because methanol is stored in 2077

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liquid form onboard, efficiency calculations must be based on liquid methanol’s calorific value of 638.3 (=676.2 − 37.9) kJ/ mol. For each mole of methanol reacted, the approximate amounts of liquid methanol energy contained in 2 moles of H2 and 1 mole of CO are 76% (=2 × 241.8/638.3) and 44% (=283/638.3), respectively.21 The resulting heat recovery via the endothermic decomposition reaction (eq 1) (90.36/638.3 = 14%) and vaporization of liquid methanol (37.9/638.3 = 6%) is 20% of the calorific value of liquid methanol. While the decomposition reaction enthalpy can only be recovered from exhaust heat, the vaporization enthalpy can also be recovered from the engine coolant because methanol’s boiling point (64.7 °C) is lower than the standard coolant temperature (90 °C). If stoichiometric aqueous methanol (w = 1) is reformed to syngas according to eq 5a, the product gas contains approximately 114% (=3 × 241.8/638.3) of the liquid methanol energy, resulting in a 14% higher calorific value. The heat recovery via the endothermic steam reforming reaction accounts for only 8% (=49.2/638.3) of liquid methanol’s calorific value. Because neither CO2 nor H2O have calorific value, they are not taken into account. Even though heat recovery due to vaporization remains unchanged (at 6%) with respect to w, vaporization of water in aqueous methanol requires additional heat input. Furthermore, the boiling point of an equimolar aqueous methanol solution (73.5 °C) is higher than that of neat methanol but still less than the standard engine coolant temperature. The theoretical reactor efficiency of neat-methanol-based syngas appears to be 7.4% higher than that of aqueous-methanol-based syngas. However, the production of neat-methanol-based syngas challenges the catalyst stability much more than the production of aqueous-methanol-based syngas. Engines fueled with a H2−CO mixture have good efficiency at low indicated mean effective pressure (IMEP) equivalent to engine torque. However, according to engine tests, a H2−CO mixture could not achieve the same maximum IMEP compared with engines fueled with gasoline or liquid methanol because the energy density of the gaseous reformate is lower than that of the liquid fuel itself.22 The system efficiency is increased at the cost of a decrease in maximum power. Although the power output of fuels with lower energy density can be improved by increasing the engine compression ratio, this alternative is limited by the combustion stability, which depends on a fuel’s motor octane number. The energy densities of different fuel options are quantified by the mixture heat value (HG), which represents the energy density of an engine’s fresh charge. The higher the mixture heat value, the higher is the maximum power produced by the engine. The mixture heat value is defined as the ratio of the engine charge’s total energy (Qt) to the engine’s displacement volume (Vd) (eq 6). The mixture heat value of a neat fuel (in units of kJ/L) can be determined from the fresh charge’s molar density (ρ) after intake valve closure (IVC), the relative air/fuel ratio (λ), the mean lower heating value (LHV), and the stoichiometric oxygen (ost), as shown in eq 6.19 These parameters will be defined and determined for any gaseous fuel mixture in the next section. HG =

Qt Vd

= ρ·

LHV [λ ·(ost /0.21) + 1]

The mean lower heating values of neat-methanol-based and stoichiometric aqueous-methanol-based syngas are equal to 255.5 [=(2 × 241.8 + 283)/3] and 181.4 (=3 × 241.8/4) kJ/ mol of fuel, respectively. The denominator of the expression for HG represents the specific molar amount of the engine charge based on 1 mole of fuel inducted (denoted below as n0T). In order to determine this amount, the stoichiometric oxygen is divided by the mole fraction of oxygen in the atmosphere (0.21), which gives the stoichiometric air. Its product with λ provides the amount of air inducted into the engine. To this amount of air, unity is added in order to account for the amount of evaporated fuel. The specific molar amounts of neatmethanol-based and stoichiometric (w = 1) aqueous-methanolbased syngas at λ = 1 are equal to 3.38 and 2.79 mol of gas/mol of fuel, respectively. The ratio of the mean LHV to the specific molar amount is the calorific value of 1 mole of air−fuel mixture, which is equal to 75.6 and 65 kJ/mol for neatmethanol-based and stoichiometric aqueous-methanol-based syngas, respectively. The standard mixture heat value (HG° ) is defined by assuming that the cylinder state at IVC is equal to the standard state (p° = 1 atm, T° = 298.15 K) and replacing the molar density at IVC (ρ) with the standard molar density (ρ° = 0.041 mol/L): HG° = ρ° ·

LHV [λ ·(ost /0.21) + 1]

The use of ρ° is a fair estimate, as the fresh charge is not throttled or supercharged. The formula for H°G no longer contains engine-specific data and enables comparison between different fuel options. Evaluation of this formula for a stoichiometric air−fuel mixture results in standard mixture heat values of 3.45 kJ/L for isooctane and 3.39 kJ/L for methanol, while the standard mixture heat values of neatmethanol-based and stoichiometric aqueous-methanol-based syngas decrease to 3.09 and 2.66 kJ/L, respectively. When λ is doubled to form a lean mixture, the decrease in the standard mixture heat value is much more pronounced for liquid fuels than for syngas. At λ = 2, HG° drops to 1.74 kJ/L for isooctane, 1.81 kJ/L for both methanol and neat-methanolbased syngas, and 1.62 kJ/L for stoichiometric aqueousmethanol-based syngas. When λ is increased further, the values of HG° for hydrogen-rich gaseous fuels decrease further but remain higher than that of gasoline or methanol at the same λ. This shows that the maximum power produced by syngas may exceed those of liquid fuels when lean (λ > 2) air−fuel mixtures are combusted.20 If the engine’s kinetic parameters are adjusted properly, the peak cylinder pressure during combustion represents the engine’s maximum power output. In an experimental analysis of an SI engine working at 1800 rpm and an absolute air pressure of 61.5 kPa, λ was increased from stoichiometric to 1.36. The peak cylinder pressure decreased by 24.9% for combustion of lean gasoline and by 18.4% for combustion of 2.5% ethanol-based syngas-blended gasoline, while the indicated thermal efficiency increased from 36% to 38.6% for lean gasoline and to 40% for lean syngas-blended combustion.23 The peak engine brake thermal efficiency of an SI engine at 1400 rpm and an absolute air pressure of 61.5 kPa increased from 30.55% at λ = 1.13 to 31.95% at λ = 1.25 when pure H2 was blended with methanol at a volume fraction of 3%. The brake mean effective pressure was increased by about 19.3% at λ = 1.39 for the same amount of H2 blending.24 Besides increasing the maximum engine brake thermal efficiency and

(6) 2078

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eq 6 and rp in eq 7. Substituting Qp + Qd for Qt in eq 7 and rearranging gives eq 8, which shows that Qd is determined by rp and Qp.

brake mean effective pressure, blending of methanol with H2 also shifted the maximum engine brake thermal efficiency to higher λ. The higher efficiency and higher engine peak pressure in lean combustion of syngas-blended gasoline and hydrogen-blended methanol can be explained by the higher flame and diffusion speed of CO and H2 compared with gasoline and methanol. The higher flame and diffusion speed initiate more complete combustion of the air−fuel mixture in the engine because CO and H2 stimulate the formation of O and OH radicals, improving the chain reaction in the fuel combustion process.25 High engine pressure may cause combustion instabilities during combustion of fuels with higher syngas or H2 blending due to a higher radical concentration. Because of the high flame speed, very low requirement of ignition energy, and small quenching distance of hydrogen, attempts to burn either hydrogen-enriched fuels or hydrogen itself are plagued by abnormal combustion of hydrogen in the form of preignition, backfiring, and knocking. These problems are exacerbated particularly when the relative air/fuel ratio is dropped to stoichiometric in order to increase the engine power.1 Furthermore, because of the high flame speed of H2, its addition tends to increase the combustion temperature, which increases NO emissions under stoichiometric SI combustion.26 On the other hand, the NO, HC, and CO emissions of an idling SI engine at 790 rpm all decreased when λ was increased from stoichiometric to 1.3 while H2 was blended with gasoline at a volume fraction of 3% simultaneously.27 Hydrogen-rich syngas mixtures may be more appropriate in lean burn applications where the combustion temperature is moderated by excess air. Such mixtures could also serve in “dual-fuel” engines that operate under compression ignition using a pilot injection of diesel fuel.28 Because syngas is generated from methanol, a dual-fuel SI engine operated with methanol and methanol-based syngas does not require storage of two different fuels. The cold start of an SI engine with methanol is very difficult to perform below 5 °C because of difficulties in preparing an ignitable gaseous air−fuel mixture due to the higher boiling point and heat of vaporization of methanol compared with gasoline.21 The production of decomposed methanol onboard the vehicle for use as a cold-start fuel is a possible solution. After heat-up of the engine and reformer is complete, an appropriate amount of produced gas can be stored in a small, moderately pressurized metal flask for the next few cold starts.29 The method of starting an engine with a battery is based on a similar strategy of storing energy from a previous steady run to be used in the next transient cold start. Mixing of liquid fuel and hydrogen-rich syngas is also a strategy to reduce emissions of diesel engines. Exhaust gas fuel reforming in diesel engines has been studied experimentally as a way to assist premixed charge compression ignition (PCCI) operation. By substitution of part of the main fuel with hydrogen-rich gas, NOx and smoke emissions can be reduced simultaneously.30 The premixed fuel ratio (rp) is the ratio of the calorific energy of the premixed fuel (syngas) (Qp) to the total fuel energy of the fresh charge (eq 7):31 rp =

Q d = Q p·

rp

(8)

Mixing of liquid fuel and methanol-based syngas can alter the power produced according to road load. The air/fuel ratio must be decreased to form a near-stoichiometric mixture for a proper load response during increasing road load. The syngas amount has to be substituted by liquid methanol or gasoline simultaneously to approach the maximum engine power produced with liquid methanol or gasoline only. Liquid fuel can be injected into the intake port or into the engine to increase the mixture heat value. In addition to power adjustment, substituting methanol or gasoline in place of syngas at higher loads also results in more stable combustion thanks to the higher motor octane numbers of both methanol and gasoline compared with hydrogen. Methanol increases the mixture’s average motor octane number, leading to stable combustion of a near-stoichiometric hydrogen−methanol−air mixture. Also, addition of alcohol blends to gasoline boosts the octane number while reducing CO and NOx emissions.32 To decrease the engine power, liquid fuel must be partially substituted with syngas. Simultaneously, the air/fuel ratio must be increased and adapted to the premixed fuel ratio to prevent engine knock. The power can be further decreased if the syngas amount is also throttled, until the lean flammability limit of hydrogen is reached. This keeps the manifold pressure constant to run the engine unthrottled with hydrogen, regulating the power output by varying the syngas flow and thereby excess air.33 An onboard decomposed methanol engine can be operated with a wide-open throttle over a large torque range at a practical maximum air excess of around 4.21 This method of power regulation eliminates pumping losses of an SI engine, although a homogeneous charge is combusted in the engine.

2. DETERMINATION OF COMBUSTION PRODUCTS AT CHEMICAL EQUILIBRIUM The theoretical complete combustion of an oxygenated fuel with excess air (λ > 1) is shown in eq 9. The relative air/fuel ratio must be higher than unity to minimize the amount of incomplete combustion products. Cc HhOo + ost ·λ(O2 + 3.76N2) h → cCO2 + H 2O + ost [(λ − 1)O2 + λ ·3.76N2] 2

(9)

The stoichiometric amount of O2 required for theoretical complete combustion of an oxygenated hydrocarbon fuel to CO2 and H2O (ost, in units of mol of O2/mol of fuel) is determined in eq 10 as a function of the fuel’s chemical composition. The amount of oxygen inducted into the engine with fresh charge is the actual oxygen (O2,in), which is equal to the product of ost and λ. The excess oxygen and unreacted nitrogen are added to the amount of CO2 and H2O to determine the total amount of complete combustion products (nC, in units of mol of gas/mol of fuel).

Qp Qt

1 − rp

(7)

The sum of Qp and the direct fuel energy (Qd) is equal to the total fuel energy Qt used to determine the mixture heat value in 2079

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Table 1. Enthalpies and Free Enthalpies of Formation and Polynomial Coefficients for the Heat Capacity H O OH N NO H2 H2O CO CO2 O2 N2 C8H18 CH3OH C

Δhf° (J/mol)

Δgf° (J/mol)

A

103·B (K−1)

106·C (K−2)

10−5·D (K2)

218000 249170 39460 472680 90291 0 −241818 −110525 −393509 0 0 −208750 −200660 0

203290 231770 34280 455510 86550 0 −228572 −137169 −394359 0 0 16260 −161960 0

2.50 2.52 2.88 2.52 3.48 2.97 3.00 3.28 5.45 3.79 3.17 4.11 2.21 1.77

0 −0.026 0.891 −0.021 0.784 0.696 2.299 0.893 1.530 0.513 0.946 70.57 12.22 0.771

0 0.0076 −0.127 0.0069 −0.154 −0.065 −0.361 −0.173 −0.295 −0.062 −0.179 −22.21 −3.450 0

0 0.108 0.556 −0.017 −0.308 0.440 0.397 −0.189 −1.728 −0.602 −0.048 0 0 −0.867

ost = c +

h o − 4 2

nC = c +

h + (4.76·λ − 1) ·ost 2

enough and the temperature of the burnt mixture is high enough to satisfy the ideal gas relationship. For an ideal gas i, the molar free enthalpy (gi) is equal to the chemical potential (μi) of the reacting ideal gas i. The molar free enthalpy of reacting gas i at standard pressure, g°i (T), is determined from the standard molar enthalpy of formation (Δhf,i° ), the standard molar entropy (sT° ° ), the change in molar enthalpy Δhi(T), and the change in molar entropy Δsi(T) as a function of the reaction temperature T (eq 14):

(10)

The standard reaction enthalpy of combustion is determined in eq 11 in terms of the standard formation enthalpies of the fuel (Δhf,CHO ° ), CO2 (Δh°f,CO2) and H2O (Δh°f,H2O). The absolute value of the standard combustion enthalpy (ΔH°c ) is the lower heating value (LHV) of the fuel, which can be inserted into eq 6 to determine the mixture heat value. ΔHc◦

=

◦ c·Δhf,CO 2

gi°(T ) = Δh°f , i + Δhi(T ) − T ·[sT°°, i + Δsi(T )]

The standard molar entropy (s°T°) is the ratio of the difference between the standard molar enthalpy of formation (Δhf°) and the standard molar free enthalpy of formation (Δgf°) to the standard temperature (T°), as shown in eq 15, where the changes in molar enthalpy and entropy are also defined.

h ◦ ◦ + ·Δhf,H − Δhf,CHO 2O 2

LHV = |ΔHc◦|

(11)

2.1. Fundamentals of the Chemical Equilibrium Calculation. The assumption of chemical equilibrium to determine the composition of the combustion products is a good approximation for performance estimates in engines. During combustion at elevated temperatures, intermediate products such as H2, CO, and NO and radicals such as H, N, OH, and O are formed along with the complete combustion products in eq 9. The products of the reaction in eq 12 are the major species formed during combustion assuming chemical equilibrium.

sT°°, i =

Δhi(T ) =

∫T ° Cp,i dTi

Δsi(T ) =

∫T °

Cp , i Ti

dTi

(15)

The molar heat capacity at constant pressure (Cp), which is required for the determination of the molar enthalpy and entropy changes, can be assumed to be constant for small temperature differences. However, a polynomial function such as in eq 16, in which R is the universal gas constant (8.3145 J mol−1 K−1), can be used to estimate the heat capacity as a function of the reaction temperature because the temperature change during combustion is very high.35

+ n H2 H 2 + nH H + nN N + nOO + nOHOH (12)

Cp , i(T ) = R ·(Ai + Bi ·T + Ci·T 2 + Di ·T −2)

The products in eq 12 are formed by means of the simultaneous multiple reactions depicted in eq 13:

T

⇒ Δhi(T ) =

1 1 CO + O2 ⇄ CO2 , H 2 + O2 ⇄ H 2O 2 2 2H ⇄ H 2 , 2O ⇄ O2 , H + OH ⇄ H 2O N2 + O2 ⇄ 2NO

T°

T

⇄ nCO2CO2 + n H2OH 2O + n N2 N2 + nO2O2 + nCOCO

2N ⇄ N2 ,

Δhf,°i − Δg f,°i T

Cc HhOo + ominλ(O2 + 3.76N2)

+ nNO NO

(14)

∫T ° Cp,i(Ti) dTi

(16)

Gasoline is a mixture of several hydrocarbons with high knock resistance. In this article, gasoline is idealized and represented by isooctane. The coefficients A, B, C, and D for carbon, methanol, and isooctane (Table 1) were obtained from several sources36−38 and are valid below 1500 K. Although the combustion temperature is much higher than 1500 K, these fuels are oxidized or dissociated before the combustion

(13)

The gas species that make up the working fluids in internal combustion engines can usually be treated as ideal gases.34 The partial pressures of the gases in the unburnt mixture are low 2080

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temperature is reached. The polynomial coefficients of the other gases are applied to the same polynomial type and were obtained by least-squares optimization of the gases’ enthalpy values from a different source.39 Values of Δhf° and Δgf° for the products and reactants during combustion are listed in Table 1.35 For each variable listed in the column headings of Table 1, the corresponding values for the gases were collected into a column vector. Instead of entering a scalar value for a single gas into any of the formulas presented above and doing the same calculation for each gas, the column vector was entered into the formulas embedded in a code written with Mathematica. The equations of mass conservation, equilibrium, and energy were determined by parameters and algebraic formulas entered into a code that determines the product distribution numerically. The mass conservation during combustion can be represented by the conservation matrix (A) depicted in Table 2. Each gas i present among the combustion products is

f g = {0, 0, 0, 0, 0, x H2 , 0, xCO , xCO2 , 0, 0, 0, 0, 0}T xCO =

1−w , 3+w

xCO2 =

w , 3+w

x H2 =

2+w 3+w (19)

The premixed fuel vector fp is the product of the premixed fuel’s mole fraction xp (0 ≤ xp ≤ 1) and the product gas vector fg (eq 20):

f p = x p· f g

(20)

The liquid fuel is composed of isooctane and methanol. The alcohol content of the liquid fuel stored in the fuel tank (0 ≤ al ≤ 1) determines the liquid fuel vector (fl) (eq 21): f l = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, al, 1 − al, 0}T

(21)

The product of the direct fuel’s mole fraction (xd = 1 − xp) and the liquid fuel vector determines the direct fuel vector fd (eq 22):

Table 2. Conservation Matrix

fd = (1 − xp) ·f l

(22)

After fp from eq 20 and fd from eq 22 are substituted into eq 18, the fuel vector f becomes a linear combination of the product gas vector and the liquid fuel vector (eq 23): f = xp·f g + (1 − xp) ·f l

The sum of the components of the fuel vector must be unity because all of the components of the reactant and product vectors are based on 1 mole of fuel. The unity condition is satisfied because the sums of the components of fl and fg equal unity, too.

represented by a column of the matrix. Each entry aki shows the number of atoms of the element listed in row k (C, H, O, N) in the formula of gas i. Mass conservation provides one equation for the conservation of each element. The four conservation equations of the elements C, H, O, and N are derived by setting the matrix product of the conservation matrix (A) and the reactant vector (n0) equal to the matrix product of A and the product vector (n). The reactant vector can be determined as a function of parameters such as the relative air/fuel ratio (λ), the water content of the methanol solution (w), the premixed fuel ratio (rp), and the alcohol content of the direct fuel (al). The product vector n is the solution of chemical equilibrium calculation and satisfies eq 17.

An 0 = An

n

∑ fg,i i=1

n

=1∧

n

∑ fl,i

=1⇒

i=1

∑ fi

=1

i=1

The mole fraction of the premixed fuel (xp) must be determined as a function of the premixed fuel ratio rp. The LHVs of the direct and premixed fuel mixtures must be calculated to determine xp from rp. The LHV of an oxygenated fuel is given by eq 11. The numbers of atoms of carbon (c) and hydrogen (h) in the formula of each gas are listed in the first row (A[1]) and second row (A[2]) of A, respectively. The standard formation enthalpy vector (Δh°f ) is shown in the first column of Table 1. The standard combustion enthalpy vector (ΔHc°) is determined by inserting the carbon and hydrogen row vectors of A into eq 11, as shown in eq 24:

(17)

◦ ΔH◦c = A[1]·Δhf,CO + 2

2.2. Determination of the Reactant Vector. The reactant vector is composed of the fuel vector (f) and the air vector (a). The fuel vector affects the air vector according to the complete combustion reaction in eq 9. The fuel is composed of gaseous premixed fuel and liquid direct fuel. Consequently, the fuel vector is represented as the vector sum of the premixed fuel vector (fp) and the direct fuel vector (fd) (eq 18). Both fp and fd must be determined to calculate the fuel and air vectors. f = f p + fd

(23)

1 ◦ A[2]·Δhf,H − Δh◦f T 2O 2

(24)

The lower heating value vector (LHV) is the negative of the standard combustion enthalpy vector because the LHV is the absolute value of the standard combustion enthalpy and the latter quantity is always negative: LHV = |ΔHc◦| ∧ ΔHc◦ < 0 ⇒ LHV = −ΔH◦c

(25)

The dot products of the lower heating value vector with the premixed fuel vector, direct fuel vector, and fuel vector determine the premixed fuel, direct fuel, and total fuel energies, respectively, based on 1 mole of fuel inducted. Multiplying these scalar products by the molar amount of the fuel inducted into the engine until IVC (nf) determines the corresponding energies Qp, Qd, and Qt consumed during an engine cycle (eq 26):

(18)

The premixed fuel is a product gas generated by the reaction of methanol with water and is composed of CO, CO2, and H2. The product gas vector (fg) (eq 19) is determined by the stoichiometric coefficients of the reaction in eq 5a: 2081

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Scheme 1. Flow Diagram of the Proposed Model

Q p = (LHV·f p) ·n f Q d = (LHV·fd) ·n f Q t = Q p + Q d = (LHV·f) ·n f

(26)

The definition of the premixed fuel ratio in eq 7 is used to derive a relationship between rp and xp. This determines xp as a function of rp and the mean LHV ratio of liquid to gaseous fuel (d), as defined in eq 27. The parameter d is a function of w and al. The value of xp determined using eq 27 can be inserted back into eq 23 to evaluate f. rp =

Qp Qt

=

LHV·f p LHV·f

where d =

⇒ xp =

rp·d 1 + rp·(d − 1)

,

LHVl LHV·f l = = f (w , al) LHVg LHV·f g

(27)

mixture. Although the diagram shows two separate tanks containing the premixed fuel and direct fuel, both fuels may also be supplied from the same source. The premixed fuel may be generated from pure methanol or a mixture of aqueous methanol and gasoline. If gasoline is in the premixed fuel, the above calculation must consider the amount of unreacted gasoline leaving the reformer. The storage of premixed and direct fuel in separate tanks in Scheme 1 should demonstrate the effects of the amount of methanol in the direct fuel and the amount of water in the premixed fuel independently. Equation 33 shows how the mixture heat value HG can be evaluated for an arbitrary fuel mixture by inserting Qt from eq 26 into eq 6 and replacing Vd in eq 6 by n0T/ρ, since the molar density is given by ρ = n0T/Vd:

The stoichiometric oxygen amount for the combustion of an oxygenated fuel is determined in eq 10. The coefficients of C, H, and O constitute a vector. Its product with the conservation matrix generates the stoichiometric oxygen vector (ost) (eq 28). If a component of ost is negative, the corresponding gas is an oxygen source. ost =

{1, 14 , − 12 , 0}A

(28)

The stoichiometric amount of oxygen required for the combustion of the fuel mixture (O2,st) is the dot product of ost and f (eq 29): O2,st = ost·f

(29)

The actual amount of oxygen inducted (O2,in) is the product of the relative air/fuel ratio (λ) and the fuel mixture’s stoichiometric oxygen amount (O2,st). The amount of nitrogen inducted into the engine (N2,in) is determined according to its fraction in the atmosphere: O2,in = λ ·O2,st ,

N2,in =

0.79 ·O2,in 0.21

HG =

HG° = ρ° ·

(30)

(31)

(32)

For a more exact calculation, the amount of exhaust gas left in the cylinder from the previous cycle could be added to the sum. The sum of the reactant vector components is equal to the molar amount of the fresh engine charge (n0T) which is also the denominator of eq 6.

LHV·f n0T

(33)

LHV·f n0T

Table 3. Product and Reactant Vectors Setting up the Conservation Equation System

14

n0T =

= ρ·

If the direct fuel is injected into the engine in the liquid phase, the volume of the liquid fuel can be neglected, and the control variable i will not include the fuel as a component in the reactant vector (i.e., i ≤ 11). 2.3. Determination of the Product Vector. The reactant vector in eq 32 can be inserted into eq 17. The components of n0 are shown explicitly in Table 3. The product vector (n) shown in Table 3 is substituted by the product of the total molar amount of combustion products (nT) and the mole fraction vector (y). The components of the mole fraction vector are the mole fractions of the combustion products (yi). The values of yi are between 0 and 1, which are easier to

The reactant vector (n0) is the sum of the fuel and air vectors (eq 32):

n0 = f + a

Vd

As in the case of eq 6, the mixture heat value can be based on the standard state by replacing the molar density ρ with the standard molar density ρ°:

The air vector based on 1 mole of fuel (a) contains only O2 and N2 as components (eq 31): a = {0, 0, 0, 0, 0, 0, 0, 0, 0, O2,in , N2,in , 0, 0, 0}T

Qt

∑ n0,i i=1

Scheme 1 shows a flow diagram of the model presented in this work. The vectors defined above are represented as gas streams in this diagram. The mixing of liquid fuel with hot reformate evaporates the liquid fuel, cools the hot reformate, and facilitates the formation of a more homogeneous fuel 2082

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reaction pressure. The equilibrium constants can be determined as functions of temperature using the least-squares method.40 The seven reactions in eq 13 provide seven equations of the above type. In addition, the five conservation equations in Table 4 are added to the equation system. The eleven components of the mole fraction vector and nT in Table 4 represent a total of 12 unknowns in total. Therefore, the equation system can be solved numerically for a given reaction temperature.41 However, the solution for multiple reactions does not lend itself to standardization that allows a computer solution, especially if the reaction temperature is to be determined by adding the energy equation to the equation system. An alternative equilibrium criterion applied to multiple reactions is the minimization of the system’s total free enthalpy. This method offers a standard solution procedure based on the method of Lagrange undetermined multipliers.42 This criterion has hardly been applied to combustion calculations where the fuel properties vary with the premixed fuel ratio and the liquid fuel composition. Moreover, the number of different gases assumed to be present in the burning mixture can be increased for a more realistic equilibrium calculation. Instead of reactions, this method assumes only the presence of different gases in the reaction environment. The free enthalpy of formation for a gas (ΔG°f ) is the free enthalpy change for the reaction in which the gas is formed from its elements. The reactants of the formation reaction are the elemental substances from which the gas is formed (C, H2, O2, N2), and the product is the gas for which ΔG°f is to be determined. The free enthalpies of each gas and its constituent elements in the formation reaction must be determined according to eq 14. Instead of evaluating ΔGf° for each gas separately by inserting the free enthalpies into eq 35, the ΔG°f values of all of the gases are embodied in the vector of the system’s free enthalpy change (ΔGf°). The definition of ΔGf° in terms of the conservation matrix’s row vectors and the free enthalpies of the elements, an additional contribution of this work, is shown in eq 37:

calculate numerically compared with the components of n, which may be any positive real number. An additional equation must be set up to determine the additional unknown nT. This equation expresses the fact that the sum of the components of the mole fraction vector is unity. The fuels and carbon are assumed to be fully consumed, so their corresponding amounts (the last three components of the product and mole fraction vectors) are set equal to 0. The matrix multiplication in Table 3 is carried out in Table 4 for the case of pure isooctane without premixed fuel to derive the conservation equations for the elements C, H, O, and N. Table 4. Conservation Equation System for Combustion of Pure Isooctane (al = 0, rp = 0, xp = 0)

The initial value of nT in the numerical calculation is set equal to the molar amount of complete combustion products nC defined in eq 10. For the fuel mixture f defined in eq 18, this amount can be determined from the dot products of the row vectors A[1] and A[2] with f, which give the molar amounts of CO2 and H2O, respectively (eq 34). Excess oxygen and unreacted nitrogen are added to this sum of the dot products, as was done in eq 10. 1 nC = A[1]·f + A[2]·f + (λ − 1)·O2,in + N2,in (34) 2 The products of the combustion reaction in eq 12 are formed through multiple reactions that are assumed to be in equilibrium. The standard free enthalpy change for a reaction, ΔG°(T), is defined as the difference between the standard free enthalpies of the products and reactants weighted by their stoichiometric coefficients. If signed stoichiometric coefficients (νi negative for reactants and positive for products) are used, ΔG°(T) can be written as the sum of the standard free enthalpies weighted with νi (eq 35). For the calculation of the standard free enthalpies g°i (T) according to eq 14, each product and reactant is assumed to be a pure substance at the standardstate pressure and at the reaction temperature T.

ΔG°f (T ) = g °(T ) − gC°(T ) ·A[1]T − −

∑ νi·gi°(T ) i=1

(35)

The value of ΔG°(T) at the end-state temperature determines the equilibrium constant K(T) for that reaction (eq 36): K (T ) = e−ΔG°(T )/ RT

2

2

·A[3]T −

g N° (T ) 2

2

2

2

·A[2]T

·A[4]T

(37)

where g°(T) is the vector of standard free enthalpies of the gases. ΔGf° in eq 37 is evaluated as a column vector on the lefthand side of Table 5. The corresponding formation reactions

n

ΔGi°(T ) =

gO° (T )

g H° (T )

Table 5. Vector of the Free Enthalpies of Formation

(36)

The equilibrium constant depends on temperature and provides a relationship among the partial pressures (pi) of the gases, assuming that they are ideal: n

⎛ p ⎞∑i=1 νi K (T ) = ∏ ⎜ i ⎟ p° ⎠ i=1 ⎝ n

The partial pressures can be replaced with the corresponding mole fractions using the fact that pi = p·yi, where p is the 2083

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⎫ ⎪ dH = dU + p dV + V dp ⎬ ⇒ dH = dQ + V dp ⎪ dW = − p dV ⎭

are depicted on the right-hand side of the table. The components of the column vector generated by eq 37 reflect the free enthalpies of formation reactions of the gases according to eq 35. The free enthalpies of formation of C, H2, O2, and N2 are zero because they are elemental substances. The condition of chemical equilibrium in the method of the Lagrange undetermined multipliers is shown in eq 38. The derivation of this equation is shown in ref 35. The components λk of the vector of undetermined multipliers λ = (λC, λH, λO, λN) are the undetermined multipliers corresponding to the elements C, H, O, and N that constitute the gases in the reaction environment. The fuels and carbon are assumed to be fully consumed according to the product vector in Table 3, and therefore, the control variable i varies from 1 to 11. ⎛ p⎞ ΔGf,°i + RT ln⎜⎜yi · ⎟⎟ + ⎝ p° ⎠

dU = dQ + dW

adiabatic ⇒ dQ = 0 ⎫ ⎬ ⇒ dH = 0 ⇒ H0 = HT isobaric ⇒ V dp = 0 ⎭ ⎪

⎪

The initial enthalpy (H0) is the sum of the initial thermal enthalpy (n0·hT0) and the initial chemical enthalpy (n0·Δhf°), while the final enthalpy (HT) is the sum of the final thermal enthalpy (n·hT) and the final chemical enthalpy (n·Δh°f ) (eq 41). The molar thermal enthalpies are based on the standard temperature T° and can be calculated using the molar heat capacities at constant pressure (Cp). If the reactants have different initial temperatures, the initial temperature T0 must be substituted with an initial temperature vector T0.

4

∑ λk ·aik = 0 k=1

(38)

H0 = n 0·(Δh°f + h T0),

Equation 38 can be solved for the mole fractions yi of the combustion products (eq 39): yi =

p° −[ΔGf,°i + (λ A)i ]/ RT ·e p

(40)

HT = n·(Δh°f + h T),

h T0 =

T0

∫T °

Cp(T ′) dT ′

T

hT =

∫T ° Cp(T′) dT′

(41)

The energy expression in terms of mole fraction vectors (eq 42) is obtained by inserting the enthalpies in eq 41 into eq 40:

(39)

When the reaction temperature and pressure are inserted into eq 39, the undetermined multipliers will remain as the only unknown parameters. The mole fractions determined in eq 39 are inserted into the conservation equations in Table 3. After the resulting 5 × 5 equation system has been solved numerically for the four undetermined multipliers and the molar amount of combustion products nT, the undetermined multipliers are determined. These parameters can be inserted back into eq 39 to evaluate the mole fraction vector. The product of nT and y is equal to the product vector. The product vector components display the molar amounts of the combustion products based on 1 mole of fuel. 2.4. Determination of the Adiabatic Flame Temperature. The final temperature of the products in an adiabatic combustion process is called the adiabatic flame temperature. This is the reaction temperature T presumed for the calculation of the product vector. Constant-volume and constant-pressure combustion are two ideal processes of an adiabatic combustion. In reality, the pressure and the volume change simultaneously during combustion in the engine’s combustion chamber. The higher the flame speed of an air−fuel mixture, the faster the combustion is finished. Thus, a high flame speed indicates that a constant-volume process describes the adiabatic flame temperature better than a constant-pressure process because the change in volume is small during a fast combustion. Because most fuels have a finite flame speed and the combustion takes tens of crank angles, the real change of state during combustion is somewhere between constantpressure combustion and constant-volume combustion. 2.4.1. Adiabatic Flame Temperature for a ConstantPressure Process. For an adiabatic constant-pressure process of a steady-state open system, the enthalpy (H) remains constant. The initial state of the system is the state of the reactants, identified by the subscript 0. The final state of the system is the state of the products, identified by the subscript T.

n0T ·[y0·(Δh°f + h T0)] = nT ·[y·(Δh°f + h T)], 14

n0T =

∑ n 0i ,

y0 = n 0/n0T

i=1

(42)

After the mole fraction vector y in eq 39 is inserted into both eq 42 and the conservation equations derived in Table 3, eq 42 must be solved simultaneously with the conservation equations. The resulting 6 × 6 equation system can be solved for the four undetermined multipliers, the molar amount of the combustion products nT, and the reaction temperature T, which is equal to the adiabatic flame temperature at constant pressure (Tp). The undetermined multipliers and the flame temperature must be inserted back into eq 39 to determine the mole fraction and the product vectors. 2.4.2. Adiabatic Flame Temperature for a ConstantVolume Process. A constant-volume change of state is called an isochoric process. During an isochoric combustion, the volume of the combustion chamber (V) remains constant. Consequently, the reaction pressure of the isochoric process (pV) has to be determined with an equation of state. The ideal gas equation is an appropriate equation of state for internal combustion. Applying the ideal gas equation to the initial state (subscript 0) and the final state (subscript V) of the adiabatic isochoric combustion allows pV to be determined (eq 43). The reaction pressure depends on the flame temperature of the adiabatic isochoric combustion (Tv). p0 ·V = n0T ·R ·T0 , pV ·V = nT ·R ·TV ⇒ pV = p0 ·

TV nT · T0 n0T (43)

The final combustion pressure pV can be inserted into eq 39 instead of the pressure p of the constant-pressure process. The mole fraction vector for the constant-volume process is then determined as a function of undetermined multipliers, the molar amount of combustion products nT, and the adiabatic flame temperature TV (eq 44): 2084

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Figure 1. Isothermal mole fractions of combustion products (y) from (left) this work and (right) Heywood, 1988.34

yi =

p° T0·n0T −[ΔGf,°i + (λ A)i ]/ RTV · ·e p0 TV ·nT

energy equation for the constant volume process (eq 47). The calculation with internal energies is reduced to the calculation with enthalpies, which were defined in eq 42.

(44)

The internal energy (U) of a closed system remains constant during an adiabatic constant-volume process. The enthalpies in eq 42 are substituted with internal energies to determine the energy equation of a closed system’s constant volume process in eq 45.

n0T ·[y0·(Δh°f + h T0) − R ·T0] = nT ·[y·(Δh°f + h T) − R ·T ] ⇒ H0 − n0T ·R ·T0 = H − nT ·R ·T

dU = dQ + dW ⎫ ⎬ ⇒ dU = dQ − p dV dW = − p dV ⎭

The mole fraction vector of the constant volume process from eq 44 can be inserted into eq 47 and into the conservation equation system in Table 3. The conservation equations and eq 47 constitute a 6 × 6 equation set to be solved for the four undetermined multipliers, the isochoric adiabatic flame temperature, and nT simultaneously. The undetermined multipliers, TV, and nT must be inserted back into eq 44 to determine y and n.

⎪

⎪

adiabatic ⇒ dQ = 0 ⎫ ⎬ ⇒ dU = 0 ⇒ U0 = UT isochoric ⇒ pdV = 0 ⎭ ⎪ ⎪

U0 = UT ⇒ n 0·(Δu°f + u T0) = n·(Δu°f + u T)

(45)

The molar thermal internal energy (uT) can be determined by integrating the differential du = CV dT, where CV is the molar heat capacity at constant volume, which is equal to the difference between the molar heat capacity at constant pressure and the universal gas constant (i.e., CV = Cp − R). The subtraction of the scalar R from the vector Cp is mathematically not correct. Therefore, in eq 46 the scalar R is substituted with the product of the scalar R and the unit vector (I) in order to subtract R from each component of the vector. u T0 =

T0

∫T °

3. RESULTS The equation systems presented and the computer model written according to the method in this work were tested by comparing the results with results of a reference. The comparison can be easily illustrated by presenting the results of this work and the results from the reference in graphical form side by side. The reference analyzes combustion of isooctane. Following the side-by-side comparison, the effects of the premixed fuel ratio and the direct fuel’s alcohol content on both the flame temperature and combustion products were analyzed. The calculation of exhaust gas content depends also on how fast the fuel is burned by the flame front and how fast the gas cools during expansion. The flame speeds of different combustible gases must be taken into account. Nevertheless, the distribution of combustion products derived from an assumption of chemical equilibrium must also be given in order to determine the exhaust gas composition. 3.1. Verification of the Presented Model by Comparison to the Results of a Reference. The mole fractions of isooctane combustion products at a pressure of 30 atm and a temperature 2750 K are depicted as functions of equivalence

(Cp(T ′) − R ·I) dT ′

T

uT =

∫T ° (Cp(T′) − R·I) dT′

⇒u T0 = h T0 − R ·(T0 − T °) ·I u T = h T − R ·(T − T °) ·I Δu°f = Δh°f − R ·T °·I

(47)

(46)

Substitution of the vectors for the internal energy of formation (Δuf°) and the initial (uT0) and final (uT) molar thermal internal energies from eq 46 into eq 45 determines the 2085

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Figure 2. Adiabatic flame temperatures Tp and TV and the isochoric combustion pressure pV from (left) this work and (right) Heywood, 1988.34

Figure 3. Effect of the premixed fuel ratio on the adiabatic flame temperature for (left) λ = 1 and (right) λ = 2.

isochoric combustion and increased to pV according to the prevailing parameters. Figure 2 shows that an increase in the equivalence ratio from zero to unity, corresponding to a decrease in the air/fuel ratio from infinity to unity, increases the adiabatic flame temperatures because the amount of excess air decreases. Increasing ϕ further (ϕ > 1) decreases the flame temperatures because of incomplete combustion of the fuel. However, the maximum temperatures are reached when the air−fuel mixture is slightly richer than stoichiometric. Both Figure 1 and Figure 2 show that the differences between the results of this work (on the left-hand sides) and those of the reference (on the right-hand sides) are negligible. 3.2. Adiabatic Flame Temperatures. The adiabatic flame temperature significantly affects both outputs of the engine: performance and exhaust gases. Methanol’s and isooctane’s adiabatic flame temperatures for a constant pressure process are shown in Figure 3 as functions of the premixed fuel ratio. The left and right panels of Figure 3 are valid for relative air/fuel ratios of 1 (stoichiometric) and 2, respectively. In both of the charts in Figure 3, the amount of alcohol in the direct fuel is shown on the primary vertical axis, where the fuel consists of direct fuel only (rp = 0). M0 and M100 indicate direct fuels containing only isooctane and methanol,

ratio (ϕ = 1/λ) in Figure 1. In the right panel of Figure 1, the results of a reference34 using a model developed at NASA43 are shown. In the left panel of Figure 1, the results of the model presented here are shown. Table 4 shows the conservation equation system for this evaluation. The model in the reference assumed the radical N not to be present in the reaction environment. For the purpose of comparison, the product vector’s fourth component was set to 0 (yN = nN = 0) in our model. An increase in the equivalence ratio (i.e., a decrease in λ) promotes combustion products (H2, CO) and radicals (H) that contain elements coming from the fuel. On the other hand, a decrease in the equivalence ratio promotes combustion products and radicals that contain elements coming from the air (O, NO, O2). The adiabatic flame temperatures at constant pressure (Tp) and constant volume (TV) according to eqs 42 and 47 are shown in Figure 2. In addition, the combustion pressure at constant volume (pV) according to eq 43 is also shown. The results of this work (Figure 2 left) are almost identical to the results of the reference (Figure 2 right). The results are valid for isooctane without premixed fuel at an initial temperature of T0 = 700 K. The combustion pressure was 10 atm for isobaric combustion. This pressure was also the initial pressure of the 2086

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by drawing a line parallel to the line of the corresponding water amount of the charge temperature at 700 K. This line should begin at the methanol amount corresponding to the increased charge temperature (e.g., 900 K). The effect of the relative air/fuel ratio on the adiabatic flame temperature is shown in Figure 4. Except for the curves marked

respectively, while M85 indicates a fuel with a methanol mole fraction of 85%. The amount of water in the premixed fuel is shown on the secondary vertical axis in these charts. The secondary vertical axis is located at rp = 1, indicating that the fuel consists of premixed fuel only. The mole fraction of the premixed fuel (xp) is determined as a function of rp in eq 27. The use of xp instead of rp could have simplified the equilibrium calculation. However, the function containing rp is inserted into the fuel vector instead of xp. The adiabatic flame temperature varies linearly with respect to rp, and the resulting curves are lines connecting the two points on the primary and secondary vertical axis of the chart. Consequently, the choice of the parameter rp instead of xp is proven to be more appropriate for presenting the results. According to the Figure 3 (right), the adiabatic flame temperature of a lean mixture (λ = 2) always increases with respect to the premixed fuel ratio. The comparison of the two charts shows that the flame temperature decreases almost linearly with increasing water content in the premixed fuel because the amount of inert CO2 in the premixed fuel increases linearly with the increase in water amount. In contrast to the water content, the direct fuel’s alcohol content does not have a linear effect on the flame temperature. The flame temperature decreases with increasing alcohol content for near-stoichiometric mixtures and increases with increasing alcohol content for lean fuel mixtures. For a given near-stoichiometric fuel mixture, there exists a water content w of the premixed fuel for which the flame temperature remains constant with respect to the premixed fuel ratio. The resulting functions are horizontal lines representing flame isotherms at this water content. For a stoichiometric M85 air−fuel mixture the flame isotherm is at 2520 K, located at w = 0.612. In order to reach the maximum power of the engine, the air/ fuel ratio must be decreased to its minimum value above stoichiometric. When the air/fuel ratio is decreased toward stoichiometric, the amount of the premixed fuel must be decreased to prevent engine knock resulting from a low motor octane number of hydrogen in the premixed fuel. If the water amount of the premixed fuel corresponds to the one of the direct fuel’s flame isotherms, the adiabatic flame temperature remains constant with respect to the premixed fuel ratio when supplementing premixed fuel with direct fuel. Therefore, the flame isotherm could help define the water amount of the premixed fuel for a given alcohol amount of the direct fuel. While the piston moves to the top dead center, the flame propagates from the plugs toward the piston and cylinder walls, increasing the pressure of both the burnt gas and unburnt charge. While the flame penetrates into the unburnt charge, the charge temperature increases because of the rise in pressure and further compression of the piston. Because the initial temperature of the charge increases during flame propagation, it will have an impact on the adiabatic flame temperature as well. Therefore, the left chart in Figure 3 contains an additional group of results valid for a charge temperature of 900 K identified by this temperature. When the premixed fuel ratio of the M100 fuel mixture is varied from 0 to unity at a charge temperature of 900 K, the flame temperature increases only 3 K when w = 0.776. The M100 fuel mixture with w = 0.776 is a flame isotherm at 700 K. Consequently, the additional results for M100 at 900 K are almost parallel to the results of the same direct fuel at 700 K. The results for a different charge temperature can be estimated

Figure 4. Effect of the relative air/fuel ratio on the adiabatic flame temperature.

with M0 and M100, all of the results in the figure are valid for fuel mixtures consisting of indirect fuel only (rp = 1). The flame temperature decreases with respect to λ for any given parameter because the amount of inert N2 and excess O2 in the combustion products increases with λ. The flame temperature for constant-volume combustion is about 250 K higher than that for constant-pressure combustion. The y axis is situated at λ = 0.9 to show the parameters. Figure 4 also shows the effect of the charge temperature and pressure for λ = 1 and λ = 1.5. For these results, λ on the x axis is just a parameter increased from 1 to 1.8 while the air/fuel ratio becomes a constant parameter. T0 and p are multiplied by λ to vary T0 from 700 to 1260 K (=1.8 × 700 K) and p from 10 to 18 atm (=1.8 × 10 atm) while holding w equal to unity. The increase in the charge pressure from 10 to 18 atm has only a marginal effect on the flame temperature. The adiabatic flame temperature increases almost linearly from 2431 to 2722 K for λ = 1 when the charge temperature is increased from 700 to 1260 K. The linearity of the flame temperature with respect to the charge temperature explains how the results for T0 = 900 K in Figure 3 (left) can be deduced from the results for T0 = 700 K. The results for different charge temperatures can be derived by drawing lines parallel to that for the corresponding water amount of the premixed fuel at 700 K because the adiabatic flame temperature increases linearly with respect to T0, w, and rp. Figure 5 shows the amounts of combustion products as a function of rp while the parameters w, λ, al, p, and T0 are held constant. The combustion products NO (Figure 5 left) and CO (Figure 5 right) decrease almost linearly with increase in the premixed fuel’s water amount, similar to the adiabatic flame temperature. The water in the premixed fuel is reduced to CO2 during reforming of methanol to H2 in the reactor. CO2 is an inert gas and decreases the adiabatic flame temperature and the amounts of both NO and CO indirectly. The shapes of the two charts are similar to that for the adiabatic flame temperature, showing that the flame temperature has a stronger effect on the combustion products compared with other parameters. 2087

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Figure 5. Effect of the premixed fuel ratio on the isobaric adiabatic combustion products for (left) NO and (right) CO.

The combustion of isooctane at λ = 1 produces 4500 ppm NO and 18 000 ppm CO in the combustion products at chemical equilibrium according to Figure 5. On the other hand, an SI engine working at 1800 rpm and an absolute air pressure of 61.5 kPa caused tailpipe emissions of 2800 ppm NO and 2300 ppm CO at λ = 1.23 The combustion products NO and CO are further reacted to complete combustion products during the expansion process after combustion in the engine. The decrease in the amount of CO during expansion from 18 000 to 2300 ppm is much more pronounced than the decrease in the amount of NO from 4500 to 2800 ppm because the rate of NO reduction to N2 is lower than the rate of oxidation of CO at temperatures during expansion. When ethanol-based 2.5% syngas was added to gasoline, NO emissions and CO emissions were higher than for gasoline combustion at the same λ, but a decrease in both emissions was possible when λ was increased to 1.35.23 However, syngas must be added to the charge to lean out the charge further. According to Figure 5, decomposed neat methanol (w = 0) causes higher CO and NO amounts in the combustion products than combustion of gasoline as well. On the other hand, the increase in syngas volume fraction in the charge of a hybrid syngas−gasoline engine from 0% to 1.84% at λ = 1.2 decreased the NO emissions in exhaust gases from 2047 to 1499 ppm in an experimental study.44 The amount of NO in the combustion products decreases to 2900 ppm from 4600 and 3600 ppm for isooctane and methanol, respectively, when an equimolar water−methanol mixture (w = 1) is reformed in the reactor and used instead of liquid fuel (rp = 1) according to Figure 5. Such substitutions of gasoline and methanol results in 37% and 19% reductions of NO in the combustion products, respectively. Thus, combustion of syngas reformed out of more aqueous methanol causes lower NO and CO emissions even at near-stoichiometric. Increasing λ further for load adjustment can decrease the emissions further. Substitution of gasoline with methanol results in a 22% reduction of NO in the combustion products assuming constant-pressure combustion. The flame isotherms for M0, M85, and M100 from the left chart in Figure 3 are also shown in both panels of Figure 5. The lines representing the flame isotherms in Figure 3 show a decrease in the amounts of NO and CO in the combustion products when the premixed fuel ratio is increased from zero to unity. When the air/fuel ratio is doubled (λ = 2), the results for different alcohol amounts in the direct fuel overlap. The direct fuel’s alcohol content does not affect the NO amount for lean

mixtures. As the fuel mixture gets leaner, the water content of the premixed fuel has a greater impact than the alcohol content of the direct fuel on the fraction of combustion products. Figure 6 shows the effect of the relative air/fuel ratio on the combustion products. The results for NO and CO are shown

Figure 6. Effect of λ on the NO and CO fractions in the combustion products.

on the primary (left) and secondary (ight) vertical axes, respectively. While the CO amounts for methanol (M100), isooctane (M0), reformed dry methanol (w = 0), and reformed stoichiometric methanol (w = 1) decrease monotonically with λ, the NO amounts have a maximum between λ = 1.2 and 1.3. The trend in the NO amount in the combustion products of M0, M100, and reformed stoichiometric methanol intersect at λ = 2, similar to the left panel in Figure 5, while the combustion products of reformed dry methanol remain higher because of its higher adiabatic flame temperature. The different flame speeds of methanol, gasoline, CO, and H2 are not taken into account in the results presented above. Hydrogen has a higher flame speed than other fuels such as gasoline and methanol, so combustion of stoichiometric syngas mixtures could be better described with constant-volume combustion. On the other hand, gasoline has a lower flame speed, and its constant-volume combustion may result in the prediction of a too high adiabatic flame temperature. Constant-pressure combustion was assumed in order to compare different fuels on the same basis. Therefore, stoichiometric mixtures with rp near unity may have a higher adiabatic flame temperature than predicted because constantvolume combustion causes a flame temperature 200−400 K 2088

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article could be interesting for decision makers in engine design in the automotive industry and for catalyst manufacturers, oil refineries, fuel distributors, and all NGOs and governments that advocate for lower-CO2-emitting vehicles.

higher than the assumed constant-pressure combustion according to Figure 2. The higher flame temperature will also have a negative impact on NO emissions. In fact, experimental works show an increase in NOx emissions if liquid methanol is substituted with neat-methanol-based syngas while the air/fuel ratio is held constant.45 Consequently, from a perspective of emission control, a high premixed fuel ratio is only applicable for lean combustion at part load, which is also a prerequisite for combustion stability. Increasing the premixed fuel ratio and relative air/fuel ratio simultaneously accomplishes a higher engine and system efficiency at part load, which decreases CO2 and other noxious emissions based on produced energy.

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AUTHOR INFORMATION

Corresponding Author

*E-mail: [email protected]. Notes

The authors declare no competing financial interest.

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ACKNOWLEDGMENTS The authors are grateful for the support of Istanbul Technical University.

4. CONCLUSION In research related to onboard fuel reforming, the purpose is to increase the thermal efficiency via recovery of the exhaust heat of an internal combustion engine and decrease emissions by enabling combustion of leaner mixtures. Hydrogen as the main product of reforming allows stable combustion of ultralean mixtures in the engine. Despite its lower charge density, hydrogen-rich syngas can be mixed with liquid gasoline or methanol to increase the energy density of the engine charge. On the other hand, the high flame speed of hydrogen causes higher NOx emissions and combustion instability when syngas is combusted with a near-stoichiometric air/fuel ratio. The method of Lagrange undetermined multipliers has been applied to the combustion of hydrogen-rich syngas premixed with a methanol−gasoline mixture. The method of calculation in this article is independent of engine parameters, valid only for chemical equilibrium conditions, and presents a well organized, well-defined, and easy to follow programming code that takes different parameters related to the fuel’s composition into account. This is possible because the programming code expresses gas, liquid fuels, air, reactants, and products as vectors. Moreover, the calculation results are expressed as vectors. Final exhaust emissions from the tailpipe and thermal efficiency can only be determined given specific geometric and cycle parameters of an engine. This method can determine the amount of water in the aqueous methanol that is needed for a given specific alcohol amount in the liquid fuel in order to maintain a stable flame temperature regardless of the premixed fuel ratio for a specified air/fuel ratio. The results of the article also indicate that with increasing amount of water in the aqueous methanol, the adiabatic flame temperature and combustion products such as CO and NO are reduced. When an equimolar water−methanol mixture is reformed to syngas, as the premixed ratio increases, the NO and CO ratio of the combustion products decreases regardless of other parameters. Combustion of syngas reformed out of an equimolar water−methanol mixture causes a decrease in the NO ratio by 20% compared with methanol and 40% compared with gasoline under equilibrium conditions, whereas the adiabatic flame temperature decreases by only 50 and 100 K, respectively. This decrease in the NO and CO ratio causes a reduction in exhaust emissions since the initial values of the NO and CO fraction among postcombustion exhaust emissions are lowered. In future research, the addition of kinetic data on combustion reactions and geometric parameters of an engine to this programming code will enable the calculation of tailpipe exhaust emissions, indicated power, and efficiency as well. This

NOMENCLATURE

Symbols

a = air vector (mol/mol of fuel) A = conservation matrix aki = entry of A at row k and column i al = mole fraction of alcohol in the direct fuel c = amount of carbon in a molecule Cp = molar heat capacity at constant pressure (J mol−1 K−1) CV = molar heat capacity at constant volume (J mol−1 K−1) d = mean LHV ratio of liquid to gaseous fuel f = fuel vector (mol/mol of fuel) fd = direct fuel vector (mol/mol of fuel) fg = product gas vector (mol/mol of fuel) fl = liquid fuel vector (mol/mol of fuel) fp = premixed fuel vector (mol/mol of fuel) g°i = molar free enthalpy of gas i based on standard pressure (J/mol) h = amount of hydrogen in a molecule H = enthalpy of the engine charge (J) HG = mixture heat value (kJ/L) H°G = standard mixture heat value (kJ/L) K(T) = equilibrium constant for a reaction LHV = lower heating value vector (kJ/mol) LHV = lower heating value (kJ/mol) LHVg = mean lower heating value of gaseous fuel (kJ/mol) LHVl = mean lower heating value of liquid fuel (kJ/mol) n = product vector (mol/mol of fuel) nC = amount of theoretical complete combustion products (mol/mol of fuel) nf = molar amount of fuel combusted during an engine cycle (mol) n0 = reactant vector (mol/mol of fuel) n0T = molar amount of engine’s fresh charge (mol/mol of fuel) nT = molar amount of combustion products (mol/mol of fuel) N2,in = amount of nitrogen inducted into the engine (mol of N2/mol of fuel) o = amount of oxygen in a molecule ost = stoichiometric oxygen vector for theoretical complete combustion (mol of O2/mol) O2,st = stoichiometric oxygen for complete combustion of the fuel mixture (mol of O2/mol of fuel) O2,in = amount of oxygen inducted into the engine (mol of O2/mol of fuel) p = reaction pressure (Pa) pi = partial pressure of gas i (Pa) 2089

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p° = standard pressure (1 atm = 101 325 Pa) pV = isochoric combustion pressure (Pa) Q = heat exchanged between the engine and its surroundings (J) Qd = direct fuel energy (J) Qp = premixed fuel energy (J) Qt = total fuel energy (J) rp = premixed fuel ratio R = universal gas constant (8.3145 J mol−1 K−1) sT0° = standard molar entropy (J mol−1 K−1) T = reaction temperature (K) T0 = temperature of unburnt charge (K) T° = standard temperature (298.15 K) Tad = adiabatic flame temperature (K) Tp = combustion temperature of an isobaric adiabatic process (K) TV = combustion temperature of an isochoric adiabatic process (K) u = molar internal energy of the gas in the engine (J/mol) U = internal energy of the gas in the engine (J) Vd = displacement volume (L) w = water-to-methanol molar ratio in aqueous methanol W = work done by the engine (J) xi = mole fraction of gas i in the product gas xp = premixed fuel’s mole fraction yi = mole fraction of gas i y = mole fraction vector of combustion products y0 = mole fraction vector of fresh engine charge yad = mole fraction vector of adiabatic combustion products ϕ = equivalence ratio Δgf° = standard molar free enthalpy of formation (J/mol) ΔGf°(T) = standard change in free enthalpy for the formation reaction of a gas (J/mol) Δh(T) = change in molar enthalpy (J/mol) ΔG°(T) = standard free enthalpy change of a reaction (J/ mol) Δh°f = standard molar enthalpy of formation (J/mol) Δh°f,CHO = standard molar enthalpy of formation of an oxygenated hydrocarbon (J/mol) ΔHc° = standard combustion enthalpy (kJ/mol) ΔH°r = standard reaction enthalpy (kJ/mol) Δs(T) = change in molar entropy (J mol−1 K−1) λ = relative air/fuel ratio λ = vector of Lagrange undetermined multipliers μ = chemical potential of a gas (J/mol) ν = signed stoichiometric coefficient of a gas in a reaction ρ = molar density of the engine charge (mol/L) ρ° = standard molar density (0.041 mol/L)

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p = variable of an isobaric process T = variable based on the reaction temperature T V = variable of an isochoric process

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Abbreviations

CNG = compressed natural gas LPG = liquefied petroleum gas IMEP = indicated mean effective pressure IVC = intake valve closure PCCI = premixed charge compression ignition SI = spark ignition STP = standard temperature (T° = 298.15 K) and pressure (p° = 1 atm) syngas = synthesis gas Indices

ad = adiabatic i = gas component i, column i of the conservation matrix A k = row number of an entry in the conservation matrix A 2090

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