JP2003073676A - Method for calculating theoretical air/fuel ratio of gasoline and method for producing gasoline with desired theoretical air/fuel ratio - Google Patents

Abstract

PROBLEM TO BE SOLVED: To provide a method for strictly calculating a theoretical air/fuel ratio of gasoline, and a method for producing gasoline having a desired theoretical air/fuel ratio obtained by the calculation. SOLUTION: The theoretical air/fuel ratio is calculated by using the expression: 138.03 (A+B/2)/Wf (wherein A and B are the coefficients of a composition of combustion gas ACO2 +BH2 O; and Wf is the weight of gasoline).

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a method for calculating a stoichiometric air-fuel ratio of gasoline and a method for producing a gasoline having a desired stoichiometric air-fuel ratio, and more particularly to a stoichiometric air-fuel ratio of gasoline for which the stoichiometric air-fuel ratio of gasoline can be determined exactly. The present invention relates to a fuel ratio calculation method and a method for producing gasoline having a desired stoichiometric air-fuel ratio based on this calculation method.

[0002]

2. Description of the Related Art Gasoline is called a hydrocarbon C _n H _m fuel and is made by combining carbon C and hydrogen H. About 20
It is a mixture of zero hydrocarbon compounds C _n H _m . By no means is gasoline a single hydrocarbon C _n H _m . In that sense, gasoline is a very complex fuel.

On the other hand, essentially a piston engine is an incomplete combustion. Since the intake and exhaust are repeated alternately, the time allowed for fuel combustion is too short, and when the flame of the combustion gas ignited by the spark plug propagates and reaches near the combustion chamber surface, the combustion chamber surface This is because the combustion flame is extinguished in the thin boundary layer and the fuel in the layer does not burn. The reason for quenching the flame in the boundary layer on the surface of the combustion chamber is that the combustion temperature of the fuel is about 2000 ° C, which is extremely high, while the temperature of the combustion chamber surface is maintained at an extremely low temperature of around 300 ° C, which was cooled by cooling water. This is because you are dripping. Due to such incomplete combustion, CO, HC of unburned fuel, and NO _X generated due to the high combustion temperature of 2000 ° C. are discharged into the exhaust gas. Purification of CO, HC and NO _{X in} exhaust gas is required.

CO and HC are converted into CO ₂ and H ₂ O by the oxidation reaction, and NO
_X becomes N ₂ and O ₂ by the reduction reaction. That is, CO, HC, and NO _X must be cleaned by simultaneously performing opposite oxidation and reduction reactions. It is possible 3
The original catalyst was developed. In order to use this three-way catalyst, it becomes necessary to operate the engine at the stoichiometric air-fuel ratio. When operated at the stoichiometric air-fuel ratio, the exhaust gas becomes a neutral atmosphere nor a reducing atmosphere in an oxidizing atmosphere, it is possible to three-way catalyst to exhibit the function CO, HC, reducing the three components of the NO _X. Therefore, the stoichiometric air-fuel ratio of a gasoline engine becomes important.

The definition of air-fuel ratio is the ratio of air to fuel weight under certain operating conditions. Therefore, when designing the engine, for example, assuming that the stoichiometric air-fuel ratio of regular gasoline is 14.7, the fuel supply amount is set so that the air-fuel ratio becomes 14.7. However, the stoichiometric air-fuel ratio is determined by the mixing ratio of about 200 kinds of hydrocarbon hydrocarbons in gasoline. When the theoretical air-fuel ratio of gasoline is 14.5, which is different from 14.7 when the engine was designed,
The O ₂ sensor is attached to the exhaust system to detect the air-fuel ratio, and always tries to control the fuel supply amount so that the theoretical air-fuel ratio of the gasoline becomes 14.5.

"The theoretical air-fuel ratio of regular gasoline is 1
Assuming that the value is 4.7 ", the value of the stoichiometric air-fuel ratio of gasoline, which is the standard for setting the engine, differs depending on the manufacturer. For example, Company A has 14.5, Company B has 14.
It is 7 and so on. Although it is narrower than 14.5-14.7, it is evidence that the theoretical air-fuel ratio of commercially available gasoline varies. Therefore, it is important to exactly determine the stoichiometric air-fuel ratio of the gasoline used. At the same time, if all theoretical air-fuel ratios of commercially available gasoline can be controlled to a constant value, it would be better than that. It is desirable to establish a calculation formula that strictly calculates the theoretical air-fuel ratio of produced gasoline.

There are two methods for measuring the air-fuel ratio during engine operation. One is the intake air-fuel ratio, which is a method of obtaining the air-fuel ratio from the intake air weight and the supply gasoline weight. The other is the exhaust gas air-fuel ratio method. The intake air-fuel ratio method is a traditional method and is not used much at present. This is because the exhaust gas analysis technology has advanced since the enforcement of the Exhaust Gas Regulations Act that started in the 1970s, and it has become possible to measure engine exhaust gas with high accuracy. The air-fuel ratio can be calculated from the measured CO, CO ₂ and HC concentration of unburned fuel.

In 1973, William H. Holl of GM USA announced the calculation method. The assumption of the calculation is the hydrogen / carbon ratio of gasoline. This hydrogen / carbon ratio refers to gasoline as one of the typical hydrocarbons C _n H _m
I will represent it. It is the m / n ratio of the hydrocarbon C _n H _m . Holl set the theoretical air-fuel ratio to 1.85. As described above, gasoline is a mixture of about 200 kinds of hydrocarbons, but generally, one kind of hydrocarbon CH _1.85 represents gasoline. Actually, such hydrocarbon does not exist. Normally, the combustion chemical reaction equation for calculating the theoretical air-fuel ratio of gasoline, which is assumed in this way, is as shown in equation (5).

CH _1.85 +1.463 (O ₂ + 3.785N ₂ ) → CO ₂ + 0.925H ₂ O + 5.537N ₂ --- (5)

Here, it is assumed that the air is (O ₂ + 3.785N ₂ ). Gasoline CH _1.85 burns with oxygen O ₂ in the air to produce CO ₂ +0.
Completely burns to 925H ₂ O. The number on the right side of oxygen O ₂ required for complete combustion is 1+ (0.925 / 2) = 1.463. Therefore, the coefficient of air (O ₂ + 3.785N ₂ ) on the left side of equation (5) is 1.463, and equation (5) holds. The stoichiometric air-fuel ratio shown by the equation (5) can be calculated by knowing the atomic weights of C and H and the molecular weights of O ₂ and N ₂ .

The weight of air is 1.463 (O ₂ +
3.785N ₂ ), the molecular weight of oxygen O ₂ is 32.00, the molecular weight of nitrogen N ₂ is
From 28.014 to 1.463 (32.00 + 3.785 × 28.014) = 201.94. Also, the gasoline weight is the same as CH _{1.85 on the} left side and carbon C
Atomic weight of 12.011, hydrogen H atomic weight of 1.008, 12.011 + 1.8
5 x 1.008 = 13.87. Therefore, the stoichiometric air-fuel ratio is 201.94 / 13.87 = 14.55. This method is a general method for obtaining the stoichiometric air-fuel ratio of gasoline.

Assuming that the hydrogen / carbon ratio is 1.95, the combustion chemical reaction formula is CH _1.95 +1.4875 (O ₂ + 3.785N ₂ ) → CO ₂ + 0.975H ₂ O + 5.678N ₂ --- ( 6) becomes, and the theoretical air-fuel ratio is calculated in the same way as in the case of equation (5).
14.69.

[0013]

That is, the conventional gasoline stoichiometric air-fuel ratio calculation method is merely the stoichiometric air-fuel ratio when the gasoline is regarded as a single hydrocarbon and the hydrogen / carbon ratio is assumed to be a certain value. There is no guarantee that the gasoline will be represented by a single hydrocarbon and at some hydrogen / carbon ratio.

Therefore, an object of the present invention is to calculate a stoichiometric air-fuel ratio of gasoline capable of rigorously determining the stoichiometric air-fuel ratio of gasoline and a method for producing a gasoline having a desired stoichiometric air-fuel ratio based on this calculation method. To provide.

[0015]

[Means for Solving the Problems] First, it is necessary to clarify the theoretical air-fuel ratio of gasoline as a reference. Gasoline is not a single substance like methane or benzene, but a hydrocarbon formed by combining carbon atom C and hydrogen atom H.
Mainly composed of C _n H _m . As described above, the carbon number is a mixture of about 200 kinds of hydrocarbon compounds ranging from 4 to 16 or so. The C has four bonds,
H has one bond. Petroleum hydrocarbons C _n H _m are classified into several systems depending on how these bonds are connected.

(1) Paraffin hydrocarbon A general term for hydrocarbons in which several Cs and Cs are linked by a single bond such as -C-C-. This hydrocarbon is generally the most abundant in crude oil. For example, isooctane (2,2,4-trimethylpentane) with an octane number of 100, which is the standard of octane number and is hard to cause knock,
Normal heptane with an octane number of 0, which is prone to knock, is also a paraffinic hydrocarbon.

(2) Aromatic (Aromatic) hydrocarbon A state in which 6 C's are cyclically bonded to each other with 3 double bonds every 3 is called a benzene nucleus. Hydrocarbons having a benzene nucleus are called aromatic hydrocarbons. Aromatic hydrocarbons are rarely contained in crude oil, but a large amount are contained when they are produced in the catalytic reforming process.

(3) Olefin-based hydrocarbon One of the bonds between C and C has two bonds at one place, and -C = C-
Hydrocarbons that are linked by bonds such as are collectively referred to.

(4) Naphthene type hydrocarbon It is a hydrocarbon in which C is cyclically connected without having a double bond. Some are more abundant and some are not, depending on the origin of the crude oil.

When considering the air-fuel ratio of such gasoline, it is easier to represent hydrocarbons as shown in Table 1. That is,

[Table 1] Can be classified as.

[0021] (A) paraffinic hydrocarbons C _n H _{(2n + 2)} stoichiometric ratio paraffinic hydrocarbons C _n H _{(2n + 2)} complete combustion chemical equation becomes (7).

C _n H _{(2n + 2)} + (1.5n + 0.5) (O ₂ + 3.785N ₂ ) → nCO ₂ + (n + 1) H ₂ O + 3.785 (1.5n + 0.5) N _2- -(7)

The left side shows the paraffinic hydrocarbon before combustion.
It shows air, and the right side shows the combustion product gas after combustion.
It As a result of combustion, carbon contained in hydrocarbons is carbon dioxide.
₂, H is steam H₂Become O. C contained in hydrocarbon is n
, H is (2n + 2), so H₂Will be (n + 1).
Therefore, the complete combustion product gas is nCO₂+ (n + 1) H₂It becomes O.
Nitrogen N contained in air₂Is an inert gas, as it is 3.785
(1.5n + 0.5) N₂Only discharged. Combustion generation on the right side of equation (7)
Oxygen contained in gas O₂Is n + (n + 1) / 2 = 1.5n + 0.5
It Therefore paraffinic hydrocarbon C_nH_{(2n + 2)}Full of
The amount of air required for combustion is O ₂From the number of (1.5n + 0.5) (O₂+3.78
5N₂) And equation (7) holds.

For complete combustion of paraffinic hydrocarbons
The ratio of air weight to fuel weight is the reason for paraffinic hydrocarbons.
The air-fuel ratio. The air weight on the left side of equation (7) is (1.5n + 0.
5) (O ₂+ 3.785N₂) Is required. Oxygen and nitrogen molecular weight
Assuming 32.00 and 28.014 respectively, the air weight is 138.03 (1.5n
+0.5). On the other hand, the fuel weight is C_nH_{(2n + 2)}From carbon,
If the atomic weights of hydrogen are 12.011 and 1.008, respectively, (14.027
n + 2.016).

Therefore, the theoretical air-fuel ratio is theoretical air-fuel ratio = 138.03 × (1.5n + 0.5) / (14.027n + 2.018)-(8). As is clear from the equation (8), the theoretical air-fuel ratio of paraffinic hydrocarbon is a function of n. That means
This means that the theoretical air-fuel ratio differs depending on the carbon number n.
Table 2 shows the theoretical air-fuel ratio for carbon numbers C1 to C16.

[0026]

[Table 2]

As is clear from Table 2, the theoretical air-fuel ratio becomes smaller as the carbon number increases.

Since the stoichiometric air-fuel ratio of gasoline is around 14.6, the stoichiometric air-fuel ratio of paraffin hydrocarbons is about 3% larger than the stoichiometric air-fuel ratio of gasoline. That is, the paraffinic hydrocarbon corrects the stoichiometric air-fuel ratio of the hydrocarbon that constitutes the gasoline and has a small stoichiometric air-fuel ratio.

(B) Aromatic hydrocarbon C _n H
Chemical equation complete combustion of _(2n-6) of the stoichiometric air-fuel ratio Aromateikku hydrocarbons C _n H _(2n-6) becomes equation (9).

C _n H _(2n-6) + (1.5n-1.5) (O ₂ + 3.785N ₂ ) → nCO ₂ + (n-3) H ₂ O + 3.785 (1.5n-1.5) N _2- -(9)

By burning, n hydrocarbons C are carbon dioxide gas nC
In O ₂ , (2n-6) H becomes steam (n-3) H ₂ O. Therefore, the combustion product gas on the right side of the equation (9) is nCO ₂ + (n-3) H ₂ O. Nitrogen N ₂ contained in air is 3.785 (1.5n-1.5) N as it is
Only _{2 is} discharged. The number of oxygen O ₂ contained in the combustion product gas is n + (n-3) /2=1.5n-1.5. Therefore, the amount of air required for complete combustion of the aromatic hydrocarbon C _n H _(2n-6) is (1.5n-1.5) (O ₂ + 3.785N ₂ ) and the equation (9) is established.

The air weight for complete combustion of aromatic hydrocarbons is calculated from (1.5n-1.5) (O ₂ + 3.785N ₂ ). From the molecular weight of oxygen and nitrogen, the weight of air is 138.03 (1.5n
-1.5). On the other hand, the fuel weight is calculated from C _n H _(2n-6) to (14.02
7n-6.048).

Therefore, the theoretical air-fuel ratio is theoretical air-fuel ratio = 138.03 × (1.5n-1.5) / (14.027n-6.048) --- (10). As is clear from the equation (10), the theoretical air-fuel ratio of the aromatic hydrocarbon also becomes a function of n, like the paraffinic hydrocarbon. Table 3 shows the theoretical air-fuel ratio for carbon numbers C6 to C12.

[0034]

[Table 3]

The theoretical air-fuel ratio of the aromatic hydrocarbon is about 10% smaller than the theoretical air-fuel ratio of gasoline, and the theoretical air-fuel ratio shows a large value as the carbon number increases. Aromatic hydrocarbons have benzene nuclei and emit harmful benzene nuclei when incomplete combustion occurs.
Therefore, the volume% of aromatic hydrocarbons in gasoline is not so large.

(C) Olefin and naphthenic hydrocarbon C _n
The stoichiometric air-fuel ratio of H _2n and the naphthenic hydrocarbon C _n H _{2n The} complete chemical reaction formula is obtained by the same formula as that of the paraffinic hydrocarbon and the aromatic hydrocarbon, and is represented by the formula (11).

C _n H _2n + 1.5n (O ₂ + 3.785N ₂ ) → nCO ₂ + nH ₂ O + 3.785 × 1.5nN ₂ --- (11)

The number of oxygen O ₂ contained in the combustion product gas on the right side of the equation (11) is n + 0.5n = 1.5n. Thus the amount of air required for complete combustion of the olefin and naphthene hydrocarbons C _n H _2n is 1.5n (O ₂ + 3.785N ₂₎ becomes (11) is established.

The air weight on the left side of the equation (11) is 1.5n (O ₂ +3.78).
5N ₂ ). From the molecular weight of oxygen and nitrogen, the weight of air is 138.03 × 1.5n. On the other hand, the fuel weight is from C _n H _2n ,
If the carbon and hydrogen atomic weights are 12.011 and 1.008 respectively, 1
It becomes 4.027n. Therefore, the theoretical air-fuel ratio is 138.03 × 1.5n / 14.02n = 14.77 --- (12). That is, the stoichiometric air-fuel ratio of olefins and naphthenic hydrocarbons is a constant value of 14.77 regardless of the carbon number n. The value is close to the theoretical air-fuel ratio of gasoline.

The relationship between the carbon number of hydrocarbons constituting the gasoline and the theoretical air-fuel ratio is shown in Table 2 for paraffinic hydrocarbons, Table 3 for aromatic hydrocarbons, and the theoretical air-fuel ratio of olefins and naphthenic hydrocarbons as carbon. It was found to show a constant value of 14.77 regardless of the number. FIG. 1 shows these relationships by the relationship between the carbon number and the theoretical air-fuel ratio.

The theoretical air-fuel ratio of regular gasoline is 14.
Since it is around 6, the theoretical air-fuel ratios of olefin and naphthenic hydrocarbons are almost similar to those of regular gasoline, paraffinic hydrocarbons are larger than regular gasoline, and aromatic hydrocarbons are small. When they are mixed at a certain ratio, the stoichiometric air-fuel ratio of regular gasoline becomes around 14.6.

The characteristics of the theoretical air-fuel ratio of paraffinic hydrocarbons, aromatic hydrocarbons, olefins and naphthenic hydrocarbons have been clarified. These properties provide powerful information for stoichiometric adjustment. For example, a method of changing the already commercialized gasoline having a theoretical air-fuel ratio of 14.5 to 14.7 can be achieved by increasing the paraffinic hydrocarbon. Further, in the case of gasoline in the process of refining, it may be possible to use it together with narrowing the addition ratio of aromatic hydrocarbons. In the opposite case, a countermeasure can be easily considered from the revealed characteristic of the theoretical air-fuel ratio.

The present invention has been made on the basis of the above-mentioned findings, and is characterized by the following.

The invention according to claim 1 is characterized in that the theoretical air-fuel ratio of gasoline is calculated by the following equation (1).

[0045]   Theoretical air-fuel ratio = 138.03 (A + B / 2) / Wf --- (1)

However, in the above equation (1),     A = α₁+ 2α₂+ 3α₃+ 4α_Four+ 5α_Five+ 6α₆+ 7α₇+ 8α₈+ 9α₉+ 10α_Ten+ 11α₁₁+ 12α₁₂ + 13α₁₃+ 14α₁₄+ 15α₁₅+ 16α₁₆+ 17α₁₇+ 18α₁₈+ 19α₁₉+ 20α₂₀+ 21α_{twenty one}+ 22α_{twenty two}+2 3α_{twenty three}+ 24α_{twenty four}+ 25α_{twenty five}+ 6β₆+ 7β₇+ 8β₈+ 9β₉+ 10β_Ten+ 11β₁₁+ 12β₁₂+ 13β₁₃+ 14β ₁₄ + 15β₁₅+ 16β₁₆+ 17β₁₇+ 18β₁₈+ 19β₁₉+ 20β₂₀+ 21β_{twenty one}+ 22β_{twenty two}+ 23β_{twenty three}+ 24β_{twenty four} + 25β_{twenty five}+ γ₁+ 2γ₂+3 (γ₃+ δ₃) +4 (γ_Four+ δ_Four) +5 (γ_Five+ δ_Five) +6 (γ₆+ δ₆) +7 (γ₇+ δ₇) + 8 (γ₈+ δ₈) +9 (γ₉+ δ₉) +10 (γ_Ten+ δ_Ten) +11 (γ₁₁+ δ₁₁) +12 (γ₁₂+ δ₁₂) +13 (γ₁₃+ δ₁₃) +14 (γ₁₄+ δ₁₄) +15 (γ₁₅+ δ₁₅) +16 (γ₁₆+ δ₁₆) +17 (γ₁₇+ δ₁₇) +18 (γ₁₈+ δ ₁₈ ) +19 (γ₁₉+ δ₁₉) +20 (γ₂₀+ δ₂₀) +21 (γ_{twenty one}+ δ_{twenty one}) +22 (γ_{twenty two}+ δ_{twenty two}) +23 (γ_{twenty three}+ δ_{twenty three} ) +24 (γ_{twenty four}+ δ_{twenty four}) +25 (γ_{twenty five}+ δ_{twenty five}) --- (2)

[0047]     B = 2α₁+ 3α₂+ 4α₃+ 5α_Four+ 6α_Five+ 7α₆+ 8α₇+ 9α₈+ 10α₉+ 11α_Ten+ 12α₁₁+ 13α ₁₂ + 14α₁₃+ 15α₁₄+ 16α₁₅+ 17α₁₆+ 18α₁₇+ 19α₁₈+ 20α₁₉+ 21α₂₀+ 22α_{twenty one}+ 23α_{twenty two} + 24α_{twenty three}+ 25α_{twenty four}+ 26α_{twenty five}+ 3β₆+ 4β₇+ 5β₈+ 6β₉+ 7β_Ten+ 8β₁₁+ 9β₁₂+ 10β₁₃+ 11β_{1 Four} + 12β₁₅+ 13β₁₆+ 14β₁₇+ 15β₁₈+ 16β₁₉+ 17β₂₀+ 18β_{twenty one}+ 19β_{twenty two}+ 20β_{twenty three}+ 21β_{twenty four}+ 22β_{twenty five}+ γ₁+ 2γ₂+3 (γ₃+ δ₃) +4 (γ_Four+ δ_Four) +5 (γ_Five+ δ_Five) +6 (γ₆+ δ₆) +7 (γ₇+ δ₇) +8 (γ₈+ δ₈) +9 (γ₉+ δ₉) +10 (γ_Ten+ δ_Ten) +11 (γ₁₁+ δ₁₁) +12 (γ₁₂+ δ₁₂) +13 (γ₁₃+ δ₁₃) +14 (γ₁₄+ δ₁₄) +15 (γ₁₅+ δ₁₅) +16 (γ₁₆+ δ₁₆) +17 (γ₁₇+ δ₁₇) +18 (γ₁₈+ δ ₁₈ ) +19 (γ₁₉+ δ₁₉) +20 (γ₂₀+ δ₂₀) +21 (γ_{twenty one}+ δ_{twenty one}) +22 (γ_{twenty two}+ δ_{twenty two}) +23 (γ_{twenty three}+ δ_{twenty three} ) +24 (γ_{twenty four}+ δ_{twenty four}) +25 (γ_{twenty five}+ δ_{twenty five}) --- (3)

[0048]     Wf = α₁CH_Four+ α₂C₂H₆+ α₃C₃H₈+ α_FourC_FourH_Ten+ α_FiveC_FiveH₁₂+ α₆C₆H₁₄+ α₇C₇H₁₆+ α₈ C₈H₁₈+ α₉C₉H₂₀+ α_TenC_TenH_{twenty two}+ α₁₁C₁₁H_{twenty four}+ α₁₂C₁₂H₂₆+ α₁₃C₁₃H₂₈+ α₁₄C₁₄H₃₀+ α ₁₅ C₁₅H₃₂+ α₁₆C₁₆H₃₄+ α₁₇C₁₇H₃₆+ α₁₈C₁₈H₃₈+ α₁₉C₁₉H₄₀+ α₂₀C₂₀H₄₂+ α_{twenty one}C_{twenty one}H ₄₄ + α_{twenty two}C_{twenty two}H₄₆+ α_{twenty three}C_{twenty three}H₄₈+ α_{twenty four}C_{twenty four}H₅₀+ α_{twenty five}C_{twenty five}H₅₂+ β₆C₆H₆+ β₇C₇H₈ + β₈C₈H_Ten + β₉C₉H₁₂+ β_TenC_TenH₁₄+ β₁₁C₁₁H₁₆+ β₁₂C₁₂H₁₈+ β₁₃C₁₃H₂₀+ β₁₄C₁₄H_{twenty two}+ β₁₅C₁₅ H_{twenty four}+ β₁₆C₁₆H₂₆+ β₁₇C₁₇H₂₈+ β₁₈C₁₈H₃₀+ β₁₉C₁₉H₃₂+ β₂₀C₂₀H₃₄+ β_{twenty one}C_{twenty one}H₃₆+ β _{twenty two} C_{twenty two}H₃₈+ β_{twenty three}C_{twenty three}H₄₀+ β_{twenty four}C_{twenty four}H₄₂+ β_{twenty five}C_{twenty five}H₄₄+ γ₁CH₂+ γ₂C₂H_Four+ (γ₃+ δ₃) C₃H₆+ ( γ_Four+ δ_Four) C_FourH₈+ (γ_Five+ δ_Five) C_FiveH_Ten+ (γ₆+ δ₆) C₆H₁₂+ (γ₇+ δ₇) C₇H₁₄+ (γ₈+ δ₈) C₈H₁₆ + (γ₉+ δ₉) C₉H₁₈+ (γ_Ten+ δ_Ten) C_TenH₂₀+ (γ₁₁+ δ₁₁) C₁₁H_{twenty two}+ (γ₁₂+ δ₁₂) C₁₂H_{twenty four}+ ( γ₁₃+ δ₁₃) C₁₃H₂₆+ (γ₁₄+ δ₁₄) C₁₄H₂₈+ (γ₁₅+ δ₁₅) C₁₅H₃₀+ (γ₁₆+ δ₁₆) C₁₆H₃₂+ ( γ₁₇+ δ₁₇) C₁₇H₃₄+ (γ₁₈+ δ₁₈) C₁₈H₃₆+ (γ₁₉+ δ₁₉) C₁₉H₃₈+ (γ₂₀+ δ₂₀) C₂₀H₄₀+ ( γ_{twenty one}+ δ_{twenty one}) C_{twenty one}H₄₂+ (γ_{twenty two}+ δ_{twenty two}) C_{twenty two}H₄₄+ (γ_{twenty three}+ δ_{twenty three}) C_{twenty three}H₄₆+ (γ_{twenty four}+ δ_{twenty four}) C_{twenty four}H₄₈+ ( γ_{twenty five}+ δ_{twenty five}) C_{twenty five}H₅₀--- (4)

However, in the above equations (2), (3) and (4), α _n : volume content of paraffinic hydrocarbon (%) β _n : volume content of aromatic hydrocarbon (%) γ _n : Capacity content of olefinic hydrocarbons (%) δ _n : Capacity content of naphthenic hydrocarbons (%)

The invention according to claim 2 is characterized in that a gasoline having a desired stoichiometric air-fuel ratio is produced by adjusting the content ratio of hydrocarbons constituting the gasoline based on the above formula (1). I have.

The invention according to claim 3 is characterized in that when the stoichiometric air-fuel ratio of gasoline is increased, the content ratio of paraffinic hydrocarbons is increased and the content ratio of aromatic hydrocarbons is decreased. It is a thing.

[0052]

BEST MODE FOR CARRYING OUT THE INVENTION Dividing gasoline into hydrocarbon systems,
When analyzed by carbon number, paraffin hydrocarbons are carbon C
Number of C1 to C25 of 25 types, Aromatic system is C6 to C
The capacity contents of 20 kinds of 25, 25 kinds of C1 to C25 for olefins, and 23 kinds of C3 to C25 for naphthenes are analyzed. Assuming that all hydrocarbons are completely combusted and carbon C becomes carbon dioxide CO ₂ and hydrogen H becomes steam H ₂ O, a general formula for obtaining the theoretical air-fuel ratio can be obtained from the chemical reaction formulas of these complete combustion. .

If the volume contents of paraffinic hydrocarbons, aromatic hydrocarbons, olefinic hydrocarbons and naphtheneic hydrocarbons are α _n , β _n , γ _n and δ _n , respectively,
The chemical reaction equation for complete combustion is as shown in equation (13).

[0054]   [Α₁CH_Four+ α₂C₂H₆+ α₃C₃H₈+ α_FourC_FourH_Ten+ α_FiveC_FiveH₁₂+ α₆C₆H₁₄+ α₇C₇H₁₆+ α₈C₈H₁₈+ α₉C₉H₂₀+ α_TenC_TenH_{twenty two}+ α₁₁C₁₁H_{twenty four}+ α₁₂C₁₂H₂₆+ α₁₃C₁₃H₂₈+ α₁₄C₁₄H₃₀+ α₁₅C₁₅H ₃₂ + α₁₆C₁₆H₃₄+ α₁₇C₁₇H₃₆+ α₁₈C₁₈H₃₈+ α₁₉C₁₉H₄₀+ α₂₀C₂₀H₄₂+ α_{twenty one}C_{twenty one}H₄₄+ α_{2 2} C_{twenty two}H₄₆+ α_{twenty three}C_{twenty three}H₄₈+ α_{twenty four}C_{twenty four}H₅₀+ α_{twenty five}C_{twenty five}H₅₂+ β6C₆H₆+ β7C₇H₈+ β8C₈H_Ten+ β9C₉H ₁₂ + β10C_TenH₁₄+ β11C₁₁H₁₆+ β12C₁₂H₁₈+ β13C₁₃H₂₀+ β14C₁₄H_{twenty two}+ β15C₁₅H_{twenty four}+ β1 6C₁₆H₂₆+ β17C₁₇H₂₈+ β18C₁₈H₃₀+ β19C₁₉H₃₂+ β20C₂₀H₃₄+ β21C_{twenty one}H₃₆+ β22C_{twenty two}H_{3 8} + β23C_{twenty three}H₄₀+ β24C_{twenty four}H₄₂+ β25C_{twenty five}H₄₄+ γ₁CH₂+ γ₂C₂H_Four+ (γ₃+ δ₃) C₃H₆+ (γ_Four+ δ_Four ) C_FourH₈+ (γ_Five+ δ_Five) C_FiveH_Ten+ (γ₆+ δ₆) C₆H₁₂+ (γ₇+ δ₇) C₇H₁₄+ (γ₈+ δ₈) C₈H₁₆+ (γ₉+ δ₉) C₉H₁₈+ (γ_Ten+ δ_Ten) C_TenH₂₀+ (γ₁₁+ δ₁₁) C₁₁H_{twenty two}+ (γ₁₂+ δ₁₂) C₁₂H_{twenty four}+ (γ₁₃+ δ ₁₃ ) C₁₃H₂₆+ (γ₁₄+ δ₁₄) C₁₄H₂₈+ (γ₁₅+ δ₁₅) C₁₅H₃₀+ (γ₁₆+ δ₁₆) C₁₆H₃₂+ (γ₁₇+ δ ₁₇ ) C₁₇H₃₄+ (γ₁₈+ δ₁₈) C₁₈H₃₆+ (γ₁₉+ δ₁₉) C₁₉H₃₈+ (γ₂₀+ δ₂₀) C₂₀H₄₀+ (γ_{twenty one}+ δ _{twenty one} ) C_{twenty one}H₄₂+ (γ_{twenty two}+ δ_{twenty two}) C_{twenty two}H₄₄+ (γ_{twenty three}+ δ_{twenty three}) C_{twenty three}H₄₆+ (γ_{twenty four}+ δ_{twenty four}) C_{twenty four}H₄₈+ (γ_{twenty five}+ δ _{twenty five} ) C_{twenty five}H₅₀] + Y (O₂+ 3.785N₂) → ACO₂+ BH₂O + 3.785yN₂                           --- (13)

In equation (13), the fuel gasoline is
On the left   α₁CH_Four+ α₂C₂H₆+ α₃C₃H₈+ α_FourC_FourH_Ten+ α_FiveC_FiveH₁₂+ α₆C₆H₁₄+ α₇C₇H₁₆+ α₈C₈H₁₈+ α ₉ C₉H₂₀+ α_TenC_TenH_{twenty two}+ α₁₁C₁₁H_{twenty four}+ α₁₂C₁₂H₂₆+ α₁₃C₁₃H₂₈+ α₁₄C₁₄H₃₀+ α₁₅C₁₅H₃₂ + α₁₆C₁₆H₃₄+ α₁₇C₁₇H₃₆+ α₁₈C₁₈H₃₈+ α₁₉C₁₉H₄₀+ α₂₀C₂₀H₄₂+ α_{twenty one}C_{twenty one}H₄₄+ α_{twenty two}C _{twenty two} H₄₆+ α_{twenty three}C_{twenty three}H₄₈+ α_{twenty four}C_{twenty four}H₅₀+ α_{twenty five}C_{twenty five}H₅₂+ β6C₆H₆+ β7C₇H₈+ β8C₈H_Ten+ β9C₉H₁₂ + β10C_TenH₁₄+ β11C₁₁H₁₆+ β12C₁₂H₁₈+ β13C₁₃H₂₀+ β14C₁₄H_{twenty two}+ β15C₁₅H_{twenty four}+ β16C ₁₆ H₂₆+ β17C₁₇H₂₈+ β18C₁₈H₃₀+ β19C₁₉H₃₂+ β20C₂₀H₃₄+ β21C_{twenty one}H₃₆+ β22C_{twenty two}H₃₈+ β23C_{twenty three}H₄₀+ β24C_{twenty four}H₄₂+ β25C_{twenty five}H₄₄+ γ₁CH₂+ γ₂C₂H_Four+ (γ₃+ δ₃) C₃H₆+ (γ_Four+ δ_Four) C _Four H₈+ (γ_Five+ δ_Five) C_FiveH_Ten+ (γ₆+ δ₆) C₆H₁₂+ (γ₇+ δ₇) C₇H₁₄+ (γ₈+ δ₈) C₈H₁₆+ (γ₉+ δ₉ ) C₉H₁₈+ (γ_Ten+ δ_Ten) C_TenH₂₀+ (γ₁₁+ δ₁₁) C₁₁H_{twenty two}+ (γ₁₂+ δ₁₂) C₁₂H_{twenty four}+ (γ₁₃+ δ₁₃) C₁₃H₂₆+ (γ₁₄+ δ₁₄) C₁₄H₂₈+ (γ₁₅+ δ₁₅) C₁₅H₃₀+ (γ₁₆+ δ₁₆) C₁₆H₃₂+ (γ₁₇+ δ₁₇) C₁₇H₃₄+ (γ₁₈+ δ₁₈) C₁₈H₃₆+ (γ₁₉+ δ₁₉) C₁₉H₃₈+ (γ₂₀+ δ₂₀) C₂₀H₄₀+ (γ_{twenty one}+ δ_{twenty one}) C_{twenty one}H₄₂+ (γ_{twenty two}+ δ_{twenty two}) C_{twenty two}H₄₄+ (γ_{twenty three}+ δ_{twenty three}) C_{twenty three}H₄₆+ (γ_{twenty four}+ δ_{twenty four}) C_{twenty four}H₄₈+ (γ_{twenty five}+ δ_{twenty five}) C_{twenty five}H₅₀  --- (14) , Y (O₂+ 3.785N₂) Completely burns its gasoline
It represents the amount of air needed to get it to work. And the right side is
Combustion product gas AC produced by complete combustion of gasoline
O₂+ BH₂O and nitrogen, an inert gas that does not participate in combustion
Gas 3.785yN₂Is.

The coefficients A and B of the combustion product gas ACO ₂ + BH ₂ O and the coefficient y of the amount of air required for complete combustion are (13)
From the formula,

[0057]   A = α₁+ 2α₂+ 3α₃+ 4α_Four+ 5α_Five+ 6α₆+ 7α₇+ 8α₈+ 9α₉+ 10α_Ten+ 11α₁₁+ 12α₁₂+1 3α₁₃+ 14α₁₄+ 15α₁₅+ 16α₁₆+ 17α₁₇+ 18α₁₈+ 19α₁₉+ 20α₂₀+ 21α_{twenty one}+ 22α_{twenty two}+23 α_{twenty three}+ 24α_{twenty four}+ 25α_{twenty five}+ 6β₆+ 7β₇+ 8β₈+ 9β₉+ 10β_Ten+ 11β₁₁+ 12β₁₂+ 13β₁₃+ 14β_{1 Four} + 15β₁₅+ 16β₁₆+ 17β₁₇+ 18β₁₈+ 19β₁₉+ 20β₂₀+ 21β_{twenty one}+ 22β_{twenty two}+ 23β_{twenty three}+ 24β_{twenty four}+ 25β_{twenty five}+ γ₁+ 2γ₂+3 (γ₃+ δ₃) +4 (γ_Four+ δ_Four) +5 (γ_Five+ δ_Five) +6 (γ₆+ δ₆) +7 (γ₇+ δ₇) +8 (γ₈+ δ₈) +9 (γ₉+ δ₉) +10 (γ_Ten+ δ_Ten) +11 (γ₁₁+ δ₁₁) +12 (γ₁₂+ δ₁₂) +13 (γ₁₃+ δ₁₃) +14 (γ₁₄+ δ₁₄) +15 (γ₁₅+ δ₁₅) +16 (γ₁₆+ δ₁₆) +17 (γ₁₇+ δ₁₇) +18 (γ₁₈+ δ ₁₈ ) +19 (γ₁₉+ δ₁₉) +20 (γ₂₀+ δ₂₀) +21 (γ_{twenty one}+ δ_{twenty one}) +22 (γ_{twenty two}+ δ_{twenty two}) +23 (γ_{twenty three}+ δ_{twenty three} ) +24 (γ_{twenty four}+ δ_{twenty four}) +25 (γ_{twenty five}+ δ_{twenty five}) --- (2)

[0058]   B = 2α₁+ 3α₂+ 4α₃+ 5α_Four+ 6α_Five+ 7α₆+ 8α₇+ 9α₈+ 10α₉+ 11α_Ten+ 12α₁₁+ 13α₁₂ + 14α₁₃+ 15α₁₄+ 16α₁₅+ 17α₁₆+ 18α₁₇+ 19α₁₈+ 20α₁₉+ 21α₂₀+ 22α_{twenty one}+ 23α_{twenty two}+2 4α_{twenty three}+ 25α_{twenty four}+ 26α_{twenty five}+ 3β₆+ 4β₇+ 5β₈+ 6β₉+ 7β_Ten+ 8β₁₁+ 9β₁₂+ 10β₁₃+ 11β₁₄+ 12β₁₅+ 13β₁₆+ 14β₁₇+ 15β₁₈+ 16β₁₉+ 17β₂₀+ 18β_{twenty one}+ 19β_{twenty two}+ 20β_{twenty three}+ 21β_{twenty four}+22 β_{twenty five}+ γ₁+ 2γ₂+3 (γ₃+ δ₃) +4 (γ_Four+ δ_Four) +5 (γ_Five+ δ_Five) +6 (γ₆+ δ₆) +7 (γ₇+ δ₇) +8 ( γ₈+ δ₈) +9 (γ₉+ δ₉) +10 (γ_Ten+ δ_Ten) +11 (γ₁₁+ δ₁₁) +12 (γ₁₂+ δ₁₂) +13 (γ₁₃+ δ ₁₃ ) +14 (γ₁₄+ δ₁₄) +15 (γ₁₅+ δ₁₅) +16 (γ₁₆+ δ₁₆) +17 (γ₁₇+ δ₁₇) +18 (γ₁₈+ δ₁₈ ) +19 (γ₁₉+ δ₁₉) +20 (γ₂₀+ δ₂₀) +21 (γ_{twenty one}+ δ_{twenty one}) +22 (γ_{twenty two}+ δ_{twenty two}) +23 (γ_{twenty three}+ δ_{twenty three}) + 24 (γ_{twenty four}+ δ_{twenty four}) +25 (γ_{twenty five}+ δ_{twenty five}) --- (3)

[0059]   y = A + B / 2 --- (15)

The weight of air can be obtained from the air amount y (O ₂ + 3.785N ₂ ) in the equations (13) and (15). Molecular weight of oxygen O ₂ 3
From 2.00 and the molecular weight of nitrogen N ₂ of 28.014, the air weight is y × (31.998 + 3.785 × 28.014) = 138.03 y = 138.03 (A + B / 2) --- (16).

The gasoline weight Wf is α_n, Β_n,
γ_n, Δ_nThe atomic weight of carbon C given by the analysis result of 12.01
1, from the atomic weight of hydrogen H 1.008, from equation (14)   Wf = α₁CH_Four+ α₂C₂H₆+ α₃C₃H₈+ α_FourC_FourH_Ten+ α_FiveC_FiveH₁₂+ α₆C₆H₁₄+ α₇C₇H₁₆+ α₈C₈ H₁₈+ α₉C₉H₂₀+ α_TenC_TenH_{twenty two}+ α₁₁C₁₁H_{twenty four}+ α₁₂C₁₂H₂₆+ α₁₃C₁₃H₂₈+ α₁₄C₁₄H₃₀+ α₁₅ C₁₅H₃₂+ α₁₆C₁₆H₃₄+ α₁₇C₁₇H₃₆+ α₁₈C₁₈H₃₈+ α₁₉C₁₉H₄₀+ α₂₀C₂₀H₄₂+ α_{twenty one}C_{twenty one}H₄₄ + α_{twenty two}C_{twenty two}H₄₆+ α_{twenty three}C_{twenty three}H₄₈+ α_{twenty four}C_{twenty four}H₅₀+ α_{twenty five}C_{twenty five}H₅₂+ β₆C₆H₆+ β₇C₇H₈ + β₈C₈H_Ten+ β₉C₉H₁₂+ β_TenC_TenH₁₄+ β₁₁C₁₁H₁₆+ β₁₂C₁₂H₁₈+ β₁₃C₁₃H₂₀+ β₁₄C₁₄H_{twenty two}+ β₁₅C₁₅H _{twenty four} + β₁₆C₁₆H₂₆+ β₁₇C₁₇H₂₈+ β₁₈C₁₈H₃₀+ β₁₉C₁₉H₃₂+ β₂₀C₂₀H₃₄+ β_{twenty one}C_{twenty one}H₃₆+ β_{2 2} C_{twenty two}H₃₈+ β_{twenty three}C_{twenty three}H₄₀+ β_{twenty four}C_{twenty four}H₄₂+ β_{twenty five}C_{twenty five}H₄₄+ γ₁CH₂+ γ₂C₂H_Four+ (γ₃+ δ₃) C₃H₆+ ( γ_Four+ δ_Four) C_FourH₈+ (γ_Five+ δ_Five) C_FiveH_Ten+ (γ₆+ δ₆) C₆H₁₂+ (γ₇+ δ₇) C₇H₁₄+ (γ₈+ δ₈) C₈H₁₆ + (γ₉+ δ₉) C₉H₁₈+ (γ_Ten+ δ_Ten) C_TenH₂₀+ (γ₁₁+ δ₁₁) C₁₁H_{twenty two}+ (γ₁₂+ δ₁₂) C₁₂H_{twenty four}+ ( γ₁₃+ δ₁₃) C₁₃H₂₆+ (γ₁₄+ δ₁₄) C₁₄H₂₈+ (γ₁₅+ δ₁₅) C₁₅H₃₀+ (γ₁₆+ δ₁₆) C₁₆H₃₂+ ( γ₁₇+ δ₁₇) C₁₇H₃₄+ (γ₁₈+ δ₁₈) C₁₈H₃₆+ (γ₁₉+ δ₁₉) C₁₉H₃₈+ (γ₂₀+ δ₂₀) C₂₀H₄₀+ ( γ_{twenty one}+ δ_{twenty one}) C_{twenty one}H₄₂+ (γ_{twenty two}+ δ_{twenty two}) C_{twenty two}H₄₄+ (γ_{twenty three}+ δ_{twenty three}) C_{twenty three}H₄₆+ (γ_{twenty four}+ δ_{twenty four}) C_{twenty four}H₄₈+ ( γ_{twenty five}+ δ_{twenty five}) C_{twenty five}H₅₀--- (4) Required from. Therefore, this theoretical air-fuel ratio is     Theoretical air-fuel ratio = 138.03 (A + B / 2) / Wf --- (1) Becomes This method is an exact solution of gasoline theoretical air-fuel ratio calculation
is there.

The content ratio of hydrocarbons constituting gasoline is
By adjusting based on the above formula (1), gasoline having a desired stoichiometric air-fuel ratio can be manufactured.

[0063]

[Example] Separate regular gasoline into each hydrocarbon system,
Table 4 shows an example of analysis by carbon number. The regular air-fuel ratio of the regular gasoline shown in Table 4 is calculated as follows by applying this general formula.

[0064]

[Table 4]

The paraffinic hydrocarbon has a carbon number C3 to C.
13 types, 13 types of aromatic compounds, 7 types of C6 to C12 types, 8 types of olefins and naphthenes of 8 types of C4 to C11 types, total 26 types of hydrocarbons are completely burned and carbon C
Is converted into carbon dioxide gas CO ₂ and hydrogen H is converted into steam H ₂ O, the combustion chemical reaction formula is obtained as shown in formula (17).

0.03C ₃ H ₈ + 2.55C ₄ H ₁₀ + 15.35C ₅ H ₁₂ + 15.90C ₆ H ₁₄ + 8.91C ₇ H ₁₆ + 3.45C ₈ H ₁₈ + 2.31C ₉ H ₂₀ + 1.78C ₁₀ H ₂₂ + 0.32C ₁₁ H ₂₄ + 0.14C ₁₂ H ₂₆ + 0.04C ₁₃ H ₂₈ + 0.62C ₆ H ₆ + 10.87C ₇ H ₈ + 4.38C ₈ H ₁₀ + 5.94C ₉ H ₁₂ + 3.63C ₁₀ H ₁₄ + 0.30C ₁₁ H ₁₆ + 0.12C ₁₂ H ₁₈ + 2.04C ₄ H ₈ + 5.97C ₅ H ₁₀ + 4.30C ₆ H ₁₂ + 5.47C ₇ H ₁₄ + 2.41C ₈ H ₁₆ + 2.69C ₉ H ₁₈ + 0.37C ₁₀ H ₂₀ + 0.10C ₁₁ H ₂₂ + y (O ₂ + 3.785N ₂ ) → ACO ₂ + BH ₂ O + 3.785yN ₂ --- (17)

Coefficients A, B, y for obtaining the theoretical air-fuel ratio
From Equation (2), A = 676.46 Equation (3), B = 649.66 From Equation (15), y = 676.46 + 649.66 / 2 = 1001.29

Therefore, the air weight is 1001.29 (O ₂ +3.7
85N ₂ ) = 1001.29 × (32.00 + 3.785 × 28.014) = 138209.05

The fuel weight Wf is From equation (4), Wf = 9434.680

Therefore, the strictly determined theoretical air-fuel ratio of the regular gasoline shown in Table 4 is theoretical air-fuel ratio = 138209.05 / 9434.680 = 14.65.

A method of correcting the stoichiometric air-fuel ratio of gasoline will be described.

The stoichiometric air-fuel ratio of regular gasoline shown in Table 4 is 14.65. The theoretical air-fuel ratio of this gasoline is 14.70
Let's think about. When increasing the air-fuel ratio, qualitatively, the paraffinic hydrocarbon having a large stoichiometric air-fuel ratio may be increased and the small aromatic hydrocarbon may be decreased.

Now, in Table 4, paraffin hydrocarbons
Increase the capacity% of C6 and C7 by 1.00% and 2.00% respectively to 15.90
To 16.91 and 8.91 to 10.91%, and conversely reduce the volume% of aromatic hydrocarbons C7 and C9 by 2.00 and 1.00%, respectively.
If 10.87 is set to 8.87 and 5.94 is set to 4.94% and the theoretical air-fuel ratio is calculated in the same manner as above, A = 667.48 B = 652.98 y = 993.97

Therefore, the air weight is 993.97 (O ₂ + 3.785N ₂ ).
= 993.97 × (32.00 + 3.785 × 28.014) = 137200.65. The gasoline weight is Wf = 9333.51.

This theoretical air-fuel ratio is A / Fc = 137200.65 / 9333.51 = 14.70 The stoichiometric air-fuel ratio can be adjusted.

In this way, by adjusting the content ratio of hydrocarbons constituting gasoline based on the above equation (1), it becomes possible to produce gasoline having a desired theoretical air-fuel ratio.

[0077]

As described above, according to the present invention, the stoichiometric air-fuel ratio of gasoline can be determined exactly,
Based on this calculation method, a useful effect such that gasoline having a desired stoichiometric air-fuel ratio can be produced is brought about.

[Brief description of drawings]

FIG. 1 is a graph showing the relationship between carbon number and stoichiometric air-fuel ratio.

Claims (3)

[Claims] 1. The theoretical air-fuel ratio of gasoline is calculated by the following equation (1).
The theoretical sky of gasoline, characterized by being calculated by
Calculation method of fuel ratio.   Theoretical air-fuel ratio = 138.03 (A + B / 2) / Wf --- (1) However, in the above formula (1),     A = α₁+ 2α₂+ 3α₃+ 4α_Four+ 5α_Five+ 6α₆+ 7α₇+ 8α₈+ 9α₉+ 10α_Ten+ 11α₁₁+ 12α₁₂ + 13α₁₃+ 14α₁₄+ 15α₁₅+ 16α₁₆+ 17α₁₇+ 18α₁₈+ 19α₁₉+ 20α₂₀+ 21α_{twenty one}+ 22α_{twenty two}+2 3α_{twenty three}+ 24α_{twenty four}+ 25α_{twenty five}+ 6β₆+ 7β₇+ 8β₈+ 9β₉+ 10β_Ten+ 11β₁₁+ 12β₁₂+ 13β₁₃+ 14β ₁₄ + 15β₁₅+ 16β₁₆+ 17β₁₇+ 18β₁₈+ 19β₁₉+ 20β₂₀+ 21β_{twenty one}+ 22β_{twenty two}+ 23β_{twenty three}+ 24β_{twenty four} + 25β_{twenty five}+ γ₁+ 2γ₂+3 (γ₃+ δ₃) +4 (γ_Four+ δ_Four) +5 (γ_Five+ δ_Five) +6 (γ₆+ δ₆) +7 (γ₇+ δ₇) + 8 (γ₈+ δ₈) +9 (γ₉+ δ₉) +10 (γ_Ten+ δ_Ten) +11 (γ₁₁+ δ₁₁) +12 (γ₁₂+ δ₁₂) +13 (γ₁₃+ δ₁₃) +14 (γ₁₄+ δ₁₄) +15 (γ₁₅+ δ₁₅) +16 (γ₁₆+ δ₁₆) +17 (γ₁₇+ δ₁₇) +18 (γ₁₈+ δ ₁₈ ) +19 (γ₁₉+ δ₁₉) +20 (γ₂₀+ δ₂₀) +21 (γ_{twenty one}+ δ_{twenty one}) +22 (γ_{twenty two}+ δ_{twenty two}) +23 (γ_{twenty three}+ δ_{twenty three} ) +24 (γ_{twenty four}+ δ_{twenty four}) +25 (γ_{twenty five}+ δ_{twenty five}) --- (2)     B = 2α₁+ 3α₂+ 4α₃+ 5α_Four+ 6α_Five+ 7α₆+ 8α₇+ 9α₈+ 10α₉+ 11α_Ten+ 12α₁₁+ 13α ₁₂ + 14α₁₃+ 15α₁₄+ 16α₁₅+ 17α₁₆+ 18α₁₇+ 19α₁₈+ 20α₁₉+ 21α₂₀+ 22α_{twenty one}+ 23α_{twenty two} + 24α_{twenty three}+ 25α_{twenty four}+ 26α_{twenty five}+ 3β₆+ 4β₇+ 5β₈+ 6β₉+ 7β_Ten+ 8β₁₁+ 9β₁₂+ 10β₁₃+ 11β_{1 Four} + 12β₁₅+ 13β₁₆+ 14β₁₇+ 15β₁₈+ 16β₁₉+ 17β₂₀+ 18β_{twenty one}+ 19β_{twenty two}+ 20β_{twenty three}+ 21β_{twenty four}+ 22β_{twenty five}+ γ₁+ 2γ₂+3 (γ₃+ δ₃) +4 (γ_Four+ δ_Four) +5 (γ_Five+ δ_Five) +6 (γ₆+ δ₆) +7 (γ₇+ δ₇) +8 (γ₈+ δ₈) +9 (γ₉+ δ₉) +10 (γ_Ten+ δ_Ten) +11 (γ₁₁+ δ₁₁) +12 (γ₁₂+ δ₁₂) +13 (γ₁₃+ δ₁₃) +14 (γ₁₄+ δ₁₄) +15 (γ₁₅+ δ₁₅) +16 (γ₁₆+ δ₁₆) +17 (γ₁₇+ δ₁₇) +18 (γ₁₈+ δ ₁₈ ) +19 (γ₁₉+ δ₁₉) +20 (γ₂₀+ δ₂₀) +21 (γ_{twenty one}+ δ_{twenty one}) +22 (γ_{twenty two}+ δ_{twenty two}) +23 (γ_{twenty three}+ δ_{twenty three} ) +24 (γ_{twenty four}+ δ_{twenty four}) +25 (γ_{twenty five}+ δ_{twenty five}) --- (3)     Wf = α₁CH_Four+ α₂C₂H₆+ α₃C₃H₈+ α_FourC_FourH_Ten+ α_FiveC_FiveH₁₂+ α₆C₆H₁₄+ α₇C₇H₁₆+ α₈ C₈H₁₈+ α₉C₉H₂₀+ α_TenC_TenH_{twenty two}+ α₁₁C₁₁H_{twenty four}+ α₁₂C₁₂H₂₆+ α₁₃C₁₃H₂₈+ α₁₄C₁₄H₃₀+ α ₁₅ C₁₅H₃₂+ α₁₆C₁₆H₃₄+ α₁₇C₁₇H₃₆+ α₁₈C₁₈H₃₈+ α₁₉C₁₉H₄₀+ α₂₀C₂₀H₄₂+ α_{twenty one}C_{twenty one}H ₄₄ + α_{twenty two}C_{twenty two}H₄₆+ α_{twenty three}C_{twenty three}H₄₈+ α_{twenty four}C_{twenty four}H₅₀+ α_{twenty five}C_{twenty five}H₅₂+ β₆C₆H₆+ β₇C₇H₈ + β₈C₈H_Ten + β₉C₉H₁₂+ β_TenC_TenH₁₄+ β₁₁C₁₁H₁₆+ β₁₂C₁₂H₁₈+ β₁₃C₁₃H₂₀+ β₁₄C₁₄H_{twenty two}+ β₁₅C₁₅ H_{twenty four}+ β₁₆C₁₆H₂₆+ β₁₇C₁₇H₂₈+ β₁₈C₁₈H₃₀+ β₁₉C₁₉H₃₂+ β₂₀C₂₀H₃₄+ β_{twenty one}C_{twenty one}H₃₆+ β _{twenty two} C_{twenty two}H₃₈+ β_{twenty three}C_{twenty three}H₄₀+ β_{twenty four}C_{twenty four}H₄₂+ β_{twenty five}C_{twenty five}H₄₄+ γ₁CH₂+ γ₂C₂H_Four+ (γ₃+ δ₃) C₃H₆+ ( γ_Four+ δ_Four) C_FourH₈+ (γ_Five+ δ_Five) C_FiveH_Ten+ (γ₆+ δ₆) C₆H₁₂+ (γ₇+ δ₇) C₇H₁₄+ (γ₈+ δ₈) C₈H₁₆ + (γ₉+ δ₉) C₉H₁₈+ (γ_Ten+ δ_Ten) C_TenH₂₀+ (γ₁₁+ δ₁₁) C₁₁H_{twenty two}+ (γ₁₂+ δ₁₂) C₁₂H_{twenty four}+ ( γ₁₃+ δ₁₃) C₁₃H₂₆+ (γ₁₄+ δ₁₄) C₁₄H₂₈+ (γ₁₅+ δ₁₅) C₁₅H₃₀+ (γ₁₆+ δ₁₆) C₁₆H₃₂+ ( γ₁₇+ δ₁₇) C₁₇H₃₄+ (γ₁₈+ δ₁₈) C₁₈H₃₆+ (γ₁₉+ δ₁₉) C₁₉H₃₈+ (γ₂₀+ δ₂₀) C₂₀H₄₀+ ( γ_{twenty one}+ δ_{twenty one}) C_{twenty one}H₄₂+ (γ_{twenty two}+ δ_{twenty two}) C_{twenty two}H₄₄+ (γ_{twenty three}+ δ_{twenty three}) C_{twenty three}H₄₆+ (γ_{twenty four}+ δ_{twenty four}) C_{twenty four}H₄₈+ ( γ_{twenty five}+ δ_{twenty five}) C_{twenty five}H₅₀                                              --- (4) However, in the above equations (2), (3) and (4), α_n: Paraffin hydrocarbon content by volume (%) β_n: Volume content of aromatic hydrocarbons (%) γ_n: Volume content of olefinic hydrocarbons (%) δ_n: Capacity content of naphthene hydrocarbons (%) 2. A method for producing gasoline, which comprises producing a gasoline having a desired stoichiometric air-fuel ratio by adjusting the content ratio of hydrocarbons constituting the gasoline based on the above formula (1). . 3. The method according to claim 2, wherein when the stoichiometric air-fuel ratio of gasoline is increased, the content ratio of paraffinic hydrocarbons is increased and the content ratio of aromatic hydrocarbons is decreased. , Method of manufacturing gasoline.