CN105868565A - Theoretical calculation method for lean combustion limit of fuel - Google Patents

Abstract

The invention provides a theoretical calculation method for the lean combustion limit of fuel. The method specifically includes the following steps that firstly, the mole fraction of each component is determined; secondly, each type of diluent gas is converted to a mole fraction of equivalent N2; thirdly, mixed gas containing fuel is approximated to mixed gas composed of fuel, O2 and N2; fourthly, normalization processing is conducted on the mole fractions of the components; fifthly, the mole fraction ratio lambda of N2 to O2 is worked out; sixthly, a non-theoretical specific general reaction equation of fuel, O2 and N2 under the oxygen-enriched condition is constructed; seventhly, the adiabatic combustion temperature of fuel under the lean combustion limit is assumed, and the mole fraction xf of fuel is obtained through reverse deduction according to an adiabatic combustion temperature calculation formula and the non-theoretical specific general reaction equation.

Description

A kind of fuel lean-burn limit theory computational methods

Technical field

The invention belongs to technical field of fuel combustion, be specifically related to a kind of fuel lean-burn limit theory computational methods.

Background technology

The high-efficiency cleaning of commercial production safety, fuel is burnt significant by the inflammability limit of fuel.On May 25th, 1812 there occurs famous Felling mine disaster near Newcastle UK, cause 92 people dead, the investigation of current mine disaster event finds to have stepped the first step that the fuel fire limit is explored by the mankind, the team of Davy jazz leader proposes the concept of inflammability limit and disclose the principle of fuel fire first, becomes the indispensable safety device in colliery as the Davy safelight lamp of one of this achievement investigated.Carry out the large scale experiment for the fuel fire limit in 20 th Century United States Bureau of Mines to study, having obtained the gaseous mixture inflammability limit that a large amount of fuel includes that hydro carbons, alcohols, ketone etc. are constituted from different diluent gas under room temperature (25 DEG C), normal pressure (1atm), its experimental data is still researcher, the commercial production business research extremely important reference material of the fuel fire limit so far.Jones summarizes and affects fire tube and measure 8 factors of the fuel fire limit, including: the flame direction of propagation, fire tube designs, the length of tube that catches fire, mixture temperature, pressure, the concentration of diluent gas and ignition energy.

In addition to coal mining enterprise, the fuel fire limit is also paid close attention to by multiple industries and is included chemistry and process industrial safety in production, cold-producing medium working medium safety, security against fire and combustion completion.As a example by cold-producing medium working medium safety, the nineties in 20th century, countries in the world government decided was phased out alkyl halide cold-producing medium, and Substitute Working Medium has quite a few to be flammable, it is therefore desirable to once reveal caused fire risk in view of working medium.And the fierce swift and violent even explosion type of the Ignition of part working medium, the most more need the inflammability limit of working medium is made systematic research.

Summary of the invention

At present the fuel fire limit typically under room temperature, normal pressure by constant volume combustion bomb or fire tube test, the accuracy of experimental result is affected the biggest by the volume of experimental facilities, planform, size, visualization constant volume combustion bomb especially typically can not be run with fire tube under high temperature (＞ 1200K), high pressure (＞ 20bar), and therefore the measuring of the fuel fire limit is greatly limited.And the inflammability limit that the various test fuel one by one of the classification of fuel is under different temperatures, pressure, workload is huge and causes the wasting of resources, it is therefore necessary to propose a kind of simple, theoretical calculation method that accuracy preferably predicts the fuel fire limit.

The present invention proposes a kind of fuel lean-burn limit theory computational methods, it is characterised in that: for calculate that pure matter fuel and one or more diluent gas constitute containing the aerial lean flammability limit of fuel mixture；Specifically include following steps:

A) determine containing component molar mark each in fuel mixture, it is A%Fuel+B1%Inert1+B2%Inert2+ respectively ... Bj%Inertj+ ...+Bp%Inertp+C%Air, i.e. include that fuel Fuel that molar fraction is A%, molar fraction are respectively B1%, B2% ... the p kind diluent gas Inert of Bj% ... Bp% and the air Air that molar fraction is C%, and A%+B1%+B2%+ ... Bj%+ ...+Bp%+C%=100%；

B) respective molar fraction is multiplied by corresponding equivalent coefficient Kk respectively by p kind diluent gas according to dilution effect, is converted into the nitrogen N of equivalent₂Molar fraction D%, i.e. D%=B1% × Kk1+B2% × Kk2+ ... Bj% × Kkj+ ...+Bp% × Kkp；Now it is respectively A%Fuel+D%N containing each component molar mark in fuel mixture₂The equivalent coefficient Kk of the various conventional diluent gas of+C%Air is as shown in the table:

C) fuel mixture will be contained be approximately by fuel-oxy-nitrogen (Fuel-O₂-N₂) gaseous mixture that constitutes, now the molar fraction of fuel Fuel is A%, oxygen O₂Molar fraction be E%=20.95% × C%, nitrogen N₂Molar fraction be F%=D%+79.05% × C%, be now respectively A%Fuel+E%O containing each component molar mark in fuel mixture₂+ F%N₂；

D) three component gaseous mixture Fuel-O approximation obtained₂-N₂Molar fraction be normalized, fuel Fuel molar fraction is G%=A%/(A%+E%+F%) %, oxygen O₂Molar fraction is H%=E%/(A%+E%+F%) %, nitrogen N₂Molar fraction be I%=F%/(A%+E%+F%) %, be now respectively G%Fuel+H%O containing each component molar mark in fuel mixture₂+ I%N₂, but now G%+H%+I%=100%；

E) nitrogen N is obtained₂With oxygen O₂The ratio λ=I%/H% of molar fraction；

F) structure single-fuel Fuel, oxygen O₂And nitrogen N₂In excess oxygen i.e. equivalent proportionNon-theory than general reactional equation:

Wherein:

$v_{\mathrm{air}}=a+\frac{b}{4}-\frac{c}{2}$

${\mathrm{\μ}}_{{\mathrm{CO}}_2}=a$

${\mathrm{\μ}}_{H_{2}O}=\frac{b}{2}$

${\mathrm{\μ}}_{N_2}=\mathrm{\λ}\×{\mathrm{\μ}}_{\mathrm{air}}+\frac{d}{2}$

${\mathrm{\μ}}_{o_2}={\mathrm{\μ}}_{\mathrm{air}}-a-\frac{b}{4}+\frac{c}{2}$

G) assume that fuel adiabatic combustion temperature under lean flammability limit is T_AFT, ambient temperature be T_o, search the enthalpy of combustion Δ H of fuel_c, the specific heat at constant pressure c of all combustion products_{p , i}, according to following adiabatic combustion temperature computing formula and above-mentioned non-theory molar fraction x that push away now fuel more counter than general reactional equation_f, obtain fuel lean flammability limit x under this gaseous mixture_L；

$T_{\mathrm{AFT}}=T_o+\frac{\mathrm{\Δ}{H}_c}{\mathrm{\Σ}{\mathrm{\μ}}_i\×c_{p,i}}$

Wherein:

T_AFT-adiabatic combustion temperature, unit K；

T_o-ambient temperature, unit K；

ΔH_cThe enthalpy of combustion of-fuel, unit J/mol；

μ_i-non-theory is than the coefficient of i-th kind of combustion product in general reactional equation；

c_{p , i}-non-theory is than the specific heat at constant pressure of i-th kind of combustion product in general reactional equation, unit J/ (mol*K).

Accompanying drawing explanation

Fig. 1 is the flowchart of the present invention.

Fig. 2 is the methane ignition administrative division map that experiment draws.

Fig. 3 is for utilizing the principle of the invention calculated methane lean flammability limit schematic diagram.

Detailed description of the invention

Describe the detailed description of the invention of the present invention below in conjunction with drawings and Examples in detail, but the present invention is not limited by described specific embodiment.The flow process that realizes of following embodiment accompanying drawing 1 the most to specifications calculates.

Embodiment 1: existing containing fuel mixture 9%CH₄+ 10%CO₂+ 81%Air, in volume is 12L constant volume combustion bomb, seeks (10%CO under identical diluting condition₂)CH₄Lean flammability limit.

Answer:

A () determines containing component molar mark each in fuel mixture: this gaseous mixture contains 9%CH₄, 10%CO₂And 81%Air；

B () utilizes the N of diluent gas₂Equivalent coefficientBy CO₂The nitrogen N of equivalent it is converted into according to dilution effect₂Molar fraction；

CO is known by above table₂N₂Equivalent coefficient Kk=1.5

CO₂Equivalent N₂Molar fraction=10% × 1.5=15%

C () will contain fuel mixture and be approximately by fuel-oxy-nitrogen (Fuel-O₂-N₂) gaseous mixture that constitutes；

CH₄Molar fraction=9%

O₂Molar fraction=81% × 20.95%=16.9695%

N₂Molar fraction=15%+81% × 79.05%=79.0305%

CH₄-O₂-N₂In the gaseous mixture constituted, each component molar mark is respectively 9%Fuel+16.9695%O₂+ 79.0305%N₂

Total molar fraction=the 9%+16.9695%+79.0305%=105% of gaseous mixture after conversion

D three component gaseous mixture Fuel-O that approximation is obtained by ()₂-N₂Molar fraction be normalized；

8.5714%+16.1614%+75.2671%=100%

E () obtains nitrogen N₂With oxygen O₂The ratio λ of molar fraction；

$\mathrm{\λ}=\frac{75.2671\%}{16.1614\%}=4.6572$

(f) structure single-fuel Fuel, oxygen O₂And nitrogen N₂In excess oxygen i.e. equivalent proportionNon-theory than general reactional equation:

Wherein:

$v_{\mathrm{air}}=1+\frac{4}{4}=2$

${\mathrm{\μ}}_{{\mathrm{CO}}_2}=1$

${\mathrm{\μ}}_{H_{2}O}=\frac{4}{2}=2$

${\mathrm{\μ}}_{N_2}=\mathrm{\λ}\×{\mathrm{\μ}}_{\mathrm{air}}=\mathrm{\λ}\×\frac{1-x_f}{4.773x_f}$

${\mathrm{\μ}}_{O_2}={\mathrm{\μ}}_{\mathrm{air}}-a-\frac{b}{4}+\frac{c}{2}=\frac{1-x_f}{4.773x_f}-2$

G () assumes CH₄Adiabatic combustion temperature under lean flammability limit is T_AFT=1450K, ambient temperature are T_o=298K, searches CH₄Enthalpy of combustion Δ H_c=802480J/mol, CO₂、H₂O、N₂And O₂Specific heat at constant pressureWithIt is respectively 54.3J/ (mol*K), 41.2J/ (mol*K), 32.7J/ (mol*K) and 34.9J/ (mol*K).

Above-mentioned condition is substituted into adiabatic combustion temperature computing formula

$T_{\mathrm{AFT}}=T_o+\frac{\mathrm{\Δ}{H}_c}{\mathrm{\Σ}{\mathrm{\μ}}_i\×c_{p,i}}$

I.e.

$1450=298+\frac{802480}{1\×54.3+2\×41.2+34.9\×(\frac{1-x_f}{4.773x_f}-2)+32.7\×\mathrm{\λ}\×\frac{1-x_f}{4.773x_f}}$

Solve equation

CH₄Molar fraction x_f=5.878%

I.e. CH₄(10%CO under this diluting condition₂) lean flammability limit x_L=x_f=5.878%

The experimental data that contrast constant volume combustion bomb obtains, now CH₄Lean flammability limit is 5.6% preferable with lean flammability limit predictive value 5.878% degree of agreement of the present invention.

Thus the present invention seen from embodiment achieves the beneficial effect of basic Accurate Prediction fuel lean flammability limit.

In like manner, fuel lean flammability limit in multicomponent diluent gas can be calculated according to the principle of the present invention, only the different component in diluent gas need to be converted into the N of respective quality mark₂?.EGR Technology measure is such as often taked to reduce NO in combustion process of the internal-combustion engine_x, and the waste gas participating in recirculation is usually CO₂、H₂O and N₂Gaseous mixture, therefore prediction fuel lean flammability limit in many components diluent gas the most necessary.Example below utilizes the principle of the invention to calculate fuel lean flammability limit in multicomponent diluent gas.

Embodiment 2: existing containing fuel mixture 7%CH₄+ 20%Inert+73%Air is in volume is 12L constant volume combustion bomb, and wherein diluent gas Inert is by 20%CO₂And 80%N₂Composition, ambient temperature and pressure are respectively 25 DEG C, 1atm, seek (20%Inert) CH under identical diluting condition₄Lean flammability limit.

Answer:

A () determines containing component molar mark each in fuel mixture；

CO in diluent gas Inert₂Molar fraction=20% × 20%=4%

N in diluent gas Inert₂Molar fraction=20% × 80%=16%

Therefore this gaseous mixture contains 7%CH₄, 4%CO₂, 16%N₂And 73%Air.

B () utilizes the N of diluent gas₂Equivalent coefficientRespectively by the CO in diluent gas Inert₂And N₂The nitrogen N of equivalent it is converted into according to dilution effect₂Molar fraction；

CO is known by above table₂N₂Equivalent coefficient Kk=1.5, N₂N₂Equivalent coefficient Kk=1

CO₂Equivalent N₂Molar fraction=4% × 1.5=6%

N₂Equivalent N₂Molar fraction=16% × 1=16%

C () will contain fuel mixture and be approximately by fuel-oxy-nitrogen (Fuel-O₂-N₂) gaseous mixture that constitutes；

CH₄Molar fraction=7%

O in gaseous mixture₂Molar fraction=73% × 20.95%=16.9695%

N in gaseous mixture₂Molar fraction=6%+16%+73% × 79.05%=79.7065%

CH₄-O₂-N₂In the gaseous mixture constituted, each component molar mark is respectively 7%Fuel+16.9695%O₂+ 79.7065%N₂

Total molar fraction=the 7%+16.9695%+79.7065%=103.676% of gaseous mixture after conversion

D three component gaseous mixture Fuel-O that approximation is obtained by ()₂-N₂Molar fraction be normalized；

6.7518%+16.3678%+76.8804%=100%

E () obtains nitrogen N₂With oxygen O₂The ratio λ of molar fraction；

$\mathrm{\λ}=\frac{76.8804\%}{16.3678\%}=4.6971$

(f) structure single-fuel Fuel, oxygen O₂And nitrogen N₂In excess oxygen i.e. equivalent proportionNon-theory than general reactional equation:

Wherein:

$v_{\mathrm{air}}=1+\frac{4}{4}=2$

${\mathrm{\μ}}_{{\mathrm{CO}}_2}=1$

${\mathrm{\μ}}_{H_{2}O}=\frac{4}{2}=2$

${\mathrm{\μ}}_{N2}=\mathrm{\λ}\×{\mathrm{\μ}}_{\mathrm{air}}=\mathrm{\λ}\×\frac{1-x_f}{4.773x_f}$

${\mathrm{\μ}}_{O_2}={\mathrm{\μ}}_{\mathrm{air}}-a-\frac{b}{4}+\frac{c}{2}=\frac{1-x_f}{4.773x_f}-2$

G () assumes CH₄Adiabatic combustion temperature under lean flammability limit is T_AFT=1400K, ambient temperature are T_o=298K, searches CH₄Enthalpy of combustion Δ H_c=802480J/mol, CO₂、H₂O、N₂And O₂Specific heat at constant pressureWithIt is respectively 54.3J/ (mol*K), 41.2J/ (mol*K), 32.7J/ (mol*K) and 34.9J/ (mol*K).

Above-mentioned condition is substituted into adiabatic combustion temperature computing formula

$T_{\mathrm{AFT}}=T_o+\frac{\mathrm{\Δ}{H}_c}{\mathrm{\Σ}{\mathrm{\μ}}_i*c_{p,i}}$

I.e.

$1400=298+\frac{802480}{1\×54.3+2\×41.2+34.9\×(\frac{1-x_f}{4.773x_f}-2)+32.7\×\mathrm{\λ}\×\frac{1-x_f}{4.773x_f}}$

Solve equation

CH₄Molar fraction x_f=5.649%

I.e. CH₄Lean flammability limit x under this diluting condition_L=x_f=5.649%

The experimental data that contrast constant volume combustion bomb obtains, now CH₄Lean flammability limit is 5.4% preferable with lean flammability limit predictive value 5.649% degree of agreement of the present invention.

Thus the present invention seen from embodiment achieves the beneficial effect of basic Accurate Prediction fuel lean flammability limit.

Accompanying drawing 2 be fire tube methane air-diluent gas of obtaining of experiment at 25 DEG C, fire area under 1atm, accompanying drawing 3 is the lean flammability limit using principle of the invention prediction, and absolute error is all within 6%

Summary, in fuel fire Limits properties, fuel lean-burn limit theory computational methods proposed by the invention are preferable with experimental result degree of agreement, achieve the beneficial effect of basic Accurate Prediction fuel lean flammability limit.

The embodiment of present invention described above, is not intended that limiting the scope of the present invention.Any amendment, equivalent and improvement etc. done within the spiritual principles of the present invention, within should be included in the claims of the present invention.

Claims (1)

1. fuel lean-burn limit theory computational methods, it is characterised in that: it is used for calculating pure matter fuel and dilutes with one or more Gas constitute containing the aerial lean flammability limit of fuel mixture；Specifically include following steps:

A) determine containing component molar mark each in fuel mixture, be A%Fuel+B1%Inert1+B2%Inert2+ respectively ... Bj%Inertj+ ...+Bp%Inertp+C%Air, i.e. includes fuel Fuel, molar fraction that molar fraction is A% It is respectively B1%, B2% ... the p kind diluent gas Inert of Bj% ... Bp% and molar fraction are the sky of C% Gas Air, and A%+B1%+B2%+ ... Bj%+ ...+Bp%+C%=100%；

B) respective molar fraction is multiplied by corresponding equivalent coefficient Kk respectively by p kind diluent gas according to dilution effect, folding It is combined into the nitrogen N of equivalent₂Molar fraction D%, i.e. D%=B1% × Kk1+B2% × Kk2+ ... Bj% × Kkj+ ...+Bp% × Kkp；Now it is respectively containing each component molar mark in fuel mixture A%Fuel+D%N₂+ C%Air；The equivalent coefficient Kk of various conventional diluent gas is as shown in the table:

C) fuel mixture will be contained be approximately by fuel-oxy-nitrogen (Fuel-O₂-N₂) gaseous mixture that constitutes, now fire The molar fraction of material Fuel is A%, oxygen O₂Molar fraction be E%=20.95% × C%, nitrogen N₂'s Molar fraction is F%=D%+79.05% × C%, is now respectively containing each component molar mark in fuel mixture A%Fuel+E%O₂+ F%N₂；

D) three component gaseous mixture Fuel-O approximation obtained₂-N₂Molar fraction be normalized, fuel Fuel Molar fraction is G%=A%/(A%+E%+F%) %, oxygen O₂Molar fraction be H%=E%/ (A%+E%+F%) %, nitrogen N₂Molar fraction be I%=F%/(A%+E%+F%) %, now contain In fuel mixture, each component molar mark is respectively G%Fuel+H%O₂+ I%N₂, but now G%+H%+I%=100%；

E) nitrogen N is obtained₂With oxygen O₂The ratio λ=I%/H% of molar fraction；

F) structure single-fuel Fuel, oxygen O₂And nitrogen N₂In excess oxygen i.e. equivalent proportionNon-theory ratio general Reactional equation:

$C_a{H}_b{O}_c{N}_d+{\mathrm{\μ}}_{\mathrm{air}}(O_2+{\mathrm{\λN}}_2)\→{\mathrm{\μ}}_{{\mathrm{CO}}_2}{\mathrm{CO}}_2+{\mathrm{\μ}}_{H_{2}O}{H}_{2}O+{\mathrm{\μ}}_{N_2}{N}_2+{\mathrm{\μ}}_{O_2}{O}_2$

Wherein:

$v_{air}=a+\frac{b}{4}-\frac{c}{2}$

${\mathrm{\μ}}_{{\mathrm{CO}}_2}=a$

${\mathrm{\μ}}_{H_{2}O}=\frac{b}{2}$

${\mathrm{\μ}}_{N_2}=\mathrm{\λ}\×{\mathrm{\μ}}_{air}+\frac{d}{2}$

${\mathrm{\μ}}_{o_2}={\mathrm{\μ}}_{air}-a-\frac{b}{4}+\frac{c}{2}$

G) assume that fuel adiabatic combustion temperature under lean flammability limit is T_AFT, ambient temperature be T_o, search the burning of fuel Enthalpy Δ H_c, the specific heat at constant pressure C of all combustion products_p,i, according to following adiabatic combustion temperature computing formula and above-mentioned Non-theory molar fraction x that push away now fuel more counter than general reactional equation_f, obtain dilute under this gaseous mixture of fuel Combustion limit x_L；

$T_{AFT}=T_o+\frac{{\mathrm{\ΔH}}_c}{\Σ{\mathrm{\μ}}_i\×c_{p,i}}$

Wherein:

T_AFTAdiabatic combustion temperature, unit K；

T_oAmbient temperature, unit K；

ΔH_cThe enthalpy of combustion of fuel, unit J/mol；

μ_iNon-theory is than the coefficient of i-th kind of combustion product in general reactional equation；

c_p,iNon-theory is than the specific heat at constant pressure of i-th kind of combustion product in general reactional equation, unit J/ (mol*K).