一維CFD模擬仿真設(shè)計

1、.專業(yè)整理 .CFD simulation in Laval nozzleSIAE 090441313AbstractWe aim to simulate the quasi one dimension flow in the Lavalnozzle based on CFDcomputationin thispaper .We considerthechange of the temperature ,the pressure ,the density and the speed of the flow to study the flow.The analytic solution ofth

2、e flow in the Laval nozzle is provided when the input velocity is supersonic.We use the Mac-Cormack Explicit Difference Scheme to slove the question.Key words :Laval nozzle ,CFD,throat narrow.ContentsAbstract .1Introduction .2Simulationofone-dimensionalsteadyflow.3. 學習幫手 .專業(yè)整理 .Basisequations .3Dime

3、nsionless .10Mac-CormackExplicitDifferenceScheme.11Boundaryconditions .13Reference .13Annex .14IntroductionLaval nozzle is the most commonly used components of rocket engines and aero-engine, constituted by two tapered tube, one shrink tube, another expansion tube.Laval nozzle is an important part o

4、f the thrust chamber. The first half of the nozzle from large to small contraction to a narrow throat to the middle. Narrow throat and then expand. 學習幫手 .專業(yè)整理 .outwards from small to big to the end. The gas in the rocket body by the front half of the high pressure into the nozzle, through the narrow

5、 throat to escape by the rear half. This architecture allows the speed of the air flow changes due to changes in the jet cross-sectional area, the airflow fromsubsonic to the speed of sound, untilacceleratedto transonic.So, people flarednozzle calledtransonicnozzle.Since itwasinvented by the Swedish

6、 Laval, also known as Laval nozzle.Analysis of the principle of the Laval nozzle. The rocketengines of the gas flow in the combustion chamber under pressure,afterthe backward movement of the nozzle into thenozzle .Atthisstage,the gas movementfollowthe principleof thefluidmoves in the tube , the smal

7、l cross-section at the flow rate large sectional large flow velocity, thus accelerating airflow.Laval nozzleWhen you reach the narrow throat, the flow rate has exceeded. 學習幫手 .專業(yè)整理 .the speed of sound. Transonic fluid movement they no longer follow the principle of cross-section at small velocity, a

8、ta flowrate ofsmallcross-section large, but on the contrarythe larger cross-sectional flow faster. The gas flow speed isfurtheracceleratedto 2-3 km / sec,equivalentto 7-8 times thespeed of sound, thuscreatinga greatthrust.The Laval nozzlefact played the role of a flow rate Enlargement Device. Infact

9、,not justrocketengines,missilenozzleis this horn shape,so the Laval nozzle weapons has a very wide range of applications.Simulationofone-dimensionalsteadyflow1.Basis equationsAs we know,Laval nozzle is a zooming nozzle flow channel to narrow further expansion.Allows the airflow to further accelerate

10、 to reach the speed of sound at the throat into a supersonic flow.Now,we want to simulate the quasi one-dimension flowing.Firstly,we will analysis on theory.The flow is isentropic,so we can apply the following equations.(1)Continuity equation:. 學習幫手 .專業(yè)整理 .In the flow, we need to consider the follow

11、ing physical quantities.The pression ,the temperature ,the speed of the fluid and the cross-section .They are respectively represented by P,T,u,A. We apply the conservation of the mass.we will obtain this equation.uA(d)( AdA )( udu )And then we getdAdud0Au(2)Equation of momentum(in the direction of

12、the axis)According to the theory of momentum:uA(u du)uAu PA (PdP)(A dA) (PdP)dA2The simplification of this equation isududP(3)Energy equation2vdhtd (h)dhudu02Ideal gas equation of statePRTMR is ideal gas constant,R=8.314J/g/K.M is the masse per mole. 學習幫手 .專業(yè)整理 .(4)The equation of ThermodynamicsT dS

13、dePdV , h e PV ;TdSdhVdP , dh C p dT ;dSC pdTV dP C pdTR dPTTTPBecause the flow is isentropic,sodS=0And we use the equation of momentum,we haveC pP （T ）R( T )1Combine with others equations,we result withu 2RTWe called u the speed of sound,we noted a.a 2RTWe apply the continuity equationdA( uAa221)We

14、 defined the Mach numberuMaIf we have the relation asA1.3980.347 tanh(0.8x4)We have the figure1. 學習幫手 .1) du u.專業(yè)整理 .SodA (M 2 AM1,supersonicIf dA0.If dA0,we have du0.M1,subsonicIf dA0.If dA0,we have du0.This is the reason why this architecture allows the speed ofthe air flow changes due to changes

15、in the jet cross-sectional area, the airflow from subsonic to the speed of sound, until accelerated to transonic. 學習幫手 .專業(yè)整理 .We have the consequence as follows11mAMP ()2( 1)PAMR Ttt12RT1M2TTt; PPt1 M2 )1 M2 )(1(1221Then we replace P and T in this equation.The consequence will become11 M 2Pt AMm2RTt

16、12)1(1M2To simplifyPt11mAM ()2( 1)RTt121 M 2In this equation,the variable is the much number,as the speed of the flow is from subsonic to supersonic ,so we can suppose that there exist a critical section where M equal to 1.ThenA11 M2112)2(1)A(1M2. 學習幫手 .專業(yè)整理 .Figure2This section is called narrow thr

17、oat.The same method,we canobtainT12T11M22P112()1P1M2121(2)11M212. 學習幫手 .專業(yè)整理 .Figure3We know the section in narrow throat is minimum.d ( m ) K (11(1) ( 1)M )(11)2( 1)1M ()2( 1) dM A1M22(1)1M2112we can judge that the function attains the maximum or not11f ( M )M ()2(1)11M222 DimensionlessCombining CF

18、D with one-dimension flow theory,we make the variables dimensionless.According to the condition initial which is given .We note. 學習幫手 .專業(yè)整理 .TT T00uu u 0xx lAAA0tt lu 0We use the variable dimensionless to represent theequations.And the equations have the following changes(1)Continuity equationvt xt

19、v A v 0A x x v v ln A v x x x (2)Equation of momentumvtv 1T T vx (x x )(3)Energy equationTv T1)T ( vvlnA()txxx. 學習幫手 .專業(yè)整理 .3.Mac-Cormack Explicit Difference SchemeThen we use the Mac-Cormack Explicit Difference Scheme,the. 學習幫手 .專業(yè)整理 .principal of this theory is using the surrounding points to pres

20、ent differential parts of a point and we consider thequestion with one dimension.The distance between two points is h.f(x)f 0( x x0 )(f1f 221f 33) 0(x2 ) 0 (xx0 )(3 ) 0 ( xx0 )x2!3!xx1x0hx3x0hh( f ) 0h2f2f1f 0(2 )0x2xh( f ) 0h2f2f3f 0(2 ) 0x2xSo we can use two points adjacent to present the differen

21、tial parts.(f) 0f1f 3x2h(2 ff1f3 2 f02) 0h2xUsing thismethod,we make an estimationand correct the error.Estimationttttt ln Ai 1nAitti 1itvi 1vit( t )ivixixivixcorrect the errort tt tt ttt(ii1)ivitxIntermediate value. 學習幫手 .專業(yè)整理 .t tttii(t) itThen the equation has the following change( ) av1 () it()

22、it t )t2ttAt the moment t ,we will know the value in the whole plan .And we defineticxiviaitmin(t 1t ,t 2t ,t3t ,t 4t ,t 5t . t tN )4.Boundary conditionsHyperbola equation has two characteristics lines.When one ofthecharacteristicslinesentertheflowzone .Weadmitaparameter to be fixed ,otherwise when

23、one of the characteristics go out the flow zone ,we admit a parameter tobe a variable depends the time.Applying this theory ,we can determine the boundary conditions.Reference：1 章利特，高鐵瑜，夏慶鋒，徐廷相 . 拉瓦爾噴管的準一維定常流動 .中國科技論文在線。2 王平，昌平，柏松 . 基于 CFD數(shù)值模擬的拉瓦爾噴管流場分析 . 航空計算技術(shù) 2012 年 7 月第 42 卷第 4 期。3 王如根，瑞賢，全通 . 基

24、于實際發(fā)動機拉瓦爾噴管的流場分析 .99. 學習幫手 .專業(yè)整理 .學術(shù)會議空軍工程學院飛機推進系統(tǒng)實驗室。4 周文祥，黃金泉，周人治 . 拉瓦爾噴管計算模型的改進及其整機仿真驗證 . 航空動力學報 2009 年 11 月第 24 卷第 11 期。5 王克印，韓星星，曉濤，耀鵬，吉潮 . 縮擴型超音速噴管的設(shè)計與仿真 . 中國工程機械學報 2011 年 9 月第 9 卷第 3 期。Annex 1:Figure 1程序：x=0:0.1:5;a=1.398+0.374*tanh(0.8*x-4);plot(x,a)Figure 2程序：gama=1.33;M=0:0.01:2;A=(1./M).*

25、(1+(gama-1).*M.2./2)./(gama+1)./2).(gama+1)./(2.*(gama-1);Xlabel(variable M);ylabel(variable A);Plot(M,A)Figure 3程序：gama=1.33;M=0:0.01:2;T=(gama+1)./2)./(1+(gama-1.).*M.2./2);P=(gama+1)./2)./(1+(gama-1).*M.2./2).(1./(gama-1);rho=(P./T).*(gama-1)./2)./(1+(gama-1).*M.2./2).(gama./(gama-1);x=T;P;rho;y=

26、M;M;plot(M,T,M,P,M,rho);subplot(221);plot(M,T);xlabel(variable M);ylabel(variable T)；subplot(222);plot(M,P);xlabel(variable M);ylabel(variable P)；subplot(2,2,3:4);plot(M,rho);xlabel(variable M);ylabel(variable rho)；模擬程序（未完成） ：M1=1.5; %input( 來流馬赫數(shù) M1=);P1=47892.4;%input(來流氣體壓強 P1=);rho1=1.222;%input

27、(來流氣體密度 rho1=);gama=1.4; %input(比熱比 gama=);. 學習幫手 .專業(yè)整理 .R=8.314; %input(氣體常量 R=);C=1.5; %input(科朗數(shù) C=);T1=293; %input(來流氣體溫度 T1=);a1=sqrt(gama*R*T1);V1=M1*a1;L=10;%input(噴管長度 L=);I=300; %input(等分步數(shù) I=);N=1000;%input(時間步數(shù) N=);d_t=0;e=0;delta_x=1/(I-1);A1=1.398-0.347*tanh(4);A=zeros(I,1);V=zeros(I,N)

28、;rho=zeros(I,N);T=zeros(I,N);ahead_V=zeros(I,N);ahead_rho=zeros(I,N);ahead_T=zeros(I,N);ahead_der_V=zeros(I,N);ahead_der_rho=zeros(I,N);ahead_der_T=zeros(I,N);der_V=zeros(I,N);der_rho=zeros(I,N);der_T=zeros(I,N);ave_der_V=zeros(I,N);ave_der_rho=zeros(I,N);ave_der_T=zeros(I,N);delta_t=zeros(I,N);a=ze

29、ros(I,N);e=zeros(I,N);forn=1:Nahead_V(1,n)=M1;ahead_rho(1,n)=1;ahead_T(1,n)=1;V(1,n)=M1;rho(1,n)=1;T(1,n)=1;endfori=1:IV(i,1)=(1.5+1.09*i*delta_x)*sqrt(T(i,1);rho(i,1)=1-0.3146*i*delta_x;T(i,1)=1-0.2314*i*delta_x;A(i,1)=(1.398+0.347*tanh(0.8*(i-1)*delta_x*L-4);endforn=1:Nfori=2:(I-1)der_V(i,n)=-V(i,

30、n)*(V(i+1,n)-V(i,n)/delta_x-1/gama*(T(i+1,n)-T(i,n)/delta_x+ T(i,n)/rho(i,n)*(rho(i+1,n)-rho(i,n)/delta_x);der_rho(i,n)=-rho(i,n)*(V(i+1,n)-V(i,n)/delta_x-V(i,n)*(rho(i+1,n)-rho(i,n)/d elta_x-rho(i,n)*V(i,n)*(log(A(i+1,1)-log(A(i,1)/delta_x;. 學習幫手 .專業(yè)整理 .der_T(i,n)=-V(i,n)*(T(i+1,n)-T(i,n)/delta_x-(

31、gama-1)*T(i,n)*(V(i+1,n)-V(i,n)/delta_x+V(i,n)*(log(A(i+1,1)-log(A(i,1)/delta_x);a(i,n)=sqrt(gama*R*T(i,n)/a1;delta_t(i,n)=C*delta_x/(V(i,n)+a(i,n);endd_t=min(delta_t(i,n);fori=2:1:(I-1)ahead_V(i,n+1)=V(i,n)+der_V(i,n)*d_t;ahead_rho(i,n+1)=rho(i,n)+der_rho(i,n)*d_t;ahead_T(i,n+1)=T(i,n)+der_T(i,n)*d_t;ahe