汽车理论习题答案

1.3、1）绘制汽车驱动力与行驶阻力平衡图。

2）求汽车的最高车速、最大爬坡度及克服该坡度时相应的附着率。 3）绘制汽车行驶加速倒数曲线，用图解积分法求汽车有Ⅱ档起步加速行驶至70km/h的车速－时间曲线，或者用计算机求汽车用Ⅱ档起步加速至70km/h的加速时间。

n_min=600; n_max=4000; m1=2000; m2=1800; mz=3880; g=9.81; r=0.367; CdA=2.77; f=0.013; nT=0.85;

ig=[5.56 2.769 1.644 1.00]; i0=5.83; If=0.218; Iw1=1.798; Iw2=3.598;

Iw=2*Iw1+4*Iw2; L=3.2; a=1.947; b=L-a; hg=0.9; g=9.8;

1）绘制汽车驱动力与行驶阻力平衡图： for i=1:3401 n(i)=599+i;

Ttq(i)=-19.313+295.27*(n(i)/1000)-165.44*(n(i)/1000)^2+40.874*(n(i)/1000)^3-3.8445*(n(i)/1000)^4; end

for j=1:4

for i=1:3401

Ft(i,j)=Ttq(i)*ig(j)*i0*nT/r; ua(i,j)=0.377*r*n(i)/(ig(j)*i0);

Fz(i,j)=CdA*ua(i,j)^2/21.15+mz*g*f; end end

plot(ua,Ft,ua,Fz) ,grid

title('汽车驱动力与行驶阻力平衡图'); xlabel('ua(km/h)'); ylabel('Ft(N)');

2）①汽车的最高车速：

②最大爬坡度： for k=1:1000

n1(k)=1800+k*0.5;

Ttq(k)=-19.313+295.27*(n1(k)/1000)-165.44*(n1(k)/1000)^2+40.874*(n1(k)/1000)^3-3.8445*(n1(k)/1000)^4; Ft(k)=Ttq(k)*ig(1)*i0*nT/r; ua(k)=0.377*r*n1(k)/(ig(1)*i0);

Fz(k)=CdA*ua(k)^2/21.15+mz*g*f; af(k)=asin((Ft(k)-Fz(k))/(mz*g)); end

for k=1:1000

if(af(k)==max(af)) i=tan(af(k));

disp('汽车最大爬坡度='); disp(i); end end

汽车最大爬坡度= 0.3522

③相应的附着率：

3）①绘制汽车行驶加速倒数曲线: n=(n_min:1:n_max); N=n./1000;

Tq=-19.313+295.27*N-165.44*N.^2+40.874*N.^3-3.8445*N.^4; Ft1=Tq*ig(1) *i0*nT/r; Ft2=Tq*ig(2) *i0*nT/r; Ft3=Tq*ig(3)*i0*nT/r; Ft4=Tq*ig(4)*i0*nT/r; ua1=0.377*n*r/ig(1) /i0; ua2=0.377*n*r/ig(2)/i0; ua3=0.377*n*r/ig(3)/i0; ua4=0.377*n*r/ig(4)/i0; Fw1= CdA*ua1.^2/21.15; Fw2= CdA*ua2.^2/21.15; Fw3= CdA*ua3.^2/21.15; Fw4= CdA*ua4.^2/21.15; Ff= mz*g*f;

deta1=1+Iw/(mz*r^2)+If*ig(1)^2*i0^2*nT/(mz*r^2); deta2=1+Iw/(mz*r^2)+If*ig(2)^2*i0^2*nT/(mz*r^2); deta3=1+Iw/(mz*r^2)+If*ig(3)^2*i0^2*nT/(mz*r^2); deta4=1+Iw/(mz*r^2)+If*ig(4)^2*i0^2*nT/(mz*r^2); a1=( Ft1- Fw1- Ff)/(deta1*mz); a2=( Ft2- Fw2- Ff)/(deta2*mz); a3=( Ft3- Fw3- Ff)/(deta3*mz); a4=( Ft4- Fw4- Ff)/(deta4*mz); x1=1./a1; x2=1./a2; x3=1./a3; x4=1./a4;

plot(ua1,x1, 'k', ua2,x2, 'b',ua3,x3, 'r', ua4,x4,'y'),grid title('加速度倒数曲线'); xlabel('ua(km/h)'); ylabel('1/a');

②汽车用Ⅱ档起步加速至70km/h的加速时间

2.7、计算与绘制题1 中货车的1）汽车功率平衡图；

2）最高档与次高档的等速百公里油耗曲线。

m1=2000; m2=1800; mz=3880; g=9.81; r=0.367; CdA=2.77; f=0.013; nT=0.85;

ig=[5.56 2.769 1.644 1.00]; i0=5.83; If=0.218; Iw1=1.798; Iw2=3.598;

n1=[815 1207 1614 2012 2603 3006 3403 3804]; Iw=2*Iw1+4*Iw2; Qid=0.299; L=3.2; a=1.947; b=L-a; hg=0.9; g=9.8;

1）汽车功率平衡图： for j=1:4

for i=1:3401 n(i)=599+i;

Ttq(i)=-19.313+295.27*(n(i)/1000)-165.44*(n(i)/1000)^2+40.874*(n(i)/1000)^3-3.8445*(n(i)/1000)^4; Pe(i)=n(i)*Ttq(i)/9549;

ua(i,j)=0.377*r*n(i)/(ig(j)*i0);

Pz(i,j)=(mz*g*f*ua(i,j)/3600.+CdA*ua(i,j)^3/76140.)/nT; end end

plot(ua,Pe,ua,Pz),grid; title('功率平衡图'); xlabel('ua(km/h)'); ylabel('Pe,Pz(kw)');

for i=1:8

Td(i)=-19.313+295.27*(n1(i)/1000.0)-165.44*(n1(i)/1000.0)^2+40.874*(n1(i)/1000.0)^3-3.8445*(n1(i)/1000.0)^4; Pd(i)=n1(i)*Td(i)/9549;

u3(i)=0.377*n1(i)*r/(ig(3)*i0); u4(i)=0.377*n1(i)*r/(ig(4)*i0); end

b(1)=0.17768*Pd(1)^4-5.8629*Pd(1)^3+72.379*Pd(1)^2-416.46*Pd(1)+1326.8; b(2)=0.043072*Pd(2)^4-2.0553*Pd(2)^3+36.657*Pd(2)^2-303.98*Pd(2)+1354.7; b(3)=0.0068164*Pd(3)^4-0.51184*Pd(3)^3+14.524*Pd(3)^2-189.75*Pd(3)+1284.4; b(4)=0.0018555*Pd(4)^4-0.18517*Pd(4)^3+7.0035*Pd(4)^2-121.59*Pd(4)+1122.9; b(5)=0.00068906*Pd(5)^4-0.091077*Pd(5)^3+4.4763*Pd(5)^2-98.893*Pd(5)+1141.0; b(6)=0.00035032*Pd(6)^4-0.05138*Pd(6)^3+2.8593*Pd(6)^2-73.714*Pd(6)+1051.2; b(7)=0.00028230*Pd(7)^4-0.047449*Pd(7)^3+2.9788*Pd(7)^2-84.478*Pd(7)+1233.9;

b(8)=-0.000038568*Pd(8)^4-0.00075215*Pd(8)^3+0.71113*Pd(8)^2-45.291*Pd(8)+1129.7; for i=1:8

Qs3(i)=Pd(i)*b(i)/(1.02*u3(i)*7.15); Qs4(i)=Pd(i)*b(i)/(1.02*u4(i)*7.15); end

plot(u3,Qs3, '*') ,grid; hold on;

plot(u4,Qs4, '>');

title('3档4档等速百公里油耗图'); xlabel('u (km/h)'); ylabel('Qs(L/100km)');

3、改变1.3题中轻型货车的主减速器传动比，作出i0为5.17、5.43、5.83、6.17、6.33时的燃油经济性——加速时间曲线，讨论不同i0值对汽车性能的影响和采用不同变速器对汽车性能的影响。 （以最高档，较高转速（n取3403r/min），最经济负荷（即90%负荷大约60Kw）行驶时的油耗）： m1=2000; m2=1800; mz=3880; g=9.81; r=0.367; CdA=2.77; f=0.013; nT=0.85;

ig=[5.56 2.769 1.644 1.00]; i0=[5.17 5.43 5.83 6.17 6.33]; If=0.218; Iw1=1.798; Iw2=3.598;

Iw=2*Iw1+4*Iw2; n=3403; Qid=0.299; Pe=60;

b=0.00028230*Pe^4-0.047449*Pe^3+2.9788*Pe^2-84.478*Pe+1233.9; for i=1:5

ua(i)=0.377*n*r/ig(4) /i0(i); Qs(i)=Pe*b/(1.02*ua(i)*7.15); end

disp('百公里消耗燃油='); disp(Qs); disp('L');

百公里消耗燃油=

26.9678 28.3240 30.4105 32.1840 33.0186 L

for q=1:5 for i=1:3401 n(i)=599+i;

Ttq(i)=-19.313+295.27*(n(i)/1000)-165.44*(n(i)/1000)^2+40.874*(n(i)/1000)^3-3.8445*(n(i)/1000)^4; end

for i=1:3401

deta=1+Iw/(mz*r^2)+If*ig(4)^2*i0(q)^2*nT/(mz*r^2); ua(i)=0.377*r*n(i)/(ig(4)*i0(q));

a(i)=(Ttq(i)*ig(4)*i0(q)*nT/r-CdA*ua(i)^2/21.15-mz*g*f)/(deta*mz); b1(i)=a(i); b(i)=1/b1(i); end

for i=1:3401

if(ua(i)<=96.6) m=i; end end

t=ua(1)*b(1); for p=2:m

t1(p)=(ua(p)-ua(p-1))*(b(p)+b(p-1))*0.5; t=t+t1(p); end tz=t/3.6;

disp('加速时间='); disp(tz); disp('s'); end

加速时间=

135.3100 121.7714 101.9921 80.7631 73.8296 s

t=[135.100 121.7714 101.9921 80.7631 73.8296] Qs=[26.9678 28.3240 30.4105 32.1840 33.0186] plot(Qs,t,'r'),grid; hold on

plot(Qs,t,'*');

title('燃油经济性—加速时间曲线'); axis([25 35 50 150]) xlabel('Qs (L)'); ylabel('t(s)');

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