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1、第二次作业 车辆工程 八班 胡文贵 2013424032解：汽车功率平衡图程序 clear all f=0.013; G=38800;Ff=G*f; CdA=2.77;u1=0:1:120; Fw=CdA.*u1.2/21.15;F=Ff+Fw;Ig=5.56 2.769 1.644 1.000 0.793;k=1:5; ngk=600 600 600 600 600; ngm=4000 4000 4000 4000 4000; r=0.367;I0=5.83;eta=0.85; ugk=0.377*r*ngk(k)./(Ig(k).*I0); ukm=0.377.*r.*ngm(k)./(I
2、g(k).*I0); pz=F.*u1./3600; for k=1:5 u=ugk(k):ukm(k); n=Ig(k)*I0.*u./r/0.377;Tq=-19.313+295.27*n/1000-165.44*(n/1000).2+40.874*(n/1000).3-3.8445*(n/1000).4; Ft=Tq.*Ig(k)*I0*eta/r; pe=Ft.*u./3600; plot(u1,pz,u,pe) hold on,grid on end2)最高档和次高档百公里油耗曲线 n=600:1:4000; m=3880;g=9.8; G=m*g; ig=5.56 2.769 1.
3、644 1.00 0.793; nT=0.85;r=0.367;f=0.013;CDA=2.77;i0=5.83; L=3.2;a=1.947;hg=0.9;If=0.218;Iw1=1.798;Iw2=3.598; n0=815 1207 1614 2012 2603 3006 3403 3804; B00=1326.8 1354.7 1284.4 1122.9 1141.0 1051.2 1233.9 1129.7; B10=-416.46 -303.98 -189.75 -121.59 -98.893 -73.714 -84.478 -45.291; B20=72.379 36.657
4、14.524 7.0035 4.4763 2.8593 2.9788 0.71113; B30=-5.8629 -2.0553 -0.51184 -0.18517 -0.091077 -0.05138 -0.047449 -0.00075215; B40=0.17768 0.043072 0.0068164 0.0018555 0.00068906 0.00035032 0.00028230 - 0.000038568; B0=spline(n0,B00,n); B1=spline(n0,B10,n); B2=spline(n0,B20,n); B3=spline(n0,B30,n); B4=
5、spline(n0,B40,n); Ff=G*f; ua4=0.377*r*n/ig(4)/i0; ua5=0.377*r*n/ig(5)/i0; Fz4=Ff+CDA*(ua4.2)/21.15; Fz5=Ff+CDA*(ua5.2)/21.15; Pe4=Fz4.*ua4./(nT*3.6*1000); Pe5=Fz5.*ua5./(nT*3.6*1000); for i=1:1:3401 b4(i)=B0(i)+B1(i)*Pe4(i)+B2(i)*Pe4(i).2+B3(i)*Pe4(i).3+B4(i)*Pe4(i).4; b5(i)=B0(i)+B1(i)*Pe5(i)+B2(i)
6、*Pe5(i).2+B3(i)*Pe5(i).3+B4(i)*Pe5(i).4; end pg=7.0; Q4=Pe4.*b4./(1.02.*ua4.*pg); Q5=Pe5.*b5./(1.02.*ua5.*pg); plot(ua4,Q4,ua5,Q5); axis(0 100 10 30); title(最高档与次高档等速百公里油耗曲线); xlabel(ua(km/h); ylabel(百公里油耗（L/100km）); gtext(4),gtext(5);六工况油耗计算 1）匀速 25km/h f=0.013; G=38800;Ff=G*f; CdA=2.77;eta=0.85;ua
7、=25; pe=(Ff*ua+CdA*ua.3/21.15)/(eta*3600)pe =4.7897 r=0.367;I0=5.83;Ig=5.56 2.769 1.644 1.000 0.793; for i=1:5n=ua*Ig(i)*I0/(0.377*r) tq=-19.313+295.27*n/1000-165.44*(n/1000).2+40.874*(n/1000).3-3.8445*(n/1000).4 P=tq*n/9550 end 利用 matlab 求出的结果 n =5.8570e+03 tq =-276.9876 P = -169.8763n =2.9169e+03
8、tq =170.4435 P =52.0596 n =1.7318e+03 tq =173.5743 P =31.4764 n =1.0534e+03 tq =151.1890 P =16.6770 n =835.3613 tq = 133.8503 P =11.7082由于国标规定，尽量用最高档行驶。而阻力功率 f=0.013; G=38800;Ff=G*f; CdA=2.77;eta=0.85;ua=25; pe=(Ff*ua+CdA*ua.3/21.15)/(eta*3600)pe =4.7897 所以时速 25km/h 时可以使用五档行驶 此时，n =835.3613 tq = 133
9、.8503 P =11.7082 利用插值法 n=835;k=(835-815)/(1207-835); B1=1326.8 -416.46 72.379 -5.8629 0.17768;B2=1354.7 -303.98 36.657 -2.0553 0.043072; for i=1:5x(i)=(k*B2(i)+B1(i)/(k+1) end p=11.7082;b=x(1)+x(2)*p+x(3)*p.2+x(4)*p.3+x(5)*p.4b =303.1827 取汽油的 pg 值为 6.69 pg=6.69;Qt=p*b/(367.1*pg); Q=p*b*50/(102*25*pg
10、) Q =10.4039ml 2）匀速 40 同理可求得用第五档匀速 n =1.3366e+03 tq = 165.1160 P = 23.1089 n=1337; k=(n-1207)/(1614-n); B2=1354.7 -303.98 36.657 -2.0553 0.043072;B3=1284.4 -189.75 14.524 -0.51184 0.0068164; for i=1:5x(i)=(k*B3(i)+B2(i)/(k+1) end p=23.1089;b=x(1)+x(2)*p+x(3)*p.2+x(4)*p.3+x(5)*p.4 b =652.0141 pg=6.69
11、;Qt=p*b/(367.1*pg); Q=p*b*250/(102*25*pg) Q =220.8055ml3）匀速 50km/h 同理可求得用第五档匀速 n =1.6707e+03 tq = 172.8685 P = 30.2424 n=1671;k=(n-1614)/(2012-n); B4=122.9 -121.59 7.0035 -0.18517 0.0018555; for i=1:5x(i)=(k*B4(i)+B3(i)/(k+1) end p = 30.2424;b=x(1)+x(2)*p+x(3)*p.2+x(4)*p.3+x(5)*p.4b =217.6389 pg=6.6
12、9;Qt=p*b/(367.1*pg); Q=p*b*250/(102*25*pg)Q =96.4554ml4）I=0.218; Iw1=1.798; Iw2=3.598;L=3.2; m=3880; r=0.367; eta=0.85; f=0.013; i0=5.83; a=1.947; hg=0.9; g=10; ig=5.56;2.769;1.644;1.00;0.793; G=38800;Ff=G*f; ua=40; for i=1:5 th(i)=1+(Iw1+Iw2)/(m*r2)+(I*i02+(ig(i)2*eta)/(m*r2); end for i=1:5pe=(Ff*u
13、a+CdA*ua.3/21.15+th(i)*m*ua0.25)/(3600*eta) end pe = 22.9607 pe = 22.4813 pe = 22.3789 pe = 22.3438 pe = 22.3361 阻力功率比第五档、第四档 25km/h 行驶时功率大故可以用第三档加速 r=0.367;I0=5.83;Ig=1.644; for i=1:16 n=u(i)*Ig*I0/(0.377*r) tq=-19.313+295.27*n/1000-165.44*(n/1000).2+40.874*(n/1000).3-3.8445*(n/1000).4 P=tq*n/9550
14、end ua=25:1:40; I=0.218; Iw1=1.798; Iw2=3.598;L=3.2; m=3880; r=0.367; eta=0.85; Cd=2.77; f=0.013; i0=5.83; a=1.947; hg=0.9; g=10; ig=5.56;2.769;1.644;1.00;0.793; for j=1:16n(j)=ua(j)*ig(3)*i0/(0.377*r) end for i=1:16 tq(i)=-19.313+295.27*n(i)/1000-165.44*(n(i)/1000).2+40.874*(n(i)/1000).3- 3.8445*(n
15、(i)/1000).4 pe(i)=tq(i)*n(i)/9550 end B5=1141.0 -98.893 4.4763 -0.091077 0.00068906;B6=1051.2 -73.714 2.8593 - 0.05138 0.00035032;B4=1122.9 -121.59 7.0035 -0.18517 0.0018555;B3=1284.4 -189.75 14.524 -0.51184 0.0068164; for i=1:16if n(i) for i=1:16if n(i)2603k(i)=(n(i)-2603)/(3006-n(i)for j=1:5x(j)=(k(i)*B6(j)+B5(j)/(k(i)+1)endb(i)=x(1)+x(2)*pe(i)+x(3)*pe(i).2+x(4)*pe(i).3+x(5)*pe(i).4elsek(i)=(n(i)-2012)/(2603-n(i)for j=1:5x(j)=(k(i)*B5(j)+B4(j)/(k(i)+1)endb(i)=x(1)+x(2)*pe(i)+x(3)*pe(i).2+x(4)*pe(i).3+x(5)*p
