传热学多媒体教学辅助系统赵镇南例题中的数值计算源程序.doc

例题中的数值计算源程序目录例题3-10：热力发动机壁面的冷却例题3-11：椭圆管矩形翅片的效率例题3-12：梯形散热肋的温度分布与肋效率例题4-11：一维对流边界条件下平壁的非稳态导热源程序例题3-10：热力发动机壁面的冷却c program for ex3-10 dimension t (65),tt (65),a (65) data dx,dy,cond,tf1,h1,tf2,h2,tf3,h3/.002,.001,4.,1600.,500., * 450.,40.,350.,1500./ data eps,N/0.01,200/ write(*,100)100 format(1x,'input cond:F6.2')c read(*,200) cond200 format(f6.2) write(*,101)dx,dy,cond,tf1,h1,tf2,h2,tf3,h3101 format(1x,'dx/dy=',2f6.3,2x,'cond=',f6.2/ * 1x,'tfi/hi:',6(f6.1,1x)) Bi1=h1*dx/cond Bi2=h2*dx/cond Bi3=h3*dx/cond it=0 MN=65c area of boundary for node points a(1)=0.25 do 51 J=2,6551 a(J)=1.0 do 52 J=2,652 a(J)=0.5 do 53 J=7,35,753 a(J)=0.5 do 54 J=8,29,754 a(J)=0.5 do 55 J=30,3255 a(J)=0.5 a(33)=0.75 a(36)=0.5 a(39)=0.5 a(42)=0.5 do 56 J=38,44,356 a(J)=0.5 a(45)=0.25 do 57 J=46,4857 a(J)=0.5 a(49)=0.75 a(51)=0.5 a(58)=0.5 a(52)=0.5 a(59)=0.25 do 58 J=60,6458 a(J)=0.5 a(65)=0.25c assign initial temp. value do 10 I=1,MN10 t(I)=750.012 do 20 J=1,MN tt(J)=t(J)20 continuec nodal eq.14 continue t(1)=(0.5*t(2)+2.*t(8)+Bi1*tf1)/(2.5+Bi1) do 2 i=2,62 t(i)=(t(i-1)+t(i+1)+8.*t(i+7)+4.*Bi1*tf1)/(10.+4.*Bi1) t(7)=(0.5*t(6)+2.*t(14)+Bi1*tf1)/(2.5+Bi1) t(8)=(t(1)+t(15)+0.5*t(9))/2.5 do 3 i=9,133 t(i)=(0.5*t(i-1)+0.5*t(i+1)+2.*t(i-7)+2.*t(i+7))/5. t(14)=(t(7)+t(21)+0.5*t(13))/2.5 t(15)=(t(8)+t(22)+0.5*t(16))/2.5 do 4 i=16,204 t(i)=(0.5*t(i-1)+0.5*t(i+1)+2.*t(i-7)+2.*t(i+7))/5. t(21)=(t(14)+t(28)+0.5*t(20))/2.5 t(22)=(t(15)+t(29)+0.5*t(23))/2.5 do 5 i=23,275 t(i)=(0.5*t(i-1)+0.5*t(i+1)+2.*t(i-7)+2.*t(i+7))/5. t(28)=(t(21)+t(35)+0.5*t(27))/2.5 t(29)=(0.5*t(30)+2.*t(22)+Bi3*tf3)/(2.5+Bi3) do 6 i=30,326 t(i)=(t(i-1)+t(i+1)+8.*t(i-7)+4.*Bi3*tf3)/(10.+4.*Bi3) t(33)=(0.25*t(32)+0.5*t(34)+t(36)+2.*t(26)+0.75*Bi3*tf3)/ * (3.75+0.75*Bi3) t(34)=(0.5*t(33)+0.5*t(35)+2.*t(27)+2.*t(37))/5. t(35)=(t(28)+t(38)+0.5*t(34))/2.5 t(36)=(t(33)+t(39)+0.5*t(37)+0.5*Bi3*tf3)/(2.5+0.5*Bi3) t(37)=(0.5*t(36)+0.5*t(38)+2.*t(34)+2.*t(40))/5. t(38)=(t(35)+t(41)+0.5*t(37))/2.5 t(39)=(t(36)+t(42)+0.5*t(40)+0.5*Bi3*tf3)/(2.5+0.5*Bi3) t(40)=(0.5*t(39)+0.5*t(41)+2.*t(37)+2.*t(43))/5. t(41)=(t(44)+t(38)+0.5*t(40))/2.5 t(42)=(t(39)+t(49)+0.5*t(43)+0.5*Bi3*tf3)/(2.5+0.5*Bi3) t(43)=(0.5*t(42)+0.5*t(44)+2.*t(40)+2.*t(50))/5. t(44)=(t(41)+t(51)+0.5*t(43))/2.5 t(45)=(0.5*t(46)+2.*t(52)+Bi3*tf3)/(2.5+Bi3) do 7 i=46,487 t(i)=(t(i-1)+t(i+1)+8.*t(i+7)+4.*Bi3*tf3)/(10.+4.*Bi3) t(49)=(0.25*t(48)+0.5*t(50)+t(42)+2.*t(56)+0.75*Bi3*tf3)/ * (3.75+0.75*Bi3) t(50)=(0.5*t(49)+0.5*t(51)+2.*t(43)+2.*t(57))/5. t(51)=(t(44)+t(58)+0.5*t(50))/2.5 t(52)=(t(45)+t(59)+0.5*t(53))/2.5 do 8 i=53,578 t(i)=(0.5*t(i-1)+0.5*t(i+1)+2.*t(i-7)+2.*t(i+7))/5. t(58)=(t(51)+t(65)+0.5*t(57))/2.5 t(59)=(0.5*t(60)+2.*t(52)+Bi2*tf2)/(2.5+Bi2) do 9 i=60,649 t(i)=(t(i-1)+t(i+1)+8.*t(i-7)+4.*Bi2*tf2)/(10.+4.*Bi2) t(65)=(0.5*t(64)+2.*t(58)+Bi2*tf2)/(2.5+Bi2) do 28 K=1,MN if(abs(tt(K)-t(k))-eps) 28,28,2928 continue goto 2729 it=it+1 if(it.gt.N) goto 30 goto 1230 write(*,104) 104 format(/1x,'showing divergence')27 write(*,105) it105 format(/1x,'number of iteration',I4) write(*,106)106 format(/1x,'nodal temp.') do 40 L=1,MN40 write(*,107)(t(L),L=1,MN)107 format(1x,7(F7.1,1x)/1x,7(F7.1,1x)/1x,7(F7.1,1x)/1x,7(F7.1,1x)/ * 1x,7(F7.1,1x)/33x,3(F7.1,1x)/33x,3(F7.1,1x)/ * 33x,3(F7.1,1x)/1x,7(F7.1,1x)/1x,7(F7.1,1x)/ * 1x,7(F7.1,1x)/) pause 202 q1=0.0 g2=0.0 q3=0.0 do 50 I=1,750 q1=q1+2.*a(I)*dx*(t(I)-tf1)*h1 do 60 K=29,3260 q2=q2+2.*a(K)*dx*(t(K)-tf3)*h3 q2=q2+2.*a(33)*dy*(t(33)-tf3)*h3 do 62 L=36,39,362 q2=q2+2.*a(L)*dy*(t(L)-tf3)*h3 do 64 M=45,4864 q2=q2+2.*a(M)*dx*(t(M)-tf3)*h3 q2=q2+2.*a(49)*dy*(t(49)-tf3)*h3 do 80 J=59,65 q3=q3+2.*a(J)*dx*(t(J)-tf2)*h280 continue write(*,108) q1,q2,q3108 format(/1x,'q1/q2/q3 equal:',3(f12.2,2x)) stop end例题3。