管壳式换热器模拟计算（课本P40 2-5题）
# include <stdio.h>
# include <math.h>
main()
{double Do=0.025，Di=0.021，L=6，Wh=29.5; Th1=280，Wc=37.5， Tc1=160，Ro=0.0005,Ri=0.0001，D1,D2，D2O1=0.85，D2O2=0.919，Tc2,Th2，
MD2O1，MD2O2，K=12.5,Cp1,Cp2，ramda1,ramda2，niu1,niu2，
yita1,yita2,a1,b1,a2,b2,Tc20,Th20,Tmc,Tmh,C,Qc,Qh,
Ho,Hi,Reo,Rei,Pro,Pri,Si,rou1,ui,Hi0,Twi0,
Ai,Twi,niuwi,yitawi,Dwi,Xwi,rouwi,Ao,
Smax,B=0.23,D=0.8,t=0.032,De,uo,rou2,
Ho0,Two0,Two,niuwo,yitawo,Dwo,Xwo,rouwo，
Ko，NTU,E,F,Cmin,Cmax，Xh,Xc，ld=44,b=0.002,Dm=0.0023，e=2.71828;
int n=0;
Tc2=180;
Th2=260; /*给Tc2，Th2赋初值*/
do{ n++;
printf("n=%d\n",n);
Th20=Th2;
Tmh=(Th1+Th20)/2;
do{Tc20=Tc2; /*Tc2的迭代*/
Tmc=(Tc1+Tc20)/2;
Cp1=(0.7072+(0.00147-0.00051*D2O1)*Tmc-0.318*D2O1)*(0.055*K+0.35) *4.18*1000;
Cp2=(0.7072+(0.00147-0.00051*D2O2)*Tmh-0.318*D2O2)*(0.055*K+0.35) *4.18*1000;
C=Wh*Cp2/(Wc*Cp1);
Tc2=Tc1+C*(Th1-Th20);
}while(fabs(Tc2-Tc20)>=0.1);
Qh=Wh*Cp2*(Th1-Th2);
Qc=Wc*Cp1*(Tc2-Tc1);
Ai=3.14*Di*L*324;
Si=0.25*3.14*Di*Di*324/2;
Xc=1+Tmc/100.0;
MD2O1=pow(D2O1,2);
D1=0.942+0.248*Xc+0.174*MD2O1+0.0841/(Xc*D2O1)-0.312*Xc/D2O1-0.55 6*exp(-Xc);
rou1=1000*D1;
ui=Wc/(rou1*Si);
b1=log((log(90.0+1.22)/log(13+1.22)))/(log((50.0+273)/(100.0+273) ));
a1=log(log(90+1.22))-b1*log(50.0+273);
niu1=exp(exp(a1+b1*log(Tmc+273)))-1.22;
yita1=niu1*rou1/1000000;
Rei=Di*ui*rou1/yita1;
ramda1=0.4213*(1-0.00054*Tmc)/D2O1/3.6;
Pri=Cp1*yita1/ramda1;
Hi0=0.027*pow(Rei,0.8)*pow(Pri,0.33)*ramda1/Di;
Twi0=Tmc+Qc/(Hi0*Ai);
do /*管壁内壁温的迭代*/
{ niuwi=pow(e,pow(e,a1+b1*log(Twi0+273)))-1.22;
Xwi=1+Twi0/100.0;
Dwi=0.942+0.248*Xwi+0.174*MD2O1+0.0841/(Xwi*D2O1)-0.312*Xwi/D2O1-0.556*exp(-Xwi);
rouwi=1000*Dwi;
yitawi=niuwi*rouwi/1000000;
Hi=Hi0*(pow((yita1/yitawi),0.14));
Twi=Tmc+Qc/(Hi*Ai);
Twi0=Twi;
}while(fabs(Twi-Twi0)>=0.5);
Ao=3.14*Do*L*324;
Smax=B*D*(1-Do/t);
De=4*(t*t-0.25*3.14*Do*Do)/(3.14*Do);
Xh=1+Tmh/100.0;
MD2O2=pow(D2O2,2);
D2=0.942+0.248*Xh+0.174*MD2O2+0.0841/(Xh*D2O2)-0.312*Xh/D2O2-0.556*ex p(-Xh);
rou2=1000*D2;
uo=Wh/(rou2*Smax);
b2=log((log(1500+1.22)/log(120+1.22)))/(log((50.0+273)/(100.0+273 )));
a2=log(log(1500+1.22))-b2*log(50.0+273);
niu2=exp(exp(a2+b2*log(Tmh+273)))-1.22;
yita2=niu2*rou2/1000000;
Reo=De*uo*rou2/yita2;
ramda2=0.4213*(1-0.00054*Tmh)/D2O2/3.6;
Pro=Cp2*yita2/ramda2;
Ho0=0.36*pow(Reo,0.55)*pow(Pro,0.33)*ramda2/De;
Two0=Tmh-Qh/(Ho0*Ao);
do /*管壁外壁温的迭代*/
{ niuwo=exp(exp(a1+b1*log(Two0+273)))-1.22;
Xwo=1+Two0/100;
Dwo=0.942+0.248*Xwo+0.174*MD2O1+0.0841/(Xwo*D2O1)-0.312*Xwo/D2O1-0.556*exp(-Xwo);
rouwo=1000*Dwo;
yitawo=niuwo*rouwo/1000000;
Ho=Ho0*(pow((yita2/yitawo),0.14));
Two=Tmh-Qh/(Ho*Ao);
Two0=Two;
}while(fabs(Two-Two0)>=0.5)；
Ko=1/((1/Hi+Ri)*Ao/Ai+1/Ho+Ro+b/ld*Do/Dm); /*以管外表面为基准，计算传热系数K*/
Cmin=Wh*Cp2;
Cmax=Wc*Cp1;
NTU=Ko*Ao/Cmin; /*传热单元数*/
F=NTU*sqrt(1+pow(Cmin/Cmax,2));
E=2/((1+Cmin/Cmax)+sqrt(1+Cmin/Cmax)*(1+exp(-F))/(1-exp(-F))); /*传热效率*/
Th2=Th1-E*(Th1-Tc1);
Tc2=Tc1+C*(Th1-Th2);
printf("Th2=%.1lf\tTc2=%.1lf\n",Th2,Tc2);
printf("Rei=%.2lf\tPri=%.2lf\tHi=%.2lf\n",Rei,Pri,Hi);
printf("Reo=%.2lf\tPro=%.2lf\tHo=%.2lf\n",Reo,Pro,Ho);
printf("Qc=%.2lf\tQh=%.2lf\n",Qc/1000,Qh/1000);
printf("Ko=%.2lf\tNTU=%.3lf\tE=%.3lf\n",Ko,NTU,E);
}while( fabs(Th20-Th2) >= 0.1); return 0；}
运行结果如下：。