公式的推导过程见传热学的书籍,假设套管长为H,外径为d,管厚δ。套管伸入大管子,大管子壁上温度为t0,温度计示数为tw,则外温是
t f = t w ch ( m H ) − t 0 ch ( m H ) − 1 t_f=\frac{t_w\ch(mH)-t_0}{\ch(mH)-1} tf=ch(mH)−1twch(mH)−t0
其中 m = h λ δ m=\sqrt{\frac{h}{\lambda \delta}} m=λδh
计算的代码如下
c
#include<stdio.h>
#include<math.h>
int main(){
double H=0.14;
double delta=0.001;
double h=29.1;
double lambda=58.2;
double mH=sqrt(h/lambda/delta)*H;
double t0=50.0;
double tw=100;
double chmH=(exp(mH)+exp(-mH))/2;
double tf=(tw*chmH-t0)/(chmH-1);
printf("外温:%lf\n",tf);
printf("误差:%lf",(tf-tw)/tf);
}
对于一般的肋片,如果顶端绝热,则顶端的过余温度为 θ = θ 0 ch ( m H ) \theta=\frac{\theta_0}{\ch(mH)} θ=ch(mH)θ0,如果给定测量误差α和来流温度tf,则θ=αtf。设 x = e m H x=e^{mH} x=emH,则 x 2 − 2 ( t f − t 0 ) x α t f + 1 = 0 x^2-\frac{2(t_f-t_0)x}{\alpha t_f}+1=0 x2−αtf2(tf−t0)x+1=0,如果已知m,可以由此解出H,代码如下
c
#include<stdio.h>
#include<math.h>
int main(){
double delta=0.001;
double h=29.1;
double lambda=58.2;
double m=sqrt(h/lambda/delta);
double t0=50.0;
double tf=104;
double alpha=0.047;
double B=2*(tf-t0)/alpha/tf;
double H=log((B+sqrt(B*B-4))/2)/m;
printf("套管高度(m):%lf\n",H);
}