clear all
clc
syms x xe k n CR T
Tin=915;
CR=1;
xe=1;
eq=(xe*k^n*((Tin-T)/CR)^n)/(1+k^n*((Tin-T)/CR)^n)==x;
eq1=subs(eq,T,888);
eq1=subs(eq1,x,5.43/100);
eq2=subs(eq,T,840);
eq2=subs(eq2,x,73/100);
sol=solve([eq1,eq2],[k,n]);

 Akzeptierte Antwort

Walter Roberson
Walter Roberson am 18 Mai 2019

1 Stimme

The equation is too complicated for MATLAB.
The solutions are:
ln = @log;
k = exp(-(ln(73/(100*xe-73))*ln(25/9)+ln(75)*ln(543/73*(100*xe-73)/(10000*xe-543)))/ln(543/73*(100*xe-73)/(10000*xe-543)))
n = -ln(543/73*(100*xe-73)/(10000*xe-543))/ln(25/9)

2 Kommentare

sudheer kumar
sudheer kumar am 18 Mai 2019
Thanks for your response.
Actually i want to change my input values so that i will get corresponding unknown values k and n.
The solution you have given will not work.
Instead of
Tin=915;
CR=1;
xe=1;
we will use the symbols Tin, CR, and xe.
Let 888 be T1 and corresponding x 5.43/100 be x1.
Let 840 be T2 and corresponding x 73/100 be x2.
ln = @log;
k = exp((-ln((-T2+Tin)/CR)*ln(x2*(x1-xe)/x1/(x2-xe))+ln(-x2/(x2-xe))*(ln((-T2+Tin)/CR)-ln((-T1+Tin)/CR)))/ln(x2*(x1-xe)/x1/(x2-xe)))
n = ln(x2*(x1-xe)/x1/(x2-xe))/(ln((-T2+Tin)/CR)-ln((-T1+Tin)/CR))
The solution I gave was fine for the question that was asked. It was after 4 am here when I answered, and my mind-reading abilities are tired.

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