I need the Coding to solve the simultaneous equations. To find T, tow, TC.

2 次查看(过去 30 天)
lc
clear all
c0=40;
cd=50;
y0=0.09;
h=0.7;
r=5;
u=0.05;
a=260;
b=0.1;
x='((a*h)/2)+((a*r*T)/3)+((a*h*T*((exp(-u*tow))+b))/3+((a*r*(T^2)*(exp(-u*tow))+b))/8)-(c0/(T^2))+((cd*((3*a*y0*(exp(-u*tow)))-2(2*T*a*b*y0*(exp(-u*tow)))-(2*T*a*(b^2))/6)';
y='1-((u*a*h*(T^2)*y0*(exp(-u*tow)))/6)-((u*a*r*(T^3)*y0*(exp(-u*tow)))/24)-(cd*u*T*y0*(exp(-u*tow))*((3*a-a*b*T)/6))';
TC=(c0/T)+(1/T)*(((a*h*(T^2))/2)+((a*r*(T^3))/6)+((((a*(y0*(exp(-u*tow)))+b)/2))*(((h*(T^3))/3)+((r*(T^4))/12))))+((cd/T)*((((T^2)*y0*(exp(-u*tow))*(3*a-a*b*T))-(a*(b^2)*(T^3)))/6))+tow

采纳的回答

Walter Roberson
Walter Roberson 2021-1-1
You cannot solve that without fixing your formulas.
syms T tow
c0=40;
cd=50;
y0=0.09;
h=0.7;
r=5;
u=0.05;
a=260;
b=0.1;
A = a; %??? maybe ?
R = r; %??? maybe ?
B = b; %??? maybe ?
x = ((a*h)/2)+((a*r*T)/3)+((a*h*T*((exp(-u*tow))+b))/3+((a*r*(T^2)*(exp(-u*tow))+b))/8)-(c0/(T^2))+((cd*((3*a*y0*(exp(-u*tow)))-2*(2*T*a*b*y0*(exp(-u*tow)))-(2*T*a*(b^2))/6);
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
12 1 0 12 1 0 12 34 5 43 21 23 4 3 4 5 43 21 0 1 2 10 12 34 5 6 543 4 5 6 543 4 5 43 2
You have two more ( than you have )
y = 1-((u*a*h*(T^2)*y0*(exp(-u*tow)))/6)-((u*a*r*(T^3)*y0*(exp(-u*tow)))/24)-(cd*u*T*y0*(exp(-u*tow))*((3*a-a*b*T)/6));
sol = solve([x,y], [T tow]);
T = sol.T;
tow = sol.tow;
TC = (c0/T)+(1/T)*(((a*h*(T^2))/2)+((a*r*(T^3))/6)+((((a*(y0*(exp(-u*tow)))+b)/2))*(((h*(T^3))/3)+((r*(T^4))/12))))+((cd/T)*((((T^2)*y0*(exp(-u*tow))*(3*a-a*b*T))-(a*(b^2)*(T^3)))/6))+tow;
disp(T)
disp(tow)
disp(TC)
vpa(T)
vpa(tow)
vpa(TC)

更多回答(1 个)

shunmugam hemalatha
Even i couldn't find answer for the same.
clc
clear all
c0=40;
cd=50;
y0=0.09;
h=0.7;
r=5;
u=0.05;
a=260;
b=0.1;
syms T tow
x='((a*h)/2)+((a*r*T)/3)+((a*h*T*((exp(-u*tow))+b))/3+((a*r*(T^2)*(exp(-u*tow))+b))/8)-(c0/(T^2))+((cd*((3*a*y0*(exp(-u*tow)))-2*(2*T*a*b*y0*(exp(-u*tow)))-(2*T*a*(b^2))/6)';
y = '(1-((u*a*h*(T^2)*y0*(exp(-u*tow)))/6)-((u*a*r*(T^3)*y0*(exp(-u*tow)))/24)-(cd*u*T*y0*(exp(-u*tow))*((3*a-a*b*T)/6))';
sol = solve([x,y], [T tow]);
T = sol.T;
tow = sol.tow;
TC = (c0/T)+(1/T)*(((a*h*(T^2))/2)+((a*r*(T^3))/6)+((((a*(y0*(exp(-u*tow)))+b)/2))*(((h*(T^3))/3)+((r*(T^4))/12))))+((cd/T)*((((T^2)*y0*(exp(-u*tow))*(3*a-a*b*T))-(a*(b^2)*(T^3)))/6))+tow;
disp(T)
disp(tow)
disp(TC)
vpa(T)
vpa(tow)
vpa(TC)
  2 个评论
Walter Roberson
Walter Roberson 2021-1-4
Using solve with quoted strings has not been valid for several releases, and has not been a good idea since 2016 (there were some bugs then that forced using quoted strings for certain differential equations... for nearly any purpose other than differential equations, using quoted strings for symbolic expressions has not been a good idea since r2014a.
I posted the version without quoted strings already and you should have built on that.

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