unable to find symbolic solution
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Dhananjay Singh
am 23 Nov. 2020
Kommentiert: Dhananjay Singh
am 23 Nov. 2020
%heat conduction with an electrical heat source in copper wire
clc; clear;
close all;
%parameters
R = 0.2; % radius in m
L = 10; % length of wire in m
ke = 60000000; % electrical conductivity of copper in S/m
k = 401; % thermal conductivity of copper in W/m.K
I = 10; % electrical current in Amp
Se = (I*I)/ke; % rate of heat production per unit volume
T0 = 20 % Temp in Celcius
%Energy Balance in cylindrical shell of thickness del(r) and length L
% Doing the energy balance and taking del(r) common and limiting it to 0, a
% linear differential equation is formed.
% SOLVING DIFFERENTIAL EQUATION
syms q(r) %q is the heat flux
ode45 = diff(r*q,r) == Se*r
cond = q(0)~=Inf;%B.C at r = 0,q~=inf is invalid so we took it's value 0 to get similar answer
qSol(r)=dsolve(ode45,cond)
%Temperature Profile
syms T(r)
ode45 = diff(T,r) ==(-Se*r/(2*T))
cond = q(R)==T0;
TSol(r)=dsolve(ode45,cond)
0 Kommentare
Akzeptierte Antwort
Stephan
am 23 Nov. 2020
1.
The condition for the second ode is invalid - i assume you meant:
cond = T(R)==T0;
instead of
cond = q(R)==T0;
2.
As you stated by yourself in the code replace:
cond = q(0)~=Inf; %B.C at r = 0,q~=inf is invalid so we took it's value 0 to get similar answer
by
cond = q(0)==0; %B.C at r = 0,q~=inf is invalid so we took it's value 0 to get similar answer
3.
And finally i recommend to replace the variable ode45 by ode, because this can cause trouble due to an inbuilt function named ode45.
Making those changes worked well:
%heat conduction with an electrical heat source in copper wire
clc; clear;
close all;
%parameters
R = 0.2; % radius in m
L = 10; % length of wire in m
ke = 60000000; % electrical conductivity of copper in S/m
k = 401; % thermal conductivity of copper in W/m.K
I = 10; % electrical current in Amp
Se = (I*I)/ke; % rate of heat production per unit volume
T0 = 20 % Temp in Celcius
%Energy Balance in cylindrical shell of thickness del(r) and length L
% Doing the energy balance and taking del(r) common and limiting it to 0, a
% linear differential equation is formed.
% SOLVING DIFFERENTIAL EQUATION
syms q(r) %q is the heat flux
ode = diff(r*q,r) == Se*r
cond = q(0)==0;%B.C at r = 0,q~=inf is invalid so we took it's value 0 to get similar answer
qSol(r)=dsolve(ode,cond)
%Temperature Profile
syms T(r)
ode = diff(T,r) ==(-Se*r/(2*T))
cond = T(R)==T0;
TSol(r)=dsolve(ode,cond)
4 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Calculus finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!