Steady state temperature of insulated rod

[jc2009]

Problem: Find the steady state temperature of a laterally insulated rod subject to the following conditions. Length of the rod 10, , left end kept at 0 and right end at 100

Solution: i got this equation from a book i don't know if this applied to this problem but i don't know what to do.
u(x,t) = v(x) + w(x,t) , v is the steady solution and w the transient putting this is the PDE of Heat , we get two equations , one of which is a function of v and this can be easily solved ( but i don't see what the equations are) .

From this point i don't know what to do

thanks

 

[coomast]

Hello jc2009,

The basic equation of heat conduction in 1 dimension is given as:
\frac{\partial u}{\partial t}=k \cdot \frac{\partial ^2 u}{\partial x^2}
Now if it can be assumed that the equation can be written as the sum of a transient and steady state part as follows:
u(x,t)=v(x)+w(x,t)
with v(x) the steady-state and w(x,t) the transient solution and u(x,t) the total solution, you can obtain two new equations by substituting this into the original PDE. This is thus:
\frac{\partial v}{\partial t}+\frac{\partial w}{\partial t}=k \cdot \left[\frac{\partial ^2 v}{\partial x^2}+ \frac{\partial ^2 w}{\partial x^2}\right]
Now the first term is zero because the steady-state is independent of time. This gives two new equations as follows:
\frac{\partial w}{\partial t}=k \cdot \frac{\partial ^2 w}{\partial x^2}
0=\frac{\partial ^2 v}{\partial x^2}
The second one gives you the steady-state solution, the first one the transient one. In your case the solution for the transient can be written as:
v(x)=A\cdot x + B
Using the boundary conditions, you get:
v(x)=10\cdot x

Hope this helps,
coomast

 

[jc2009]

Hello jc2009,

The basic equation of heat conduction in 1 dimension is given as:
\frac{\partial u}{\partial t}=k \cdot \frac{\partial ^2 u}{\partial x^2}
Now if it can be assumed that the equation can be written as the sum of a transient and steady state part as follows:
u(x,t)=v(x)+w(x,t)
with v(x) the steady-state and w(x,t) the transient solution and u(x,t) the total solution, you can obtain two new equations by substituting this into the original PDE. This is thus:
\frac{\partial v}{\partial t}+\frac{\partial w}{\partial t}=k \cdot \left[\frac{\partial ^2 v}{\partial x^2}+ \frac{\partial ^2 w}{\partial x^2}\right]
Now the first term is zero because the steady-state is independent of time. This gives two new equations as follows:
\frac{\partial w}{\partial t}=k \cdot \frac{\partial ^2 w}{\partial x^2}
0=\frac{\partial ^2 v}{\partial x^2}
The second one gives you the steady-state solution, the first one the transient one. In your case the solution for the transient can be written as:
v(x)=A\cdot x + B
Using the boundary conditions, you get:
v(x)=10\cdot x

Hope this helps,
coomast

excellent answer, very detailed, thank you coomast

 

[jc2009]

Hello jc2009,

The basic equation of heat conduction in 1 dimension is given as:
\frac{\partial u}{\partial t}=k \cdot \frac{\partial ^2 u}{\partial x^2}
Now if it can be assumed that the equation can be written as the sum of a transient and steady state part as follows:
u(x,t)=v(x)+w(x,t)
with v(x) the steady-state and w(x,t) the transient solution and u(x,t) the total solution, you can obtain two new equations by substituting this into the original PDE. This is thus:
\frac{\partial v}{\partial t}+\frac{\partial w}{\partial t}=k \cdot \left[\frac{\partial ^2 v}{\partial x^2}+ \frac{\partial ^2 w}{\partial x^2}\right]
Now the first term is zero because the steady-state is independent of time. This gives two new equations as follows:
\frac{\partial w}{\partial t}=k \cdot \frac{\partial ^2 w}{\partial x^2}
0=\frac{\partial ^2 v}{\partial x^2}
The second one gives you the steady-state solution, the first one the transient one. In your case the solution for the transient can be written as:
v(x)=A\cdot x + B
Using the boundary conditions, you get:
v(x)=10\cdot x

Hope this helps,
coomast

In this part :

The second one gives you the steady-state solution, the first one the transient one. In your case the solution for the transient can be written as:
v(x)=A\cdot x + B
Using the boundary conditions, you get:
v(x)=10\cdot x

I think you meant to say "THe solution for the steady state,(not the transient one) can be written as:
v(x)=A\cdot x + B
Using the boundary conditions, you get:
v(x)=10\cdot x

because you if you integrate 2 times the v(x) part using the boundary condition you get
u'' = 0 , Integrate [ u '' ] = u' = A
Integrate [u ' ] = Ax+B ----> v(x) = 10x , or this is the transient one ?
Also how exactly you got 10x? , is this because of the length of the rod ? , how do you know A = 10 , and B=0 in Ax+B ?

 

[coomast]

Hello jc2009,

Indeed I meant the steady-state solution :-)

The boundary conditions are:
v=0 at x=0
v=100 at x=10

Putting this into the solution v=A*x+B gives you the system to solve:

100=A*10+B
0=A*0+B

From which you get: v=10*x

best regards,

coomast