What Is the Steady-State Solution for This Heat Conduction Problem?

Click For Summary
SUMMARY

The discussion focuses on finding the steady-state solution for the heat conduction equation ∂²u/∂x² = ∂u/∂t, specifically under the boundary conditions u(0,t) = 0 and u(L,t) = 0. The solution involves separating variables, leading to the ordinary differential equations X'' + βX = 0 and T + ∂²βT' = 0. The user successfully derives the form X(x) = C₁cos(√(β)x) and identifies that C₂ must equal zero due to the boundary condition at x = 0. The challenge lies in applying the correct method to reach the steady-state solution, with the user questioning the necessity of separation of variables versus direct integration.

PREREQUISITES
  • Understanding of the heat conduction equation and its boundary conditions.
  • Familiarity with separation of variables technique in solving partial differential equations.
  • Knowledge of ordinary differential equations, specifically second-order linear equations.
  • Basic concepts of steady-state solutions in thermal analysis.
NEXT STEPS
  • Study the method of separation of variables in greater detail.
  • Learn about boundary value problems in the context of partial differential equations.
  • Explore the implications of steady-state solutions in heat conduction scenarios.
  • Review examples of solving the heat equation with different boundary conditions.
USEFUL FOR

Students and professionals in engineering, particularly those specializing in thermal dynamics and heat transfer, as well as anyone studying partial differential equations and their applications in real-world problems.

Jamin2112
Messages
973
Reaction score
12

Homework Statement




In each of Problems 1 through 8 find the steady-state solution of the heat conduction equation ∂2uxx=ut that satisfieds that given set of boundary problems.

...

3. ux(0,t)=0, u(L,t)=0

Homework Equations



Assume u(x,t)=X(x)T(t)


The Attempt at a Solution



2X'' T = X T'

X'' / X = (1/∂2) T' / T = -ß

X'' + ßX = 0,
T + ∂2ßT' = 0.

Since ux(0,t)=0 and u(x,t)=X(x)T(t),

X'(0)T(t)=0 ---> X'(0)=0.

Solving X'' + ßX = 0 like a I would any differential equation,

X(x) = C1cos(√(ß)x) + C2sin(√(ß)x)

--->

X'(x) = -√(ß)C1sin(√(ß)x) + √(ß)C2(cos(√(ß)x)

--->

X'(0)=0= 0 + √(ß)C2 ---> C2=0

--->

X(x) = C1cos(√(ß)x).

Also, u(L,t)=0 ---> X(L)T(t)=0 ---> X(L)=0.

I'm getting somewhere, right?

Since we're talking about a steady-state solution being reached, some function of t and possibly x will disappear, leaving us with just a function v(x) that shows the temperature at any place in the rod.

u(x,t) = v(x) + w(x,t),

limt-->∞ u(x,t) = v(x)


...


Anyhow, I'm sort of stuck. Problem 1 was was easy because it gave me u(0,t)=T1, u(L,t)=T2, just like the examples in the chapter; but this one is throwing me off with the ux(0,t). I'm having trouble putting the pieces together.

Please help, geniuses.
 
Physics news on Phys.org
At steady state, \partial T/\partial t=0. Why not just integrate \partial^2 T/\partial x^2=0 twice and apply the boundary conditions? Or are you required to use separation of variables?
 
Mapes said:
At steady state, \partial T/\partial t=0. Why not just integrate \partial^2 T/\partial x^2=0 twice and apply the boundary conditions? Or are you required to use separation of variables?

Hmmmmm ... In the textbook example it never has us integrate T''(t)=0 when solving these problems. It does say, however:

Since v(x) must satisfy the equation of the heat equation ß2uxx=ut, we have

v''(x)=0, 0<x<L.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
Replies
7
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
11
Views
2K
Replies
2
Views
1K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K