Download presentation
Presentation is loading. Please wait.
Published byKristin Axelsson Modified over 6 years ago
1
The Heat Equation for Two-Dimensional Objects
2
Example A thin iron square 5 m by 4 m is kept at 100 π C at the bottom 5 m side, and kept at 0 π C at the other three sides. Find the steady-state solution. Solution Temperature now depends on three variables, π₯,π¦,π‘: so we write π=π π₯,π¦,π‘ . It satisfies the heat equation: Two-dimensional heat equation π π₯π₯ + π π¦π¦ = 1 πΌ π π‘ Definition A steady-state solution is a solution to a partial differential equation with the property that π π‘ =0 --hence it depends only on π₯ and π¦. If this is the case, then we have The Dirichlet Problem π π₯π₯ + π π¦π¦ =0 An interesting consequence: the steady-state solution does not depend on the material constant πΌ.
3
Example A thin iron square 5 m by 4 m is kept at 100 π C at the bottom 5 m side, and kept at 0 π C at the other three sides. Find the steady-state solution. Solution π=π π₯,π¦ π π₯π₯ + π π¦π¦ =0 Bottom: π π₯,0 =100 Left & Right: π 0,π¦ =π 5,π¦ =0 Top: π π₯,4 =0
4
Solution The Dirichlet problem can be solved by separation of variables. π π₯π₯ + π π¦π¦ =0 π π₯,π¦ =π π₯ π π¦ π π₯,0 =100 π β²β² π₯ π π¦ +π π₯ π β²β² π¦ =0 π 0,π¦ =π 5,π¦ =0 π β²β² π₯ π π₯ =β π β²β² π¦ π π¦ βπ= π π₯,4 =0 π β²β² π₯ +ππ π₯ =0 π β²β² π¦ βππ π¦ =0 π 0,π¦ =0 βΉ π 0 π π¦ =0 βΉ π 0 =0 π 5,π¦ =0 βΉ π 5 π π¦ =0 βΉ π 5 =0 π= π π = π 2 π 2 25 π π₯ = π π π₯ = sin ππ 5 π₯
5
Solution Letβs discuss the situation weβre in. We guessed π π₯,π¦ =π π₯ π π¦ π π₯π₯ + π π¦π¦ =0 In order to satisfy the the left and right boundary conditions π 0,π¦ =π 5,π¦ =0 π π₯,0 =100 We were forced to take π= π π = π 2 π , and π π₯ = sin ππ 5 π₯ . We had no choice in the matter. π 0,π¦ =π 5,π¦ =0 Now weβre faced with an upsetting situation: π π₯,4 =0 π 0 = 100 π π₯ , π 4 =0 π β²β² π¦ βππ π¦ =0 We cannot possibly satisfy π 0 = 100 π π₯ because a constant cannot be equal to a non-constant. So we will ignore that boundary condition, and deal with it later with superposition. For reasons that will be clear later, we will instead demand that π 0 =1 and π 4 =0. Thus, we will have satisfied all our boundary conditions except π π₯,0 =100. π 0 =1, π 4 =0
6
Solution π π = π 2 π 2 25 π π₯π₯ + π π¦π¦ =0 π β²β² π¦ βππ π¦ =0 π 0 =1, π 4 =0 π π₯,0 =100 π 2 β π 2 π =0 π 0,π¦ =π 5,π¦ =0 π π₯,4 =0 π=Β± ππ 5 π π π¦ = πΆ 1 π ππ 5 π¦ + πΆ 2 π β ππ 5 π¦ π π π₯,π¦ = π π π₯ π π π¦ πΆ 1 + πΆ 2 =1 π π π₯,π¦ = sin ππ 5 π₯ πΌ π π ππ 5 π¦ + π½ π π β ππ 5 π¦ πΆ 1 π 4ππ 5 + πΆ 2 π β 4ππ 5 =0 Where πΌ π = 1 1β π 8ππ 5 , π½ π = π 8ππ 5 π 8ππ 5 β1 π π₯,π¦ = π=1 β π π sin ππ 5 π₯ πΌ π π ππ 5 π¦ + π½ π π β ππ 5 π¦ π π π¦ = 1 1β π 8ππ 5 π ππ 5 π¦ + π 8ππ 5 π 8ππ 5 β1 π β ππ 5 π¦
7
Solution π π₯π₯ + π π¦π¦ =0 π π₯,π¦ = π=1 β π π sin ππ 5 π₯ πΌ π π ππ 5 π¦ + π½ π π β ππ 5 π¦ π π₯,0 =100 Where πΌ π = 1 1β π 8ππ 5 , π½ π = π 8ππ 5 π 8ππ 5 β1 π 0,π¦ =π 5,π¦ =0 π π₯,4 =0 Letβs satisfy the boundary condition π π₯,0 =100 π π₯,0 = π=1 β π π sin ππ 5 π₯ πΌ π + π½ π =100 Remember when we had that boundary condition π 0 = 100 π π₯ that we couldnβt solve? But, miraculously, πΌ π + π½ π =1. Why? So we arbitrarily made it π 0 =1 instead? Now we are reaping the benefits of that decision. π=1 β π π sin ππ 5 π₯ =100, 0β€π₯β€5 (Odd 10-periodic function)
8
Solution π π₯π₯ + π π¦π¦ =0 π π₯,π¦ = π=1 β π π sin ππ 5 π₯ πΌ π π ππ 5 π¦ + π½ π π β ππ 5 π¦ π π₯,0 =100 Where πΌ π = 1 1β π 8ππ 5 , π½ π = π 8ππ 5 π 8ππ 5 β1 π 0,π¦ =π 5,π¦ =0 π π₯,4 =0 Itβs interesting to note that The center 2.5 , 2 of the slab is π C
9
Discussion (Dealing with other boundary conditions)
The technique we just used works for the following situation: f(x) You might ask, what about other boundary conditions? No problem!
10
General Boundary Conditions
π 2 π₯ π 1 π¦ π 2 π¦ π 1 π₯ Just solve four easier problems: π 2 π₯ π 1 π¦ π 2 π¦ π 1 π₯ And add your four solutions together to get the solution to the original problem
11
Discussion We now have a satisfactory technique for finding a steady-state solution for any boundary conditions. But weβre not just interested in steady state solutions. Consider, for example, the following problem:
12
Example A 5mΓ4m steel slab is initially at 500 π C throughout when it is subjected to the following boundary conditions: 500 π C Will the center ever cool to 160 π C? If so, when? Use πΌ=9.88Γ 10 β5 . Solution Temperature depends on π₯,π¦,π‘: π π₯,π¦,π‘ . Letβs write π π₯,π¦,π‘ = π 0 π₯,π¦ + π π₯,π¦,π‘ where π 0 π₯,π¦ is the steady-state solution. From previous work, we find that the steady state solution π 0 π₯,π¦ is: π 0 π₯,π¦ = π=1 β π π sin ππ 5 π₯ πΌ π π ππ 5 π¦ + π½ π π β ππ 5 π¦ Where πΌ π = 1 1β π 8ππ 5 , π½ π = π 8ππ 5 π 8ππ 5 β1 π π = π π₯ sin 2ππ 10 π₯ π π₯ = β€π₯β€5 β500 5β€π₯β€10
13
Solution Letβs write π π₯,π¦,π‘ = π 0 π₯,π¦ + π π₯,π¦,π‘ where π 0 π₯,π¦ is the steady-state solution. π 0 π₯,π¦ = π=1 β π π sin ππ 5 π₯ πΌ π π ππ 5 π¦ + π½ π π β ππ 5 π¦ Substitute π π₯,π¦,π‘ = π 0 π₯,π¦ + π π₯,π¦,π‘ into the 2D-heat equation π π₯π₯ + π π¦π¦ = 1 πΌ π π‘ Where πΌ π = 1 1β π 8ππ 5 , π½ π = π 8ππ 5 π 8ππ 5 β1 π 0 π₯π₯ + π 0 π¦π¦ + π π₯π₯ + π π¦π¦ = 1 πΌ π 0 π‘ + 1 πΌ π π‘ π π = π π₯ sin 2ππ 10 π₯ π π₯π₯ + π π¦π¦ = 1 πΌ π π‘ π π₯ = β€π₯β€5 β500 5β€π₯β€10 Since π 0 satisfies the inhomogeneous boundary conditions, π has to satisfy homogeneous boundary conditions. π π₯,π¦,0 =500 since the slab starts at 500 π C π π₯,π¦,0 + π 0 π₯,π¦ =500 π π₯,π¦,0 =500β π 0 π₯,π¦
14
Solution π π₯,π¦,π‘ =π π₯ π π¦ β(π‘) π π₯π₯ + π π¦π¦ = 1 πΌ π π‘ π β²β² π₯ π π¦ β π‘ +π π₯ π β²β² π¦ β π‘ = 1 πΌ π π₯ π π¦ β β² π‘ π π₯,π¦,0 =500β π 0 π₯,π¦ π β²β² π₯ π π₯ + π β²β² π¦ π π¦ = 1 πΌ β β² π‘ β π‘ =βπ 0 ππ π‘βπ πππ’πππππππ π β²β² π₯ π π₯ + π β²β² π¦ π π¦ =βπ π₯=0,π₯=5,π¦=0,π¦=4 1 πΌ β β² π‘ β π‘ =βπ π β²β² π₯ π π₯ =βπ π π,π = π π + π π = π 2 π π 2 π π β²β² π¦ π π¦ =βπ β π,π β² π‘ +πΌ π π,π β π,π π‘ =0 β π,π π‘ = π βπΌ π π,π π‘ π+π=π π π₯,π¦,π‘ = π=1 β π=1 β π π,π π π π₯ π π π¦ β π,π π‘ π β²β² π₯ +ππ π₯ =0, homog. at 0 and 5 π β²β² π¦ +ππ π¦ =0, homog at 0 and 4 π= π π = π 2 π π π π₯ = sin ππ 5 π₯ π= π π = π 2 π π π π¦ = sin ππ 4 π¦
15
Solution π π₯,π¦,0 =500β π 0 π₯,π¦ π π₯,π¦,π‘ = π=1 β π=1 β π π,π π π π₯ π π π¦ β π,π π‘ π=1 β π=1 β π π,π sin πππ₯ 5 sin πππ¦ =500β π 0 π₯,π¦ βFix π¦β π=1 β π=1 β π π,π sin ππ π¦ sin πππ₯ 5 =500β π 0 π₯, π¦ 0
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.