Baixe trabalho metodos numericos e outras Exercícios em PDF para Métodos Numéricos em Engenharia, somente na Docsity!
Use o método ADF para determinar a temperatura interna de uma placa, sabendo que Tⱼᵢ = 0°C. Utilize
∆t=10s, ∆x=10cm, k=0,835 cm²/s.
Fase 01
(-λT¹ᵢⱼ₊₁+2(1+λ)T¹іј-λT¹ᵢⱼ₊₁=λT°ᵢ₊₁ⱼ+2(1-λ)T°ᵢⱼ+λT°ᵢ₊₁ⱼ)
Fase 02
(-λT²ᵢⱼ₊₁+2(1+λ)T²іј-λT²ᵢⱼ₊₁=λT¹ᵢ₊₁ⱼ+2(1-λ)T¹ᵢⱼ+λT¹ᵢ₊₁ⱼ) ∆x= 10
λ= 0,0835 k= 0,835 2(1+λ)= 2,167 2(1-λ)= 1,833 ∆t= 10
1º Iteração: Fase 1 i=
(-λT¹₁₀+2(1+λ)T¹₁₁-λT¹₁₂=λT°₀₁+2(1-λ)T°₁₁+λT°₂₁)
(-λT¹₁₁+2(1+λ)T¹₁₂-λT¹₁₃=λT°₀₂+2(1-λ)T°₁₂+λT°₂₂)
(-λT¹₁₂+2(1+λ)T¹₁₃-λT¹₁₄=λT°₀₃+2(1-λ)T°₁₃+λT°₂₃)
2,167 -0,0835 0 T¹₁₁ 6,2625 3,
T¹₁₂
0 -0,0835 2,167 T¹₁₃ 14,6125 6,
T¹₁₁= 3,0160 T¹₁₂= 3,2708 T¹₁₃= 6,
i=
(-λT¹₂₀+2(1+λ)T¹₂₁-λT¹₂₂=λT¹₁₁+2(1-λ)T°₂₁+λT°₃₁)
(-λT¹₂₁+2(1+λ)T¹₂₂-λT¹₂₃=λT¹₁₂+2(1-λ)T°₂₂+λT°₃₂)
(-λT¹₂₂+2(1+λ)T¹₂₃-λT¹₂₄=λT¹₁₃+2(1-λ)T°₂₃+λT°₃₃)
2,167 -0,0835 0 T¹₂₁ 0,2518 0,
-0,0835 2,167 -0,0835 T¹₂₂ 0,2731 0,
0 -0,0835 2,167 T¹₂₃ 8,9236 4,
T¹₂₁= 0,1274 T¹₂₂= 0,2900 T¹₂₃= 4,
i=
(-λT¹₃₀+2(1+λ)T¹₃₁-λT¹₃₂=λT¹₂₁+2(1-λ)T°₃₁+λT°₄₁)
(-λT¹₃₁+2(1+λ)T¹₃₂-λT¹₃₃=λT¹₂₂+2(1-λ)T°₃₂+λT°₄₂)
(-λT¹₃₂+2(1+λ)T¹₃₃-λT¹₃₄=λT¹₂₃+2(1-λ)T°₃₃+λT°₄₃)
2,167 -0,0835 0 T¹₃₁ 4,1856 2,
T¹₃₂
0 -0,0835 2,167 T¹₃₃ 12,8698 6,
T¹₃₁= 2,0181 T¹₃₂= 2,2477 T¹₃₃= 6,
Fim da 1º fase
1º Iteração: Fase 2 i=
(-λT²₀₁+2(1+λ)T²₁₁-λT²₂₁=λT¹₁₀+2(1-λ)T¹₁₁+λT¹₁₂)
(-λT²₁₁+2(1+λ)T²₂₁-λT²₃₁=λT¹₂₀+2(1-λ)T¹₂₁+λT¹₂₂)
(-λT²₂₁+2(1+λ)T²₃₁-λT²₄₁=λT¹₃₀+2(1-λ)T¹₃₁+λT¹₃₂)
2,167 -0,0835 0 T²₁₁ 12,0639 5,
-0,0835 2,167 -0,0835 T²₂₁ 0,2577 0,
0 -0,0835 2,167 T²₃₁ 8,0619 3,
T²₁₁= 5,5855 T²₂₁= 0,4782 T²₃₁= 3,
i=
(-λT²₀₂+2(1+λ)T²₁₂-λT²₂₂=λT²₁₁+2(1-λ)T¹₁₂+λT¹₁₃)
(-λT²₁₂+2(1+λ)T²₂₂-λT²₃₂=λT²₂₁+2(1-λ)T¹₂₂+λT¹₂₃)
(-λT²₂₂+2(1+λ)T²₃₂-λT²₄₂=λT²₃₁+2(1-λ)T¹₃₂+λT¹₃₃)
2,167 -0,0835 0 T²₁₂ 13,0833 6,
T²₂₂
0 -0,0835 2,167 T²₃₂ 8,9667 4,
T²₁₂= 6,0685 T²₂₂= 0,8038 T²₃₂= 4,
i=
(-λT²₀₃+2(1+λ)T²₁₃-λT²₂₃=λT²₁₂+2(1-λ)T¹₁₃+λT¹₁₄)
(-λT²₁₃+2(1+λ)T²₂₃-λT²₃₃=λT²₂₂+2(1-λ)T¹₂₃+λT¹₂₄)
(-λT²₃₂+2(1+λ)T²₃₃-λT²₄₃=λT²₃₂+2(1-λ)T¹₃₃+λT¹₃₄)
2,167 -0,0835 0 T²₁₃ 27,4769 12,
-0,0835 2,167 -0,0835 T²₂₃ 15,9429 8,
0 -0,0835 2,167 T²₃₃ 23,7576 11,
T²₁₃= 12,9992 T²₂₃= 8,2928 T²₃₃= 11,
Fim da 2º Fase e 1º Iteração
2º Iteração: Fase 1 i=
(-λT³₁₀+2(1+λ)T³₁₁-λT³₁₂=λT²₀₁+2(1-λ)T²₁₁+λT²₂₁) 16,
(-λT³₁₁+2(1+λ)T³₁₂-λT³₁₃=λT²₀₂+2(1-λ)T²₁₂+λT²₂₂) 17,
(-λT³₁₂+2(1+λ)T³₁₃-λT³₁₄=λT²₀₃+2(1-λ)T²₁₃+λT²₂₃) 30,
2,167 -0,0835 0 T³₁₁ 16,5407 7,
-0,0835 2,167 -0,0835 T³₁₂ 17,4532 8,
0 -0,0835 2,167 T³₁₃ 30,7825 14,
T³₁₁= 7,9768 T³₁₂= 8,9220 T³₁₃= 14,
- i=
- (-λT³₂₀+2(1+λ)T³₂₁-λT³₂₂=λT³₁₁+2(1-λ)T²₂₁+λT²₃₁) 1,
- (-λT³₂₁+2(1+λ)T³₂₂-λT³₂₃=λT³₁₂+2(1-λ)T²₂₂+λT²₃₂) 2,
- (-λT³₂₂+2(1+λ)T³₂₃-λT³₂₄=λT³₁₃+2(1-λ)T²₂₃+λT²₃₃) 17, - 2,167 -0,0835 0 T³₂₁ 1,8548 0,
- -0,0835 2,167 -0,0835 T³₂₂ 2,5664 1, - 0 -0,0835 2,167 T³₂₃ 17,3576 8, - T³₂₁= 0,9149 T³₂₂= 1,5305 T³₂₃= 8,
- i=
- (-λT³₃₀+2(1+λ)T³₃₁-λT³₃₂=λT³₂₁+2(1-λ)T²₃₁+λT²₄₁) 11,
- (-λT³₃₁+2(1+λ)T³₃₂-λT³₃₃=λT³₂₂+2(1-λ)T²₃₂+λT²₄₂) 11,
- (-λT³₃₂+2(1+λ)T³₃₃-λT³₃₄=λT³₂₃+2(1-λ)T²₃₃+λT²₄₃) 25, - 2,167 -0,0835 - 11,1045 5,
- -0,0835 2,167 -0,0835 T³₃₂ 11,9442 6, - 0 -0,0835 2, - 25,5303 12, T³₃₃ - T³₃₁ 5,3626 T³₃₂ 6,1817 T³₃₃ 12,
- 2º Iteração: Fase 2 i= Fim da 1º fase - (-λT⁴₀₁+2(1+λ)T⁴₁₁-λT⁴₂₁=λt³₁₀+2(1-λ)T³₁₁+λT³₁₂) 15, - (-λT⁴₁₁+2(1+λ)T⁴₂₁-λT⁴₃₁=λT³₂₀+2(1-λ)T³₂₁+λT³₂₂) 1, - (-λT⁴₂₁+2(1+λ)T⁴₃₁-λT⁴₄₁=λT³₃₀+2(1-λ)T³₃₁+λT³₃₂) 10, - 2,167 -0,0835 0 T⁴₁₁ 15,3664 7, - -0,0835 2,167 -0,0835 T⁴₂₁ 1,8048 1, - 0 -0,0835 2,167 T⁴₃₁ 10,3457 4, - T⁴₁₁= 7,1410 T⁴₂₁= 1,2939 T⁴₃₁= 4,
- i=
- (-λT⁴₀₂+2(1+λ)T⁴₁₂-λT⁴₂₂=λT⁴₁₁+2(1-λ)T³₁₂+λT³₁₃) 18, - 3, (-λT⁴₁₂+2(1+λ)T⁴₂₂-λT⁴₃₂=λT⁴₂₁+2(1-λ)T³₂₂+λT³₂₃)
- (-λT⁴₂₂+2(1+λ)T⁴₃₂-λT⁴₄₂=λT⁴₃₁+2(1-λ)T³₃₂+λT³₃₃) 12, - 2,167 -0,0835 0 T⁴₁₂ 18,1652 8,
- -0,0835 2,167 -0,0835 T⁴₂₂ 3,5872 2, - 0 -0,0835 2,167 T⁴₃₂ 12,7374 5, - T⁴₁₂= 8,4679 T⁴₂₂= 2,2114 T⁴₃₂= 5,
- i=
- (-λT⁴₀₃+2(1+λ)T⁴₁₃-λT⁴₂₃=λT⁴₁₂+2(1-λ)T³₁₃+λT³₁₄) 35,
- (-λT⁴₁₃+2(1+λ)T⁴₂₃-λT⁴₃₃=λT⁴₂₂+2(1-λ)T³₂₃+λT³₂₄) 23,
- (-λT⁴₃₂+2(1+λ)T⁴₃₃-λT⁴₄₃=λT⁴₃₂+2(1-λ)T³₃₃+λT³₃₄) 30, - 2,167 -0,0835 0 T⁴₁₃ 35,7253 16,
- -0,0835 2,167 -0,0835 T⁴₂₃ 23,3250 11, - 0 -0,0835 2,167 T⁴₃₃ 30,8799 14, - 16, T⁴₁₃= - 11, T⁴₂₃= - 14, T⁴₃₃=
- 3º Iteração: Fase 1 i= Fim da 2º Fase e 2º Iteração - (-λT⁵₁₀+2(1+λ)T⁵₁₁-λT⁵₁₂=λT⁴₀₁+2(1-λ)T⁴₁₁+λT⁴₂₁) 19, - (-λT⁵₁₁+2(1+λ)T ⁵₁₂-λT⁵₁₃=λT⁴₀₂+2(1-λ)T⁴₁₂+λT⁴₂₂) 21, - (-λT⁵₁₂+2(1+λ)T⁵₁₃-λT⁵₁₄=λT³₀₃+2(1-λ)T³₁₃+λT³₂₃) 38, - 2,167 -0,0835 0 T⁵₁₁ 19,4599 9, - -0,0835 2,167 -0,0835 T⁵₁₂ 21,9687 11, - 0 -0,0835 2,167 T⁵₁₃ 38,3285 18, - T⁵₁₁ 9,4116 T⁵₁₂ 11,1987 T⁵₁₃ 18,
- i=
- (-λT⁵₂₀+2(1+λ)T⁵₂₁-λT⁵₂₂=λT⁵₁₁+2(1-λ)T⁴₂₁+λT⁴₃₁) 3,
- (-λT⁵₂₁+2(1+λ)T⁵₂₂-λT⁵₂₃=λT⁵₁₂+2(1-λ)T⁴₂₂+λT⁴₃₂) 5,
- (-λT⁵₂₂+2(1+λ)T⁵₂₃-λT⁵₂₄=λT⁵₁₃+2(1-λ)T⁴₂₃+λT⁴₃₃) 24, - 2,167 -0,0835 0 T⁵₂₁ 3,5604 1,
- -0,0835 2,167 -0, - 5,4866 3, T⁵₂₂ - 0 -0,0835 2,167 T⁵₂₃ 24,7074 11, - T⁵₂₁ 1,7603 T⁵₂₂ 3,0436 T⁵₂₃ 11,
- i=
- (-λT⁵₃₀+2(1+λ)T⁵₃₁-λT⁵₃₂=λT⁵₂₁+2(1-λ)T⁴₃₁+λT⁴₄₁) 13,
- (-λT⁵₃₁+2(1+λ)T⁵₃₂-λT⁵₃₃=λT⁵₂₂+2(1-λ)T⁴₃₂+λT⁴₄₂) 15,
- (-λT⁵₃₂+2(1+λ)T⁵₃₃-λT⁵₃₄=λT⁵₂₃+2(1-λ)T⁴₃₃+λT⁴₄₃) 32, - 2,167 -0,0835 0 T⁵₃₁ 13,1645 6,
- -0,0835 2,167 -0,0835 T⁵₃₂ 15,3595 7, - 0 -0,0835 2,167 T⁵₃₃ 32,1036 15, - T⁵₃₁ 6,3800 T⁵₃₂ 7,9164 T⁵₃₃ 15,
- 3º Iteração: Fase 2 i= Fim da 1º fase - 18, (-λT⁶₀₁+2(1+λ)T⁶₁₁-λT⁶₂₁=λt⁵₁₀+2(1-λ)T⁵₁₁+λT⁵₁₂) - (-λT⁶₁₁+2(1+λ)T⁶₂₁-λT⁶₃₁=λT⁵₂₀+2(1-λ)T⁵₂₁+λT⁵₂₂) 3, - 12, (-λT⁶₂₁+2(1+λ)T⁶₃₁-λT⁶₄₁=λT⁵₃₀+2(1-λ)T⁵₃₁+λT⁵₃₂) - 2,167 -0,0835 - 18,1866 8, T⁶₁₁ - -0,0835 2,167 -0,0835 T⁶₂₁ 3,4808 2, - 0 -0,0835 2,167 T⁶₃₁ 12,3556 5, - T⁶₁₁ 8,4756 T⁶₂₁ 2,1557 T⁶₃₁ 5,
- i=
- (-λT⁶₀₂+2(1+λ)T⁶₁₂-λT⁶₂₂=λT⁶₁₁+2(1-λ)T⁵₁₂+λT⁵₁₃) 22,
- (-λT⁶₁₂+2(1+λ)T⁶₂₂-λT⁶₃₂=λT⁶₂₁+2(1-λ)T⁵₂₂+λT⁵₂₃) 6,
- (-λT⁶₂₂+2(1+λ)T⁶₃₂-λT⁶₄₂=λT⁶₃₁+2(1-λ)T⁵₃₂+λT⁵₃₃) 16, - 2,167 -0,0835 0 T⁶₁₂ 22,7478 10,
- -0,0835 2,167 -0,0835 T⁶₂₂ 6,7207 3, - 0 -0,0835 2,167 T⁶₃₂ 16,2562 7, - T⁶₁₂ 10,6440 T⁶₂₂ 3,8062 T⁶₃₂ 7,
- i=
- (-λT⁶₀₃+2(1+λ)T⁶₁₃-λT⁶₂₃=λT⁶₁₂+2(1-λ)T⁵₁₃+λT⁵₁₄) 42, - 29, (-λT⁶₁₃+2(1+λ)T⁶₂₃-λT⁶₃₃=λT⁶₂₂+2(1-λ)T³₂₃+λT³₂₄)
- (-λT⁶₃₂+2(1+λ)T⁶₃₃-λT⁶₄₃=λT⁶₃₂+2(1-λ)T³₃₃+λT³₃₄) 36,
