【数学建模】高温作业专用服装设计(2018A)隐式差分推导

发布于:2024-07-15 ⋅ 阅读:(296) ⋅ 点赞:(0)

为方便计算,对区域进行离散化处理,采用隐式差分格式进行离散计算。隐式差分格式如图:
请添加图片描述

每层材料内部

对第 j j j层材料:

其中, λ j \lambda_j λj​表示第 j j j层的热扩散率, c j c_j cj​表示第 j j j层的比热容, ρ j \rho_j ρj​表示第 j j j层的材料密度。

∂ T j ∂ t = λ j c j ρ j ∂ 2 T j ∂ x 2 \dfrac{\partial T_j}{\partial t} =\dfrac{\lambda_j}{c_j \rho_j}\dfrac{\partial^2 T_j}{\partial x^2} ∂t∂Tj​​=cj​ρj​λj​​∂x2∂2Tj​​

对方程左端采用向后差分格式,方程右端采用中心差分格式,有:

T j ( x i , t k + 1 ) − T j ( x i , t k ) Δ t = λ j c j ρ j T j ( x i + 1 , t k + 1 ) − 2 T j ( x i , t k + 1 ) + T j ( x i − 1 , t k + 1 ) ( Δ x ) 2 \dfrac{T_j(x_i,t_{k+1})-T_j(x_i,t_{k})}{\Delta t} =\dfrac{\lambda_j}{c_j \rho_j} \dfrac{T_j(x_{i+1},t_{k+1})-2T_j(x_i,t_{k+1}) + T_j(x_{i-1},t_{k+1})}{(\Delta x)^2} ΔtTj​(xi​,tk+1​)−Tj​(xi​,tk​)​=cj​ρj​λj​​(Δx)2Tj​(xi+1​,tk+1​)−2Tj​(xi​,tk+1​)+Tj​(xi−1​,tk+1​)​

整理得:
− T j ( x i , t k ) = λ j Δ t c j ρ j ( Δ x ) 2 F ( X ) − T j ( x i , t k + 1 ) -T_j(x_i,t_{k}) = \dfrac{\lambda_j \Delta t}{c_j \rho_j (\Delta x)^2} F(X) - T_j(x_i,t_{k+1}) −Tj​(xi​,tk​)=cj​ρj​(Δx)2λj​Δt​F(X)−Tj​(xi​,tk+1​)

F ( X ) = T j ( x i + 1 , t k + 1 ) − 2 T j ( x i , t k + 1 ) + T j ( x i − 1 , t k + 1 ) F(X) = T_j(x_{i+1},t_{k+1})-2T_j(x_i,t_{k+1}) + T_j(x_{i-1},t_{k+1}) F(X)=Tj​(xi+1​,tk+1​)−2Tj​(xi​,tk+1​)+Tj​(xi−1​,tk+1​)
令 F ( X ) = r j F(X) = r_j F(X)=rj​

整理得:

T j ( x i , t k ) = − r j T j ( x i + 1 , t k + 1 ) − r j T j ( x i − 1 , t k + 1 ) + ( 1 + 2 r j ) T j ( x i , t k + 1 ) T_j(x_i,t_k) = -r_jT_j(x_{i+1},t_{k+1})-r_jT_j(x_{i-1},t_{k+1})+(1+2r_j)T_j(x_i,t_{k+1}) Tj​(xi​,tk​)=−rj​Tj​(xi+1​,tk+1​)−rj​Tj​(xi−1​,tk+1​)+(1+2rj​)Tj​(xi​,tk+1​)

左边界条件

T j ( x , 0 ) = 37 ° T_j(x,0) = 37\degree Tj​(x,0)=37°
此前求得的左边界条件:

− k 1 ∂ T 1 ∂ x + k o u t T 1 ∣ x = 0 = k o u t T o u t -k_1 \dfrac{\partial T_1}{\partial x} + k_{out}T_1|_{x=0} = k_{out}T_{out} −k1​∂x∂T1​​+kout​T1​∣x=0​=kout​Tout​

对其进行离散化处理:

− k 1 T ( 2 , k ) − T ( 0 , k ) 2 Δ x = k o u t ( T o u t − T ( 1 , k ) ) -k_1 \dfrac{T_(2,k) - T(0,k)}{2\Delta x} = k_{out}(T_{out} - T(1,k)) −k1​2ΔxT(​2,k)−T(0,k)​=kout​(Tout​−T(1,k))

整理可得:

T ( 0 , k ) = 2 Δ x k o u t T o u t k 1 − 2 Δ x k o u t T ( 1 , k ) k 1 + T ( 2 , k ) T(0,k) = \dfrac{2\Delta xk_{out}T_{out}}{k_1} - \dfrac{2\Delta xk_{out}T(1,k)}{k_1} + T(2,k) T(0,k)=k1​2Δxkout​Tout​​−k1​2Δxkout​T(1,k)​+T(2,k)

根据每层材料内部,有

T j ( 1 , k − 1 ) = − r 1 T j ( 2 , k ) − r 1 T j ( 0 , k ) + ( 1 + 2 r 1 ) T j ( 1 , k ) T_j(1,k-1) = -r_1T_j(2,k)-r_1T_j(0,k)+(1+2r_1)T_j(1,k) Tj​(1,k−1)=−r1​Tj​(2,k)−r1​Tj​(0,k)+(1+2r1​)Tj​(1,k)

联立得:

T j ( 1 , k − 1 ) = − r j T j ( 2 , k ) − r j ( 2 Δ x k o u t T o u t k 1 − 2 Δ x k o u t T ( 1 , k ) k 1 + T ( 2 , k ) ) + ( 1 + 2 r j ) T j ( 1 , k ) T_j(1,k-1) = -r_jT_j(2,k)-r_j(\dfrac{2\Delta xk_{out}T_{out}}{k_1} - \dfrac{2\Delta xk_{out}T(1,k)}{k_1} + T(2,k))+(1+2r_j)T_j(1,k) Tj​(1,k−1)=−rj​Tj​(2,k)−rj​(k1​2Δxkout​Tout​​−k1​2Δxkout​T(1,k)​+T(2,k))+(1+2rj​)Tj​(1,k)

T j ( 1 , k − 1 ) + r 1 2 Δ x k o u t T o u t k 1 = − 2 r 1 T j ( 2 , k ) + ( 1 + 2 r 1 + r 1 2 Δ x k o u t k 1 ) T j ( 1 , k ) T_j(1,k-1)+r_1\dfrac{2\Delta xk_{out}T_{out}}{k_1} = -2r_1T_j(2,k)+(1+2r_1+r_1\dfrac{2\Delta xk_{out}}{k_1})T_j(1,k) Tj​(1,k−1)+r1​k1​2Δxkout​Tout​​=−2r1​Tj​(2,k)+(1+2r1​+r1​k1​2Δxkout​​)Tj​(1,k)

右边界条件

此前求得的右边界条件:

k 4 ∂ T 1 ∂ x + k o u t T 4 ∣ x = x s k i n = k s k i n T s k i n k_4 \dfrac{\partial T_1}{\partial x} + k_{out}T_4|_{x=x_{skin}} = k_{skin}T_{skin} k4​∂x∂T1​​+kout​T4​∣x=xskin​​=kskin​Tskin​

对其进行离散化处理:

k 4 T ( N + 1 , k ) − T ( N − 1 , k ) 2 Δ x = k s k i n ( T s k i n − T ( N , k ) ) k_4 \dfrac{T_(N+1,k) - T(N-1,k)}{2\Delta x} = k_{skin}(T_{skin} - T(N,k)) k4​2ΔxT(​N+1,k)−T(N−1,k)​=kskin​(Tskin​−T(N,k))

整理可得:

T ( N − 1 , k ) = T ( N + 1 , k ) − 2 Δ x k s k i n T s k i n k 4 + 2 Δ x k s k i n T ( N , k ) k 4 T(N-1,k) = T(N+1,k) -\dfrac{2\Delta xk_{skin}T_{skin}}{k_4} + \dfrac{2\Delta xk_{skin}T(N,k)}{k_4} T(N−1,k)=T(N+1,k)−k4​2Δxkskin​Tskin​​+k4​2Δxkskin​T(N,k)​
根据每层材料内部,有

T j ( N , k − 1 ) = − r 1 T j ( N + 1 , k ) − r 1 T j ( N − 1 , k ) + ( 1 + 2 r 1 ) T j ( N , k ) T_j(N,k-1) = -r_1T_j(N+1,k)-r_1T_j(N-1,k)+(1+2r_1)T_j(N,k) Tj​(N,k−1)=−r1​Tj​(N+1,k)−r1​Tj​(N−1,k)+(1+2r1​)Tj​(N,k)

联立得:

T j ( N , k − 1 ) = − r j T j ( N + 1 , k ) − r j ( T ( N + 1 , k ) − 2 Δ x k s k i n T s k i n k 4 + 2 Δ x k s k i n T ( N , k ) k 4 ) + ( 1 + 2 r ) T j ( N , k ) T_j(N,k-1) = -r_jT_j(N+1,k)-r_j(T(N+1,k) -\dfrac{2\Delta xk_{skin}T_{skin}}{k_4} + \dfrac{2\Delta xk_{skin}T(N,k)}{k_4})+(1+2r)T_j(N,k) Tj​(N,k−1)=−rj​Tj​(N+1,k)−rj​(T(N+1,k)−k4​2Δxkskin​Tskin​​+k4​2Δxkskin​T(N,k)​)+(1+2r)Tj​(N,k)

T j ( N , k − 1 ) − 2 Δ x k s k i n T s k i n k 4 = − 2 r j T j ( N + 1 , k ) + ( 1 + 2 r j − r j 2 Δ x k s k i n k 4 ) T j ( N , k ) T_j(N,k-1)-\dfrac{2\Delta xk_{skin}T_{skin}}{k_4} = -2r_jT_j(N+1,k)+(1+2r_j-r_j\dfrac{2\Delta xk_{skin}}{k_4})T_j(N,k) Tj​(N,k−1)−k4​2Δxkskin​Tskin​​=−2rj​Tj​(N+1,k)+(1+2rj​−rj​k4​2Δxkskin​​)Tj​(N,k)

N = L N=L N=L

本文参考自【数学建模】2018年国赛A题详解(一)

更多建模相关信息看Github仓库(包括真题练习)

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