DIRFERENCESOLUTLONTOTHEQUESTONSOFPLAIN (3) Toward boundary node 0 which has third boundary conditions as in Fig 7-3b, we need to make difference equation about unknown To o In order to eliminate the temperature T of the void node in the equation, we can use the boundary conditions aT (7o-T) Teis a known temperature of the outside boundary medium, from the difference formula we can get T B 7o-72) amendatory difference equation; ac ch Get T, then substitute it into formul 1), we can get the 4+ 2hB n-T2-27;-7 ChB T (3) When the boundary plumbs y axes we can get similar amendatory difference equation like above (4) Toward the boundary node which has fourth boundary conditions, if the objects are completely tangent, their temperature 21 field is continue
21 （3）Toward boundary node 0 which has third boundary conditions as in Fig 7-3b，we need to make difference equation about unknown 。In order to eliminate the temperature of the void node in the equation,we can use the boundary conditions： T0 T1 ( ) 0 0 T Te x T = − − ( ) 2 0 1 3 T Te h T T = − − − is a known temperature of the outside boundary medium,from the difference formula we can get: Te Te h T T T T h 2 2 2 4 0 − 2 − 3 − 4 = − + （4）Toward the boundary node which has fourth boundary conditions ,if the objects are completely tangent,their temperature field is continue. （3） Get T1 ，then substitute it into formula（1），we can get the amendatory difference equation ： When the boundary plumbs y axes ,we can get similar amendatory difference equation like above
平二的盖分 (3)对于具有第三类边界条件的边界结点0,如图73b,也须 立出相应于未知值T的差分方程。为了消去该方程中的虚结点 温度T,可利用边界条件得: OT' (T-T) 其中T为边界以外的介质的已知温度。应用差分公式,可得: T 2h=-2(-T) 解出7,代入(1)式,即得修正的差分方程: 2hB 4+ n-T2-27;-7 ChB T (3) 当边界垂直于y轴时,也可导出与上式相似的修正差分方程 (4)对于具有第四类边界条件的边界结点,在完全接触的情 况下,由于两个接触体的温度场是连续的, 22
22 （3）对于具有第三类边界条件的边界结点0，如图7-3b，也须 立出相应于未知值 的差分方程。为了消去该方程中的虚结点 温度 ，可利用边界条件得： T0 T1 ( ) 0 0 T Te x T = − − ( ) 2 0 1 3 T Te h T T = − − − 其中 Te 为边界以外的介质的已知温度。应用差分公式，可得： 解出 T1 ，代入（1）式，即得修正的差分方程： 当边界垂直于 y 轴时，也可导出与上式相似的修正差分方程。 （4）对于具有第四类边界条件的边界结点，在完全接触的情 况下，由于两个接触体的温度场是连续的， Te h T T T T h 2 2 2 4 0 − 2 − 3 − 4 = − + （3）
So when they have the same heat nature constant the boundary node is the same as inside node. If they are not completely tangent or they have different heat nature constant the question is very complex and we will not discuss it here Example: Suppose a square sheet( Fig. 7-4)is 8 meters in 35302520 40 length, 6 meters in width Its right boundary is adiabatic boundary 32-a The temperature of other three 6r 24 boundaries is signed at every node (unit is C), try to find the 18 21/ 121 temperature fromt toT of nodes in the sheet Fig. 7-4 Solution: use a 4x3 grid h=2 m establish difference equations of nodes from a to f according to formula (1) 4-T-T-35-32=0 23 4T-T-T-16-24=0
23 So when they have the same heat nature constant,the boundary node is the same as inside node.If they are not completely tangent ,or they have different heat nature constant ,the question is very complex ,and we will not discuss it here. Example : Suppose a square sheet(Fig.7-4) is 8 meters in length ,6 meters in width .Its right boundary is adiabatic boundary. The temperature of other three boundaries is signed at every node. （unit is ℃），try to find the temperature from to of nodes in the sheet . Ta Ti 6m 8m a b c d e f g i 40 35 30 25 20 16 14 12 10 32 24 18 Establish difference equations of nodes from a to f according to formula (1): Solution：use a 4x3 grid ,h=2 m 4 16 24 0 4 35 32 0 − − − − = − − − − = b a d a b c T T T T T T Fig.7-4
平二的盖分 因此只要两个接触体具有相同的热性常数,这个边界结点就和 内结点完全一样。如果接触不完全,或者两个接触体具有不同 的热性常数,则问题比较复杂,这里不进行讨论。 例:设有矩形薄板,如图7-4, 35302520 长8米,宽6米,右边界为绝热0 边界,其余三边界上的已知结 32 点温度标在各结点上(单位为 ℃),试求板内的结点温度x2441 至T 8 21/ 121 解:用4×3的网格,h=2米。 8m 按照(1)式立出结点a至f处的差分方程: 图7-4 47-Tb-T-35-32=0 4T-T-T-16-24=0 24
24 因此只要两个接触体具有相同的热性常数，这个边界结点就和 内结点完全一样。如果接触不完全，或者两个接触体具有不同 的热性常数，则问题比较复杂，这里不进行讨论。 例：设有矩形薄板，如图7-4， 长8米，宽6米，右边界为绝热 边界，其余三边界上的已知结 点温度标在各结点上（单位为 ℃），试求板内的结点温度 至 。 Ta Ti 6m 8m a b c d e f g i 40 35 30 25 20 16 14 12 10 32 24 18 按照（1）式立出结点a至f处的差分方程： 解：用 43 的网格， h = 2 米。 4 16 24 0 4 35 32 0 − − − − = − − − − = b a d a b c T T T T T T 图7-4
4T-T--1-30=0 47-T,-T-T-14=0 47-7-T-T-25=0 4T-7-T-T-12=0 Establish difference equations of nodes g and i according to formula(3) 4T-T-2T-20=0 47-T。-2T7-10=0 ) Solve the 8 equations above together, we get(unit is C T=28.51,T=22.03,T=24.99,T=1961 T=21.84,T=1741,Tg=1997,T=1620 When the temperature has curve or declining boundary, irregular inside nodes will happen near the boundary. Like node o in the Fig. 7-5a
25 4 12 0 4 25 0 4 14 0 4 30 0 − − − − = − − − − = − − − − = − − − − = f d e i e c f g d b c f c a d e T T T T T T T T T T T T T T T T Establish difference equations of nodes g and i according to formula (3): 4 2 10 0 4 2 20 0 − − − = − − − = i g f g i e T T T T T T Solve the 8 equations above together,we get(unit is ℃）： 21.84, 17.41, 19.97, 16.20 28.51, 22.03, 24.99, 19.61 = = = = = = = = e f g i a b c d T T T T T T T T When the temperature has curve or declining boundary,irregular inside nodes will happen near the boundary. Like node 0 in the Fig.7-5a