温动向边签解法 可见,导热系数表示“在单位温度梯度下通过等温面单位面积 的热流速度”。 由(1)和(3)可见,热流密度的大小 热流密度在坐标轴上的投影 x=— 一 可见:热流密度在任一方向的分量,等于导热系数乘以 温度在该方向的递减率
16 n T q = 由（1）和（3）可见，热流密度的大小 可见，导热系数表示“在单位温度梯度下通过等温面单位面积 的热流速度” 。 热流密度在坐标轴上的投影 可见：热流密度在任一方向的分量，等于导热系数乘以 温度在该方向的递减率。 z T q y T q x T q z y x = − = − = −
86-2 The Differential Equation of Heat Conduction The principle of heat quantity equilibrium: Within any period of time, the heat quantity accumulated in any minute part of the object equals the heat quantity conducted into this minute part plus the heat quantity supplied by internal heat source Take a minute hexahedron dxdydz as shown in the above figure. Suppose that the temperature of this hexahedron rises from T to T+ OT dt he heat quantity accumulated by temperature is Cpdxdydz aT - dt, where p is the density of the object, C is the heat quantity needed when the temperature of the object with a unit mass rise one degree-mspecific thermal capability 17
17 The principle of heat quantity equilibrium: Within any period of time, the heat quantity accumulated in any minute part of the object equals the heat quantity conducted into this minute part plus the heat quantity supplied by internal heat source. §6-2 The Differential Equation of Heat Conduction dx x q q x x + qx x y z Take a minute hexahedron dxdydz as shown in the above figure. Suppose that the temperature of this hexahedron rises from T to . The heat quantity accumulated by temperature is , where is the density of the object, C is the heat quantity needed when the temperature of the object with a unit mass rise one degree——specific thermal capability. dt t T T + dt t T C dxdydz
温动向边签解法 §6-2热传导微分方程 热量平衡原理:在任意一段时间内,物体的任一微小部 分所积蓄的热量,等于传入该微小部分的热量加上内部热源 所供给的热量 q 上 取图示微小六面体 dxdydz。假定该六面体的温度在d时间内 由7升高到7+0d。由温度所积蓄的热量是 Cpdxdydz--dt ot 其中是物体的密度,C是单位质量的物体升高一度时所需的 热量。比热容。 18
18 热量平衡原理：在任意一段时间内，物体的任一微小部 分所积蓄的热量，等于传入该微小部分的热量加上内部热源 所供给的热量。 §6-2 热传导微分方程 dx x q q x x + x q x y z 取图示微小六面体dxdydz。假定该六面体的温度在dt时间内 由T 升高到 。由温度所积蓄的热量是 ， 其中 是物体的密度，C 是单位质量的物体升高一度时所需的 热量——比热容。 dt t T T + dt t T C dxdydz
Within the same period of time dt, the heat quantity q dydzdt is conducted into the hexahedron from left, and the heat quantity 9x +a- dx )dydsdtis conducted out the hexahedron through right. Hence, the net heat quantity conducted into is ogx dxdydzdt Introduce q,=-14 into it. We can see that a-T The net heat quantity conducted into it from left and rignt is: or dt O-T ne net heat quantity conducted into it from top and bottom is: oy dydedxdt The net heat quantity conducted into it from front and back is: a O-T dzdxdyat Hence, the total net heat quantity conducted into the hexahedron is O-T 0-T0T ay O2 ) dxdydzdt which can be abbreviated as 况V2 dxdydzat 19
19 Within the same period of time dt, the heat quantity qxdydzdt is conducted into the hexahedron from left, and the heat quantity is conducted out the hexahedron through right. Hence, the net heat quantity conducted into is dx dydzdt x q q x x ( ) + dxdydzdt x qx − x T qx Introduce = − into it . We can see that dxdydzdt x T 2 2 dydzdxdt y T 2 2 dzdxdydt z T 2 2 The net heat quantity conducted into it from left and right is: The net heat quantity conducted into it from top and bottomis: The net heat quantity conducted into it from front and back is: dxdydzdt z T y T x T ( ) 2 2 2 2 2 2 + + Hence, the total net heat quantity conducted into the hexahedron is: Tdxdydzdt 2 which can be abbreviated as:
温动向边签解法 在同一段时间d内,由六面体左面传入热量 g, dydzdt, 由右面传出热量(g+x)dvda。因此,传入的净热量为 O ag x dxdydzdt ax OT 将=代入可见: aT 由左右两面传入的净热量为 dxdvdzat 由上下两面传入的净热量为 O- dydzdx di 由前后两面传入的净热量为: Oe2 dedxdvdt 因此,传入六面体的总净热重为xc0o7,oT1dhb 简记为: 20
20 在同一段时间dt内，由六面体左面传入热量qxdydzdt， 由右面传出热量 dx dydzdt 。因此，传入的净热量为 x q q x x ( ) + dxdydzdt x qx − x T qx 将 = − 代入可见： dxdydzdt x T 2 2 dydzdxdt y T 2 2 由左右两面传入的净热量为 由上下两面传入的净热量为 由前后两面传入的净热量为： 因此，传入六面体的总净热量为： 简记为： dzdxdydt z T 2 2 dxdydzdt z T y T x T ( ) 2 2 2 2 2 2 + + Tdxdydzdt 2