一道流体力学题 Two flat plates are inclined to one another at an angle 2α.A liquid is forced through a small gap at the apex at a steady flow rate of q m3/s.m-width.In the sketch,identify first the coordinate system that you would use.For suc

一道流体力学题 Two flat plates are inclined to one another at an angle 2α.A liquid is forced through a small gap at the apex at a steady flow rate of q m3/s.m-width.In the sketch,identify first the coordinate system that you would use.For suc
一道流体力学题
Two flat plates are inclined to one another at an angle 2α.A liquid is forced through a small gap at the apex at a steady flow rate of q m3/s.m-width.In the sketch,identify first the coordinate system that you would use.For such a system,only one velocity component exists.Analyze first the continuity equation,which tells us partly how the velocity field varies with the coordinates.Substitute this result into the equations of motion and simplify.What additional simplification can be obtained if the Reynolds number and α are small?Solve such an equation to obtain the velocity profile and the pressure drop.
\x05In the same system,the walls are slightly soluble.Calculate the net amount dissolved.

一道流体力学题 Two flat plates are inclined to one another at an angle 2α.A liquid is forced through a small gap at the apex at a steady flow rate of q m3/s.m-width.In the sketch,identify first the coordinate system that you would use.For suc
两块平板成一个2α的夹角 流体以恒定流量q由顶点通过狭缝受迫流动
这里q是单位时间单位宽度体积流量
在草图中先确认坐标系 我比较倾向于以原点作为两平板顶点 以x轴正方向作为流体流动方向 两平板分别在第一和第四区间形成α夹角
只有一个速度分量存在 即速度矢量平行于x轴
先分析连续性方程 连续性方程部分地告诉了我们速度场如何随坐标变化
连续性方程q=vA 其中v为流速 A为过流面积 连续性方程实质就是质量守恒定律方程
由于流量q已知 过流面积A由坐标决定 因而流速v的变化就随坐标而变化
把结果（即由连续性方程求得的流速分布）代入运动方程并化简
所谓运动方程就是动量方程和能量方程
当雷诺数和角度α都较小时 流动可以看作层流 进而可以应用牛顿内摩擦定律进一步化简方程
根据动量方程F=d(mv) 其中F为内摩擦力 由牛顿内摩擦定律求得 m为单位时间质量流量 m=ρq ρ为流体密度 v为流速
求出各断面流速v进而求得流速分布
再根据能量方程（伯努利方程）p1/(ρg)+z1+(v1^2)/(2g)=p1/(ρg)+z1+(v1^2)/(2g)+H 其中H为沿程阻力损失 H=λ(l/d)(v^2)/(2g) 由于是层流 λ=64/Re 于是可以求出各断面压强p进而求得压降
在该系统中 如果壁面是微溶的 要求计算净溶解量的话
在温度给定并且知道压强分布的条件下 可以确定各断面的溶解度
然后根据流量和溶解度 即可求得溶解量