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均匀静止理想流体小振幅波运动方程

均匀静止理想流体小振幅波运动方程

作者: itkkanae | 来源:发表于2020-04-20 20:39 被阅读0次

    还是熟悉的图,假设各时空压强满足p=p(x,y,z,t)

    根据压强分析上图所示微元受力情况,首先分析x方向上AB面所受压力:
    \overset{\rightarrow}{F_{A}}=dydz(p-\frac{\partial p}{\partial x}\frac{1}{2}dx)\overset{\rightarrow}{i}

    \overset{\rightarrow}{F_{B}}=-dydz(p+\frac{\partial p}{\partial x}\frac{1}{2}dx)\overset{\rightarrow}{i}

    得到x方向上所受到的力:
    \overset{\rightarrow}{F_{x}}=-dxdydz\frac{\partial p}{\partial x}\overset{\rightarrow}{i}=-dV\frac{\partial p}{\partial x}\overset{\rightarrow}{i}

    个方向上受到的合力:
    \overset{\rightarrow}{F}=-dV(\frac{\partial p}{\partial x}\overset{\rightarrow}{i}+\frac{\partial p}{\partial y}\overset{\rightarrow}{i}+\frac{\partial p}{\partial z}\overset{\rightarrow}{i})=-dVgradp=-dV\bigtriangledown p

    代入牛二定律得:
    m\overset{\rightarrow}{a}=-dV\bigtriangledown p

    \rho\frac{d\overset{\rightarrow}{u}}{dt}=-\bigtriangledown p

    根据均匀静止理想流体小振幅波的条件p_{0}\rho_{0}为常数,u_{0}为0:
    p=p_{0}+\Delta p

    u=u_{0}+\mu

    \rho=\rho_{0}+\Delta\rho

    代入条件得:
    (\rho_{0}+\Delta\rho)\frac{d\overset{\rightarrow}{\mu}}{dt}=-\bigtriangledown\Delta p

    (\rho_{0}+\Delta\rho)(\frac{\partial\overset{\rightarrow}{\mu}}{\partial t}+\frac{\partial\overset{\rightarrow}{\mu}}{\partial x}\frac{dx}{dt}+\frac{\partial\overset{\rightarrow}{\mu}}{\partial y}\frac{dy}{dt}+\frac{\partial\overset{\rightarrow}{\mu}}{\partial z}\frac{dz}{dt})=-\bigtriangledown\Delta p

    由小振幅波的条件可知,\Delta\rho\frac{\partial\overset{\rightarrow}{\mu}}{\partial t}\frac{\partial\overset{\rightarrow}{\mu}}{\partial x}\frac{dx}{dt}等均为一阶小量,方程最后化为:
    \rho_{0}\frac{\partial\overset{\rightarrow}{\mu}}{\partial t}+...=-\bigtriangledown\Delta p

    其中省略部分包含二阶和三阶小量,相较于一阶小量\rho_{0}\frac{\partial\overset{\rightarrow}{\mu}}{\partial t}可以忽略不计,因此运动方程化为:
    \rho_{0}\frac{\partial\overset{\rightarrow}{\mu}}{\partial t}=-\bigtriangledown\Delta p

    另外,对方程两端时间积分得欧拉方程(流体力学):

    \overset{\rightarrow}{\mu}=-\frac{1}{\rho_{0}}\int\bigtriangledown\Delta pdt

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