1

If $(u(y),0,0)$ are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the $X$-direction, then $u(y)$ is a

$(a)$ linear function of $y$.

$(b)$ quadratic function of $y$.

$(c)$ cubic function of $y$.

$(d)$ constant.

I started with Euler dynamical equation as follows : $$\frac{D \vec{q}}{Dt}=\vec{F}-\frac{1}{\rho}(\nabla p)$$ where $\vec{q}=(u(y),0,0)$ and $\vec{F}$ is the external force, $\nabla p$ being pressure gradient. Now as per question we have along $X$-direction that $$\frac{\partial u(y)}{\partial t}+u(y)\frac{\partial u(y)}{\partial x}+0.\frac{\partial u(y)}{\partial y}+0.\frac{\partial u(y)}{\partial z}=0-\frac{1}{\rho}\frac{\partial p}{\partial x} \\ \implies \frac{\partial u(y)}{\partial t}=-\frac{1}{\rho}\frac{\partial p}{\partial x} \\ \implies u(y)=-\int \bigg(\frac{1}{\rho}\frac{\partial p}{\partial x}\bigg)dt+C$$ $C$ being arbitrary constant. But from here I am unable to conclude whether this is linear or quadratic or cubic or constant. Is my thinking correct here? Any help is appreciated.

am_11235...
  • 2,142

1 Answers1

1

The Navier-Stokes equations apply as the fluid is Newtonian and, hence, viscous.

Neglecting body force terms other than gravity which can be absorbed into the pressure field, the $x$-component of velocity satisfies

$$\tag{*}\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z} = - \frac{1}{\rho} \frac{\partial p}{\partial x} + \frac{\mu}{\rho}\nabla^2 u,$$

where $\mu$ is the viscosity.

The flow is steady, fully developed ($u$ depends only on $y$), and unidirectional. Consequently the partial derivatives of $u$ with respect to $t$, $x$, and $z$ vanish and $v = w = 0$. Thus, (*) reduces to

$$\tag{**} \frac{d ^2u}{dy^2} = \frac{1}{\mu}\frac{\partial p}{\partial x} = \frac{G}{\mu}$$

The pressure gradient $G$ is independent of $y$. This follows from the $y$-component of the Navier-Stokes equations which reduces to

$$\frac{\partial p}{\partial y} = 0$$

Integrating both sides of (**) twice with respect to $y$ we get the quadratic function

$$u(y) = \frac{G}{2\mu}y^2 + C_1y + C_2,$$

wher $C_1$ and $C_2$ are integration constants.

RRL
  • 90,707