Chapter 4, Problem 4.34P

To determine

(a)

Whether the flow field is a valid solution to continuity and Navier Stokes.

Answer to Problem 4.34P

The Flow field satisfy three dimensional incompressible continuity equation and Navier Stokes

Explanation of Solution

Given information:

The Flow field is a vector representing a three dimensional incompressible flow field Here, The velocity flow field vector is , velocity in $x$ direction is $Kx$, velocity in $y$ direction is $Ky$, velocity in $z$ direction is $-2Kz$.

Write the expression for three dimensional incompressible continuity Equation.

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$ ... (I)

Here, the velocity of fluid along $x,y$ and $z$ directions are $u,v$ and $w$ respectively.

Calculation:

Substitute $Kx$ for $u$, $Ky$ for $v$, $-2Kz$ for $w$ in Equation (I).

$\begin{array}{l}\frac{\partial }{\partial x}\left(Kx\right)+\frac{\partial }{\partial y}\left(Ky\right)+\frac{\partial }{\partial z}\left(-2Kz\right)=0\\ K+K-2K=0\\ 2K-2K=0\end{array}$

Conclusion:

The Flow field Satisfy the three dimensional continuity equation and therefore it will satisfy Navier Stokes Equation.

To determine

(b)

The pressure field $p(x,y,z)$ for .

Answer to Problem 4.34P

The pressure field $p\left(x,y,z\right)=-\rho gz-\frac{\rho K^2}{2}\left(x^2+y^2+z^2\right)$.

Explanation of Solution

Write the expression for incompressible Navier stoke equation in $x$ direction.

$\rho \left(u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial x}+u\frac{\partial w}{\partial x}\right)=-\frac{\partial p}{\partial x}+v\left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}+\frac{{\partial }^{2}w}{\partial z^2}\right)$ ... (II)

Write the expression for incompressible Navier stoke equation in $y$ direction.

$\rho \left(u\frac{\partial u}{\partial y}+v\frac{\partial v}{\partial y}+u\frac{\partial w}{\partial y}\right)=-\frac{\partial p}{\partial y}+v\left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}+\frac{{\partial }^{2}w}{\partial z^2}\right)$ ... (III)

Write the expression for incompressible Navier stoke equation in $y$ direction.

$\rho \left(u\frac{\partial u}{\partial z}+v\frac{\partial v}{\partial z}+u\frac{\partial w}{\partial z}\right)=\rho \left(-g\right)-\frac{\partial p}{\partial y}+v\left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}+\frac{{\partial }^{2}w}{\partial z^2}\right)$ ... (IV)

Write the total pressure of the field.

$p\left(x,y,z\right)=p_x+p_y+p_z$ ... (V)

Calculation:

Substitute $Kx$ for $u$, $Ky$ for $v$, $-2Kz$ for $w$ in Equation (II).

$\begin{array}{l}\left[\rho \left(\begin{array}{l}\left(Kx\right)\frac{\partial \left(Kx\right)}{\partial x}+\left(Ky\right)\frac{\partial \left(Ky\right)}{\partial x}\end{array}\right)\right]\end{array}$

+ −2Kz ∂ −2Kz ∂x

= − ∂p ∂x +v ∂ 2 Kx ∂ x 2 + ∂ 2 Ky ∂ y 2 + ∂ 2 −2Kz ∂ z 2 ρ Kxk+ Ky0+ −2Kz0=−∂p∂x+v0+0+0ρK2x+0+0=−∂p∂xρK2x∂x=−∂p Further

$\rho \left(K^{2}x\right)=\frac{-\partial p}{\partial x}$

Integrate the above equation.

$\begin{array}{l}{\displaystyle \int \rho \left(K^{2}x\right)}={\displaystyle \int \left(\frac{-\partial p}{\partial x}\right)}\\ \frac{\rho \left(K^2{x}^2\right)}{2}=-p_x\\ p_x=\frac{-\rho \left(K^2{x}^2\right)}{2}\end{array}$ ... (VI)

Substitute $Kx$ for $u$, $Ky$ for $v$, $-2Kz$ for $w$ in Equation (III) for pressure field in $y$ direction.

$\begin{array}{l}\left[\rho \left(\begin{array}{l}\left(Kx\right)\frac{\partial \left(Kx\right)}{\partial y}+\left(Ky\right)\frac{\partial \left(Ky\right)}{\partial y}\end{array}\right)\right]\end{array}$

+ −2Kz ∂ −2Kz ∂y

= − ∂p ∂y +v ∂ 2 Kx ∂ x 2 + ∂ 2 Ky ∂ y 2 + ∂ 2 −2Kz ∂ z 2 ρ Kx0+ KyK+ −2Kz0=−∂p∂y+v0+0+0ρ0+K2y+0=−∂p∂yρK2y∂y=−∂p $\rho \left(K^{2}y\right)=\frac{-\partial p}{\partial y}$

Integrate the above equation.

$\begin{array}{l}{\displaystyle \int \rho \left(K^{2}y\right)}={\displaystyle \int \left(\frac{-\partial p}{\partial y}\right)}\\ \frac{\rho \left(K^2{y}^2\right)}{2}=-p_y\\ p_y=\frac{-\rho \left(K^2{y}^2\right)}{2}\end{array}$ ... (VII)

Substitute $Kx$ for $u$, $Ky$ for $v$, $-2Kz$ for $w$ in Equation (IV) for pressure field in $z$ direction.

$\begin{array}{l}\rho \left(\begin{array}{l}\left(Kx\right)\frac{\partial \left(Kx\right)}{\partial z}+\left(Ky\right)\frac{\partial \left(Ky\right)}{\partial z}\\ +\left(-2Kz\right)\frac{\partial \left(-2Kz\right)}{\partial z}\end{array}\right)=\rho \left(-g\right)-\frac{\partial p}{\partial z}+v\left(\begin{array}{l}\frac{{\partial }^2\left(Kx\right)}{\partial x^2}\\ +\frac{{\partial }^2\left(Ky\right)}{\partial y^2}+\frac{{\partial }^2\left(-2Kz\right)}{\partial z^2}\end{array}\right)\\ \rho \left(\left(Kx\right)0+\left(Ky\right)0+\left(-2Kz\right)-2K\right)=\rho \left(-g\right)-\frac{\partial p}{\partial z}+v\left(0+0+0\right)\\ \rho \left(0+0+4K^{2}z\right)=\rho \left(-g\right)-\frac{\partial p}{\partial z}\\ \rho \left(4K^{2}z\right)=\rho \left(-g\right)-\frac{\partial p}{\partial z}\end{array}$ $\begin{array}{l}\rho \left(4K^{2}z\right)+\rho \left(-g\right)=-\frac{\partial p}{\partial z}\\ \partial p=-\left(\rho g+4\rho K^{2}z\right)\partial z\end{array}$

Integrate the above Equation.

$\begin{array}{l}{\displaystyle \int \partial p}={\displaystyle \int -\left(\rho g+4\rho K^{2}z\right)\partial z}\\ p_z=-\left[\rho gz+4\rho K^2\left(\frac{z^2}{2}\right)\right]\\ p_z=-\left[\rho gz+2\rho K^2{z}^2\right]\end{array}$ ... (VIII)

Substitute $\frac{-\rho \left(K^2{x}^2\right)}{2}$ for $p_x$, $\frac{-\rho \left(K^2{y}^2\right)}{2}$ for $p_y$, $-\left[\rho gz+2\rho K^2{z}^2\right]$ for $p_z$ in Equation (V)

$\begin{array}{c}p\left(x,y,z\right)=\frac{-\rho \left(K^2{x}^2\right)}{2}+\frac{-\rho \left(K^2{y}^2\right)}{2}+-\left[\rho gz+2\rho K^2{z}^2\right]\\ =\frac{-\rho \left(K^2{x}^2\right)}{2}+\frac{-\rho \left(K^2{y}^2\right)}{2}-\rho gz-2\rho K^2{z}^2\\ =\frac{-\rho \left(K^2{x}^2\right)}{2}+\frac{-\rho \left(K^2{y}^2\right)}{2}-\rho gz-2\rho K^2{z}^2\times \left(\frac{2}{2}\right)\\ =-\rho gz-\frac{\rho K^2}{2}\left(x^2+y^2+4z^2\right)\end{array}$

Conclusion:

The pressure field $p(x,y,z)$ for is $-\rho gz-\frac{\rho K^2}{2}\left(x^2+y^2+4z^2\right)$.

(c)

To determine

Whether the flow is rotational or irrotational.

Answer to Problem 4.34P

The flow is irrotational in nature

Explanation of Solution

Given information:

The Flow field is a vector representing a three dimensional incompressible flow field Here, $\text{V}$ is flow field vector, $Kx$ is velocity in $x$ direction, $Ky$ is velocity in $y$ direction, $-2Kz$ is velocity in $z$ direction

Write the expression for curl of the velocity field.

... (IX)

Here the velocity in $x$ direction is $u$, the velocity in the $y$ direction is $v$, velocity in the $z$ direction is $w$.

Calculation:

Substitute $Kx$ for $u$, $Ky$ for $v$, $-2Kz$ for $w$ in Equation (IX).

Conclusion:

Since $\nabla \times \text{V=}\text{0}$, the flow is irrotational.