Chapter 4, Problem 120P

A steady, two-dimensional velocity field in the ay-plane is given by $\stackrel{\to }{V}=\left(a+bx\right)\stackrel{\to }{i}+\left(c+dy\right)\stackrel{\to }{j}+0\stackrel{\to }{k}$
(a) What are the primary dimensions (m. L,t. T,.. .) of coefficients a, b. c, and ci?
(b) What relationship between the coefficients is necessary in order for this flow to be incompressible?
(c) What relationship between the coefficients is necessary in order for this flow to be irrotational?
(d) Write the strain rate tensor for this (e) For the simplified case of d = -b, derive an equation for the streamlines of this tiow, namely, = function(x, a, b, c).

To determine

(a)

The primary dimensions ( $m$, $L$, $t$, $T$ ) of coefficients $a$, $b$, $c$, and $d$.

Answer to Problem 120P

The primary dimension for $a$ is $\left[L{T}^{-1}\right]$, the primary dimension for $b$ is $\left[T^{-1}\right]$, the primary dimension for $c$ is $$

  $\left[L{T}^{-1}\right]$, and the primary dimension for $d$ is $\left[T^{-1}\right]$.

Explanation of Solution

Given information:

The flow is steady and two-dimensional.

The velocity field in the $xy$ -plane is $\overrightarrow{V}=\left(a+bx\right)\overrightarrow{i}+\left(c+dy\right)\overrightarrow{j}+0\overrightarrow{k}$. ....... (I)

Here, $a$, $b$, $c$, and $d$ are coefficients.

Write the expression of the comparison for $a$ for the velocity.

  $\overrightarrow{V}=a$   ....... (II)

Here, the velocity vector is $\overrightarrow{V}$, the coefficient is $a$, the distance is $D$, and the time is $T$.

Write the expression for the velocity

  $\overrightarrow{V}=\frac{D}{T}$

Here, the distance is $D$ and the time is $T$.

Substitute $\frac{D}{T}$ for $\overrightarrow{V}$ in Equation (II).

  $a=\frac{D}{T}$

Substitute $\text{L}$ for $D$, and $\text{T}$ for $T$ in the Equation (II).

  $a=\left[{\text{LT}}^{-1}\right]$

Here, the length is $\text{L}$, and the time is $\text{T}$.

Write the expression of the comparison for $b$.

  $\overrightarrow{V}=bx$   ....... (III)

Here, the coefficient is $b$.

Substitute $\frac{D}{T}$ for $\overrightarrow{V}$ and $D$ for $x$ in the Equation (III).

  $\frac{D}{T}=bD$

Substitute $\text{L}$ for $D$, and $\text{T}$ for $T$ in the Equation (III).

  $\begin{array}{l}\left[{\text{LT}}^{-1}\right]=b\left[\text{L}\right]\\ b=\left[{\text{T}}^{-1}\right]\end{array}$

Write the expression of the comparison for $c$.

  $\begin{array}{c}\overrightarrow{V}=c\\ \overrightarrow{V}=\frac{D}{T}\end{array}$   ....... (IV)

Here, the coefficient is $c$.

Substitute $\text{L}$ for $D$, and $\text{T}$ for $T$ in the Equation (I

V).

  $c=\left[{\text{LT}}^{-1}\right]$

Write the expression of the comparison for $d$.

  $\overrightarrow{V}=dy$   ....... (V)

Substitute $\frac{D}{T}$ for $\overrightarrow{V}$ and $D$ for $y$ in the Equation (V).

  $\frac{D}{T}=dD$

Substitute $\text{L}$ for $D$, and $\text{T}$ for $T$ in the Equation (V)

  $\begin{array}{l}\left[{\text{LT}}^{-1}\right]=d\left[\text{L}\right]\\ d=\left[{\text{T}}^{-1}\right]\end{array}$

Here, the coefficient is $d$.

Conclusion:

The primary dimension for $a$ is $\left[{\text{LT}}^{-1}\right]$, the primary dimension for $b$ is $\left[{\text{T}}^{-1}\right]$, the primary dimension for $c$ is $$

  $\left[{\text{LT}}^{-1}\right]$, and the primary dimension for $d$ is $\left[{\text{T}}^{-1}\right]$.

To determine

(b)

The relationship between the coefficients in order for the given flow to be incompressible.

Answer to Problem 120P

The relation between the coefficients is $b+d=0.$

Explanation of Solution

Write the expression for the flow to be incompressible.

  $\nabla \overrightarrow{V}=0$   ....... (VI)

Here, the del operator is $\nabla$, and the velocity vector is $\overrightarrow{V}$.

Write the expression for del operator.

  $\nabla =\left(\overrightarrow{i}\frac{\partial }{\partial x}+\overrightarrow{j}\frac{\partial }{\partial y}+\overrightarrow{k}\frac{\partial }{\partial z}\right)$

Here, the vector along $x$ -direction is $\overrightarrow{i}$, the derivative in $x$ -direction is $\frac{\partial }{\partial x}$, the vector in $y$ -direction is $\overrightarrow{j}$, the derivative in $y$ -direction is $\frac{\partial }{\partial y}$, the vector in $z$ -direction is $\overrightarrow{k}$, and the derivative in $z$ -direction is $\frac{\partial }{\partial z}$.

Substitute $\left(\overrightarrow{i}\frac{\partial }{\partial x}+\overrightarrow{j}\frac{\partial }{\partial y}+\overrightarrow{k}\frac{\partial }{\partial z}\right)$ for $\nabla$, and $\left(a+bx\right)\overrightarrow{i}+\left(c+dy\right)\overrightarrow{j}+0\overrightarrow{k}$ for $\overrightarrow{V}$ in the Equation (VI).

  $\begin{array}{c}\left(\overrightarrow{i}\frac{\partial }{\partial x}+\overrightarrow{j}\frac{\partial }{\partial y}+\overrightarrow{k}\frac{\partial }{\partial z}\right)\left(\left(a+bx\right)\overrightarrow{i}+\left(c+dy\right)\overrightarrow{j}+0\overrightarrow{k}\right)=0\\ \left(\overrightarrow{i}\frac{\partial }{\partial x}+\overrightarrow{j}\frac{\partial }{\partial y}+\overrightarrow{k}\frac{\partial }{\partial z}\right)\left(\left(a+bx\right)\overrightarrow{i}+\left(c+dy\right)\overrightarrow{j}\right)=0\\ b+d=0\end{array}$

Conclusion:

The relation between the coefficients is $b+d=0$

To determine

(c)

The relationship between the coefficients in order for the given flow to be irrotational.

Answer to Problem 120P

The relation between the coefficients for the flow being irrotational is 0.

Explanation of Solution

Write the expression for the flow to be irrotational.

  $\nabla \times \overrightarrow{V}=0$   ....... (VII)

Here, the del operator is $\nabla$, and the velocity vector is $\overrightarrow{V}$.

Substitute $\left(\overrightarrow{i}\frac{\partial }{\partial x}+\overrightarrow{j}\frac{\partial }{\partial y}+\overrightarrow{k}\frac{\partial }{\partial z}\right)$ for $\nabla$, and $\left(a+bx\right)\overrightarrow{i}+\left(c+dy\right)\overrightarrow{j}+0\overrightarrow{k}$ for $\overrightarrow{V}$ in the Equation (III).

  $\left(\left(\frac{\partial w}{\partial y}-\frac{\partial v}{\partial z}\right)\overrightarrow{i}-\left(\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}\right)\overrightarrow{j}+\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\overrightarrow{k}\right)=0$   ....... (VIII)

Here, the $y$ -component of velocity is $v$, the $x$ -component of velocity is $u$, the $z$ -component of velocity is $w$, the derivative in $x$ -direction is $\frac{\partial }{\partial x}$, the derivative in $y$ -direction is $\frac{\partial }{\partial y}$, and the derivative in $z$ -direction is $\frac{\partial }{\partial z}$.

Substitute $\left(a+bx\right)$ for $u$, $\left(c+dy\right)$ for $v$, and $0$ for $w$.

  $\begin{array}{c}\left(\left(\frac{\partial \left(0\right)}{\partial y}-\frac{\partial \left(c+dy\right)}{\partial z}\right)\overrightarrow{i}-\left(\frac{\partial \left(0\right)}{\partial x}-\frac{\partial \left(a+bx\right)}{\partial z}\right)\overrightarrow{j}+\left(\frac{\partial \left(c+dy\right)}{\partial x}-\frac{\partial \left(a+bx\right)}{\partial y}\right)\overrightarrow{k}\right)=0\\ =0\end{array}$

Conclusion:

The relation between the coefficients for the flow being irrotational is 0.

To determine

(d)

The strain rate tensor for the given flow.

Answer to Problem 120P

The strain rate tensor for two dimensional flow is $\left(\begin{array}{cc}b& 0\\ 0& d\end{array}\right)$

Explanation of Solution

Write the expression for the strain tensor for two-dimensional flow.

  $\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{1}{2}\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)\\ \frac{1}{2}\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)& \frac{\partial v}{\partial y}\end{array}\right)$   ....... (IX)

Here, the derivative in $x$ -direction is $\frac{\partial }{\partial x}$, the $x$ -component of velocity is $u$, the derivative in $y$ -direction is $\frac{\partial }{\partial y}$, and the $y$ -component of velocity is $v$.

Substitute $\left(a+bx\right)$ for $u$, and $\left(c+dy\right)$ for $v$ in the Equation (V).

  $\begin{array}{c}\left(\begin{array}{cc}\frac{\partial \left(a+bx\right)}{\partial x}& \frac{1}{2}\left(\frac{\partial \left(a+bx\right)}{\partial y}+\frac{\partial \left(c+dy\right)}{\partial x}\right)\\ \frac{1}{2}\left(\frac{\partial \left(a+bx\right)}{\partial y}+\frac{\partial \left(c+dy\right)}{\partial x}\right)& \frac{\partial \left(c+dy\right)}{\partial y}\end{array}\right)\\ =\left(\begin{array}{cc}b& 0\\ 0& d\end{array}\right)\end{array}$

Conclusion:

The strain rate tensor for two dimensional flow is $\left(\begin{array}{cc}b& 0\\ 0& d\end{array}\right)$

To determine

(e)

An equation for the streamlines of the given flow, namely, $y=\text{fun}\left(x,a,b,c\right)$, for the simplified case of $d=-b$.

Answer to Problem 120P

The equation for the streamlines for $d=b$ is $\frac{1}{b}\left(c+\frac{1}{\left(a+bx\right)k}\right)$.

Explanation of Solution

Write the expression for the streamline equation.

  $\frac{dy}{dx}=\frac{v}{u}$   ....... (X)

Here, the differential with respect to $x$ -component is $\frac{dy}{dx}$, the $y$ -component of velocity is $v$, the $x$ -component of velocity is $u$, and the function is $y$.

Substitute $\left(a+bx\right)$ for $u$ and $\left(c+dy\right)$ for $v$ in the Equation (VI).

  $\frac{dy}{dx}=\frac{\left(c+dy\right)}{\left(a+bx\right)}$   ....... (XI)

Substitute $-b$ for $d$ in the Equation (VII).

  $\begin{array}{l}\frac{dy}{dx}=\frac{\left(c-by\right)}{\left(a+bx\right)}\\ \frac{dy}{\left(c-by\right)}=\frac{dx}{\left(a+bx\right)}\end{array}$

Integrate

  $\begin{array}{l}{\displaystyle \int \frac{dy}{\left(c-by\right)}}={\displaystyle \int \frac{dx}{\left(a+bx\right)}}\\ -\frac{\ln \left(c-by\right)}{b}=\frac{\ln \left(a+bx\right)}{b}+\ln k\\ \frac{-1}{\left(c-by\right)}=\left(a+bx\right)k\\ y=\frac{1}{b}\left(c+\frac{1}{\left(a+bx\right)k}\right)\end{array}$

Here, the integration constant is $k$.

Conclusion:

The equation for the streamlines for $d=b$ is $\frac{1}{b}\left(c+\frac{1}{\left(a+bx\right)k}\right)$.