Chapter 4, Problem 66P

Consider steady, incompressible, two-dimensional shear flow for which the velocity field is $\stackrel{\to }{V}=\left(u,v\right)=\left(a+by\right)\stackrel{\to }{i}+0\stackrel{\to }{j}$ where a and b are constants. Sketched in Fig. P4-66 is a small rectangular fluid particle of dimensions dx and dy at time t. The fluid particle moves and deforms with the flow such that at a later time (t + dt), the particle is no longer rectangular, as also shown in the figure. The initial location of each corner of the fluid particle is labeled in Fig. P4-66. The lower-left corner is at (x,y) at time t, where the x-component of velocity is u = a +by. At the later time, this corner moves to (x + u dt, y), or $\left(x+\left(a+by\right)dt,y\right)$
(a) In similar fashion, calculate the location of each of the other three corners of the fluid particle at time t+dt.
(b) From the fundamental definition of linear strain rate (the rate of increase in length per unit length), calculate linear strain rates ${\epsilon }_{xx}$ and ${\epsilon }_{yy}$ .
(c) Compare your results with those obtained from the equations for ${\epsilon }_{xx}$ and ${\epsilon }_{yy}$ in Cartesian coordinates. i.e.,
   ${\epsilon }_{xx}=\frac{\partial u}{\partial x}{\epsilon }_{yy}=\frac{\partial u}{\partial y}$

To determine

(a)

The location of each of the other three corners of the fluid particle at time $t+dt$.

Answer to Problem 66P

The location of the lower left corner after time $t+dt$ is $\left[\left(x+\left(a+by\right)dt\right),y\right]$.

The location of the lower right corner after time $t+dt$ is $\left[\left(x+\left(a+by\right)dt+dx\right),y\right]$.

The location of the upper left corner after time $t+dt$ is $\left[\left(x+\left(a+b\left(y+dy\right)\right)dt\right),\left(y+dy\right)\right]$.

The location of the upper right corner after time $t+dt$ is $\left[\left(x+dx+\left(a+b\left(y+dy\right)\right)dt+dx\right),\left(y+dy\right)\right]$.

Explanation of Solution

Given information:

Two-dimensional shear flow, flow is incompressible, the velocity field is $\left(a+by\right)\overrightarrow{i}+0\overrightarrow{j}$, dimension of the fluid particle at the initial time is $dx,$ and $dy$ and initial x component of the velocity is $a+by$.

Write the expression for the two-dimensional velocity field in the vector form.

  $\overrightarrow{V}=\left(a+by\right)\overrightarrow{i}+0\overrightarrow{j}$   ...... (I)

Here, the constants are $a$ and $b$, the distance in $x$ direction is $x$ and the distance in $y$ direction is $y$.

The following figure shows the position of the corners at time $t$ and $t+dt$.

  

Figure-(1)

Here, the length of the lower edge at time $t$ is $dx$, length of the left edge at time $t$ is $dy$, length of the lower edge at time $t+dt$ is $n,$ and the length of the right edge at time $t+dt$ is $m$.

Write the expression for location of the lower left corner after time $t+dt$.

  $A=\left[\left(x+udt\right),y\right]$   ...... (II)

Write the expression for location of the lower right corner after time $t+dt$.

  $B=\left[\left(x+udt+dx\right),y\right]$   ...... (III)

Write the expression for location of the upper left corner after time $t+dt$.

  $C=\left[\left(x+u^{\prime }dt\right),\left(y+dy\right)\right]$   ...... (IV)

Write the expression for location of the upper right corner after time $t+dt$.

  $D=\left[\left(x+dx+u^{\prime }dt+dx\right),\left(y+dy\right)\right]$   ...... (V)

Write the expression for velocity along x direction.

  $u=a+by$   ...... (VI)

Calculation:

Substitute $a+by$ for $u$ in Equation (II).

  $A=\left[\left(x+\left(a+by\right)dt\right),y\right]$

Substitute $a+by$ for $u$ in Equation (III).

  $B=\left[\left(x+\left(a+by\right)dt+dx\right),y\right]$

Substitute $a+b\left(y+dy\right)$ for $u^{\prime }$ in Equation (IV).

  $C=\left[\left(x+\left(a+b\left(y+dy\right)\right)dt\right),\left(y+dy\right)\right]$

Substitute $a+b\left(y+dy\right)$ for $u^{\prime }$ in Equation (V).

  $D=\left[\left(x+dx+\left(a+b\left(y+dy\right)\right)dt+dx\right),\left(y+dy\right)\right]$

Conclusion:

The location of the lower left corner after time $t+dt$ is $\left[\left(x+\left(a+by\right)dt\right),y\right]$.

The location of the lower right corner after time $t+dt$ is $\left[\left(x+\left(a+by\right)dt+dx\right),y\right]$.

The location of the upper left corner after time $t+dt$ is $\left[\left(x+\left(a+b\left(y+dy\right)\right)dt\right),\left(y+dy\right)\right]$.

The location of the upper right corner after time $t+dt$ is $\left[\left(x+dx+\left(a+b\left(y+dy\right)\right)dt+dx\right),\left(y+dy\right)\right]$.

To determine

(b)

The linear strain rates.

Answer to Problem 66P

The linear strain rate along x axis is $0$.

The linear strain rate along y axis is $0$.

Explanation of Solution

Write the expression for the strain rate along x direction.

  ${\epsilon }_{xx}=\frac{1}{dt}\left[\frac{n-dx}{dx}\right]$   ...... (VII)

Write the expression for the strain rate along y direction.

  ${\epsilon }_{yy}=\frac{1}{dt}\left[\frac{m-dy}{dy}\right]$  ...... (VIII)

Write the expression for the length of the lower edge at time $t+dt$.

  $n=\left[x+udt+dx\right]-\left[x+udt\right]$   ...... (IX)

Write the expression for the length of the lower edge at time $t+dt$.

  $\begin{array}{l}m=\left[y+dy\right]-y\\ m=dy\end{array}$   ...... (X)

Calculation:

Substitute $a+by$ for $u$ in Equation (IX).

  $\begin{array}{c}n=\left[x+\left(a+by\right)dt+dx\right]-\left[x+\left(a+by\right)dt\right]\\ =\left[x+\left(a+by\right)dt+dx\right]-x-\left(a+by\right)dt\\ =dx\end{array}$

Substitute $dx$ for $n$ in Equation (VII).

  $\begin{array}{c}{\epsilon }_{xx}=\frac{1}{dt}\left[\frac{dx-dx}{dx}\right]\\ =\frac{1}{dt}\left[\frac{0}{dx}\right]\\ =0\end{array}$

Substitute $dy$ for $m$ in Equation (VIII).

  $\begin{array}{c}{\epsilon }_{yy}=\frac{1}{dt}\left[\frac{dy-dy}{dy}\right]\\ =\frac{1}{dt}\left[\frac{0}{dy}\right]\\ =0\end{array}$

Conclusion:

The linear strain rate along x axis is $0$.

The linear strain rate along y axis is $0$.

To determine

(c)

The linear strain rates in Cartesian coordinates.

Comparison of the linear strain rate by fundamental principal to the linear strain rates in Cartesian coordinates.

Answer to Problem 66P

The linear strain rate in Cartesian coordinates along x axis is $0$.

The linear strain rate in Cartesian coordinates along y axis is $0$.

The linear strain rate by fundamental principal and the linear strain rates in Cartesian coordinates are same

Explanation of Solution

Given information:

Linear strain along x axis is $\frac{\partial u}{\partial x}$ and linear strain along y axis is $\frac{\partial v}{\partial y}$.

Write the expression for the velocity along y direction.

  $v=0$   ...... (XI)

Write the expression for the linear strain rate along x direction in Cartesian coordination.

  ${\epsilon }_{xx}=\frac{\partial u}{\partial x}$   ...... (XII)

Write the expression for the linear strain rate along y direction in Cartesian coordination.

  ${\epsilon }_{yy}=\frac{\partial v}{\partial y}$   ...... (XIII)

Calculation:

Substitute $a+by$ for $u$ in Equation (XII).

  $\begin{array}{c}{\epsilon }_{xx}=\frac{\partial }{\partial x}\left(a+by\right)\\ =0\end{array}$

Substitute $0$ for $v$ in Equation (XIII).

  $\begin{array}{c}{\epsilon }_{yy}=\frac{\partial }{\partial y}\left(0\right)\\ =0\end{array}$

Conclusion:

The linear strain rate in Cartesian coordinates along x axis is $0$.

The linear strain rate in Cartesian coordinates along y axis is $0$.

The linear strain rate by fundamental principal and the linear strain rates in Cartesian coordinates are the same.