Chapter 4, Problem 53P

To determine

The expression for the linear strain rate in the x direction.

Answer to Problem 53P

The expression for the linear strain rate in x direction is ${\epsilon }_{xx}=b$.

Explanation of Solution

Given information:

The fluid particle is located on the centre line.

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

  $\overrightarrow{V}=\left(U_0+bx\right)\overrightarrow{i}-by\overrightarrow{j}$   ...... (I)

Here, the horizontal speed is $U_0$, the constant is $b,$ and the distance in $x$ direction is $x$ and the distance in $y$ direction is $y$.

Write the expression for the velocity component along x direction.

  $u=U_0+bx$  ...... (II)

Here, the variable is $x$ in x direction.

Write the expression for the velocity component along x direction.

  $v=-by$   ...... (III)

Here, the variable is $y$ in y direction.

Write the expression for the velocity in x direction in differential form.

  $\frac{dx}{dt}=u$   ...... (IV)

Write the expression for the initial length.

  $\zeta =x_B-x_A$  ...... (V)

Here, the initial location of A is $x_A$ and the initial location of B is $x_B$.

Write the expression for the final length.

  $\zeta +\Delta \zeta =x_{B^{\prime }}-x_{A^{\prime }}$   ...... (VI)

Here, the final location of A is $x_{A^{\prime }}$, the final location of B is $x_{B^{\prime }}$.the change in length is $\Delta \zeta$.

Write the expression for the change in lengths.

  $\Delta \zeta =\left(\zeta +\Delta \zeta \right)-\zeta$   ...... (VII)

Write the expression for the strain rate in x direction.

  ${\epsilon }_{xx}=\frac{d}{dt}\left(\frac{\Delta \zeta }{\zeta }\right)$   ...... (VIII)

Write the expression for $e^{bt}$ when $t$ tends to zero.

  $e^{bt}=1+bt$  ...... (IX)

Calculation:

Substitute $U_0+bx$ for $u$ in Equation (IV).

  $\begin{array}{l}\frac{dx}{dt}=U_0+bx\\ \frac{dx}{U_0+bx}=dt\end{array}$   ...... (X)

Integrate the Equation (X).

  $\begin{array}{l}{\displaystyle \int \frac{dx}{U_0+bx}}={\displaystyle \int dt}\\ \frac{1}{b}\ln \left(U_0+bx\right)=t-\ln C_1\\ \ln {\left(U_0+bx\right)}^{\frac{1}{b}}+\ln C_1=t\\ t=C_1\ln {\left(U_0+bx\right)}^{\frac{1}{b}}\end{array}$  ...... (XI)

Substitute $x_A$ for $x$ and $0$ for $t$ in Equation (XI).

  $\begin{array}{l}0=C_1\ln {\left(U_0+b{x}_A\right)}^{\frac{1}{b}}\\ \ln \left(1\right)=\ln \left[C_1{\left(U_0+b{x}_A\right)}^{\frac{1}{b}}\right]\\ C_1{\left(U_0+b{x}_A\right)}^{\frac{1}{b}}=1\\ C_1=\frac{1}{{\left(U_0+b{x}_A\right)}^{\frac{1}{b}}}\end{array}$

Substitute $\frac{1}{{\left(U_0+b{x}_A\right)}^{\frac{1}{b}}}$ for $C_1$ in Equation (XI).

  $\begin{array}{l}t=\ln \left(\frac{1}{{\left(U_0+b{x}_A\right)}^{\frac{1}{b}}}{\left(U_0+bx\right)}^{\frac{1}{b}}\right)\\ t=\frac{1}{b}\ln \left(\frac{U_0+bx}{U_0+b{x}_A}\right)\\ bt=\ln \left(\frac{U_0+bx}{U_0+b{x}_A}\right)\\ e^{bt}=\left(\frac{U_0+bx}{U_0+b{x}_A}\right)\end{array}$

  $\begin{array}{l}{e}^{bt}\left(U_0+b{x}_A\right)=\left(U_0+bx\right)\\ x=\frac{1}{b}\left[\left(U_0+b{x}_A\right)e^{bt}-U_0\right]\\ x_{A^{\prime }}=\frac{1}{b}\left[\left(U_0+b{x}_A\right)e^{bt}-U_0\right]\\ x_{B^{\prime }}=\frac{1}{b}\left[\left(U_0+b{x}_B\right)e^{bt}-U_0\right]\end{array}$

Substitute $x_{B^{\prime }}-x_{A^{\prime }}$ for $\zeta +\Delta \zeta$ and $x_B-x_A$ for $\zeta$ in Equation (VII).

  $\Delta \zeta =\left(x_{B^{\prime }}-x_{A^{\prime }}\right)-\left(x_B-x_A\right)$   ...... (XII)

Substitute $\frac{1}{b}\left[\left(U_0+b{x}_B\right)e^{bt}-U_0\right]$ for $x_{B^{\prime }}$ and $\frac{1}{b}\left[\left(U_0+b{x}_A\right)e^{bt}-U_0\right]$ for $x_{A^{\prime }}$ in Equation (XII).

  $\begin{array}{c}\Delta \zeta =\left(\frac{1}{b}\left[\left(U_0+b{x}_B\right)e^{bt}-U_0\right]-\frac{1}{b}\left[\left(U_0+b{x}_A\right)e^{bt}-U_0\right]\right)-\left(x_B-x_A\right)\\ =\left(\frac{1}{b}\left[\left(U_0+b{x}_B\right)e^{bt}-U_0-\left(U_0+b{x}_A\right)e^{bt}+U_0\right]\right)-\left(x_B-x_A\right)\\ =\left(x_B-x_A\right)e^{bt}-\left(x_B-x_A\right)\\ =\left(x_B-x_A\right)\left(e^{bt}-1\right)\end{array}$

Substitute $\left(x_B-x_A\right)\left(e^{bt}-1\right)$ for $\Delta \zeta$ and $x_B-x_A$ for $\zeta$ in Equation (VIII).

  $\begin{array}{c}{\epsilon }_{xx}=\frac{d}{dt}\left(\frac{\left(x_B-x_A\right)\left(e^{bt}-1\right)}{\left(x_B-x_A\right)}\right)\\ {\epsilon }_{xx}=\frac{d}{dt}\left(e^{bt}-1\right)\end{array}$   ...... (XIII)

Substitute $1+bt$ for $e^{bt}$ in Equation (XIII).

  $\begin{array}{c}{\epsilon }_{xx}=\frac{d}{dt}\left(1+bt-1\right)\\ =\frac{d}{dt}\left(bt\right)\\ =b\end{array}$

Conclusion:

The expression for strain rate in x direction is ${\epsilon }_{xx}=b$.