Chapter 4, Problem 58P

Using the results of Prob. 4—57 and the fundamental definition of linear strain rate (the rate of increase in length per unit length), develop an expression for the linear strain rate in the y-direction $\left({\epsilon }_{yy}\right)$ of fluid particles moving down the channel. Compare your result to the general expression for ${\epsilon }_{yy}$ in terms of the velocity field. ${\epsilon }_{yy}=\partial v/\partial y$. (Hint: Take the limit as time $t\to 0$ . You may need to apply a truncated series expansion for $e^{-bt}$ .)

To determine

The expression for the linear strain rate in y direction.

Answer to Problem 58P

The expression for the linear strain rate in y direction is ${\epsilon }_{yy}=-b$.

Explanation of Solution

Given information:

The general strain in y direction is $\frac{\partial v}{\partial y}$.

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 y direction in differential form.

  $\frac{dy}{dt}=v$   ...... (IV)

Write the expression for the initial length.

  $\eta =y_B-y_A$  ...... (V)

Here, the initial location of A is $y_A$ and the initial location of B is $y_B$.

Write the expression for the final length.

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

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

Write the expression for the change in lengths.

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

Write the expression for the linear strain rate in y direction.

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

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

  $e^{-bt}=1-bt$   ...... (IX)

Calculation:

Substitute $-by$ for $v$ in Equation (IV).

  $\begin{array}{l}\frac{dy}{dt}=-by\\ \frac{dy}{-by}=dt\end{array}$   ...... (X)

Integrate the Equation (X).

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

Substitute $y_A$ for $y$ and $0$ for $t$ in Equation (XI).

  $\begin{array}{l}0=\ln \left(\frac{C_1}{{\left(y_A\right)}^{\frac{1}{b}}}\right)\\ \ln \left(1\right)=\ln \left(\frac{C_1}{{\left(y_A\right)}^{\frac{1}{b}}}\right)\\ \frac{C_1}{{\left(y_A\right)}^{\frac{1}{b}}}=1\\ C_1={\left(y_A\right)}^{\frac{1}{b}}\end{array}$

Substitute ${\left(y_A\right)}^{\frac{1}{b}}$ for $C_1$ in Equation (XI).

  $\begin{array}{l}t=\ln \left(\frac{{\left(y_A\right)}^{\frac{1}{b}}}{y^{\frac{1}{b}}}\right)\\ t=\frac{1}{b}\ln \left(\frac{y_A}{y}\right)\\ bt=\ln \left(\frac{y_A}{y}\right)\\ e^{bt}=\frac{y_A}{y}\end{array}$

  $\begin{array}{c}y=\frac{y_A}{e^{bt}}\\ y=y_A{e}^{-bt}\end{array}$

Substitute $y_B$ for $y$ and $0$ for $t$ in Equation (XI).

  $\begin{array}{l}0=\ln \left(\frac{C_1}{{\left(y_B\right)}^{\frac{1}{b}}}\right)\\ \ln \left(1\right)=\ln \left(\frac{C_1}{{\left(y_B\right)}^{\frac{1}{b}}}\right)\\ \frac{C_1}{{\left(y_B\right)}^{\frac{1}{b}}}=1\\ C_1={\left(y_B\right)}^{\frac{1}{b}}\end{array}$

Substitute ${\left(y_B\right)}^{\frac{1}{b}}$ for $C_1$ in Equation (XI).

  $\begin{array}{l}t=\ln \left(\frac{{\left(y_B\right)}^{\frac{1}{b}}}{y^{\frac{1}{b}}}\right)\\ t=\frac{1}{b}\ln \left(\frac{y_B}{y}\right)\\ bt=\ln \left(\frac{y_B}{y}\right)\\ e^{bt}=\frac{y_B}{y}\end{array}$

  $\begin{array}{c}y=\frac{y_B}{e^{bt}}\\ y=y_B{e}^{-bt}\end{array}$

Substitute $y_A{e}^{-bt}$ for $y_{A^{\prime }}$ and $y_B{e}^{-bt}$ for $y_{B^{\prime }}$ in Equation (VI).

  $\eta +\Delta \eta =\left(y_B{e}^{-bt}\right)-\left(y_A{e}^{-bt}\right)$

Substitute $y_{B^{\prime }}-y_{A^{\prime }}$ for $\eta +\Delta \eta$ and $y_B-y_A$ for $\eta$ in Equation (VII).

  $\Delta \eta =\left(y_{B^{\prime }}-y_{A^{\prime }}\right)-\left(y_B-y_A\right)$   ...... (XII)

Substitute $\left(y_B{e}^{-bt}\right)-\left(y_A{e}^{-bt}\right)$ for $\left(y_{B^{\prime }}-y_{A^{\prime }}\right)$ in Equation (XII).

  $\begin{array}{c}\Delta \eta =\left(y_B{e}^{-bt}-y_A{e}^{-bt}\right)-\left(y_B-y_A\right)\\ =\left(y_B-y_A\right)e^{-bt}-\left(y_B-y_A\right)\\ =\left(y_B-y_A\right)\left(e^{-bt}-1\right)\end{array}$

Substitute $\left(y_B-y_A\right)\left(e^{-bt}-1\right)$ for $\Delta \eta$ and $\left(y_B-y_A\right)$ for $\eta$ in Equation (VIII).

  $\begin{array}{l}{\epsilon }_{yy}=\frac{d}{dt}\left(\frac{\left(y_B-y_A\right)\left(e^{-bt}-1\right)}{\left(y_B-y_A\right)}\right)\\ {\epsilon }_{yy}=\frac{d}{dt}\left(e^{-bt}-1\right)\end{array}$   ...... (XIII)

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

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

Conclusion:

The expression for the linear strain rate in y direction is ${\epsilon }_{yy}=-b$.