Chapter 4, Problem 52P

To determine

The expression for the change in length of the line segment.

Answer to Problem 52P

The expression for the change in length of the line segment is $\left(x_B-x_A\right)\left(e^{bt}-1\right)$.

Explanation of Solution

Given information:

The flow is symmetric about x axis.

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)

Calculation:

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

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

Integrate the Equation (VIII).

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

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

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

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

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

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

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

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

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 (X).

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

Conclusion:

The expression for the change in length of the line segment is $\left(x_B-x_A\right)\left(e^{bt}-1\right)$.