Chapter 4, Problem 55P

Converging duct flow is modeled by the steady, two-dimensional velocity field of Prob. 4-16. As vertical line segment AB moves downstream it shrinks from length $\eta$ to length $\eta -\Delta \eta$ as sketched in Fig. P4-57. Generate an analytical expression for the change in length of the line segment. $\Delta \eta$ Note that the change in length. $\Delta \eta$ , is negative. (Hint: Use the result of Prob. 4-54.)

To determine

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

Answer to Problem 55P

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

Explanation of Solution

Given information:

Initially fluid particle is located at $x_A$ along x axis, and at $y_A$ along y direction.

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)

Calculation:

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

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

Integrate the Equation (VIII).

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

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

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

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

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

  $\begin{array}{c}y=\frac{y_A}{e^{bt}}\\ y=y_A{e}^{-bt}\\ y_{A^{\prime }}=y_A{e}^{-bt}\\ y_{B^{\prime }}=y_B{e}^{-bt}\end{array}$

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

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

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

  $\begin{array}{c}\Delta \eta =\left(\frac{1}{b}\left[\left(U_0+b{y}_B\right)e^{bt}-U_0\right]-\frac{1}{b}\left[\left(U_0+b{y}_A\right)e^{bt}-U_0\right]\right)-\left(y_B-y_A\right)\\ =\left(\frac{1}{b}\left[\left(U_0+b{y}_B\right)e^{bt}-U_0-\left(U_0+b{y}_A\right)e^{bt}+U_0\right]\right)-\left(y_B-y_A\right)\\ =\left(y_B-y_A\right)e^{bt}-\left(y_B-y_A\right)\\ =\left(y_B-y_A\right)\left(e^{bt}-1\right)\end{array}$

Conclusion:

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