Chapter 4, Problem 23P

For the velocity field of Prob. 4−22, calculate the fluid acceleration along the diffuser centerline as a function of x and the given parameters. For $L=1.56\text{m}$ , $u_{\text{entrance}}=24.3\text{m/s}$ , and $u_{\text{exit}}=16.8\text{m/s}$ , calculate the acceleration at $x=0$ and $x=1.0$ m.

To determine

The acceleration at $x=0\text{m}$.

The acceleration at $x=1\text{m}$.

Answer to Problem 23P

The acceleration at $x=0\text{m}$ is $0\text{m}/{\text{s}}^{\text{2}}$.

The acceleration at $x=1\text{m}$ is $-130.773\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given information:

The length of the diffuser is $1.56\text{m}$, the velocity at entrance is $24.3\text{m}/\text{s},$ and the velocity at exit is $16.8\text{m}/\text{s}$.

Write the expression for general parabolic equation in $\text{x}$ direction.

   $u=a{\left(\text{x}-h\right)}^2+k$ ...... (I)

Here, acceleration of fluid from diffuser is $a$, the length in $\text{x}$ direction is $\text{x,}$ and the total height is $h$ and the constant is $k$.

Write the expression for acceleration of the fluid along $\text{x}$ direction.

   $a_{\text{x}}=u\frac{\partial u}{\partial \text{x}}$ ...... (II)

Here, the partial derivative of velocity with respect to $\text{x}$ is $\partial u/\partial \text{x}$.

Calculation:

Substitute initial boundary condition $0$ for $\text{x}$, $0$ for $h$ and $u_{entrance}$ for $u$ in Equation (I).

   $\begin{array}{l}{u}_{entrance}=a{\left(0-0\right)}^2+k\\ k=u_{entrance}\end{array}$

Substitute final boundary condition $L$ for $\text{x}$, $0$ for $h$, $u_{entrance}$ for $k$ and $u_{exit}$ for $u$ in Equation (I).

   $\begin{array}{l}{u}_{exit}=a{\left(L-0\right)}^2+u_{entrance}\\ a=\frac{u_{exit}-u_{entrance}}{L^2}\end{array}$

Substitute $\frac{u_{exit}-u_{entrance}}{L^2}$ for $a$, $0$ for $h$ and $u_{entrance}$ for $k$ in Equation (I).

   $u=\frac{u_{exit}-u_{entrance}}{L^2}{\text{x}}^2+u_{entrance}$ ...... (III)

Substitute $\frac{u_{exit}-u_{entrance}}{L^2}{\text{x}}^2+u_{entrance}$ for $u$ in Equation (II).

   $\begin{array}{c}{a}_{\text{x}}=\left(\frac{u_{exit}-u_{entrance}}{L^2}{\text{x}}^2+u_{entrance}\right)\frac{\partial \left(\frac{u_{exit}-u_{entrance}}{L^2}{\text{x}}^2+u_{entrance}\right)}{\partial \text{x}}\\ =2u_{entrance}\frac{u_{exit}-u_{entrance}}{L^2}\text{x}+2\frac{{\left(u_{exit}-u_{entrance}\right)}^2}{L^4}{\text{x}}^3\end{array}$ ...... (IV)

Substitute $0$ for $\text{x}$ in Equation (IV).

   $\begin{array}{c}{a}_x=2u_{entrance}\frac{u_{exit}-u_{entrance}}{L^2}\left(0\right)+2\frac{{\left(u_{exit}-u_{entrance}\right)}^2}{L^4}\left(0\right)\\ =0+0\\ =0\end{array}$

Substitute $1\text{m}$

   $\text{x}$, $16.8\text{m}/\text{s}$ for $u_{exit}$, $1.56\text{m}$ for $L$ and $24.3\text{m}/\text{s}$ for $u_{entrance}$ in Equation (IV).

   $\begin{array}{c}{a}_x=\left[\begin{array}{l}2\left(24.3\text{m}/\text{s}\right)\frac{\left(16.8\text{m}/\text{s}\right)-\left(24.3\text{m}/\text{s}\right)}{{\left(1.56\text{m}\right)}^2}\left(1\right)\\ +2\frac{{\left(16.8\text{m}/\text{s}-24.3\text{m}/\text{s}\right)}^2}{{\left(1.56\text{m}\right)}^4}\left(1\right)\end{array}\right]\\ =-149.77\text{m}/{\text{s}}^{\text{2}}+18.99\text{m}/{\text{s}}^{\text{2}}\\ =-130.773\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

The acceleration at $x=0\text{m}$ is $0\text{m}/{\text{s}}^{\text{2}}$.

The acceleration at $x=1\text{m}$ is $-130.773\text{m}/{\text{s}}^{\text{2}}$.