Chapter 3, Problem 3.23P

Prove that the flow field specified in Example 2.1 is not incompressible; i.e., it is a compressible flow as stated without proof in Example 2.1.

To determine

To prove:

The flow in example 2.1 is compressible.

Explanation of Solution

Calculation:

Velocity field for the subsonic flow is expressed as follows:

$\begin{array}{l}u=V_{\infty }\left(1+\frac{h}{\beta }\frac{2\pi }{l}\cos \left(\frac{2\pi x}{l}\right)e^{-\frac{2\pi \beta y}{l}}\right)\mathrm{......}\left(1\right)\\ v=-V_{\infty }h\frac{2\pi }{l}\sin \left(\frac{2\pi x}{l}\right)e^{-\frac{2\pi \beta y}{l}}\mathrm{......}\left(2\right)\end{array}$

Differentiate the equation (1) with respect to x as follows:

$\frac{\partial u}{\partial x}=-V_{\infty }\frac{h}{\beta }{\left(\frac{2\pi }{l}\right)}^2\sin \left(\frac{2\pi x}{l}\right)e^{-\frac{2\pi \beta y}{l}}$

Differentiate the equation (2) with respect toy as follows:

$\frac{\partial v}{\partial x}=V_{\infty }\beta h{\left(\frac{2\pi }{l}\right)}^2\sin \left(\frac{2\pi x}{l}\right)e^{-\frac{2\pi \beta y}{l}}$

The divergence of a vector field is calculated as follows:

$\begin{array}{l}\nabla \cdot V=-V_{\infty }\frac{h}{\beta }{\left(\frac{2\pi }{l}\right)}^2\sin \left(\frac{2\pi x}{l}\right)e^{-\frac{2\pi \beta y}{l}}+V_{\infty }\beta h{\left(\frac{2\pi }{l}\right)}^2\sin \left(\frac{2\pi x}{l}\right)e^{-\frac{2\pi \beta y}{l}}\\ \nabla \cdot V=V_{\infty }h\left(\beta -\frac{1}{\beta }\right){\left(\frac{2\pi }{l}\right)}^2\sin \left(\frac{2\pi x}{l}\right)e^{-\frac{2\pi \beta y}{l}}\end{array}$

Rate of change of volume is calculated as follows:

$\begin{array}{l}\nabla \cdot V=V_{\infty }h\left(\beta -\frac{1}{\beta }\right){\left(\frac{2\pi }{l}\right)}^2\sin \left(\frac{2\pi }{4}\right)e^{-\frac{2\pi \beta y}{l}}\\ \nabla \cdot V=V_{\infty }h\left(\beta -\frac{1}{\beta }\right){\left(\frac{2\pi }{l}\right)}^2{e}^{-\frac{2\pi \beta y}{l}}\end{array}$

Substitute the values in the above equation as follows:

$\begin{array}{l}\nabla \cdot V=V_{\infty }h\left(\beta -\frac{1}{\beta }\right){\left(\frac{2\pi }{l}\right)}^2\sin \left(\frac{2\pi }{4}\right)e^{-\frac{2\pi \beta y}{l}}\\ \nabla \cdot V=240\times 0.01\left(0.714-\frac{1}{0.714}\right){\left(\frac{2\pi }{1}\right)}^2{e}^{-\frac{2\pi \times 0.714}{1}}\\ \nabla \cdot V=-0.686\times 2.4\times 397.47\times 0.01\\ \nabla \cdot V=-0.7327{\text{s}}^{\text{-1}}\end{array}$

Apply the control volume equation as follows:

$\begin{array}{l}\frac{1}{\rho }\left(\frac{\partial \rho }{\partial t}\right)=-\nabla \cdot V\\ \frac{1}{\rho }\left(\frac{\partial \rho }{\partial t}\right)=-\left(-0.7327\right)\\ \frac{1}{\rho }\left(\frac{\partial \rho }{\partial t}\right)=0.7327\\ \text{or,}\\ \frac{1}{\rho }\left(\frac{\partial \rho }{\partial t}\right)=73.27\%\end{array}$

This shows that the given flow is compressible by 73.37%.