Velocity potential and stream function for a source flow is given as \(\phi=\frac{\Lambda }{2\pi}ln\;r\) and \(\psi=\frac{\Lambda }{2\pi}\;\theta\).

 \[\phi=\frac{\Lambda }{2\pi}ln\;r
\\\frac{\partial \phi}{\partial r}=\frac{\Lambda }{2\pi}\frac{1}{r}\;,\;\frac{\partial^2 \phi}{\partial r^2}=-\frac{\Lambda }{2\pi}\frac{1}{r^{2}}
\\\frac{\partial \phi}{\partial \theta}=0\;;\;\frac{\partial^2 \phi}{\partial \theta^2}=0\]

Therefore Laplace equation \[\frac{1}{r}\frac{\partial }{\partial r}\left ( r\; \frac{\partial \phi}{\partial r} \right )+\frac{1}{r^{2}}\frac{\partial^2 \phi}{\partial \theta^2}=0+0=0\]

is satisfied.

\[\psi=\frac{\Lambda }{2\pi}\theta\]\[\frac{\partial \psi}{\partial r}=0\;;\;\frac{\partial^2 \psi}{\partial r^2}=0
\\\frac{\partial \psi}{\partial \theta}=\frac{\Lambda }{2\pi}\;;\;\frac{\partial^2 \psi}{\partial \theta^2}=0\]

Therefore Laplace equation \[\frac{1}{r}\frac{\partial }{\partial r}\left ( r\;\frac{\partial \psi}{\partial r} \right )+\frac{1}{r^{2}}\frac{\partial^2 \psi}{\partial \theta^2}=\frac{1}{r}\frac{\partial (0)}{\partial r}+\frac{1}{r^{2}}\left ( 0 \right )=0+0=0\]

is satisfied.

We  need to find here the velocity of the airplane.\[P_{0}=P_{\infty}+\frac{1}{2}\rho V_{\infty}^{2}
\\\Rightarrow V_{\infty}=\sqrt{\frac{2\left ( P_{0}-P_{\infty} \right )}{\rho }}
\\=\sqrt{\frac{2\left ( 1.07-1.01 \right )\times 10^{5}}{1.23}}
\\=98.8\;m/s\]

Venturi has a throat area ratio of 0.85.We need to calculate here the maximum vacuum obtainable in the reservoir when the venturi is placed in an airstream of \(90\;m/s\).

\[P_{1}-P_{2}=\frac{1}{2}\rho V_{1}^{2}\left [ \left ( \frac{A_{1}}{A_{2}} \right )^{2}-1 \right ]
\\=\frac{1}{2}\left ( 1.23 \right )\left ( 90 \right )^{2}\left [ \left ( \frac{1}{0.85} \right )^{2}-1 \right ]
\\=1913\;N/m^{2}\]

We denote the pressure distributions over the upper and lower walls by \(p_{u}(x)\) and \(p_{l}(x)\) respectively.

Figure : Wind Tunnel

\(p=p_{\infty}\) and \(v=0\) (Assuming faces \(ai\) and \(bh\) are far enough upstream and downstream of the model such that \(p=p_{\infty}\) and \(v=0\))

\[L = – \mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}\limits_s
{\left( {\rho \overrightarrow {v} \cdot \overrightarrow {ds} } \right)} v \; – \iint\limits_{abhi} {\left( {p\overrightarrow {ds} } \right)}y\]

Here first integral = 0 , because \(\vec{v}\cdot \vec{ds}=0  \) or because \(v=0\).

Therefore \[{L^\prime } = – \mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}\limits_{abhi}
{\left( {p\overrightarrow {ds} } \right)} y = – \left[ {\int_a^b {{p_u}dx \;- \int\limits_i^h {{p_l}dx} } } \right]\]

\[\Rightarrow{L^\prime } = \left[ {\int_i^h {{p_l}dx \;- \int\limits_a^b {{p_u}dx} } } \right] = 0\]

\(x\) and \(y\) components of velocity are given by \(u=cx\) and \(v=-cy\).We need to calculate the stream function and velocity potential.Here

\[u=cx
\\\frac{\partial \psi }{\partial y}=cx
\\\Rightarrow \partial \psi=cx\partial y
\\\Rightarrow \psi=cxy+f(x)
\\v=-cy=-\frac{\partial \psi}{\partial x}
\\\Rightarrow \psi=cxy+f(y)\]

Here f(x) and f(y) are constants:

\[\psi=cxy+const.
\\u=cx=\frac{d\phi}{dx}
\\\Rightarrow d\phi=cxdx
\\\Rightarrow \phi=cx^{2}+f(y)
\\v=-cy=\frac{\partial \phi}{\partial y}:\phi=-cy^{2}+f(x)\]

Therefore \[f(y)=-cy^{2}\;and \;f(x)=cx^{2}\]

and \[\phi=c\left ( x^{2}-y^{2} \right )
\\\psi=cxy+const.\]

Differentiating the above equation of \(\psi\) with respect to ‘\( x\)’, holding \(\psi=constant\)

\[\frac{\partial \psi}{\partial x}=cx\frac{dy}{dx}+cy\]

or \[\left ( \frac{dy}{dx} \right )_{\psi=constant}=\left ( \frac{-y}{x} \right )\]

Differentiating the above equation of \(\phi\) with respect to ‘\(x\)’,holding \(\phi=constant\)

\[0=2cx-2cy\left ( \frac{dy}{dx} \right )\]

or \[\left ( \frac{dy}{dx} \right )_{\phi=constant}=\left ( \frac{x}{y} \right )\]

On comparing the above equations we get \[\left ( \frac{dy}{dx} \right )_{\psi=constant}=\frac{-1}{\left (\frac{dy}{dx} \right )_{\phi=constant}}\]

Therefore the lines of constant \(\phi\) are perpendicular to the lines of constant \(\psi\).

Length of the pipe is bent into ‘U’ shape.Here inside diameter = \(0.5 \;m\) and air density = \(1.23\;kg/m^{3}\).

Figure

‘-R’ exerted by the pipe, reverses the flow velocity.Cross sectional area =\( \frac{\pi\;d^{2}}{4}\)

\[A=\frac{\pi d^{2}}{4}
\\=\frac{\pi (0.5)^{2}}{4}
\\=0.196\;m^{2}\]

Mass flow entering the pipe is \[\dot{m}=\rho _{1}A_{1}V_{1}=1.23\times 0.196\times 100
\\=24.11\;kg/sec
\\R=-\mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}
 \left ( \rho V\cdot ds \right )V
\\V_{2}=-V_{1}
\\R=-\left [ -\dot{m} V_{1}+\dot{m}V_{2}\right ]=\dot{m}\left ( V_{1}-V_{2} \right )
\\=\dot{m}\left [ V_{1}-\left ( -V_{1} \right ) \right ]=\dot{m}2V_{1}
\\R=24.11\times 2\times 100
\\=4822\;N\]

a) Let for a vortex flow \(u=\frac{cy}{x^{2}+y^{2}}\) and \(v=\frac{-cx}{x^{2}+y^{2}}\).In polar coordinates the time rate of change of a fluid element per unit volume is given as \[\nabla\cdot \vec{V}=\frac{1}{r}\frac{\partial }{\partial r}\left ( rV_{r} \right )+\frac{1}{r}\frac{\partial V_{\theta}}{\partial \theta}\]

The velocity component  is transformed by using \(x=rcos\theta\) and \(y=rsin\theta\).

\[V_{r}=ucos\theta+vsin\theta
\\V_{\theta}=-usin\theta+vcos\theta\]

\[u=\frac{cy}{x^{2}+y^{2}}=\frac{crsin\theta}{r^{2}}=\frac{c sin\theta}{r}
\\v=\frac{-cx}{x^{2}+y^{2}}=\frac{crcos\theta}{r^{2}}=\frac{-c cos\theta}{r}
\\V_{r}=\frac{c}{r}cos\theta sin\theta-\frac{c}{r}cos\theta sin\theta=0
\\V_{\theta}=\frac{-c}{r}sin^{2}\theta-\frac{c}{r}cos^{2}\theta=\frac{-c}{r}\]

\[\nabla\cdot \vec{V}=\frac{1}{r}\frac{\partial }{\partial r}\left ( 0 \right )+\frac{1}{r}\frac{\partial }{\partial \theta}\left ( \frac{-c}{r} \right )=0+0=0\]

b) The vorticity

Vorticity is given as \[\nabla \times \vec{V}=\vec{e_{z}}\left [ \frac{\partial }{\partial r}\left ( \frac{-c}{r} \right )-\frac{c}{r^{2}}-\frac{1}{r}\frac{\partial (0)}{\partial \theta} \right ]
\\=\vec{e_{z}}\left [ \frac{c}{r^{2}}-\frac{c}{r^{2}} -0\right ]
\\=0\]

Here \(\nabla \times \vec{V}=0\) except at origin, where r=0.The flow field is singular at the origin.

Here \(u=cx\) and \(v=-cy\).Since \[vdx – udy=0\]

Therefore \[\frac{dy}{dx}=\frac{v}{u}=\frac{-y}{x}
\\\Rightarrow \frac{dy}{y}=\frac{-dx}{x}\]

On integrating \[ln(y)=xln(x)+C_{1}
\\\Rightarrow y=\frac{C_{2}}{x}\]

These streamlines are in hyperbola shape.

Figure

Here \(x\) and \(y\) components of velocities are given.We need to find equation of streamlines. \[vdx-udy=0
\\\frac{dy}{dx}=\frac{v}{u}=\frac{-cx}{x^{2}+y^{2}}\times \frac{x^{2}+y^{2}}{cy}=\frac{-x}{y}
\\\Rightarrow \frac{dy}{dx}=\frac{-x}{y}
\\\Rightarrow ydy=-xdx\]

On integrating we get \[\Rightarrow \frac{y^{2}}{2}=\frac{-x^{2}}{2}+C
\\\Rightarrow x^{2}+y^{2}=4C
\\\Rightarrow x^{2}+y^{2}=C_{1}\]

Therefore streamlines are concentric with their centres at the origin.

Here \(x\) component of velocity is \(u=\frac{cx}{x^{2}+y^{2}}\) and \(y\) component of velocity is \(v=\frac{cy}{x^{2}+y^{2}}\) \[vdx-udy=0\]
\(\Rightarrow vdx=udy
\\\Rightarrow \frac{dy}{dx}=\frac{v}{u}=\frac{\frac{cy}{x^{2}+y^{2}}}{\frac{cx}{x^{2}+y^{2}}}=\frac{y}{x}
\\\Rightarrow \frac{dy}{dx}=\frac{y}{x}=\frac{dy}{y}=\frac{dx}{x}\)

On integrating

\(ln(y)=ln(x)+C
\\\Rightarrow ln\left ( \frac{y}{x} \right )=C
\\\Rightarrow y=e^{C}x
\\\Rightarrow y=xC_{1}\)

Therefore streamlines are straight lines originating from the origin.