pressure coefficient is given as  $${C_p} = 1 – {\left( {\frac{v}{{{v_\infty }}}} \right)^2}$$ $$= 1 – {\left( {\frac{{220}}{{100}}} \right)^2}$$ $$= 1 – 4.84$$ $$=  – 3.84$$

(a) For inviscid and incompressible flow coefficient of pressure is given as  $${C_p} = 1 – {\left( {\frac{v}{{{v_\infty }}}} \right)^2}$$ $$v = \sqrt {v_\infty ^2\left( {1 – {C_p}} \right)}$$ $$= \sqrt {{{\left( {50} \right)}^2}\left( {1 – \left( { – 3.84} \right)} \right)}$$ $$= \sqrt {2500\left( {1 + 3.84} \right)}$$ $$= 110\,{\rm{ft/sec}}$$ also (b) for ${{v_\infty }}$ = 200 ft/sec  $${C_p} = 1 – {\left( {\frac{v}{{{v_\infty }}}} \right)^2}$$ $$v = \sqrt {v_\infty ^2\left( {1 – {C_p}} \right)}$$ $$= \sqrt {{{\left( {200} \right)}^2}\left( {1 – \left( { – 3.84} \right)} \right)}$$ $$= \sqrt {40000\left( {1 + 3.84} \right)}$$ $$= 440\,{\rm{ft/sec}}$$

Considering the flow to be inviscid and incompressible, Pressure coefficient ${{\rm{C}}_{\rm{p}}} = 1 – {\left( {\frac{v}{{{v_\infty }}}} \right)^2}$ $$= 1 – {\left( {\frac{{150}}{{100}}} \right)^2}$$ $$=  – 1.25$$

For a uniform flow velocity potential $\phi  = {V_\infty }x$ $$\frac{{\partial \phi }}{{\partial x}} = {V_\infty },\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} = 0$$ $$\frac{{\partial \phi }}{{\partial y}} = 0\,;\,\frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0$$ Laplace’s equation $$\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0 + 0 = 0$$ is satisfied. For a uniform flow stream function $\psi  = Vy\,;$ $$\frac{{\partial \psi }}{{\partial x}} = 0\,,\,\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} = 0$$ $$\frac{{\partial \psi }}{{\partial y}} = V,\frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0$$ Therefore, Laplace’s equation = $$\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0 + 0 = 0$$ is satisfied.

For a source flow , velocity potential $\phi  = \frac{\Lambda }{{2\pi }}\ln r$ $$\frac{{\partial \phi }}{{\partial r}} = \frac{\Lambda }{{2\pi }}\frac{1}{r}\,,\,\frac{{{\partial ^2}\phi }}{{\partial {r^2}}} =  – \frac{\Lambda }{{2\pi }}\frac{1}{{{r^2}}}$$ $$\frac{{\partial \phi }}{{\partial \theta }} = 0\,,\,\frac{{{\partial ^2}\phi }}{{\partial {\theta ^2}}} = 0$$ Laplace’s equation =  $$= \frac{1}{r} \cdot \frac{\partial }{{\partial r}}\left( {r\frac{{\partial \phi }}{{\partial r}}} \right) + \frac{1}{{{r^2}}} + \frac{1}{{{r^2}}}\frac{{{\partial ^2}\phi }}{{\partial {\theta ^2}}}$$ $$= \frac{1}{r}\frac{\partial }{{\partial r}}\left[ {\frac{\Lambda }{{\partial \pi }}} \right] + 0 = 0$$ is satisfied.For a source flow stream function  $$\psi  = \frac{\Lambda }{2} = \theta \,;\,\frac{{\partial \psi }}{{\partial r}} = 0\,,\,\frac{{{\partial ^2}\psi }}{{\partial {r^2}}} = 0$$ $$\frac{{\partial \psi }}{{\partial \theta }} = \frac{\Lambda }{{2\pi }}\,,\,\frac{{{\partial ^2}\psi }}{{\partial {\phi ^2}}}\, = \,0$$ Therefore , Laplace’s equation  $$\frac{1}{r}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial \psi }}{{\partial r}}} \right) + \frac{1}{{{r^2}}}\frac{{{\partial ^2}\psi }}{{\partial {\theta ^2}}}$$ $$= \frac{1}{r}\frac{\partial }{{\partial r}}\left( 0 \right) + \frac{1}{{{r^2}}}\left( 0 \right) = 0$$ is satisfied.

For a uniform flow $${V_\infty } = u = {\rm{constant}}\,\,{\rm{;}}\,\,\mathop V\limits^ \to   = {V_\infty }\mathop i\limits^ \to$$ Divergence of velocity field $$\nabla  \cdot \mathop V\limits^ \to   = \frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}}$$ $$= 0 + 0 + 0$$ $$= 0$$ Therefore, it shows that it is a physically possible incompressible flow.

Also, curl of velocity field  $$\nabla  \times \mathop V\limits^ \to   = \left| {\begin{array}{*{20}{c}}{\mathop i\limits^ \to  }&{\mathop j\limits^ \to  }&{\mathop k\limits^ \to  }\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\u&v&w\end{array}} \right| = \mathop i\limits^ \to  \left( {0 – 0} \right) – \mathop j\limits^ \to  \left( {0 – \frac{{\partial u}}{{\partial x}}} \right) + \mathop k\limits^ \to  \left( {0 – \frac{{\partial u}}{{\partial y}}} \right)$$ $$\nabla  \times \mathop V\limits^ \to   = 0$$ This shows that the flow is irrotational.

For a source flow, velocity $\mathop V\limits^ \to   = {V_r}\mathop {{e_r}}\limits^ \to   = \frac{\Lambda }{{2\pi r}}\mathop {{e_r}}\limits^ \to$In polar co-ordinates :, the curl of velocity field will be  $$\nabla  \cdot \mathop V\limits^ \to   = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {r{V_r}} \right) + \frac{1}{r}\frac{{\partial {V_\theta }}}{{\partial \theta }}$$ $$\nabla  \cdot \mathop V\limits^ \to   = \frac{1}{r}\frac{\partial }{{\partial r}}\left[ {r\frac{\Lambda }{{2\pi r}}} \right] + \frac{1}{r}\frac{{\partial \left( 0 \right)}}{{\partial \theta }}$$ $$\nabla  \cdot \mathop V\limits^ \to   = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {\frac{\Lambda }{{2\pi }}} \right) + 0 = 0$$ Therefore, the flow is a physically possible incompressible flow, except at origin where r =0.

Also, $$\nabla  \times \mathop V\limits^ \to   = \frac{1}{r}\left| {\begin{array}{*{20}{c}}{\mathop {{e_r}}\limits^ \to  }&{r\mathop {{e_\theta }}\limits^ \to  }&{\mathop {{e_z}}\limits^ \to  }\\{\frac{\partial }{{\partial r}}}&{\frac{\partial }{{\partial \theta }}}&{\frac{\partial }{{\partial z}}}\\{{V_r}}&{r{V_\theta }}&{{V_z}}\end{array}} \right| = \frac{1}{r}\left| {\begin{array}{*{20}{c}}{\mathop {{e_r}}\limits^ \to  }&{r\mathop {{e_\theta }}\limits^ \to  }&{\mathop {{e_z}}\limits^ \to  }\\{\frac{\partial }{{\partial r}}}&{\frac{\partial }{{\partial \theta }}}&{\frac{\partial }{{\partial z}}}\\{\frac{\Lambda }{{2\pi r}}}&0&0\end{array}} \right|$$ $$\nabla  \times \mathop V\limits^ \to   =  – r\mathop {{e_\theta }}\limits^ \to  \left( {\frac{{\partial 0}}{{\partial r}} – \frac{{\partial \Lambda /2\pi r}}{{\partial z}}} \right) + \mathop {{e_z}}\limits^ \to  \left( {\frac{{\partial 0}}{{\partial r}} – \frac{{\partial \Lambda /2\pi r}}{{\partial \theta }}} \right) = 0$$ Therefore, $\nabla  \times \mathop V\limits^ \to   = 0$ (everywhere in the flow).