- Worldtech
- Earth
- Email is hidden.
503
Points
Questions
112
Answers
70
-
On changing the equation to polar co-ordinates \[\begin{array}{l}x = r\cos \theta \\y = r\sin \theta \\{V_r} = u\cos \theta + v\sin \theta \\{V_\theta } = – u\sin \theta + v\cos \theta \end{array}\]\[\begin{array}{l}u = \frac{{4y}}{{({x^2} + {y^2})}} = \frac{{4r\sin \theta }}{{{r^2}}} = \frac{{4\sin \theta }}{r}\\v = \frac{{ – 4x}}{{({x^2} + {y^2})}} = \frac{{4r\cos \theta }}{{{r^2}}} = \frac{{ – 4\cos \theta }}{r}\\{V_r} = \frac{4}{r}\cos \theta \sin \theta – \frac{4}{r}\cos \theta \sin \theta = 0\\{V_\theta } = \frac{{ – 4}}{r}{\sin ^2}\theta – \frac{4}{r}{\cos ^2}\theta = \frac{{ – 4}}{r}\end{array}\]
Time rate of change of volume of a fluid element per unit volume is given as\[\begin{array}{l}\nabla .\mathop V\limits^ \to = \frac{1}{r}\frac{{\partial r{V_r}}}{{\partial r}} + \frac{1}{r}\frac{{\partial {V_\theta }}}{{\partial \theta }}\\\nabla .\mathop V\limits^ \to = \frac{1}{r}\frac{{\partial (0)}}{{\partial r}} + \frac{1}{r}\frac{{\partial \left( { – c/r} \right)}}{{\partial \theta }} = 0 + 0 = 0\end{array}\] (b) Vorticity = \[\begin{array}{l}\nabla \times \mathop V\limits^ \to = {e_z}\left[ {\frac{{\partial \left( { – c/r} \right)}}{{\partial r}} – \frac{c}{{{r^2}}} – \frac{1}{r}\frac{{\partial \left( 0 \right)}}{{\partial \theta }}} \right]\\ = {e_z}\left[ {\frac{c}{{{r^2}}} – \frac{c}{{{r^2}}} – 0} \right]\\ = 0\end{array}\]\(\nabla \times \mathop V\limits^ \to = 0\) except at origin, since it is singular at origin.
- 1257 views
- 1 answers
- 0 votes
-
Let \(x = r\cos \theta ,\,y = r\sin \theta \), therefore \({x^2} + {y^2} = {r^2}\). In polar- coordinates \({V_r} = u\cos \theta + v\sin \theta \) and \({V_\theta } = – u\sin \theta + v\cos \theta \)
Hence,
\(u = \frac{{5y}}{{{x^2} + {y^2}}} = \frac{{5r\sin \theta }}{{{r^2}}} = \frac{{5\sin \theta }}{r}\) , \(v = – \frac{{5x}}{{{x^2} + {y^2}}} = – \frac{{5r\cos \theta }}{{{r^2}}} = – \frac{{5\cos \theta }}{r}\)
\({V_r} = \frac{{5\sin \theta }}{r}\left( {\cos \theta } \right) + \left( { – \frac{{5\cos \theta }}{r}} \right)\sin \theta = 0\) , \({V_\theta } = – \frac{{5\sin \theta }}{r}\sin \theta + \left( { – \frac{{5\cos \theta }}{r}} \right)\cos \theta = – \frac{5}{r}\)
\(V.ds = \left( {{V_r}{e_r} + {V_\theta }{e_\theta }} \right).\left( {dr{e_r} + rd\theta {e_\theta }} \right)\)\( = \left( {{V_r}dr + r{V_\theta }d\theta } \right)\)\( = 0 + r\left( { – \frac{5}{r}} \right)d\theta \)\( = – 5d\theta \)
Therefore, \(\tau = – \oint_c {V.ds} = – \int\limits_0^{2\pi } { – 5d\theta } = 5\int\limits_0^{2\pi } {d\theta } \)\( = 5 \times 2\pi \ = \,10\,{m^2}/s\).Here, value of circulation, is independent of diameter of circular path.
- 1704 views
- 1 answers
- 0 votes
-
Location of center of pressure is given as \[{x_{cp}} = \left( {\frac{c}{4}} \right) – \frac{{M_{c/4}^\prime }}{{{L^{^\prime }}}}\]
Where\({x_{cp}}\) is the distance of center of pressure from the leading edge of airfoil, and ‘c’ is the chord length .\(M_{c/4}^\prime \) is moment per unit span about quarter-chord point and \({L^\prime }\) is the lift per unit span;\[\begin{array}{l}{x_{cp}} = \frac{c}{4} – \frac{{M_{c/4}^\prime }}{{{L^\prime }}}\\\frac{{{x_{cp}}}}{c} = \frac{1}{4} – \frac{{\left( {M_{c/4}^\prime /{q_\infty }{c^2}} \right)}}{{\left( {{L^\prime }/{q_\infty }c} \right)}}\end{array}\]\[\begin{array}{l} = \frac{1}{4} – \left( {\frac{{{c_{m,c/4}}}}{{{c_l}}}} \right)\\ = \frac{1}{4} – \left( {\frac{{ – 0.08}}{{0.80}}} \right)\\ = \frac{1}{4} + 0.1\\ = 0.25 + 0.1\\ = 0.35\end{array}\]Therefore, location of center of pressure is at 0.35 with respect to chord length.
- 1123 views
- 1 answers
- 0 votes
-
pressure coefficient is given as \[ {C_p} = 1 – {\left( {\frac{v}{{{v_\infty }}}} \right)^2}\]\[ = 1 – {\left( {\frac{{220}}{{100}}} \right)^2}\]\[ = 1 – 4.84\]\[ = – 3.84\]
- 836 views
- 1 answers
- 0 votes
-
(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 \[{C_p} = 1 – {\left( {\frac{v}{{{v_\infty }}}} \right)^2}\]\[v = \sqrt {v_\infty ^2\left( {1 – {C_p}} \right)} \]\[ = \sqrt {{{\left( {200} \right)}^2}\left( {1 – \left( { – 3.84} \right)} \right)} \]\[ = \sqrt {40000\left( {1 + 3.84} \right)} \]\[ = 440\,{\rm{ft/sec}}\]
- 992 views
- 1 answers
- 0 votes
-
Considering the flow to be inviscid and incompressible, Pressure coefficient \({{\rm{C}}_{\rm{p}}} = 1 – {\left( {\frac{v}{{{v_\infty }}}} \right)^2}\)\[ = 1 – {\left( {\frac{{150}}{{100}}} \right)^2}\]\[ = – 1.25\]
- 818 views
- 1 answers
- 0 votes
-
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.
- 2251 views
- 1 answers
- 0 votes
-
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.
- 1195 views
- 1 answers
- 0 votes
-
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).
- 1332 views
- 1 answers
- 0 votes
-
Lift per unit span is given as \({L^\prime} = {\rho _\infty }{V_\infty }\tau \)\[\tau = \frac{{{L^\prime}}}{{{\rho _\infty }{V_\infty }}} = \frac{8}{{\left( {1.23} \right)\left( {50} \right)}} = 0.13\,{m^2}/s\]
- 1095 views
- 1 answers
- 0 votes