Find the velocity, pressure and temperature at, \((0.07\,m, 0.07\,m)\).

Velocity potential of a subsonic compressible flow is \[\phi(x,y) = V_{\infty}x + \frac{5}{\sqrt{1 – M_{\infty}^{2}}}e^{-2\pi\sqrt{1-M_{\infty}^{2}}y}sin2\pi x\]

Freestream velocity, pressure and temperature are \(200\,m/s\), \(1\,atm\) and \(300\,K\). Find the velocity, pressure and temperature at, \((0.07\,m, 0.07\,m)\).

Asked on 13th January 2022 in Aerodynamics.
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    Here,\[ \phi \left ( x,y \right )=V_{\infty}x+\frac{5}{\sqrt{1-M_{\infty}^{2}}}e^{-2\pi\sqrt{1-M_{\infty}^{2}}y}sin2\pi x\]\[u=\frac{\partial \phi}{\partial x}=\frac{\partial }{\partial x}\left \{V_{\infty}x+\frac{5}{\sqrt{1-M_{\infty}^{2}}}e^{-2\pi\sqrt{1-M_{\infty}^{2}}y}sin2\pi x \right \} \]\[=V_{\infty}+\left \{ \frac{5}{\sqrt{1-M_{\infty}^{2}}}e^{-2\pi\sqrt{1-M_{\infty}^{2}}y}cos2\pi x \right \}2\pi\]\[=V_{\infty}+\frac{5\left ( 2\pi \right )}{\sqrt{1-M_{\infty}^{2}}}e^{-2\pi\sqrt{1-M_{\infty}^{2}}y}cos\left(2\pi x\right)\]\[v=\frac{\partial \phi}{\partial y}=\frac{\partial }{\partial y}\left \{V_{\infty}x+\frac{5}{\sqrt{1-M_{\infty}^{2}}}e^{-2\pi\sqrt{1-M_{\infty}^{2}}y}sin2\pi x \right \} \]\[=-\frac{5}{\sqrt{1-M_{\infty}^{2}}}\left (2\pi \sqrt{1-M_{\infty}^{2}} \right )e^{-2\pi\sqrt{1-M_{\infty}^{2}}y}sin2\pi x\]\[=-5\left ( 2\pi \right )e^{-2\pi\sqrt{1-M_{\infty}^{2}}y}sin\left ( 2\pi x \right )\]Speed of sound, \(a_{\infty} = \sqrt{\gamma RT}=\sqrt{1.4 \times 287\times 300}=347.189\,m/s\)

    Mach number, \[M_{\infty} = \frac{V_{\infty}}{a_{\infty}}=\frac{200}{347.189}=0.576\]Therefore, at \(\left ( x,y \right ) = \left ( 0.07\,m, 0.07\,m \right )\)

    \[u=200+\frac{5\left ( 2\pi \right )}{\sqrt{1-\left ( 0.576 \right )^{2}}}e^{-2\pi\sqrt{1-\left ( 0.576 \right )^{2}}\left ( 0.07 \right )}cos\left(2\pi \left ( 0.07 \right )\right)\]\[=224.272\,m/s\]\[v=-5\left( 2\pi \right )e^{-2\pi\sqrt{1-\left(0.576\right)^{2}}\left( 0.07 \right )}sin\left ( 2\pi \left ( 0.07 \right )\right )\]\[=-9.337\,m/s\]Velocity, \[V=\sqrt{u^{2}+v^{2}}=\sqrt{\left(224.272\right)^{2}+\left ( -9.337 \right )^{2}}=224.47\,m/s\]From isentropic flow properties, for, \[M_{\infty}=0.576,\frac{T_{\infty}}{T_{0}}=0.93777382\]\[T_{0}=\frac{300}{0.93777382}=319.9066\,K\]Speed of sound,\[ a_{0}=\sqrt{\left ( 1.4 \right )\left ( 287 \right )\left ( 319.9066 \right )}=358.5226\,K\]\[a^{2}=a_{0}^{2}-\frac{\gamma -1}{2}\left ( V^{2} \right )\]\[a^{2}=\left(358.5226\right)^{2}-\frac{1.4-1}{2}\left ( 224.47 \right )^{2}\]\[\Rightarrow a = 344.182\,m/s\]Mach number, \[M = \frac{V}{a}=\frac{224.47}{344.182}=0.6522\]From isentropic flow properties, for, \[M = 0.576, \frac{p_{\infty}}{p_{0}}=0.7986, \frac{T_{\infty}}{T_{0}}=0.93777\]From isentropic flow properties, for, \[M = 0.6522, \frac{p}{p_{0}}=0.7514, \frac{T}{T_{0}}=0.9216\]Therefore, at \(\left ( 0.07\,m,0.07\,m \right )\)

    \[p = \left ( \frac{p}{p_{0}} \right )\left ( \frac{p_{0}}{p_{\infty}} \right )p_{\infty}=\left ( 0.7514 \right )\left ( \frac{1}{0.7986} \right )\left ( 1\,atm \right )=0.941\,atm\]

    \[T = \left ( \frac{T}{T_{0}} \right )\left ( \frac{T_{0}}{T_{\infty}} \right )T_{\infty}=\left ( 0.9216 \right )\left ( \frac{1}{0.93777} \right )\left ( 300\, K \right )= 294.83\,K\]

    Answered on 21st January 2022.
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