# 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

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.