Pressure coefficient is given as \[C_{p}=\frac{p_{1} – p_{\infty }}{q_{\infty }}\]This is used for incompressible to compressible flow. However, for incompressible flow, this expression can be used in terms of velocity alone, on applying the Bernoulli’s equation. On using Bernoulli’s equation for two points in a flow, \[p_{1}+\frac{1}{2}\rho V_{1}^{2} = p_{\infty } + \frac{1}{2}\rho v_{\infty }^{2}\]\[p_{1}-p_{\infty} = \frac{1}{2} \rho \left ( V_{\infty}^{2}-V_{1}^{2} \right )\]Considering \(p_{1}\) as the pressure at a point on the wing of the aircraft, and \(p_{\infty}\) as freestream pressure, coefficient of pressure is,\[C_{p} = \frac{p_{1} – p_{\infty }}{q_{\infty}}=\frac{\frac{1}{2}\rho \left ( V_{\infty}^{2}-V_{1}^{2} \right )}{\frac{1}{2}\rho V_{\infty}^{2}}=1-\left ( \frac{V_{1}}{V_{\infty}} \right )^{2}\]\[\Rightarrow C_{p} = 1 – \left ( \frac{140}{100} \right )^{2}=-0.96\]

Velocity components are \(u=\left ( 6x^{2}y+t \right )\); \(v=\left ( 3yx+t^{2}+3 \right )\); \(w=\left ( 3+3t^{2}y \right)\)
Velocity vector is given as,
\[\vec{V}=u\hat{i}+v\hat{j}+w\hat{k}\]At point (\(3,2,4)\) and at \(t=5\), velocity components will be,
\[u=\left ( 6x^{2}y+t \right )=\left ( 6\left ( 3 \right )^{2}2+5 \right )=113\]\[v=\left ( 3yx+t^{2}+3 \right )=\left ( \left ( 3\times 2\times 3 \right )+25+3 \right )=\left ( 18+25+3 \right )=46\]\[w=\left ( 3+3t^{2}y \right )=\left ( 3+\left ( 3\left ( 5 \right )^{2}2 \right ) \right )=\left ( 3+150 \right )=153\]
Therefore, velocity vector is,
\[\vec{V}=113\hat{i}+46\hat{j}+153\hat{k}\]Velocity magnitude is, \[V=\sqrt{u^{2}+v^{2}+w^{2}}\]\[=\sqrt{113^{2}+46^{2}+153^{2}}=195.69\,m/s\]

The stream function for the flow field is \[\psi =2x^{2}-y^{2}\]Velocity component \(u\) and \(v\) of the two-dimensional flow field is \[u=\frac{\partial \psi }{\partial y},v=-\frac{\partial \psi }{\partial x}\]Therefore,\[u=\frac{\partial \psi }{\partial y}=\frac{\partial }{\partial y}\left ( 2x^{2}-y^{2} \right )
=\frac{\partial }{\partial y}\left ( 2x^{2} \right )-\frac{\partial }{\partial y}\left ( y^{2} \right )=0-2y=-2y\]\[v=-\frac{\partial \psi }{\partial x}
=-\frac{\partial }{\partial x}\left ( 2x^{2}-y^{2} \right )
=-\frac{\partial }{\partial x}\left ( 2x^{2} \right )+\frac{\partial }{\partial x}\left ( y^{2} \right )=-4x+0=-4x\] Therefore, at point \((2,3)\),\[u=-2y=-2\left ( 3 \right )=-6\\v=-4x=-4\left ( 2 \right )=-8\]Velocity, \(V=-6i-8j\)
Magnitude of velocity = \[\sqrt{u^{2}+v^{2}}=\sqrt{\left ( -6 \right )^{2}+\left ( -8\right)^{2}}=\sqrt{36+64}=10\]

Flow passing through a shock wave

Specific heat at constant pressure,
\[c_{p}=\frac{\gamma R}{\gamma -1}=\frac{\left ( 1.4 \right )\left ( 287 \right )}{\left ( 1.4-1 \right )}=1004.5\,J/\left  (kg.K \right )\]
Specific heat at condtant volume,
\[c_{v}=\frac{R}{\gamma -1}=\frac{287}{0.4}=717.5\,J/kg.K\]
Enthalpy change across the shock wave,
\[h_{2}-h_{1}=c_{p}\left ( T_{2}-T_{1} \right )=1004.5\left ( 700-300 \right )=4.018\times 10^{5}\,J/kg\]
Internal energy change across the shock wave,
\[e_{2}-e_{1}=c_{v}\left ( T_{2}-T_{1} \right )=717.5\left ( 700-300 \right )=2.87\times 10^{5}\,J/kg\]
Entopy change across the shock wave,
\[s_{2}-s_{1}=c_{p}ln\left ( \frac{T_{2}}{T_{1}} \right )
-Rln\left ( \frac{p_{2}}{p_{1}} \right )
=1004.5ln\left ( \frac{700}{300} \right )-287ln\left ( \frac{9}{1} \right )=220.51\,J/\left (kg.K \right )\]

Prandtl-Meyer expansion waves

Downstream Mach number can be calculated from Prandtl-Meyer function
\[\nu (M)=\sqrt{\frac{\gamma +1}{\gamma -1}}tan^{-1}\sqrt{\frac{\gamma -1}{\gamma +1}\left ( M^{2}-1 \right )}-tan^{-1}\sqrt{M^{2}-1}\]

For a Mach number of \(2.2\), \(\nu \left ( M_{1} \right )=31.73^{^{\circ}}\)

\[\theta =\nu \left ( M_{2} \right )-\nu \left ( M_{1} \right )\]

\[\Rightarrow \nu \left ( M_{2} \right )=\theta + \nu \left ( M_{1} \right ) = 25^{\circ}+31.73^{\circ}=56.73^{\circ}\]

for, \(\nu \left ( M_{2} \right )=56.73^{^{\circ}}\), Mach number \(M_{2}=3.389\)

From Isentropic flow properties for, \(M_{1}=2.2\)

\[\frac{p_{01}}{p_{1}}=10.69\] and
\[\frac{T_{01}}{T_{1}}=1.968\]

For, \(M_{2}=3.389\)
\[\frac{p_{02}}{p_{2}}=65.0798\] and
\[\frac{T_{02}}{T_{2}}=3.297\]

Total pressure and total temperature is constant through the expansion wave, so

\(p_{01}=p_{02}\) and \(T_{01}=T_{02}\)

Therefore,
\[p_{2}=\left ( \frac{p_{2}}{p_{02}} \right )\left ( \frac{p_{01}}{p_{1}} \right )p_{1}=\left ( \frac{1}{65.0798} \right )\left ( 10.69 \right )0.8=0.1314\,atm\]

\[T_{2}=\left ( \frac{T_{2}}{T_{02}} \right )\left ( \frac{T_{01}}{T_{1}} \right )T_{1}=\left ( \frac{1}{3.297} \right )\left ( 1.968 \right )360=214.886\,K\]

\[p_{2}=\rho _{2}RT_{2}\]

\[\Rightarrow 0.1314=\rho _{2}\left ( 287 \right )\left ( 214.886\,K \right )\]

\[\Rightarrow \rho _{2}=\frac{0.1314}{\left ( 287 \right )\left ( 214.886 \right )}=2.13\times 10^{-6}\,kg/m^{3}\]

Since,
\[p_{01}=p_{02}\Rightarrow p_{02}=\left ( \frac{p_{01}}{p_{1}} \right )\left ( p_{1} \right )=\left ( 10.69 \right )\left ( 0.8 \right )=8.552\,atm\]

Also,

\[T_{01}=T_{02}\Rightarrow T_{02}=\left ( \frac{T_{01}}{T_{1}} \right )\left ( T_{1} \right )=\left ( 1.968 \right )\left ( 360 \right )=708.48\,K\]

Mach angle is given by \(\mu = Sin^{-1}\left ( \frac{1}{M} \right )\)

Therefore Mach angle \(\mu _{1}\) made by upstream Mach wave for \(M_{1}=2.2\) is \(27.04^{\circ}\)

Mach angle \(\mu _{2}\) made by downstream Mach wave for\(M_{2}=3.389\) is \(17.16^{\circ}\)

and, angle made by forward Machline is \(27.04^{\circ}\) and angle made by the rearward Machline is \(17.16^{\circ}-25^{\circ}=-7.84^{\circ}\)

Oblique shock wave

Normal component of the upstream Mach number to the oblique shock wave,

\[M_{n1}=M_{1}Sin\beta =4\left ( Sin30^{\circ} \right )=2\]

From Normal shock wave properties table, for \(M_{n1}=2\)

\(\frac{p_{2}}{p_{1}}=4.5\), \(\frac{T_{2}}{T_{1}}=1.687\),  \(\frac{p_{02}}{p_{01}}=0.7209\), \(M_{n2}=0.5774\)

At an altitude of \(9000\, m \) , \(p_{1}=3.08\times 10^{4} N/m^{2} \), \(T_{1}=229.74K\)

Therefore,

\[P_{2}=\left ( \frac{P_{2}}{P_{1}} \right )P_{1}=\left ( 4.5 \right )\left ( 3.08\times 10^{4} \right)= 13.86\times 10^{4} N/m^{2}\]

\[T_{2}=\left ( \frac{T_{2}}{T_{1}} \right )T_{1}=\left ( 1.687 \right )229.74=387.571K\]

From the \(\theta -\beta -M\) diagram , for, wave angle of \(\beta =30^{\circ}\) and Mach number of \(4\), deflection angle, \(\theta =17.7^{\circ}\).

Therefore,

\[M_{2}=\frac{M_{n2}}{Sin\left ( \beta -\theta \right )}=\frac{0.5774}{Sin\left ( 30^{\circ} -17.7^{\circ}\right )}=2.710\]

From the table of isentropic flow properties , for

\[M_{1}=4,\frac{p_{01}}{p_{1}}=151.8,\frac{T_{01}}{T_{1}}=4.2\]

Total pressure,

\[p_{02}=\left ( \frac{p_{02}}{p_{01}} \right )\left ( \frac{p_{01}}{p_{1}} \right )p_{1}=\left ( 0.0709 \right )\left ( 151.8 \right )\left ( 3.08\times 10^{4} \right )=331488.696N/m^{2}\]

Total temperature,

\[T_{02}=T_{01}=\left ( \frac{T_{01}}{T_{1}} \right )T_{1}=4.2\times 229.74K=964.908K\]

Satellite orbit

\[\frac{{{r_{\max }}}}{{{r_{\min }}}} = \frac{{1 + e}}{{1 – e}} = \frac{{1 + 0.0014}}{{1 – 0.0014}} = 1.0028\]

\[ \Rightarrow {r_{\max }} = {r_{\min }}\left( {1.0028} \right)\]

\[{r_{\min }} = {\rm{radius}}\,{\rm{of}}\,{\rm{earth}}\,{\rm{ + }}\,{\rm{altitude}}\,{\rm{above}}\,{\rm{earth’s}}\,{\rm{surface}}\]

\[ \Rightarrow {{\rm{r}}_{\min }} = 6.4 \times {10^6} + 0.4 \times {10^6} = 6.8 \times {10^6}\,m\]

\[\Rightarrow {r_{\max }} = 6.8 \times {10^6}\left( {1.0028} \right) = 681904\,m = 6.81904 \times {10^6}\,m\]

Therefore satellite’s altitude at apogee is

\[6.81904 \times {10^6} – 6.4 \times {10^6} = 419040\,m = 419.04\,km\]

Time period of the satellite is given as

\[{T^2} = \frac{{4{\pi ^2}}}{{{k^2}}}{a^3}\]

\[{k^2} = G{M_{earth}} = 3.986 \times {10^{14}}\,{m^3}/{s^2}\]

\[ \Rightarrow a = {\rm{semi}}\,{\rm{major}}\,{\rm{axis}}\,{\rm{ = }}\frac{{{r_{\max }} + {r_{\min }}}}{2} = \frac{{6.81904 \times {{10}^6} + 6.8 \times {{10}^6}}}{2} = 6.80952 \times {10^6}\,m\]

\[{T^2} = \frac{{4{\pi ^2}}}{{3.986 \times {{10}^{14}}}} \times {\left( {6.80952 \times {{10}^6}} \right)^3}\]

\[ \Rightarrow T = 5592.242\,s = 1.5534\,hr\]

\[{V_{perigee}} = {r_{\min }}\dot \theta \]

\[\dot \theta = \frac{h}{{{{\left( {{r_{\min }}} \right)}^2}}}\]

\[{h^2} = {r_{\min }}\left( {1 + e} \right){k^2}\]

\[ \Rightarrow {h^2} = \left( {6.8 \times {{10}^6}} \right)\left( {1 + 0.0014} \right)3.986 \times {10^{14}}\]

\[ \Rightarrow h = 5.20987 \times {10^{10}}\]

Therefore,

\[\dot \theta = \frac{{5.20987 \times {{10}^{10}}}}{{{{\left( {6.8 \times {{10}^6}} \right)}^2}}} = 0.0011267\,rad/s\]

\[{V_{perigee}} = \left( {6.8 \times {{10}^6} \times 0.0011267} \right) = 7661.56\,m/s = 7.66156\,km/s\]

Burnout velocity: It is the velocity of the rocket when propellants stop burning. It is the maximum velocity achieved by a rocket.

Escape velocity : It is the minimum velocity required for a body to escape from a gravitational center of attraction. Kinetic energy of the body is equal to the potential energy and in the absence of any frictional resistance the body will escape from the planet. The escape velocity from earth is \(11.2\, km/s\). Burnout velocity for a rocket is given by

\[{V_b} = {g_0}{I_{sp}}\ln \left( {\frac{{{M_i}}}{{{M_f}}}} \right)\]

\[ \Rightarrow 11.2 \times 1000 = \left( {9.8} \right)\left( {390} \right)\ln \left( {\frac{{{M_i}}}{{{M_f}}}} \right)\]

\[ \Rightarrow \left( {\frac{{{M_i}}}{{{M_f}}}} \right) = 18.74\]

Also,

\[{M_i} = \left( {{M_p} + {M_f}} \right)\]

\[ \Rightarrow {M_f} = \left( {{M_i} – {M_p}} \right)\]

\[ \Rightarrow \frac{{{M_i}}}{{{M_f}}} = \frac{{{M_i}}}{{{M_i} – {M_p}}} = \frac{1}{{1 – \left( {\frac{{{M_p}}}{{{M_i}}}} \right)}}\]

\[ \Rightarrow 18.74 = \frac{1}{{1 – \left( {\frac{{{M_p}}}{{{M_i}}}} \right)}}\]

\[ \Rightarrow 1 – \left( {\frac{{{M_p}}}{{{M_i}}}} \right) = \frac{1}{{18.74}}\]

\[ \Rightarrow \frac{{{M_p}}}{{{M_i}}} = 0.947\]

Static margin is the distance between the center of gravity and neutral point of an aircraft. It is expressed as percentage in terms of mean aerodynamic chord of the wing. Neutral point of an aircraft is the point where aircraft is neutrally stable, when it is disturbed from its trim angle of attack longitudinally.

Neutral point is given as

\[{h_n} = {h_{ac,wb}} + {V_H}\frac{{{a_t}}}{a}\left( {1 – \frac{{\partial \varepsilon }}{{\partial \alpha }}} \right)\]

\[ \Rightarrow {h_n} = 0.3 + \left( {0.352} \right)\frac{{0.12}}{{0.085}}\left( {1 – 0.36} \right)\]

\[ \Rightarrow {h_n} = 0.3 + \left( {0.352} \right)\left( {1.41} \right)\left( {0.64} \right) = 0.618\]

Static margin, \({h_n} – h = 0.618 – 0.38 = 0.238\,{\rm{chord}}\,{\rm{length}}\)

Airplane’s turn radius is given by

\[R = \frac{{V_\infty ^2}}{{g\sqrt {{n^2} – 1} }}\]

and turn rate is \[\frac{{{V_\infty }}}{R}\]

Here, \(n\) = load factor = \((lift/weight)=L/W\)

Airplane velocity = \(403.2\,km/h = 112\,m/s\)

\[L = \frac{1}{2}{\rho _\infty }V_\infty ^2S{C_L} = \frac{1}{2} \times 1.225 \times {\left( {112} \right)^2} \times 48 \times 1.3 = 479431.68\,N\]

Therefore,\[n = \frac{{479431.68}}{{100000}} = 4.79\]

Turn radius, \[R = \frac{{V_\infty ^2}}{{g\sqrt {{{n}^2} – 1} }} = \frac{{{{\left( {112} \right)}^2}}}{{9.8\sqrt {{{\left( {4.79} \right)}^2} – 1} }} = 273.244\,m\]

Turn rate,\[\frac{{{V_\infty }}}{R} = \frac{{112}}{{273.244}} = 0.41\,rad/s\]