We have

$$q=T\left ( s_{2} -s_{1}\right )$$
$\\=\left ( 175+273 \right )\left ( 6.622-7.055 \right ) \\=-193.983 \\W=q-\left ( u_{2}-u_{1} \right ) \\=\left ( -193.984-\left ( 2579-2606 \right ) \right )\times 2 \\= -500.94\;kJ$

Therefore work required for the process is approximately  -501 kJ.

$Q/^{\circ}C=4.5\;kJ \\Q/^{\circ}F=\frac{Q/^{\circ}C}{1.8} \\=\frac{4.5}{1.8} \\=2.5\;kJ$

Therefore amount of heat lost is 2.5 kJ.

mu = 1; % Doublet strength
% GRID:
x = -5:.02:5;
y = -5:.02:5;
for m = 1:length(x)
for n = 1:length(y)
xx(m,n) = x(m); yy(m,n) = y(n);
% Velocity potential function:
phi_Doublet(m,n) = mu * x(m)/(x(m)^2+(y(n)+.01)^2);
% Stream function:
psi_Doublet(m,n) = - mu * y(n)/(x(m)^2+(y(n)+.01)^2);
end
end
% Plots
% Doublet at origin of coordinate system:n
figure(4)
contour(xx,yy,psi_Doublet,[-4:0.25:4],'k'),hold on
contour(xx,yy,phi_Doublet,[-20:0.5:20],'r')
legend('streamlines' ,'potential')
title('Doublet at origin')
axis image,hold off

Doublet at origin

 

 

 

Mach number is defined as the ratio of velocity of the object to the velocity of sound $M = \frac{V}{a}$.Mach number is generally associated with flying vehicles. Mach number is a very useful quantity for estimation of compressible flow aerodynamics, particularly high speed aerodynamics,where there is a creation of normal and oblique shock waves. Flying vehicles such as airplanes generally fly in Mach number of 0.8. Supersonic aircrafts like fighter aircrafts fly in Mach number  around of 2.

Re-entry vehicles like space shuttle have Mach number above 25 during its entry to earth. Speed of sound varies with altitude as it is changing with air temperature. So at different altitude on earths atmosphere Mach number is different even if flight speed is same. Also speed of sound is dependent on type of medium. So on mars where there is a different atmosphere than on earth Mach number for the flight vehicle would be different even if it is flying at same flight speed as on earth. $$a = \sqrt {\gamma RT}$$

$$M = \frac{V}{{\sqrt {\gamma RT} }}$$

We need to obtain $\nu \left( {{M_2}} \right)$ from Prandtl-Meyer expansion function.

$\theta = \nu \left( {{M_2}} \right) – \nu \left( {{M_1}} \right)$

We need to calculate $\nu \left( {{M_1}} \right)$ from Appendix C “Prandtl-Meyer expansion function and Mach angle table” for

${M_1} = 2,{\nu _1} = {26.38^ \circ }$

On substituting this value

$\begin{array}{l}{3^ \circ } = \nu \left( {{M_2}} \right) – \nu \left( {{M_1}} \right)\\{3^ \circ } = \nu \left( {{M_2}} \right) – {26.38^ \circ }\\\nu \left( {{M_2}} \right) = {26.38^ \circ } + {3^ \circ }\\ = {29.38^ \circ }\end{array}$

We need to obtain ${M_2}$ from Appendix C “Prandtl-Meyer expansion wave and Mach angle table”, for

$\begin{array}{l}\nu \left( {{M_2}} \right) = {29.38^ \circ }\\{M_2} = 2.1\end{array}$

So, average Mach number across AB is ${2.1}$

 

 

The supersonic flow over a flat plate at an angle of attack in a free stream with given Mach number, M_∞ is sketched below

The flow direction downstream of the leading edge is given by line ab. The flow direction is below the horizontal (below the direction of M_∞) because lift is produced on the flat plate, and due to overall momentum considerations, the downstream flow will be inclined slightly downward.

Line ab is a slip line. Since flows over top and bottom of plate have gone through shock waves of different strengths entropy in region 4 and 5 are different.

The boundary conditions that must hold across the slip line is constant pressure that P₄ will be equal to P₅ and strengths of trailing edge shock waves and expansion waves is fixed by this boundary condition.

We can use iterative approach in order to calculate flow direction and strength of trailing edge shocks and expansion waves.

1. a) We have to assume a value of ∅.
2. b) In order to obtain P₄, we need to calculate strength of trailing edge shock by using local deflection angle (α – ∅).
3. c) In order to obtain P₅, we need to calculate strength of trailing edge expansion wave for a local expansion angle (α – ∅).
4. d) If P₄ and P₅ are different than we have to assume a new value of ‘∅’.
5. e) When P₄ = P_{5 ,} than condition is satisfied and iteration is said to be converged, trailing edge flow is determined.

The flow over a cylinder is considered to be two dimensional, whereas flow over a sphere can be considered as three dimensional. The flow over a cylinder is moving upward and downward, however in sphere apart from upward and downward motion, it is also moving sideways. Since it has an extra direction, there is a lower speed when flowing over sphere. Therefore free stream mach number is higher in order to have a sonic velocity over the sphere, means a larger critical mach number.

Given NACA 2213 at wing root and NACA 2205 at wing tip

We have to consider NACA 2213

Root airfoil chord = 8.33 ft.

C_d = 0.006

Re = 9 × 10⁶

Altitude = 18000 ft.

Density at this altitude = 1.3553×10^-3 slug/ft³

(a) We have to find velocity of aircraft

Here, Considering dynamic viscosity and temperature at sea level dynamic viscosity at 18000 ft. is calculated as

$\begin{array}{l}\frac{{{\mu _2}}}{{{\mu _1}}} = \sqrt {\frac{{{T_2}}}{{{T_1}}}} \\{\mu _2} = \sqrt {\frac{{454.55}}{{518.69}}} \times 3.737 \times {10^{ – 7}}\\ = 3.498 \times {10^{ – 7}}\end{array}$

Reynold’s number is given as

$\begin{array}{l}{{\mathop{\rm Re}\nolimits} _c} = \frac{{{\rho _\infty }{v_\infty }c}}{{{\mu _\infty }}}\\ \Rightarrow \frac{{1.3553 \times {{10}^{ – 3}} \times {v_\infty } \times 8.33}}{{3.498 \times {{10}^{ – 7}}}} = 9 \times {10^6}\\ \Rightarrow {v_\infty } = 278.86ft/\sec \end{array}$

(b)

Since, the flow is turbulent, so skin friction drag coefficient

$\begin{array}{l}{c_f} = \frac{{0.074}}{{{\mathop{\rm Re}\nolimits} _c^{1/5}}}\\ = 0.003\end{array}$

Total profile drag coefficient

Pressure drag coefficient + skin friction drag coefficient

Therefore, pressure drag coefficient = 0.006 – 0.003 = 0.003

Percentage = 

$\begin{array}{l}\frac{{0.003}}{{0.006}} \times 100\\ = \,50\,\% \end{array}$

 

We need to obtain drag ratio per unit span for cylinder and wedge.

Drag coefficient for cylinder  ${c_d} = 4/3$

Obtaining drag per unit span for the cylinder with ${c_d}$ based on frontal area we have

$\begin{array}{l}{\left( {{D^\prime}} \right)_{cyl}} = {q_\infty }s{c_d}\\ = {q_\infty }d(1) \times \frac{4}{3}\end{array}$

Obtaining drag per unit span for the dimensional wedge airfoil we have

$\begin{array}{l}{\left( {{D^\prime}} \right)_w} = \left( {{p_2} – {p_3}} \right)t\\\frac{{\left( D \right)_{cyl}^\prime}}{{{{\left( D \right)}^\prime }_w}} = \frac{{\frac{4}{3}\left( d \right){q_\infty }}}{{\left( {{p_2} – {p_3}} \right)t}}\end{array}$

Here,

$\begin{array}{l}t = d,{q_\infty } = \frac{\gamma }{2}{p_1}M_1^2\\\frac{{{{\left( {{D^\prime}} \right)}_{cyl}}}}{{{{\left( {{D^\prime}} \right)}_w}}} = \frac{{\frac{4}{3}\left( {\frac{\gamma }{2}} \right)M_1^2}}{{\left( {\frac{{{P_2}}}{{{P_1}}} – \frac{{{P_3}}}{{{P_1}}}} \right)}}\\ = \frac{{\frac{2}{3}\gamma M_1^2}}{{\left( {\frac{{{P_2}}}{{{P_1}}} – \frac{{{P_3}}}{{{P_1}}}} \right)}}\end{array}$

Now we need to obtain the static pressure ratio ${P_2}/{P_1}$

$\begin{array}{l}{M_1} = 5,\theta = {5^ \circ },\beta = {15.1^ \circ }\\{M_{n,1}} = {M_1}\sin \beta \\ = 5\sin \left( {{{15.1}^ \circ }} \right)\\ = 1.303\end{array}$

From Appendix A, Isentropic flow properties table for ${M_{n,1}} = 1.302,{P_2}/{P_1} = 1.805$

$\begin{array}{l}{M_2} = \frac{{{M_{n,2}}}}{{\sin \left( {\beta – \theta } \right)}}\\ = \frac{{0.786}}{{\sin \left( {15.1 – 5} \right)}}\\ = 4.48\end{array}$

Now we need to obtain parameters from Appendix C, Prandtl-Meyer function and Mach angle table.

For ${M_2} = 4.48$ ,

$\begin{array}{l}{\nu _2} = 71.83\\{\nu _3} = {\nu _2} + \theta \\ = {71.83^ \circ } + {10^ \circ }\\ = {81.38^ \circ }\\\ {M_3} = 5.6\end{array}$

From Appendix A, Isentropic flow properties table for

$\begin{array}{l}{M_1} = 5,\frac{{{P_{01}}}}{{{P_1}}} = 529.1\\{M_3} = 5.6,\frac{{{P_{03}}}}{{{P_3}}} = 1037\end{array}$

From Appendix B, Normal shock wave properties table for

${M_{n,1}} = 1.303,\frac{{{P_{02}}}}{{{P_{01}}}} = 0.9794$

Obtaining static pressure ratio ${P_3}/{P_1}$

$\begin{array}{l}\frac{{{P_3}}}{{{P_1}}} = \frac{{{P_3}}}{{{P_{03}}}} \times \frac{{{P_{03}}}}{{{P_{02}}}} \times \frac{{{P_{02}}}}{{{P_{01}}}} \times \frac{{{P_{01}}}}{{{P_1}}}\\ = \left( {\frac{1}{{1037}}} \right)\left( 1 \right)\left( {0.9794} \right)\left( {529.1} \right)\\ = 0.5\end{array}$

Now we need to find the required drag ratio for cylinder and wedge.

$\begin{array}{l}\frac{{{{\left( {{D^\prime}} \right)}_{cyl}}}}{{{{\left( {{D^\prime}} \right)}_w}}} = \frac{{\left( {\frac{2}{3}\gamma M_1^2} \right)}}{{\left( {\frac{{{P_2}}}{{{P_1}}} – \frac{{{P_3}}}{{{P_1}}}} \right)}}\\ = \frac{{\frac{2}{3} \times 1.4 \times {5^2}}}{{\left( {1.805 – 0.5} \right)}}\\ = 17.9\end{array}$

Therefore required drag ratio is 17.9 .

This shows that blunt bodies create a large amount of drag than a sharp-nosed slender body in a supersonic flow.

 

$u = 4x = \frac{{\partial \psi }}{{\partial y}}$ 

 

$\Rightarrow \partial \psi  = 4x\partial y$

 

$\Rightarrow \int {\partial \psi  = \int {4x\partial y} }$

 

$\Rightarrow \psi  = 4xy + f\left( x \right)$

 

                                      $v =  – 4y$ = -$\frac{{ \partial \psi }}{{\partial x}}$

 

$\Rightarrow \frac{{\partial \psi }}{{\partial x}} = 4y$

 

$\Rightarrow \partial \psi  = 4y\partial x$

 

$\Rightarrow \int {\partial \psi  = \int {4y\partial x} }$

 

$\Rightarrow \psi  = 4xy + f\left( y \right)$

 

On comparing these two equations for $\psi$, stream functions is $\psi  = 4xy + {\rm{constant}}$.

 

Also                               $u = 4x = \frac{{\partial \phi }}{{\partial x}}$

 

$\Rightarrow 4x\partial x = \partial \phi$

 

$\Rightarrow \int {\partial \phi  = \int {4x\partial x} }$

 

$\Rightarrow \phi  = 4{x^2} + f\left( y \right)$

 

                                      $v =  – 4y = \frac{{\partial \phi }}{{\partial y}}$

 

$\Rightarrow \frac{{\partial \phi }}{{\partial y}} =  – 4y$

 

$\Rightarrow \int {\partial \phi  = \int { – 4y\partial y} }$

 

$\Rightarrow \phi  =  – 4{y^2} + f\left( x \right)$

 

On comparing the above two equations, $f\left( y \right) =  – 4{y^2}$ and $f(x) = 4{x^2}$, velocity potential is $\phi  = 4\left( {{x^2} – {y^2}} \right)$.

 

Now

 

$\psi  = 4xy + {\rm{constant}}$, differentiating with respect to x, holding $\psi  = {\rm{constant}}$

 

$0 = 4x\frac{{dy}}{{dx}} + 4y$

 

$\Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\psi  = {\rm{constant}}}} =  – \frac{y}{x}$

 

Differentiating $\phi  = 4\left( {{x^2} – {y^2}} \right)$ with respect to x, holding $\phi  = {\rm{constant}}$ 

 

$0 = 2 \times 4x – 2 \times 4 \times y\frac{{dy}}{{dx}}$

 

$\Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\phi  = {\rm{constant}}}} = \frac{x}{y}$

On comparing the above, we get $${\left( {\frac{{dy}}{{dx}}} \right)_{\psi  = {\rm{constant}}}} =  – {\frac{1}{{\left( {\frac{{dy}}{{dx}}} \right)}}_{\phi  = {\rm{constant}}}}$$ This shows that lines of constant $\phi$ are perpendicular to lines of constant $\psi$ .