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\title{Problems of Chapter 8 \\(Linearized flows)}
\author{Yi-Chao XIE}
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\maketitle
\begin{center}
 \textcolor{red}{\bf {Due November $3^{rd}$, 2023}}
 \end{center}
%9.1 Show that this nonlinear equation is valid for transonic flow with small perturbations:
%
%\begin{center}
%$(1-M_{\infty}^2)\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}+\frac{\partial^2 \phi}{\partial z^2}=M^2_{\infty}[\frac{\gamma+1}{V_{\infty}}\frac{\partial \phi}{\partial x}]\frac{\partial^2 \phi}{\partial x^2}$
%\end{center}
%
%Solution:\\

9.2 The low-speed lift coefficient for a NACA 2412 airfoil at an angle of attack of $4^o$ is 0.65. Using the Prandtl -Glauert rule, calculate the lift coefficient for $M_{\infty}=0.7$.\\
%\textcolor{blue}{Solution:\\
% According to the Prandtl-Glauert rule for subsonic flow $$C_l=\frac{C_{l0}}{\sqrt{1-M^2_\infty}}$$
% Thus, $$C_l=\frac{0.65}{\sqrt{1-0.7^2}}=0.91$$}

\vspace{10cm}
9.3 In low-speed flow, the pressure coefficient at a point on an airfoil is -0.9. Calculate the value of $C_p$ at the same point for $M_{\infty}=0.6$ by means of:\\
(a) The Prandtl-Glauert rule\\
(b) Laitone's correction\\
(c) The Karman-Tsien rule\\

%\textcolor{blue}{Solution:\\
%(a) According to the Prandtl-Glauert rule for subsonic flow $$C_p=\frac{C_{p0}}{\sqrt{1-M^2_\infty}}$$
% Thus $$C_p=\frac{-0.9}{\sqrt{1-0.6^2}}=-1.25$$.}
% 
%\textcolor{blue}{ (b) According to the Laitone's rule for subsonic flow $$C_p=\frac{C_{p0}}{\sqrt{1-M^2_\infty}+[M^2_\infty(1+\frac{\gamma-1}{2}M^2_\infty)/(2\sqrt{1-M^2_\infty})]C_{p0}}$$
%Thus $$C_p=\frac{-0.9}{\sqrt{1-0.6^2}+[0.6^2\times(1+\frac{1.4-1}{2}\times0.6^2)/(2\sqrt{1-0.6^2})]\times(-0.9)}=-1.544$$.}
%
%\textcolor{blue}{ (c) According to the Karman-Tsien's rule for subsonic flow $$C_p=\frac{C_{p0}}{\sqrt{1-M^2_\infty}+\frac{M^2_\infty}{\sqrt{1+M^2_\infty}}\frac{C_{p0}}{2}}$$
%Thus $$C_p=\frac{-0.9}{\sqrt{1-0.6^2}+(0.6^2/\sqrt{1+0.6^2)}\times(-0.45)}=-1.2676$$}.
 \vspace{10cm}
 

9.4 Consider a flat plat with chord length $c$ at an angle of attack $\alpha$ to a supersonic free stream of Mach number $M_{\infty}$. Let $L$ and $D$ be the lift and drag per unit span, $S=c(1)$. Using linearized theory, derive the following expression for the lift and drag coefficients (where $C_L\equiv\frac{L}{\frac{1}{2}\rho_{\infty}V^2_{\infty}S}$ and $C_D\equiv\frac{D}{\frac{1}{2}\rho_{\infty}V^2_{\infty}S}$):\\
(a)$C_L=\frac{4\alpha}{\sqrt{M^2_{\infty}-1}}$\\
(b)$C_D=\frac{4\alpha^2}{\sqrt{M^2_{\infty}-1}}$\\

%\textcolor{blue}{Solution:\\ for supersonic flow.\\The respective lift and drag force for a plate with angle of attack $\alpha$ is $$L=(p_2\cos\alpha-p_1\cos\alpha)S$$ and $$D=(p_2\sin\alpha-p_1\sin\alpha)S$$ Thus the lift and drag coefficient are, respectively, $$C_L=\frac{(p_2\cos\alpha-p_1\cos\alpha)S}{\frac{1}{2}\rho_\infty V^2_\infty S}=(C_{p_2}-C_{p_1})\cos\alpha$$ and $$C_D=\frac{(p_2\sin\alpha-p_1\sin\alpha)S}{{\frac{1}{2}\rho_\infty V^2_\infty S}}=(C_{p_2}-C_{p_1})\sin\alpha$$ Consider linearised flow, i.e., $\alpha$ is small, we have $\cos\alpha=1$ and $\sin\alpha=\alpha$, we have $$C_L=C_{p_2}-C_{p_1}$$ and$$C_D=(C_{p_2}-C_{p_1})\alpha$$ According to the Prandtl Glauert rule, $C_p=\frac{2\theta}{\sqrt{M_\infty^2-1}}$, $C_{p_1}=-\frac{2\alpha}{\sqrt{M_\infty^2-1}}$ and $C_{p_2}=\frac{2\alpha}{\sqrt{M_\infty^2-1}}$, we have $$C_l=\frac{4\alpha}{\sqrt{M^2_{\infty}-1}}\qquad  and \qquad C_D=\frac{4\alpha^2}{\sqrt{M^2_{\infty}-1}}$$}
\vspace{10cm}
9.5 For the flat plate in Problem 9.4, the quarter-chord point is located, by definition, at a distance equal to $c/4$ from the leading edge. Using linearised theory, derive the following expression for the moment coefficient about the quarter-chord point for supersonic flow:
\begin{center}
$C_{M_{c/4}}=\frac{-\alpha}{\sqrt{M^2_{\infty}-1}}$
\end{center}
where $C_{M_{c/4}}\equiv M_{c/4}/\frac{1}{2}\rho_{\infty}V^2_{\infty}Sc$, and as usual in aeronautical practice, a positive moment by convention is in the direction of the increasing angle of attack.

% \textcolor{blue}{Solution:\\ The moment at $\frac{c}{4}$ can be calculated as following:\\ $$M_{c/4}=(-p_1\frac{c}{4}\times 1+p_2\frac{c}{4}\times 1)\frac{c}{8}+(p_1\frac{3c}{4}\times 1-p_2\frac{3c}{4}\times 1)\frac{3c}{8}=(p_1-p_2)\frac{c^2}{4}$$ Hence, the moment coefficient about the quard-chord point is $$ C_{M_{c/4}}=\frac{(p_1-p_2)\frac{c^2}{4}}{\frac{1}{2}\rho_\infty V_\infty^2 c\times1\times c}$$ $$\qquad\quad=(C_{p_1}-C_{p_2})\times\frac{1}{4} $$ According to the Prandtl-Glauert rule, $C_{p_1}=-\frac{2\alpha}{\sqrt{M_\infty^2-1}}$ and $C_{p_2}=\frac{2\alpha}{\sqrt{M_\infty^2-1}}$, we have $$C_{M_{c/4}}=\frac{-\alpha}{\sqrt{M^2_{\infty}-1}}$$ }
\vspace{10cm}
%9.6 Consider a flat plate at an angle of attack of $4^o$.\\
%(a) Calculate $C_L$ and $C_{M_{c/4}}$ for $M_{\infty}=0.03$ (essentially incompressible flow).(Hint: Consult a book, such as ref 104 for the aerodynamic properties of a flat plate using incompressible flow thin airfoil theory.)\\
%(b) Apply the Prandtl-Glauert rule to the result of part (a), and calculate $C_L$ and $C_{M_{c/4}}$ for $M_{\infty}=0.6.$


%Solution:\\

9.7 Consider a diamond-shaped airfoil such as that sketched in Fig 4.35. The half angle is $\epsilon$, thickness is $t$, and chord length is $c$. For supersonic flow, use linearized theory to derive the following expression for $C_D$ at $\alpha=0$.
\begin{center}
$C_D=\frac{4}{\sqrt{M^2_{\infty}-1}}(\frac{t}{c})^2$
\end{center}

%\textcolor{blue}{Solution:\\ The drag on the diamond airfoil with unit span is $$D=(p_1\sin\epsilon+p_3\sin\epsilon-p_4\sin\epsilon-p_2\sin\epsilon)\times l\times 1=(p_1+p_3-p_4-p_2)\frac{t}{2}$$ with $l$ being the length of one side of the wing.\\ The drag coefficient is then $$C_D=\frac{D}{\frac{1}{2}r\rho_\infty V^2_\infty S}=\frac{(p_1+p_3-p_4-p_2)\frac{t}{2}}{\frac{1}{2}r\rho_\infty V^2_\infty c\times 1}=(C_{p_1}+C_{p_3}-C_{p_4}-C_{p_2})\frac{t}{2c}$$ Recall with linearised flow, we have $$C_{p_1}=\frac{2\epsilon}{\sqrt{M^2_\infty-1}},\quad C_{p_2}=-\frac{2\epsilon}{\sqrt{M^2_\infty-1}}, \quad C_{p_3}=\frac{2\epsilon}{\sqrt{M^2_\infty-1}}\quad and \quad C_{p_4}=-\frac{2\epsilon}{\sqrt{M^2_\infty-1}}$$ Thus, $$C_D=\frac{4\epsilon}{\sqrt{M^2_\infty-1}}\frac{t}{c}$$ From the geometrical relation, we know that $\epsilon=\tan\epsilon=\frac{t}{c}$, hence we have $$C_D=\frac{4}{\sqrt{M^2_\infty-1}}(\frac{t}{c})^2$$}

%9.8 Supersonic linearized theory predicts that, for a thin airfoil of arbitrary shape and thickness at angle of attack $\alpha$, $C_L=4\alpha/\sqrt{M^2_{\infty}-1}$, independent of the shape and thickness. Prove this.
%
%Solution:\\
%
%9.9 Repeat Problem 4.17, except using linearized theory. Plot the linearized results on top of the same graphs produced for Prob 4.17 in order to assess the differences between linear theory (which is approximate) and shock-expansion theory (which is exact). From this comparison, over what angle of attack range would you feel comfortable in applying linear theory?
%
%Solution:\\
%
%9.10 Linear supersonic theory predicts that the curve of wave drag versus Mach number has a minimum point at a certain value of $M_{\infty}>1$.\\
%(a) Calculate this value of $M_{\infty}$.\\
%(b) Does it make physical sense for the wave drag to have a minimum value at some supersonic value of $M_{\infty}$ above 1? Explain. What does this say about the validity of linear theory for certain Mach number ranges?

%Solution:\\
\vspace{10cm}

9.11 At $\alpha=0^o$, the minimum pressure coefficient for a NACA 0009 airfoil in low-speed flow is -0.25. Calculate the critical Mach number for this airfoil using\\
(a) The Prandtl-Glauert rule.\\
(b) The (more accurate) Karman-Tsien rule.

%\textcolor{blue}{Solution: The critical pressure coefficient is given by \begin {equation}C_{p_{cr}}=\frac{1}{\gamma M^2_\infty} [(\frac{1+\frac{\gamma-1}{2}M^2_\infty}{1+\frac{\gamma-1}{2}})^{\frac{\gamma}{\gamma-1}}-1]\end{equation} (a) The Prandtl-Glauert correction for an airfoil is given by \begin{equation}C_p=\frac{C_{p0}}{\sqrt{1-M^2_\infty}}\end{equation} When the flow reaches the critical Mach number, we have $C_p=C_{p_{cr}}$. Thus, \begin{equation} \frac{C_{p0}}{\sqrt{1-M^2_{cr}}}=\frac{1}{\gamma M^2_{cr}} [(\frac{1+\frac{\gamma-1}{2}M^2_{cr}}{1+\frac{\gamma-1}{2}})^{\frac{\gamma}{\gamma-1}}-1]\end{equation} (b) The Karman-Tsien correction states \begin{equation}C_p=\frac{C_{p0}}{\sqrt{1-M^2_\infty}+\frac{M^2_\infty}{\sqrt{1+M^2_\infty}}\frac{C_{p0}}{2}}\end{equation} Let equation (4) equals to equation (1), we have \begin{equation} \frac{1}{\gamma M^2_{cr}} [(\frac{1+\frac{\gamma-1}{2}M^2_{cr}}{1+\frac{\gamma-1}{2}})^{\frac{\gamma}{\gamma-1}}-1]=\frac{C_{p0}}{\sqrt{1-M^2_{cr}}+\frac{M^2_{cr}}{\sqrt{1+M^2_{cr}}}\frac{C_{p0}}{2}}\end{equation} Substitute $C_{p0}=-0.25$ into equation (3) and (5), solve the resultant equation with $M_{cr}$, one can have $M_{cr}=0.752$ for Prandtl-Glauert rule and $M_cr=0.778$ for the Karman-Tsien rule.  Equations (3) and (5) are solved graphically. See the graph below.}
%\begin{center}
%\includegraphics[width=0.5\textwidth] {9.11}
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