Chapter 1, Problem 1.2P

Starting with Equations $\left(1.7\right),\left(1.8\right)$, and $\left(1.11\right)$, derive in detail Equations $\left(1.15\right),\left(1.16\right)$, and $\left(1.17\right)$.

To determine

Derive the equation of force and moment coefficients with the help of given equations.

Explanation of Solution

Given:

The equation of normal force is,

${\text{N}}^{\text{'}}=-{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u\cos \theta +{\tau }_u\sin \theta \right)}d{s}_u+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_l\cos \theta -{\tau }_l\sin \theta \right)}d{s}_l$

The equation of axial force is,

${\text{A}}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left(-p_u\sin \theta +{\tau }_u\cos \theta \right)}d{s}_u+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_l\sin \theta +{\tau }_l\cos \theta \right)}d{s}_l$

The equation of the moment about leading edge per unit span is,

$\begin{array}{l}{M}_{LE}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left[\left(p_u\cos \theta +{\tau }_u\sin \theta \right)x-\left(p_u\sin \theta -{\tau }_u\cos \theta \right)y\right]}d{s}_u\\ +{\displaystyle \underset{LE}{\overset{TE}{\int }}\left[\left(-p_l\cos \theta +{\tau }_l\sin \theta \right)x+\left(p_l\sin \theta +{\tau }_l\cos \theta \right)y\right]}d{s}_l\end{array}$

Here, p_u and p_l is the pressure upper and below the surface of the body and ${\tau }_{u}and{\tau }_l$ is the shear stressupper and below the surface of the body respectively.

Calculation:

The calculation for coefficient of normal force (1.15) is as follows:

The equation of normal force is ${\text{N}}^{\text{'}}=-{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u\cos \theta +{\tau }_u\sin \theta \right)}d{s}_u+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_l\cos \theta -{\tau }_l\sin \theta \right)}d{s}_l$

Fig:

Nomenclature for the integration of pressure and shear stress distributions over a two-dimensional body surface.

From figure:

$\begin{array}{l}ds\cos \theta =dx\\ ds\sin \theta =-dy\end{array}$

Then, normal force per unit span is

$\begin{array}{l}{\text{N}}^{\text{'}}=-{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u-p_l\right)}dx+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}dy\\ {\text{N}}^{\text{'}}=-{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(\left(p_u-p_{\infty }\right)-\left(p_l-p_{\infty }\right)\right)}dx+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}dy\end{array}$

Now, divide by $q_{\infty }S=q_{\infty }c\left(1\right)$

$\frac{{\text{N}}^{\text{'}}}{q_{\infty }c}=-\frac{1}{c}{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(\left(\frac{p_u-p_{\infty }}{q_{\infty }}\right)-\left(\frac{p_l-p_{\infty }}{q_{\infty }}\right)\right)}dx+\frac{1}{c}{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(\frac{{\tau }_u+{\tau }_l}{q_{\infty }}\right)}dy$

Here,

Normal force coefficient $c_n=\frac{{\text{N}}^{\text{'}}}{q_{\infty }c}$

Pressure coefficient $c_p=\frac{p-p_{\infty }}{q_{\infty }}$

Skin friction coefficient $c_f=\frac{\tau }{q_{\infty }}$

Thus, the equation can be written as:

$\begin{array}{l}{c}_n=\frac{1}{c}{\displaystyle \underset{0}{\overset{c}{\int }}\left(c_{p_l}-c_{p_u}\right)}dx+\frac{1}{c}{\displaystyle \underset{0}{\overset{c}{\int }}\left(c_f{}_u+c_f{}_l\right)}dyor,\\ c_n=\frac{1}{c}\left[{\displaystyle \underset{0}{\overset{c}{\int }}\left(c_{p_l}-c_{p_u}\right)}dx+{\displaystyle \underset{0}{\overset{c}{\int }}\left(c_{f_u}+c_f{}_l\right)}dy\right]\mathrm{.................}(eq1.15)\end{array}$

Now the calculation for coefficient of axial force (1.16) is as follows:

The equation of axial force is ${\text{A}}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left(-p_u\sin \theta +{\tau }_u\cos \theta \right)}d{s}_u+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_l\sin \theta +{\tau }_l\cos \theta \right)}d{s}_l$

From figure:

$\begin{array}{l}ds\cos \theta =dx\\ ds\sin \theta =-dy\end{array}$

Then,

$\begin{array}{l}{\text{A}}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u-p_l\right)}dy+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}dx\\ {\text{A}}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left(\left(p_u-p_{\infty }\right)-\left(p_l-p_{\infty }\right)\right)}dy+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}dx\end{array}$

Divide by $q_{\infty }S=q_{\infty }c\left(1\right)$

$\frac{{\text{A}}^{\text{'}}}{q_{\infty }c}=\frac{1}{c}{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(\left(\frac{p_u-p_{\infty }}{q_{\infty }}\right)-\left(\frac{p_l-p_{\infty }}{q_{\infty }}\right)\right)}dy+\frac{1}{c}{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(\frac{{\tau }_u+{\tau }_l}{q_{\infty }}\right)}dx$

Here,

Axial force coefficient $c_a=\frac{{\text{A}}^{\text{'}}}{q_{\infty }c}$

Pressure coefficient $c_p=\frac{p-p_{\infty }}{q_{\infty }}$

Skin friction coefficient $c_f=\frac{\tau }{q_{\infty }}$

Then, the above equation can be written as:

$c_a=\frac{1}{c}\left[{\displaystyle \underset{0}{\overset{c}{\int }}\left(c_{p_u}-c_{p_l}\right)}dy+{\displaystyle \underset{0}{\overset{c}{\int }}\left(c_f{}_u+c_f{}_l\right)}dy\right]\mathrm{......................................}(eq1.16)$

And, the calculation for coefficient of moment force (1.17) is as follows:

The equation of the moment about leading edge per unit span is,

$\begin{array}{l}{M}_{LE}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u-p_l\right)}xdx-{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}xdy+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u-p_l\right)}ydy+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}ydx\\ M_{LE}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left[\left(p_u-p_{\infty }\right)-\left(p_l-p_{\infty }\right)\right]}xdx-{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}xdy+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left[\left(p_u-p_{\infty }\right)-\left(p_l-p_{\infty }\right)\right]}ydy+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}ydx\end{array}$

Divide by $q_{\infty }{c}^2$ :

$\begin{array}{l}\frac{M_{LE}^{\text{'}}}{q_{\infty }{c}^2}=\frac{1}{c^2}{\displaystyle \underset{LE}{\overset{TE}{\int }}\left[\left(\frac{p_u-p_{\infty }}{q_{\infty }}\right)-\left(\frac{p_l-p_{\infty }}{q_{\infty }}\right)\right]}xdx-\frac{1}{c^2}{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(\frac{{\tau }_u+{\tau }_l}{q_{\infty }}\right)}xdy\\ +\frac{1}{c^2}{\displaystyle \underset{LE}{\overset{TE}{\int }}\left[\left(\frac{p_u-p_{\infty }}{q_{\infty }}\right)-\left(\frac{p_l-p_{\infty }}{q_{\infty }}\right)\right]}ydy+\frac{1}{c^2}{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(\frac{{\tau }_u+{\tau }_l}{q_{\infty }}\right)}ydx\end{array}$

Here,

Coefficient of moment $c_m=\frac{{\text{M}}_{LE}^{\text{'}}}{q_{\infty }{c}^2}$

Pressure coefficient $c_p=\frac{p-p_{\infty }}{q_{\infty }}$

Skin friction coefficient $c_f=\frac{\tau }{q_{\infty }}$

Then, the above equation can be written as:

$\begin{array}{l}{c}_{M_{LE}^{\text{'}}}=\frac{1}{c^2}{\displaystyle \underset{0}{\overset{c}{\int }}\left[c_{p_u}-c_{p_l}\right]}xdx-\frac{1}{c^2}{\displaystyle \underset{0}{\overset{c}{\int }}\left(c_f{}_u+c_{f_l}\right)}xdy\\ +\frac{1}{c^2}{\displaystyle \underset{0}{\overset{c}{\int }}\left[c_{p_u}-c_{p_l}\right]}ydy+\frac{1}{c^2}{\displaystyle \underset{0}{\overset{c}{\int }}\left(c_f{}_u+c_{f_l}\right)}ydx\mathrm{.................}(eq1.17)\end{array}$