Chapter 4, Problem 4.9P

Starting with Equations $\left(\text{4}.\text{35}\right)$ and $\left(\text{4}.\text{43}\right)$, derive Equation $\left(\text{4}.\text{62}\right)$.

To determine

To derive the equation $c_{m,le}=-\frac{\pi }{2}\left(A_0+A_1-\frac{A_2}{2}\right)$ using the given equations.

Answer to Problem 4.9P

The moment coefficient about the leading edge $c_{m,le}=-\frac{\pi }{2}\left(A_0+A_1-\frac{A_2}{2}\right)$

Explanation of Solution

Given equation:

  $\text{γ}\left(\text{θ}\right){\text{=2V}}_{\infty }\left({\text{A}}_{\text{0}}\frac{\left(\text{1+cosθ}\right)}{\text{sinθ}}\text{+}{\displaystyle \sum _{\text{n=1}}^{\infty }{\text{A}}_{\text{n}}\text{sin nθ}}\right)$ and,

The total moment about the leading edge, $M_{LE}^{\text{'}}=-{\rho }_{\infty }{V}_{\infty }{\displaystyle \underset{0}{\overset{c}{\int }}\xi \gamma \left(\xi \right)}d\xi$

Calculation:

The moment coefficient about the leading edge can be written as,

  $\begin{array}{l}{c}_{m,le}=\frac{M_{LE}^{\text{'}}}{q_{\infty }{c}^2}or,\\ q_{\infty }{c}^2\times c_{m,le}=M_{LE}^{\text{'}}\end{array}$

Substitute the value of $M_{LE}^{\text{'}}$ in the above equation,

  $\begin{array}{l}{q}_{\infty }{c}^2\times c_{m,le}=-{\rho }_{\infty }{V}_{\infty }{\displaystyle \underset{0}{\overset{c}{\int }}\xi \gamma \left(\xi \right)}d\xi or,\\ \frac{{\rho }_{\infty }{V}_{\infty }^2{c}^2}{2}\times c_{m,le}=-{\rho }_{\infty }{V}_{\infty }{\displaystyle \underset{0}{\overset{c}{\int }}\xi \gamma \left(\xi \right)}d\xi \\ \frac{V_{\infty }{c}^2}{2}\times c_{m,le}=-{\displaystyle \underset{0}{\overset{c}{\int }}\xi \gamma \left(\xi \right)}d\xi \_\_\_\_\_\_(1)\end{array}$

Here, $\text{ξ =}\frac{\text{c}}{\text{2}}\left(\text{1-cosθ}\right),\text{dξ =}\frac{\text{c}}{\text{2}}\text{sinθ}\text{.dθ}\text{and,}$

  $\text{γ}\left(\text{θ}\right){\text{=2V}}_{\infty }\left({\text{A}}_{\text{0}}\frac{\left(\text{1+cosθ}\right)}{\text{sinθ}}\text{+}{\displaystyle \sum _{\text{n=1}}^{\infty }{\text{A}}_{\text{n}}\text{sin nθ}}\right)$

Substituting all these values in equation (1),

  $\begin{array}{l}\frac{V_{\infty }{c}^2}{2}\times c_{m,le}=-{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{\text{c}}{\text{2}}\left(\text{1-cosθ}\right){\text{.2V}}_{\infty }\left({\text{A}}_{\text{0}}\frac{\left(\text{1+cosθ}\right)}{\text{sinθ}}\text{+}{\displaystyle \sum _{\text{n=1}}^{\infty }{\text{A}}_{\text{n}}\text{sin nθ}}\right)}\frac{\text{c}}{\text{2}}\text{sinθ}\text{.dθ}\\ \frac{V_{\infty }{c}^2}{2}\times c_{m,le}=-\frac{{\text{c}}^2{V}_{\infty }}{\text{2}}{\displaystyle \underset{0}{\overset{\pi }{\int }}\left[{\text{A}}_{\text{0}}\frac{\left({\text{1- cos}}^2\text{θ}\right)}{\text{sinθ}}\text{+}{\displaystyle \sum _{\text{n=1}}^{\infty }{\text{A}}_{\text{n}}\left(\text{1-cosθ}\right)\text{sin nθ}}\right]}\text{ sinθ}\text{.dθ}\\ c_{m,le}={\displaystyle \underset{0}{\overset{\pi }{\int }}{\text{A}}_{\text{0}}\left({\text{1- cos}}^2\text{θ}\right)\text{dθ +}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle \sum _{\text{n=1}}^{\infty }{\text{A}}_{\text{n}}\left(\text{1-cosθ}\right)\text{sin nθ sinθ}\text{.dθ}}}}\end{array}$

By integrating the above equation, the equation obtained is as follows:

  $\begin{array}{l}{c}_{m,le}=-\frac{\text{π}}{\text{2}}{\text{A}}_{\text{0}}-\frac{\text{π}}{\text{2}}{\text{A}}_{\text{1}}+\frac{\text{π}}{\text{4}}{\text{A}}_{\text{2}}\\ c_{m,le}=-\frac{\text{π}}{\text{2}}\left({\text{A}}_{\text{0}}{\text{+A}}_{\text{1}}-\frac{{\text{A}}_{\text{2}}}{\text{2}}\right)\end{array}$