Chapter 1, Problem 1.4P

Consider an infinitely thin flat plate with a 1 m chord at an angle of attack of $\text{1}0°$ in a supersonic flow. The pressure and shear stress distributions on the upper and lower surfaces are given by $p_u=4\times {10}^4{\left(x-1\right)}^2+5.4\times {10}^4,p_l=2\times {10}^4{\left(x-1\right)}^2+1.73\times {10}^5,{\tau }_u=288x^{-0.2}$, and ${\tau }_l=731x^{-0.2}$, respectively, where x is the distance from the leading edge in meters and p and r are in newtons per square meter. Calculate the normal and axial forces, the lift and drag, moments about the leading edge. and moments about the quarter chord, all per unit span. Also, calculate the location of the center of pressure.

To determine

The normal, axial, lift and drag forces, moment about the leading and quarter chord and the centre pressure.

Answer to Problem 1.4P

Normal force ${\text{N}}^{\text{'}}\text{= 11}\text{.2}\times {\text{10}}^4{\rm N}$

Axial force ${\text{A}}^{\text{'}}\text{=}1274{\rm N}$

Lift force $\text{L= 11}\text{.01}\times {\text{10}}^4{\rm N}$

Drag force $\text{D =}\text{2}\text{.07}\times {\text{10}}^4{\rm N}$

Moment about the leading ${{\rm M}}_{LE}^{\text{'}}\text{= -5}\text{.78}\times {\text{10}}^4\text{Ν-m}$

Moment about quarter chord ${{\rm M}}_{c/4}^{\text{'}}\text{= -3}\text{.02}\times {\text{10}}^4\text{Ν-m}$

Centre of pressure $x_{cp}=0.516\text{m}$

Explanation of Solution

Given:

Pressure on upper surface ${\text{p}}_u\text{= 4}\times {10}^4{\left(x-1\right)}^2\text{+5}\text{.4}\times {10}^4{\text{Ν/m}}^2$

Pressure on lower surface ${\text{p}}_l\text{= 2}\times {10}^4{\left(x-1\right)}^2\text{+1}\text{.73}\times {10}^5{\text{Ν/m}}^2$

Shear stress on upper surface ${\tau }_u{\text{= 228x}}^{-0.2}{\text{Ν/m}}^2$

Shear stress on lower surface ${\tau }_l{\text{= 731x}}^{-0.2}{\text{Ν/m}}^2$

Length of chord $\text{c = 1 m}$

Angle of attack $\alpha {\text{ = 10}}^{°}$

Calculation:

The normal force per unit span is,

$N^{\prime }=-{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u+p_l\right)}\cos \theta dx+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u-{\tau }_l\right)\sin \theta }dy$

Since the plate is thin $\theta =0,$

${\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u-{\tau }_l\right)}\sin \theta dy=0$

Then, normal force per unit span is,

$\begin{array}{l}{N}^{\text{'}}=-{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u-p_l\right)}dx\\ N^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_l-p_u\right)}dx\\ N^{\text{'}}={\displaystyle \underset{0}{\overset{1}{\int }}\left[2\times {10}^4{\left(x-1\right)}^2+1.73\times {10}^5-\left[4\times {10}^4{\left(x-1\right)}^2+5.4\times {10}^4\right]\right]}dx\\ N^{\text{'}}={\displaystyle \underset{0}{\overset{1}{\int }}\left[2\times {10}^4{\left(x-1\right)}^2+1.73\times {10}^5-4\times {10}^4{\left(x-1\right)}^2-5.4\times {10}^4\right]}dx\\ N^{\text{'}}={\displaystyle \underset{0}{\overset{1}{\int }}\left[-2\times {10}^4\left(x^2+1-2x\right)+11.9\times {10}^4\right]dx}\end{array}$

$\begin{array}{l}{N}^{\text{'}}=-2\times {10}^4{\left[\frac{x^3}{3}+x-x^2\right]}_0^1+11.9\times {10}^4{\left[x\right]}_0^1\\ N^{\text{'}}=-\frac{2}{3}\times {10}^4+11.9\times {10}^4\\ N^{\text{'}}=11.2\times {10}^4{\rm N}\end{array}$

Now, axial force,

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

Since the plate is thin $\theta =0,$

${\displaystyle \underset{LE}{\overset{TE}{\int }}\left(-p_u+p_l\right)}\sin \theta dy=0$

Then, axial force per unit span is,

$\begin{array}{l}{A}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_l+{\tau }_u\right)}dx\\ A^{\text{'}}={\displaystyle \underset{0}{\overset{1}{\int }}\left[731x^{-0.2}+288x^{-0.2}\right]}dx\\ A^{\text{'}}={\left[1274x^{0.8}\right]}_0^1\\ A^{\text{'}}=1274{\rm N}\end{array}$

Lift and drag force:

Lift $L=N^{\text{'}}\cos \alpha -A^{\text{'}}\sin \alpha$

$\begin{array}{l}L=11.2\times {10}^4\cos {10}^{°}-1274\sin {10}^{°}\\ L=11.01\times {10}^4{\rm N}\end{array}$

Drag $D=N^{\text{'}}\sin \alpha +A^{\text{'}}\cos \alpha$

$\begin{array}{l}D=11.2\times {10}^4\sin {10}^{°}+1274\cos {10}^{°}\\ D=2.07\times {10}^4{\rm N}\end{array}$

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

$M_{LE}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u-p_l\right)}xdx-{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u-{\tau }_l\right)}xdy+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u-p_l\right)}ydy+{\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}ydy$

Here, p_u and p_l is the pressure upper and below the plate, respectively.

Since the plate is thin,

${\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u-p_l\right)}ydy=0$

Since shear stresses are also negligible, therefore,

${\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u-{\tau }_l\right)}xdy={\displaystyle \underset{LE}{\overset{TE}{\int }}\left({\tau }_u+{\tau }_l\right)}ydy=0$

Then, from above equation the moment about leading edge per unit span is,

$\begin{array}{l}{M}_{LE}^{\text{'}}={\displaystyle \underset{LE}{\overset{TE}{\int }}\left(p_u-p_l\right)}xdx\\ M_{LE}^{\text{'}}={\displaystyle \underset{0}{\overset{1}{\int }}\left[2\times {10}^4{\left(x-1\right)}^2-1.19\times {10}^5\right]}xdx\\ M_{LE}^{\text{'}}=2\times {10}^4{\left[\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}\right]}_0^1-\left[0.595\times {10}^5{\left(x\right)}_0^1\right]\\ M_{LE}^{\text{'}}=-5.78\times {10}^4\text{N-m}\end{array}$

Again, the moment about the quarter chord per unit span is,

$\begin{array}{l}{M}_{c/4}^{\text{'}}=M_{LE}^{\text{'}}+L^{\text{'}}\left(\frac{c}{4}\right)\\ M_{c/4}^{\text{'}}=-5.78\times {10}^4+11.01\times {10}^4\left(\frac{1}{4}\right)\\ M_{c/4}^{\text{'}}=-3.02\times {10}^{4}N/m\end{array}$

The equation of location of centre of pressure of the plate is,

$\begin{array}{l}{x}_{cp}=-\frac{{\text{M}}_{\text{LE}}^{\text{'}}}{{\text{N}}^{\text{'}}}\\ x_{cp}=-\frac{\left[-5.78\times {10}^4\right]}{11.2\times {10}^4}\\ x_{cp}=0.516\text{m}\end{array}$