Question 8.3: Consider the infinitesimally thin airfoil which has the shap......

For the free-stream flow,

\nu_{\infty}=28.000^{\circ} \frac{p_{\infty}}{p_{t 1}}=0.11653 \quad \theta_{\infty}=0^{\circ}

Since the turning angles are small, it will be assumed that both the acceleration of the flow over the upper surface and the deceleration of the flow over the lower surface are isentropic processes. Note that the expansion waves on the upper surface diverge as the flow accelerates, but the compression waves of the lower surface coalesce. Since the flow is isentropic, we can use equations (8.59a) and (8.59b).

\nu-\theta=R   (8.59a)

\nu+\theta=Q   (8.59b)

Furthermore, the stagnation pressure is constant throughout the flow field and equal to p_{t 1} (which is the value for the free-stream flow).

In going from the free-stream (region \infty in Fig. 8.9) to the first segment on the upper surface (region u a ), we move along a right-running characteristic to cross the left-running Mach wave shown in the figure. Thus,

\nu+\theta=Q

or

d \nu=-d \theta

so

\begin{aligned} \nu_{u a t} & =\nu_{\infty}-\left(\theta_{u a}-\theta_{\infty}\right) \\ & =28.000^{\circ}-\left(-1.145^{\circ}\right)=29.145^{\circ} \end{aligned}

and

M_{u a}=2.1018

Using Table 8.1, p_{\text {ua }} / p_{t 1}=0.1091

Similarly, in going from the free stream to the first segment on the lower surface (region la) we move along a left-running characteristic to cross the right-running Mach wave shown in the figure. Thus,

\nu-\theta=R

or

d \nu=d \theta

so

\begin{aligned} \nu_{l a} & =\nu_{\infty}+\left(\theta_{l a}-\theta_{\infty}\right) \\ & =28.000^{\circ}+\left(-1.145^{\circ}\right)=26.855^{\circ} \end{aligned}

Summarizing, table 1

Let us now calculate the lift coefficient and the drag coefficient for the airfoil. Note that we have not been given the free-stream pressure (or, equivalently, the altitude at which the airfoil is flying) or the chord length of the airfoil. But that is not critical, since we seek the force coefficients.

C_{l}=\frac{l}{\frac{1}{2} \rho_{\infty} U_{\infty}^{2} c}

Referring to equation (8.46),

\frac{0.5 \rho U^2}{p_{t 1}}=\frac{\frac{1}{2}(p / R T)(\gamma / \gamma) U^2}{p_{t 1}}=\frac{1}{2} \frac{\gamma p}{p_{t 1}} \frac{U^2}{\gamma R T}=\frac{\gamma M^2}{2} \frac{p}{p_{t 1}}     (8.46)

it is evident that, for a perfect gas,

q_{\infty}=\frac{1}{2} \rho_{\infty} U_{\infty}^{2}=\frac{\gamma}{2} p_{\infty} M_{\infty}^{2}    (8.60)

Thus,

C_{l}=\frac{l}{(\gamma / 2) p_{\infty} M_{\infty}^{2} c}     (8.61)

Referring again to Fig. 8.9, the incremental lift force acting on any segment (i.e., the i th segment) is

d l_{i}=\left(p_{l i}-p_{u i}\right) d s_{i} \cos \theta_{i}=\left(p_{l i}-p_{u i}\right) d x_{i}

Similarly, the incremental drag force for any segment is

d d_{i}=\left(p_{l i}-p_{u i}\right) d s_{i} \sin \theta_{i}=\left(p_{l i}-p_{u i}\right) d x_{i} \tan \theta_{i}

table 2

Finally,

\begin{aligned} C_{l} & =\frac{\sum d l_{i}}{(\gamma / 2) p_{\infty} M_{\infty}^{2} c}=\frac{0.6784 c}{0.7(4.24) c}=0.2286 \\ C_{d} & =\frac{\sum d d_{i}}{(\gamma / 2) p_{\infty} M_{\infty}^{2} c}=\frac{0.0896 c}{0.7(4.24) c}=0.0302 \\ \frac{l}{d} & =\frac{C_{l}}{C_{d}}=7.57 \end{aligned}