Question 10.3: For purposes of discussion, the flow field has been divided ......

Since the surface of region 1 is parallel to the free stream, the flow does not turn in going from the free-stream conditions (\infty) to region 1 . Thus, the properties in region 1 are the same as in the free stream. The pressure coefficient on the airfoil surface bounding region 1 is zero. Thus,

M_{1}=2.0 \quad \nu_{1}=26.380^{\circ} \quad \theta_{1}=0^{\circ} \quad C_{p 1}=0.0

Furthermore, since the flow is not decelerated in going to region 1, a Mach wave (and not a shock wave) is shown as generated at the leading edge of the upper surface. Since the Mach wave is of infinitesimal strength, it has no effect on the flow. However, for completeness, let us calculate the angle between the Mach wave and the free-stream direction. The Mach angle is

\mu_{a}=\sin ^{-1} \frac{1}{M_{\infty}}=30^{\circ}

Since the surface of the airfoil in region 2 “turns away” from the flow in region 1 , the flow accelerates isentropically in going from region 1 to region 2. To cross the left-running Mach waves dividing region 1 from region 2, we move along right-running characteristics. Therefore,

d \nu=-d \theta

Since the flow direction in region 2 is

\theta_{2}=-20^{\circ}

\nu_{2} is

\nu_{2}=\nu_{1}-\Delta \theta=46.380^{\circ}

Therefore,

M_{2}=2.83

To calculate the pressure coefficient for region 2 ,

C_{p 2}=\frac{p_{2}-p_{\infty}}{\frac{1}{2} \rho_{\infty} U_{\infty}^{2}}=\frac{p_{2}-p_{\infty}}{(\gamma / 2) p_{\infty} M_{\infty}^{2}}

Thus,

C_{p 2}=\left(\frac{p_{2}}{p_{\infty}}-1\right) \frac{2}{\gamma M_{\infty}^{2}}

Since the flow over the upper surface of the airfoil is isentropic,

p_{t \infty}=p_{t 1}=p_{t 2}

Therefore,

C_{p 2}=\left(\frac{p_{2}}{p_{t 2}} \frac{p_{t \infty}}{p_{\infty}}-1\right) \frac{2}{\gamma M_{\infty}^{2}}

Using Table 8.1. or equation (8.36),

\frac{p_{t 1}}{p}=\left(1+\frac{\gamma-1}{2} M^2\right)^{\gamma /(\gamma-1)}    (8.36)

to calculate the pressure ratios given the values for M_{\infty} and for M_{2},

C_{p 2}=\left(\frac{0.0352}{0.1278}-1\right) \frac{2}{1.4(4)}=-0.2588

The fluid particles passing from the free stream to region 3 are turned by the shock wave through an angle of 20^{\circ}. The shock wave decelerates the flow and the pressure in region 3 is relatively high. To calculate the pressure coefficient for region 3 , we must determine the pressure increase across the shock wave. Since we know that M_{\infty}=2.0 and \theta=20^{\circ}, we can use Fig. 8.12 \mathrm{~b} to find the value of C_{p 3} directly; it is 0.66 . As an alternative procedure, we can use Fig. 8.12 to find the shock-wave angle, \theta_{w}, which is equal to \theta_{w}=53.5^{\circ}. Therefore, as discussed in Section 8.5, we can use M_{\infty} \sin \theta (instead of M_{\infty} ), which is 1.608 , and the correlations of Table 8.3 to calculate the pressure increase across the oblique shock wave:

\frac{p_{3}}{p_{\infty}}=2.848

Thus,

C_{p 3}=\left(\frac{p_{3}}{p_{\infty}}-1\right) \frac{2}{\gamma M_{\infty}^{2}}=0.66

Using Fig. 8.12c, we find that M_{3}=1.20.

Having determined the flow in region 3 ,

M_{3}=1.20 \quad \nu_{3}=3.558^{\circ} \quad \theta_{3}=-20^{\circ}

one can determine the flow in region 4 using the Prandtl-Meyer relations. One crosses the right-running Mach waves dividing regions 3 and 4 on leftrunning characteristics. Thus,

d \nu=d \theta

Since \theta_{4}=0^{\circ}, d \theta=+20^{\circ} and

\nu_{4}=23.558^{\circ}

so

M_{4}=1.90

Note that because of the dissipative effect of the shock wave, the Mach number in region 4 (whose surface is parallel to the free stream) is less than the free-stream Mach number.

Whereas the flow from region 3 to region 4 is isentropic, and

p_{t 3}=p_{t 4}

the presence of the shock wave causes

p_{t 3}<p_{t \infty}

To calculate

\begin{aligned} C_{p 4} & =\left(\frac{p_{4}}{p_{\infty}}-1\right) \frac{2}{\gamma M_{\infty}^{2}} \\ & =\left(\frac{p_{4}}{p_{3}} \frac{p_{3}}{p_{\infty}}-1\right) \frac{2}{\gamma M_{\infty}^{2}} \end{aligned}

the ratio

\frac{p_{4}}{p_{3}}=\frac{p_{4}}{p_{t 4}} \frac{p_{t 3}}{p_{3}}

can be determined since both M_{3} and M_{4} are known. The ratio p_{3} / p_{\infty} has already been found to be 2.848 . Thus,

C_{p 4}=\left[\frac{0.1492}{0.4124}(2.848)-1.0\right] \frac{2}{1.4(4.0)}=0.0108

We can calculate the section lift coefficient using equation (10.24).

C_l=\frac{\sum p \cos \theta\left(0.5 c / \cos \delta_w\right)}{(\gamma / 2) p_{\infty} M_{\infty}^2 c}    (10.24a)

C_l=\frac{1}{2 \cos \delta_w} \Sigma C_p \cos \theta    (10.24b)

Similarly, using equation (10.25)

C_d=\frac{\Sigma p \sin \theta\left(0.5 c / \cos \delta_w\right)}{(\gamma / 2) p_{\infty} M_{\infty}^2 c}     (10.25a)
or
C_d=\frac{1}{2 \cos \delta_w} \Sigma C_p \sin \theta    (10.25b)

to calculate the section wave-drag coefficient,

C_{d}=\frac{1}{2 \cos 10^{\circ}}\left(C_{p 3} \sin 20^{\circ}-C_{p 2} \sin 20^{\circ}\right)

Thus,

C_{d}=0.1595

The lift/drag ratio in our example is

\frac{l}{d}=2.782

for this airfoil section.

In some cases, it is of interest to locate the leading and trailing Mach waves of the Prandtl-Meyer expansion fans at b and d. Thus, using the subscripts l and t to indicate leading and trailing Mach waves, respectively, we have

\begin{aligned} & \mu_{l b}=\sin ^{-1} \frac{1}{M_{1}}=30^{\circ} \\ & \mu_{t b}=\sin ^{-1} \frac{1}{M_{2}}=20.7^{\circ} \\ & \mu_{l d}=\sin ^{-1} \frac{1}{M_{3}}=56.4^{\circ} \\ & \mu_{t d}=\sin ^{-1} \frac{1}{M_{4}}=31.8^{\circ} \end{aligned}

Each Mach angle is shown in Fig. 10.5.

To calculate the pitching moment about the midchord point, we substitute the values we have found for the pressure coefficients into equation (10.27)

C_{m_{0.5 c}}=\left(-C_{p 1}+C_{p 2}\right. \left.+C_{p 3}-C_{p 4}\right) \frac{1}{8} +\left(C_{p 1}-C_{p 2}-C_{p 3}+C_{p 4}\right) \frac{\tan ^2 \delta_w}{8}       (10.27)

and get

C_{m_{0.5 c}}=0.04728