Question 12.1: Neglecting the effects of skin friction and using the modifi......

Because the sphere is a blunt configuration, the pressure forces are the principal component of the drag force at high Reynolds numbers. Thus, using the coordinate system shown in Fig. 12.10, the drag on the sphere due to the pressure is

D=\int p[2 \pi y(d s)] \sin \theta_b=\int p 2 \pi y d y

where the incremental surface area on which the pressure acts is [2 \pi y(d s)]; refer to the shaded region on the forebody in Fig. 12.10. As discussed in Chapter 5 , the net resultant force in any direction due to a constant pressure acting over a closed surface is zero. Thus, the pressure coefficient can be used in our expression for the drag:

D=q_{\infty} \int C_p 2 \pi  y  d y      (12.18)

Using modified Newtonian theory to define the pressure, as given by equation (12.17),

C_p=C_{p, 12} \sin ^2 \theta_b=C_{p, 12} \cos ^2 \phi    (12.17)

with the coordinates shown in Fig. 12.10,

C_p=C_{p, 12} \sin ^2 \theta_b=C_{p, 12}\left(\frac{d y}{d s}\right)^2=C_{p, t 2} \frac{\left(y^{\prime}\right)^2}{1+\left(y^{\prime}\right)^2}     (12.19)

where y^{\prime}=d y / d x. Combining equations (12.18) and (12.19), the expression for the drag becomes

D=q_{\infty} C_{p, 2} \int_0^R \frac{\left(y^{\prime}\right)^2}{1+\left(y^{\prime}\right)^2} 2 \pi y y^{\prime} d x    (12.20)

The limits of the integration are 0 \leq x \leq R, since, according to the Newtonian flow model, the pressure coefficient on the leeward side (i.e., the crosshatched region of the sphere in Fig. 12.10) is zero.

For the sphere,

y=\left(2 x R-x^2\right)^{0.5}

and

\frac{d y}{d x}=\frac{R-x}{\left(2 x R-x^2\right)^{0.5}}

Substituting these expressions into equation (12.20),

D=2 \pi q_{\infty} C_{p, t 2} \int_0^R \frac{\left[\frac{R-x}{\left(2 x R-x^2\right)^{0.5}}\right]^3}{1+\frac{(R-x)^2}{2 x R-x^2}}\left(2 x R-x^2\right)^{0.5} d x

Simplifying and integrating yields

\begin{aligned} D & =\frac{2 \pi q_{\infty} C_{p, t 2}}{R^2}\left[R^3 x-\frac{3}{2} x^2 R^2+x^3 R-\frac{x^4}{4}\right]_0^R \\ & =\frac{C_{p, t 2}}{2} q_{\infty} \pi R^2 \end{aligned}

Thus,

C_D=\frac{D}{q_{\infty} \pi R^2}=\frac{C_{p, t 2}}{2}     (12.21)