Question 5.9: A fluid of constant density ρ flows over a solid body of arb...
A fluid of constant density \rho flows over a solid body of arbitrary shape with a uniform free stream velocity U_{\infty}, as shown in Fig. 5.11. The velocity distribution at the downstream of the body is given by \frac{u}{U_{\infty}}=\sin \left(\frac{\pi}{2} \frac{y}{a}\right) . The width of flow perpendicular to the plane of the figure is w. Assuming that the flow is symmetrical with respect to the centre line, find the total drag force on the solid body exerted by the fluid.
Choose a CV ABCD as shown in Fig. 5.11(a), such that AD and BC are streamlines, i.e., there are no fluxes across these two surfaces.
Writing the conservation of mass for the CV results in
0=0+\int_{C S} \rho(\vec{V} \cdot \hat{n}) d A0=0+\int_{A B} \rho(\vec{V} \cdot \hat{n}) d A+\int_{C D} \rho(\vec{V} \cdot \hat{n}) d A
0=0+2 \rho \int_{0}^{h}\left(-U_{\infty}\right) w d y+\int_{0}^{a} u w d y
U_{\infty} h=\int_{0}^{a} u d y (5.25)
Applying the x component of the linear momentum conservation equation, we have
\sum F_{x}=\int_{A B} \rho u(\vec{V} . \hat{n}) d A+\int_{C D} \rho u(\vec{V} . \hat{n}) d A=2 \rho \int_{0}^{h} U_{\infty}\left(-U_{\infty}\right) w d y+2 \rho \int_{0}^{a} u(u) w d y
=-2 \rho U_{\infty}^{2} w h+2 \rho w \int_{0}^{a} u^{2} d y
=2 \rho w\left[\int_{0}^{a} u^{2} d y-U_{\infty} \int_{0}^{a} u d y\right] (5.26)
Substituting, \frac{u}{U_{\infty}}=\sin \left(\frac{\pi}{2} \frac{y}{a}\right) in Eq. (5.26), and letting \theta=\frac{\pi}{2} \frac{y}{a} , we obtain
\sum F_{X}=2 \rho w U_{\infty}^{2} \cdot \frac{2 a}{\pi}\left[\int_{0}^{\frac{\pi}{2}} \sin ^{2} \theta d \theta-\int_{0}^{\frac{\pi}{2}} \sin \theta d \theta\right]=\frac{4 a \rho w U_{\infty}^{2}}{\pi}\left[\frac{\theta}{2}-\frac{\sin 2 \theta}{4}+\cos \theta\right]_{0}^{\frac{\pi}{2}}
=\frac{4 a \rho w U_{\infty}^{2}}{\pi}\left[\frac{\pi}{4}-1\right]
This is the force exerted by the solid body on the fluid. By Newton’s third law, the drag force exerted by the fluid on the solid body will be \frac{4 a \rho w U_{\infty}^{2}}{\pi}\left[1-\frac{\pi}{4}\right] .
