4. (20 points) One way of reducing the skin-friction drag is to use boundary-layer suction to delay the transition to turbulence. Shown in figure 2 is the flow over a porous flat plate with a constant wall-suction velocity, $V_w$. Far downstream, the boundary-layer will eventually stop growing and the velocity profile will not change. (a) (10 points) Far downstream, where the velocity profile does not change along the plate, establish the governing equations (continuity and momentum conservation equations) and boundary conditions. You have to simplify the equations with proper assumptions and scaling analysis. And solve them to find the streamwise velocity ($u$) profiles. (b) (5 points) Find the skin-friction coefficient, $C_f = frac{ au_w}{(0.5 ho u_infty^2)}$. (c) (5 points) Determine the $y$ location (let's say this location as $delta_{99}$) where the streamwise velocity is 99% of the free-stream velocity ($u_infty$).
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These equations are: u = u(x, t) p = p(x, t) We can solve these equations for u(x, t) using the Laplace transforms: u(x, t) = Laplace(u(x, 0), u(x, t)) p(x, t) = Laplace(p(x, 0), p(x, t)) Show more…
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Sri K.
Numerical results of the Blasius solution to the Prandtl boundary-layer equations are presented in Table $9.1 .$ Consider steady, incompressible flow of standard air over a flat plate at freestream speed $U=5 \mathrm{m} / \mathrm{s}$. At $x=20 \mathrm{cm},$ estimate the distance from the surface at which $u=0.95 U$. Evaluate the slope of the streamline through this point. Obtain an algebraic expression for the local skin friction, $\tau_{w}(x)$. Obtain an algebraic expression for the total skin friction drag force on the plate. Evaluate the momentum thickness at $L=1 \mathrm{m}$.
Suppose the radius at $A$ is $R$ and it decreases uniformaly to $r$ at $B$ where $S=\pi R^{2}$ and $s=\pi r^{2} .$ Assume also that the semi vectical angle at 0 is $\alpha$. Then $$ \frac{R}{L_{2}}=\frac{r}{L_{1}}=\frac{y}{x} $$ So $$ y=r+\frac{R-r}{L_{2}-L_{1}}\left(x-L_{1)}\right. $$ where $y$ is the radius at the point $P$ distant $x$ from the vertex $O$. Suppose the velocity with which the liquid flows out is $V$ at $A, v$ at $B$ and $u$ at $P$. Then by the equation of continuity $\pi R^{2} V=\pi r^{2} v=\pi y^{2} u$ The velocity $v$ of efflux is given by $$ v=\sqrt{2 g h} $$ and Bernoulli's theorem gives $$ p_{p}+\frac{1}{2} \rho u^{2}=p_{0}+\frac{1}{2} \rho v^{2} $$ where $p_{p}$ is the pressure at $P$ and $p_{0}$ is the atmospheric pressure which is the pressure just outside of $B$. The force on the nozzle tending to pull it out is then $F=\int\left(p_{\rho}-p_{0}\right) \sin \theta 2 \pi y d s$ We have subtracted $p_{0}$ which is the force due to atmosphenic pressure the factor $\sin \theta$ gives horizontal component of the force and $d s$ is the length of the element of nozzle surface, $d s=d x \sec \theta$ and $$ \tan \theta=\frac{R-r}{L_{2}-L_{1}} $$ Thus $$ F=\int_{L}^{L_{2}} \frac{1}{2}\left(v^{2}-u^{2}\right) \rho 2 \pi y \frac{R-r}{L_{2}-L_{1}} d x $$$$ \begin{aligned} &=\pi \rho \int_{r}^{R} v^{2}\left(1-\frac{r^{4}}{y^{4}}\right) y d y \\ &=\pi \rho v^{2} \frac{1}{2}\left(R^{2}-r^{2}+\frac{r^{4}}{R^{2}}-r^{2}\right)=\rho g h\left(\frac{\pi\left(R^{2}-r^{2}\right)^{2}}{R^{2}}\right) \end{aligned} $$ $=\rho g h(S-s)^{2} / S=6.02 \mathrm{~N}$ on putting the values. Note : If we try to calculate $F$ from the momentum change of the liquid flowing out wi will be wrong even as regards the sign of the force. There is of course the effect of pressure at $S$ and $s$ but quantitative derivation of $F$ fron Newton's law is difficult.
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