Blasius Flow Past a Flat Plate

Abstract

After a pastoral interlude revisiting singular perturbation theory by determining higher order terms in the Method of Matched Asymptotic Expansions and comparing these results with those obtained by the method of multiple scales for the constant coefficient second-order ordinary differential equation treated in Chapter , steady-state Blasius boundary-layer flow of a viscous fluid streaming uniformly past a flat plate is investigated. This is accomplished by applying singular perturbation theory techniques to the governing partial differential equation satisfied by the relevant stream function governing that flow. The solution of the resulting boundary-layer equation to lowest order requires a similarity solution argument. When the drag on the plate is calculated for that lowest order solution it is nonzero resolving D’Alembert’s paradox in this instance. In order to make an interpretation of these results the second order term in the outer free stream solution is calculated and compared with that deduced for inviscid flow past an effective body consisting of the flat plate plus the boundary layer. The problems involve two extensions: The first extending the singular perturbation techniques developed in the pastoral interlude for a constant coefficient example to a variable coefficient one and the second extending the Blasius flow methodology employed in the chapter for a uniform stream flow situation to a variable flow one.

Problems

14.1.

Consider the following singularly perturbed boundary value problem for $y = y(x;\varepsilon )$:

$$\begin{aligned} \varepsilon \frac{d^2y}{dx^2} + (1+x)\frac{dy}{dx} + y = 0, \, 0< x< 1, \, 0< \varepsilon<< 1; \end{aligned}$$

$$\begin{aligned} y(0;\varepsilon ) = 0, \, y(1;\varepsilon ) = 1. \end{aligned}$$

• (a) Given that the boundary layer occurs at $x = 0$ and has thickness $\delta (\varepsilon ) = \varepsilon$, find the first two terms in the outer and inner expansions for $y(x;\varepsilon )$ employing the intermediate limit technique of matched asymptotic expansions designating them by

  $$\begin{aligned} y_0(x) + \varepsilon y_1(x) \text { and } Y_0(\xi ) + \varepsilon Y_1(\xi ) \text { where } \xi = \frac{x}{\varepsilon }, \end{aligned}$$

  respectively.

• (b) Use these results to construct a two-term uniformly valid additive composite defined by

  $$\begin{aligned} y_u^{(1)}\left( x,\frac{x}{\varepsilon };\varepsilon \right) = y_0(x) + \varepsilon y_1(x) + Y_0\left( \frac{x}{\varepsilon }\right) + \varepsilon Y_1\left( \frac{x}{\varepsilon }\right) - y_0(0) - y_0^\prime (0)x - \varepsilon y_1(0) \end{aligned}$$

  and represent it in the form

  $$\begin{aligned} y_u^{(1)}\left( x,\frac{x}{\varepsilon };\varepsilon \right) = y_{u_0}^{(1)}\left( x,\frac{x}{\varepsilon }\right) + \varepsilon y_{u_1}^{(1)}\left( x,\frac{x}{\varepsilon }\right) . \end{aligned}$$

  Compare $y_{u_0}^{(1)}$ to the one-term uniformly valid additive composite

  $$\begin{aligned} y_u^{(0)}\left( x,\frac{x}{\varepsilon }\right) = y_0(x) + \varepsilon Y_0\left( x,\frac{x}{\varepsilon }\right) - y_0(0) \end{aligned}$$

  and note in this context that the term $\varepsilon \xi ^2e^{-\xi } = (x^2/\varepsilon )e^{-x/\varepsilon }$ belongs in $y_{u_0}^{(1)}$ rather than in $\varepsilon y_{u_1}^{(1)}$.

• (c) Find a one-term uniformly valid asymptotic representation for $y(x;\varepsilon )$ by the following generalized method of multiple scales. Define

  $$\begin{aligned} y(x;\varepsilon ) = f(x,\zeta ;\varepsilon ) = f_0(x,\zeta ) + \varepsilon f_1(x,\zeta ) + O(\varepsilon ^2) \text { as } \varepsilon \rightarrow 0, \, \zeta = \frac{g(x)}{\varepsilon } > 0, \, 0< x < 1; \end{aligned}$$

  where g(x) is such that $g(x) \sim x$ as $x\rightarrow 0$ and the partial differential equation satisfied by $f_0$ with respect to $\zeta$ is of the same form as the ordinary differential equation satisfied by $Y_0$ with respect to $\xi$. Once g(x) has been determined, calculate $f_0(x,\zeta )$ by proceeding exactly as in the regular method of multiple scales, with $\zeta$ assuming the same role as $\xi$, and compare it for $\zeta = g(x)/\varepsilon$ to $y_{u_0}^{(1)}(x, x/\varepsilon )$ from part (b). Hint: $g(x) = x + x^2/2$ and observe that unlike the regular method of multiple scales the change of variables involves an extra term proportional to $g^{\prime \prime }(x)$. Further $y_{u_0}^{(1)}(x, x/\varepsilon )$ will bear the same partial sum relationship to $f_0(x,[x+x^2/2]/\varepsilon )$ as it did to $f_0(x, x/\varepsilon )$ in the regular method of multiple scales. That is $e^{-x^2/(2\varepsilon )} \sim 1-x^2/(2\varepsilon )$.

14.2.

Consider Blasius-type flow past an infinite flat plate where the velocity profile satisfies the free-stream condition

$$\begin{aligned} \varvec{v}^*= u^*\varvec{i} + v^*\varvec{j} \sim \frac{U_0}{\ell }(x^*\varvec{i} - y^*\varvec{j}) \text { as } y^*\rightarrow \infty , \, x^*> 0. \end{aligned}$$

This implies that the velocity components satisfy the far-field conditions

$$\begin{aligned} u^*\sim \frac{U_0x^*}{\ell }, \, v^*\sim -\frac{U_0y^*}{\ell } \text { as } y^*\rightarrow \infty , \, x^*> 0; \, \, u \sim x, \, v \sim -y \text { as } y \rightarrow \infty , \, x > 0; \end{aligned}$$

in dimensional and nondimensional variables, respectively, instead of (14.2.1d) and (14.2.2d) while all the other equations and boundary conditions of systems (14.2.1) and (14.2.2) remain the same.

• (a) Defining the stream function

  $$\begin{aligned} \psi = \psi (x, y;\varepsilon ) \in C^2[\{(x, y): x > 0, y \ge 0\}] \text { such that } u = \frac{\partial \psi }{\partial y} \text { and } v = -\frac{\partial \psi }{\partial x}, \end{aligned}$$

  conclude that (14.3.1d) is replaced by

  $$\begin{aligned} \psi (x, y;\varepsilon ) \sim xy \text { as } y \rightarrow \infty \end{aligned}$$

  while (14.3.1a, b, c) remain the same.

• (b) Deduce that the outer or free-stream solution to lowest order of (14.3.4) is then given by

  $$\begin{aligned} \psi _0(x, y) = \lim _{\varepsilon \rightarrow 0}\psi (x, y;\varepsilon ) = xy \text { for } y \text { in the free stream.} \end{aligned}$$

• (c) Introducing the boundary-layer variables of system (14.3.7) concludes that the lowest order inner or boundary-layer solution

  $$\begin{aligned} \varphi _0(x,\eta ) = \lim _{\varepsilon \rightarrow 0}\varphi (x,\eta ;\varepsilon ) \end{aligned}$$

  then satisfies the matching condition

  $$\begin{aligned} \varphi _{0_\eta }(x,\eta ) \sim \psi _{0_y}(x, 0) = x \text { as } \eta \rightarrow \infty \end{aligned}$$

  rather than (14.3.7c) while (14.3.7a, 14.3.7b) remain the same.

• (d) Hence conclude that (14.3.7a) in conjunction with the matching condition of part (c) implies

  $$\begin{aligned} \varphi _{0_{\eta \eta \eta }} + \varphi _{0_x}\varphi _{0_{\eta \eta }} - \varphi _{0_\eta }\varphi _{0_{x\eta }} = g(x) = -\psi _{0_y}(x, 0)\psi _{0_{yx}}(x, 0) = -x \end{aligned}$$

  rather than (14.3.8). Thus denoting $\varphi _0$ by $\varphi$ for ease of exposition the problem for it reduces to

  $$\begin{aligned} \varphi _{\eta \eta \eta } + \varphi _x\varphi _{\eta \eta } - \varphi _{\eta }\varphi _{x\eta } = -x; \, 0< \eta < \infty ; \end{aligned}$$
  $$\begin{aligned} \varphi (x, 0) = \varphi _\eta (x, 0) = 0; \, \varphi _\eta (x,\eta ) \sim x \text { as } \eta \rightarrow \infty . \end{aligned}$$

• (e) Consider the transformation

  $$\begin{aligned} x^\prime = \frac{x}{c}, \, \eta ^\prime = \eta , \, \varphi ^\prime = \frac{\varphi }{c} \text { for } c > 0. \end{aligned}$$

  Show that then the system of part (d) becomes for $\varphi ^\prime = \varphi ^\prime (x^\prime ,\eta ^\prime )$:

  $$\begin{aligned} \varphi _{\eta ^\prime \eta ^\prime \eta ^\prime }^\prime + \varphi _{x^\prime }^\prime \varphi _{\eta ^\prime \eta ^\prime }^\prime - \varphi _{\eta ^\prime }^\prime \varphi _{x^\prime \eta ^\prime }^\prime = -x^\prime ; \, 0< \eta ^\prime < \infty ; \end{aligned}$$
  $$\begin{aligned} \varphi ^\prime (x^\prime , 0) = \varphi _{\eta ^\prime }(x^\prime , 0) = 0; \, \varphi _{\eta ^\prime }^\prime (x^\prime ,\eta ^\prime ) \sim x^\prime \text { as } \eta ^\prime \rightarrow \infty . \end{aligned}$$

  In other words that system is invariant under this transformation.

• (f) Since the transformed system of part (e) is identical in form to the original system of part (d) this tells one that the original system has a similarity solution such that $\varphi /x$ must be a function of $\eta$ which will convert its partial differential equation into an ordinary differential equation. Hence assume

  $$\begin{aligned} \frac{\varphi (x,\eta )}{x} = f(\eta ) \text { or } \varphi (x,\eta ) = xf(\eta ) \end{aligned}$$

  and show that $f(\eta )$ then satisfies

  $$\begin{aligned} f^{\prime \prime \prime } + f f^{\prime \prime } - (f^{\prime })^2 + 1 = 0, \, 0< \eta < \infty ; \end{aligned}$$
  $$\begin{aligned} f(0) = f^\prime (0) = 0, \, f^\prime (\eta ) \sim 1 \text { as } \eta \rightarrow \infty . \end{aligned}$$

• (g) Noting that

  $$\begin{aligned} \psi (x, y;\varepsilon ) \sim \varepsilon ^{1/2}\varphi _0\left( x,\frac{y}{\varepsilon ^{1/2}}\right) = \varepsilon ^{1/2}xf\left( \frac{y}{\varepsilon ^{1/2}}\right) \text { as } \varepsilon \rightarrow 0 \text { for } y \text { in the boundary layer}, \end{aligned}$$

  where

  $$\begin{aligned} \varepsilon = \frac{1}{R}, \,\, f^{\prime \prime }(0) = 1.2326, \,\, \eta - f(\eta ) \sim \beta _1 = 0.6480 \text { as } \eta \rightarrow \infty ; \end{aligned}$$

  compute the following parameters of the boundary layer:

  $$\begin{aligned} \delta _1(x) = \int _{0}^{\infty }{\left[ 1 - \frac{u(x, y)}{U(x)}\right] \, dy} \equiv \frac{\beta _1}{\sqrt{R}} = \frac{0.6480}{\sqrt{R}} \text { with } U(x) = x; \end{aligned}$$
  $$\begin{aligned} C_D = \frac{1}{R}\left[ \frac{\partial u(x, 0)}{\partial y} + \frac{\partial v(x, 0)}{\partial x}\right] = \frac{1}{R}\frac{\partial u(x, 0)}{\partial y} = \frac{x f^{\prime \prime }(0)}{\sqrt{R}}; \end{aligned}$$
  $$\begin{aligned} D = \int _{0}^{1}{C_D\, dx} = \frac{0.6163}{\sqrt{R}}. \end{aligned}$$