Chapter 7 Boundary layers

The first step is to develop a scaling estimate for each of the terms in these equations. Here our objective is to identify how each term in (7.1.2) scales with the parameters of the problem, viz., $\rho$, $\mu$, $U$ and $L$, and then balance the magnitudes of the dominant terms.

Each dependent and independent variable is assigned a constant magnitude in terms of the parameters governing the problem. For example, the magnitude of $u$ is taken to be $U$, expressed as $u\sim U$. Similarly, the scale for $x$ is taken to be $L$, i.e. $x\sim L$. We do not yet know the magnitudes for $v$, $y$ and $p$, so we define symbols $V$, $\delta$ and $P$, respectively, to represent them. The magnitudes of partial derivatives are estimated using ratios of the variables in the derivative, e.g., ${\displaystyle \frac{\partial u}{\partial x}}\sim {\displaystyle \frac{U}{L}}$. Our task is to express these undefined symbols in terms of the known parameters using dominant balances.

As an illustration, consider the mass conservation equation (7.1a). The magnitude of the two terms in this equation are presented in the second row of Table 7.1. Balancing the magnitudes of the two terms yields the magnitude for $v$ as

${\displaystyle \frac{U}{L}}={\displaystyle \frac{V}{\delta }}⟹V={\displaystyle \frac{U\delta }{L}}.$

(7.3)

Note that this is consistent with our estimates in the simple picture in §7.1.1.

Table 7.1 also shows the magnitude estimates for the momentum conservation equations in (7.1.2). Using (7.3), it is evident that in (7.1b)

${\displaystyle \frac{\rho U^2}{L}}={\displaystyle \frac{\rho UV}{\delta }},$

(7.4)

implying that both the terms $\rho u{\displaystyle \frac{\partial u}{\partial x}}$ and $\rho v{\displaystyle \frac{\partial u}{\partial y}}$ are of equal magnitude. Furthermore, given that the pressure field is established to enforce incompressibility, its magnitude is set up to balance the strongest term in the equations. This implies that the $\rho u{\displaystyle \frac{\partial u}{\partial x}}$ is of a magnitude comparable to ${\displaystyle \frac{\partial p}{\partial x}}$, or

${\displaystyle \frac{P}{L}}={\displaystyle \frac{\rho U^2}{L}}⟹P=\rho U^2.$

(7.5)

Between the two viscous terms, $\mu {\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}$ and $\mu {\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}$, which scale as $\mu U/L^2$ and $\mu U/{\delta }^2$, respectively, because $\delta \ll L$ it is evident that, $\mu {\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}\ll \mu {\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}$. Thus, the term $\mu {\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}$ is perhaps negligible compared to at least one other term in (7.1b). Neglecting the $\mu {\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}$ also will render the dominant balance equivalent to that of an inviscid fluid, which is contrary to our original intention of studying the influence of viscosity. Therefore, balancing the dominant of the two viscous terms $\mu {\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}$ with the inertial terms yields

${\displaystyle \frac{\rho U^2}{L}}={\displaystyle \frac{\mu U}{{\delta }^2}}⟹\delta =\sqrt{{\displaystyle \frac{\nu L}{U}}}.$

(7.6)

This expression is eerily similar to one derived in §7.1.1, except that $x$ is replaced by $L$. We can now verify our assumption

${\displaystyle \frac{\delta }{L}}\ll 1⟺\sqrt{{\displaystyle \frac{\nu }{UL}}}={\text{Re}}^{-1/2}\ll 1⟺\text{Re}\gg 1,$

(7.7)

where we defne the Reynolds number $\text{Re}=UL/\nu$.

We now turn our attention to (7.1c). Substituting $V$ from (7.3) and $P$ from (7.5) into (7.1c) yields the magnitude of every term in (7.1c). In light of the assumption $\delta \ll L$ yields the most dominant term in that equation to be ${\displaystyle \frac{\partial p}{\partial y}}$, which scales as ${\displaystyle \frac{\rho U^2}{L}}\times \sqrt{\text{Re}}$, while all the other terms are negligible in comparison. See Table 7.1 for details.

With these estimates, (7.1.2) simplify to

	${\displaystyle \frac{\partial u}{\partial x}}+{\displaystyle \frac{\partial v}{\partial y}}$	$=0,$		(7.8a)
	$\rho \left(u{\displaystyle \frac{\partial u}{\partial x}}+v{\displaystyle \frac{\partial u}{\partial y}}\right)$	$=-{\displaystyle \frac{\partial p}{\partial x}}+\mu {\displaystyle \frac{{\partial }^{2}u}{\partial y^2}},$		(7.8b)
	$0$	$=-{\displaystyle \frac{\partial p}{\partial y}},$		(7.8c)

which are subject to boundary conditions given by (7.1.2). The set of equations (7.1.2) are known as the Prandtl boundary-layer equations after Ludwig Prandtl, who first identified the presence of boundary layers in high Reynolds number flow of viscous fluids, and the approximation leading to them is called the Prandtl boundary-layer approximation.

For flow past a flat plate, the flow is uniform outside the boundary layer, and consequently the pressure is uniform there. Combined with (7.8c), this implies that the pressure is uniform everywhere and ${\displaystyle \frac{\partial p}{\partial x}}=0$. Thus, for the case of flow past a flat plate, (7.1.2) simplify further to

	${\displaystyle \frac{\partial u}{\partial x}}+{\displaystyle \frac{\partial v}{\partial y}}$	$=0,$		(7.9a)
	$u{\displaystyle \frac{\partial u}{\partial x}}+v{\displaystyle \frac{\partial u}{\partial y}}$	$=\nu {\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}.$		(7.9b)

The scale invariance of the geometry of the flow past a long plate, approximated as semi-infinite, is evident. Upon zooming out, the plate appears semi-infinite and the uniform flow remains uniform. The governing dynamics in the boundary layer are also scale invariant. To see this, rescale the variables in (7.1.2) and (7.1.2) as

$u=\alpha \tilde{u},v=\beta \tilde{v},x=\gamma \tilde{x}\text{and}y=\delta \tilde{y},$

(7.10)

where $\alpha$, $\beta$, $\gamma$ and $\delta$ are positive transformation factors. The rescaled equations (7.1.2) and (7.1.2) are

	${\displaystyle \frac{\alpha }{\gamma }}{\displaystyle \frac{\partial \tilde{u}}{\partial \tilde{x}}}+{\displaystyle \frac{\beta }{\delta }}{\displaystyle \frac{\partial \tilde{v}}{\partial \tilde{y}}}$	$=0,$		(7.11a)
	${\displaystyle \frac{{\alpha }^2}{\gamma }}\tilde{u}{\displaystyle \frac{\partial \tilde{u}}{\partial \tilde{x}}}+{\displaystyle \frac{\alpha \beta }{\delta }}\tilde{v}{\displaystyle \frac{\partial \tilde{u}}{\partial \tilde{y}}}$	$={\displaystyle \frac{\alpha }{{\delta }^2}}\nu {\displaystyle \frac{{\partial }^2\tilde{u}}{\partial {\tilde{y}}^2}},$		(7.11b)

and

	$\alpha \tilde{u}(\alpha \tilde{x}\ge 0,\delta \tilde{y}=0)=0,$	$⟹$	$\tilde{u}(\tilde{x}\ge 0,\tilde{y}=0)=0$			(7.12a)
	$\beta \tilde{v}(\alpha \tilde{x}\ge 0,\delta \tilde{y}=0)=0,$	$⟹$	$\tilde{v}(\tilde{x}\ge 0,\tilde{y}=0)=0$			(7.12b)
	$\alpha \tilde{u}(\alpha \tilde{x},\delta \tilde{y}\to \mathrm{\infty })=U.$	$⟹$	$\tilde{u}(\tilde{x},\tilde{y}\to \mathrm{\infty })={\displaystyle \frac{U}{\alpha }}$			(7.12c)

Equation (7.1.3) and (7.1.3) are identical to (7.1.2) and (7.1.2) if

${\displaystyle \frac{\beta }{\delta }}={\displaystyle \frac{\alpha }{\gamma }},{\displaystyle \frac{{\alpha }^2}{\gamma }}={\displaystyle \frac{\alpha }{{\delta }^2}},\text{and}\alpha =1.$

(7.13)

These relations between the scaling factors are satisfied if $\alpha =1$, $\delta =\sqrt{\gamma }$ and $\beta =1/\sqrt{\gamma }$, irrespective of the value of $\gamma$. This demonstrates the invariance of the governing equations, and thereby the flow, to the (anisotropic) scaling of the independent and dependent variables.

The scale invariance implies that the flow velocity at points related by the transformation is also related. In particular, If the functional form of $u(x,y)$ is given by $u(x,y)=F(x,y)$, then an indetical form descrives the velocity in the transformed coordinates, i.e., $\tilde{u}(\tilde{x},\tilde{y})=F(\tilde{x},\tilde{y})$. Given the relation between $u$ and $\tilde{u}$ from (7.10),

$F(x,y)=\alpha F(\tilde{x},\tilde{y})=\alpha F({\displaystyle \frac{x}{\gamma }},{\displaystyle \frac{y}{\delta }}).$

(7.14)

Expressing $\alpha$ and $\delta$ in terms of $\gamma$ and $F$ in terms of $u$ yields

$u(x,y)=u({\displaystyle \frac{x}{\gamma }},{\displaystyle \frac{y}{\sqrt{\gamma }}}).$

(7.15)

Choosing $\gamma =x$ reduces this relation to a form useful for our interpretation as

$u(x,y)=u(1,{\displaystyle \frac{y}{\sqrt{x}}}).$

(7.16)

This equation states that the profile of $u$ versus $y$ at any location $x$ is related to that at $x=1$ (arbitrary units) by a simple rescaling of $y$ by $\sqrt{x}$. Similarly,

$v(x,y)={\displaystyle \frac{1}{\sqrt{x}}}v(1,{\displaystyle \frac{y}{\sqrt{x}}}).$

(7.17)

In this way, we expect the velocity profile to not be independently dependent on $x$ and $y$, but merely on the combination $y/\sqrt{x}$ alone.

Once the possibility of self-similarity is established, the following approach can be used to simplify the governing equations. First, a precursory method incorporates the dependence of the dimensional parameters into the scaling invariance. At this stage, we simply replace derivatives in (7.1.2) with ratios, and replace the dependent and independent variables with the scaling estimates in terms of known problem parameters, very similar to scaling estimates in §7.1.2. In this analysis, armed by the knowledge of scale invariance, we recognize that $x$ itself represents the scale for the independent variable $x$, and $\delta (x)$ for $y$. The scales for $u$ and $v$ are $U$ and $U\delta (x)/x$, respectively, in view of (7.9a). The dominant balance between terms in (7.9b) is

${\displaystyle \frac{U^2}{x}}=2\nu {\displaystyle \frac{U}{\delta {(x)}^2}}⟹\delta (x)=\sqrt{{\displaystyle \frac{2\nu x}{U}}},$

(7.18)

in agreement with the intuition presented in §7.1.1, except for the factor of 2. (The explanation for this factor being an algebraic convenience will appear soon.) This approach also explains the appearance of $L$ in the expression for $\delta$ in (7.6) instead of $x$ as in (7.18). Because of the scale-invariance, the plate length $L$ is not the appropriate scale for $x$, but $x$ itself is. Apart from this difference, the balance of terms that led to (7.18) and (7.6) are identical. Equation (7.18) now incorporates the dimensional dependence on the parameters $\nu$ and $U$.

We proceed to determine the self-similar flow profile by incorporating the results (7.18) and (7.16) by asserting that

$u=U{f}^{\prime }(\xi ),\text{where}\xi ={\displaystyle \frac{y}{\delta (x)}}.$

(7.19)

Here the variables $\xi$ and $f(\xi )$ are dimensionless and ${}^{\prime }$ denotes differentiation with respect to the argument. There exist multiple equivalent formulations, which represent the same flow, from which we choose one. To facilitate the subsequent analysis, note that

${\displaystyle \frac{d\delta }{dx}}=\sqrt{{\displaystyle \frac{\nu }{2Ux}}}={\displaystyle \frac{\delta }{2x}}\text{so}{\displaystyle \frac{\partial \xi }{\partial x}}=-{\displaystyle \frac{y}{{\delta }^2}}{\delta }^{\prime }(x)=-{\displaystyle \frac{\xi }{2x}}\text{and}{\displaystyle \frac{\partial \xi }{\partial y}}={\displaystyle \frac{1}{\delta }}.$

(7.20)

Using (7.20),

	${\displaystyle \frac{\partial u}{\partial x}}$	$=U{f}^{\prime \prime }(\xi ){\displaystyle \frac{\partial \xi }{\partial x}}=-{\displaystyle \frac{U}{2x}}\xi f^{\prime \prime }(\xi ),$		(7.21a)
	${\displaystyle \frac{\partial u}{\partial y}}$	$=U{f}^{\prime \prime }(\xi ){\displaystyle \frac{\partial \xi }{\partial y}}={\displaystyle \frac{U}{\delta }}{f}^{\prime \prime }(\xi ),$		(7.21b)
	${\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}$	$={\displaystyle \frac{U}{{\delta }^2}}{f}^{\prime \prime \prime }(\xi ),$		(7.21c)
	${\displaystyle \frac{\partial v}{\partial y}}$	$={\displaystyle \frac{1}{\delta }}{\displaystyle \frac{\partial v}{\partial \xi }}.$		(7.21d)

Substitution into (7.9a) yields

${\displaystyle \frac{\partial v}{\partial \xi }}=U{\displaystyle \frac{\delta }{2x}}\xi f^{\prime \prime }(\xi ),⟹v=U{\displaystyle \frac{\delta }{2x}}\left(\xi f^{\prime }(\xi )-f(\xi )\right).$

(7.22)

Substituting (7.19) and (7.22) in (7.9b) yields

${\displaystyle \frac{U^2}{2x}}\left(-\xi f^{\prime }(\xi )f^{\prime \prime }(\xi )+\left(\xi f^{\prime }(\xi )-f(\xi )\right)f^{\prime \prime }(\xi )\right)=\nu {\displaystyle \frac{U}{{\delta }^2}}{f}^{\prime \prime \prime }(\xi ).$

(7.23)

This equation reveals the utility of the dominant balance using scaling. Note that in (7.18) we balanced the terms $U^2/x$ and $\nu U/{\delta }^2$, which are also the dimensional pre-factors of the terms in (7.23). The additional factor of 2 in (7.18) is simply to account for the factor of 2 that arises in (7.23) – this factor is not necessary for deriving the self-similar flow profile. The essential reason underlying the utility of the method to derive (7.18) is that, according to the chain rule applied in (7.20) and (7.1.3), differentiation with respect to $x$ yields a dimensional factor of $1/x$, and with respect to $y$ yields $1/\delta$. These were precisely the rules we used to derive (7.18).

In view of (7.18), equation (7.23) simplifies to

$f^{\prime \prime \prime }(\xi )+f(\xi )f^{\prime \prime }(\xi )=0.$

(7.24)

The boundary conditions (7.1.2) simplify to

	$f^{\prime }(\xi =0)=0,$		(7.25a)
	$f(\xi =0)=0,$		(7.25b)
	$f^{\prime }(\xi \to \mathrm{\infty })=1.$		(7.25c)

Equations (7.24-7.1.3) are called the Blasius equations for the self-similar profile. They constitute an ordinary differential boundary value problem, which cannot be solved in closed form analytically but can be easily solved numerically. The solution is shown in Figure 7.3 as the dashed curve, which agrees very well with the numerically computed profiles.