# Show that for some $f$ it holds that $\dot{\gamma}_0 = \dfrac{U^2\rho}{\eta}f\bigg(\dfrac{Ux\rho}{\eta}\bigg)$

I'm not a physicist and I'm having some trouble understanding the following problem:

We model the ground by a horizontal flat plate (standing still) with the air or water flowing over it, assuming that the flow is uniform before it reaches the plate. We choose coordinates such that the plate is in the half plane $(x\geq 0, -\infty < y <\infty, z= 0)$. Since the $y$-coordinate is irrelevant, we shall omit it. In front of the plate edge, where $x\leq 0$, the flow is uniform with velocity $U$ in the positive $x$-direction. The fluid mass density $\rho$ with $[\rho] = ML^{-3}$ and viscosity $\eta$ with $[\eta] = ML^{-1}T^{-1}$. Close to the plate the friction between plate and flow decelerates the flow. This friction causes the flow velocity near the plate to drop. For increasing $x$, the boundary condition will approach the no-slip condition with the velocity at the plate exactly vanishing. For large $x$ values, the velocity profile will become independent of $x$ and approach a stationary profile, independent of $x$ and thus $t$. We want to know how fast this convergence takes place as a function of time $t$ and thus of position $x$. In the figure above the velocity profiles above the plate are sketched for different $x$-positions. Let the velocity of the flow be denoted by $(u,w)$ with $u$ the velocity in the $x$-direction and $w$ the velocity in the normal $z$-direction. The so-called shear rate is the variation of the horizontal velocity $u$ in the normal direction. This is commonly denoted as $\dot{\gamma}$: $$\dot{\gamma}(x,z) = \dfrac{\partial u}{\partial z}(x,z).$$ Its value at the plate is denoted as $$\dot{\gamma}_0(x) = \dot{\gamma}(x,0).$$ This quantity only depends on the distance $x$ from the edge of the plate, and it is this dependence that we want to investigate. In the figure an angle $\phi(x)$ is indicated. It is related to $\dot{\gamma}(x)$ via $\dot{\gamma}_0(x) = \tan\phi(x)$. Far from the plate, where $z$ is large, the flow is little influenced by the presence of the plate, and so there we may take $(u,w)= (U,0)$. In addition to the dependence on the distance $x$ from the edge, the shear rate also will depend on the viscosity, the velocity $U$, and the density $\rho$. In the steady state we can generally write $$\dot{\gamma}_0 =\dot{\gamma}_0(x;\eta,U,\rho).$$ In the following steps we want to find this relationship as precisely as possible.

a) Determine two dimensionless variables from the set $\dot{\gamma}_0, x, \eta, U$, and $\rho$.

b) Show that for some function $f$ it holds that $$\dot{\gamma}_0 = \dfrac{U^2\rho}{\eta}f\bigg(\dfrac{Ux\rho}{\eta}\bigg).$$

I think I found two dimensionless quantities: $x^* = x\dfrac{U\rho}{\eta}$ and $\eta^* = \eta\dfrac{1}{xU\rho}$. When I looked for these quantities I ignored $\dot{\gamma}_0$ because I don't understand how that function is defined.

Question: The value of the shear-rate at the plate, $\dot{\gamma}_0(x)$ is first given as a function that only depends on $x$. Later however, $\dot{\gamma}_0$ is said to depend on the viscosity, the velocity and the density as well and supposedly can be written as: $$\dot{\gamma}_0(x;\eta,U,\rho).$$

How does this make sense? Does it or does it not depend on $x$ alone? Furthermore, how would you solve $b$?

Thanks and happy easter.

Does it or does it not depend on $x$ alone?

When your text states that $\dot\gamma_0$ 'only depends on the distance $x$ from the edge of the plate', that implies 'from all the variables describing the fluid domain' (that is $x$, $y$ and $z$). This quantity is only defined on the plate, this is a boundary term as opposed to, say, velocity components $u$, $v$ which are defined at any point in the fluid volume. Of course, the explicit expression for $\dot\gamma_0$ could contain other quantities characterizing the problem as a whole: fluid density, viscosity etc. That is why the $x$ variable is separated by semicolon from the rest: it is the only spatial variable here.

I think I found two dimensionless quantities: $x^* = x\dfrac{U\rho}{\eta}$ and $\eta^* = \eta\dfrac{1}{xU\rho}$. When I looked for these quantities I ignored $\dot\gamma_0$ because I don't understand how that function is defined.

Your two dimensionless quantities $x^{*}$, $\eta^{*}$ are inverse of one another, so they are not independent. The five dimensionful quantities have dimensions in units of $M$, $L$ and $T$, so to construct the second independent dimensionless quantity (the first could be either of $x^{*}$, $\eta^{*}$) you must include $\dot\gamma_0$ ($5 = 3 + 2$).

Even if you do not understand definition of $\dot\gamma_0$ you could find its dimension: it is a derivative of velocity component w.r.t coordinate so its dimension is $$[\dot\gamma_0]=\left[\frac{\partial u}{\partial z}\right]=\frac{[u]}{[z]}=\frac{L T^{-1}}{L} =T^{-1}.$$ Incidentally, the (b) part of your question gives a hint on how to construct that second quantity: rewrite the equation $$\dot{\gamma}_0 = \dfrac{U^2\rho}{\eta}f\bigg(\dfrac{Ux\rho}{\eta}\bigg) \tag{*}$$ so that the right hand side contains only $f(x^{*})$. Since it is a dimensionless quantity the left hand side would be dimensionless and would be the required second quantity (let us call it $\Gamma$).

Edit: Since it seems to be not obvious, the answer for the second dimensionless quantity: $$\Gamma = \frac{\dot\gamma_0 \eta}{U^2 \rho} .$$

Now you have conjectured functional dependence: $$\dot{\gamma}_0 =\dot{\gamma}_0(x;\eta,U,\rho),$$ a dimensionless variable $x^{*}$ from arguments of r.h.s and another one, $\Gamma$ from $\dot\gamma_0$, $U$, $\rho$ and $\eta$. So this allows us to apply Buckingham $\pi$ theorem by stating that $\Gamma$ must be a function of $x^{*}$: $$\Gamma = f(x^{*}).$$

After simple transformation you would obtain the statement of your (b) part.

• Thanks for your reply! However, I don't really understand what you mean when you say "Even if you do not understand the definition of $\dot{\gamma}$ you could find its dimension: it is a derivative of velocity component w.r.t. coordinate so its dimension is $T^{-1}$." Does that mean that $[\dot{\gamma}_0] = \dfrac{\partial L/T}{\partial L}$? Apr 1, 2018 at 17:52
• Furthermore, I don't understand what you mean with $\Gamma$ from $\dot{\gamma}$ . What would $\Gamma$ be in $\Gamma = f(x^*)$? Apr 1, 2018 at 17:53
• @titusAdam: to your first comment: I have now written it explicitly. $\partial a/\partial b$ has the dimension of $a/b$, since derivative is just limit of ratios. Apr 1, 2018 at 19:18
• to second comment: $\Gamma$ is dimensionless quantity of the form $\dot\gamma_0 U^a \eta^b\rho^c$, where $a$, $b$ and $c$ you have to determine. The important part: there is no $x$ in $\Gamma$ and it is linear in $\dot\gamma_0$. Apr 1, 2018 at 19:24
• @titusAdam: anyway, I've written it explicitly. Apr 1, 2018 at 19:31