Find a general nondimensional relation for the fluid velocity induced by a moving plate in a Couette flow (cf. Fig. 9.7 of Chap. 9).
Find a general nondimensional relation for the fluid velocity induced by a moving plate in a Couette flow (cf. Fig. 9.7 of Chap. 9).
Following the five-step recipe, SCALE, let us first specify a general functional relationship for the velocity in terms of variables that we expect to influence it. For example, the velocity \nu_{x} in the fluid may depend on the position y between the plates as well as the velocity U_{0} of the moving plate and the gap distance h. Although we would also expect the fluid properties, viscosity and density, to likewise play a role, for simplicity let us consider only the following: Step 1: \nu_{x}=f(U_{0}, h, y). Step 2: Next, we must consider appropriate fundamental dimensions, which, in general, are L, T, M, and \Theta , but for this isothermal problem, length, time, and mass will suffice. Moreover, we must identify dimensions of each variable, namely^{3}
\left[\nu _{x}\right]=\frac{L}{T}=L^{1}T^{-1} M^{0} , \left[h\right]=L=L^{1}T^{0}M^{0},\left[U_{0}\right]=\frac{L}{T}=L^{1}T^{-1}M^{0} , \left[y\right]=L=L^{1} T^{0}M^{0}.
Step 3: This is the most important and indeed the most challenging step: assign appropriate scales. Because each variable in the basic equation depends only on length and time, we only need two scales. Clearly, a convenient and natural intrinsic length scale is the separation distance h between the plates. Picking a timescale is much different. Generally, one tries to find a quantity having dimensions of time based on variables in the list. In this case, we see that dividing a length by a velocity will yield a time. Hence, let
L_{s}=h and T_{s}=\left(\frac{h}{U_{0}} \right).
Step 4: Following the formula in the above recipe, we now list the computed non-dimensional Pi variables for each of our four variables in the original function:
^{3} Note that we use [x] to denote the dimension of x; this is not to be confused with the use of brackets to denote a matrix.
\pi _{1}=\frac{\nu _{x}}{(h)^{1}(h/U_{0})^{-1}(1)^{0}(1)^{0}}=\frac{\nu _{x}}{U_{0}}, \pi _{3}=\frac{h}{(h)^{1}(h/U_{0})^{0}}=\frac{h}{h}=1,\pi _{2}=\frac{U_{0}}{(h)^{1}(h/U_{0})^{-1}}=\frac{U_{0}}{U_{0}} =1, \pi _{4}=\frac{y}{(h)^{1}(h/U_{0})^{0}}=\frac{y}{h}.
Step 5: Therefore, the final step is to express our relation of interest,
x_{1}=f(x_{2},x_{3},x_{4})\Leftrightarrow \nu _{x}=f(U_{0},h,y)in terms of Pi-groups:
\pi _{1}=g(\pi _{2},\pi _{3},\pi _{4})\Leftrightarrow \frac{\nu _{x}}{U_{0}}=g\left(1,1,\frac{y}{h} \right).Hence, according to the Buckingham Pi Theorem, we merely need to relate two nondimensional parameters functionally, not four dimensional parameters:
\frac{\nu _{x}}{U_{0}}=g\left(\frac{y}{h} \right).Note: The values of unity in the function [e.g., g(1, 1)] simply imply constants in the general function and thus do not need to be written explicitly. This clearly reduces the experimental need. Rather than performing experiments wherein we measure velocities for multiple combinations of U_{0} and h at multiple values of y, we merely need to relate two nondimensional quantities for any y. This is yet another example where theory tells us what to measure and why (i.e., how to interpret the data). Indeed, if we look back to Example 9.2, we find that a Navier–Stokes solution revealed that \nu_{x}=U_{0}(y/h), which is recovered by the Buckingham Pi result if the function g is simply linear in y/h Clearly, this theorem can aid in the experimental identification of various functional rela-tions of interest.