Question 6.7: A fully developed laminar incompressible flow between two fl...
A fully developed laminar incompressible flow between two flat plates with one plate moving with a uniform velocity U , with respect to other is known as Couette flow. In a Couette flow, the velocity u , at a point depends on its location y (measured perpendicularly from one of the plates), the distance of separation h , between the plates, the relative velocity U , between the plates, the pressures gradient \mathrm{d} p / \mathrm{d} x imposed on the flow, and the viscosity \mu of the fluid. Find a relation in dimensionless form to express u in terms of the independent variables as described above.
The Buckingham’s \pi theorem is used for this purpose. The variables describing a Couette flow are u, U, y h, \mathrm{~d} p / \mathrm{d} x \text { and } \mu . Therefore, m (the total no. of variables) = 6.
n (the number of fundamental dimensions in which the six variables are expressed) = 3 (M, L and T)
Hence no. of independent p terms is 6-3=3
To determine these \pi terms, U, h and \mu are taken as repeating variables. Then,
\pi_{1}=U^{a} h^{b} \mu^{c} u\pi_{2}=U^{a} h^{b} \mu^{c} y
\pi_{3}=U^{a} h^{b} \mu^{c} d p / d x
The above three equations can be expressed in terms of the fundamental dimensions of each variable as
\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right)^{c} \mathrm{LT}^{-1} (6.32)
\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right)^{c} \mathrm{~L} (6.33)
\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right)^{c} \mathrm{ML}^{-2} \mathrm{~T}^{-2} (6.34)
Equating the exponents of M, L and T on both sides of the above equations we get the following:
From Eq. (6.32): c = 0
a+b-c+1=0
-a-c-1=0
which give a = –1, b = 0 and c = 0
Therefore, \pi_{1}=\frac{u}{U}
From equation (6.33): c = 0
a+b-c+1=0
-c-a=0
Which give a = 0, b = –1 and c = 0
Therefore, \pi_{2}=\frac{y}{h}
It is known from one of the corolaries of the \pi theorem, as discussed earlier, that if any two physical quantities defining a problem have the same dimensions, the ratio of the quantities is a \pi term. Therefore, there is no need of evaluating the terms \pi_{1} \text { and } \pi_{2} through a routine application of \pi theorem as done here; instead they can be written
straight forward as \pi_{1}=u / U \text { and } \pi_{2}=y / h .
From Eq. (6.34)
c + 1 = 0
a+b-c-2=0
-a-c-2=0
Which give a = –1, b = 2 and c = –1
Therefore, \pi_{3}=\frac{h^{2}}{\mu U} \frac{\mathrm{d} p}{\mathrm{~d} x}
Hence, the governing relation amongst the different variables of a couette flow in dimensionless form is
f\left(\frac{u}{U}, \frac{y}{h}, \frac{h^{2}}{\mu U} \frac{\mathrm{d} p}{\mathrm{~d} x}\right)=0Or \frac{u}{U}=F\left(\frac{y}{h}, \frac{h^{2}}{\mu U} \frac{\mathrm{d} p}{\mathrm{~d} x}\right) (6.35)
It is interesting to note, in this context, that from the exact solution of Navier Stokes equation, the expression of velocity profile in case of a couette flow has been derived in Chapter 8 (Sec. 8.3.2) and is given by Eq. (6.34) as
\frac{u}{U}=y / h-\left(\frac{h^{2}}{2 \mu U} \frac{\mathrm{d} p}{\mathrm{~d} x}\right) \frac{y}{h}\left(1-\frac{y}{h}\right)However, \pi theorem can never determine this explicit functional form of the relation between the variables.