Consider steady, viscous, fully developed flow of a fluid through a circular pipe of internal diameter D. We know that the velocity profile forms a paraboloid about the pipe centre line, given by : $V=-C\left(r^2-\frac{D^2}{4}\right)$ m/s, where C is a constant. The rate of kinetic energy (in J/s) at the control surface A-B, as shown in the figure, is proportional to Dⁿ. The value of n is ________.

The correct answer is 8.

Problem Analysis:
We are given a steady, viscous, fully developed flow of a fluid through a circular pipe of internal diameter D (and radius R = D/2). The velocity profile in terms of radial coordinate r is parabolic and given by:

$V=-C(r^2-\frac{D^2}{4})$

where C is a constant. We need to find the power index n such that the rate of kinetic energy (kinetic energy flux) crossing the control surface A-B shown in the figure is proportional to Dⁿ.

Step 1: Express velocity V in terms of radius R
Since the pipe radius is $R=\frac{D}{2}$ , we can write:

$R^2=\frac{D^2}{4}$

Substituting this into the velocity profile:

$V=-C(r^2-R^2)=C(R^2-r^2)$

Step 2: Formulate the rate of kinetic energy (kinetic energy flux)
Consider an elemental circular ring of radius r and thickness dr on the cross-section A-B. The area of this elemental ring is:

$dA=2\pi rdr$

The mass flow rate through this ring is:

$d\dot{m}=\rho VdA=\rho V\left(2\pi rdr\right)$

The rate of kinetic energy (K.E.) associated with this mass flow is:

$d(K.E.)=\frac{1}{2}d\dot{m}{V}^2=\frac{1}{2}\left(\rho V2\pi rdr\right)V^2=\rho \pi V^{3}rdr$

Step 3: Integrate over the entire pipe cross-section
Integrating from r = 0 to r = R:

$K.E.={\int }_0^R\rho \pi V^{3}rdr$

Substituting the expression for V:

$K.E.={\int }_0^R\rho \pi {\left[C\right(R^2-r^2\left)\right]}^{3}rdr$

$K.E.=\mathrm{��}\pi C^3{\int }_0^R{(R^2-r^2)}^{3}rdr$

Step 4: Solve the integral
Using u-substitution, let:

$u=R^2-r^2$

Then:

$du=-2rdr\Rightarrow rdr=-\frac{du}{2}$

Changing the limits of integration:
When r = 0, u = R².
When r = R, u = 0.
Substituting these into the integral:

$K.E.=\rho \pi C^3{\int }_{R^2}^0{u}^3(-\frac{du}{2})$

Using the minus sign to flip the integration limits:

$K.E.=\frac{1}{2}\rho \pi C^3{\int }_0^{R^2}{u}^{3}du$

$K.E.=\frac{1}{2}\rho \pi C^3{\left[\frac{u^4}{4}\right]}_0^{R^2}=\frac{1}{2}\rho \pi C^3\left(\frac{R^8}{4}\right)=\frac{\rho \pi C^3}{8}{R}^8$

Step 5: Determine proportionality in terms of D
Substituting $R=\frac{D}{2}$ :

$K.E.=\frac{\rho \pi C^3}{8}{(\frac{D}{2}}^{})8=\frac{\rho \pi C^3}{2048}{D}^8$

This clearly shows that:

$K.E.\propto D^8$

Comparing this to the given relation $D^n$, we find:

$n=8$