Question

Axial velocity profile $u(r)$ for an axisymmetric flow through a circular tube of radius $R$ is given as, \[ \frac{u(r)}{U} = \left(1 - \frac{r}{R}\right)^{1/n} \] where $U$ is the centerline velocity. If $V$ refers to the area-averaged velocity (volume flow rate per unit area), then the ratio $V/U$ for $n = 1$ (rounded off to two decimal places) is $________________$.

Hint:

For average velocity, always use $V = Q/A$ with $Q = \int u \, dA$. In axisymmetric cases, switch to polar coordinates ($2\pi r dr$).

Answer 1

Correct Answer: 0.32

Step 1: Write given velocity profile.
For $n=1$: \[ u(r) = U \left(1 - \frac{r}{R}\right) \]
Step 2: Definition of average velocity.
The average velocity is: \[ V = \frac{\int_A u(r)\, dA}{A} \] For a circular pipe (axisymmetric flow): \[ V = \frac{\int_0^R u(r) \, 2\pi r \, dr}{\pi R^2} \]
Step 3: Substitute velocity expression.
\[ V = \frac{2}{R^2}\int_0^R U\left(1 - \frac{r}{R}\right) r \, dr \]
Step 4: Simplify integral.
\[ V = \frac{2U}{R^2} \int_0^R \left(r - \frac{r^2}{R}\right) dr \] \[ = \frac{2U}{R^2} \left[ \frac{r^2}{2} - \frac{r^3}{3R} \right]_0^R \] \[ = \frac{2U}{R^2} \left( \frac{R^2}{2} - \frac{R^2}{3} \right) \] \[ = \frac{2U}{R^2} . \frac{R^2}{6} = \frac{U}{3} \]

Step 5: Compute ratio.
\[ \frac{V}{U} = \frac{1}{3} = 0.33 \] Final Answer: \[ \boxed{0.33} \]