Question

A fluid flowing steadily in a circular pipe of radius \( R \) has a velocity that is everywhere parallel to the axis (centerline) of the pipe. The velocity distribution along the radial direction is \[ V_r = U \left( 1 - \frac{r^2}{R^2} \right) \] where \( r \) is the radial distance as measured from the pipe axis and \( U \) is the maximum velocity at \( r = 0 \). The average velocity of the fluid in the pipe is

• A. \( \frac{U}{2} \)
• B. \( \frac{U}{3} \)
• C. \( \frac{U}{4} \)
• D. \( \frac{5}{6} U \)

Hint:

To find the average velocity of a fluid in a pipe, integrate the velocity profile over the cross-sectional area of the pipe.

Answer 1

The Correct Option is A

To find the average velocity of the fluid, we need to calculate the average value of the velocity \( V_r \) over the entire cross-sectional area of the pipe. The average velocity is given by: \[ V_{\text{avg}} = \frac{1}{A} \int_0^R V_r(r) \, 2\pi r \, dr \] Where \( A = \pi R^2 \) is the area of the pipe cross-section. Substitute \( V_r = U \left( 1 - \frac{r^2}{R^2} \right) \) into the integral: \[ V_{\text{avg}} = \frac{1}{\pi R^2} \int_0^R U \left( 1 - \frac{r^2}{R^2} \right) 2\pi r \, dr \] Simplifying the expression: \[ V_{\text{avg}} = \frac{2U}{R^2} \int_0^R \left( r - \frac{r^3}{R^2} \right) dr \] Now, integrate: \[ V_{\text{avg}} = \frac{2U}{R^2} \left[ \frac{r^2}{2} - \frac{r^4}{4R^2} \right]_0^R \] \[ V_{\text{avg}} = \frac{2U}{R^2} \left( \frac{R^2}{2} - \frac{R^4}{4R^2} \right) \] \[ V_{\text{avg}} = \frac{2U}{R^2} \left( \frac{R^2}{2} - \frac{R^2}{4} \right) = \frac{2U}{R^2} \times \frac{R^2}{4} = \frac{U}{2} \] Thus, the average velocity of the fluid is \( \frac{U}{2} \), which corresponds to option (A).
Final Answer: (A) \( \frac{U}{2} \)