Question

Maximum velocity in laminar flow in a pipe is

• A. Equal to the average velocity
• B. Twice the average velocity
• C. Half the average velocity
• D. Four times the average velocity

Hint:

Remember:
- Circular pipe laminar flow: $u_{\text{max}} = 2 \cdot u_{\text{avg}}$
- Parallel flat plates laminar flow: $u_{\text{max}} = 1.5 \cdot u_{\text{avg}}$

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This fluid mechanics question asks for the exact relationship between the maximum flow velocity and the average flow velocity in a fully developed laminar pipe flow.

Step 2: Key Formula or Approach:
For steady, fully developed laminar flow of a viscous, incompressible fluid in a circular pipe (Hagen-Poiseuille flow), the velocity profile is parabolic:
\[ u(r) = u_{\text{max}} \cdot \left[ 1 - \left(\frac{r}{R}\right)^2 \right] \]
The average velocity ($u_{\text{avg}}$) is determined by integrating this velocity profile over the cross-sectional area of the pipe.

Step 3: Detailed Explanation:

• The total volumetric flow rate ($Q$) is the integral of the local velocity over the pipe cross-section:
\[ Q = \int_0^R u(r) \cdot (2\pi r \cdot dr) \]
\[ Q = 2\pi \cdot u_{\text{max}} \int_0^R \left( r - \frac{r^3}{R^2} \right) dr \]
\[ Q = 2\pi \cdot u_{\text{max}} \cdot \left[ \frac{R^2}{2} - \frac{R^4}{4 R^2} \right] = \frac{\pi R^2 \cdot u_{\text{max}}}{2} \]

• The average velocity is defined as the flow rate divided by the total cross-sectional area ($A = \pi R^2$):
\[ u_{\text{avg}} = \frac{Q}{A} = \frac{\frac{\pi R^2 \cdot u_{\text{max}}}{2}}{\pi R^2} = \frac{u_{\text{max}}}{2} \]

• Rearranging this gives:
\[ u_{\text{max}} = 2 \cdot u_{\text{avg}} \]

Step 4: Final Answer:
The maximum velocity in laminar flow in a pipe is twice the average velocity.