Question

The Reynolds number for flow of oil through a 1 cm diameter pipe is 1000. The kinematic viscosity, $\nu = 1.0 \times 10^{-6} \text{ m}^2/\text{s}$. What is the velocity at a point 0.25 cm away from the wall?

• A. 0.1 m/s
• B. 0.15 m/s
• C. 0.2 m/s
• D. 0.3 m/s

Hint:

At exactly halfway between the pipe center and the wall ($r/R = 0.5$), the local velocity is always exactly $75\%$ of the maximum velocity ($u_{\text{max}}$) for laminar pipe flow.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This fluid dynamics problem requires us to determine the local velocity of fluid flow in a pipe at a specified radial distance from the wall.

Step 2: Key Formula or Approach:
First, we compute the average velocity ($V$) of the flow using the Reynolds number formula:
\[ Re = \frac{V \cdot d}{\nu} \]
Since $Re = 1000 < 2300$, the flow is laminar. For fully developed laminar flow in a pipe, the velocity profile is parabolic:
\[ u(r) = u_{\text{max}} \left[ 1 - \left(\frac{r}{R}\right)^2 \right] \]
where:
$R$ is the radius of the pipe,
$r$ is the radial distance from the center of the pipe, and
$u_{\text{max}}$ is the centerline maximum velocity, which is exactly twice the average velocity ($u_{\text{max}} = 2 V$).

Step 3: Detailed Explanation:

• Identify the given parameters:
- Diameter, $d = 1 \text{ cm} = 0.01 \text{ m} \implies$ Radius, $R = 0.5 \text{ cm} = 0.005 \text{ m}$
- Kinematic viscosity, $\nu = 1.0 \times 10^{-6} \text{ m}^2/\text{s}$
- Reynolds number, $Re = 1000$

• Calculate average velocity ($V$):
\[ V = \frac{Re \cdot \nu}{d} = \frac{1000 \times 1.0 \times 10^{-6}}{0.01} = 0.1 \text{ m/s} \]

• Calculate maximum velocity ($u_{\text{max}}$):
\[ u_{\text{max}} = 2 \cdot V = 2 \times 0.1 = 0.2 \text{ m/s} \]

• The point is at distance $y = 0.25 \text{ cm}$ from the wall.
The radial position from the center ($r$) is:
\[ r = R - y = 0.5 - 0.25 = 0.25 \text{ cm} \]

• Substitute $r$ and $R$ into the parabolic velocity profile:
\[ \frac{r}{R} = \frac{0.25}{0.5} = 0.5 \]
\[ u(r) = 0.2 \left[ 1 - (0.5)^2 \right] = 0.2 \left[ 1 - 0.25 \right] = 0.2 \times 0.75 = 0.15 \text{ m/s} \]

Step 4: Final Answer:
The velocity of the oil at $0.25 \text{ cm}$ from the wall is $0.15 \text{ m/s}$.