Question

In a laminar flow of a Newtonian fluid through a circular pipe of radius 5 cm, the maximum velocity is found to be 2 m/s. The velocity (in m/s) at a radial distance of 2.50 cm from the axis of the pipe is

• A. 1.00
• B. 1.25
• C. 1.50
• D. 1.75

Hint:

Use the parabolic Hagen-Poiseuille profile \(u(r)=u_{max}[1-(r/R)^2]\) with \(r/R=0.5\).

Answer 1

The Correct Option is C

Step 1: Recall the velocity profile for laminar (Hagen-Poiseuille) flow in a pipe.
For steady laminar flow of a Newtonian fluid in a circular pipe of radius \(R\), the velocity varies parabolically across the cross-section:
\[ u(r) = u_{max}\left[1-\left(\frac{r}{R}\right)^2\right] \]
where \(u_{max}\) is the velocity at the pipe axis (\(r=0\)), and \(u(r)\) is the velocity at radial distance \(r\) from the axis.

Step 2: Substitute the given values.
Here \(R=5\) cm, \(u_{max}=2\) m/s, and we need \(u\) at \(r=2.50\) cm.
\[ \frac{r}{R} = \frac{2.50}{5} = 0.5 \]
\[ \left(\frac{r}{R}\right)^2 = 0.25 \]

Step 3: Compute the velocity.
\[ u(2.5) = 2\times(1-0.25) = 2\times0.75 = 1.5 \text{ m/s} \]

Step 4: Rule out the other options.
Option (A) 1.00 would need \((r/R)^2=0.5\), i.e. \(r\approx3.54\) cm, not 2.5 cm. Option (B) 1.25 would need \((r/R)^2=0.375\), i.e. \(r\approx3.06\) cm, not 2.5 cm. Option (D) 1.75 would need \((r/R)^2=0.125\), i.e. \(r\approx1.77\) cm, not 2.5 cm. Only option (C) matches the actual given radial distance of 2.5 cm exactly.

Final Answer:
\[ \boxed{u=1.50 \text{ m/s}} \]