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: Derive the profile from the governing pressure-gradient equation, instead of quoting the normalized profile directly.
For steady laminar flow in a circular pipe driven by a constant pressure gradient $-\frac{dp}{dx}$, the velocity profile obtained from the Navier-Stokes / Hagen-Poiseuille analysis is
\[ u(r) = \frac{1}{4\mu}\left(-\frac{dp}{dx}\right)\left(R^2-r^2\right) \]
where $\mu$ is the fluid viscosity and $R$ is the pipe radius.

Step 2: Find the maximum velocity from this expression.
The velocity is largest at the pipe axis, $r=0$:
\[ u_{max} = \frac{1}{4\mu}\left(-\frac{dp}{dx}\right)R^2 \]

Step 3: Form the ratio $u(r)/u_{max}$ to eliminate the unknown pressure gradient and viscosity.
\[ \frac{u(r)}{u_{max}} = \frac{R^2-r^2}{R^2} = 1-\left(\frac{r}{R}\right)^2 \]
This is the same parabolic shape used before, but now it comes directly from the flow-driving equation rather than being recalled as a ready-made formula.

Step 4: Substitute the numbers.
$R=5$ cm, $r=2.5$ cm, so $r/R=0.5$ and $(r/R)^2=0.25$.
\[ \frac{u(2.5)}{2} = 1-0.25=0.75 \]
\[ u(2.5) = 2\times0.75 = 1.5 \text{ m/s} \]

Step 5: Confirm against the options.
This gives 1.50 m/s, matching option (C). Since the velocity falls off with the SQUARE of $r/R$, halfway to the wall ($r/R=0.5$) the velocity has fallen by only $0.25\times100\%=25\%$ from $u_{max}$, not by half, which is why options like 1.00 (a 50% drop) are wrong.

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