Question

The viscosity of an incompressible Newtonian fluid is measured using a capillary tube of diameter \(0.5\ \text{mm}\) and length \(1.5\ \text{m}\). The fluid flow is laminar, steady and fully developed. For a flow rate of \(1\ \text{cm}^3\text{s}^{-1}\), the pressure drop across the length of the tube is \(1\ \text{MPa}\). If the viscosity of the fluid is \(k \times 10^{-3}\ \text{Pa}\cdot\text{s}\), the value of \(k\) is ____________________ (rounded off to two decimal places).

Hint:

For laminar, fully developed flow in a circular tube, use Hagen–Poiseuille: \(Q = \dfrac{\pi D^4}{128\mu L}\Delta P\).
Always convert mm \(\rightarrow\) m, cm\(^3\) \(\rightarrow\) m\(^3\), and MPa \(\rightarrow\) Pa before substituting.
If an answer is given as \(k\times10^{-3}\), compute \(\mu\) in Pa·s and then set \(k=\mu/(10^{-3})\).

Answer 1

Correct Answer: 1.01

Step 1: Use the Hagen–Poiseuille relation for fully developed laminar flow in a circular tube: \[ Q \;=\; \frac{\pi\,D^{4}}{128\,\mu\,L}\,\Delta P \quad\Rightarrow\quad \mu \;=\; \frac{\pi\,D^{4}\,\Delta P}{128\,L\,Q}. \] Step 2: Convert all quantities to SI units and substitute. \[ D = 0.5\ \text{mm} = 5.0\times 10^{-4}\ \text{m},\quad L = 1.5\ \text{m},\] \[Q = 1\ \text{cm}^3\!/\text{s} = 1.0\times 10^{-6}\ \text{m}^3/\text{s},\quad \Delta P = 1\ \text{MPa} = 1.0\times 10^{6}\ \text{Pa}. \] \[ \mu \;=\; \frac{\pi (5.0\times 10^{-4})^{4} (1.0\times 10^{6})}{128\,(1.5)\,(1.0\times 10^{-6})} \;=\; 1.02265\times 10^{-3}\ \text{Pa}\cdot\text{s}. \] Step 3: Express \(\mu\) as \(k\times 10^{-3}\ \text{Pa}\cdot\text{s}\) and round \(k\) to two decimals: \[ k \;=\; 1.02265 \;\approx\; \boxed{1.02}. \]