Question

The apparent wall shear stress in a 0.6 m long pipe line carrying refined oil is 12.5 Pa. If the pressure drop along the length is 300 Pa and flow rate is 0.25 m\(^3\)/s, the absolute viscosity of oil in \(10^{-3}\) Pa·s is \underline{\hspace{3cm}}.

Hint:

In laminar pipe flow, viscosity can be back-calculated using Hagen–Poiseuille's law when discharge, pressure drop, and dimensions are known.

Answer 1

Step 1: Relation between wall shear stress and pressure drop. For a cylindrical pipe in steady laminar flow: \[ \tau_w = \frac{\Delta P \cdot R}{2L} \] where: - \(\tau_w\) = wall shear stress (Pa), - \(\Delta P\) = pressure drop across length \(L\) (Pa), - \(R\) = pipe radius (m).

Step 2: Solve for pipe radius. \[ 12.5 = \frac{300 \cdot R}{2 \times 0.6} \] \[ 12.5 = \frac{300R}{1.2} \] \[ R = \frac{12.5 \times 1.2}{300} = 0.05 \, m \] Thus, diameter = 0.1 m.

Step 3: Hagen–Poiseuille equation for volumetric flow. \[ Q = \frac{\pi R^4}{8 \mu L} \Delta P \] Rearranging to find viscosity: \[ \mu = \frac{\pi R^4 \Delta P}{8 Q L} \]

Step 4: Substitute values. \[ R = 0.05, L = 0.6, Q = 0.25, \Delta P = 300 \] \[ \mu = \frac{3.1416 \times (0.05)^4 \times 300}{8 \times 0.25 \times 0.6} \] \[ = \frac{3.1416 \times 6.25 \times 10^{-6} \times 300}{1.2} \] \[ = \frac{5.89 \times 10^{-3}}{1.2} = 4.91 \times 10^{-3} \, Pa·s \]

Step 5: Convert to requested units. Since answer required in multiples of \(10^{-3}\) Pa·s: \[ \mu = 4.91 \, (10^{-3} \, Pa·s) \] \[ \boxed{4.91} \]