Laminar oil flow in a pipe

Given data

$\rho = 850\ \text{kg/m}^3$, $\mu = 0.08\ \text{Pa s}$, $D = 0.05\ \text{m}$, $R = 0.025\ \text{m}$, $L = 100\ \text{m}$, $Q = 1.5\times10^{-3}\ \text{m}^3/\text{s}$.

Mean velocity and Reynolds number

$$\begin{aligned} A &= \frac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}\ \text{m}^2 \\ \bar V &= \frac{Q}{A} = \frac{1.5\times10^{-3}}{1.9635\times10^{-3}} = 0.764\ \text{m/s} \\ Re &= \frac{\rho \bar V D}{\mu} = \frac{850 \times 0.764 \times 0.05}{0.08} = 406 \end{aligned}$$

$Re = 406 < 2000$, so the flow is laminar.

(a) Maximum velocity

$$u_{max} = 2\bar V = 2 \times 0.764 = 1.528\ \text{m/s}$$

(b) Pressure drop and head loss

From Hagen-Poiseuille, $\Delta p = \dfrac{32\mu \bar V L}{D^2}$:

$$\Delta p = \frac{32 \times 0.08 \times 0.764 \times 100}{(0.05)^2} = 78\,228\ \text{Pa} \approx 78.2\ \text{kPa}$$ $$h_f = \frac{\Delta p}{\rho g} = \frac{78\,228}{850 \times 9.81} = 9.38\ \text{m of oil}$$

(c) Wall shear stress

$$\tau_w = \frac{\Delta p D}{4L} = \frac{78\,228 \times 0.05}{4 \times 100} = 9.78\ \text{Pa}$$

(d) Friction factor

$$f = \frac{64}{Re} = \frac{64}{406} = 0.158$$

Check: $h_f = f\dfrac{L}{D}\dfrac{\bar V^2}{2g} = 0.158 \times 2000 \times 0.02975 = 9.38\ \text{m}$. This agrees.

(e) Radius where $u = \bar V$

Since $u = u_{max}(1 - r^2/R^2) = \bar V = u_{max}/2$:

$$r = \frac{R}{\sqrt2} = \frac{25}{1.414} = 17.7\ \text{mm}$$

(f) Velocity at $r = 15\ \text{mm}$

$$u = 1.528\left[1 - \left(\frac{15}{25}\right)^2\right] = 1.528 \times 0.64 = 0.978\ \text{m/s}$$

(g) Power to overcome friction

$$P = Q\,\Delta p = 1.5\times10^{-3} \times 78\,228 = 117.3\ \text{W}$$

Answer: $Re = 406$ (laminar); $u_{max} = 1.53\ \text{m/s}$; $\Delta p = 78.2\ \text{kPa}$ ($h_f = 9.38\ \text{m}$); $\tau_w = 9.78\ \text{Pa}$; $f = 0.158$; $r = 17.7\ \text{mm}$; $u = 0.978\ \text{m/s}$; $P = 117\ \text{W}$