Laminar oil flow in inclined pipe

Given. $\rho = 900$ kg/m³, $\nu = 0.0002$ m²/s so $\mu = \rho\nu = 0.18$ Pa s; $d = 0.06$ m, $L = 10$ m, inclination 40°. At section 1: $p_1 = 350\,000$ Pa, $z_1 = 0$. At section 2: $p_2 = 250\,000$ Pa, $z_2 = L\sin 40^\circ = 6.428$ m.

Direction of flow

Flow goes from high to low piezometric pressure $p^* = p + \rho g z$, not from high to low $p$.

$$p_1^* = 350\,000 + 0 = 350\,000\ \text{Pa}, \qquad p_2^* = 250\,000 + 900(9.81)(6.428) = 306752\ \text{Pa}$$

Since $p_1^* > p_2^*$, the flow is from 1 to 2, i.e. upward, as required. The driving pressure drop is

$$\Delta p^* = p_1^* - p_2^* = 43248\ \text{Pa}$$

Flow rate (Hagen-Poiseuille, laminar)

$$Q = \frac{\pi d^4\,\Delta p^*}{128\,\mu L} = \frac{\pi (0.06)^4 (43248)}{128(0.18)(10)} = 0.007643\ \text{m}^3/\text{s} = 7.64\ \text{l/s}$$

Check that the flow is laminar

$$V = \frac{Q}{\pi d^2/4} = 2.703\ \text{m/s},\qquad Re = \frac{Vd}{\nu} = \frac{2.703(0.06)}{0.0002} = 811 < 2000$$

So the laminar assumption is valid.

Answer: The flow is upward; $Q = 0.007643$ m³/s ($7.64$ l/s), $Re \approx 811$.