Pressure to drive oil up inclined pipe

Data. $\rho = 850$ kg/m³ (s.g. 0.85), $\mu = 0.01$ Pa s, $D = 0.12$ m, $L = 50$ m, slope 20°, outlet pressure 1 bar $= 10^5$ Pa, $Q = 0.25$ m³/min $= 0.004167$ m³/s. The oil is viscous, so Hagen-Poiseuille (laminar) flow is assumed.

(i) Pressure needed if there were no slope

Mean velocity and Reynolds number:

$$V = \frac{Q}{\pi D^2/4} = 0.3684\ \text{m/s}, \qquad Re = \frac{\rho V D}{\mu} = 3758$$

(This is above 2000, so the flow is really in the critical zone. The laminar formula is used because the oil is described as viscous and the flow is steady. A turbulent estimate would give a friction drop of only about 1 kPa, so the friction loss is small in either case.)

Friction pressure drop (Hagen-Poiseuille):

$$\Delta p_f = \frac{128\mu L Q}{\pi D^4} = \frac{128(0.01)(50)(0.004167)}{\pi(0.12)^4} = 409.3\ \text{Pa}$$

Pressure at the inlet without slope: $P = 10^5 + 409.3 = 100409.3$ Pa $\approx 100.41$ kPa.

Pressure needed on the actual slope

Extra pressure to lift the oil through the height $L\sin 20^\circ = 17.10$ m:

$$\Delta p_z = \rho g L \sin 20^\circ = 850(9.81)(50)(0.3420) = 142597\ \text{Pa}$$ $$P = 10^5 + 409.3 + 142597 = 243006\ \text{Pa} \approx 243.0\ \text{kPa}$$

(ii) Number of 8 cm pipes in parallel

At the same driving pressure the friction drop per pipe is the same, and in laminar flow $Q \propto D^4$ (the slope term is the same for any pipe size). Flow per 8 cm pipe compared with the 12 cm pipe:

$$\frac{Q_8}{Q_{12}} = \left(\frac{8}{12}\right)^4 = 0.1975$$ $$n = \frac{Q_{12}}{Q_8} = \left(\frac{12}{8}\right)^4 = 5.0625 \;\Rightarrow\; n = 6 \text{ pipes (rounded up)}$$

Check with 6 pipes: each carries $Q/6 = 0.000694$ m³/s, $V = 0.138$ m/s, $Re = 939$ (laminar), and the friction drop is $345.4$ Pa, which is less than 409.3 Pa, so the driving pressure is not exceeded.

Answer: (i) $P \approx 100.41$ kPa without slope ($\Delta p_f = 409$ Pa); on the 20° slope $P \approx 243.0$ kPa; (ii) 6 pipes of 8 cm diameter.