Oil through two parallel pipes

Data. $Q = 12$ l/s $= 0.012$ m³/s, $D_1 = 0.12$ m, $D_2 = 0.10$ m, $L = 1000$ m each, s.g. 0.85 ($\rho = 850$ kg/m³), $k = 0.26$ mm, $\nu = 9$ cm²/s $= 9\times10^{-4}$ m²/s.

The kinematic viscosity is very high, so the Reynolds numbers are small. Assume laminar flow in both pipes and verify afterwards.

Flow in each pipe

In parallel, both pipes have the same head loss. For laminar flow $h_f = \dfrac{128\nu LQ}{\pi gD^4}$, so $Q \propto D^4$:

$$\frac{Q_1}{Q_2} = \left(\frac{0.12}{0.10}\right)^4 = 2.0736$$ $$Q_1 = 0.012\left(\frac{2.0736}{1 + 2.0736}\right) = 0.008096\ \text{m}^3/\text{s} = 8.096\ \text{l/s},\qquad Q_2 = 0.003904\ \text{m}^3/\text{s} = 3.904\ \text{l/s}$$

Check for laminar flow

$$V_1 = \frac{Q_1}{\pi(0.12)^2/4} = 0.7158\ \text{m/s},\ Re_1 = \frac{V_1D_1}{\nu} = 95.4 \qquad V_2 = 0.4971\ \text{m/s},\ Re_2 = 55.2$$

Both are well below 2000, so the flow is laminar and the roughness (0.26 mm) has no effect.

Head loss and pump power

$$h_f = \frac{128\nu LQ_1}{\pi gD_1^4} = \frac{128(9\times10^{-4})(1000)(0.008096)}{\pi(9.81)(0.12)^4} = 145.94\ \text{m of oil}$$

The same value is obtained from pipe 2 ($145.94$ m). Pump head $= 145.94$ m (equal to the friction head, with no static lift).

Power delivered to the oil:

$$P = \rho g Q h_f = 850(9.81)(0.012)(145.94) = 14603\ \text{W} \approx 14.60\ \text{kW}$$

Answer: $Q_1 = 8.10$ l/s (12 cm pipe), $Q_2 = 3.90$ l/s (10 cm pipe); pump power $\approx 14.6$ kW (head 145.9 m of oil).