Discharge in 20 km pipe between reservoirs

Data. $D = 0.02$ m, $L = 20\,000$ m, $H = 5$ m, $\nu = 1.02\times10^{-6}$ m²/s. The pipe is very long and thin, so the velocity will be small; try laminar flow first.

Laminar flow assumption

The available head is used up in friction (velocity head and minor losses are negligible against 5 m over 20 km). Hagen-Poiseuille:

$$h_f = \frac{32\nu LV}{gD^2} = H \;\Rightarrow\; V = \frac{gD^2H}{32\nu L} = \frac{9.81(0.02)^2(5)}{32(1.02\times10^{-6})(20\,000)} = 0.0301\ \text{m/s}$$

Check

$$Re = \frac{VD}{\nu} = \frac{0.0301(0.02)}{1.02\times10^{-6}} = 589 < 2000$$

The flow is laminar, so the assumption is right. (Velocity head $V^2/2g = 4.6\times10^{-5}$ m is negligible against 5 m.)

Discharge

$$Q = \frac{\pi}{4}D^2V = \frac{\pi}{4}(0.02)^2(0.0301) = 0.0000094\ \text{m}^3/\text{s} = 0.00944\ \text{l/s} \approx 34.0\ \text{l/hour}$$

Answer: $Q \approx 9.44\times10^{-6}$ m³/s (about 0.0094 l/s, 34 l/h); laminar flow with $Re \approx 589$.