Rate of level change in standpipes

Data. Two open standpipes, $D = 0.75$ m, joined at the bottom by a pipe $d = 75$ mm, $L = 4$ m, $e = 0.3$ mm. At the instant considered the water in the left standpipe is $H = 2.5$ m above that in the right. Water at 20 °C ($\nu = 10^{-6}$ m²/s).

Velocity in the connecting pipe

The head difference is used up in the entrance loss (0.5), friction, and the exit loss (1.0) at the other standpipe:

$$H = \left(0.5 + f\frac{L}{d} + 1.0\right)\frac{V^2}{2g}, \qquad \frac{e}{d} = 0.0040$$

$f$ depends on $Re$ (Colebrook-White), so iterate: assume $f \approx 0.029$, $V \approx 4$ m/s, $Re \approx 3\times10^5$; the Colebrook equation then gives

$$f = 0.0288,\qquad f\frac{L}{d} = 1.535,\qquad 2.5 = 3.035\,\frac{V^2}{2g} \;\Rightarrow\; V = 4.020\ \text{m/s},\quad Re = 301490$$

Discharge

$$Q = \frac{\pi}{4}d^2V = \frac{\pi}{4}(0.075)^2(4.020) = 0.01776\ \text{m}^3/\text{s} = 17.76\ \text{l/s}$$

Rate of change of level

Area of the standpipe $A_s = \dfrac{\pi}{4}(0.75)^2 = 0.4418$ m². The left standpipe loses water at the rate $Q$, so

$$\frac{dh}{dt} = -\frac{Q}{A_s} = -\frac{0.01776}{0.4418} = -0.0402\ \text{m/s} = -40.2\ \text{mm/s}$$

Answer: The water level in the left standpipe falls at about 0.040 m/s ($\approx 40$ mm/s) at this instant, while the right one rises at the same rate; the head difference therefore decreases by twice this rate, 0.080 m/s, and the flow slows down gradually.