Flow rate 2 s after pump failure

Data. Static lift $H_s = 12$ m. Pipe 1: $L_1 = 30$ m, $D_1 = 0.225$ m, $f = 0.028$. Pipe 2: $L_2 = 100$ m, $D_2 = 0.15$ m, $f = 0.032$. Steady flow $Q_0 = 115$ l/s. After the pump stops instantaneously, the water column keeps moving for a short time, decelerating under the static head and friction. Treat the water as incompressible and the pipes as rigid (rigid-column theory).

Constants

• Areas: $A_1 = 0.03976$ m², $A_2 = 0.01767$ m².
• Inertia term: $I = \sum\dfrac{L}{A} = \dfrac{30}{0.03976} + \dfrac{100}{0.01767} = 6413.4\ \text{m}^{-1}$
• Friction coefficients, $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$: $r_1 = 120.4$, $r_2 = 3481.9$, so $r = r_1 + r_2 = 3602.3$ s²/m⁵ (check: the steady friction head at 115 l/s is $r Q_0^2 = 47.64$ m).

Equation of motion

After the pump fails, two effects decelerate the whole water column: the static head $H_s$ (pushing the water back) and friction (opposing the forward motion). For the pipes in series:

$$\frac{\sum (L/A)}{g}\frac{dQ}{dt} = -\left(H_s + rQ^2\right) \;\Rightarrow\; \frac{dQ}{dt} = -\frac{g}{I}\left(H_s + rQ^2\right)$$

Separate the variables and integrate from $Q_0$:

$$\int_{Q_0}^{Q}\frac{dQ}{H_s + rQ^2} = -\frac{g}{I}t \;\Rightarrow\; \frac{1}{\sqrt{H_sr}}\left[\tan^{-1}\left(Q\sqrt{\frac{r}{H_s}}\right) - \tan^{-1}\left(Q_0\sqrt{\frac{r}{H_s}}\right)\right] = -\frac{g}{I}t$$ $$Q = \sqrt{\frac{H_s}{r}}\tan\left[\tan^{-1}\left(Q_0\sqrt{\frac{r}{H_s}}\right) - \frac{g\sqrt{H_sr}}{I}\,t\right]$$

Numbers

$$\sqrt{\frac{H_s}{r}} = \sqrt{\frac{12}{3602.3}} = 0.05772,\qquad Q_0\sqrt{\frac{r}{H_s}} = 0.115\sqrt{\frac{3602.3}{12}} = 1.9925,\qquad \tan^{-1}(1.9925) = 1.1056\ \text{rad}$$ $$\frac{g\sqrt{H_sr}}{I} = \frac{9.81\sqrt{12(3602.3)}}{6413.4} = 0.31803\ \text{s}^{-1}$$

At $t = 2$ s:

$$Q = 0.05772\tan\left(1.1056 - 0.31803(2)\right) = 0.05772\tan(0.4696) = 29.29\times10^{-3}\ \text{m}^3/\text{s}$$

(The flow stops completely at $t = 1.1056/0.31803 = 3.48$ s.)

Answer: The flow rate 2 s after the power failure is about $29.3$ l/s (still forward), down from 115 l/s.