Why can the water column be treated as a decelerating rigid body here?

Because elasticity of water and pipe is disregarded

2075 Bhadra · 8 marks

Duration of flow after pump stops

Short answer

With rigid-column theory, the deceleration equation is integrated from initial flow to zero. The flow into the tank continues for 2.62 s after power failure.

A pump draws water from a reservoir and delivers it through a pipe 150 mm diameter, 90 m long, to a tank in which the free surface level is 8 m higher than that in the reservoir. The flow rate is steady at 0.05 m³/s until a power failure causes the pump to stop. Neglecting minor losses in the pipe and in the pump and assuming that the pump stops instantaneously, determine for how long flow into the tank continues after the power failure. The friction factor f may be taken as constant at 0.028 and elastic effects in the water or pipe material may be disregarded.

Data. $D = 0.15$ m, $L = 90$ m, static lift $H_s = 8$ m, $f = 0.028$ , steady $Q_0 = 0.05$ m³/s. The pump stops instantaneously; minor losses and elasticity are neglected, so the water column decelerates as a rigid body under the static head (back-pressure) and friction.

Constants

A = \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2,\qquad I = \frac{L}{A} = \frac{90}{0.01767} = 5093.0\ \text{m}^{-1}

r = \frac{8fL}{\pi^2gD^5} = \frac{8(0.028)(90)}{\pi^2(9.81)(0.15)^5} = 2742.0\ \text{s}^2/\text{m}^5 \quad(\text{steady friction head } rQ_0^2 = 6.85\ \text{m})

Equation of motion

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

The flow into the tank stops when $Q = 0$ . Integrating from $Q_0$ to 0:

t = \frac{I}{g}\int_0^{Q_0}\frac{dQ}{H_s + rQ^2} = \frac{I}{g\sqrt{H_sr}}\tan^{-1}\left(Q_0\sqrt{\frac{r}{H_s}}\right)

Q_0\sqrt{\frac{r}{H_s}} = 0.05\sqrt{\frac{2742.0}{8}} = 0.9257,\qquad \tan^{-1}(0.9257) = 0.7468\ \text{rad}

t = \frac{5093.0}{9.81\sqrt{8(2742.0)}}(0.7468) = \frac{0.7468}{0.28528} = 2.62\ \text{s}

Answer: The flow into the tank continues for about $2.62$ s (about 2.6 s) after the power failure.

✓ Answer checked by Claude Opus 5.5 (high effort)

Flow rate 2 s after pump failure · Hydraulics, chapter 4
Siphon with 90 mm pipe, number and length · Hydraulics, chapter 2
Pump power and pipe length with Colebrook · Hydraulics, chapter 2
Pipe outlet minimum water level, 50 lps · Irrigation and Drainage Engineering, chapter 7
Reservoir level change, water balance, 1600 ha · Engineering Hydrology, chapter 1 · asked 2×
Reservoir level change, water balance, 15.5 km2 · Engineering Hydrology, chapter 1 · asked 2×
Settling tank 20 m long, 3.5 m deep; 0.02 and 0.01 mm · Water Supply Engineering, chapter 6 · asked 3×

All Hydraulics chapter 4 questions →

Duration of flow after pump stops

Constants

Equation of motion

Related questions