The wellbore

The wellbore has a tributary of 2 m³ per hour. When there is no tapping, there are a stable 28 liters of water in the well. The pump suction basket is at the bottom of the well. At 14.00, the water was pumped out at a rate of 0.5 liters of water every second. What time will the water in the well be depleted?

Result

t = hh:mm (Correct answer is: INF)

Solution:

$$\begin{gathered} Q=2000\text{ l/h } \\ {Q}_1=Q\mathrm{/}3600=2000\mathrm{/}3600={\displaystyle \frac{5}{9}}\doteq 0.5556\text{ l/s } \\ {Q}_2=0.5\text{ l/s } \\ {Q}_2<Q_1 \\ V=28\text{ l } \\ x\cdot Q_2=V+Q_1\cdot x \\ t=\mathrm{\infty } \end{gathered}$$

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