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:

$Q=2000 \ \text{l/h} \ \\ Q_{1}=Q/3600=2000/3600 \doteq \dfrac{ 5 }{ 9 } \doteq 0.5556 \ \text{l/s} \ \\ Q_{2}=0.5 \ \text{l/s} \ \\ \ \\ Q_{2} < Q_{1} \ \\ \ \\ V=28 \ \text{l} \ \\ \ \\ x \cdot \ Q_{2}=V + Q_{1} \cdot \ x \ \\ \ \\ t=INF$

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