# 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:

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

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