Question

if a 3 m deep well is half filled with water, then how much time needed to raise 2 kg water with a 10 kw pump whose efficiency is 70%?

Answer 1

Answer on Question #50804-Physics-Mechanics-Kinematics-Dynamics

If a $H = 3$ m deep well is half filled with water, then how much time needed to raise $m = 2$ kg water with a $P = 10$ kW pump whose efficiency is $\eta = 70\%$?

Solution

To up the mass $dm = \rho Adh$ of water from the depth $h$, where $A$ is area, we need energy

$$dE = dmgh = \rho Adhgh = (\rho gA)hdh.$$

To fulfill the well we need

$$E = \int_{0}^{H} (\rho gA)hdh = (\rho gA) \frac{h^{2}}{2}_{0}^{H} = (\rho gA) \frac{H^{2}}{2} = \frac{mgH}{2}.$$

The efficient power is

$$\eta P = \frac{E}{t}.$$

Thus

$$t = \frac{E}{\eta P} = \frac{mgH}{2\eta P} = \frac{2 \cdot 9.8 \cdot 3}{2 \cdot 0.7 \cdot 10000} = 4.2 \text{ ms}.$$

Answer: 4.2 ms.

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