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I was wondering if water exits from the pump and reach the ground, is the pressure of the water at the point of the ground be 0?

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I was wondering if water exits from the pump and reach the ground, is the pressure of the water at the point of the ground be 0?

No, it is not. See below.


Pressure of water on ground is combination of the hydrostatic pressure and of hydrodynamic pressure due change of its kinetic state.

The former is determined by water layer thickness and is rather minimal, 1 cm is equivalent 0.001 atm.

The latter is determined by relation of the total force causing acceleration of water to stop its vertical motion.

$$\int p \cdot dA = \int a_z \cdot dm $$

Note that relation of this ground water pressure and the pump pressure is very indirect, via the water speed, the water volumetric flow and the stream shape.

As water, leaving a pump, has lost the pump hydrostatic pressure, falling by the free fall. The ground pressure is given by characteristics of this fall.

If the stream is about compact, the pressure at the stream impact zone can be very roughly estimated for the vertical stream as $$p=\frac 12 \cdot \rho v^2=500 \cdot v^2$$ where v is water speed in m/s.

The speed can be determined from the diameter of the pump outlet pipe D[m], water flow Q [m3/s] and the height of the fall H [m].

The initial speed is $$v1=\frac {4Q}{\pi \cdot D^2}$$

The kinetic energy of the water stream gain the potential energy by falling. $$\frac12 \cdot m \cdot v_2^2=\frac 12 \cdot m \cdot v_1^2 + m\cdot g\cdot H$$ $$v2=\sqrt {\left(\frac{4Q}{\pi D^2}\right)^2 + 5H}$$

The impact area $$A=\frac Q{v_2}=\frac Q{\sqrt {\left(\frac{4Q}{\pi D^2}\right)^2 + 5H}}$$

The impact pressure estimation(for vertical and compact stream) is

$$p=500 \cdot \left(\left(\frac{4Q}{\pi D^2}\right)^2 + 5H\right)$$

Note that this is rough estimation of average pressure. In reality, due turbulences, real pressure is turbulent and has fast space and time variability.

For general angles, the pressure will be smaller, due smaller vertical velocity component and lower hydrodynamic drag of skewed surface.

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  • $\begingroup$ is there any possible way if i know the initial velocity of water exiting the pump and flow rate of the pump, to find out the water pressure exiting the pump? I have tried to use the bernoulli's equation, however, i dun really understand how to deal with the P1 and P2 $\endgroup$ – Struggle May 8 at 7:09
  • $\begingroup$ @Struggle See the answer update. Water exiting the pump has no pressure, while it's outlet is free and impact has not accepted yet $\endgroup$ – Poutnik May 8 at 7:47
  • $\begingroup$ The pressure toward the outlet more or less linerly decreases to zero, dealing with pipe surface drag forces. that is the requirement of the water flow continuity. Leaving water then has no pressure until the impact. $\endgroup$ – Poutnik May 8 at 7:52
  • $\begingroup$ I am afraid you want what you should not want. Water leaving the pipe at the outlet has zero pressure. Water in the pipe near outlet as pressure just to deal with the friction, or it has dynamic pressure needed for water acceleration in case of variability of the flow. $\endgroup$ – Poutnik May 8 at 8:07
  • $\begingroup$ What then does mean the title of the question ? "What is the pressure of water after entering atmosphere?" "I was wondering if water exits from the pump and reach the ground, is the pressure of the water at the point of the ground be 0?" $\endgroup$ – Poutnik May 8 at 8:19

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