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If a solenoid valve is placed right after water tank's water tap does the water pressure changes if the water tank is lifted to 3 meters?
The context is this: the electric valve is functional only if the water pressure surpasses 0.6 bar.

And a second question: from other sources I found out that 1000 liters of water create a 1 bar pressure. Does this pressure is the same at the water tap exit?

enter image description here

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  • $\begingroup$ 1000 liters of water in a puddle on a floor exerts very little hydrostatic pressure, but that same amount of water in a tall column (e.g., a water tower) will exert a great deal of pressure. It's the height of the water column that matters - you can't convert a volume of water into an estimate of pressure without knowing something about the geometry. Not sure if that 1000L=1 bar rule of thumb is for the specific device pictured, but it's not a general rule. $\endgroup$ – Nuclear Wang Jun 17 '19 at 14:20
  • $\begingroup$ @NuclearWang from other sources I found out that 1 meter depth of water can exert 0.98 bar. In this case, the minimum length of the tube before the tap should be 10 meters to reach 1 bar. $\endgroup$ – George I. Jun 17 '19 at 15:45
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If you mount the tank at a height 3 meters, but have the tap at 0 meters then you will have more than enough pressure.

The valve might operate correctly while the tank and tap are as shown in your image but only while the tank is more than 2/3 full. However, once the level starts to drop from full, the operation may become unreliable.

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As mentioned in the other answer and a comment, it is the height of the water column above the tap that determines the pressure. (That means if you lift the whole tank, including the valve, the pressure will not change.) The shape of the water doesn't matter, so you could have the cubic tank with a thin tube of, say, 3 m length downwards, and the pressure would correspond to a height of $3 \text{ m} + (\text{height in the tank})$.

Specifically, in a body of water, at a depth $d$ below the surface, the pressure is $$p=\rho\cdot g \cdot d\,$$ where $\rho=1000\text{ kg}/\text{m}^3$ is the density and $g=9.81\text{ m}/\text{s}^2$ is gravitational acceleration. (These are values for pure (drinking) water and at sea level. For salt water, $\rho$ is higher, and higher up on a mountain $g$ is lower, but you can probably ignore that. You can also approximate $g=10\text{ m}/\text{s}^2$ for easier calculation.)

Hence, one meter of water gives $p\approx1000\text{ kg}/\text{m}^3 \cdot 1\text{ m}\cdot 10\text{ m}/\text{s}^2=10\,000 \text{ N}/\text{m}^2=0.1 \text{ bar}$. So for $0.6 \text{ bar}$, you would need $6\text {m}$ of water.

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