What pressure value should be used for fluid flow calculations in a pipe?

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For fluid flow calculations in a pipe, the pressure value used should account for both atmospheric pressure and the pressure exerted by the water column. When water is drawn from a depth of 12 feet, the downward pressure is approximately 5 psi, while atmospheric pressure is 14.7 psi. The effective upward pressure to consider for calculations is the difference, resulting in about 9.7 psi available to lift the water. It's important to note that as the water rises, the pressure at the top can change, affecting flow rates. In hydrostatic equilibrium, the upward atmospheric pressure balances the downward pressures, confirming the relationship between pressures at different points in the system.
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Homework Statement


I found an equation used to calculate the flow of a fluid through a pipe. But I'm having some difficulty with the pressure value I should use in the calculations. It's the flow rate of water moving by atmospheric force. For example if you have a well and you suck up water through a tube, And the water is 12feet down which would be around 5psi, The atmospheric force is 14.7psi. So would I use 9.7psi to determine the upward force because the difference in pressure between 14.7 psi upward and 5 psi downward force.


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The Attempt at a Solution

 
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The pump at the top need to reduce the pressure in order for the water to flow upwards. The pressure at the bottom will then be in effect atmospheric pressure (if the ground water is in contact with the outside atmosphere). The pressure at the top will then be whatever is delivered by the pump to the column of water - which then need to be at most 9.7 psi for the water to just reach the top. The problem with hand operated pumps is that the pressure on top rises as the water rises in the pipe (the partial vacuum is reduced by compressing the air on top of the rising column).
 
What if there's a situation where the air at the top does not compress to create pressure. If the reduced pressure stays the same. No matter how high the water level got in the pipe. Would you need less atmospheric pressure to force the water up that high.
 
If the pressure at the top changes the flow rate will change. The maximum height that the water can be lifted to is limited to around 10.3 meters or 33.8 feet no matter how effective the vacuum is at the top.
 
Ok. But if the water is 12 feet high creating 5.19psi down force pressure does it cancel out that amount of pressure out of the atmospheric pressure pushing the water upward. So instead of 14.7psi upward force there would be 9.505psi upward force. Is this right?
 
It the water column is in hydrostatic equilibrium (stationary) the "upwards" pressure (atmospheric) and "downwards" pressure (pressure on top of the liquid and pressure caused by the weight of the liquid) will be equal. This is described by the relationship

p_{bottom} = p_{top} + \rho g h

where p_bottom is atmospheric pressure in this case.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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