Calculate Water to Increase Cylinder Pressure from 50 to 150 PSI

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To calculate the amount of water needed to raise the pressure in a cylinder from 50 PSI to 150 PSI, the relationship between pressure, height, and density must be considered. The discussion highlights that while the initial assumption is to triple the water volume to achieve a threefold increase in pressure, this approach does not account for the incompressibility of water. It is emphasized that water behaves differently than an ideal gas, necessitating the inclusion of water's compressibility factor in calculations. The density of water at 50 PSI is noted to be just over 1 g/cm3, and the total volume of water in the pipeline is approximately 57500 cubic feet. Accurate calculations require a deeper understanding of fluid dynamics beyond simple volume adjustments.
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If I have a cylinder with water at 50 PSI, is there any way to calculate how much water it would take to raise the pressure to 150 PSI?
 
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Pressure = height x density.
 
Let me add some additional information. This is for a pipeline with roughly 3.6 million pounds of water and 57500 cubic feet of area. The density is just over 1 g/cm3 at 50 PSI. Height will remain constant as will the volume. So by my thinking, I will have to add 3 times the water to increase the density by 3 times and thus the pressure 3 times. But that is contrary to my experiences with raising the PSI of pipelines in this manner.

Any thoughts?
 
You are thinking of an ideal gas. Water is no such thing. You need a figure for the compessiblity of water.
 
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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