# How do I Calculate Pressure at Multiple Points in Time while Filling a Container with Water?

Say I have a container completely full of water with a volume of 3000 gallons and I'm filling this container with water at a rate of 8.2 gallons per minute, it's already full at 200 PSI, how do I calculate how much pressure I've gotten up to at 10 sec, 20 sec, etc?

• Are you sure you aren't just spilling water onto the ground at a rate of 8.2 gallons per minute? – J. Ari Dec 27 '19 at 18:00
• No sir, I'm adding water to this container thus increasing its pressure. I'm pressurizing this container with more water. – Sramirez Dec 27 '19 at 18:02
• If the container is completely full you cannot add more water, water is incompressible. It would depend on what the pump can handle. Is it moving 8.2 gallons/minute at 200 psi? – jko Dec 27 '19 at 19:39
• Nothing is completely incompressible. Liquids are relatively incompressible when compared to gasses (but probably pretty spongy compared to neutronium). Water has a bulk modulus of $3.12 \times 10^5 \mathrm{PSI}$, or about $2.15 \mathrm{GPa}$. That works out to roughly a 104PSI pressure rise for each gallon increase, assuming that the container is perfectly rigid. I suspect, however, that for any reasonably economical tank, the volume of the tank would also change with pressure, and would perhaps contribute more to the volume accommodated per unit of pressure rise than the water itself. – TimWescott Dec 27 '19 at 20:23
• So, you need to tell us about the tank -- what's it made of, how big is it, are any of your loved ones in the path of destruction when it bursts, etc. – TimWescott Dec 27 '19 at 20:24

we need to assume some parameters to start building a model.

let's say your container is a steel sphere with thickness, t = 0.25" to make things a bit easier with a volume of 3000 gallons. So converting to cubic ft, we get $$V= 3000*0.133=401ft^3 \\ R=\sqrt{3/4*V/\pi} =4.574'$$

And we calculate the stress of this steel container for 200psi.

$$\sigma = \text{200*area of sphere/ t*circumfrence }=\frac{200*(4.54*12)^2*\pi}{ t*2\pi 4.574*12 }\\=100*4.574*12*4=21955psi$$

Every 10 seconds you are increasing the volume of the container (water is incompressible) by a ratio of $$(8.2/6)/3000 =0.00045$$ this will increase the radius of the container by the $$R_{expander}= R_{Initial}*(1+\sqrt{0.00045}) \$$

And stress will increase on the container by a ratio of $$\ R_{expanded}^2/R_{expanded}^3.$$

We can calculate the new stress using $$E*\epsilon=\sigma .$$

And then plug in the new stress in the formula and calculate the pressure. I suspect the container will explode soon, considering steel's huge Young modules.

Basically your container will act like a balloon we fill with water. I