# Determining pressure as a function of time for water filled balloon

Looking to determine pressure as a function of time for a water filled "balloon" at extremely high pressures (~90MPa) with a fixed flow rate out. This is to determine the time it takes to hit a certain pressure $$P_f$$ from an initial pressure $$P_0$$. I have recognized the importance of creating an equation of state.

$$\frac{\delta(P)}{\delta(t)} = \frac{\delta(P)}{\delta(\rho)} * \frac{\delta(\rho)}{\delta(t)}$$

$$\frac{\delta(P)}{\delta(\rho)}$$ may be found with a compressed water table.

$$\frac{\delta(\rho)}{\delta(t)} = \frac{\delta(\rho)}{\delta(m)} * \frac{\delta(m)}{\delta(t)}$$

Where $$\frac{\delta(m)}{\delta(t)}$$ is simply the fixed mass flow rate.

Now $$\frac{\delta(\rho)}{\delta(m)}$$ is where I begin to get difficulties. Volume will change as a function of pressure, ie. $$V=f(P)$$ and mass will change as a function of density and volume, ie. $$m=f(\rho,V)$$.

Simply, it should functionally look like something along these lines: $$\Delta(\rho (\frac{kg}{m^3}) ) = (\rho (\frac{kg}{m^3}) )_0 - \Delta(m (\frac{kg}{s}) ) + \Delta(V (\frac{m^3}{s}) )$$

The change in volume with pressure is a problem I have saved for later, so for now, assume that equation can be readily expressed.

Assume all initial conditions are known and the system is isothermal.

Is this a correct way to express this problem?

Thanks, Brad

• Is the "balloon" some type of steel container? Do you know the function for volume expansion of the container as a function of pressure? Your posting is a bit short on details, making it difficult to answer. – David White Jan 18 at 18:55
• I'm not sure the material is relevant. Whether it's plastic or steel, the principals remain the same. Assume steel if it helps. I have stated that the change in volume wrt. pressure is a separate question and can be assumed known for this problem. I'm really more looking at whether this is the right track, and how to model the change in density wrt. mass if the change in volume wrt pressure is known. The biggest question I had is how to create those expressions, as volume depends on pressure, pressure depends on density, and density depends on volume. – Brad Thomas Jan 18 at 19:01
• I can't solve your problem but I can add something to think about. Make sure that you have the same number of independent equations as you have variables. If you do, you either need to use substitution or some form of matrix manipulation to work this problem. – David White Jan 18 at 19:04
• Brad, I will further qualify my last sentence from the previous comment. Yes, I realize that matrix mathematics works only for linear equations. Despite this, there are routines that linearize equations around a guessed starting point, and use trial-and-error techniques to iteratively converge to the correct answer. Such techniques work with systems of non-linear equations, and they are very useful where substitution becomes overly cumbersome. – David White Jan 18 at 19:14
• Is it the balloon stretching that is forcing the liquid out of the balloon (and the liquid is basically incompressible) or is it the balloon that is more rigid, and the compression of the liquid is being released? – Chet Miller Jan 18 at 20:38