Assumptions:
- The hole is in the region below the air pocket (so water, not air, is leaking)
- Air pocket volume is $V_p$ when the pressure is $P$
- Isothermal process (slow expansion: temperature constant)
- Volume of container doesn't change with pressure (probably not true… - this will underestimate the leakage rate)
you can write the rate of change of the volume of the air pocket as a function of pressure:
$$\frac{PV}{T} = const\\
P_1V_1 = P_2V_2$$
Differentiating $PV = constant$:
$$P dV + V dP = 0$$
Dividing by $dT$ and rearranging:
$$\frac{dV}{dt}=-\frac{V}{P}\frac{dP}{dt}$$
From this you compute the flow rate from the pressure change. As you can see, the smaller the volume $V$, the smaller the volume change $dV$ that you can calculate for a given change in pressure.
This leaves you with the problem of calibrating the air pocket. This is best done by having an air filled capillary somewhere near the top of your system: you will be able to see the liquid rise in this capillary as the system is pressurized, and from the rate at which it drops you can determine the leakage rate immediately - if you know the diameter of the capillary, no other math is required…
Do note that a heat exchanger is likely to expand when pressurized - you should be able to determine how much this affects the result by having a small calibrated plunger (ideally the same size as the capillary) with which you can inject a small known amount of additional liquid into the system. If the liquid plus heat exchanger were truly incompressible (constant volume, constant density) then the air in the capillary should rise as you push the plunger down. When this does not happen (say the capillary rises half as much) then you know what volume of liquid corresponds to what volume change in the capillary, and this gives you the calibration from capillary volume to liquid volume.
I answer my own question and give a good thanks to DavidPh, who has not really gave the answer, but in fact, it was impossible for him to give it. Here is "why":
I'm French, so I've many fire hydrant data but from France. And when applying them to the formulas, the result was wrong...
In fact, the problem is not the formula but the way we measure the pressure and from vocabulary confusion.
In France, firefighter consider that a fire hydrant must provide a flow rate of 60m3 per hour, so 1000 liters per minute, at a pressure of 1 bar.
In order to check that, here is how we do:
- Put a manometer and a valve at hydrant output
- Open water
- Close slowly the valve in order to increase pressure
- When pressure is at 1 bar, measure flow rate which must be higher than 1000 liters per minute
This mean we change the diameter, in order to get 1 bar. So the formular cant' applied as i fact, we don't know the diameter we have.
This explain also why , in the USA, some fire hydrants flow 7000 liters per minutes when in France they flow only 2000. But in the USA, they flow at a low dynamic pressure when in France the flow is measured at 1 bar dynamic pressure.
Best regards to you all
Peter
Best Answer
The flow rate is dependent on the pressure, and the pressure is dependent on the height of water above the hole.
Over a short enough period of time, the change in water level can be ignored and the flow rate can be considered constant. At an infinitesimally short scale we get the concept of instantaneous flow, which again can be applied over a short time scale to closely approximate the change in water volume during that time.
Let's say your water level starts at 100mm and your needle pokes a hole that at that pressure loses 1mm/s of water. (I'm phrasing the measurements as such so that we can work with pressure instead of volume.) You poke one hole in the bottom of balloon A and at the same time poke two holes in the bottom of balloon B. Over the first second, the water level in A drops by 1mm and the water level in B drops by 2mm, a difference of 1mm.
Since the pressure is decreasing proportionally to the water height, the flow rate in the next second is reduced. Between t=1 and t=2, balloon A loses .99mm and balloon B loses .98mm, so the total water level difference is 2.98mm-1.99mm = 0.99mm. Continuing such numerical analysis considering only pressure implies that the two balloons asymptotically approach zero flow as the empty and thus empty out at the same time - an infinite time in the future.
Obviously the infinity there indicates that we've left something out, and that missing factor is the consideration of what causes the pressure in the first place. In the typical case of a rigid container, it's gravity: and even when the water level approaches zero height, the flow rate is non-zero due to the actual weight of the water. When the water level is higher pressure dominates flow rate, but the water weight has its own effect throughout the process. So rather than asymptotically stopping, the flow continues at least above a minimum rate.
The end result considering both pressure and gravity (or tension in a balloon, or a pump) is that the balloon with more holes empties quicker, because the flow rate for a given water level is higher with more drain holes and the balloon with more holes reaches each successive level sooner. The balloon with fewer holes falls behind and never catches up.