How Much Water is Leaking
fOOnOOn
bar in a closed fire hydrant network.
After 30 minutes without any water discharge, the pressure drops down
to 11.2 bar. Then a jockey pump starts repressurizing the network to
12 bar again.
Can someone please tell me how much water is leaking in the 30 minutes
period! or at least give me the way to calculate the leakage!
Thanks
Bruce Rosen
and the section "Density and change in Pressure"
me
wrote:
Trick question, no. You said without any water discharge. This means
zero leakage, no?
Olin Perry Norton
Dear fOOnOOn,
I will first calculate the answer to your problem, and then I will
explain why I don't think
the answer is really very meaningful.
First, the answer. What we need to know is the compressibility of
water -- how much it
changes in volume in response to changes in pressure.
(Alternatively, we could use the
bulk modulus of elasticity, which is the inverse of the
compressibility.) We could quibble about
whether to use the isothermal or the isentropic compressibility, but
it really won't make much
difference. I think that a 30 minute time period is enough time for
temperatures to equilibrate,
so I'd use the isothermal compressibility.
Wikipedia
(http://en.wikipedia.org/wiki/Properties_of_water#Compressibility) gives
values in the ballpark of 4e-10 to 5e-10 (in units of inverse Pa)
depending on temperature and
pressure.
Let's use 4.5e-10 inverse Pa as our compressibility.
Your pressure change is
(12 bar) - (11.2 bar) = (0.8 bar)
1 bar is 100 kPa, so 0.8 bar is 80 kPa.
Fractional change in volume is thus
(8e4 Pa) * (4.5e-10 1/Pa) = 3.6e-5
Change in volume is thus
(3.6e-5) * (2100 cubic meters) = 7.56e-2 cubic meters
If this were a homework problem, this would be the "answer".
In real life, things get more complicated.
The problem is that the compressibility of water is so small that other
effects can easily be of the same, or greater, importance.
Consider the effects of gas (air) bubbles. The compressibility of
gases is much
greater than that of water, so that even a small air pocket or
bubble can contribute
a lot more to the total compressibility than the water does.
It is easy to show that for an ideal gas the isothermal
compressibility is the
inverse of the absolute pressure. Your average pressure is 11.6 bar.
Assuming that is
a gage pressure, the average absolute pressure is 12.6 bar, so an
average
compressibility for a gas over this pressure range is 1/(1.26e6 Pa)
= 0.79e-6 1/Pa.
The fractional change in volume of a gas will then be
(8e4 Pa) * (7.9e-7 1/Pa) = 6.32e-2
Now, let's suppose that your system contains 1 cubic meter of
trapped air.
The volume change will be
(1 cubic meter) * (6.32e-2) = 6.32e-2 cubic meters.
Recall that the volume change of the 2100 cubic meters of water was
7.56e-2 cubic meters. One cubic meter of air trapped in the system
gives roughly the same amount of volume change as 2100 cubic meters
of water.
To use the answer I computed, you must be sure that there is
no air trapped in your system.
Another thing you need to consider is that, as the pressure inside a
pipe
increases, the pipe diameter increases, thus increasing the volume
inside
the pipe.
Olin Perry Norton
fOOnOOn
Thank you all for responding. I would like to add that it is not a
trick question! I should have said in my original post "... without
any intentional water discharge ..."
In principle, I want to consider the value calculated by Olin as an
answer, keeping in mind the issues he raised regarding its accuracy.
On the other hand, one of our colleagues suggested a practical way to
measure the amount of leakage: After the pressure reaches the highest
12 bar, he proposed to manually bleed some water at a much faster
rate, a minute or so, until the pressure drops down to 11.2, and
measure its volume.
This we will do later on because the system is now sectionalized into
more than 10 segments for individual hydro-testing and repair. Once
done, I will hopefully post the answer. By then however, the system,
after working on it, would most probably be much tighter than before.
Thanks again
Olin Perry Norton
I like your colleague's practical suggestion.
Please let us know what the final result is. I'm curious to know.
Olin Perry Norton
fOOnOOn
<mylastn...@icet.msstate.invalid> wrote:
> On 2/21/2010 8:47 PM, fOOnOOn wrote:
>
>
>
>
>
> > On Feb 14, 4:23 pm, fOOnOOn<foon...@gmail.com> wrote:
>
> >> bar in a closed fire hydrant network.
>
> >> to 11.2 bar. Then a jockey pump starts repressurizing the network to
> >> 12 bar again.
>
> >> period! or at least give me the way to calculate the leakage!
>
> >> Thanks
>
> > Thank you all for responding. I would like to add that it is not a
> > trick question! I should have said in my original post "... without
>
> > In principle, I want to consider the value calculated by Olin as an
> > answer, keeping in mind the issues he raised regarding its accuracy.
>
> > On the other hand, one of our colleagues suggested a practical way to
> > measure the amount of leakage: After the pressure reaches the highest
> > 12 bar, he proposed to manually bleed somewaterat amuchfaster
> > rate, a minute or so, until the pressure drops down to 11.2, and
> > measure its volume.
>
> > This we will do later on because the system is now sectionalized into
> > more than 10 segments for individual hydro-testing and repair. Once
> > done, I will hopefully post the answer. By then however, the system,
>
> > Thanks again
>
> I like your colleague's practical suggestion.
>
> Please let us know what the final result is. I'm curious to know.
>
>
> - Show quoted text -
After finding and repairing a number of leaks, the system now takes
2:18h to drop from 12 to 11.2 bars. We bled 900 Liter from the network
through the method described earlier.