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design/size venturi tube (carburetor type operation)

  • Thread starter Thread starter Ilario
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Well, the fact is that it is precisely the company that produces purifiers, they have always had some softeners (poorly called purifiers) and water treatment systems, but now we are increasing the flow rates and the range and the venturi system there have never been "logics" but copies from the competition without knowing " why".
reverse osmosis has nothing to do with it, those work with osmotic membranes, not resins and membranes do not regenerate, but change.
 
a couple of perhaps trivial considerations, but not knowing the application. . .
the first, definitely practical... the system works until it works! in the sense that if there is no fluid passage or if this is very different from the project data, the depression does not create and the brine is not aspired. In fact, being the pressure system, you have a "travaded" of water in the brine container, unless this is conveniently pressurized. which could also be a half idea...
the second, just for info... but is a homogeneization of the brine required in the water or "what is there?" because this makes the difference between a tube inside the other and a real carburetor
Moreover, how much can the flow range vary? of a few percent or from 0 to 100% open, with constant stenchiometric ratio?
 
a couple of perhaps trivial considerations, but not knowing the application. . .
the first, definitely practical... the system works until it works! in the sense that if there is no fluid passage or if this is very different from the project data, the depression does not create and the brine is not aspired
In fact, being the pressure system, you have a "travaded" of water in the brine container, unless this is conveniently pressurized. which could also be a half idea...
to this could be obvious by inserting a sort of "pin valve" in the suction duct (accorgimento that in some way is already contemplated in such systems)

the second, just for info... but is a homogeneization of the brine required in the water or "what is there?" because this makes the difference between a tube inside the other and a real carburetor
Moreover, how much can the flow range vary? of a few percent or from 0 to 100% open, with constant stenchiometric ratio?
all just and absolutely quotable:smile:

greetings
Mar
 
to this could be obvious by inserting a sort of "pin valve" in the suction duct (accorgimento that in some way is already contemplated in such systems)
true, but add additional load losses and detachment problems if the system is rarely used.
if you use salt water with risk of deposits, each moving part is a seamless grenade.. .
 
a couple of perhaps trivial considerations, but not knowing the application. . .
the first, definitely practical... the system works until it works! in the sense that if there is no fluid passage or if this is very different from the project data, the depression does not create and the brine is not aspired. In fact, being the pressure system, you have a "travaded" of water in the brine container, unless this is conveniently pressurized. which could also be a half idea...
the second, just for info... but is a homogeneization of the brine required in the water or "what is there?" because this makes the difference between a tube inside the other and a real carburetor
Moreover, how much can the flow range vary? of a few percent or from 0 to 100% open, with constant stenchiometric ratio?
for the first part, yes, I am aware of it, but since this type of equipment should be placed in a domestic water system or in bars, restaurants, hotels, etc... the incoming pressure is the pressure provided by the aqueducts, so everything is dimensioned with the nominal pressure of 3bar and minimum of 2bar that you, can also fall further, but considering however the load losses upstream of the entrance into this venturi, and the passages of water can be neglected, it is not a big problem. However, it is always recommended to put a by-pass before the plant to avoid problems in case the softener should be excluded from the plant. for the travase from the container (which cannot be put into pressure because those who use it often do not know how it works and the only thing that is obliged to do is replenish after tot days the salt that has been consumed inside the container) is provided a safety float inside that serves to other similar purposes, but can also help in this case.

the "stechiometric ratio" in this case is not fundamental, the important is the sizing of the quantity of aspired brine according to the amount of resin to be regenerated, only that for quantity of small resin (6-8 litres) serve non-excessive flow rates otherwise the contact time between brine and resin would not be enough to regenerate, while for quantity of large resin they would serve wider flow because otherwise it would not be sufficient force to lift the resin bed. for this reason different injectors must be sized according to the size of the softener.

However today I am collecting all the articles and I am grouping them so as to have everything possible to make comparisons with counter tests so that then I can have a prontuary for future dimensioning. :
 
but since this type of equipment should be placed in a domestic water system or in bars, restaurants, hotels, etc...
Haargh!
don't tell me it's for those "macchinette" shit to produce "pure" water to drink (and "mineral" say..). they cost a barrel, they do not serve anything and "rovinano" water?:biggrin: (I hate the restaurants you need those carafes; I want bottled water closed with label! as it seems to be by law).
Forgive me if you build them, but it has been tried several times (and confirm it to me of the manufacturers/sellers I know) that it is all "fuff" that often shatters the commercial scam (not to mention the sales systems. .

I'm sorry again, but it's my considerations. .

However the technical side can be interesting.

greetings
Marco:smile:
 
no no, it's not about:), we don't do those, as you say, scams, even if those who produce them have had the clinical eye for "good ignorance" and however they have a name that in the field means quality and are making money with "smoke".

are water softeners, all exercises like bars, restaurants, hotels, etc... they have to preserve from limestone plants such as coffee machine, dishwashers and sinks, ice makers and all that makes intensive use of drinking water (in addition I think it is coming out a law that obliges commercial exercises of this type to mount them).
in houses many people use it to preserve the piping, taps, showers, etc. put it downstream of the water meter to have the centralization of the softening water.
I assure you that these have nothing to do with the "caraffine" :).
 
Unfortunately for now I am in affairs with other kinds of facets, but I was preparing a spreadsheet with the various formulas for the verification of the various cases... quiet that as soon as I am new to you.... so surely I will intrigue to something that will go wrong when I come back on :)))))
 
if the formula is:

p1/rho + c^2/2 + gz1 = p2/rho + c2^2/2 + gz2

Then:

30591.5/1000 + 7.8^2/2 + 9.81*z1 = x/1000 + y2/2 + 9.81*z2

....
then:

30591.5/1000 + 7.8^2/2 = x/1000 + y^2/2
Today I came back up (until they let me be... until it lasts....), I ask something I have a doubt:
being the relationship of bernoulli an equation, it is possible that being the fluid always the same and incomprehensible (water) can I remove from the equation "rho"? (excuse the banality, but I haven't applied maths :smile:).
and for the same principle can I also eliminate the "/2" of the "c1^2/2" and "c2^2/2"?
same thing also for "g" as for "z1" and "z2" given the horizontality?
In short, for the equation are data negligible in my case?

Thanks :redface:
 
I did a check and I think I said shit behind the other, except for the last part about "g", "z1" and "z2"?

Do you confirm? (you really miss ignorance, when I went to school I was number one in mathematics, but training makes the difference.... :redface:)
 
I confirm... g*z1 and g*z2 you can delete it.
The rest is better to leave it
Right.

in this case the density would seem "numerically" influential, being for water equal to 1.
but leave it for "counting" of the measuring unit.

greetings
Mar
 
Right.

in this case the density would seem "numerically" influential, being for water equal to 1.
but leave it for "counting" of the measuring unit.

greetings
Mar
However, I have tried to maintain the measuring units of yes, so I had to convert 1 g/cm^3 of water to 1000 kg/m^3 so the data, for the units of measurement I have adopted, I need it.
 
However, I have tried to maintain the measuring units of yes, so I had to convert 1 g/cm^3 of water to 1000 kg/m^3 so the data, for the units of measurement I have adopted, I need it.
I told you to keep it.
but "the entity" does not change, they are equivalences. 1000 gr are equal to one kg, right?

greetings
Marco:smile:
 
at the end of the fair, the equation compares energies to two points, so the units of measurement must be consistent and tending to get joules in both members of the equation
In fact, this is not true, as it is "divided" by the flow (which is constant), then the two members will give j/(kg/sec) and that is m^2/s^2
If I didn't tell the fuck about five minus a quarter!
 
small update:
with bernouli, the data in my possession and fixing the depression to 0.5 calculating 0 the atmospheric pressure the accounts seem to come back and it seems that I approached resolution of a part of the problem.. .
I'm just a little screwed because the problem as a whole is more complicated than expected, many more variables have come into play than I thought... I hope you will come out, even if for now I have some doubts... anyway for everything you have recommended to me right now I thank you because it was very helpful.
If you want more specific, I will gladly place you:).

p.s.: taking a step back, if you also want to calculate the mixing concentration (say the stechiometric ratio) knowing for example that it will have to be 10%, how should I behave?
 
Here I am. Now I'm bumping into load losses.
we say that at this moment I am interested mainly to the localized one since the path is very short, the diameters under examination are very small and everything goes into a very smooth plastic material. but I have a very strong narrowing of section, step from a diameter 12mm approximately to a diameter of 1.2mm.

to find the load loss I applied the formula
y=k*v^2/2gwhere:
k= 0.5 (section tightening)
v= 192.8 m/sec (value previously derived from bernoulli equation)

I have therefore:
= 0.5*192.8^2/2g = 947.3m

Well... but 947.3m at this point I need? in the sense; how can I determine the difference in scope I have from the beginning of the "conduct" at the end having a value in meters?
 

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