WATER FLOW AT THE END OF A MECHANICAL DEVICEMENT...HOW MUCH?

[Saver2008]

Hi!

I have the following problema (see attached file)
On figure 1.0, you can see the system that is formed by two ítems, one of the is like a cap where it is going to be opened due to wáter comes to the system (see figure 2.0)
In external part of cap, it is a pressure of 2 kg/cm2g that has to push the cap when wáter @ 20000 kg/h / 3 kg/cm2g / 50°C. If I consider the continuity equation
Q=VA
It is supposed that all fluid that comes into the pipe, at the end of the pipde will exit the same quantity of fluid.....an example: if I have a straight pipe of 2"Ø, if at one end comes in wáter at 1000 kg/h, at the other end will exit the same 1000 kg/h.

But in reality, in my devicement this is not going to happen. I need to calculate with that área (see figure 3.0) how much wáter will exit? I believe It should be less tan 20000 kg/h but I do not know how to demonstrate it.

Hope you can help me

If you need more data please let me know

Rolander

 

[bimr]

Hard to tell if you have a circular or square orifice.

Pressure drop in valves such as your device is determined by experiment for each valve.

There is no method to calculate precisely the pressure drop and flow through the outlet device. The results will just be a rough estimate. The most critical factors are the orifice size and internal flow path.

You can use one of the online calculators, but the accuracy will suffer.

http://www.spray.com/sprayware/pressure_drop_calculator.aspx

 

[LittleInch]

As said, it's not easy to see if your plug is round or square, but unless there is some serious misalignment of the lug as it moves off the seat, then flow should be equal all the way around the plug.

E.g if this is a square plug, but only open at the top then flow through the two areas will be 10,000kgs/h each - +/- 1000 kg/hr

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Saver2008]

Hi!

bimr and LittleInch thank you very much for your answers.
@bimr and @ LittleInch
Attached you will find more details of dimensions, section cross áreas and cases. It is supposed that when you have the cap or plug totally opened, the Exit Q of wáter will be 20000 kg/hr.
As I close the cap or the plug, it is supposed that Exit Q of wáter will be decreasing, This is what I think that happens. For example, in your kitchen when you open a Little bit the wáter valve for washing dishes or whatever, you will see that not all the total flow of wáter exits. You need to open it completely so you can have more wáter and at the end the total flow of wáter will be exited.

Now for this example, I know that when we have case 3, the Exit Q of wáter will be less tan case 1....are you agreed?

But If I want to calculate, it is supposed that I have the following equation: Continuity Equation---> Q = V A....I know A and I can know V but If I use that.....the initial wáter Q flow ( 20,000 kg/h) will be exiting at the end (20,000 kg/h) because the nature of Continuity Equation.

This is when I get confused....what do you think?

Thank you both of you

Rolander

 

[LittleInch]

Hmmm,

The problem here is that you're stating one thing as a fixed amount then saying it can vary.

You either have a certain pressure and then a certain and perhaps variable pressure drop across your strange plug thing and then the flow rate can vary, or you have a fixed flow rate in which the pressure can increase to overcome any such additional pressure drop. You can't have both. In your tap example, you have a fixed pressure which you then adjust the flow by varying the pressure drop across the tap. All you need to know the flow is the inlet and outlet pressure and the valve CV at different opening points. I think that what you need to do for your plug thing. The main way to discover valve Cv is to do some experiments.

The simplistic nature of this thing and your seeming complete lack of understanding about fluid flow leads me to think you're a student??

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[pumpingtips]

Rolander,
You are focusing too much on this "continuity" idea. As little inch says, it is all about discharge coefficients.

You wont achieve the flow rate with the known / fixed pressures on inlet AND outlet sides. You must either 1/ accept that the plunger pressure is not "2" to achieve the flow rate, or 2/ accept that the plunger pressure is "2" that will most likely result in a different flow rate (could be higher or lower than 20000kg/h).

I think that you are searching for "the flow rate that will be achieved in this system with a pressure drop of (3-2) = 1 kg/cm2 over the "nozzle". You will have to estimate that by trying to select a discharge coeficient that best fits the shape of the discharge.