meccanicamg
Guest
the area in square meters and not square millimeters.
TECNOMODEL
Guest
the area served me only to find the equivalent ø, that I inserted it in meters.
If I express the area in meters then I get a diameter in meters, it should not change anything.
Right?
If I express the area in meters then I get a diameter in meters, it should not change anything.
Right?
TECNOMODEL
Guest
I did not understand what you are referring to, I was applying the formula indicated by mechanicsmg.......
Stan9411
Guest
I apologized not to have read the previous posts, I am accustomed to q = a*sqrt( 2p/rho) which is a volumetric flow.... its is premultiplicated for rho because it is the expression of a mass flow
TECNOMODEL
Guest
so you to get a multiply flow "only" the area *sqrt?
what units do you use?
Stan9411
Guest
the root comes out from the reversal of the bernoulal theorem, assuming initial speed nothing, and obviously represents a speed. multiplying through the passage area you get the volumetric flow, multiplying by the density you get the mass flow. this obviously from the ideal point of view. in reality at the exit of the hole you will have a contracted vein, hence the need for contraction coefficients that are experimental and usually tabulated according to other parameters of your system. Mechanicalmg skipped the multiplication by density by incorporating everything into one equation that from the mass flow by result. the units are those of the s.i.
ingmex
Guest
but is this orifice the only thing in the plant? I think it's kind of weird if it wasn't. the orifice at that point you have to consider it as a localized pressure drop. .evaluate the flow of the circuit considering the other losses and then calculate the relative speed.
TECNOMODEL
Guest
in reality it is not in a plant but inside a pump.
is used to bring a quantity of oil for lubrication by taking it from a pressure duct.
is used to bring a quantity of oil for lubrication by taking it from a pressure duct.
ingmex
Guest
Okay maybe I get it.. you have a pressure oil and you want to take a known amount. the problem is that you also need to know the pressure to the other end of the circuit. This oil goes inside the pump, and there will be load losses.. in a well-known pressure tank?
TECNOMODEL
Guest
the oil goes inside the pump in a pressure zone 0 (or better, at ambient pressure).
from here comes out through the drainage outlet and goes into the reservoir, always at room pressure.
from here comes out through the drainage outlet and goes into the reservoir, always at room pressure.
ingmex
Guest
If you do the calculations like: https://it.wikipedia.org/wiki/orifizio_tarato, get a range of about 0.3 m3/s. as cd I used 0.61 (https://neutrium.net/fluid_flow/discharge-coefficient-for-nozzles-and-orifices/) and b<1 considered null.
TECNOMODEL
Guest
That's exactly the value I don't get.
I have a drainage of the whole pump of 10/12 l/min, the flow of that orifice should be a small part of these 10/12.
I have a drainage of the whole pump of 10/12 l/min, the flow of that orifice should be a small part of these 10/12.
ingmex
Guest
the problem is that this calculation presupposes a practically infinite source at costly pressure. If you can't provide that flow to that pressure, the upstream pressure will flow. I don't think you have a system that can pump 0.3 m3/s of oil to 79 bar..
or, downstream of the orifice generates a counter-pressure
or, downstream of the orifice generates a counter-pressure
crix79
Guest
Hello, I have the problem of calculating the diameter of the orifice to reduce the pressure as I can do
meccanicamg
Guest
If you do everything in excel you find the solution alone. If you want lower pressure you need to increase the pass area.
crix79
Guest
Do I have to use this formula? and try to change values until the q flow meets me?
I have known:flow q: 3570,88 lt/min
delta p: 2 bar
diam tube: 159 mm
fluid density; 1,33 water
meccanicamg
Guest
if you read everything from the first post, you will find all your answers, including formulas. It's just a matter of wanting.