Hydraulic restrictors 2

[Michiel1984]

How do I have to calculate the pressure drop over a restrictor? If I calculate it with the formula stated below, the pressure drop will be huge!!!

Q=Cf*A*SQRT((2*differential pressure)/Density)

A=0.7 mm^2
Q=1.66 GPM
Density=0.997 g/ml

The units have to be calculated in cubic meters and liters (liters/second). The Cf is mostly in between 0.6 and 0.9, but I don't know what the right value of it is. If somebody does know how I can determine the value of it, it would be really great!

Thank you in advance

Ps: the system pressure is 3000 PSI (20,7 MPa)

 

[FredGarvin]

That is a .94 mm orifice that you're trying to get 1.7 GPM through (at 3000 psi). I would think you'd see a pretty healthy drop across it. Is this a Lee plug of some kind?

 

[Michiel1984]

Yes this is indeed a Lee plug with a Lohm rate of 550. According to the formulaes given at their site, the pressure drop is about 2000 PSI. If the Lohm rate increases to 1000 and the flow rate remeans the same, the differential pressure is about 8000 PSI. That is not possible since the system pressure is only 3000 PSI. I'm desperate....!

 

[israelkk]

I still don't undestand what you are after. The pressure drop is physics if you try to move such a high flow rate through such a small orifice you will have a large pressure drop. Why can't you use a larger orifice? Are you trying to design a valve and the sealing force getting too high? Can you give more info?

 

[Michiel1984]

This orifice is used in a already designed hydraulic system of the Dash 8 Aircraft. I'm trying to calculate the actuating times of the Main landing gear actuator. The normal flowrate in hydraulic lines is 2,91 gal/min. I want to calculate the flow rate after the flow is restricted if the flow rate in the hydraulic lines becomes 1,14 gal/min.

If the flow rate is 2,91 gal/min the actuator needs 4 seconds for a full working stroke (that is a given value). The volume of the actuator is 0,418 liters. I can calculate the flow rate after the flow is restricted, namely:

0,418/4=0.1045 l/s = 1.66 gal/min.

So the restrictor restricts a flow of 2.91 gal/min to a 1.66 gal/min.

My question: If the flow becomes 1,14 gal/min before the restrictor, what flow is permitted after the restrictor?

I tried to calculate this by means of the differential pressure.

 

[israelkk]

What is the needed pressure inside the actuator to be able to open/close it against aerodynamic and other loads? The flow will be the result of the differential pressure between the pressure before the restrictor and after it. The 2.91 gal/min has no importance only the differential pressures over the restrictor. If the system pressure is 3000psi there is also a pressure drop to create the 2.91 gal/min therefore, the pressure before the restrictor will be less than the 3000 psi.

 

[israelkk]

To clarify, there is no way that the flow rate before the restrictor and behind are different. If the flow after the restrictor for example is 1.66 gal/min the flow before it is the same (conservation of mass law). The 2.91 may repesent the maximum flow that the system can handle.

 

[Michiel1984]

The pressure needed to open the actuator is about 1000 PSI.

 

[Michiel1984]

But if I'm not wrong, an orifice decreases the hydraulic flow? To control the speed of the actuator and to prevent cavitation in the system?

 

[israelkk]

If the pressure needed to move the actuator is 1000psi than the maximum differential pressure over the restrictor is 2000psi therefore, the maximum flow rate through the restrictor is the calculated flow rate when Differential pressure=2000psi. Just calculate this flow rate and see if the filling time is 4 seconds.

 

[Michiel1984]

Thank you...It is really logicall indeed. But than you assume the flow rate is always the same. It is really hard for me to believe that, because I did see with my own eyes, the actuating times are different, if the input flow rate differs. What to do with it?

 

[israelkk]

My previous response was an approximate analysis and it give an intelligent estimate. If you want a more accurate analysis you need to solve the actuator and landing gear dyanamic system including the aerodynamic effect altogether with the flow rate. Basically you need to write the differential equations of the whole system to solve it accurately assuming you know the aerodynamic model.

 

[Michiel1984]

That's what I thought...What a pity! Those aerodynamic models aren't available for me! So I think, I can't solve the problem with the means I have!

 

[LHA]

To answer your question about the discharge coefficient:

The coefficient varies due to the type of orifice. A sharper-edged orifice will give closer to 0.6, a well-rounded edge will give as high as 0.98.

It also varies with the length of the the constriction, relative to its diameter. Bump it down roughly 0.1 with L=0.5D, up 0.15 for L=2 or 3 D.

The value of C is really an engineering judgement, there are numerous references to back you up.

 

[btrueblood]

I guess I'm still confused over what you're after. If you put a smaller restriction orifice in the line, the flow is reduced, and the actuation time for the landing gear will be increased. If you increase the orifice size, the flow rate will increase and the actuation time will decrease. The flow will likely cavitate at the orifice irregardless of the orifice size - cavitation will be driven by the differential pressure available. Yes, the orifice is sized originally to prevent cavitation or loss of pressure at the pump; but decreasing the orifice size from the original shouldn't cause a problem. Aero loads will impact the extension/retraction time, but orifice will still be a dominant driver (i.e. smaller orifice will create longer stroke times, assuming all other variables are held constant).

 

[aviat]

I will try and answer some of the questions.
One of many formulas for pressure drop across an orifice is:
DP = C * (Q/A)^2. Q is flow, A is area of orifice and constant C varies with fluid viscosity and the shape of the orifice. DP is affected by flow and for a given orifice; an increase in pressure drop from inlet to outlet is always accompanied by an increase in flow.

Changing your Lee plug LOHM from 550 to 1000 has to be accompanied with an increase in flow. This is shown on Lees site as LOHM versus hole diameter. The formula that is hard to read is L = 0.76/d^2. That's why you get an extremely large DP when trying to use same flow rate. I get a DP of approx 2000 using their formulas.

Cylinder rod speed is a function of flow and piston area. Formulas for speed are:
Metric v = Q/A*16.7 where v is rod speed in cm/s, Q is flow in l/min and A is area in cm^2.
Imperial v = Q/A*19.25 where v is rod speed in ft/min, Q is flow in gal/min and A is area in in^2.

In your restrictor you have an inlet flow of 2.91 GPM and an outlet of 1.66 GPM and the difference of 1.25 GPM is dropped across as heat (using 1.5 HP). Hydraulic fluid heats when flowing though a restriction as the pressure energy upstream of the restriction is converted into thermal energy.

You want to know the flow after the restrictor when input is only 1.14 GPM. Because the required flow to pass test is 1.66 GPM and you have less, there is nothing you can change with the restrictor, even removing it (not recommended), to obtain desired flow. Your problem is upstream of the restrictor, either another component that has too much restriction, filter(s) blocked etc. or your pump is not providing enough flow. If you are using a mule you may need a larger one.

Is this a Q400?

Cheers!

 

[Michiel1984]

Thank you for your answers IHA and aviat.

According to the formulaes of the LEE company, I also get a differential pressure of 2000 PSI. I don't get this since the system pressure is 3000 PSI. Why would you use such a loss in energy in a system?

The LEE jets used are of the type:

JEHA1872550L Used in the "Up" port of the cylinder
FCFA2810100D Used in the "Down" port of the cylinder.

The FCFA control valve does not restrict the flow if power is applied to the "Down" port. Only the other way around.

I tried to calculate the actuating times with the formulaes you gave me Aviat. If I can calculate the pressure drop if the inlet flow is 2,91 GPM (outlet 1.66GPM), is it then possible to calculate the flow after restriction, if the inlet flow is 1,14 GPM instead of 2,91GPM. I thought this was possible by calculating the pressure drop.Because of the relations stated below:

Q=SQRT(P)
P=Q^2

P stands for differential pressure
Q for flow rate

I hope you know where I am after? If not please ask me because I'm really desperate to calculate these actuating times.

Michiel

 

[Michiel1984]

For implementation, the hydraulic system already excists. It is the hydraulic system of a Dash 8 aircraft

 

[aviat]

The JEHA is a High Watt Jet with a LOHM of 550, hole dia of 0.037 and restricts both ways. The FCFA is a directional flow control with a LOHM of 1000, hole dia of 0.027.
The FCFA will restrict flow a lot more than the JEHA and it would depend if it is metering into or out of the down line and if the down line is on rod or cap end of cylinder.

You can't use the formula you have because it doesn't account for Cv, which you could calculate. DP changes with output flow and you have neither of the variables.

The easiest way is to determine output flow is to click on the yellow icon on Lees site that says ASK LEE. If you can't get help there you can remove thr restrictor and flow test it.

I hope you are aware that changes to a landing gear constitutes a major alteration.

 

[Michiel1984]

Thank you,

I indeed askek LEE. They called me on the phone. I told them what my problem was, but even they could not calculate this. I think I have to give up with the calculations. It is really hard. And I am not going to use other restrictors or control valves in the hydraulic system. But thank you for your time and effort with helping me. I really apreciate it.

Best Regards,

Michiel