Hydraulic restrictors
Hydraulic restrictors
(OP)
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)
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)





RE: Hydraulic restrictors
RE: Hydraulic restrictors
RE: Hydraulic restrictors
RE: Hydraulic restrictors
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.
RE: Hydraulic restrictors
RE: Hydraulic restrictors
RE: Hydraulic restrictors
RE: Hydraulic restrictors
RE: Hydraulic restrictors
RE: Hydraulic restrictors
RE: Hydraulic restrictors
RE: Hydraulic restrictors
RE: Hydraulic restrictors
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.
RE: Hydraulic restrictors
RE: Hydraulic restrictors
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!
RE: Hydraulic restrictors
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
RE: Hydraulic restrictors
RE: Hydraulic restrictors
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.
RE: Hydraulic restrictors
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
RE: Hydraulic restrictors
1. The landing gear geometry
2. a model of the aerodynamic loads (worst case, typical, and optimal case etc.)
3. The inernal resistance of the landing gear such as friction etc.
How are you expect to select the desired restrictor?
The calculations are not so complicated, all you need is to combine the hydraulic model with the dynamic model of the landing gear deployment which will result in a set of differential equation which can quite easily be numerically solved for the set of conditions.
I have done and still doing such analysis from time to time for pneumatic, electric and spring actuated systems using Mathcad Octave/Matlab.
RE: Hydraulic restrictors
Flow is reduced from 6,5 gal to 1,5 (Factor 4.3333)
This means the actuating time is increased wiht 4.3333. I don't think this is right because of the non-lineair components in the system. But it is a good approximation I think (if the factor not becomes to high)
However, I think such analysis are real fun to find out.