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Flow rate compensation: Pressure and or Temperature

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sqaurecut

Mechanical
Oct 13, 2011
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Hi,

I have a test setup where I measure the flow rate of hydraulic oil at high pressure (before a pressure relief valve). The pressure ranges from 100-350 bar. On another test setup this flow is measured at the low pressure side (i.e after the pressure relief valve).
The data from these two tests are slightly different and increases with pressure. I suspect it is due to compression of oil at high pressure. The oil I am using is (grade 32)

I want to able to convert the data measured in one setup to the other setup. (i.e To cover up the difference) Is there is an easy way to do this? or some kind of rule of thumbs regarding this? Any suggestions?
 
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Need more information. Are these different systems? Same system, measuring before and after the same relief valve? Can you provide system setup schematics?

Ted
 
Thanks for the reply TED!

The system is exactly the same in all aspects except for the place where the flow is being measured.

It is not possible for me to provide a schematic right now.But I will try to explain the setup a bit more in detail.

First of all it is an open circuit (as opposed to a closed circuits- so we have a tank/oil reservoir in the system). A hydraulic pump is delivering/pumping oil to a load (which is a hydraulic motor). In between these two machine (pump and motor) we have a safety valve (a pressure relief valve). So this safety valve connects the line providing flow to the motor and the tank.

I am actually measuring the flow through the safety valve AFTER it opens. In one of the test setup this measurement is done before this safety valve while in the other measurement is done after the valve.

And these two values are slightly different.

 
Fixed displacement pump? What is driving the pump? Any directional valve between the pump and motor?

Are you saying the outlet of the relief valve is connected to the motor input? Or does the outlet of the relief valve connect to the return to the tank?

What is the difference in the measured values, before the relief valve compared to after the relief valve?

Ted
 
If the motor is still rotating when both tests are performed, there will be a difference in flow measurement.
before RV = pump output
after RV = RV bleed amount
If it's stalled then
before rv = pump output (presuming fixed disp gear pump)
after rv = RV bleed amount i.e pump output - motor leakage
so both cases in my imaginery circuit give different flow measurements.
 
The bulk modulus of mineral oil dictates that the compression will be negligible at these pressures.

Internal leakage in the motor, the valve and the pump will increase with pressure, so are not comparable over the pressure range.

The dynamic response of the valve will vary with pressure, don't assume that all of the oil will go through the valve.
 
sqaurecut,

Considering a common relief valve connection, there are three flow paths. Incoming flow, Qin. Outgoing flow, Qout. Bypassing flow, Qb. All flow leaving the valve must equal the flow coming in. Qout + Qb = Qin Bypass flow will include valve leakage if the valve is not leak free.

When the relief valve is closed and not leaking, Qb = 0 and Qout = Qin at all pressures and temperatures up to the valve cracking pressure.

When pressure rises up to the valve cracking pressure or higher bypass flow, Qb, will be >0. Bypass flow rate is dependant upon the flow characteristics of the valve which relate bypass flow rate to pressure causing the valve to open. Measuring one of three variables, say Qb, will not tell you anything about the other two, Qin and Qout, unless you know, for instance, that Qin is constant over the pressure range you measure.

At dead head, Qout = 0, unless the motor leaks. Motor leakage will depend on pressure and temperature and motor clearances. Assume no motor leakage, then Qb = Qin. Otherwise, Qout + Qb = Qin, and knowing only Qb, for instance, tells you nothing about Qin and Qout. Unless you know that Qin is constant for your measuring conditions.

Ted
 
"Motor leakage will depend on pressure, temperature and motor clearances."
I agree. The losses will not be the same even if the pumps are theoretically identical.

The downstream oil will be hotter because you are turning pressure into heat as it goes through the valve, therefor thermal expansion of the oil will add to your error. It is approximately .00043 cubic inches per cubic inch of volume for 1°F. My guess is that one rig flow more than the other, so it will generate more heat.

"The bulk modulus of mineral oil dictates that the compression will be negligible at these pressures."
True if there is negligible air in the oil. One rig may have aeration issues.

ISZ
 
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