Measuring Pressure Gradient Across Valve 1

[TomFin]

System consists of a cylinder (piston), pressurizing fluid through a mechanical flow control valve in the restricted direction, (metered flow in 1 direction, free flow in oppositte direction), pouring into a resevoir exposed to atm press.
The system was tapped with two pressure gages immediately before and after the metered control valve. Operating the piston gives me a gage reading of 1000psi just before the valve and 0psi just after the valve. I'm assuming the 0psi is due to zero back pressure (gage) at the resevoir therefore the pressure head is in the form of (V^2)/2g.
Comments appreciated.
Now if I apply resistance to the outlet hence backpressure the gage after the valve reads 800 psi. So a 200 psi (1000-800) drop occurs across the valve according to the gages.
Without the backpressure one might interpret the pressure drop as 1000psi. With the backpressure, one might interpret the drop as 200psi. If a cap was used on the outlet to provide infinite resitance both pressure gages would read the same 1000psi, right? So with an added resistance to the outlet (backpressure) I'm assuming the difference in gage readings give me the pressure gradient across the flow valve. The various gradients just represent different flows with delta 0psi corresponding to 0 flow. Is this correct?

 

[JimCasey]

Sounds pretty good to me. Make sure your pressure readings are free-stream and not so close downstream as to get the vena contracta, which would show an artificially low pressure due to high velocity.

 

[BigInch]

Using the Bernoulli equation, you can see you have an overdetermined system, i.e. not enough unknowns. If you know the inlet pressure and the flow, you must calculate (or physically adjust) the outlet pressure. If you know the outlet pressure and flow, you must calculate the inlet pressure. If you know both inlet and outlet pressures, you must calculate flow. Here you are simply physically adjusting the outlet pressure (backpressure) to correspond with your gauge observations and the mathematical results predicted by the Bernoulli equation given a known inlet pressure, a known flow and an unknown outlet pressure. In this case the known flow could be determined from the maximum possible flow across the valve, Q = Cv x dP^2/SG^0.5 You have thus determined that the outlet pressure must be 800 psi (or less) to match the inlet pressure and this valve's maximum flowrate for that system. You have also reached the valve's maximum flow capacity at your particular %open when you have a 200 psi differential pressure across the valve and you have found that any more pressure drop will not accelerate the flow nor increase the actual pressure drop across the valve any more than 200 psi you already have.