Relationship between valve opening degree and flow, pressure, and velocity

My understanding isn’t necessarily comprehensive; it’s just my own opinion. You mean that the second idea is basically not in line with the facts. I think the core of Bernoulli’s equation is energy conservation; however, when you reduce the valve opening, the resistance throughout the system increases and energy is reduced, so Bernoulli’s equation can no longer be applied.

Could it be said from this perspective that this is why valves with wait-hundred characteristic curves exist?

This post was last edited by HEJIYUER on 2016-4-12 00:15. An automatic valve is used to \"compensate\" for changes in pressure drop along the pipeline. By changing the opening degree of the control valve (i.e., the valve’s own resistance coefficient), the control parameter is brought back to the set value. The process parameters change first, and then the control valve \"follows up\" – it’s a matter of sequence. The Bernoulli equation calculates the pressure distribution in a steady state, while control valves are used to \"correct\" deviations in process parameters. According to the design of the fuel station, a pressure-controlled PIC (or PCV) is used to stabilize the oil pressure at the outlet, and it is desired that the outlet pressure remain stable—at your set value—regardless of changes in user traffic. 1. If the flow rate required by the machine increases (as demanded by the machine), the outlet pressure will decrease. At this point, the PIC will automatically increase the opening of the control valve, reducing its resistance coefficient and thereby decreasing the total resistance loss in the pipeline, so as to meet the requirements regarding the main pipe resistance at an increased flow rate. This process continues until a new equilibrium is reached between the flow rate and the outlet pressure (calculated using Bernoulli’s equation), after which the control valve no longer needs to be opened further. 2. If the oil filter of the machine becomes clogged, the oil pressure at the outlet will increase. In such a situation, the PIC will close the valve to reduce the pressure; since the PIC monitors the outlet pressure regardless of the flow rate, this is a dangerous condition. 3. If I use flow FIC as the outlet control, the valve will close down at value 1, because its setpoint is the flow rate, in order to maintain stability in flow. However, in the case of 2, the control valve will automatically open wider. Analyzing points 1, 2, and 3: at the fuel station, what we control is the pressure, while the flow rate is determined by the machine itself, which decides how much flow is needed; this is achieved by adjusting the resistance coefficient of the valve. In Bernoulli’s equation, flow rate and pressure loss are two variables; you can only choose one parameter to control (pressure or flow rate), leaving the other one uncontrolled. The control of centrifugal pumps is the same; it is determined by the pump’s operating curve.

Thank you so much; you were extremely detailed. At a gas station, it is the changes in the flow rate required by the machinery that cause variations in the pressure in the main pipes; the control valve then automatically adjusts its opening degree to maintain stable pressure. Another thing I’m not quite sure about is whether or not orifice plates really have the effect of increasing pressure. Usually, an orifice plate is installed when the pressure in the lubricating oil branch line is insufficient. However, as I understand it, an orifice plate increases the back pressure; it can only raise the pressure before the orifice plate, which makes the pressure gauge before it show that the pressure is sufficient. Meanwhile, the pressure in the lubricating oil branch line behind the orifice plate does not increase – in fact, it decreases slightly due to pressure loss. Are there still some principles of orifice plates that I don’t know about?

I majored in mechanical engineering; I’m working on the oil supply system for centrifugal compressors. There are some valve pressure issues that I’ve been thinking about for a long time without finding any solution. I guess I haven’t mastered fluid dynamics well; I’m seeking advice from everyone here. 1. As the valve opening increases, the work required to operate the valve decreases; in other words, the pressure drop across the valve reduces. If the pressure before the valve remains constant, both the flow rate and pressure after the valve will increase. 2. As the valve opening increases, the flow rate increases as well, and the velocity rises; the static pressure of the oil is converted into dynamic pressure. The decrease in static pressure (lower pressure at higher flow velocities) is derived from Bernoulli’s equation for fluids. 3. The two conclusions are opposite; the first point is definitely correct. I have two ideas regarding the mistake about the second point. The first case is that before and after the valve opening increases, it belongs to two separate systems, so the Bernoulli equation cannot be applied. The Bernoulli equation holds only when measuring the conditions before and after the valve at the same time; when the flow rate is the same and the cross-sectional area remains constant, the flow velocity is also the same, and thus the dynamic pressure is identical. However, the static pressure decreases due to losses associated with the valve. The second idea is that the Bernoulli equation can be applied before and after a change in the valve opening degree. When the valve is opened wider, the flow rate increases, the velocity rises, and the pressure drops. However, at the same time, as the valve is opened wider, the pressure loss decreases, causing the pressure to rise. Between these increases and decreases, it is the increase in pressure that plays a decisive role (I’m not aware of the exact formula, but I guess the pressure drop caused by the increased velocity is similar to a higher-order infinitesimal of the pressure loss; therefore, it is the pressure loss that determines whether the pressure rises or falls, with the ultimate result being an increase in pressure). There are no objections to the above explanation. 4. The pressure at the pump outlet is determined by the system resistance. If the shut-off valve at the pump outlet is closed, the pressure at the pump outlet will rise. But at the same time, the energy required to overcome the resistance of the valve also increases. So, how does the pressure behind the valve change? Comments: For centrifugal pumps, the pump outlet head is determined by both the pump’s characteristic curve and the pipeline curve, and the head includes dynamic pressure and static pressure. Closing the shut-off valve reduces the flow rate, increases the total head, decreases the dynamic head, and raises the static head. When the valve is closed, the valve pressure loss increases and the pressure downstream of the valve decreases. You said that the pump outlet pressure is determined by the system resistance; this refers to positive-displacement pumps, not centrifugal pumps.

My understanding is that installing an orifice plate increases the pipe resistance, which in turn raises the outlet pressure of the positive displacement pump, that is, the pressure in the main pipe. For centrifugal pumps, although they can also increase the pressure at the pump outlet, the increase is not significant, and this is determined by the characteristics of the centrifugal pump’s performance curve. For any type of pump, the orifice plate should be placed behind the user in order to increase the user’s flow rate by raising P1, subject to the limitations imposed by the pump’s capacity.