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Physics of Water Pressure Regulators/Reducing Valves

by mindarson
Tags: physics, pressure, valves, water
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mindarson
#1
Nov20-13, 02:16 PM
P: 56
My brother is a professional plumber and I'm a physics major, so he often comes to me for explanations of some of the physics underlying his job. I'm familiar with the basics of fluid mechanics, e.g. hydrostatic principles, continuity equation, Bernoulli's law, Poiseuille's law. But often I struggle to help him because I am ignorant of how real-world water distribution works.

So I have a few questions, mostly regarding water pressure regulators/reducing valves (don't know if those two are the same). If you're kind enough to answer, please don't get TOO technical. Please stick to the simplest possible regulator that can be used to answer my questions.

In fact, here is a diagram of the simple regulator I'm using to think about it:



Here is my basic understanding of how these regulators work. As downstream demand changes, this changes the pressure in the pipes. But the pressure in the pipes ideally would stay the same. The regulator's operation allows demand to be met while counteracting accompanying pressure changes to keep pressure within acceptable limits.

As for how this is accomplished, when someone opens a tap, pressure (downstream from the regulator) drops. So the regulator needs to increase the pressure. This happens because, since the pressure dropped, the spring force downward is now able to push the valve down, which actually enlarges the opening, allowing more water flow, which increases the pressure. This happens until the pressure increases enough so that the spring force is now counterbalanced by the upward pressure, and the valve stays in place, i.e. constant pressure. And similarly for an increase in pressure, but everything is reversed.

Finally, all of this can be calibrated to a desired constant pressure by adjusting the initial force of the spring.

Here are my questions.

1. When the tap is opened - e.g. someone is using the faucet - the water pressure in the piping drops, correct? Is this due to Bernoulli's principle, i.e. as the water's velocity increases, its pressure decreases? In other words, after opening a tap, we would expect a drop in the pressure in the pipes, but the reason we don't observe a drop is because the regulator is in place to counteract that drop?

2. My basic understanding is that

open tap > lower downstream pressure > enlarged opening > more flow > increased downstream pressure

is this correct?

3. In the literature, when 'flow' is mentioned - as in, increased flow leads to increased pressure - how exactly is this flow defined? Is it flow rate, in units of volume per second? Or is it something else?

4. My biggest question is the physics of why increased 'flow' results in increased pressure. I know that, according to Poiseuille's law, flow rate and pressure difference across a section of flow are directly proportional to each other. Is that what's happening here? I'm more used to thinking of an increase in pressure difference as giving rise to an increase in flow rate, but it could work the other way around, couldn't it?

5. To what extent does Bernoulli's principle apply to this situation? I did read in one of the texts I found that Bernoulli's principle does not apply to flow around a valve, which is what I'm considering here. At the same time, I feel like it must enter into the relationship between flow and pressure in some way.

Thanks for any help anyone can give in understanding these issues!
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Travis_King
#2
Nov20-13, 04:20 PM
P: 818
Ok, so typically (and in the case where you have a faucet downstream of the valve) what you would call a regulator takes a high pressure stream of fluid and reduces it's pressure so that it can be used at a manageable pressure (or velocity). In the case of a faucet, say you have a water main that is pressurized to 100 psig, and at the same elevation you have a faucet valve that is closed. The pressure just before that valve will also be at 100 psig. When the valve opens to atmosphere, the pressure immediately after the valve drops to 1 ATM and the water velocity will increase until the pressure loss across that valve is sufficient enough to drop the line pressure. This might mean a LOT of fluid, depending on the faucet. Not only that, but the unecessarily large amount of water flowing through the rest of the pipes to meet this demand will cause losses in your system which will reduce the pressure for all other users.

A regulator basically causes a pressure drop across the valve surfaces (based on the principals you outlined in your post) in order to drop the pressure from high to low. Since the pressure downstream of the regulator valve is lower, the losses required to bring the system into equilibrium are lower (and therefore so is velocity, and thus the volume, of water required).

The biggest thing to take from this is that the pressure in the line at the valve is 0 psig when emptying into an open container. The regulator doesn't increase pressure at all, it does the exact opposite. So to talk to point 2, I'd say:

Open Tap -> Pressure at outlet falls to 0 psig -> Opening enlarges -> More flow -> Increased fluid velocity -> Increased friction losses across valve -> At a certain velocity/capacity, the friction losses are sufficient to bring the system back into equilibrium.

3. Flow is volume per unit time for fluids (gallon per minute, cubic feet per minute, cubic meters per hour, etc, etc). Some gasses use mass flow rate.

4. Increased flow results in decreased downstream line pressure. It does give you a higher back pressure (like for pumps and such), but when you are looking downstream, higher flow means higher friction losses, which means a lower downstream pressure.

5. Bernoulli isn't much help in these situations. Valves are a lot more complicated that the flow types assumed by the bernoulli equation. (with that said, it's not entirely correct to say it doesn't apply at all, just that it's not practical, and you can't apply it as is). edit: It may actually be correct to say that it doesn't apply. I thought I read somewhere that with a bunch of tweaking and added terms it could be a useful approximation for energy conservation, but the general idea is that this works for inviscid, steady flow (which flow around a valve poppet surely isn't) where inertial and gravitational effects dominate (which isn't really/necessarily the case in valves).

Also, this is a better picture for you to visualize it: http://www.ctgclean.com/tech-blog/wp...e-1024x752.jpg

Check this out
mindarson
#3
Nov20-13, 04:56 PM
P: 56
Thanks for your thorough reply! I'm still thinking on a lot of it, but there is one big question I have about what you're telling me. I think I'm misunderstanding something fundamental about the mechanism. If increased flow results in decreased downstream pressure, would that not mean that decreased pressure leads to a further decrease in pressure? Suppose a tap is opened, then downstream pressure drops, as you say. Then the diaphragm/spring force acting downward would overwhelm the pressure force upward, and the spring/valve would drop. This would enlarge the opening, thus increasing the flow, thus decreasing the pressure.

Is it because increased flow results in increased friction losses, which in turn results in decreased pressure? Am I understanding that correctly?

I guess I was under the impression that the purpose of a regulator is to maintain a certain pressure in the piping, against drops in that pressure. But if the above is the case, how could it possibly do that?

Travis_King
#4
Nov20-13, 06:04 PM
P: 818
Physics of Water Pressure Regulators/Reducing Valves

There are many types of valves, but a regulator's job is to take a high pressure stream and reduce it's pressure to a low pressure stream which cannot safely or properly use the high pressure stream.

The fundamental point to grasp is this:
If you placed a pressure gauge just prior to the faucet valve, that pressure at the valve is only high when the valve is closed. When the valve is open, the pressure at the valve drops to 0 psig (atmosphere, or ~14.7 psia at sea level). This cannot be increased by any means. In order to conserve energy, there needs to be friction losses in the pipe so that the water pressure at that outlet is 0 psig. This means that the water will tend to increase it's velocity (by increasing the volume of water going through the pipe).

Suppose a tap is opened, then downstream pressure drops, as you say. Then the diaphragm/spring force acting downward would overwhelm the pressure force upward, and the spring/valve would drop. This would enlarge the opening, thus increasing the flow, thus decreasing the pressure.
When the outlet pressure is low, the spring will keep the valve wide open. The regulator has it's own Cv (basically, friction/pressure loss), so it will cause it's own pressure drop, reducing the amount of flow required in order to maintain the equilibrium. As fluid passes through the pipes and ultimately the tap, it will create friction losses which are equivalent to a backpressure being put on the system (from the perspective of the regulator). When enough fluid is passing through the system that that back pressure reaches the set point, the regulator will do it's job and maintain that flow.

If you were to close the tap slightly, you would be creating an additional friction loss, thus back pressure. This additional back pressure reduces the amount of fluid required to flow through the valve in order to maintain the set point (hence, when you close valves, the volume is lower). The local velocity of the fluid at the slightly closed tap may be higher (due to lower flow area) but don't be fooled, the volume is actually lower than a wide open tap.

As you continue to close the tap, the back pressure (friction loss) from the fluid passing over that tap's valve internals increases, until the tap is closed. As it does so, the volume required to flow through the regulator becomes less and less, until the regulator finally shuts when the tap is closed and the downstream pressure rises as the fluid has no place to go; that is, when the pressure on the diaphragm reaches the set point, so that the spring cannot overcome it and open the valve seat.


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