In a closed system does pressure vary with elevation?

[timsea81]

In a closed piping system (such as a chilled or hot water system in a tall building), is the water pressure at the top less than that at the bottom?

Bernoulli's equation would lead you to believe that it is, but I cannot find anything explicitly stating that this applies to closed systems as it does for open systems like plumbing fixtures or sprinklers. I have found a lot saying that in closed systems like these the pump is sized for friction losses only without mentioning a need to make sure you have enough to have some minimal pressure at the top. This leads me to believe the opposite, that the pressure at the top should be equal to that at the bottom, so I am confused.

For example, say you have a pump in the basement of a tall building that circulates water to the top floor. Say the theory is true, that the pressure at the top will be less than in the basement. If you have 100 psi leaving the pump you might have only 70 at the top, then it would increase in pressure on the way back down and be 100 minus friction losses on the inlet of the pump. The pump would only need to make up those friction losses and bring it back to 100 psi. On the other hand, if you only had 30 psi in the basement you would have 0 at the top and the system wouldn't work right, so in sizing the pump you would also have to verify that the pressure at the top is greater than zero (otherwise your flow would stop, right?).

I've looked through my Fluid Dynamics and HVAC textbooks, ASHRAE papers and all over the web trying to figure this seemingly simple question out and I'm completely confused why I can't find an explanation. If you know the answer please let me know what I need to do to look up to find a full explanation of the physics of the situation. Thanks.

 

[timsea81]

Okay I think typing this out helped me understand this a little better. If the system is full and not flowing, you have 0 psig at the top and some higher pressure, say 30 psig, at the bottom. You don't need to size a pump for that 30 psi because you get it for free as a result of the weight of the water. If 30 psig is what you get from the weight of the water it would be impossible to get less than that (minus friction losses) in the basement, so you don't have to do any pumping in the basement to stay above atmospheric pressure on the top floor. The pump just raises the pressure to make up friction losses, and gravity does the rest, which is why everyone says you don't take into account elevation when sizing a pump for a closed system. However, the pressure is still 1 psi per 2.31 feet of elevation change less as you go up the building.

Correct?

 

[SpectraCat]

Okay I think typing this out helped me understand this a little better. If the system is full and not flowing, you have 0 psig at the top and some higher pressure, say 30 psig, at the bottom. You don't need to size a pump for that 30 psi because you get it for free as a result of the weight of the water. If 30 psig is what you get from the weight of the water it would be impossible to get less than that (minus friction losses) in the basement, so you don't have to do any pumping in the basement to stay above atmospheric pressure on the top floor. The pump just raises the pressure to make up friction losses, and gravity does the rest, which is why everyone says you don't take into account elevation when sizing a pump for a closed system. However, the pressure is still 1 psi per 2.31 feet of elevation change less as you go up the building.

Correct?

I think that's basically correct. Another way to think about it is, since it is a closed loop with water flowing in one direction, you always have one side where the water is flowing up fighting gravity, and one where it is flowing down, assisted by gravity, so the two basically balance out. I think it's similar to operating an elevator with a counterweight .. you don't need a very big motor, because whichever is higher (the weight or the elevator car) is doing almost all of the work to pull the other one up.

 

[timsea81]

Awesome! I think I understand it now. Now can you tell me where I can find this explanation in a Physics book?

 

[firavia]

the pressure at the highest point would be low relatively to the lowest one when the water is flowing, but as spectracat said they are balanced cause when you want to calculate the whole pressure loss in the pipe you will add the friction loss + the loss due to height - the assistance of gravity and if it is a closed loop thn the 2 last elements cancels out cause they are obviously geometrically in y-axis equal.

 

[Drakkith]

I don't believe Bernoulli's principle is referring to pressure due to gravity.

In a closed system the pressure due to the weight of the water will be greater at the bottom of the system than at the top if you have one long pipe where the wieght of the water on top adds to the weight further down.

I THINK you can use this instead. (I'm not experienced with fluids, so I may be wrong)

Liquid pressure or pressure at depth
Used with liquid columns of constant density or at a depth within a substance (example: pressure at 20 km depth in the Earth).

P = ρgh
Where:

P is Pressure
g is gravity at the surface of overlaying material
ρ is density of liquid or overlaying material
h is height of liquid or depth within a substance

 

[cjl]

Drakkith is correct - this has nothing to do with bernoulli. This is simply hydrostatic pressure, in which P = ρgh (assuming an incompressible fluid).

 

[timsea81]

If it's hydrostatic pressure it does have to do with Bernoulli. Bernoulli's equation states that the sum of the static, dynamic, and hydrostatic pressures are constant throughout the streamline. In a closed system where velocity (and therefore dynamic pressure) is constant, a rise in hydrostatic press must correspond with a decrease in static pressure, so as ρgh goes up, the static pressure goes down.

 

[Drakkith]

If it's hydrostatic pressure it does have to do with Bernoulli. Bernoulli's equation states that the sum of the static, dynamic, and hydrostatic pressures are constant throughout the streamline. In a closed system where velocity (and therefore dynamic pressure) is constant, a rise in hydrostatic press must correspond with a decrease in static pressure, so as ρgh goes up, the static pressure goes down.

Sure, but I was not sure if the OP's question was actually about Bernoulli's laws or not. I don't know if its merely pressure do to gravity, or if you actually have to take all of Bernoulli's laws into account.

 

[sleeper365]

I cannot speak to the mathematical facets of your question, however I can tell you from life experiences in plumbing that pressure within a closed pipe system used to supply water through several floors is stepped up for losses via a recirculating pump, this pump is widely regarded as an instant hot water circulation pump, but it is a necessity to maintain pressure as well when you go up so many floors. Just as water systems use larger sizes to reduce to smaller sizes to keep water volumes steady with increases in demands like extra bathrooms and kitchenettes added to the systems, pressures do fall in any system that gravity can be applied to. The second law of thermo also tells us that length of tube is going to affect pressures in hydraulic systems. It is true that closed systems like those in refrigeration do not experience a sufficient loss due to their limited travel and diameter, but the negligible loss does exist. The one place I am certain proves this is long hydraulic lines, they can get to a point where they are so long they fail to operate the system, this is because the volume of fluid increases with the length, and therefore changes the area part of the equation.

 

[cjl]

If it's hydrostatic pressure it does have to do with Bernoulli. Bernoulli's equation states that the sum of the static, dynamic, and hydrostatic pressures are constant throughout the streamline. In a closed system where velocity (and therefore dynamic pressure) is constant, a rise in hydrostatic press must correspond with a decrease in static pressure, so as ρgh goes up, the static pressure goes down.

True enough, but the simple phenomenon of hydrostatic pressure isn't really Bernoulli per se, and the most common form of the Bernoulli equation (at least that I've seen) tends to neglect the hydrostatic pressure changes due to differences in elevation.

 

[russ_watters]

The most common form isn't the general form and really is only most common when talking about air. Bernoulli did indeed include hydrostatic pressure in his equation: http://en.wikipedia.org/wiki/Bernoulli's_principle#Incompressible_flow_equation

 

[arildno]

Bernoulli's equation, in its standard form, is simply an integrated form of F=ma (along a streamline) for (locally) stationary inviscid flow.

It is of little use for a closed pipe system in which viscous effects will soon predominate when the fluid is moving about.

There are probably lots of engineering formulae that are quasi-Bernoullian in which the effects of viscosity are taken care of, but that is another matter.

 

[timsea81]

Bernoulli's equation, in its standard form, is simply an integrated form of F=ma (along a streamline) for (locally) stationary inviscid flow.

It is of little use for a closed pipe system in which viscous effects will soon predominate when the fluid is moving about.

There are probably lots of engineering formulae that are quasi-Bernoullian in which the effects of viscosity are taken care of, but that is another matter.

Actually that is exactly the matter, if the Bernoulli equation is not valid then what is? Also it was my understanding that not all moving flow was viscous (which must be the case since there is a velocity component to Bernoulli's equation), and if the Reynolds number is high enough you can assume it is inviscid because the friction effects are negligible. In the system I am trying to understand the friction forces are not negligible, but that is why there is a pump to recirculate the water: the pump is sized for the head loss due to friction. I guess that would make it one of the quasi-Bernoullian equations you mention, where head loss is added to one side of the equation. So my question is, in this case will pressure on the top floor be about 1 psi per 2.31 feet less than at the bottom, as it would be if the system were open and/or static? Or does moving the water in a closed loop have some effect where the pressure evens out throughout the system and no longer varies with elevation?