Temperature Pressure Relationship for Water

If it is temperature related the data logger would spot it and track it as a trend. If not, then the data logger would probably show no change.

If no change on the data logger, then the pressure must be spiking as you suspect.

You need a water hammer preventer.

What is happening is the difference between the thermal expansion of water and steel pipes as the temperature changes. The back flow preventer does not allow a little water to flow back out. So you need a small water hammer stopper. This is a buyable item, but can be made with a short length of pipe filled with air with a vertical run to trap the air. Only problem is the air gradually leaves by dissolution and osidation. The water hammer stoppers have a membrane to hold the air.

Do not think about the temperature-pressure relationship, there may be some other different causes ;

1-Your very safety backflow preventer can be older and there may be water leakage from preventer or any other joints...

2-If there are any air pockets at your system, the air is slowly disolved in water, a few hours later, water absorbs oxygen and the pressure will drop...

Because, there is a little change of pressure-temperature and volume for liquids... I think your problem causes from oxygen dissolution into water...

Check these situations.

Cem ATASER

Mechanical Engineer

Adana - TURKIYE

Hi Bill,

I am no thermodynamics engineer but from the little that I know about expansion with temperature the pressure would be controlled by four factors.

1/. Difference in temperature

2/. Expansion of the steel pipes

3/. Expansion of the water

4/. Volume of air trapped in the system.

So given the above variables we know that the temperature variation is 15ºC and the thermal expansion coefficient of steel would be

E_S = ΔT x τ_S

E_S = 15 x 9.28 x 10^-6

E_S = 1.392 x 10^-4

Now the expansion of the water is more complex as the coefficient of expansion varies with temperature as follows

τ_W17 = 1.7392 x 10^-4

τ_W32 = 3.1984 x 10^-4

and if you wok out it gives you an increase of

E_W = 3.113 x 10^-3

So the effective expansion of the water would be the expansion of the water less the expansion of the steel

E_T = E_W – E_S

E_T = 3.113 x 10^-3 – 0.1392 x 10^-3

E_T = 2.9783 x 10^-3

E_T ≈ 0.3%

For this example we can assume that water is an incompressible liquid and therefore there must be some air in the system otherwise the pipes would burst. The question is therefore how much air would there be in the system so that a 0.3% increase in the volume of the water would explain the increase of pressure we see. We know that the initial volume of gas V1 is related to the final volume as follows

V₂ = V₁- 0.3%

And if we ignore the thermal expansion of the gas trapped in the water then

P₁ V₁ = P₂ V₂

And substituting we get

P₁ V₁ = P₂(V₁ - 0.3%)

V₁ P₁/P₂ = V₁ – 0.3%

P₁/P₂ = (V₁ – 0.03%)/V₁

P₁/P₂ = 1 - .03%/V₁

.03%/V1 = 1 - P1/P2

1/V₁ = (1 – P₁/P₂)/0.3%

V₁ = 0.3%/(1 – P₁/P₂)

V₁ = 0.03%/(1 – 120/375)

V₁ = 0.3%/(1 – 0.32)

V₁ = 0.3%/0.68

V₁ = 0.441% ≈ 0.5%

So the question is how accurate is your estimate of the volume of air in the water being between 1% and 10%? If the air volume is in fact as low as 0.5% then the temperature variation you have quoted could indeed cause the type of pressure fluctuations you are seeing.

P.S. I apologize for including every little step in the process but I haven't done any thermal calculations for about 30 years and wanted to make sure that you could follow my calculations and correct any errors I have made.

You have more patience than I -- thanks for the analysis.

Bill

"Can anyone explain in plain English what the pressure rise of water will be in a closed vessel with no entrained air per degree Kelvin rise in temperature? I'm not concerned about nonlinearities because the temperature range is 290^oK to 305^oK (17^oC to 32^oC or 60^oF to 90^oF). "

Thermodynamically, for liquid water there will be no pressure change related to the change in temperature for that range. a good model for this reasoning is dP/dT|v = α_p/k_T. for water: α_p = 207K^-1 and k_T = 49000000bar^-1

solving for dP gives dP =(α_p/k_T)*dT. since α_p/k_T is such a small term (4.22*10^-5) you would need a very large dT to see a change in dP.

so that rules out the temperature dependency for pressure based on having only liquid water in your system. Im still thinking about the other scenerio with entrainment.

There may be zero free air in the system. dissolved air is part of the fluid and is not compressible.

The fact that you have a sealed system with the back flow preventers in place with no compressible component means that even a small change in water volume greater than the expansion of the steel over the same range can lead to high pressure differentials.

The solution is either a compressible section (water hammer preventer) or a capillary back flow tube that bypasses the back flow preventer.

The only other explanation is the water main pressure rises, which is not likely as it is probably stand pipe controlled.

Bill,

I have gone a bit further and factored in the increase in temperature of the air trapped in the system and the result was that if there was a trapped air component (not dissolved air) of 0.452% in the system and no leakage back past the non return valve then you would get the pressure rise you have with an increase in temperature increase you have. So my original calculation that if you have around 0.5% of the total volume of the system being trapped air then you would indeed see the pressure rise you are getting due to the thermal expansion of the water in the system.

In answer to ChemE119 about the pressure increase in the system due to thermal expansion with no entrained air. For this type of problem you can treat the water as uncompressible so therefore if there is nothing that can be compressed the pressure rise will be controlled by the tensile qualities and strength of the piping. The result would be a massive increase in pressure that would cause the pipes to expand until the elastic limit was reached at which point deformation would take place. If the temperature continued to rise it would result in a catastrophic failure of the piping. That's why there are pressure relief valves on water heaters. The short answer is that the pressure rise is controlled by the mechanical properties of the pipe and therefore can't easily be stated. The other point is that you cant ignore the change in thermal expansion coefficient as if varies considerably. The chart below shows the expansion coefficient of water from 0º to 100ºC. Note that as the water cools and approaches 0ºC the coefficient becomes negative.

If you are a masochist and wish to follow the calculation for the volume of trapped air through here it is with every algebraic step included. Best of luck.

The truth of the matter is the long pipe runs will stretch a little and relieve the prssure so it will not go to breakage as it does when water freezes and expands by ~10% and will break even thick pipes.

There is a limit. If the pipes are thich enough the water will fail to freeze. A refecnce to the phase diagram fro water will tell you why. Hellish high pressure thuough.

http://www.lsbu.ac.uk/water/phase.html

"If you are a masochist and wish to follow the calculation for the volume of trapped air through here it is with every algebraic step included. Best of luck."

You're worried about me being a masochist??? I didn't do as much work as you before I asked CR4 people for help . . .

Bill,

I am a firm believer in the adage that a picture is worth a thousand words so I have produced a plot of the expected head pressure verses temperature it the problem of the rising head pressure is caused by thermal expansion.

Something to note is the fact that the pressure increase with temperature is non linear. In fact if you look at the equations you see that the pressure is inversely proportional to the original volume of the trapped air minus the volume that the water has expanded by. As a result there will be a point when the expansion of the water equals the volume of trapped air and when this happens the bottom line of the equation will be zero and the pressure will be off the scale.

The practical use of the graph would be to compare it with a plot of the actual pressures you are experiencing against the temperature and compare it with the one above. If the relationship of temperature and pressure doesn't resemble this graph then the problem you are experiencing isn't temperature related. I would be interested knowing what was the final result of your investigation to see if it is anything like my calculations.

Yeah, but the amount of trapped air is significant. The sprinkler riser pipes comprise about 5% of the volume of the system, and there is trapped air in each one. The only way temperature can cause a rise in pressure is for the pressure to get high enough to virtually eliminate the volume of air, which can't be caused by temperature with 5% of the volume being air.

Several people mentioned putting in an expansion tank (water hammer preventer, etc., (a rose is a rose, is a rose, is a rose . . .)) to overcome the problem, which will work, and it is acceptable by our fire marshall and is allowable under local and state codes.

Thanks again, Bill

I assume the backflow prevention valves are doing a really good job. This is cheaper to fix than to study. Guess at a bladder tank size and put it in. If the first one isnt enough put in another .....so forth and so on. After doing all of the calc work a consultant will tell you to do the same thing and charge you 10 times what it took to do the job.

easy task to add all the runs of assorted pipe of various diameters and compute the gallons. Then assume zero trapped air and callculate how much that water will expand over the temp range = max bladder size.

I would expect a 250 CC bladder would suffice for this. As pressure rises the trapped air goes into solution and loses it's effect in preventing the problem. That means it will change the above curve by flattening and the slope will change when the last air is gone into solution when it will get very much steep

Since this is a sprinkler system I am not sure about the legality and safety of installing an anti hammer device with a diaphragm. There are several thing I think could be important:

1/. What would happen in the event of a fire that engulfed the anti hammer device?

2/. Could failure of the anti hammer device effect the performance of the system?

3/. Could it cope with the fire brigade coming along and connecting their fire tenders to pump extra water in and boos the sprinkler header pressure and flow rates?

I expect this has been solved many times in the field. These

A search shows 55,000 hits on this.

http://www.google.ca/search?hl=en&q=%22sprinkler+system%22+%2B%22backflow+prevention%22&btnG=Search&meta=

A water hammer device is just a small dead end with some trapped gas and it is inside a steel shell with a pressure valve (a schrader type, as on tires) so you can center the membrane.

Backflow protection and anti hammer are two separate problems. Normally with a sprinkler system you never need to worry about hammer since once a sprinkler is set off the only way to turn off the flow is at the source which is usually a header tank. My point is that sprinkler systems need to be failsafe and immune to damage by fire. That's one of the reasons the pipes are steel as copper and plastic can fail in a fire and render the system useless.

Of course they are, I never said otherwise.

The problem here is what to do to stop the pressure excursions since the law says you have to have a BF preventer.

One way is some type of expansion area, as in anti water hammer device. The other is an overpressure vent to drain and another is a vent to by pass the BF preventer.

Probably an overpressure vent is the thing you need.

The trouble with sprinkler systems is that they need to be failsafe or at least operation in the event of a failure. Some years ago I worked on a system in a high rise building that was over 1,000 feet from basement to roof. As you can expect trying to cope with those sort of header pressures is a nightmare. Here is a simplified diagram of the system.

As you can see the solution was to use three separate sprinkler header tanks that were filled from three separate cross linked pumps in the basement. Each of the sprinkler systems had a booster pump that could be started in the event of a fire to supplement the flow of water to the system. The electricity for the pumps was supplied from the mains with a diesel generator as backup and the cabling was all fireproof. There were two fires in the building during the time I worked there and on both occasions the fire was well and truly out by the time the fire brigade arrive on site.

The idea of using header tanks is one solution but I doubt it is practical in this situation. Tthe question of how to control the spikes in pressure in a fail safe manner still remains. The only thing I can think of is as you stated some sort of hammer trap. To make the system failsafe you would need to install some sort of flow restriction device that would allow the static pressure to flow either way to the hammer trap but that would restrict the flow so in the event of a failure it would stop the flow of water. Also sprinkler systems usually have connections to enable the fire brigade to connect one of their tenders to the system as a jacking pump, like the boost pumps in the system above. This is so they can supplement the pressure and flow in the sprinkler system in the event of a fire so there may be all sorts of pressure fluctuations while the system is operating.

All this of course is extraneous as the original question asked whether the increase in pressure could be due to either the thermal expansion of the water or hammer coming from the mains?

Wow! Thank all of you for your analyses and input. There is, without a doubt, dissolved air and trapped air in the system. Whether or not air gets entrained by some weird process before the water reaches the end of the line where this anomoly is being seen is another story.

The backflow preventers (as well as the sprinkler system and the rest of the building) are relatively new, but I haven't yet figured out why old ones (or new ones for that matter) would have anything to do with it.

Other buildings of near the same age and physically near the one in question do not exhibit such water pressure anomolies and are different only by being somewhere other than at the end of a water main. The building next door has temperature fluctuations over the same range.

Since this is not a common problem, common sense tells me that pressure rise is not due to temperature, and it would be since most places are subjected to temperature fluctuations in the range I'm concerned with, and some are subjected to much, much greater fluctuations. Common sense tells me that it is due mainly (only?) to water hammer spikes greater than maximum recorded pressure and of short duration which do not occur at night when I know nearby processes which might cause them do not happen.

Thanks again, Bill

The more I think about it the more I am inclined, like you, to believe the problem is with water hammer rather than temperature. If the mains were terminated like this

rather than being connected in a T you could very high hammer pressures. This sort of connection can actually amplify the pressure wave rather than just reflecting it back down the pipe. The result is that the poor old person at the end of the pipe cops all the hammer and nobody else gets any.

I think he has a sustained high pressure caused by too good a back flow seal trapping water and as the temperature cucles = a high pressure.

. Water hammer creates a transient very high pressure = the knocking sound

Aurizon and masu I think we're all on the same page now. Good point about the age of backflow preventers -- if they're old, they may leak more due to corrosion. Unfortunately, the ones we're dealing with are fairly new and probably work well.

The sprinkler system tap to the main is literally at the end of the main water supply to the building, which is a larger diameter pipe (6") than the sprinkler main, which is 4". The main at the road is 8".

I have just learned (morning 11/27/06) that when a water utility company switches from one source to another (daily) pressure spikes as high as 175 psi have been measured on 12" mains within a mile of the facility, and has caused blow-outs in the main. That's another problem which has to be solved.

Thanks again, Bill