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WATER PUMPS, TANKS, TESTS, WELLS, REPAIRS
WATER CONTAMINANT LEVELS
WATER HAMMER NOISE DIAGNOSE & CURE
WATER ODORS, CAUSE CURE
WATER PUMP REPAIR GUIDE
WATER PRESSURE LOSS DIAGNOSIS & REPAIR
WATER PUMP SHORT CYCLING
WATER SOFTENERS & CONDITIONERS
WATER TANK REPAIR PROCEDURES
WATER TANK: USES, TROUBLESHOOTING
WATER TESTS, CONTAMINANTS, TREATMENT
WATER TREATMENT EQUIPMENT CHOICES
WELLS CISTERNS & SPRINGS
WELL FLOW RATE
WELL WATER PRESSURE DIAGNOSIS
WELL YIELD IMPROVEMENT
WINTERIZE A BUILDING
This article describes Water Tank Pressure Calculations - the effects of temperature, air charge, tank size, pump pressure settings on water system performance, using Boyle's Law, Charle's Law, and the Combined Gas Law in a building water supply system where a private well is the water source.
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This article examines the effects of ambient air temperature changes, water temperatures, and the water tank air charge or tank pressures on water pump and water tank operations and performance. If your water pump, well, or water tank don't seem to be working right, this article gives some useful and basic underlying theory that might help understand what's going on. Practical advice about water pump, well, and water tank diagnosis and repair are provided in a series of articles organized at WATER PUMPS & TANKS & WELLS and linked-to by links at the left of these pages.
The effects of temperature on water pressure tank volume and water delivery are very small and are generally ignored in most calculations regarding the behavior of water pressure tanks. We visited this topic and documented the underlying principles and calculations at the request of several readers.
The amount of usable water volume that is delivered by a water pressure tank is essentially equal to the volume of expanding air inside the tank while water is leaving the tank. (Water is not very compressible so all of the volume change is due to the air pressure.) Our discussion of Boyle's Law (below) explains that P1V1 = P2V2
or stating Boyle's gas law in plain English, that the multiple of Pressure x Volume remains constant. In other words, as water leaves the tank and thus air pressure inside the water tank decreases, the volume of air expanding inside the tank increases accordingly. So if P1V1 represents the starting pressure and volume, P2V2 will represent the ending pressure and volume, and while the P's and V's themselves change, the multiple of P1V1 will always equal P2V2 (unless air also escapes from the tank along with water).
A reader wrote that letting air OUT of her water tank improved the pump performance, leading her to pose that the tank had previously contained too much air volume (or pressure) due to changes in ambient air around the tank in warm weather. Let's consider what happens to air pressure in a water tank under varying conditions. While experts agree that the effects of temperature on the operating cycle of a water pressure tank are generally negligible, we agreed to look at the question more technically and in detail.
A bladder-type water pressure tank has the air pumped into the tank, usually from the top at a schrader valve, and that air never escapes unless the tank or its water-bladder has a leak. This tank pre-charge is set at the factory but might be adjusted by a plumber depending on the operating pressure range of a particular water pump and control.
The volume of the water tank is fixed. Here I'm using an idealized 30 gallon tank.
As water is sent by the pump into the bladder in the tank bottom, the bladder swells and the air above it in the tank is compressed. The volume occupied by the air decreases. The pressure of the air increases (up to the cut-off pressure set by the pump control). The volume occupied by the water (in the bladder) increases by the same amount as the air volume decreases.
Sketch courtesy of Carson Dunlop Associates.
How Much Does Water Tank Air Volume Vary?
For our theoretical, simplified water tank and treating air as an ideal gas:
Suppose we have a water tank that is 30 gallons of cubic space and our pump control is set to cut-in at 33 psi And cut out at 50 psi.
Suppose inside the tank is a water bladder that can hold between 0 gallons and 10 gallons of water. I am cheating a little on choice of these numbers just to make the math easy as you'll see below.
30 gallons (total tank size) is 4 cubic feet of space. (http://www.western-water.com/Acre-Foot_formula.htm)
With the water bladder full we have:
We can re-examine the calculation of variation in water tank air volume in more detail using Boyle's Law.
Boyles Law: P1V1 = P2V2 at constant temperature tells us how the volume of a gas changes with changes in pressure, or vice versa, how the pressure of gas changes if we change its volume.
Turning on the water somewhere in the system, we run all of the water out of the tank. (This is not what happens in normal water pressure tank operation. Normally the water pump comes on before all of the water has been pushed out of the water tank. We are using all of the water in this example for simplicity and to describe the absolute maximum amount of water that could be taken out of a water pressure tank.
What happens in the water tank as we run out its water completely? The water volume in the tank goes from 10 gallons to zero. The air volume in the tank goes from 20 gallons to 30 gallons. (Or if you prefer we can convert all of these gallons to cu. ft. - the equations don't care)
What happens to the air pressure during this change? We can work in gallons or cu. ft. of volume.
20g x 50 psi = 30g x ? PSI solving for PSI = (20x50)/30 = 33 psi
Now in the tank we have 30 gallons of air (or 4 cubic feet of air) at 33 psi. and we have 0 gallons of water with a tank bottom pressure at the water outlet = the air pressure in the tank = 33 psi.
At this point the pump comes back on and pushes water back into the tank. As water enters the tank bladder it expands the bladder and takes up room in the tank, compressing the air in the tank from 33 psi. back up to 50 psi (the pump cut-off point) and back to our original 20 gallons of air at 50 psi and occupying 2.67 cubic feet and 10 gallons of water being pressed-on with 50 psi of force and occupying 1.33 cubic feet.
A strictly accurate use of Boyle's law would also consider that the pressures we see on a water tank pressure gauge are not absolute pressure which is what Boyle works with but rather, relative pressure - that is, when a water tank pressure gauge reads "0" psi, it has been calibrated for sea level on earth, where the normal air pressure is about 14.7 psi. (This is the "weight" of the atmosphere pressing down at the earth's surface, per square inch.) So to comply with a strictly correct use of Boyle's law, we'd add 14.7 psi to our water tank pressure gauge readings to get absolute pressure.
Let's see what happens when we use absolute pressures and then repeat the Boyle's Law calculation performed above:
For a water tank whose pump turns on at gauge pressure of 30 psi and off at gauge pressure of 50 psi, the absolute pressures (adding 14.7 psi for one atmosphere at sea level) will be 44.7 psi and 64.7 psi respectively.
Using our example of pretending that we can take all of the water out of the water tank (0 or now 14.7 psi when the tank is "empty" of water) and that we then pump water back in to a pressure of 50 psi, and using a water tank with a 10 gallon internal volume which will be occupied by some volume of water and air when the pump cycles off:
Tank Volume = 10 gallons = Water Volume + Air Volume
Tank Volume = 10 gallons = 0 gals Water + 10 gals Air (at 0 gauge pressure or 14.7 absolute pressure) when the tank is empty
Let's now pump in water until our water tank pressure reaches 50 psi (gauge) or 64.7 psi (absolute pressure)
Tank Volume = 10 gallons = W gals Water + A gals Air
Using basic algebra, W gals Water = 10 gallons - A gals Air
The air volume in the tank has changed to 10 gallons at 14.7 psi to A gallons at 50 psi.
Using P1V1 = P2V2, we have (14.7 psi x 10 gals) = (50 psi x A gals) (just looking at the change in air volume)
(14.7 x 10) / 50 = A gals of Air or 2.94 gallons. Our air, compressed from 10 gallons at 14.7 psi to the new pressure of 50 psi will now occupy just about 3 gallons of volume.
Finally! How much space will be occupied by water in our 10 gallon tank?
10 gallons of air compressed down to 3 gallons of air, letting in (10-3) or 7 gallons of water. We have 7 gallons of water in the 10 gallon water tank when the pump shut off at 50 psi.
Now let's adjust our calculations to account for the normal operating range of a water pressure tank and water pump that turns on at 20 psi and off at 50 psi.:
At 50 psi (gauge) in our 10 gallon water tank we have 7 gallons of water and 3 gallons of air.
At 0 psi (gauge) in our 10 gallon water tank we have 0 gallons of water and 10 gallons of air (at 0 gauge or 14.7 psi absolute pressure)
At 20 psi what are our water and air volumes?
Air: (14.7 psi x 10 g) = (20 psi x A gals) or (14.7 x 10)/20 = 7.35 gals of air which leaves (10 - 7.35) = 2.65 gals of water. So at 20 psi we have 7.35 gallons of air and 2.65 gallons of water in the pressure tank, and at this point the pump comes back on to send in more water.
So what is the draw down volume of water in a 20-50 psi 10 gallon water pressure tank? We subtract the 20 psi water volume (2.65) from the 50 psi water volume (7.00) and we get (7.00 - 2.65 = 4.35). So the draw down volume of water in a 10 gallon water pressure tank operating between 20 psi and 50 psi will be just 4.35 gallons!
What is the actual water volume draw down during a typical water pump and water tank operating cycle?
One can conclude a few useful things from this calculation
Charle's Law Describes How Water Tank Air Pressure Changes With Shifts in Air Temperature in the Water Tank
What about the effects of changes in ambient temperature around a water storage/pressure tank? Suppose we wonder if seasonal temperature changes might cause important changes in in-water-tank pressure. For simplicity we'll work with a fixed volume full tank of just air. We'll change the temperature of the tank and its air from 60 deg.F. to 90 deg.F.
As most people would guess from practical experience, raising the temperature of a container of air while keeping the container size fixed will increase the pressure of air in the container.
Charles' Law: V1/T1=V2/T2 says that if we raise the temperature of a cubic foot of air, the air will want to occupy a larger volume.
30gal of air (or 4 cu. ft.) of air at 33 psi and at 60 degF is changed to 90 degF (by the warming ambient air) while the container size is kept constant.
Charles would say: 4cu.ft./60degF = NEWCUFT/90degF or 0.066 = NEWCUFT/90 0.066 x 90 = NEWCUFT or about 6 cu. ft.
This is 2 cu. ft. more than where we started, or a 50% increase in the starting volume of air.
Technical note: when working with the gas laws and temperature, all temperatures have to be converted to Kelvins or "K" - their Kelvin equivalent before the law can be applied. The Kelvin temperature scale relates temperature to absolute zero, where 0 Kelvins = absolute zero. We can convert Fahrenheit to Kelvins using this simple formula: K = (degF + 459.67) / 1.8.
We can convert Kelvins to Fahrenheit using this formula: degF = (K x 1.8) - 459.67. And for folks who work with Celsius, we can convert Celsius to Kelvin with this formula: K = degC + 273.15, or we can conver4t from Kelvin back to degrees of temperature in Celsius with this formula: degC = K-273.15. (However for temperature intervals one degree Kelvin = one degree Celsius.) Also, 0 degrees Kelvin = -273.15 degC = -459.67 degF.
But inside of our water tank the total space available for air is fixed by the physical size of the tank itself. If we're talking about air inside a water tank, the volume taken up by the air can't get bigger than the tank itself. Charles and Boyle considered individually don't tell us what happens to the air pressure and volume inside of a water tank as temperatures change.
To figure this out we need to look at pressure and temperature simultaneously using the
One would conclude from this calculation that the effects of normal temperature changes in the environment of a water pressure tank will not have a significant effect on the in-tank pressure.
Note 1: A thoughtful reader pointed out that the measurement of temperature in these basic formulas must be counted from absolute zero, (-273 degrees Centigrade / -460 degrees Fahrenheit). Our references for Charle's Law confirm this requirement. Another simpler calculation of the effects of temperature change on water pressure tank internal pressures was suggested by Mr. Pryor. The difference between the starting and ending temperatures in degF. is divided by the starting temperature converted to Kelvins: a temperature change from 60 degF to 90 degF is (90-60)/(460+60) = 5.77% which is indeed only a small increase in pressure inside the water tank.
All of These Laws, Boyle's, Charle's, and the Combined Gas Law, are Wrong about Water Tank Air and Water Pressures
In THEORY our hypothesis that warming weather could cause water tank and pump short cycling since if we heated up the air in the tank it would occupy more and more space until there was little or no water in the tank. The pump would have to keep turning on and off when we were running the water.
But this is NOT the case when we consider that we're letting water back into the tank along with our air in there. Our theory is wrong because we forgot something: the air in the tank is cooled by the incoming water.
In a real, in-use water tank, the incoming water from almost any water supply is well below ambient air temperature, and from a well the water is often sitting around underground at 40 degF. which would easily cool the water tank and thus the air it contains. This is why in humid climates we see condensate on the outside of the tank. The incoming water is cooling off the water tank! The water would have to sit idly in the water tank for a long time before the tank would warm up to ambient air temperatures outside the tank.
So we have two explanations for the absence of much effect on water pressure tank performance when the ambient temperatures around the water tank change: the combined gas laws showed about a 6% pressure change in a typical water pressure tank with a 30 degF. temperature increase in the tank, and when we add the cooling effects of cool water entering the tank in most environments this pressure change is reduced further.
Only in locations where the water entering a water pressure tank is at high temperatures would the water temperature affect tank pressures noticeably. (We might see these effects in some parts of the Southwestern U.S. where groundwater or municipal water temperatures are high or where a well is tapping thermal springs.
In conclusion and despite the help from Boyle and Charles and the Combined Gas Law, the cooling effect of low-temperature incoming water which is almost always well below ambient air temperature will offset any the effects of seasonal changes of air temperature around the water tank (except maybe if the pressure tank is in hot direct sunlight which I'll skip for now).
In fact, if anything, the air in the water tank is cooled to below room temperature by the incoming water, which in turn would tend to lower, not raise, the air pressure in the system. Only if the incoming water were at or above ambient air temperatures would we be likely to see an effect on water pump and tank behavior from seasonal ambient air temperature changes.
Since We Broke the Gas Laws, What Explains Why Letting Air OUT of a Water Tank would Improve Water Pump Short Cycling?
SO the most likely explanation for the improvement that we saw when you released air from your captive-air water pressure tank is that
Using these simple gas law formulas allows us to easily calculate the effect of a change in pressure, temperature, or volume of any gas in a closed container.
Readers of this document should also see WATER TANK BLADDER PRESSURE ADJUSTMENT where we describe adjusting air pressure in a bladder type water tank to factory specs, and also see Water pump and pressure tank repair diagnosis & cost an specific case which offers an example of diagnosis of loss of water pressure, loss of water, and analyzes the actual repair cost. The illustration at page top is courtesy of Carson Dunlop, Inc. in Toronto.
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Frequently Asked Questions (FAQs)
Some of the FAQs discussed below are adapted from information provided by the Watts Regulator Company in a 1973 publication.
Question: How much air pressure is needed in the water pressure tank?
if I have a 525 gallon tank how much air pressure do I need? - Charles Collins
Reply: the air pressure needed is not a function of water tank size but rather of the pre-charge air pressure to match the pressure control switch cut-in / cut-out pressure settings.
Re Mr. Collins question: the air pressure needed in the water tank is not a function of the tank size. If your water tank is actually 525 gallons, or if your tank is just 30 gallons, the pre-charge air pressure needed is set by the cut-in and cut-out pressures of your well pump.
Question: how much air volume is needed to provide 1100 gpm of water flow?
how much Air Volume for 1100 gpm water - Musyafa
Reply: air volume addresses the pressure tank draw-down cycle time
Question: what is a pressure-heat rupture
Can you define pressure-heat rupture as it pertains to a hot water tank and can you tell me why they occur? - Anon.
A pressure-heat rupture is a sudden bursting of a hot water tank that occurs because of the combination of high pressure and high temperature superheated water in the tank, usually along with a structural weakness in the tank itself caused often by corrosion or scale. According to the Watts Regulator Company, heat itself can have a weakening effect on a hot water storage tank.
Please see BLEVE EXPLOSIONS for details about boiling liquid vapor explosions and water heater explosions.
Question: what is thermal expansion
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