Boyle's Law

Boyle's Law: this article describes and defines Boyle's Law with examples of using Boyle's Ideal Gas Law to explain what happens to air in a water storage tank, LP gas in a gas tank, oil & fumes in an oil storage tank, or air conditioning /heat pump refrigerant liquid & gas volumes inside of an air conditioning or heat pump system.
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Boyle's Ideal Gas Law explains the relationship between gas pressure and volume:
P_{1}V_{1} = P_{2}V_{2} all
(P=Pressure V=Volume) at constant temperature
Or stated in simple words, in a closed container containing a gas (air, LP gas, refrigerant, or other), the pressure of the gas times the volume of the gas equals and remains a constant (as long as temperature is also held constant).
For example, in a closed container of refrigerant, if we doubled the volume of the container the pressure of the gas contained therein would be halved from the original container.
Continuing a discussion we began at WATER TANK PRESSURE CALCULATIONS, we can reexamine the calculation of variation in water tank air volume in more detail using Boyle's Law.
Boyles Law: P_{1}V_{1} = P_{2}V_{2} 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.
[Note: We're discussing relative pressure changes within the water tank and piping system; we are not considering the effects of the atmosphere (which is another 14.7 psi at sea level) In a fullyenclosed water tank and piping system the pressures produced by the water supply or water pump should not be significantly affected by atmospheric pressure. ]
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 cutoff 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 pressedon 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.
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 P_{1}V_{1} = P_{2}V_{2}, 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.
10 gallons of air compressed down to 3 gallons of air, letting in (103) 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 2050 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!
26 August 2015 Bill said:
Correct me if I am wrong please. For your calculations using 50 psi and 20 psi, aren't they gauge readings which would also need 14.7 psi added to bring them to absolute pressure?
In that case, I get 7.73 gallons at 50 psi gauge and 5.76 gallons at 20 psi gauge giving a difference of only 1.96 gallons instead of 4.41 gallons.
If you really meant 20 psi absolute, then the gauge pressure would only be 5.3 psi which I think is not enough pressure in the house.
Bill
Thanks so much for careful reading and questioning. You're right that if we are discussing absolute pressure we need to add the atmospheric pressure to the gauge pressure. I appreciate the thoughtful editing and that you took the time to comment. Working together makes us smarter.
In the article above we comment that
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.
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