Schematic of a bladder type captive air water pressure tank (C) Carson Dunlop AssociatesCharles' Law
Charles' Gas Law explained, practical applications for building air conditioning, heat pumps, oil storage tanks, water pressure tanks, LP & natural gas systems
     

  • CHARLES' LAW - CONTENTS: definition of Charles Law with examples of using Charles 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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Charles' Law: this article describes and defines Charles Law, explaining the role of temperature in gas pressures, including examples of using Charles 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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Definition & Practical Applications of Charles' Gas Law

Charle's Law explains the relationship between the volume of a gas and its change with temperature. Simply put, Charles' law explains how and how much the volume of any gas increases with increases in temperature or decreases when the temperature drops.

V1/T1=V2/T2

(V= volume, T= Temperature)

Charles 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.

 

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