We have a container that has a fixed volume. For example, let’s say it’s 1 cubic foot. So at 0 psig, there is a certain number of gas molecules contained within that container and a certain number of collisions with the container walls occurring.

Now, let’s take that container, and we’re going to double the amount of molecules inside without changing its size at all. We know that the pressure increased, but what did it take to do this? Energy. Adding those additional gas molecules required that we add energy to force the extra gas into the container, which resulted in an increase in pressure. The thing to remember now is the law of conservation of energy — energy cannot be created or destroyed, it simply changes form.

Since heat energy is simply another form of energy, it stands to reason that adding or removing heat energy from our system will affect the energy level of the gas molecules and, ultimately, the pressure exerted by them.

Let’s return to our sample container of 1 cubic foot internal volume. We’re going to expend enough energy to put enough molecules into this container to raise the pressure to 100 psig at a temperature of 70°F. If we add more energy, not in the form of compressing more gas but in the form of heat energy, what will happen to the pressure in the container?

The heat energy is going to “excite” the molecules in the gas, increasing the number and force of the collisions that are occurring — the basis of the existing pressure. Since we’re adding energy, the pressure will rise, and it will do so in a predictable and consistent way. The reverse is also true — if we remove energy, the pressure will drop in the same predictable and consistent way.

This is why we need to understand the gas laws as technicians. They allow us to predict and understand the pressure change caused by adding or removing heat energy from a sealed, pressurized system.

PRACTICAL APPLICATION

Now that we understand how heat energy affects the pressure within a sealed system, we can apply this knowledge to pressure testing. A large number of factors are making proper leak testing at installation more important than ever, and manufacturers are demanding more detailed leak testing procedures. Add to that the fact that our tools are more refined than old school analog gauges, and a leak of even 0.5 psi over a several hour period of time is easily something a technician can spot.

Let’s take a look at an imaginary but fairly realistic scenario to see how this works and what it means on the ground and in the field.

Consider a new construction split system. Tonnage isn’t super important to this, but we just made the last brazed joint, it’s the end of a long day in the 90° heat, and a nasty thunderstorm is brewing. Let’s get this thing pressurized and get home. Run the pressure up to 350 psig of nitrogen, and get out of here. When we show up in the morning, it’s 65°, and we find that the pressure has dropped almost 16 psig. That might make us a little nervous. We checked all of our joints with a mirror and soap bubbles, but we don’t see any leaks… where did the pressure go?

Before we get excited, let’s look at how the temperature change affected the pressure within this sealed system. We pressurized to 350 psig at 90°, and it’s now only 65°. With the gas law equations, we can identify what the pressure in the system should be to eliminate wasting time looking for leaks that aren’t actually there. This is an expression of the gas laws known as Gay Lussac’s Law. In this, the system volume is a constant and can be disregarded. For our purposes, the copper piping we use to build systems is unchangeable, so we’ll use this equation.

The first step is to change the equation around to isolate the variable that we wish to get — P2= T2 (P1/T1).

Now, we have a simple equation that we can plug our numbers into and get the answer, right? Not quite yet. We have one more step before we get the calculators out. We need to convert the pressure and temperature values that we have to absolute pressure and temperature readings. To do this, add 14.7 to the pressure and 492 to the temperature.

Now, our numbers look like this:
T1 = 582°R (Rankine)
P1 = 364.7 psia
T2 = 557°R
Now, let’s solve:
P2 = 557 (364.7/ 582)
P2 = 557 (0.6266)
P2 = 349.03

But wait, our system dropped to 334 psig, so we must have a leak…
We forgot one vital step — we need to convert our P2 reading back from absolute pressure to gauge pressure. To do this, we subtract 14.7. 349.03 – 14.7 = 334.33 psig.

This says that the pressure loss within the system was due only to the temperature change and was not due to a leak.

Time to get the vacuum pump out, and finish this job up.

Publication date: 4/3/2017

Want more HVAC industry news and information? Join The NEWS on Facebook, Twitter, and LinkedIn today!