Tag: pressure

Low pressures are often measured in inches of water column or “WC. Like most units of measure, it has a very simple origin, in a water manometer 1″ of water column is literally the amount of force it takes to raise the column of water by 1”. While some water manometers (water tube) are still in use the vast majority are either dial or digital gauges that still use the same scale.

1 psi is equal to 27.71 inches of water column; this is why water column is most often used to measure pressures under 1 psi. These low pressures are most often read using a manometer or a magnahelic gauge.

When we measure water column with our tools it is calibrated at atmospheric pressure or the gauge scale instead of the absolute scale. This means that for a manometer or magnahelic to be properly used they  MUST be recalibrated before each (many auto calibrate to zero) to compensate for changes in elevation and barometric pressure. At altitudes over 2000′ above sea you will also need to follow manufacturer recommendations to adjust the gas valve and even change orifice sizes in some cases due to the effect the lower atmospheric pressure has on the gas.

Gas pressure is usually measured in “wc, most commonly we set single stage appliances to 3.5” wc on natural gas and 11″wc on propane. This varies based on manufacture specs, combustion analysis and meter clocking tests. Always read the manufacturer specs.

We also use “WC to check air static pressure on systems. Static pressure is pressure that is exerted in all directions in a contained space, it is not the directional force of the air.

We use a manometer or a magnahelic and measure the negative air pressure in the system return side before the blower (and after the filter whenever possible) and the positive pressure supply air side directly after the blower. By calculating the differential you come up with the total external static in water column. For example, if the return static is -0.3″wc  and the supply static is +0.2″wc the total static is 0.5” wc.

Many manometers and all magnahelic gauges (to my knowledge) have two ports so you can read the differential pressure all at once. This also comes in handy when reading/testing differential pressure on many furnace air pressure switches to ensure they make and break at the proper pressure.

— Bryan Orr

This article is written by my buddy and Canadian Supertech Tim Tanguay. Thanks Tim!


This P/E chart shows R410a at 100°F Saturated Condensing Temp, 10°F SC
40°F Saturated Suction Temp, 20°F SH at the compressor.
The green highlighted thumb shape is the saturation zone. Everything that occurs in the saturation zone is a latent (change of state) process.
Everything that occurs to the right (superheating) and left (subcooling) is a sensible process.
Go to the movies in your mind, imagine that you are one pound of 410A. We commence our journey at the rightmost point on the upper orange highlighted line.
At this point, you have just left the compressor. You are a superheated vapor, with a temperature of 145°F. You enter the condenser and start rejecting heat to the atmosphere. After rejecting 45°F of sensible heat (desuperheating), you hit the saturated condensing zone (100°F) and you turn into a drop of liquid. As you march your way along the condenser (follow the line left), you reject latent energy but stay the same temperature. As your latent energy decreases you become more liquid until finally, you are a solid column of liquid and you exit the saturation zone to the left of the thumb. You then give up another 10°F of sensible heat to the air and become a 90°F subcooled liquid.
You approach the sight glass as a 90°F subcooled liquid under approximately 350 PSIG of pressure. As you pass the sight glass, you fart a few bubbles just to mess with the refrigeration mechanic observing the process. You squeeze your way through the tiny orifice in the metering device and emerge into the evaporator, solidly back into the saturation zone. You find yourself as a 40°F saturated liquid at 125 PSIG (approx 78% liquid, 22% vapor, indicated along the constant quality lines).
 Now you make your way along the bottom line towards the right side of the thumb, you absorb heat energy from the warm return air rushing over the copper and aluminum evaporator fins. The heat you absorb boils you dry. You are naught but a vapor, and as such, the energy from the return air increases your sensible heat by 10°F. You emerge from the evaporator as a 50°F superheated vapor. As your journey progresses
towards the suction inlet of the compressor, you pick up another 10°F of sensible heat.
You enter the suction port of the compressor as a 60°F, superheated vapor. The compressor puts you through a strenuous workout, squeezes you into a smaller volume and in the process increases your temperature by about 85°F.
You emerge as a superheated 145°F vapor. The process begins anew.
A few things to look at. The numbers on the top represent enthalpy energy, as BTU’s per pound.  In this particular example, the sensible portions of the condenser account for approx 20% (eyeball estimate) of the total heat rejected in the condenser. The other 80% of the process is latent.
On the right-hand side of the PE diagram, you have specific volume, represented as curved dotted lines. As SST decreases, specific volume increases and vapor density decreases. This fact alone is why refrigeration compressors need to be physically larger. As specific volume increases, the volumetric efficiency of compressors decrease.   Lower SST’s (suction saturation temp) require larger compressor displacement because they need to move more gas to obtain the required mass flow. In AC and refrigeration, the mass flow of refrigerant through the system ultimately determines your system capacity.
At 40F, the latent heat of vaporization of 410A is approx 75 BTU/LB. Compare that to water, which has a latent heat of vaporization of approx 970 BTU’S per pound at 212°F/14.69 PSIA and you begin to realize why dehydration of a system takes so darn long.  It takes a LOT of energy to boil water off.
In the evaporator, about 10% of the process is sensible.  This is why a unit that is short on refrigerant isn’t able to cool properly. The refrigerant boils off leaving a large portion of the coil to collect sensible heat (higher superheat). The amount of heat that sensible processes remove from the air stream is relatively tiny, thus we lose capacity.
So too with things like water. The sensible heats involved with changing temperature are minuscule when compared to the amount of heat required to change state. Universally, latent changes require orders of magnitude more energy than sensible changes.
— Tim

The gas laws. We all learned about them in school and promptly forgot all about them. I really think that we need to dig our books out, dust that information off and work to understand and apply it.

Many will say that nitrogen pressure doesn’t change with pressure like other gasses. This is false but read on.

Let’s start by looking at the pressure a little differently. Pressure is a measure of the force exerted by a gas within a container. It exerts pressure because the individual molecules of the gas are colliding with the walls of the container. Those collisions are happening because each molecule has a specific amount of energy. So, in this way, we can view pressure as a measure of the amount of energy contained within our container of gas. That might sound complicated, so let’s kind of unpack it and see if we can understand it better.

We have a container that has a fixed volume, for example, 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 number of molecules inside that container 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 that extra gas into the container. The addition of energy to force additional molecules into the container resulting in an increase in pressure. The thing to remember now is the law of conservation of energy. Energy isn’t created nor destroyed, it simply changes form.

Since heat energy is simply another form of energy so 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 that are the basis of pressure existing. Since we’re adding energy, the pressure will rise and it will rise in a predictable and consistent way. The reverse is also true if we remove energy, the pressure will drop in the same consistent and predictable
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 in the field.

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 when it’s 65°F and 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 with 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°F and it’s now 65°F. With the gas law equations, we can know what the pressure in the system should be and eliminate time wasted 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 for change the equation around to isolate the answer we wish to get.
P2= T2 (P1/T1)

Now, we have a simple equation 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 valves that we have to absolute pressure and temperature readings, so add 14.7 to the pressure and 459.76° (Rankine scale) to the temperature to get to absolute scales

Now, our numbers look like this:
T1 = 549.67°R (Rankine)
P1 = 364.7 psia
T2 = 524.67°R
NOW, let’s solve.
P2 = 524.67 (364.7/ 549.67)
P2 = 524.67 (0.6635)
P2 = 348.11

But wait, our system dropped to 334 psig, so we have a leak…
We forgot one VITAL step. We need to convert our P2 reading back to gauge pressure.
349.03 – 14.7
333.41 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.

In summary, every gas responds to the gas laws in the same way. We use nitrogen because it is readily available (the air is mostly made of nitrogen), dry and it doesn’t readily combine with other molecules under normal circumstances.

It does change pressure with temperature and all you need to do to find out how much it will change is by changing both the before and after temperatures to absolute scales (Rankine for Fahrenheit or Kelvin for Celcius) and convert your before and after pressure readings from gauge pressure (PSIG) to absolute pressure (PSIA). Once you have your solution you can convert back to Celcius or Fahrenheit

— Jeremy Smith CMS

P.S. – I made a little before and after calculator HERE

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