What should it be? I hear that question quite often. But the reality is, there is no simple answer. If you look over my posts, you’ll see I rely heavily on AHRI, design data, and pressure-enthalpy charts because manufacturers can build just about whatever they want so long as it functions as stated and meets regulations.
There are some typical aspects of air-conditioning design that are dictated by materials of construction and good practice such as line size and sub-cooling. For instance, lines are sized to keep pressure drop low (larger lines) but promote oil return (smaller lines). The sub-cooling will probably fall around 9 – 15 degrees on a split system because they can’t control how much liquid line (added pressure drop) the installer will add. But on a package unit, where they control the entire design, it may operate just fine with 5 degrees of sub-cooling, which reduces the total charge and cost of the unit.
Here I’m going to outline a few aspects of typical air-cooled single stage air-conditioners that won’t get you into too much trouble. (heh, heh, heh).
EER rating: This is actually a pretty useful piece of information if you find it on the unit. The EER is the ratio of the Btu per hour to the Watt per hour energy it takes to achieve capacity. If you know it’s a 5000 Btu/hour unit, and has an EER of 3, then 5000/3 = 1666.7 Watts are needed, divide by 120 volts, 1666.7/120 = 13.9 amps. You now know the total amp draw of the unit.
The EER was tested when the condenser was seeing 95 F air, and the evaporator was getting 80 F air at 50% relative humidity. If your conditions are different you will have a different current draw, but it will be close.
The EER is also NOT the SEER, seasonal energy efficiency rating. The SEER is an average Btu/Watt hour for an air-conditioning season of 125 8 hour days of who knows where. The SEER is used to sell units and appease regulators.
AMPS: For the compressor allow 1 horse power per ton (12000 Btu/Hour). Multiply the horse power by 746 Watts, then divide by the appropriate single phase voltage (120 or 220 volts). For three phase, multiply the volts by 1.27 first, then divide. (it’s 1.73 X .92 X .8 for the factor, efficiency and power factor.)
For total unit, 1.5 horse power per ton of cooling may be more appropriate.
Superheat: System with a thermal expansion valve, TXV, will run 7-10 degrees of suction super-heat. Units with electronic expansion valves, EXV, will run about 4-5 degrees.
If the unit has a capillary tube or fixed orifice it can be anywhere between 5 and 45 degrees depending on the operating conditions. Use the manufacturer tables or a generic superheat chart at the very least.
Sub-cooling: The sub-cooling varies with design but will fall somewhere between 5 and 15 degrees. Check manufacturer data. Usually on one of the panels of the unit you will find a chart like the one below. Use your phone to snap a picture and begin a library of these charts by model, you’ll soon be prepared for the unit that the chart has faded.
Condenser/discharge pressure: For air-cooled units, take the outdoor air temperature and add 30 degrees. Say its 90 outside, add 30, to get 120 degrees, look this valve up in the pressure temperature chart for the refrigerant in the system. For R-22 its 260 psig.
For water cooled units, measure the inlet and exit water temperatures and add them together, divide by 2, then add 10 degrees. For instance, the inlet water is 68 F, and the exit water temperature is 81 F, 68+81 = 149, divide by 2, 149/2 = 74.5, then add 10 to get 84.5 degrees, for R-22 that’s about 155 psig.
Suction/low-side pressure: Measure the air leaving the evaporator. Take several measurements and average (add them up and divide by the number of readings you took). Subtract 15 degrees from the average and look up the pressure from the temperature you got. Say the average was 57 degrees, subtract 15 degrees to get 42 degrees. For R-22 that pressure would be about 69 psig.
Air flow: Air flow should be about 350 – 450 cubic foot per minute per ton of cooling, with 400 cfm per ton being the most common rating when the fan is set to run on medium-high speed (blue tap). So, a 3-ton system will come set up to deliver 1200 cfm at about .3 inches of static pressure.
All that said, none of this will hold true on an inverter drive system unless its operating at 100% speed.
The system refrigerant charge is either over or under charged if it isn’t correctly charged.
For a system undercharge, depending on the severity, you will have most or all the following symptoms:
1. The saturated suction temperature is less than expected after converting the suction pressure to its saturated temperature with the PT chart. For instance, common coil temperature for A/C would be 45 degrees F. On a system using R-22, your gauge reads 50 psig, you’re only running a 36 F coil.
2. Your system has a high suction superheat. After measuring the suction line surface temperature to be about 58 degrees, subtract the saturation temperature found in symptom one. In this example 58 – 36 = 22 degrees of superheat. Too high for a low saturated temperature.
3. Low liquid sub-cooling. Find the saturation temperature of the condenser pressure and subtract that number from the actual liquid line temperature. Less than 5 is low, could be zero.
4. Low saturated condenser temperature. Its 85 F outside so you add 30 degrees to get 115 F. Looking up that pressure on the PT chart you read 242 psig, but your gauge reads 190 psig, which is 98 degrees F, low.
5. Low system capacity. Unit won’t cool the room or takes a very long time to do it. Perhaps only works at night.
6. High compressor discharge superheat. Measuring the discharge line about 6 inches from the compressor you see 138 F, subtracting the saturated condensing temperature found above, 138 – 98 = 40 F. That’s above the 35 recommended.
7. Compressor body is warm, not sweating, or even hot.
8. The system could be cycling on the over temperature protection or low pressure cut-out.
9. Oil may smell burnt if running in this condition long enough.
10. Evaporator frosting or ice on the suction line for a very low charge.
Symptoms of overcharge, depending on the severity, you could have most or all the following:
1. High saturated condensing temperature. When you converted your condensing pressure to temperature it was 40 degrees higher than the ambient. Too high.
2. High saturated suction temperature. You were expecting a 45-degree coil, but after converting the pressure with a PT chart you find its running 50 degrees.
3. Low system capacity, here again the unit will not do the job because over or under charge, the system doesn’t function correctly.
4. Compressor amps will be high. This is usually from liquid getting back to the compressor.
5. Liquid sub-cooling (the actual liquid line temperature minus the condenser saturation temperature) will be greater than 15 degrees.
6. Low suction superheat. More noticeable on a system with a fixed metering device it will also cause a drop for a TXV if its severe enough.
7. Compressor may be noisy and vibrate.
8. Compressor starts with a lunge due to partial seizing caused by liquid damage and oil loss.
The above numbers were for example purposes and not from a real unit although the limits mentioned are good practice.
If you approach the diagnostic steps of troubleshooting in the same fashion every time you will be way less likely miss something. It may seem that collecting all that information is wasting time, but trust me, it will save time, money, frustration, and possible embarrassment.
First, make sure you have all the basic data.
A. Who is your contact at the job?
B. What kind of unit is it? Size/make/model?
C. How long has the problem persisted?
D. Where is the unit location? Roof/basement/attic/crawl space?
E. What is the specific complaint?
With this information, you will be better prepared to investigate the problem. It would be prudent to obtain a copy of the unit manufacturer information prior going into the field. (There may not be any cell service or you may not be allowed to use your phone.)
Once you’re on the job site, do a general inspection with the power off. Be on the lookout for:
1. Missing insulation. Can lead to unnecessary load on the unit. Missing insulation on TXV sensing bulb causes hunting.
2. Kinked or damaged lines. Poor line conditions can cause higher pressure drop or even restrict refrigerant flow.
3. Oil leaks. Units need oil to work properly for sure, but an oil leak is also a refrigerant leak.
4. Damage to condensing coils. Flattened coils reduce condenser capacity 20% is too much. Missing fins derogate condenser efficiency. Half the fins missing will affect capacity by about 20%.
5. Water leaks from line, unit, or ducts. Water will slowly destroy a unit is left free to roam. Water in the unit can lead to mold issues or worse.
6. Dirty filter/s? Dirty or plugged air filters in the unit reduce capacity and can lead to unit freeze up.
7. Bent, cracked or missing fan blades. This condition reduces the air-flow, capacity, cause vibration. A fan blade can damage a condenser beyond repair.
8. Missing unit panels. Missing panels or large gaping holes allow air to bypass the coils. Bypassing condenser coils can be enough to trigger high head pressure cut-outs.
9. Wiring condition, low and high voltage. Cracked frayed wiring is dangerous and usually means high heat. Loose connections can lead to this.
10. Damaged duct work or disconnected supply runs. The air has to get there to work.
11. Are the supply and return open and un-obstructed? Closing of half the supply is usually enough to freeze a unit.
12. Do you see ice on anything? Unit icing is low refrigerant, low air-flow, or a restriction. This has to be addressed first. Completely thaw out the unit before proceeding. Don’t immediately add gas, the coil is cold, suction will be low. Unit can be hiding a problem, check the static pressure from return to supply across the evaporator > then .15 inches of water column means the evaporator is dirty.
Any one of the above items could lead to improper operation of the unit. To what degree can vary greatly.
Now power up the unit and make a quick round to see if you hear any problems, like rattles, banging, vibration, things out of balance. (If the system won’t power up you need to check power and controls to determine why.)
Once the unit is operating gather the following data:
1. Indoor air wet and dry bulb temperatures. WB_____ DB_____
2. Outdoor ambient air dry bulb temperature. OA_____
3. Supply airflow.
4. Temperature of return air. RA____
5. Temperature of supply air. SA____
6. Temperature of air discharged from the condenser. CDA____
7. Suction pressure. SP____
8. Look up the saturated suction temperature. SST____
9. Condenser/discharge pressure. CP___
10. Look up the saturated discharge temperature. SCT____
11. Suction line surface temperature 6 inches before compressor. SLT____
12. Discharge line surface temperature 6 inches from compressor. CDT_____
13. Liquid line surface temperature just before the metering device. LLT____
14. Check the temperature difference of any liquid line filter dryers. Tin____ – Tout____ = Tdiff____
15. Amp/volts draw for the compressor. CA____ CV____
16. Amp/volts draw for the condenser fan. OFA____ OFV____
17. Amp/volts draw for the indoor fan. IFA____ IFV____
18. Check voltage of low-voltage supply. LVV____
Perform the following procedures/calculations:
1. Calculate the air-handler temperature rise (subtract the exit air from the inlet should see 20 – 25 degrees F).
2. Calculate the air handler BTUH. (Plot the indoor air wet and dry bulb versus the unit discharge conditions on a psychrometric chart and calculate the unit load. If the unit is seeing 80 F and 50% RH there is 6.7 Btu/lb of air, when the RH rises to 75%, the load on the unit doubles. The unit must overcome some of this load to become stable. Total load is the enthalpy difference multiplied by the cfm and 4.5 factor).
1. Calculated the condenser temperature rise (subtract the outdoor ambient from the exit of the fan should see 30 – 35 degrees F).
2. Calculate the condenser heat-rejection. (Multiply temperature rise by cfm and 1.08. Should be around 30% more than nameplate i.e. 24,000 Btu A/C should be rejecting about 31200 Btu from the condenser when fully loaded).
3. Determine if the unit air-flow is within the range of 350-450 cfm per ton of cooling. (This is best done with an electronic meter, but a pitot tube and calculation work fine).
4. Calculate suction superheat (Subtract saturation temperature from actual line temperature. TXV should be about 10, fixed metering use chart).
1. Calculate sub-cooling (Subtract the saturation temperature from the liquid line temperature. Should see 9-15 degrees or use charging chart if available).
2. Calculate the compressor superheat (subtract the saturation temperature from the line temperature about 6 inches from the compressor. Check against manufacture data but much higher than 35 needs looked into).
3. Check the that the voltage is not off by more than 10% (if 120 is typical line voltage then range is 108 – 132).
4. Check current draw against literature. If not available, the following is usually okay: For the compressor allow 1 horse power per ton (12000 Btu/Hour). Multiply the horse power by 746 Watts, then divide by the appropriate single phase voltage (120 or 220 volts). For three phase, multiply the volts by 1.27 first, then divide. (it’s 1.73 X .92 X .8 for the factor, efficiency and power factor.)
5. For total unit condenser and air-handler, 1.5 horse power per ton of cooling may be more appropriate for checking amps. (Package unit).
6. Filter driers with a temperature difference of 5 degrees F or higher are plugged.
7. If all this is in line with expected results, plot the system on a pressure enthalpy diagram for more detailed look at the performance.
So, the above procedure could be tailored to meet your needs, but I wouldn’t leave out too much of it. Though, not always obvious, after all these checks, the problem will emerge.
When collecting real world data to analyze system performance nothing beats a good old fashioned pencil and paper.
You can use a form to keep it organized, or just a list if that works for you. Below I provided a pictogram for up to three stage system.
This can be used for up to three stage refrigeration cycle by marking N/A (not applicable) if the system doesn’t have a particular feature.
The PRESS is short for pressure, and TEMP is short for temperature. WB and DB, are wet bulb and dry bulb temperatures of the air entering and leaving the evaporator.
Each valve has an O or C beside it to indicate whether it’s open or closed.
I’ve posted the Excel file so it can be printed on one page or altered to meet your needs.
Hope you find it helpful!
There are numerous trouble shooting charts and diagrams to help diagnose system problems. Probably one of the most familiar ones I’ve seen, posted numerous places is below.
The chart points out one thing clearly, you can’t diagnose a problem without multiple pieces of information about the system. I’m not sure who really gets credit for developing this chart, but the first place I came across it was in a very old Carrier training manual, and my guess is, they should get credit for it.
Another type of performance chart is a unit specific charging chart. Using field acquired pressure and temperature of the liquid line, the chart shows you what to do to get the unit performing at design. Below is an example of that type of chart.
Both approaches seem simple enough to use, but neither will work if the system is dirty or improperly installed.
With almost any set of charging instructions you will find the following two lines;
1. Never charge liquid into the suction port of the compressor.
2. Add small amounts of refrigerant at a time.
Why do you suppose that is?
Liquid into the compressor will damage it. It also will cause significant swing to the system pressures and temperatures. It can also quickly overcharge the system provided the compressor can take it.
As for small amounts of charge at a time, that is because when the charge is incorrect, the system is unstable. It will take as much as 20 minutes for the change you made to be realized in the readings you are observing.
To illustrate the system performance, I recorded a dataset for an air-cooled unit cooling a water stream to 45 degrees. Below is the system performance from starting, and running to set point.
Nice stable performance, the suction pressure follows the water temperature right to the set-point. Discharge pressure slowly lowers as the system is cooled, and the pressure ratio is essentially constant.
This system, by the way, used a capillary tube as a metering device. But you will see similar characteristics, such steady motor current, in a properly charged system.
Now, let’s look at a system trying to perform with about a 20% undercharge.
As might be apparent, system stability, or lack thereof, is questionable.
The water temperature warms up some during the initial operation. The discharge pressure climbs as the system can remove all the excess heat from the unit and the process.
Pressure ratio is all over the board for almost two hours, and suction pressure climbs until the system reaches a somewhat stable operation at 4:17.
So, when you are trying to interpret the information from your gauges and temperature probes, keep in mind, the system is always trying to adapt to the process conditions and its problems all at the same time.
So, use some patients and this knowledge to proceed with care when using charts to diagnose or charge a system.
Here is a spread sheet that will help plot single stage cycles for several refrigerants. It does work on Office Mobile, but won't work on Google Sheets. The tool uses lookup functions that Sheets doesn't seem to have, Enjoy! Find any errors ley me know, and the sheet isn't protected so any change you make is changed. stick to only changing the blue cells and the sheet should be fine,
With a little system information, it possible to estimate the refrigerant charge required using the pressure enthalpy diagram. Below, I plotted a R-22 system cycle based for split system air-conditioner operating at a saturated coil of 45 degrees Fahrenheit, and a saturated condenser operating at 120 degrees Fahrenheit.
The table is the data used to make the plot. The temperature lines are in 10 degree increments.
This is a 3-ton split system with a 3/8” liquid line and a ¾” suction line 50-foot long. The discharge line is ¾” and 2-foot long. There are no other components that add volume to the system. (There is a volume associated with suction gas cooled compressors but it relatively small compared to system charge.)
The net refrigeration effect, or evaporator cooling, is the difference in specific enthalpy of the evaporator, or, 175.41- 109.3 = 66.11 Btu per pound of refrigerant. Three tons of cooling is 600 Btu/minute. Divide 600 by 66.11 and you get the refrigerant flow of 600 / 66.11 = 9 pounds per minute of refrigerant flow required to meet the capacity.
Now, the dynamic condition of the refrigerant isn’t all liquid of nine pounds. The refrigerant is moving at a given velocity so that 9 pounds gets through evaporator and the condenser in one minute, not all at once.
let’s examine where the liquid is. Below is a sketch of the system, and marked in red are the points in the system that are 100% liquid.
Notice all the 100% liquid is in the last part of the condenser and the liquid line running to the expansion valve.
The rest of the system is either a mixture of liquid and gas, or 100% gas.
The evaporator gets a stream of liquid from the expansion valve at a temperature of 44.507 degrees F and a quality of .23, which means, the fluid is 23% vapor and 77% liquid.
When the refrigerant leaves the evaporator, it will be a superheated gas with no liquid remaining, this is a quality of 1, or 100% gas.
To determine how much of the refrigerant is liquid, and how much is gas, you can construct a triangle like the one below.
The total liquid available is the area under the triangle which is found by ½ X (1-.23) X .77 = .296 percent, round up to 30%, is liquid at any given moment in the evaporator. The rest is saturated vapor, or gas.
The energy can only be removed by boiling liquid refrigerant so, if 30 percent of the content is liquid, only 30 percent of the 9 pounds per minute is in the evaporator, or, .3 X 9 = 2.7 pounds of liquid at any moment in the evaporator.
The specific volume of the refrigerant liquid in the evaporator is .012738 cubic foot per pound so, the liquid occupies .012738 X 2.7 = .0343 physical cubic foot.
We know 30% of that is liquid and the other 70% is gas occupying the remaining space in the evaporator. It’s safe to say that the amount of space the gas occupies must be minimized because the gas doesn’t remove any heat to speak of. The gas and liquid are at the same temperature and pressure so the relation (P1 X V1)/T1 = (P2 X V2)/T2, The P & T’s cancel out and you have V1 = V2, so the gas is approximately occupying .0343 cubic foot also.
The amount of refrigerant gas would then be .0343 multiplied by the vapor density, 1.6421, or, .0343 X 1.6421 = .046 pounds of vapor in the evaporator.
Total volume of the evaporator, .0686, would have the equivalent volume of 127 foot of 3/8 copper line.
The total evaporator charge during operation is 2.75 pounds.
Next determine how much refrigerant is in the suction line.
Information about tube dimensions and applications of copper tubing: https://www.copper.org/applications/plumbing/cth/
¾” line has .00252 cubic foot of contents per foot. The line is 50-foot-long, so the entire volume of the suction line is .00252 X 50 = .126 cubic foot.
To know how much refrigerant is in the line, divide the volume of the line by the specific volume of the refrigerant at 55 degrees, .126/.6287 = .2 pounds in the suction line.
So, the low side of the system has, 2.75 + .2 = 2.77 pounds of refrigerant.
Moving to the high-pressure side of the refrigerant circuit we address the small amount of discharge line the same way as the suction line except we use the specific volume at 256 degrees, which is .30017. Two foot of line multiplied by the internal volume, 2 X .00252 = .005, now divide the volume by the specific volume, .005/.30017 = .017 pounds in the discharge line.
The latent process of the condenser must remove the same energy as the evaporator, the condenser also must make at least 2.7 pounds of liquid per cycle in its cycle to keep the balance of energy happy.
The specific volume of the refrigerant liquid in the condenser is .014364 cubic foot per pound so, the liquid occupies .014364 X 2.7 = .0388 physical cubic foot.
The amount of refrigerant gas would then be .0388 multiplied by the vapor density, 4.941, or, .0343 X 4.941 = .19 pounds of vapor in the latent area of the condenser.
The total refrigerant held up in the condenser is then, .19 + 2.7 = 2.89 pounds.
For the de-superheating region of the condenser we know the low side charge must be de-superheated as it continues to the condenser so, we need estimate how much line it will take.
Total heat transfer is equal to the total area in square foot multiplied by heat transfer coefficient and the temperature difference. The formula can be re-arranged to find the total area, and then the length of tubing.
Btu/hour/ (Transfer Coefficient X Temperature difference) = Area in square foot.
Our Btu/hour = 9 pounds X (209.43 -180.20) X 60 =15,784.2
Divided by (220 Btu/hour for copper X a temperature difference of 35) (air side) = 2.4 square foot.
3/8 copper pipe has an internal area of .0821 foot per foot, so 2.4/.0821 = 29 foot of pipe
The internal volume of 3/8 copper is .00054 cubic foot, so total area is, .00054 X 29 = .015 cubic foot
The density of the discharge gas is 3.3314 pounds per cubic foot multiplied by the volume of .015, you get .05 pounds in the de-superheating region.
The sub-cooling process, 112 – 109.3 = 2.7 Btu/pound to cool the condensed liquid below its saturation point.
Find the Btu/hour, 9 pounds X 2.7 X 60 = 1,458
Find the area needed, 1,458/ (220 X 35) = .189
Find the length of pipe, .189/.0821 = 2.3
Find internal area, 2.3 X .00054 = .0012
The density is 69.618 pounds per cubic foot multiplied by .0012 you get .086 pounds of liquid being sub-cooled.
All we need to complete the total is the refrigerant in the liquid line. 3/8” line has an internal volume of .00054 cubic foot per foot. Multiply by the 50 foot of line, 50 X .00054 = .027 cubic foot in the liquid line. The amount of refrigerant is found by multiplying the volume of the line by the specific density of the refrigerant at 110 degrees. The chart only give us specific volume, but no problem, the specific density is just the reciprocal of the volume, so 1/.014364 = 69.61 pounds per cubic foot of liquid at those conditions.
So, 69.61 X .027 = 1.879 pounds in the liquid line.
Evaporator and suction – 2.77 pounds.
Discharge Line - .017 pounds.
De-superheating – .05
Condensing – 2.89
Sub-cooling – .086
Liquid line – 1.879
Total estimated charge – 7.7 pounds.
This is just an estimate of the operating charge of a direct expansion air cooled system. The charge should be added at 80% of the estimated value then adjusted to meet the sub-cooling value of 10 degrees.
I worked for over thirty years in the HVACR industry. I have designed, installed, serviced, and trouble shot units of various types throughout the years. The posts here are information based on that experience, I hope you find them useful. If you have a different experience, please comment.