Estimating Refrigerant Charge with the Pressure Enthalpy Diagram

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.
Tally-ho:
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.