# Refrigerant Piping – part2

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In this post, we will be continuing our discussion on how to size refrigerant piping. This training is designed to be as simple and practical as possible, providing you with the necessary knowledge to size refrigerant piping correctly and accurately. We will go through the important aspects of this process including determining the pipe size, pressure drop, and other factors. With this comprehensive training, you will come away with the confidence to size refrigerant piping in any situation.

## Sizing Refrigerant Lines

The ASHRAE Handbook HVAC Systems And Equipment Chapters 41 and 2 include sizing for suction, discharge, and liquid lines for frequently used refrigerants. The changes in saturated suction temperature (SST) is 0.5, 1, and 2°F (0.28, 0.56, and 1.7°C) for the suction and discharge lines and 1°F (0.56°C) for liquid lines. This data is based on a condensing temperature of 105°F (40.6°C) for water-cooled equipment and must be adjusted for other temperatures, such as those of air-cooled equipment (typically 120 to 125°F [48.9 to 51.7°C]). Also, the tables assume a length of 100 feet (30.5 m) of equivalent pipe length, but the actual pressure drop can be deduced from the tables’ equations according to the actual length of the application.

Saturated suction temperature is based upon the pressure leaving the evaporator and represents the refrigerant temperature as a gas without superheat. The actual refrigerant temperature leaving the evaporator will be higher than this. The difference between the two temperatures is called superheat.

### Equivalent Length for Refrigerant Lines

Following tables provide information for estimating equivalent lengths. The actual equivalent length is estimated by calculating the path length in feet (meters) that the piping will follow and adding the pressure drops of the fittings and/or accessories along that length. The tables provide pressure drops in equivalent feet of straight pipe for fittings and accessories.

For example, in “Equivalent Length for Fittings“, we see that a 7/8-inch (22 mm) long radius elbow has a pressure drop equivalent to 1.4 feet (0.43 m) of straight copper pipe.

### How to Determine Equivalent Length

Calculate the equivalent length of the liquid line for the following condensing unit with DX air-handling unit:

The liquid line is composed of the following elements:

• •22 ft (6.7 m) of 1-3/8 inch (35 mm) piping
• 1 filter drier
• 1 sight glass
• 1 globe type isolating valve

To determine the equivalent length for the refrigerant accessories use Table 1 and Table 2).

## How to Size Liquid Lines

Size the refrigerant liquid lines and determine the sub-cooling required to avoid flashing at the TX valve for the condensing unit with DX air-handling unit shown in the previous example. The system:

• Uses R-410A
• Has copper pipes
• Evaporator operates at 40°F (4.4°C)
• Condenser operates at 120°F (48.9°C)
• Capacity is 60 tons (211 kW)
• Liquid line equivalent is 113.6 ft (34.64 m)
• Has a 20 ft (6.1 m) riser with the evaporator above the condenser

The first step in sizing liquid line is to estimate the size of the pipes needed for the system. This is followed by calculating the actual temperature differential (∆T) between the equipment and the space being served. The actual piping pressure drop must then be calculated and the total pressure drop determined. The saturated pressure of R-410A at the TX Valve, the saturation temperature at the TX Valve, and the sub-cooling required for saturated liquid at the TX Valve must also be determined. Finally, the required sub-cooling for proper operation must be calculated. Following these steps will ensure that the system is designed correctly and will operate efficiently.

### Step 1 – Estimate Pipe Size

To figure out the liquid line pipe size for a 60 ton unit, consult the following table. As per the table, a 1-3/8 inch (35 mm) pipe would be suitable for an 79.7 ton (280 kW) unit. Please note that the table conditions (equivalent length and condensing temperature) differ from the design conditions.

### Step 2 – Calculate Actual ∆T

We can calculate the saturation temperature difference based upon the design conditions:

{:[DeltaT_(“Actual “)=DeltaT_(“Table “)[(” Actual Length “)/(” Table Length “)][(” Actual Capacity “)/(” Table Capacity “)]^(1.8)],[DeltaT_(“Actual “)=1^(@)F[(113.6ft)/(100.0ft)][(60.0” Tons “)/(79.7” Tons “)]^(1.8)=0.68^(@)F],[{: Delta DeltaT_(“Actual “)=0.56^(@)C[(34.64(” “m))/(30.48(” “m))][(211(” “kW))/(280(” “kW))]^(1.8)=0.39^(@)C]]:}

### Step 3 – Calculate Actual Piping Pressure Drop

According to Table 3, the pressure drop for 1°F (0.56°C) saturation temperature drop with a 100 ft equivalent length is 4.75 PSI (32.75 kPa). The actual piping pressure drop is determined using the equation:

{:[” Pressure “” Drop “_(“Actual “)=” Pressure Drop “_(“Table “)[(DeltaT_(“Actual “))/(DeltaT_(“Table “))]],[[” Pressure Drop “p_(“Actual “)=32.75kPaquad[(0.39^(@)C)/(0.56^(@)C)]=22.81kPa]],[]:}

### Step 4 – Calculate Total Pressure Drop

Next to determine the Total pressure drop, we use Table 4, and recall that the riser is 20 ft. For R-410A the pressure drop is 0.43 PSI per ft (9.73 kPa/m).

” Pressure Drop from the Riser “=” Pressure Drop “xx(” Refrigerant Pressure Drop “)/(ft)
{:[” Pressure Drop from the Riser “=20.0ftxx(0.43PSI)/(ft)=8.6PSI],[[” Pressure Drop from the Riser “=6.1(” “m)xx(9.73kPa)/(m)=259.35kPa]]:}

Total Pressure Drop = Actual Pressure Drop + Riser Pressure Drop

Total Pressure Drop = 3.23 PSI + 8.6 PSI = 11.83 PSI

Total Pressure Drop = 59.35 kPa + 22.81 kPa = 82.16 kPa

### Step 5 – Determine the Saturated Pressure of R-410A at the TX Valve

Using refrigerant property tables which can be found in HVAC-ENG.COM or references such as ASHRAE, the saturated pressure for R-410A at 120°F is 433 PSIA (absolute) (2985 kPaA). To calculate the saturation pressure at the TX valve, we take the saturated pressure of R-410A at 120°F and subtract the total pressure drop.

Saturated PressureTX Valve = Saturated Pressure120°F – Total Pressure Drop
Saturated PressureTX Valve = 433.0 PSIA – 11.83 PSIA = 421.17 PSIA
(Saturated PressureTX Valve = 2985.0 kPa – 82.15 lPa = 2902.85 kPa)

### Step 6 – Determine the Saturation Temperature at the TXValve

Referring back to the Refrigeration property tables, the saturation temperature at the TX valve can be interpolated using the saturation pressure at the TX valve (421 PSIA). The saturation temperature at the TX valve is found to be 117.8°F.

### Step 7- Determine The Sub-cooling Required for Saturated Liquid at the TX Valve

The sub-cooling require to have saturated liquid at the TX valve can be found by:

Subcooling = Actual Saturation Temperature – Saturation TemperatureTX Valve
Subcooling = 120.0°F – 117.8°F = 2.2°F

### Step 8- Determine the Required Sub-cooling for Proper Operation

2.2°F is the amount of sub-cooling required to have saturated liquid refrigerant at the TX valve. Anything less, and the refrigerant will start to flash and the TX valve will not operate properly. For TX valves to operate properly and avoid diaphragm fluttering, there should be an additional 4°F of subcooling at the TX Valve.

Subcooling Requirement = TX Valve Temperature + Minimum System Temperature
Subcooling Requirement = 2.2°F + 4.0°F = 6.2°F

In the following posts, we will be discussing topics related to refrigerant oil, suction line sizing, oil return in suction and discharge risers, thermal expansion valves, hot gas bypass, hot gas bypass line sizing, hot gas bypass valves, how to size a hot gas bypass line, installation details, pump down, piping insulation, refrigerant line installation, low ambient operation, fan cycling and fan speed control, condenser flood back design, safety, and the environment. All of these topics are essential to understanding the different aspects of refrigerant piping and will ensure a comprehensive knowledge of the subject.

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