How To Size Liquid Lines - Daikin R-22 Refrigerant Piping Design Manual

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
Step 1 – Estimate Pipe Size
To determine the liquid line pipe size for a 60 ton unit, use
Table 9 in Appendix 2. According to the table, a 1-3/8 inch (35
mm) pipe will work for a 79.7 ton (280 kW) unit. Note, the table
conditions (equivalent length and condensing temperature) are
different than the design conditions.
Step 2 – Calculate Actual ∆T
Using Note #5 in the table, we can calculate the saturation
Liquid Line — Step 2
temperature difference based upon the design conditions:
Actual Length
∆T = ∆T
Table Length
Actual Table
Liquid Line — Step 2
113.6 ft
∆T = 1°F
Actual 100.0 ft
Actual Length
∆T = ∆T
34.64 m
∆T Actual = 0.56°C Table Table Length
Actual
30.48 m
113.6 ft
∆T = 1°F
Actual
100.0 ft
Step 3 – Calculate Actual Piping Pressure Drop
34.64 m
Liquid Line — Step 3
∆T = 0.56°C
Actual
30.48 m
According to Table 9, the pressure drop for 1°F (0.56°C)
saturation temperature drop with a 100 ft equivalent length is
Pressure Drop = Pressure Drop
4.75 PSI (32.75 kPa).
Actual
The actual piping pressure drop is determined using the
Liquid Line — Step 3
equation: Pressure Drop = 4.75 PSI
Actual
Pressure Drop = Pressure Drop
Pressure Drop Actual = 32.75 kPa
Actual
Pressure Drop = 4.75 PSI
Actual
Liquid Line — Step 4
Pressure Drop = 32.75 kPa
Actual
Pressure Drop from the Riser = Pressure Drop ×
Liquid Line — Step 4
Pressure Drop from the Riser = 20.0 ft ×
Pressure Drop from the Riser = Pressure Drop ×
AG 31-011 • REFRIGERANT PIPING DESIGN
Pressure Drop from the Riser = 6.1 m ×
Pressure Drop from the Riser = 20.0 ft ×
1.8
Actual Capacity
Table Capacity
1.8
60.0 Tons
= 0.68°F
79.7 Tons
1.8
Actual Capacity
1.8
211 kW
Table Capacity
= 0.39°C
280 kW
1.8
60.0 Tons
= 0.68°F
79.7 Tons
1.8
211 kW
= 0.39°C
280 kW
∆T
Actual
Table ∆T
Table
0.68°F
= 3.23 PSI
1°F
∆T
Actual
0.39°C
Table ∆T
= 22.81 kPa
Table
0.56°C
0.68°F
= 3.23 PSI
1°F
0.39°C
= 22.81 kPa
0.56°C
Refrigerant Pressure Drop
ft
0.43 PSI
= 8.6 PSI
ft
Refrigerant Pressure Drop
9.73 kPa
ft
= 259.35 kPa
m
0.43 PSI
= 8.6 PSI
Liquid Line — Step 3
Pressure Drop = Pressure Drop
Actual
Pressure Drop = 4.75 PSI
Actual
Pressure Drop = 32.75 kPa
Step 4 – Calculate Total Pressure Drop Actual
Next to determine the Total pressure drop, we use
page 10, and recall that the riser is 20 ft. For R-410A the
pressure drop is 0.43 PSI per ft (9.73 kPa/m).
Liquid Line — Step 4
Pressure Drop from the Riser = Pressure Drop ×
Pressure Drop from the Riser = 20.0 ft ×
Pressure Drop from the Riser = 6.1 m ×
Total Pressure Drop = Actual Pressure Drop + Riser Pressure Drop
Total Pressure Drop = Actual Pressure Drop + Riser Pressure Drop
Total Pressure Drop = 3.23 PSI + 8.6 PSI = 11.83 PSI
Total Pressure Drop = 3.23 PSI + 8.6 PSI = 11.83 PSI
(Total Pressure Drop = 59.35 kPa + 22.81 kPa = 82.16 kPa
(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
Liquid Line — Step 5
Using refrigerant property tables which can be found in
Appendix 2 of Daikin Applied's Refrigerant Application Guide
Saturated Pressure = Saturated Pressure
(AG 31-007, see www.DaikinApplied.com) the saturated
TX Valve
pressure for R-410A at 120°F is 433 PSIA (absolute) (2985
Saturated Pressure = 433.0 PSIA + 11.83 PSIA = 421.17 PSIA
TX Valve
kPaA). To calculate the saturation pressure at the TX valve, we
(Saturated Pressure = 2985.0 kPa + 82.15 kPa = 2902.85 kPa)
take the saturated pressure of R-410A at 120°F and subtract
TX Valve
the total pressure drop.
Saturated Pressure = Saturated Pressure
TX Valve
Pressure Drop
Liquid Line — Step 7
Saturated Pressure = 433.0 PSIA – 11.83 PSIA = 421.17 PSIA
TX Valve
Subcooling = Actual Saturation Temperature – Saturation Temperature
(Saturated Pressure = 2985.0 kPa – 82.15 lPa = 2902.85 kPa)
TX Valve
Subcooling = 120.0°F – 117.8°F = 2.2°F
Step 6 – Determine the Saturation Temperature at the TX
Valve
Liquid Line — Step 8
Referring back to the Refrigeration property tables in
Subcooling Requirement = TX Valve Temperature + Minimum System Temperature
Application Guide 31-007, the saturation temperature at the
TX valve can be interpolated using the saturation pressure at Subcooling Requirement = 2.2°F + 4.0°F = 6.2°F
the TX valve (421 PSIA). The saturation temperature at the TX
valve is found to be 117.8°F
18
∆T
Actual
Table ∆T
Table
0.68°F
= 3.23 PSI
1°F
0.39°C
= 22.81 kPa
0.56°C
Table 2 on
Refrigerant Pressure Drop
ft
0.43 PSI
= 8.6 PSI
ft
9.73 kPa
= 259.35 kPa
m
– Total Pressure Drop
120°F
– Total
120°F
TX Valve
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