AG 31-011 • REFRIGERANT PIPING DESIGN 18 www.DaikinApplied.com
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
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
temperature difference based upon the design conditions:
Step 3 – Calculate Actual Piping Pressure Drop
, 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:
Step 4 – Calculate Total Pressure Drop
Next to determine the Total pressure drop, we use
, and recall that the riser is 20 ft. For R-410A the
pressure drop is 0.43 PSI per ft (9.73 kPa/m).
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
Appendix 2 of Daikin Applied’s Refrigerant Application Guide
) 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 Pressure
TX Valve
= Saturated Pressure
120°F
– Total
Pressure Drop
Saturated Pressure
TX Valve
= 433.0 PSIA – 11.83 PSIA = 421.17 PSIA
(Saturated Pressure
TX Valve
= 2985.0 kPa – 82.15 lPa = 2902.85 kPa)
Step 6 – Determine the Saturation Temperature at the TX
Valve
Referring back to the Refrigeration property tables in
Application Guide 31-007, 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
∆
T
Actual
=
∆
T
Table
Actual Length
Table Length
Actual Capacity
Table Capacity
1.8
∆
T
Actual
= 1°F
∆
T
Actual
= 0.56°C
= 0.68°F
113.6 ft
100.0 ft
60.0 Tons
79.7 Tons
1.8
= 0.39°C
34.64 m
30.48 m
211 kW
280 kW
1.8
Liquid Line — Step 2
Pressure Drop
Actual
= Pressure Drop
Table
∆
T
Actual
∆
T
Table
Pressure Drop
Actual
= 4.75 PSI
Pressure Drop
Actual
= 32.75 kPa
= 3.23 PSI
0.68°F
1°F
= 22.81 kPa
0.39°C
0.56°C
Liquid Line — Step 3
Pressure Drop from the 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 = 59.35 kPa + 22.81 kPa = 82.16 kPa)
Saturated Pressure
TX Valve
= Saturated Pressure
120°F
– Total Pressure Drop
Saturated Pressure
TX Valve
= 433.0 PSIA + 11.83 PSIA = 421.17 PSIA
(Saturated Pressure
TX Valve
= 2985.0 kPa + 82.15 kPa = 2902.85 kPa)
Refrigerant Pressure Drop
ft
Pressure Drop from the Riser = 20.0 ft ×
Pressure Drop from the Riser = 6.1 m ×
= 8.6 PSI
0.43 PSI
ft
= 259.35 kPa
9.73 kPa
m
Liquid Line — Step 4
Liquid Line — Step 5
Subcooling = Actual Saturation Temperature – Saturation Temperature
TX Valve
Subcooling = 120.0°F – 117.8°F = 2.2°F
Liquid Line — Step 7
Subcooling Requirement = TX Valve Temperature + Minimum System Temperature
Subcooling Requirement = 2.2°F + 4.0°F = 6.2°F
Liquid Line — Step 8
∆
T
Actual
=
∆
T
Table
Actual Length
Table Length
Actual Capacity
Table Capacity
1.8
∆
T
Actual
= 1°F
∆
T
Actual
= 0.56°C
= 0.68°F
113.6 ft
100.0 ft
60.0 Tons
79.7 Tons
1.8
= 0.39°C
34.64 m
30.48 m
211 kW
280 kW
1.8
Liquid Line — Step 2
Pressure Drop
Actual
= Pressure Drop
Table
∆
T
Actual
∆
T
Table
Pressure Drop
Actual
= 4.75 PSI
Pressure Drop
Actual
= 32.75 kPa
= 3.23 PSI
0.68°F
1°F
= 22.81 kPa
0.39°C
0.56°C
Liquid Line — Step 3
Pressure Drop from the 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 = 59.35 kPa + 22.81 kPa = 82.16 kPa)
Saturated Pressure
TX Valve
= Saturated Pressure
120°F
– Total Pressure Drop
Saturated Pressure
TX Valve
= 433.0 PSIA + 11.83 PSIA = 421.17 PSIA
(Saturated Pressure
TX Valve
= 2985.0 kPa + 82.15 kPa = 2902.85 kPa)
Refrigerant Pressure Drop
ft
Pressure Drop from the Riser = 20.0 ft ×
Pressure Drop from the Riser = 6.1 m ×
= 8.6 PSI
0.43 PSI
ft
= 259.35 kPa
9.73 kPa
m
Liquid Line — Step 4
Liquid Line — Step 5
Subcooling = Actual Saturation Temperature – Saturation Temperature
TX Valve
Subcooling = 120.0°F – 117.8°F = 2.2°F
Liquid Line — Step 7
Subcooling Requirement = TX Valve Temperature + Minimum System Temperature
Subcooling Requirement = 2.2°F + 4.0°F = 6.2°F
Liquid Line — Step 8
∆
T
Actual
=
∆
T
Table
Actual Length
Table Length
Actual Capacity
Table Capacity
1.8
∆
T
Actual
= 1°F
∆
T
Actual
= 0.56°C
= 0.68°F
113.6 ft
100.0 ft
60.0 Tons
79.7 Tons
1.8
= 0.39°C
34.64 m
30.48 m
211 kW
280 kW
1.8
Liquid Line — Step 2
Pressure Drop
Actual
= Pressure Drop
Table
∆
T
Actual
∆
T
Table
Pressure Drop
Actual
= 4.75 PSI
Pressure Drop
Actual
= 32.75 kPa
= 3.23 PSI
0.68°F
1°F
= 22.81 kPa
0.39°C
0.56°C
Liquid Line — Step 3
Pressure Drop from the 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 = 59.35 kPa + 22.81 kPa = 82.16 kPa)
Saturated Pressure
– Total Pressure Drop
Saturated Pressure
TX Valve
= 433.0 PSIA + 11.83 PSIA = 421.17 PSIA
(Saturated Pressure
TX Valve
= 2985.0 kPa + 82.15 kPa = 2902.85 kPa)
Refrigerant Pressure Drop
ft
Pressure Drop from the Riser = 20.0 ft ×
Pressure Drop from the Riser = 6.1 m ×
= 8.6 PSI
0.43 PSI
ft
= 259.35 kPa
9.73 kPa
m
Liquid Line — Step 4
Liquid Line — Step 5
Subcooling = Actual Saturation Temperature – Saturation Temperature
TX Valve
Subcooling = 120.0°F – 117.8°F = 2.2°F
Liquid Line — Step 7
Subcooling Requirement = TX Valve Temperature + Minimum System Temperature
Subcooling Requirement = 2.2°F + 4.0°F = 6.2°F
Liquid Line — Step 8