www.DaikinApplied.com 23
AG 31-011 • REFRIGERANT PIPING DESIGN
How to Size Suction Lines
Size the suction line with a single pipe riser and determine the
pressure drop for the following air-cooled chiller with remote
evaporator:
• Uses R-134a
• Has Type L copper pipe
• Evaporator operates at 40°F (4.4°C) Saturated Suction
Temperature (SST)
• Superheat is 10°F (5.6°C)
• Condenser operates at 120°F (48.9°C)
•
Capacity is two 50 tons (176 kW) circuits with up to 20%
turn down
• Suction line equivalent length for the horizontal runs is:
—
Bottom 10 ft (30 m)
—
Top 12 ft (3.7 m)
• Suction line equivalent length for a single pipe riser is 42
ft (12.8 m)
Step 1- Estimate Suction Line Size
To determine the correct suction line size to operate the
system at minimum capacity with a single pipe riser use
in Appendix 2. According to the table, a 3-1/8 inch (79mm)
pipe will work for 57.1 tons (200.8kW) unit. Note, the table
conditions (equivalent length and condensing temperature) are
different than the design conditions.
Step 2 – Correct for Actual Operating Conditions
Sizing the pipe for full load requires a correction for the 120°F
actual condenser temperature. Referring to the correction
Actual Capacity = Table Capacity × 0.902
Actual Capacity = 57.1 Tons × 0.902 = 51.5 Tons
Step 3 – Calculate the Actual ∆T
, calculate the saturation
temperature difference based upon the actual design conditions:
Step 4 – Calculate the Actual Pressure Drop
shows the pressure drop for 40°F (4.4oC)
saturation temperature change with a 100 ft (30.5m) equivalent
length is 1.93 PSI (13.3 kPa).
A 3-1/8" pipe has 1.2°F temperature drop and a 1.16 PSI
pressure drop which is acceptable for suction pipe.
Single Pipe
Suction Line
Riser
∆
T
Actual
=
∆
T
Table
Actual Length
Table Length
Actual Capacity
Table Capacity
1.8
∆
T
Actual
= 2°F
∆
T
Actual
= 1.1°C
= 1.2°F
64.0 ft
100.0 ft
50.0 Tons
51.5 Tons
1.8
= 0.67°C
19.5 m
30.5 m
176 kW
181 kW
1.8
Suction Line — Step 3
∆
T
Actual
= 2°F
∆
T
Actual
= 1.1°C
= 0.42°F
22.0 ft
100.0 ft
50.0 Tons
51.5 Tons
1.8
= 0.23°C
6.7 m
30.5 m
176 kW
181 kW
1.8
∆
T
Actual
=
∆
T
Table
Actual Length
Table Length
Actual Capacity
Table Capacity
1.8
∆
T
Actual
= 2°F
∆
T
Actual
= 1.1°C
= 1.84°F
42.0 ft
100.0 ft
50.0 Tons
32.3 Tons
1.8
= 1.01°C
12.8 m
30.5 m
175.9 kW
113.6 kW
1.8
Suction Line — Step 6
Pressure Drop
Actual
= Pressure Drop
Table
∆
T
Actual
∆
T
Table
Pressure Drop
Actual
= 1.93 PSI
Pressure Drop
Actual
= 13.3 kPa
= 1.16 PSI
1.2°F
2°F
= 8.1 kPa
0.67°C
1.1°C
Suction Line — Step 4
Pressure Drop
Actual Vertical
= Pressure Drop
Table
∆
T
Actual Vertical
∆
T
Table
Pressure Drop
Actual Vertical
= 1.93 PSI
Pressure Drop
Actual Vertical
= 13.3 kPa
= 1.78 PSI
1.84°F
2°F
= 12.21 kPa
1.01°C
1.1°C
Pressure Drop
Actual Horizontal
= 1.93 PSI
Pressure Drop
Actual Horizontal
= 13.3 kPa
= 0.41 PSI
0.42°F
2°F
= 2.78 kPa
0.23°C
1.1°C
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
= 2°F
∆
T
Actual
= 1.1°C
= 1.2°F
64.0 ft
100.0 ft
50.0 Tons
51.5 Tons
1.8
= 0.67°C
19.5 m
30.5 m
176 kW
181 kW
1.8
Suction Line — Step 3
∆
T
Actual
= 2°F
∆
T
Actual
= 1.1°C
= 0.42°F
22.0 ft
100.0 ft
50.0 Tons
51.5 Tons
1.8
= 0.23°C
6.7 m
30.5 m
176 kW
181 kW
1.8
∆
T
Actual
=
∆
T
Table
Actual Length
Table Length
Actual Capacity
Table Capacity
1.8
∆
T
Actual
= 2°F
∆
T
Actual
= 1.1°C
= 1.84°F
42.0 ft
100.0 ft
50.0 Tons
32.3 Tons
1.8
= 1.01°C
12.8 m
30.5 m
175.9 kW
113.6 kW
1.8
Suction Line — Step 6
Pressure Drop
Actual
= Pressure Drop
Table
∆
T
Actual
∆
T
Table
Pressure Drop
Actual
= 1.93 PSI
Pressure Drop
Actual
= 13.3 kPa
= 1.16 PSI
1.2°F
2°F
= 8.1 kPa
0.67°C
1.1°C
Suction Line — Step 4
Pressure Drop
Actual Vertical
= Pressure Drop
Table
∆
T
Actual Vertical
∆
T
Table
Pressure Drop
Actual Vertical
= 1.93 PSI
Pressure Drop
Actual Vertical
= 13.3 kPa
= 1.78 PSI
1.84°F
2°F
= 12.21 kPa
1.01°C
1.1°C
Pressure Drop
Actual Horizontal
= 1.93 PSI
Pressure Drop
Actual Horizontal
= 13.3 kPa
= 0.41 PSI
0.42°F
2°F
= 2.78 kPa
0.23°C
1.1°C
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