Lennox HS29-072 Installation Instructions Manual page 12

Figure 10 illustrates the relationship between liquid line siz
ing, pressure drop per 100 feet, velocity range, and ton
nage. Remember, when using liquid line solenoid valves,
velocities should not exceed 300 fpm (1.5 m/s). Find the
cooling capacity on the left side of the chart in figure 10,
then proceed right to the smallest tube size that will not ex
ceed 300 fpm (1.5 m/s) velocity.
Table 3
Equivalent Length in Feet of Straight Pipe for
Valves and Fittings
Line
Solenoid/
Size Angle
Globe
Valve
O.D.
Valve
in.
3/8 7 4
1/2 9 5
5/8 12 6
3/4 14 7
7/8 15 8
1 1/8 22 12
1 3/8 28 15
1 5/8 35 17
2 1/8 45 22
2 5/8 51 26
NOTE − Long radius elbow. Multiply factor by 1.5 for short radius el
bow equivalent length.
Equipment that is above five tons in capacity typically oper
ates at a saturated condensing temperature of 125°F
(52°C) which corresponds to an operating pressure of 280
psig (1930 kPa). This equipment is designed to hold a
charge allowing 10°F (6°C) subcooling at 95°F (53°C) am
bient. Use the condensing temperature and the subcooling
to calculate the maximum allowable pressure drop as de
tailed below.
NOTE − 95°F (53°C) ambient is an arbitrary temperature
chosen to represent typical summer operating conditions
and the maximum allowable pressure drop. This tempera
ture (and the corresponding subcooling) will vary with re
gional climate.
Example − Calculating maximum allowable pressure
drop: Find the maximum allowable liquid line pressure
drop of a unit operating at 10°F (6°C) subcooling, 125°F
(52°C) condensing temperature and operating pressure of
280 psig (1931 kPa). Subtract 10°F (6°C) subcooling tem
perature from 125°F (52°C) condensing temperature to
equal 115°F (46°C) subcooled liquid temperature. This
corresponds with operating pressure of 245 psig (1689
kPa), which is the point at which flash gas will begin to form.
Subtract 245 psig (1689 kPa) subcooled pressure from 280
psig (1931 kPa) condensing pressure to find a maximum
allowable pressure drop of 35 psig (241 kPa).
To calculate the actual pressure drop in the liquid line, cal
culate the pressure drop due to friction and the pressure
drop due to vertical lift, then add the two:
90° 45°
Long* Long* Tee Tee
Line Branch
Radius Radius
Elbow Elbow
0.8 0.3 0.5 1.5
0.9 0.4 0.6 2.0
1.0 0.5 0.8 2.5
1.3 0.6 0.9 3.0
1.5 0.7 1.0 3.5
1.8 0.9 1.5 4.5
2.4 1.2 1.8 6.0
2.8 1.4 2.0 7.0
3.9 1.8 3.0 10
4.6 2.2 3.5 12
Pressure drop due to friction
+
Pressure drop due to vertical lift
You must consider the pressure drop due to friction in the
pipe, fittings, and field installed accessories such as driers,
solenoid valves, or other devices. Pressure drop ratings for
different pipe sizes are listed in figure 10. Pressure drop
ratings of field installed devices are typically supplied by
the manufacturer.
Pressure drop due to vertical lift (1/2 pound per foot) is
typically high and can be a limiting factor in the design of
the system.
The liquid refrigerant pressure must be sufficient to pro
duce the required flow through the expansion device. Liq
uid refrigerant (free of flash gas) should be delivered to the
expansion valve at a minimum of 175 psig (1207 kPa) to
ensure that the 100 psig (690 kPa) necessary to produce
full refrigerant flow is at the rated capacity.
Example − Liquid Line Pipe Sizing
Given: 10 ton condensing unit on ground level with a
10 ton evaporator on the third level above ground and a to
tal of 96 linear feet of piping. The unit is charged with 10°F
(6°C) subcooling at 125°F (52°C) condensing temperature
and an operating pressure of 280 psig (1930 kPa). Refer to
figure 11.
Find: Select tube size from figure 10.
Liquid Line Sizing Example
solution:
pressure drop
cannot exceed
35 psig.
3 ft.
10 ton
condensing unit
two 90° long radius elbows @ 5/8 in. o.d. = 1ft. equiv. ft. ea.
linear length + equivalent length of fittings
total friction losses =
total friction losses + lift losses + filter/drier
filter drop = 1 psig (by manufacturer)
lift losses = 40 ft. x 1/ 2 psig per foot = 20 psig
total pressure drop = 20 psig + 4.17 psig + 1 psig = 25.17 psig
Answer: 5/8 in. o.d. copper tubing can be used. Pressure loss does not exceed maximum
allowable pressure drop (6_F to 7_F subcooling will be available at the expansion valve)
and velocity is acceptable.
Solution: For a 10−ton system, select a 5/8 inch O.D. line
with 4.25 psig (29 kPa) per 100 feet drop (per figure 10).
Now, calculate pressure drop due to friction and liquid lift to
determine if this is a good selection.
Page 12
Pressure drop in
=
the liquid line
53 FT.
filter/drier
given; 10 ton evaporator
10 ton condensing unit
40 ft.
with 10°f subcooling at 125°F
length of line − 96 ft.
find: liquid line size
select a proposed tubing
size: 5/8 in. copper
total equivalent length =
total equivalent length = 98 ft.
4.25 psig
x 98 ft. = 4.17psig
100 ft.
total pressure drop=
Figure 11
10 ton
evaporator