APPENDIX A
3-1-2010
Page A-10
STEP NO. 10 – DETERMINE ATOMIZER PRESSURE
LOSS (Continued)
10b – Size Piping and Calculate Pipe Pressure Loss
(PPL) in Branch Line B:
Linear length B1 and B2 plus equivalent lengths of all
fittings in each pipe line multiplied by pressure loss (see
Table 6) equals Pipe Pressure Loss (PPL).
Example:
3/4 in. Schedule 40 piping:
Length B1, 9.25 ft (2.819 m) + B2, 6.59 ft (2.009 m) =
15.84 ft (4.828 m)
Equivalent length of 3/4 in. elbow = 2 ft (0.61 m) (see
Table 5)
One elbow in Branch Line B = 1 x 2 ft (0.61 m) (equiva-
lent length per elbow) = 2 ft (0.61 m).
Equivalent length of 3/4 in. tee = 4 ft (1.22 m) (see Table
5)
One tee in Branch Line B = 1 x 4 ft (1.22 m) (equivalent
length per side tee) = 4 ft (1.22 m).
15.84 ft (4.828 m) + 2 ft (0.61 m) + 4 ft (1.22 m) = 21.84 ft
(6.658 m) total equivalent length of pipe.
Select the correct pressure from (Table 6) for the number
of atomizers flowing through this pipe section. Since there
are two atomizers the pressure loss factor would be
0.51192 psi/ft (0.11572 bar/m) (see Table 6).
Pressure loss for Branch Line B = 21.84 ft (6.658 m) of
3/4 in. pipe x 0.51192 psi/ft (0.11572 bar/m) = 11.18 psi
(0.770 bar).
10c – Size Piping and Calculate Pipe Pressure Loss
(PPL) in Branch Line C:
Linear length C1 plus equivalent lengths of all fittings in
each pipe line multiplied by pressure loss (see Table 6)
equals Pipe Pressure Loss (PPL).
Example:
3/4 in. Schedule 40 piping:
Length C1, 2.92 ft (0.890 m) =2.92 ft (0.890 m)
Equivalent length of 3/4 in. tee = 4 ft (1.22 m) (see Table
5)
One tee in Branch Line C = 1 x 4 ft (1.22 m) (equivalent
length per side tee) = 4 ft (1.22 m).
2.92 ft (0.890 m) + 4 ft (1.22 m) = 6.92 ft (2.109 m) total
equivalent length of pipe
Select the correct pressure from (Table 6) for the number
of atomizers flowing through this pipe section. Since there
is only one atomizer the pressure loss factor would be
0.12798 psi/ft (0.02893 bar/m) (see Table 6).
Pressure loss for Branch Line C = 6.92 ft (2.109 m) of 3/4
in. pipe x 0.12798 psi/ft (0.02893 bar/m) = 0.89 psi (0.061
bar).
10d – Size Piping and Calculate Pipe Pressure Loss
(PPL) in Branch Line D:
Linear length D1 and D2 plus equivalent lengths of all
fittings in each pipe line multiplied by pressure loss (see
Table 6) equals Pipe Pressure Loss (PPL).
Example:
3/4 in. Schedule 40 piping:
Length D1, 17 ft (5.182 m) + D2, 2.92 ft (0.890 m) =
19.92 ft (6.072 m)
Equivalent length of 3/4 in. elbow = 2 ft (0.61 m) (see
Table 5)
One elbow in Branch Line D = 1 x 2 ft (0.61 m) (equiva-
lent length per elbow) = 2 ft (0.61 m)
19.92 ft (6.072 m) + 2 ft (0.61 m) = 21.92 ft (6.681 m)
total equivalent length of pipe
Select the correct pressure from (Table 6) for the number
of atomizers flowing through this pipe section. Since there
is only one atomizer the pressure loss factor would be
0.12798 psi/ft (0.02893 bar/m) (see Table 6).
Pressure loss for Branch Line D = 21.92 ft (6.681 m) of
3/4 in. pipe x 0.12798 psi/ft (0.02893 bar/m) = 2.81 psi
(0.193 bar)
10e – Size Piping and Calculate Pipe Pressure Loss
(PPL) in Branch Line E:
Linear length E1 and E2 plus equivalent lengths of all
fittings in each pipe line multiplied by pressure loss (see
Table 6) equals Pipe Pressure Loss (PPL).
Example:
3/4 in. Schedule 40 piping:
Length E1, 11.67 ft (3.557 m) + E2, 6.59 ft (2.009 m) =
18.26 ft (5.566 m)
Equivalent length of 3/4 in. elbow = 2 ft (0.61 m) (see
Table 5)
One elbow in Branch Line E = 1 x 2 ft (0.61 m) (equiva-
lent length per elbow) = 2 ft (0.61 m).
Equivalent length of 3/4 in. tee = 4 ft (1.22 m) (see Table
5)
One tee in Branch Line E = 1 x 4 ft (1.22 m) (equivalent
length per side tee) = 4 ft (1.22 m).
18.26 ft (5.566 m) + 2 ft (0.61 m) + 4 ft (1.22 m) = 24.26 ft
(7.394 m) total equivalent length of pipe.
Select the correct pressure from (Table 6) for the number
of atomizers flowing through this pipe section. Since there
are two atomizers the pressure loss factor would be
0.51192 psi/ft (0.11572 bar/m) (see Table 6).
Pressure loss for Branch Line E = 24.26 ft (7.394 m) of
3/4 in. pipe x 0.51192 psi/ft (0.11572 bar/m) = 12.42 psi
(0.856 bar).
Selector Valves
Summary of Contents for ANSUL AQUASONIC
Page 84: ...SECTION V 3 1 2010 Page 5 14 NOTES Installation ...
Page 88: ...SECTION VII 3 1 2010 Page 7 2 Inspection NOTES ...
Page 94: ...SECTION VIII 3 1 2010 Page 8 6 NOTES Maintenance ...
Page 116: ...APPENDIX C 3 1 2010 Page C 2 Main and Reserve Systems NOTES ...
Page 134: ...APPENDIX D 3 1 2010 Page D 18 008567 Typical Examples ...
Page 135: ...APPENDIX D 3 1 2010 Page D 19 008568 Typical Examples ...
Page 136: ...APPENDIX D 3 1 2010 Page D 20 008569 Typical Examples ...
Page 137: ...APPENDIX D 3 1 2010 Page D 21 008570 Typical Examples ...
Page 138: ...APPENDIX D 3 1 2010 Page D 22 008571 Typical Examples ...
Page 139: ...APPENDIX D 3 1 2010 Page D 23 008572 Typical Examples ...
Page 140: ...APPENDIX D 3 1 2010 Page D 24 008573 Typical Examples ...
Page 141: ...APPENDIX D 3 1 2010 Page D 25 008574 Typical Examples ...
